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Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Question 1.
Define chemistry.
Answer:
Chemistry is the study of matter, its physical and chemical properties and the physical and chemical changes it undergoes under different conditions.

Question 2.
Why is chemistry called a central science?
Answer:

1. Knowledge of chemistry is required in the studies of physics, biological sciences, applied sciences, and earth and space sciences.
2. Chemistry is involved in every aspect of day-to-day life, i.e. the air we breathe, the food we eat, the fluids we drink, our clothing, transportation and fuel supplies, etc.

Hence, chemistry is called a central science.

Question 3.
Give reason: Although chemistry has ancient roots, it has developed as a modern science.
Answer:
Technological development in sophisticated instruments have expanded knowledge of chemistry which, now, has been used in applied sciences such as medicine, dentistry, engineering, agriculture and in daily home use products. Hence, due to development and advancement in science and technology, chemistry has developed as modem science.

Question 4.
How is chemistry traditionally classified?
Answer:
Chemistry is traditionally classified into five branches:

• Organic chemistry
• Inorganic chemistry
• Physical chemistry
• Biochemistry
• Analytical chemistry

Question 5.
Explain the following terms:
i. Organic chemistry
ii. Inorganic chemistry
iii. Physical chemistry
Answer:
i. Organic chemistry: It deals with properties and reactions of compounds of carbon.
ii. Inorganic chemistry: It deals with the study of all the compounds which are not organic.
iii. Physical chemistry: It deals with the study of properties of matter, the energy changes and the theories, laws and principles that explain the transformation of matter from one form to another. It also provides basic framework for all the other branches of chemistry.

Question 6.
Distinguish between
i. Mixtures and pure substances
ii. Mixtures and compounds
Answer:
i.

	Mixtures	Pure substances
a.	Mixtures have no definite chemical composition.	Pure substances have a definite chemical composition.
b.	Mixtures have no definite properties.	Pure substances always have the same properties regardless of their origin.
e.g.	Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.	Pure metal, distilled water, etc.

ii.

	Mixtures	Compounds
a.	Mixtures have no definite chemical composition.	Compounds are made up of two or more elements in fixed proportion.
b.	The constituents of a mixture can be easily separated by physical method.	The constituents of a compound cannot be easily separated by physical method.
e.g.	Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.	Water, table salt, sugar, etc.

Question 7.
What is the difference between element and compound?
Answer:
Elements cannot be broken down into simpler substances while compounds can be broken down into simpler substances by chemical changes.

Question 8.
Explain: States of matter
Answer:
There are three different states of matter as follows:

1. Solid: Particles are held tightly in perfect order. They have definite shape and volume.
2. Liquid: Particles are close to each other but can move around within the liquid.
3. Gas: Particles are far apart as compared to that of solid and liquid.

These three states of matter can be interconverted by changing the conditions of temperature and pressure.

Question 9.
Explain: Physical and chemical properties
Answer:
i. Physical properties: These are properties which can be measured or observed without changing the identity or the composition of the substance. e.g. Colour, odour, melting point, boiling point, density, etc.

ii. Chemical properties: These are properties in which substances undergo change in chemical composition. e.g. Coal bums in air to produce carbon dioxide, magnesium wire bums in air in the presence of oxygen to form magnesium oxide, etc.

Question 10.
How are properties of matter measured?
Answer:

• Measurement involves comparing a property of matter with some fixed standard which is reproducible and unchanging.
• Properties such as mass, length, area, volume, time, etc. are quantitative in nature and can be measured.
• A quantitative measurement is represented by a number followed by units in which it is measured.
• These units are arbitrarily chosen on the basis of universally accepted standards. e.g. Length of class room can be expressed as 10 m. Here, 10 is the number and ‘m’ is the unit ‘metre’ in which the length is measured.

Question 11.
Define: Units
Answer:
The arbitrarily decided and universally accepted standards are called units.
e.g. Metre (m), kilogram (kg).

Question 12.
What are the various systems in which units are expressed?
Answer:
Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.

Question 13.
What are SI units? Name the fundamental SI units.
Answer:
SI Units: In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI system (abbreviated from its French name).
The seven fundamental SI units are as given below:

No.	Base physical quantity	SI unit	Symbol
i.	Length	Metre	k
ii.	Mass	Kilogram	kg
iii.	Time	Second	s
iv.	Temperature	Kelvin	K
v.	Amount of substance	Mole	mol
vi.	Electric current	Ampere	A
vii.	Luminous intensity	Candela	cd

[Note: Units for other quantities such as speed, volume, density, etc. can be derived from fundamental SI units.]

Question 14.
What is the basic unit of mass in the SI system?
Answer:
The basic unit of mass in the SI system is kilogram (kg).

Question 15.
Name the following:
i. Full form of CGS unit system
ii. Full form of FPS unit system
iii. The SI unit of length
iv. Symbol used for Candela unit
v. SI unit of electric current
vi. SI unit of electric current
Answer:
i. Centimetre Gram Second
ii. Foot Pound Second
iii. Metre (m)
iv. Cd
v. Kelvin (K)
vi. Ampere (A)

Question 16.
Give reason: The mass of a body is more fundamental property than its weight.
Answer:

• Mass is an inherent property of matter and is the measure of the quantity of matter of a body.
• The mass of a body does not vary with respect to its position.
• On the other hand, the weight of a body is a result of the mass and gravitational attraction
• Weight varies because the gravitational attraction of the earth for a body varies with the distance from the centre of the earth.

Hence, the mass of a body is more fundamental property than its weight.

Question 17.
How is gram related to the SI unit kilogram?
Answer:
The SI unit kilogram (kg) is related to gram (g) as 1 kg = 1000 g= 10³ g.
[Note: ‘Gram’ is used for weighing small quantities of chemicals in the laboratories.
Other commonly used quantity is ‘milligram’. 1 kg = 1000 g = 10⁶ mg]

Question 18.
Why are fractional units of the SI units of length often used? Give two examples of the fractional units of length. How are they related to the SI unit of length?
Answer:
i. Some properties such as the atomic radius, bond length, wavelength of electromagnetic radiation, etc. are very small and therefore, fractional units of the SI unit of length are often used to express these properties.
ii. Fractional units of length: Nanometre (nm), picometre (pm), etc.
iii. Nanometre (nm) and picometre (pm) are related to the SI unit of length (m) as follows:
1 nm = 10^-9 m, 1 pm = 10^-12 m

Question 19.
Define: Volume
Answer:
Volume is the amount of space occupied by a three-dimensional object. It does not depend on shape.

Question 20.
State the common unit used for the measurement of volume of liquids and gases.
Answer:
The common unit used for the measurement of volume of liquids and gases is litre (L).

Question 21.
How is the SI unit of volume expressed?
Answer:
The SI unit of volume is expressed as (metre)³ or m³.

Question 22.
Name some glassware that are used to measure the volume of liquids and solutions.
Answer:

• Graduated cylinder
• Burette
• Pipette

Question 23.
What is a volumetric flask used for in laboratory?
Answer:
A volumetric flask is used to prepare a known volume of a solution in laboratory.

Question 24.
What is density of a substance? How is it measured?
Answer:
Density:

• Density of a substance is its mass per unit volume. It is the characteristic property of any substance.
• It is determined in the laboratory by measuring both the mass and the volume of a sample.
• The density is calculated by dividing mass by volume.

Question 25.
How is the SI unit of density derived? State CGS unit of density.
Answer:
i. The SI unit of density is derived as follows:
Density = \(\frac{\text { SI unit mass }}{\text { SI unit volume }}\)
= \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
= kg m^-3

ii. CGS unit of density: g cm^-3
[Note: The CGS unit, g cm^-3 is equivalent to \(\frac{\mathrm{g}}{\mathrm{mL}}\) or g mL^-1.]

Question 26.
State three common scales of temperature measurement.
Answer:

1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)

Question 27.
State the temperatures in Fahrenheit scale that corresponds to 0 °C and 100 °C.
Answer:
The temperature that corresponds to 0 °C is 32 °F and the temperature that corresponds to 100 °C is 212 °F.

Question 28.
Write the expression showing the relationship between:
i. Degree Fahrenheit and Degree Celsius
ii. Kelvin and Degree Celsius
Answer:
i. The relationship between degree Fahrenheit and degree Celsius is expressed as,
°F = \(\frac {9}{5}\) (°C) + 32
ii. The relationship between Kelvin and degree Celsius is expressed as,
K = °C + 273.15

Question 29.
Convert the following degree Fahrenheit temperature to degree Celsius.
i. 50 °F ii. 10 °F
Answer:
Given: Temperature in degree Fahrenheit = 50 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 50 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
50 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(50-32) \times 5}{9}\)
= 10 °C

ii. Given: Temperature in degree Fahrenheit = 10 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 10 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
10 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(10-32) \times 5}{9}\)
= -12.2 °C
Ans: i. The temperature 50 °F corresponds to 10 °C.
ii. The temperature 10 °F corresponds to -12.2 °C.

Question 30.
What is a chemical combination?
Answer:

• The process in which the elements combine with each other to form compounds is called chemical combination.
• The process of chemical combination is governed by five basic laws which were discovered before the knowledge of molecular formulae.

Question 31.
State and explain the law of definite proportions.
Answer:
Law of definite proportions:
i. The law states that “A given compound always contains exactly the same proportion of elements by weight”.
ii. French chemist, Joseph Proust worked with two samples of cupric carbonate; one of which was naturally occurring cupric carbonate and other was synthetic sample. He found the composition of elements present in both the samples was same as shown below:

Cupric carbonate	% of copper	% of carbon	% of oxygen
Natural sample	51.35	9.74	38.91
Synthetic sample	51.35	9.74	38.91

iii. Thus, irrespective of the source, a given compound always contains same elements in the same proportion.

Question 32.
State and explain the law of multiple proportions.
Answer:
Law of multiple proportions:
i. John Dalton (British scientist) proposed the law of multiple proportions in 1803.
ii. It has been observed that two or more elements may combine to form more than one compound.
iii. The law states that, “ When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.
e.g. Hydrogen and oxygen combine to form two compounds, water and hydrogen peroxide.

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.

Question 33.
Show that NO and NO₂ satisfy the law of multiple proportions.
Answer:
Nitrogen and oxygen combine to form two compounds, nitric oxide (NO) and nitrogen dioxide (NO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of nitrogen (14 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 34.
Show that carbon monoxide and carbon dioxide satisfy the law of multiple proportions.
Answer:
Chemical reaction of carbon with oxygen gives two compounds, carbon monoxide (CO) and carbon dioxide (CO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of carbon (12 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 35.
Show that SO₂ and SO₃ satisfy the law of multiple proportions.
Answer:
Chemical reaction of sulphur with oxygen gives two compounds, sulphur dioxide (SO₂) and sulphur trioxide (SO₃).

Here, the two masses of oxygen (32 g and 48 g) which combine with the fixed mass of sulphur (32 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 32 : 48 or 2 : 3.
This is in accordance with the law of multiple proportions.

Question 36.
State and explain Gay Lussac’s law of gaseous volume with two examples.
Answer:
Gay Lussac’s law:
i. Gay Lussac proposed the law of gaseous volume in 1808.
ii. Gay Lussac’s law states that, “ When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
e.g. a. Under identical conditions of temperature and pressure, 100 mL of hydrogen gas combine with 50 mL of oxygen gas to produce 100 mL of water vapour.

Thus, the simple ratio of volumes is 2 : 1 : 2.

b. Under identical conditions of temperature and pressure, 1 L of nitrogen gas combine with 3 L of hydrogen gas to produce 2 L of ammonia gas.

Thus, the simple ratio of volumes is 1 : 3 : 2.

Question 37.
Give two examples which support the Gay Lussac’s law of gaseous volume.
Answer:
i. Under identical conditions of temperature and pressure, 1 L of hydrogen gas reacts with 1 L of chlorine gas to produce 2 L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2
This is in accordance with Gay Lussac’s law.

ii. Under identical conditions of temperature and pressure, 200 mL sulphur dioxide combine with 100 mL oxygen to form 200 mL sulphur trioxide.

Thus, the ratio of volumes is 2 : 1 : 2.
This is in accordance with Gay Lussac’s law.

Question 38.
Match the following:

	Law		Statement
i.	Law of definite proportions	a.	When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
ii.	Gay Lussac’s law	b.	Equal volumes of all gases at the same temperature and pressure contain equal number of molecules
iii.	Law of multiple proportions	c.	When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
iv.	Avogadro’s law	d.	A given compound always contains exactly the same proportion of elements by weight

Answer:
i – d,
ii – c,
iii – a,
iv – b

Question 39.
32 g of oxygen reacts with some carbon to make 56 grams of carbon monoxide. Find out how much mass must have been used.
Answer:
Given: Mass of oxygen (reactant) = 32 g, mass of carbon monoxide (product) = 56 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 16 g oxygen to form 28 g of carbon monoxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 16 = 32 g) of oxygen to give (2 × 28 = 56 g) carbon monoxide.
Ans: Mass of carbon used = 24 g

Question 40.
Calculate the mass of sulphur trioxide produced by burning 64 g of sulphur in excess of oxygen. (Average atomic mass: S = 32 u, O = 16 u).
Solution:
Given: Mass of sulphur (reactant) = 64 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 48 g oxygen to form 80 g of sulphur trioxide as follows:

Hence, (2 × 32 = 64 g) of sulphur will combine with (2 × 48 = 96 g) of oxygen to give (2 × 80 = 160 g) sulphur trioxide.
Ans: Mass of sulphur trioxide produced = 160 g

Question 41.
Explain Dalton’s atomic theory.
Answer:
John Dalton published “A New System of chemical philosophy” in the year of 1808. He proposed the following features, which later became famous as Dalton’s atomic theory.

• Matter consists of tiny, indivisible particles called atoms.
• All the atoms of a given elements have identical properties including mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.

Dalton’s atomic theory could explain all the laws of chemical combination.

Question 42.
Give reason: Dalton’s atomic theory explains the law of conservation of mass.
Answer:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• According to Dalton’s atomic theory, chemical reactions involve only the reorganization of atoms. Therefore, the total number of atoms in the reactants and products should be same and mass is conserved during a reaction.

Hence, Dalton’s atomic theory explains the law of conservation of mass.

Question 43.
Give reason: Dalton’s atomic theory explains the law of multiple proportion.
Answer:

• The law of multiple proportion states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
• According to Dalton’s atomic theory, compounds are formed when atoms of different elements combine in fixed ratio.

Hence, Dalton’s atomic theory explains the law of multiple proportion.

Question 44.
Define: Atomic mass unit (amu).
Answer:
Atomic mass unit or amu is defined as a mass exactly equal to one twelth of the mass of one carbon-12 atom.

Question 45.
How is relative atomic mass of an atom measured?
Answer:

• The mass of a single atom is extremely small, i.e. the mass of a hydrogen atom is 1.6736 × 10^-24 g. Hence, it is not possible to weigh a single atom.
• In the present system, mass of an atom is determined relative to the mass of an atom of carbon-12 as the standard. This was decided in 1961 by international agreement.
• The atomic mass of carbon-12 is assigned as 12.00000 atomic mass unit (amu).
• The masses of all other elements are determined relative to the mass of an atom of carbon-12 (C-12).
• The atomic masses are expressed in amu which is exactly equal to one twelth of the mass of one carbon-12 atom.
• The value of 1 amu is equal to 1.6605 × 10^-24 g.

Question 46.
What is meant by Unified Mass unit?
Answer:

• Presently, instead of amu, Unified Mass has now been accepted as the unit of atomic mass.
• It is called Dalton and its symbol is ‘u’ or ‘Da’.

Question 47.
What is average atomic mass?
Answer:
The atomic mass of an element which exists as mixture of two or more isotopes is the average of atomic masses of its isotopes. This is called average atomic mass.

Question 48.
Define: Molecular mass
Answer:
Molecular mass of a substance is the sum of average atomic masses of the atoms of the elements which constitute the molecule.
OR
Molecular mass of a substance is the mass of one molecule of that substance relative to the mass of one carbon-12 atom.

Question 49.
How is molecular mass of a substance calculated? Give example.
Answer:
Molecular mass is calculated by multiplying average atomic mass of each element by the number of its atoms and adding them together.
e.g. Molecular mass of carbon dioxide (CO₂) is calculated as follows:
Molecular mass of CO₂ = (1 × average atomic mass of C) + (2 × average atomic mass of O)
= (1 × 12.0 u) + (2 × 16.0 u)
= 44.0 u

Question 50.
Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g. What is the atomic mass of hydrogen in u?
Solution:
Given: Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g.
To find: Atomic mass of hydrogen in u
Calculation: 1.66056 × 10^-24 g = 1 u
∴ 1.6736 × 10^-24 g = x
x = \(\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g} / \mathrm{u}}\) = 1.008u
Ans: The atomic mass of hydrogen in u = 1.008 u

Question 51.
The mass of an atom of one carbon atom is 12.011 u. What is the mass of 20 atoms of the same isotope?
Solution:
Mass of l atom of carbon = 12.011 u
∴ Mass of 20 atoms of same carbon isotope = 20 × 12.011 u = 240.220 u
Ans: The mass of 20 atoms of same carbon isotope = 240.220 u

Question 52.
Calculate the average atomic mass of neon using the following data:

Isotope	Atomic mass	Natural Abundance
20Ne	19.9924 u	90.92%
21Ne	20.9940 u	0.26 %
22Ne	21.9914 u	8.82 %

Solution:
Average atomic mass of Neon (Ne)

Ans: Average atomic mass of neon = 20.1707 u

Question 53.
Calculate the average atomic mass of argon from the following data:

Isotope	Isotopic mass (g mol-1)	Abundance
36Ar	35.96755	0.337%
38Ar	37.96272	0.063%
40Ar	39.9624	99.600%

Solution:
Average atomic mass of argon (Ar)

Ans: Average atomic mass of argon = 39.974 g mol^-1

Question 54.
Calculate the molecular mass of the following in u:
i. H₂O ii. C₆H₅Cl iii. H₂SO₄
Solution:
i. Molecular mass of H₂O = (2 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 1.0u) + (1 × 16.0 u)
= 18 u

ii. Molecular mass of C₆H₅Cl = (6 × Average atomic mass of C) + (5 × Average atomic mass of H) + (1 × Average atomic mass of Cl)
= (6 × 12.0 u) + (5 × 1.0 u) + (1 × 35.5 u)
= 112.5 u

iii. Molecular mass of H₂SO₄ = (2 × Average atomic mass of H) + (1 × Average atomic mass of S) + (4 × Average atomic mass of O)
= (2 × 1.0 u) + (1 × 32.0 u) + (1 × 16.0 u)
= 98 u
Ans: i. The molecular mass of H₂O = 18 u
ii. The molecular mass of C₆H₅Cl = 112.5 u
iii. The molecular mass of H₂SO₄ = 98 u

Question 55.
Find the mass of 1 molecule of oxygen (O₂) in amu (u) and in grams.
Solution:
Molecular mass of O₂ = 2 × 16 u
∴ Mass of 1 molecule = 32 u
∴ Mass of 1 molecule of O₂= 32 × 1.66056 × 10^-24 g = 53.1379 × 10^-24 g
Ans: Mass of 1 molecule in amu = 32 u
Mass of 1 molecule in grams = 53.1379 × 10^-24 g

Question 56.
Find the formula mass of
i. NaCl ii. Cu(NO₃)₂
Solution:
i. Formula mass of NaCl
= Average atomic mass of Na + Average atomic mass of Cl
= 23.0 u + 35.5 u = 58.5 u

ii. Formula mass of Cu(NO₃)₂
= Average atomic mass of Cu + 2 × (Average atomic mass of N + Average atomic mass of three O)
= 63.5 + 2 × [14 + (3 × 16)] = 187.5 u
Ans: i. Formula mass of NaCl = 58.5 u
ii. Formula mass of Cu(NO₃)₂ = 187.5 u

Question 57.
Find the formula mass of
i. KCl
ii. AgCl
Atomic mass of K = 39 u, Ag =108 u and Cl = 35.5 u.
Solution:
i. Formula mass of KCl
= Average atomic mass of K + Average atomic mass of Cl
= 39 u + 35.5 u = 74.5 u

ii. Formula mass of AgCl
= Average atomic mass of Ag + Average atomic mass of Cl
= 108 + 35.5 = 143.5 u
Ans: i. Formula mass of KCl = 74.5 u
ii. Formula mass of AgCl = 143.5 u

Question 58.
Calculate the number of moles and molecules of urea present in 5.6 g of urea.
Solution:
Given: Mass of urea = 5.6 g
To find: The number of moles and molecules of urea
Formulae: i. Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Mass of urea = 5.6 g
Molecular mass of urea, NH₂CONH₂
= (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O)
= (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) = 60 u
∴ Molar mass of urea = 60 g mol^-1
∴ Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{5.6 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.09333 mol

[Calculation using log table:
\(\frac{5.6}{60}\)
= Antilog₁₀ [log₁₀ (5.6) – log₁₀ (60)]
= Antilog₁₀ [0.7482 – 1.7782]
= Antilog₁₀ [latex]\overline{2} .9700[/latex]
= 0.09333]

Now,
Number of molecules of urea
= Number of moles × Avogadro’s constant
= 0.09333 mol × 6.022 × 10²³ molecules/mol
= 0.5616 × 10²³ molecules (by using log table)
= 5.616 × 10²² molecules
Ans: Number of moles of urea = 0.0933 mol
Number of molecules of urea = 5.616 × 10²² molecules

[Calculation using log table:
0.09333 × 6.022
= Antilog₁₀ [log₁₀ (0.09333) + log₁₀ (6.022)]
= Antilog₁₀ [\(\overline{2} .9698\) + 0.7797]
= Antilog₁₀ [latex]\overline{1} .7495[/latex]
= 0.5616]

Question 59.
Calculate the number of atoms in each of the following:
i. 64 u of oxygen (O)
ii. 42 g of nitrogen (N)
Solution:
i. 64 u of oxygen (O) = x atoms
Atomic mass of oxygen (O) = 16 u
∴ Mass of one oxygen atom = 16 u
∴ x = \(\frac{64 \mathrm{u}}{16 \mathrm{u}}\) = 4 atoms

ii. 42 g of nitrogen (N)
Atomic mass of nitrogen = 14 u
∴ Molar mass of nitrogen = 14 g mol^-1
Now,
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{42 \mathrm{~g}}{14 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3 mol
Now,
Number of atoms = Number of moles × Avogadro’s constant
= 3 mol × 6.022 × 10²³ atoms/mol
= 18.07 × 10²³ atoms
= 1.807 × 10²⁴ atoms
Ans: i. Number of oxygen atoms in 64 u = 4 atoms
ii. Number of nitrogen atoms in 42 g = 1.807 × 10²⁴ atoms

Question 60.
Calculate the number of atoms in each of the following.
i. 52 moles of Argon (Ar)
ii. 52 u of Helium (He)
iii. 52 g of Helium (He)
Solution:
i. 52 moles of Argon
1 mole Argon atoms = 6.022 × 10²³ atoms of Ar
∴ Number of atoms = 52 mol × 6.022 × 10²³ atoms/mol
= 313.144 × 10²³ atoms of Argon

ii. 52 g of He
Molar mass of He = mass of 1 atom of He = 4.0 u
4.0 u = 1 He
∴ 52 u = x
∴ x = 52 u × \(\frac{1 \text { atom of He }}{4.0 \mathrm{u}}\) = 13 atoms of He

iii. 52 g of He
Molar mass of He = 4.0 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{52 \mathrm{~g}}{4.0 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 13 mol
Number of atoms of He = Number of moles × Avogadro’s constant
= 13 mol × 6.022 × 10²³ atoms/mol
= 78.286 × 10²³ atoms of He
Ans. i. Number of argon atoms in 52 moles = 313.144 × 10²³ atoms of Argon
ii. Number of helium atoms in 52 u = 13 atoms of He
iii. Number of helium atoms in 52 g = 78.286 × 10²³ atoms of He

Question 61.
Calculate the number of atoms of ‘C’, ‘H’ and ‘O’ in 72.5 g of isopropanol, C₃H₇OH (molar mass 60 g mol^-1).
Solution:
Mass of isopropanol(C₃H₇OH) = 72.5 g
The number of atoms of C, H, O
Calculation: Molecular formula of isopropanol, is C₃H₇OH.
Molar mass of C₃H₇OH = 60 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{72.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 1.208 mol
∴ Moles of isopropanol = 1.21 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of isopropanol has total 12 atoms, out of which 8 atoms are of H, 3 of C and 1 of O.
∴ Number of C atoms in 72.5 g isopropanol = (3 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 2.182 × 10²⁴ atoms of carbon
∴ Number of ‘H’ atoms in 72.5 g isopropanol = (8 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 5.819 × 10²⁴ atoms of hydrogen
∴ Number of ‘O’ atoms in 72.5 g isopropanol = (1 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 7.274 × 10²³ atoms of oxygen
Ans. 72.5g of isopropanol contain 2.182 ×10²⁴ atoms of H and 7.274 × 10²³ atoms of O.

Question 62.
Calculate the number of moles and molecules of ammonia (NH₃) gas in a volume 67.2 dm³ of it measured at STP.
Solution:
Given: Volume of ammonia at STP = 67.2 dm³
To find: Number of moles and molecules of ammonia
Formulae: i. Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
ii. Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Number of moles (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Number of moles of NH₃ = \(\frac{67.2 \mathrm{dm}^{3}}{22.4 \mathrm{dm}^{3} \mathrm{~mol}^{-1}}\)
Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
3.0 mol × 6022 × 10²³ molecules mol^-1
= 18.066 × 10²³ molecules
Ans: Number of moles of ammonia = 3.0 mol
Number of molecules of ammonia = 18.066 × 10²³ molecules

Question 63.
3.40 g of ammonia at STP occupies volume of 4.48 dm³. Calculate molar mass of ammonia.
Solution:
Given: Mass of ammonia = 3.40 g
Volume at STP = 4.48 dm³
To Find: Molar mass of ammonia
Calculation: Let ‘x’ grams be the molar mass of NH₃.
Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Volume occupied by 3.40 g of NH₃ at S.T.P = 4.48 dm³
Volume occupied by ‘x’ g of NH3 at S.T.P = 22.4 dm³
∴ x = \(\frac{22.4 \times 3.40}{4.48}\) = 17.0 g mol^-1
Ans: Molar mass of ammonia is 17.0 g mol^-1.

Question 64.
Veg puffs from a particular bakery have an average mass of 27.0 g, whereas egg puffs from the same bakery have an average mass of 40 g.
i. Suppose a person buys 1 kg of veg puff from the bakery. Calculate the number of veg puffs he receives.
ii. Determine the mass of egg puffs (in kg) that will contain the same number of eggs puffs as in one kilogram of veg puffs.
Solution:
i. Mass of a veg puff = 27.0 g = 0.027 kg
∴ Number of veg puffs in 1 kg = 1 / 0.027 = 37
ii. One kilogram of veg puffs contains 37 veg puffs.
Mass of 37 egg puffs = 37 × 0.040 = 1.48 kg
Ans: i. 37 veg puffs in 1 kg of puff.
ii. Mass of 37 egg puffs is 1.48 kg

Multiple Choice Questions:

1. The branch of chemistry which deals with carbon compounds is called ……………. chemistry.
(A) organic
(B) inorganic
(C) carbon
(D) bio
Answer:
(A) organic

2. A/an is a simple combination of two or more substances in which the constituent substances retain their separate identities.
(A) compound
(B) mixture
(C) element
(D) All of these
Answer:
(B) mixture

3. Which one of the following is NOT a mixture?
(A) Paint
(B) Gasoline
(C) Liquefied Petroleum Gas (LPG)
(D) Distilled water
Answer:
(D) Distilled water

4. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with ………………
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of reciprocal proportion
Answer:
(C) law of conservation of mass

5. A sample of calcium carbonate (CaCO₃) has the following percentage composition: Ca = 40 %; C = 12 %; O = 48 %. If the law of definite proportions is true, then the weight of calcium in 4 g of a sample of calcium
carbonate from another source will be ……………..
(A) 0.016 g
(B) 0.16 g
(C) 1.6 g
(D) 16 g
Answer:
(C) 1.6 g

6. Two elements, A and B, combine to form two compounds in which ‘a’ g of A combines with ‘b₁’ and ‘b₂’g of B respectively. According to law of multiple proportion ………………
(A) b₁ = b₂
(B) b₁ and b₂ bear a simple whole number ratio
(C) a and b₁ bear a whole number ratio
(D) no relation exists between b₁ and b₂
Answer:
(B) b₁ and b₂ bear a simple whole number ratio

7. At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour. This is in accordance with ……………….
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of gaseous volumes
Answer:
(D) law of gaseous volumes

8. One mole of oxygen molecule weighs …………….
(A) 8 g
(B) 32 g
(C) 16 g
(D) 6.022 × 10²³ g
Answer:
(B) 32 g

9. The mass of 0.002 mol of glucose (C₆H₁₂O₆) is ………………
(A) 0.20 g
(B) 0.36 g
(C) 0.50 g
(D) 1.80 g
Answer:
(B) 0.36 g

10. Which of the following is CORRECT?
(A) 1 mole of oxygen atoms contains 6.0221367 × 10²³ atoms of oxygen.
(B) 1 mole of water molecules contains 6.0221367 × 10²³ molecules of water.
(C) 1 mole of sodium chloride contains 6.0221367 × 10²³ formula units of NaCl.
(D) All of these
Answer:
(D) All of these

11. 180 g of glucose (C₆H₁₂O₆) contains ……………. carbon atoms.
(A) 1.8 × 10²³
(B) 1.8 × 10²⁴
(C) 3.6 × 10²³
(D) 3.6 × 10²⁴
Answer:
(C) 3.6 × 10²³

12. The number of molecules present in 8 g of oxygen gas is …………….
(A) 6.022 × 10²³
(B) 3.011 × 10²³
(C) 12.044 × 10²³
(D) 1.505 × 10²³
Answer:
(D) 1.505 × 10²³

13. The number of molecules in 22.4 cm³of ozone gas at STP is ……………….
(A) 6.022 × 10²⁰
(B) 6.022 × 10²³
(C) 22.4 × 10²⁰
(D) 22.4 × 10²³
Answer:
(A) 6.022 × 10²⁰

14. 11.2 cm³ of hydrogen gas at STP, contains …………….. moles.
(A) 0.0005
(B) 0.01
(C) 0.029
(D) 0.5
Answer:
(A) 0.0005

15. The mass of 224 mL of hydrogen gas at STP is
(A) 0.02 g
(B) 0.224 g
(C) 2.24 g
(D) 20.0 g
Answer:
(A) 0.02 g

16. 4.4 g of an unknown gas occupies 2.24 L of volume under STP conditions. The gas may be ………………
(A) CO₂
(B) CO
(C) O₂
(D) SO₂
Answer:
(A) CO₂

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 5 Subsidiary Books Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 5 Subsidiary Books

Objective Type Questions & Answers

1. Answer the following questions in one sentence.

Question 1.
Which transactions are recorded in the Return Outward Book?
Answer:
Return of goods purchased on credit to the suppliers is only recorded in the Return Outward Book.

Question 2.
Which transactions are recorded in the Return Inward Book?
Answer:
Goods sold on credit when returned by the customers due to certain reasons are only recorded in the Return Inward Book.

Question 3.
Who is a payee in case of cheque?
Answer:
A person to whom the amount of cheque is payable or paid is called a payee.

Question 4.
What is a dishonour of cheque?
Answer:
When the bank refuses to pay the amount of cheque on its presentation to the payee for certain reasons, the cheque is said to be dishonoured.

Question 5.
What is Pass Book?
Answer:
A bank passbook is a small booklet in which the details of all ledger entries in respect of banking transactions appearing in the books of the bank are entered for the knowledge of the account holder.

Question 6.
What is a Bank Book?
Answer:
A book maintained by businessmen to record only banking transactions is called Bank Book.

Question 7.
What do you mean by crossed cheque?
Answer:
A cheque on which two parallel transverse lines are drawn on its face at the left-hand top corner with or without any word is called crossed cheque.

Question 8.
What is meant by Bank Overdraft?
Answer:
The amount withdrawn by the account holder from his current account in excess of the balance standing in that account up to the specified limit is known as ‘Bank overdraft’.

Question 9.
What do you mean by analytical petty cash book?
Answer:
A petty cash book that had many sub-columns on the payment side for recording expenses that are repetitive in nature is called an analytical/columnar petty cash book.

Question 10.
What are opening entries?
Answer:
Entries passed at the beginning of the financial year to bring the assets and liabilities in the books of account are known as opening entries.

Question 11.
What are closing entries?
Answer:
Entries passed at the end of the financial year to close the accounts of expenses and incomes are called closing entries.

Question 12.
What are transfer entries?
Answer:
Entries passed to transfer the balance of one account to another account are known as transfer entries.

Question 13.
What are adjustment entries?
Answer:
The accounting entries are passed into the books of accounts to bring certain items which do not appear in the trial balance.
e.g. depreciation, further bad debts, closing stock, incomes receivable, etc. are called adjustment entries.

Question 14.
What is a cheque?
Answer:
A cheque is a written order given by the account holder to the bank to pay the sum of money specified there into himself or to the person in whose favour the cheque is issued.

Question 15.
What is a ‘Pay-in slip’?
Answer:
Pay-in-slip is a document used by the account holder for depositing cash as well as cheque into the bank.

Question 16.
What is a ‘Debit Note’?
Answer:
A document that is prepared to give debit if less debit is given earlier or to cancel the wrong credit or extra credit given is called a debit note.

Question 17.
What is a ‘Credit Note’?
Answer:
A document that is prepared to give credit if less credit is given earlier or to cancel the wrong debit or extra debit given is called a credit note.

2. Give a word/term or phrase for each of the following statements:

Question 1.
Signing on the backside of the cheque to transfer its ownership.
Answer:
Endorsement of Cheque

Question 2.
An extract of customer’s account maintained by the bank.
Answer:
Pass Book

Question 3.
A type of cheque whose payment is done across the counter of the bank.
Answer:
Bearer Cheque

Question 4.
Bank on whom a cheque is drawn.
Answer:
Drawee

Question 5.
Document used for depositing cash or cheque into the bank.
Answer:
Pay-in-slip

Question 6.
Subsidiary book in which only credit sale of goods is recorded.
Answer:
Sales Book

Question 7.
Subsidiary book in which return of goods purchased on credit is recorded.
Answer:
Purchase Return

Question 8.
Source document on the basis of which Purchase Book is prepared.
Answer:
Inward Invoice

Question 9.
Source document on the basis of which Sales Book is prepared.
Answer:
Outward Invoice

Question 10.
Source document on the basis of which Purchase Return Book is prepared.
Answer:
Debit Note

Question 11.
Source document on the basis of which Sales Return Book is prepared.
Answer:
Credit Note

Question 12.
Accounting entries passed in the journal proper in the beginning of the financial year.
Answer:
Opening Entries

Question 13.
Accounting entries are passed in the journal proper at the end of the financial year.
Answer:
Closing Entries

Question 14.
Bank’s refusal to pay the amount of cheque to the payee.
Answer:
Dishonour of Cheque

Question 15.
Book in which small payments are recorded.
Answer:
Petty Cash Book

Question 16.
Type of bank account in which money is deposited for a fixed period of time.
Answer:
Fixed Deposit Account

Question 17.
A written order is issued by the account holder to the bank to pay a certain amount from his account.
Answer:
Cheque

Question 18.
A cheque on which two parallel transverse lines are drawn on its face at the left-hand top corner with or without certain words.
Answer:
Crossed Cheque

Question 19.
A person who endorses the cheque.
Answer:
Endorser

Question 20.
A person in whose favour the cheque is endorsed.
Answer:
Endorsee

3. State whether the following statements are True or False with reasons:

Question 1.
The total of the payment side can be greater than the total of the receipt side in the case of the bank column.
Answer:
This statement is True.
It is an overdraft on the current account.

Question 2.
A cheque received and deposited into the bank on the same day is a contra entry.
Answer:
This statement is False.
A cheque received and deposited on the same day is not a contra entry.

Question 3.
Interest on overdraft charged by the bank is recorded in the cash column.
Answer:
This statement is False.
Interest on overdraft charged by the bank is recorded in the bank column.

Question 4.
It is not necessary to record dishonour of cheques in the cash book.
Answer:
This statement is False.
It is necessary to record dishonour of cheques in the cash book.

Question 5.
Businessman generally operates savings bank account.
Answer:
This statement is False.
Businessman generally operates a current account with the bank.

4. Do you agree with the following statements.

Question 1.
Issue of Bearer cheque record on payments side of cash book in Bank Column.
Answer:
Agree

Question 2.
Subsidiary books are the books of secondary entry.
Answer:
Disagree

Question 3.
Cash Discount is recorded in the cash book with cash and Bank columns.
Answer:
Disagree

Question 4.
Credit sale of goods to Mr. Harsh ₹ 12,000 at 5% cash discount recorded on the receipts side of cash book.
Answer:
Disagree

Question 5.
All Business transactions are recorded in the cash book only.
Answer:
Disagree

Question 6.
Subsidiary books are useful for small trades only.
Answer:
Disagree

Question 7.
The big amount of payment of the organization is recorded in Petty Cash Book.
Answer:
Disagree

Question 8.
Sale of the old computer on credit recorded in the sales journal.
Answer:
Disagree

Question 9.
Cashbook cash column debit side is always greater than credit side.
Answer:
Disagree

Question 10.
When goods are returned by the buyer to the seller, Seller sends a credit note.
Answer:
Agree

5. Complete the following sentences:

Question 1.
Cash book is a ___________ book.
Answer:
Subsidiary

Question 2.
Only ___________ transactions are recorded in Simple Cash Book.
Answer:
Cash

Question 3.
Purchase book always shows ___________ balance.
Answer:
Debit

Question 4.
Sales book shows ___________ balance.
Answer:
Credit

Question 5.
Cash column of cash book can never have a ___________ balance.
Answer:
Credit

Question 6.
The source document for recording sales book is ___________
Answer:
Outward Invoice

Question 7.
___________ columns of a cash book are never balanced.
Answer:
discount

Question 8.
Credit purchase of goods are recorded in ___________
Answer:
purchase book

Question 9.
A person who draws a cheque is called ___________
Answer:
drawer

Question 10.
All expenses are recorded on the ___________ side of Cashbook.
Answer:
payment

Question 11.
When a bank refuses to make the payment of the cheque, it is said to be ___________
Answer:
dishonoured

Question 12.
The person who keeps a record of small payments is known as ___________
Answer:
petty cashier

Question 13.
In ___________ small cash payment of expenses are recorded.
Answer:
petty cash book

Question 14.
In ___________ petty cash book, on payment side there are many sub columns.
Answer:
analytical

Question 15.
___________ advance is given to the petty cashier.
Answer:
Petty

Question 16.
___________ means fixed amount of petty advance.
Answer:
Imprest

Question 17.
___________ petty cash book is not popular in the business world.
Answer:
Simple

Question 18.
Petty cash book is balanced at the end of every ___________
Answer:
month

Question 19.
Only ___________ discount is recorded in the books of accounts.
Answer:
cash

Question 20.
Cash book is book of ___________ entry.
Answer:
prime

Question 21.
___________ book is an extract of customer’s account in the ledger of bank.
Answer:
pass

Question 22.
The debit side of the passbook is called ___________ side.
Answer:
receipt

Question 23.
To draw 2 parallel lines on the face of the cheque is called ___________ of the cheque.
Answer:
crossing

Question 24.
Discount received is recorded on ___________ side of cash book.
Answer:
credit

Question 25.
The documents from which the returns to suppliers are recorded is known as ___________
Answer:
debit note

Question 26.
Sales Return book is also called as ___________
Answer:
return inward book

Question 27.
Purchase of goods on credit basis are recorded in ___________
Answer:
purchase book

Question 28.
Account in which there is restriction on withdrawals are called ___________ account.
Answer:
savings

Question 29.
___________ Entry is recorded on both the sides of cash with and bank columns.
Answer:
Contra

Question 30.
In sale journal goods sold on ___________ are recorded.
Answer:
Credit

Question 31.
Machinery purchase on credit will be recorded in ___________ subsidiary book.
Answer:
Journal Proper

Question 32.
Return of goods sold on credit by the customers will be entered in ___________ journal.
Answer:
Sales Return

Question 33.
Cash discount is recorded in ___________ book.
Answer:
Journal Proper

Question 34.
___________ Note is prepared when goods are returned to supplier.
Answer:
Debit

Question 35.
___________ is used to deposit cash on cheque in the bank.
Answer:
Pay-in-slip

Question 36.
___________ Cheque is more safer for payments.
Answer:
Cross Account Payee

6. Correct the following sentences and rewrite them same.

Question 1.
Credit sale of goods recorded in cash book receipts side.
Answer:
Cash sale of goods recorded in cash book receipts side.

Question 2.
Journal proper book is used to record cash transactions.
Answer:
Cashbook is used to record cash transactions.

Question 3.
Pay-in-slip is us to withdraw money from Bank.
Answer:
Pay-in-slip is us to deposit the money into the Bank.

Question 4.
The dividend received is a loss for the business.
Answer:
The dividend received is profit for the business.

7. Complete the following Table.

Question 1.

Answer:
62,500

Question 2.

Answer:
3,500

Question 3.

Answer:
2,05,000

Question 4.

Answer:
26,500

Question 5.

Answer:
1,25,000

Question 6.

Answer:
3,50,000

Question 7.

Answer:
4,50,000

Question 8.

Answer:
550

Question 9.

Answer:
1,990

Solved Problems

Simple/Single Column Cash Book

Question 1.
Prepare a simple cash book from the following information:
2019 January
1 Mr. Ashvin started the business with cash of ₹ 95,000.
3 Purchased furniture for office use ₹ 17,500.
4 Purchased goods worth ₹ 11,000.
7 Purchased machinery for ₹ 15,000.
10 Sold goods of ₹ 14,300 to Sandeep Traders for cash.
13 Received from Suyash ₹ 9,000.
15 Withdrew ₹ 2,500 from business for personal use.
19 Borrowed loan from Mr. Tilak ₹ 40,000.
22 Purchased goods of ₹ 14,000 at 5% trade discount.
26 Paid salary to staff ₹ 18,000.
29 Paid carriage on goods purchased ₹ 1,400.
30 Paid electricity bill ₹ 4,980.
31 Deposited into bank ₹ 7,000.
Solution:
In the books of Mr. Ashvin

Working Notes:
1. 22nd Jan.:
Net Purchase Price = Purchase Price – 5% Trade Discount
= 14,000 – 5% of 14,000
= 14,000 – 700
= ₹ 13,300

2. Balance of Cash on 31st Jan. = Total of Receipt side – Total of Payment side
= 1,58,300 – 90,680
= ₹ 67,620

Question 2.
Prepare a simple cash book of Mr. Shriram from the following details:
2019 Feb.
1 Balance of cash ₹ 65,500.
2 Brought additional capital ₹ 12,000.
4 Received from Nilesh on account ₹ 3,800.
7 Purchased goods from United Ltd. ₹ 16,000 and paid half the amount immediately.
9 Received commission from Aarti Traders ₹ 2,650.
10 Paid to Prithviraj Traders on account ₹ 17,800.
13 Purchased stationery for office use ₹ 2,460.
17 Received rent ₹ 4,900.
20 Paid ₹ 1,750 for insurance premium.
22 Sold goods for cash ₹ 13,800.
25 Purchased 10% debentures of ₹ 20,000
27 Paid telephone bill ₹ 3,000 and electricity bill ₹ 2,340.
28 Deposited into the bank all cash in excess of ₹ 11,000.
Solution:
In the books of Mr. Shriram

Working Note:
Dates: 28th February 2011:
Excess Cash Deposited = Total of Receipt Side – Total of Payment side including Closing Cash balance
= 1,02,650 – 66,350
= ₹ 36,300

Two Column Cash Book

Question 3.
Enter the following transaction in a cash book with cash and bank columns for the month of April 2019:
2019 April
1 Started business with cash ₹ 30,000.
2 Opened a current account with the bank and deposited ₹ 9,000.
5 Received ₹ 4,850 for Cash Sales and discount allowed ₹ 50.
6 Purchased goods of ₹ 3,000 @ 10% cash discount and half the amount was paid by cash and remaining by cheque.
9 Received a crossed cheque for ₹ 7,775 from Megha Associates and discount allowed ₹ 25.
12 Purchased machinery of ₹ 13,300 and paid for installation ₹ 700.
15 Received a bearer cheque of ₹ 735 from Rahul in full settlement of his account ₹ 800.
17 Deposited into the bank ₹ 2,000.
20 Transferred ₹ 8,000 from Savings Account to Current Account.
22 Received a hearer cheque from Vinod ₹ 2,230.
24 Deposited the cheque received from Vinod into the bank.
29 Bank informed that cheque received from Vinod is dishonoured.
30 Paid for printing bill book ₹ 660.
31 Withdrew from the bank for office use ₹ 1,000.
Solution:
In the books of ___________

Working Notes:
1. Transaction dated 6th April 2019.
Cash Discount = 10% of Goods Purchased
= 10% of 3,000
= ₹ 300
Net Cash Payable = 3,000 – 300 = ₹ 2,700
Amount paid in cash = \(\frac{1}{2}\) of 2,700 = ₹ 1,350
Amount paid by cheque = \(\frac{1}{2}\) of 2,700 = ₹ 1,350

2. Dated: 12th April, 2019.
Cost of Machinery = Purchase Price + Installation Charges
= 13,300 + 700
= ₹ 14,000

3. Transaction dated: 29th April, 2019.
Entry for dishonour of Vinod’s Cheque:
Earlier Entries:

Question 4.
Enter the following transaction of Hariram Bros, in a cash book with cash and bank column for the month of May 2019.
2019 May
1 Cash Balance ₹ 33,000
Bank Balance ₹ 8,000
3 Received from Prerna cash ₹ 2,750 and a bearer cheque for ₹ 645.
6 Paid to Arjun ₹ 5,300 by cheque and discount received ₹ 125.
8 Cash sales ₹ 5,645 and discount allowed ₹ 55.
10 Cheque received on 3 May 2011 deposited into the bank for collection.
12 Deposited into bank ₹ 4,000.
15 Cheque received from Prerna returned dishonoured.
18 Purchased goods from Indrajit ₹ 4,000 at an 8% trade discount and paid half the amount immediately.
20 Bank paid insurance premium under our standing instruction ₹ 1,950 and collected interest on investment ₹ 3,650.
22 Cheque issued to Arjun was dishonoured.
24 Lucky Stores directly deposited into our bank ₹ 9,500.
25 Paid salaries by cheque ₹ 5,700.
26 Withdrew by cheque ₹ 2,000 for office use and ₹ 1,500 for personal use.
31 Deposited into the bank all cash in excess of ₹ 5,000.
Solution:
In the books of Hariram Bros.

Working Notes:
1. Transaction dated 3rd May 2019:
Receipt of bearer cheque of ₹ 645 is debited to Cash A/c
Hence, Cash A/c is debited by 2,750 + 645 = ₹ 3,395

2. Transaction dated 10th May 2019:
Entry for deposit of cheque ₹ 645
Bank A/c ……………. Dr. 645
To Cash A/c 645

3. Transaction dated 15th May 2019:
Entry for dishonour of Prerna’s cheque.

When reverse entries are reconciled Cash A/c’s gets cancelled.
? Entry for dishonour of cheque is
Prerna’s A/c…………..Dr. 645
To Bank A/c 645

4. Transaction dated 18th May, 2019:
Net Price of goods purchased = Purchase Price – Trade discount
= 4,000 – 8% of 4,000
= 4,000 – 320
= ₹ 3,680
Amount Paid = \(\frac{1}{2}\) 3,680 = ₹ 1,840

5. Date: 31st May 2019:
Excess Cash deposited = Total of a debit column of cash – Total of credit column of cash including the closing balance
= 44,040 – 11,485
= ₹ 32,555

Question 5.
Enter the following transaction of Mr. Chavan in a cash book with cash and bank column for the month of April 2019.
2019 April
1 Cash Balance ₹ 56,000
Bank Overdraft ₹ 11,000
3 Purchased goods for ₹ 13,000 for cash at 2% cash discount and amount paid by cheque.
6 Received a bearer cheque for ₹ 13,250 in full settlement of ₹ 13,500 from Govind Traders.
9 Purchased 100 shares of Amar Ltd. of ₹ 100 each at ₹ 110 each and paid by cheque.
11 Sold goods of ₹ 7,000 at a 5% cash discount to Pramod and he paid half the amount immediately.
14 Deposited into bank ₹ 11,000.
17 Received a crossed cheque for ₹ 18,000 from Gajanan Traders.
20 Bank paid our telephone bill ₹ 3,230.
21 Bank charged ₹ 540 as interest on overdraft.
22 Paid by cheque to Urmila ₹ 8,000.
25 Deposited into the bank the cheque received from Govind Traders.
27 Received a bearer cheque for ₹ 3,460 for rent which was deposited into the bank.
29 Bank informed that the cheque received for rent was dishonoured.
30 Paid life insurance premium of Mr. Chavan ₹ 4,250 by cash and electricity bill ₹ 7,400 by cheque.
Solution:
In the books of Mr. Chavan

Working Notes:
1. Transaction dated 3rd April 2019:
Net payment made on cash purchases = Purchase Price – Cash discount
= 13,000 – 2% on 13,000
= 13,000 – 260
= ₹ 12,740

2. Transaction dated 9th April 2019:
Amount paid by cheque for purchase of shares = No. of shares purchased × Market value
= 100 × 110
= ₹ 11,000

3. Closing balance of Cash book
Cash balance = 72,575 – 28,500 = ₹ 44,075
Bank overdraft balance = 57,370 (Cr. column) – 45,710 (Dr. column)
Overdraft = ₹ 11,660

Petty Cash Book

Question 6.
Enter the Following transaction in a petty cash book having analysis columns for the month of October 2019.
2019 Oct.
1 Received cash from head cashier ₹ 1,350.
3 Paid packing charges ₹ 102.
6 Paid for postage ₹ 43.
9 Purchased 3 office files of ₹ 25 each.
13 Gave a tip to watchman ₹ 20.
18 Gave advance to clerk Mr. Shrikant ₹ 280.
22 Paid for printing ₹ 162.
25 Paid for advertisement ₹ 274.
27 Paid cleaning and washing charges ₹ 46.
29 Gave donation for Diwali celebration ₹ 150.
31 Purchased revenue stamps ₹ 65.
Solution:
Analytical Petty Cash Book of ___________

Question 7.
Enter the following transaction in an analytical petty cash book under the imprest system for the month of February 2019.
2019 Feb.
1 Received cheque from Head Cashier ₹ 2,700.
4 Purchased pencils for ₹ 45, inkpot ₹ 25, and papers ₹ 55.
6 Paid for repairs ₹ 190.
10 Paid taxi fare to Manager ₹ 168.
13 Paid subscription for newspaper ₹ 212.
17 Paid for refreshments to customers ₹ 175.
19 Paid to Ranjeet in settlement of his account ₹ 230.
21 Paid wages to casual labour ₹ 220.
24 Paid electricity bill ₹ 525.
26 Paid for carriage ₹ 105.
27 Purchase a wooden chair for ₹ 275.
Solution:
Analytical Petty Cash Book of ___________

Question 8.
Mr. Shrinath maintains a columnar petty cash book on the Imprest system. The Imprest amount is ₹ 850. The following information, show how his Petty Cash Book would appear for the week ended 7th September. 2019.
2019 Sept.
1 Balance in hand ₹ 125.
Postage ₹ 28.
Stationery ₹ 35.
Refreshment expenses ₹ 20.
3 Travelling and Conveyance ₹ 34.
Miscellaneous Expenses ₹ 5.
Refreshment expenses ₹ 25.
4 Repairs charges ₹ 170.
Paid for Postage ₹ 21.
5 Refreshment expenses ₹ 22.
Travelling expenses ₹ 32.
Stationery ₹ 47.
6 Refreshment expenses ₹ 19.
7 Miscellaneous Expenses ₹ 12.
Paid for Postage ₹ 7.
Repair charges ₹ 75.
Solution:
Analytical Petty Cash Book of Mr. Shrinath

Purchase Book, Sales Book, Purchase Return Book, and Sales Return Book

Question 9.
Prepare Purchase Book and Sales Book, M/s. S.K. Traders dealer in ready-made garments, Mumbai.
2019 April
1 Purchased goods from M/s. Jalna Traders, Delhi for ₹ 10,000 at 10% T.D.
4 Sold goods to M/s Kalpana Stores, Ahmedabad for ₹ 15,000 at 5% T.D.
10 Bought goods from M/s. Raj Enterprise, Patna for ₹ 5,000.
15 Sold goods to M/s Sagar Traders, Kanpur for ₹ 20,000 at 20% T.D.
25 Bought goods from M/s. Rakesh Stores, Banglore for ₹ 12,000 @ 5% T.D.
Rate of GST applicable on ready made garments are CGST @ 9%, SGST @ 9% and IGST @ 18%.
Solution:
In the books of M/s. S.K. Traders, Mumbai
Purchase Book (Analytical)

Sales Book (Analytical)

Note: In the case of inter-state sales IGST is used.

Journal Proper

Question 10.
Journalise the following transactions in the Journal Proper of Mr. Mahadev.
1 Pass opening entries: Debtors ₹ 20,000; Creditors ₹ 30,000; Computer ₹ 50,000 and Capital ₹ 40,000.
2 Pass closing entries: Purchases ₹ 10,000; Sales ₹ 25,000; Salaries ₹ 20,000 and Carriage ₹ 2,000.
3 Pass adjustment entries: Outstanding Wages ₹ 1,000 and Prepaid Insurance ₹ 500.
4 Pass transfer entries: Gross Profit ₹ 15,000; Net loss ₹ 2,000 and Drawings ₹ 4,000
5 Bought Machinery on credit from M/s. Tech ltd. for ₹ 50,000.
6 Allowed cash discount of ₹ 1,000 to Ajay.
Solution:
In the books of Mr. Mahadev
Journal Proper

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 8 Plant Tissues and Anatomy Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 8 Plant Tissues and Anatomy

Question 1.
How plant tissues are classified on the basis of their ability to divide?
Answer:
Plant tissues are classified into meristematic tissues and permanent tissues based on their ability to divide.

Question 2.
Identify the labels i, ii and iii in the given figure of meristematic tissue and write its characteristics.

Answer:
1. Cell wall
2. Nucleus
3. Cytoplasm
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Question 3.
With the help of neat and labelled diagram explain the classification of meristematic tissue based on its position.
Answer:
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 4.
Complete the given table representing types of meristematic tissue based on its function.
Answer:

Types of meristematic tissue	Function
1. Protoderm	It is found in young growing region of a plant forming a protective covering like epidermis around the various organs.
2. Procambium	It is involved in developing primary vascular tissue.
3. Ground meristem	It forms structures like cortex, endodermis, pericycle, medullary rays, pith.

Question 5.
Which are the simple permanent tissues in plants?
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 6.
Complete the given chart by giving characteristics of following tissues:
Answer:

Question 7.
Name the type of tissue in the given figure, identify labels ‘a’ and ‘b’ and write its characteristics.

Answer:
1. The given figure represents simple permanent tissue i.e. Parenchyma.
2. a: Vacuole, b: Intercellular air spaces.

Characteristics of parenchyma: Parenchyma:

1. It is a type of simple permanent tissue.
2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
3. Cell wall is composed of cellulose.
4. Cells are living with prominent nucleus and cytoplasm with large vacuole.
5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
7. This is less specialized permanent tissue.
8. Occurrence:
   These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
9. Functions:
   These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
    secondary growth.

Question 8.
Identify the type of tissue shown in the given figure and write its characteristics.

Answer:
The given figure represents Collenchyma tissue.
Characteristics of Collenchyma:

1. It is a simple permanent tissue made up of living cells.
2. The cell wall is cellulosic but shows uneven deposition of cellulose and pectin especially at comers.
3. The walls may show presence of pits.
4. Cells are similar like parenchyma, containing cytoplasm, nucleus and vacuoles but small in size and without intercellular spaces. Thus, the cells appear to be compactly packed.
5. The cells are either circular, oval or angular in transverse section.

Function:
Collenchyma is a living mechanical tissue and serves different functions in plants.
a. It gives mechanical strength to young stem and parts like petiole of leaf.
b. It allows bending and pulling action in plant parts and also prevents tearing of leaf.
c. It also allows growth and elongation of organs.
d. Collenchyma is usually absent in monocots and roots of dicot plant.

Question 9.
With the help of neat and labelled diagrams explain the Sclerenchyma Tissue.

Answer:
Sclerenchyma Tissue:

1. It is simple permanent tissue made up of compactly arranged thick walled dead cells.
2. The cells are living at the time of production but at maturity they become dead.
3. Cells are devoid of cytoplasm.
4. Their walls are thickened due to uniform deposition of lignin.
5. Cells remain interconnected through several pits.

Types of Sclerenchyma:
Sclerenchyma cells are categorized into two types on the basis of their size and shape as
fibres and sclereids:
a. Fibres:
Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique. They provide mechanical strength.

b. Sclereids:
Sclereids are usually broad, with blunt end walls.
These occur singly or in loose groups and their pits are deep branched and straight.
These are developed due to secondary thickening of parenchyma cells and provides stiffness only.

Functions:
a. This tissue functions as the main mechanical tissue.
b. It permits bending, shearing and pulling.
c. It gives rigidity to leaves and prevents it from falling.
d. It also gives rigidity to epicarps and seeds.

Question 10.
Give a brief account of water-conducting tissues in higher plants.
Answer:

1. Xylem is the water-conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

2. Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

3. Vessels:
a. Vessels are longer than tracheids with perforated or dissolved ends and formed by union of several vessels end to end.
b. These are involved in conduction of water and minerals.
c. Their lumen is wider than tracheids and the thickening is due to lignin and similar to tracheids.
d. In monocots, vessels are rounded where as they are angular in dicot angiosperms.
e. The first formed xylem vessels (protoxylem) are small and have either annular or spiral thickenings while latter formed xylem vessels are larger (metaxylem) and have reticulate or pitted thickenings.
f. When protoxylem is arranged towards pith and metaxylem towards periphery it is called as endarch
e. g. in stem and when the position is reversed as in the roots is called as exarch.

4. Xylem parenchyma:
a. Xylem parenchyma cells are small associated with tracheids and vessels.
b. This is the only living tissue among this complex tissue.
c. The function is to store food (starch) and sometimes tannins.
d. Xylem parenchyma are involved in lateral or radial conduction of water or sap.

5. Xylem fibres:
a. Xylem fibres are sclerenchymatous cells and serve mainly mechanical support. These are called wood fibres.
b. These are also elongated, narrow and spindle shaped.
c. Cells are tapering at both the ends and their walls are lignified.

Question 11.
Draw neat and labelled diagram of xylem tissue and vascular bundle.
Answer:

Question 12.
Match the following.

Column I	Column II
1. Protoxylem	(a) Xylem with larger vessels
2. Endarch Xylem	(b) Protoxylem arranged towards pith
3. Metaxylem	(c) Metaxylem arranged towards pith
4. Exarch xylem	(d) First formed xylem vessels

Answer:

Column I	Column II
1. Protoxylem	(d) First formed xylem vessels
2. Endarch Xylem	(b) Protoxylem arranged towards pith
3. Metaxylem	(a) Xylem with larger vessels
4. Exarch xylem	(c) Metaxylem arranged towards pith

Question 13.
Describe the structure of phloem.

Answer:
Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Question 14.
Draw a diagram of phloem.
Answer:

Question 15.
Name the types of tissue systems in plants.
Answer:
The types of tissue systems in plants are epidermal tissue system, ground tissue system and vascular tissue system.

Question 16.
Write a short note on Epidermis.
Answer:
Epidermis:

1. It is the outermost protective cell layer made up of compactly arranged cells without intercellular spaces.
2. Cells show presence of central large vacuole, thin cytoplasm and a nucleus.
3. The outer side of the epidermis is often covered with a waxy thick layer called the cuticle which prevents the loss of water.
4. Root epidermis (Epiblema) has root hairs. These are unicellular, elongated and involved in absorption of sap from the soil.
5. In stem, epidermal hairs are called trichomes. These are generally multicellular, branched or unbranched, stiff or soft or even secretory. These help in preventing water loss due to transpiration.

Question 17.
Draw a diagram representing epidermal tissue system.
Answer:

Question 18.
Write a short note on Structure of stomata.
Answer:
Structure of stomata:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question 19.
Write the information related to diagrams given below.
Answer:
1. The given diagram represents stoma in dicot leaf.
[Note: We have given additional label of ‘chloroplast ’for better understanding of students]
2. Structure of stomata:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question 20.
Explain the term ground tissue.
Answer:
Ground tissue:

1. All the plant tissues excluding epidermal and vascular tissue is ground tissue.
2. It is made up of simple permanent tissue e.g. parenchyma.
3. It is present in cortex, pericycle, pith and medullary rays in the primary stem and root.
4. Collenchyma and sclerenchyma in the hypodermis and chloroplasts containing mesophyll tissue in leaves is also ground tissue.

Question 21.
Describe various types of vascular bundles.
Answer:
Vascular bundles occur in the form of distinct patches of the complex tissue viz. Xylem and Phloem. On the basis of their arrangement in the plant body they are classified as follows:
1. Radial vascular bundles:
When the complex tissues (xylem and phloem) are situated separately on separate radius as separate bundle, vascular bundle is called Radial vascular bundle. This is a common feature of roots.

2. Conjoint vascular bundles:
When the complex tissue (xylem and phloem) is collectively present as neighbours of each other on the same radius, vascular bundle is called Conjoint vascular bundle.
They are of two types:
a. Collateral vascular bundle:
In this type of vascular bundle, xylem lies inwards and the phloem lies outwards.
These bundles may be further of open type (secondary growth takes place) containing cambium in between xylem and phloem and closed type if cambium is not present (secondary growth absent).
b. Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

3. Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question 22.
Explain how formation of cambial ring occurs in dicot stem.
Answer:

1. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
2. With the onset of favourable season, meristematic cells of intrafascicular cambium become active.
3. Simultaneously, the ray parenchyma cells, both fusiform initials and ray initials become meristematic. This is known as dedifferentiation.
4. These form patch of cambial cells (meristematic cells) in between the adjacent bundles and produce interfascicular cambium.
5. Both intrafascicular and interfascicular cambium join and form a complete ring, known as the cambial ring. This is possible because they lie in one plane.

Question 23.
‘Secondary growth is observed in most of the dicot and gymnospermic root.’ State whether the given statement is true or false and justify your answer.
Answer:

1. The given statement is true.
2. Secondary growth is observed in most of the dicot and gymnospermic root by producing secondary vascular tissue and periderm.
3. Secondary growth is produced by vascular cambium and cork cambium respectively.
4. Conjunctive parenchyma cells present on the inner edges of primary phloem bundles become meristematic.
5. These cells add secondary xylem and secondary phloem on the inner and outer side respectively which results in secondary growth.

Question 24.
Differentiate between heartwood and sap wood.
Answer:

Heartwood	Sap wood
1. It is central region of secondary xylem (wood).	It is the peripheral region of secondary xylem (wood).
2. It is darker in colour due to deposition of oils, gums, resins, tannins, etc	It is lighter in colour and without any depositions.
3. It is non- functional part of secondary xylem.	It is functional part of secondary xylem.
4. It is resistant to pathogens.	It is more susceptible to pathogens
5. It is not involved in conduction of sap.	It is involved in conduction of sap.
6. It is also called as duramen.	It is also called as alburnum.

Question 25.
What are tyloses?
Answer:
Tyloses:
1. Tracheary elements of heartwood are plugged by in-growth of adjacent parenchyma cells are known as tyloses.
2. Tyloses are fdled by oils, gums, resins, tannins called as extractives.

Question 26.
Explain how periderm is formed?
Answer:
Formation of periderm:
As the stem increase in diameter due to activity of vascular cambium, the outer cortical and epidermal layer get ruptured. Thus, it becomes necessaiy to replace these cells by new cells.

1. Phellogen (cork cambium) develops in extrastelar region (cortex region) of the stem.
2. The outer cortical cells of cortex become meristematic and produce a layer of thin walled, rectangular cells. These cells cut off new cells on both sides.
3. The cells produced on outer side develop phellem (cork), whereas on the inner side produce phelloderm (secondary cortex).
4. The cork is impervious in nature and does not allow entry of water due to suberized walls. Secondary cortex is parenchymatous in nature.
5. Phellogen, phellem and phelloderm constitute periderm.

Question 27.
Explain the given terms:
1. Bark
2. Lenticels
3. Anomalous secondary growth
Answer:
1. Bark:
a. Bark is non-technical term referring to all cell types found external to vascular cambium including secondary phloem.
b. Bark of early season is soft and of the late season is hard.

2. Lenticels:
a. Lenticels are aerating pores present as raised scars on the surface of bark.
b. These are portions of periderm, where phellogen activity is more.
c. Lenticels are meant for gaseous and water vapour exchange.

3. Anomalous secondary growth:
a. Monocot stems lack cambium hence secondary growth does not take place.
b. However, accessory cambium development in plants like, Dracaena, Agave, Palms and root of sweet potato shows presence of secondary growth. This is called as anomalous secondary growth.

Question 28.
With the help of neat and labelled diagram explain the anatomy of dicot root.
Answer:

The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Question 29.
With the help of neat and labelled diagram explain the anatomy of monocot root.
Answer:
The transverse section of a typical monocotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
Endodermis: It is innermost layer of cortex. The cells of endodermis are thick walled except the passage cells which lie just opposite to the protoxylem.
Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Pericycle is present below the endodermis.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Polyarch condition of xylem is observed, (xylem bundles are more than six).
Pith: Pith is large and well developed.
Secondary growth does not occur due to absence of cambium.

Question 30.
What is polyarch condition of root?
Answer:
Polyarch condition is the one in which roots possess more than six xylem bundles.

Question 31.
Explain in detail anatomy of sunflower stem.
Answer:
A transverse section of sunflower (dicot) stem shows the following structures:
1. Epidermis: It is a single, outermost layer with multicellular outgrowth called trichomes. A layer of cuticle
is usually present towards the outer surface of epidermis.

2. Cortex: Cortex is situated below the epidermis and is usually differentiated into three regions namely, hypodermis, general cortex and endodermis.
a. Hypodermis: It is situated just below the epidermis and is made of 3-5 layers of collenchymatous cells. Intercellular spaces are absent.
b. General cortex: It is made up of several layers of large parenchymatous cells with intercellular spaces.
c. Endodermis: It is an innermost layer of cortex which is made up of barrel shaped cells. It is also called starch sheath, as it is rich in starch grain.

3. Stele: It is differentiated into pericycle, vascular bundles and pith.
a. Pericycle: It is the outermost layer of vascular system situated between the endodermis and vascular bundles. In sunflower, it is multi-layered and also called hard bast.
b. Vascular bundles: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
c. Pith: It is situated in the centre of the young stem and is made up of large-sized parenchymatous cells with conspicuous intercellular spaces.

Question 32.
With the help of neat and labelled diagram explain the anatomy of maize stem.
Answer:

A transverse section of maize (monocot) stem shows the following structures:

1. Epidermis: It is single layered and without trichomes.
2. Hypodermis: It is sclerenchymatous.
3. Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
4. Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
5. Pith: Pith is absent.

Question 33.
With the help of a neat and labelled diagram, describe the internal structure of dorsiventral leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question 34.
With the help of a neat labelled diagram, describe the anatomy of isobilateral leaf.
Answer:
The parts of isobilateral leaf are as follows:

1. Epidermis:
It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 35.
Compare between dorsiventral and isobilateral leaf.
Answer:

Dorsiventral leaf	Isobilateral leaf
1. Dorsiventral Leaf is very common in dicotyledonous plants.	Isobilateral leaf is very common in monocotyledonous plants.
2. In this mesophyll tissue is differentiated into palisade and spongy parenchyma.	In this mesophyll tissue is not differentiated into palisade and spongy parenchyma.
3. The leaves are commonly horizontal in orientation with distinct upper and lower surfaces. The upper surface which faces the sun is darker than the lower surface.	In this leaf both the surfaces are equally illuminated as both the surface can face the sun, and show similar structure. The two surfaces are equally green.
4. Stomata is absent on the upper surface of these leaves.	Stomata is present on both the upper and lower surfaces of these leaves.

Question 36.
Distinguish between anatomy of dicot root and monocot root.
Answer:

Anatomy of dicot root	Anatomy of monocot root
1. Pith is narrow.	Pith is large and well developed.
2. Diarch, triarch or tetrarch condition can be observed. (Xylem bundles vary from two to six number)	Polyarch condition is observed, (xylem bundles are more than six)
3. Cambium is formed in later stage between xylem and phloem which causes secondary growth.	Secondary growth is absent.

Question 37.
Distinguish between anatomy of dicot stem and monocot stem.
Answer:

Anatomy of Dicot stem	Anatomy of Monocot stem
1. Epidermis shows presence of multicellular trichomes.	Epidermis is without trichomes.
2. Hypodermis is made up of collenchymatous cells.	Hypodermis is made up of sclerenchymatous cells.
3. Medullary rays are present between vascular bundles.	Medullary rays are absent.
4. Vascular bundles are arranged in the form of a ring.	Vascular bundles are scattered in the ground tissue.
5. It is conjoint, collateral and open (Cambium present)	They are conjoint, collateral and closed (cambium is absent).
6. Vascular bundle is not surrounded by a sclerenchymatous bundle sheath.	Vascular bundle is surrounded by a sclerenchymatous bundle sheath.
7. Secondary growth takes place due to presence of cambium.	Secondary growth does not occur due to absence of cambium.
8. Pith is present.	Pith is absent.

Question 38.
Apply Your Knowledge

Question 1.
Which plant part would show the following:

1. Radial vascular bundles.
2. Large and well-developed pith.
3. Differentiation of mesophyll into palisade and spongy tissue.
4. Presence of stomata on both upper and lower epidermis.

Answer:

1. Root
2. Monocot root and Dicot stem,
3. Dicot leaf
4. Monocot leaf

Question 2.
When a tree is debarked, which tissues are removed?
Answer:
The bark is made up of tissues like cork, cork cambium and secondary cortex, which are removed when a tree is debarked.

Question 3.
While eating fruits like pear or guava, it feels gritty. What gives stiffness to these fruits?
Answer:
Sclereids are found in pulp of fruits like pear and guava which gives them stiffness and thus we feel gritty while eating these fruits.

Question 39.
Quick Review

Question 40.
Exercise

Question 1.
Define tissue.
Answer:
A group of cells having essentially a common function and origin is called as tissue.

Question 2.
Classify the meristematic tissue based on its origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Question 3.
Explain in detail classification of meristematic tissue based on its position.
Answer:
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 4.
Give any two examples of secondary meristematic tissue.
Answer:
Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Question 5.
Draw a diagram of meristematic cells.
Answer:
1. Cell wall
2. Nucleus
3. Cytoplasm
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Question 6.
Write a short note on tracheids.
Answer:
Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

Question 7.
Describe parenchyma in detail.
Answer:
Cell is the component that brings about important processes in the living organisms.

Question 8.
Describe the structure of xylem in detail.
Answer:
1. Xylem is the water conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

2. Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

3. Vessels:
a. Vessels are longer than tracheids with perforated or dissolved ends and formed by union of several vessels end to end.
b. These are involved in conduction of water and minerals.
c. Their lumen is wider than tracheids and the thickening is due to lignin and similar to tracheids.
d. In monocots, vessels are rounded where as they are angular in dicot angiosperms.
e. The first formed xylem vessels (protoxylem) are small and have either annular or spiral thickenings while latter formed xylem vessels are larger (metaxylem) and have reticulate or pitted thickenings.
f. When protoxylem is arranged towards pith and metaxylem towards periphery it is called as endarch
e. g. in stem and when the position is reversed as in the roots is called as exarch.

4. Xylem parenchyma:
a. Xylem parenchyma cells are small associated with tracheids and vessels.
b. This is the only living tissue among this complex tissue.
c. The function is to store food (starch) and sometimes tannins.
d. Xylem parenchyma are involved in lateral or radial conduction of water or sap.

5. Xylem fibres:
a. Xylem fibres are sclerenchymatous cells and serve mainly mechanical support. These are called wood fibres.
b. These are also elongated, narrow and spindle shaped.
c. Cells are tapering at both the ends and their walls are lignified.

Question 9.
What are Sclerenchyma fibres?
Answer:
a. Fibres:
Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique. They provide mechanical strength.

Question 10.
Write the functions of parenchyma cells.
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 11.
Write function of collenchyma tissue.
Answer:
Function:
Collenchyma is a living mechanical tissue and serves different functions in plants.
a. It gives mechanical strength to young stem and parts like petiole of leaf.
b. It allows bending and pulling action in plant parts and also prevents tearing of leaf.
c. It also allows growth and elongation of organs.

Question 12.
Which are the different types of tracheids based on the types of thickenings on their walls?
Answer:
The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

Question 13.
Death of companion cell causes death of sieve tube cells and vice versa. Justify.
Answer:
Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

Question 14.
Which are the three types of simple permanent tissues?
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 15.
Write the functions of sclerenchyma tissue.
Answer:
Functions:
a. This tissue functions as the main mechanical tissue.
b. It permits bending, shearing and pulling.
c. It gives rigidity to leaves and prevents it from falling.
d. It also gives rigidity to epicarps and seeds.

Question 16.
What are the components of xylem?
Answer:
1. Xylem is the water conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 17.
Name the living component of xylem.
Answer:
This is the only living tissue among this complex tissue.

Question 18.
Name the dead component of phloem.
Answer:
Phloem fibres are the only dead tissue among this unit.

Question 19.
What is closed vascular bundle?
Answer:
When cambium is not present between xylem and phloem, it is known as closed vascular bundle.

Question 20.
Describe two types of conjoint vascular bundles.
Answer:
Conjoint vascular bundles:
When the complex tissue (xylem and phloem) is collectively present as neighbours of each other on the same radius, vascular bundle is called Conjoint vascular bundle.
They are of two types:
a. Collateral vascular bundle:
In this type of vascular bundle, xylem lies inwards and the phloem lies outwards.
These bundles may be further of open type (secondary growth takes place) containing cambium in between xylem and phloem and closed type if cambium is not present (secondary growth absent).
b. Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 21.
Write the function of trichomes.
Answer:
In stem, epidermal hairs are called trichomes. These are generally multicellular, branched or unbranched, stiff or soft or even secretory. These help in preventing water loss due to transpiration.

Question 22.
What are bicollateral vascular bundle?
Answer:
Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 23.
Name the tissue that are not included in ground tissue.
Answer:
All the plant tissues excluding epidermal and vascular tissue is ground tissue.

Question 24.
Which type of conjoint – vascular bundles are found in members of Cucurbitaceae family?
Answer:
Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 25.
What is concentric vascular bundle?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question 26.
Define intrafascicular cambium.
Answer:
The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.

Question 27.
What is the difference between spring wood and autumn wood?
Answer:
During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

Question 28.
Explain how growth rings are formed in trees?
Answer:
1. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.
2. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.
3. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

Question 29.
Which tissues are together called as periderm?
Answer:
Phellogen, phellem and phelloderm constitute periderm.

Question 30.
What is bark?
Answer:
Bark:
a. Bark is non-technical term referring to all cell types found external to vascular cambium including secondary phloem.
b. Bark of early season is soft and of the late season is hard.

Question 31.
What is the function of lenticels?
Answer:
Lenticels are meant for gaseous and water vapour exchange.

Question 32.
Explain the term anomalous secondary growth.
Answer:
Anomalous secondary growth:
a. Monocot stems lack cambium hence secondary growth does not take place.
b. However, accessory cambium development in plants like, Dracaena, Agave, Palms and root of sweet potato shows presence of secondary growth. This is called as anomalous secondary growth.

Question 33.
Explain in detail anatomical structure of a dicot stem.
Answer:
A transverse section of sunflower (dicot) stem shows the following structures:
1. Epidermis: It is a single, outermost layer with multicellular outgrowth called trichomes. A layer of cuticle
is usually present towards the outer surface of epidermis.

2. Cortex: Cortex is situated below the epidermis and is usually differentiated into three regions namely, hypodermis, general cortex and endodermis.
a. Hypodermis: It is situated just below the epidermis and is made of 3-5 layers of collenchymatous cells. Intercellular spaces are absent.
b. General cortex: It is made up of several layers of large parenchymatous cells with intercellular spaces.
c. Endodermis: It is an innermost layer of cortex which is made up of barrel shaped cells. It is also called starch sheath, as it is rich in starch grain.

3. Stele: It is differentiated into pericycle, vascular bundles and pith.
a. Pericycle: It is the outermost layer of vascular system situated between the endodermis and vascular bundles. In sunflower, it is multi-layered and also called hard bast.
b. Vascular bundles: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
c. Pith: It is situated in the centre of the young stem and is made up of large-sized parenchymatous cells with conspicuous intercellular spaces.

Question 34.
Draw neat and labelled diagrams of dicot and monocot root and differentiate between their anatomical characters.
Answer:
The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Question 35.
Which type of vascular bundles are observed in isobilateral leaf?
Answer:
Vascular bundle:
These are conjoint, collateral and closed.

Question 36.
Describe the internal structure of a leaf in which mesophyll is differentiated in palisade and spongy parenchyma.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question 37.
Multiple Choice Questions:

Question 1.
Meristematic tissues are found
(A) only in stems of the plants
(B) in both roots and stems
(C) in all growing tips of the plant body
(D) only in roots of the plants
Answer:
(C) in all growing tips of the plant body

Question 2.
The tissue responsible for translocation of food material is _________
(A) xylem
(B) cambium
(C) parenchyma
(D) phloem
Answer:
(D) phloem

Question 3.
________ are used in making ropes and rough clothes.
(A) Phloem parenchyma
(B) Trachieds
(C) Phloem fibres
(D) Sieve tube elements
Answer:
(C) Phloem fibres

Question 4.
_______ are the only dead tissue among the phloem.
(A) Phloem parenchyma
(B) Sieve tubes
(C) Companion cells
(D) Phloem fibres
Answer:
(D) Phloem fibres

Question 5.
Phloem was named as _______ by Haberlandt as similar to xylem.
(A) Bast
(B) Leptome
(C) Wood fibres
(D) Casparian
Answer:
(B) Leptome

Question 6.
The sieve tube cell is connected to companion cell through phloem parenchyma by
(A) Plasmodesmata
(B) Interfascicular cambium
(C) Pericycle
(D) Hypodermis
Answer:
(A) Plasmodesmata

Question 7.
Which of the following tissues is with dead thick-walled cells without intercellular spaces?
(A) parenchyma
(B) collenchyma
(C) sclerenchyma
(D) phloem
Answer:
(C) sclerenchyma

Question 8.
The tissue which is present in between xylem and phloem of stem is called
(A) apical meristem
(B) pericycle
(C) vascular cambium
(D) cork cambium
Answer:
(C) vascular cambium

Question 9.
In stem, epidermal hairs are called as
(A) Cuticles
(B) Casparian strip
(C) Trichomes
(D) Companion cells
Answer:
(C) Trichomes

Question 10.
_______ forms the outer covering of plant body and is derived from protodenn or dermatogen.
(A) Ground tissue system
(B) Interfascicular cambium
(C) Vascular tissue system
(D) Epidermal tissue system
Answer:
(D) Epidermal tissue system

Question 11.
_______ play a vital role in exchange of gases and water vapour.
(A) Vascular bundles
(B) Stomata
(C) Ground tissues
(D) Trichomes
Answer:
(B) Stomata

Question 12.
In which of the following leaf possesses dumbbell shaped guard cell’?
(A) Pisum sativum
(B) Wheat
(C) Datura
(D) Sunflower
Answer:
(B) Wheat

Question 13.
Which of the following is NOT a characteristic of spring wood?
(A) Tracheids with wide lumen
(B) Less number of fibres
(C) Narrow xylem band
(D) Lighter colour
Answer:
(C) Narrow xylem band

Question 14.
Periderm consists of
(A) Phellogen
(B) Phellem
(C) Phelloderm
(D) All of these
Answer:
(D) All of these

Question 15.
Which of the following is essential for secondary growth?
(A) Xylem
(B) Pith
(C) Phloem
(D) Cambium
Answer:
(D) Cambium

Question 16.
Vascular bundles of dicot root are
(A) radial exarch
(B) radial endarch
(C) conjoint exarch
(D) conjoint endarch
Answer:
(A) radial exarch

Question 17.
In which of the following characters, a monocot root differs from dicot root?
(A) Open vascular bundle
(B) Large pith
(C) Radial vascular bundles
(D) Scattered vascular bundles
Answer:
(B) Large pith

Question 18.
Which of the following plant shows isobilateral leaves?
(A) Hibiscus
(B) Maize
(C) Mangifera indica
(D) Sunflower
Answer:
(B) Maize

Question 19.
Secondary growth does not occur in
(A) Maize stem
(B) Mango leaf
(C) Carina root
(D) All of these
Answer:
(D) All of these

Question 20.
Stele of a dicot stem consists of all the given below, EXCEPT
(A) Pericycle
(B) Cortex
(C) Vascular bundles
(D) Pith
Answer:
(B) Cortex

Question 21.
Hypodermis is collenchymatous in
(A) monocot stem
(B) dicot stem
(C) monocot root
(D) both (A) and (B)
Answer:
(B) dicot stem

Question 22.
Lysigenous cavity filled with water is present in
(A) dicot stem
(B) monocot stem
(C) monocot root
(D) dicot root
Answer:
(B) monocot stem

Question 23.
The vascular bundles in a dicot stem are
(A) collateral and open
(B) radial
(C) bicollateral and open
(D) collateral and closed
Answer:
(A) collateral and open

Question 38.
Competitive Corner

Question 1.
Phloem in gymnosperms lacks:
(A) companion cells only
(B) both sieve tubes and companion cells
(C) albuminous cells and sieve cells
(D) sieve tubes only
Answer:
(B) both sieve tubes and companion cells

Question 2.
Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following:
(A) Shrinkage of air spaces in spongy mesophyll
(B) Tyloses in vessels
(C) Closure of stomata
(D) Flaccidity of bulliform cells
Hint: Grass leaves curl inwards to the minimize water loss.
Answer:
(D) Flaccidity of bulliform cells

Question 3.
Which of the statements given below is NOT true about formation of ‘annual rings’ in trees?
(A) Activity of cambium depends upon variation in climate.
(B) Annual rings are not prominent in trees of temperate region.
(C) Annual ring is a combination of spring wood and autumn wood produced in a year.
(D) Differential activity of cambium causes light and dark bands of tissue – early and late wood respectively.
Hint: Annual rings are formed due to activity of cambium. The activity of cambium is under the control of many physiological and environmental factors. In temperate regions, the climatic conditions are not uniform throughout the year due to this, annual rings are formed.
Answer:
(B) Annual rings are not prominent in trees of temperate region.

Question 4.
Regeneration of damaged growing grass following grazing is largely due to:
(A) secondary meristem
(B) lateral meristem
(C) apical meristem
(D) intercalary meristem
Hint: Intercalary meristems occur in grasses at the base of intemode, which regenerates the grass damaged due to grazing.
Answer:
(D) intercalary meristem

Question 5.
In the dicot root, the vascular cambium originates from:
(A) intrafascicular and interfascicular tissue in a ring
(B) tissue located below the phloem bundles and a portion of pericycle tissue above protoxylem
(C) cortical region
(D) parenchyma between endodermis and pericycle
Answer:
(B) tissue located below the phloem bundles and a portion of pericycle tissue above protoxylem

Question 6.
Casparian strips occur in
(A) Cortex
(B) Pericycle
(C) Epidermis
(D) Endodermis
Answer:
(D) Endodermis

Question 7.
Plants having little or no secondary growth are
(A) Conifers
(B) Deciduous angiosperms
(C) Grasses
(D) Cycads
Hint: Secondary growth takes place in stems and roots of dicotyledons and gymnosperms, but does not occur in monocotyledons.
Answer:
(C) Grasses

Question 8.
Secondary xylem and phloem in dicot stem are produced by
(A) Phellogen
(B) Vascular cambium
(C) Apical meristems
(D) Axillary meristems
Answer:
(B) Vascular cambium

Question 9.
The vascular cambium normally gives rise to
(A) Phelloderm
(B) Primary phloem
(C) Secondary xylem
(D) Periderm
Answer:
(C) Secondary xylem

Question 10.
Identify the wrong statement in context of heartwood.
(A) Organic compounds are deposited in it
(B) It is highly durable
(C) It conducts water and minerals efficiently
(D) It comprises dead elements with highly lignified walls
Hint: In old trees, secondary xylem (wood) becomes physiologically non active. It does not conduct water and becomes dark due to organic deposits (tannins, resins, oils, aromatic substances, etc.) It comprises of dead elements and called as heart wood. It is non conductive, hard, durable and resistant to microbes and insects.
Answer:
(C) It conducts water and minerals efficiently

Question 11.
Which of the following is made up of dead cell?
(A) Xylem parenchyma
(B) Collenchyma
(C) Phellem
(D) Phloem
Hint: Cork cambium (phellogen) cuts off cells on both the sides. The outer cells differentiate into cork or phellem. The cork is impervious to water due to suberin deposition in the cell wall.
Answer:
(C) Phellem

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry

Question 1.
Define the following:
a. Atom economy.
Answer:
Atom economy : Atom economy is a measure of the amount of atoms from the starting materials that are present
in the final product at the end of chemical process.

Question 2.
How will you prevent the generation of waste or by-products?
Answer:
To prevent generating waste, there is the need to develop the zero waste technology (ZWT). ZWT in a chemical synthesis should result in waste product being zero or minimum. To use the waste product of one system as the raw material for other system is also the aim of ZWT.

For example :

• Cement and brick industry can use the bottom ash of thermal power station as the raw material.
• Thermal power station can use the effluent coming out from cleansing of machinery parts as coolant water.

Question 3.
(1) Calculate the atom economy of the following:
(At mass of C = 12, 11 = 1 ,0 = 16)

Answer:

Formula weight of ethanol = 46
ethene 28
water= 18
% atom economy = \(\frac{28}{46}\) x 100 = 60.9%

(2) Calculate the atom economy of fermentation of sugar (glucose) to ethanol.

Answer:

Formula wt of glucose = 180
Formula Wi of ethanol =46
Relative massiwt of desired useful product in thc equation = 2 x 46 =92
% Atom economy = 92/180 x 100 = 51.1%

Question 4.
ExpLain less hazardous chemical synthesis with suitable example.
Answer:

1. To avoid formation of hazardous waste from chemical processes, the chemical reactions and synthesis routes should be designed to be as safe as possible.
2. Earlier used insecticide DDT (Dichlorodiphenyltrichloroethanc) was found to be harmful for human beings. So DDT has been replaced by benzenc hexachioride (BHC) as an insecticide, one of the y-isomer (gamma) of BHC is called gammexane or lindane.

Question 5.
How will you develop products that are less toxic or which require less toxic raw materials ?
Answer:

• There is a need to design safer chemicals to prevent the workers in chemical industries from being exposed to toxic environment.
• Adipic acid is extensively used in polymer industry. In synthesis of adipic acid, benzene is used as the starting material, but benzene is carcinogenic and being volatile organic compound (VOC) it pollutes the air and environment.
• To overcome this health hazard Green technology developed by Drath and Frost, adipic acid is enzymatically synthesised from glucose.

Question 6.
How to apply the principle of green chemistry to achieve energy efficiency?
Answer:

• Energy requirements during chemical synthesis is huge. To minimize the energy use it is better to carry out reactions at room temperature and pressure.
• This can be achieved by applying the principle of green chemistry i.e. use of catalyst, use of micro-organisms or biocatalyst and use of renewable materials, etc.
• The use of less energy can be achieved by improving the technology of heating system, use of microwave, etc.

Question 7.
Explain the use of renewable feed stocks.
Answer:

• Industries use a lot of non-renewable feed stocks like petroleum. These resources are depleting fast and the future generation will be deprived. The excessive use of these resources have also put a burden on the environment.
• If renewable resources like agricultural or biological products are used, this will ensure the sharing of resources by future generations. This practice will also not put a burden on the environment.
• The products and waste produced are generally biodegradable and environmental friendly hence leading to a sustainable future.

Question 8.
Explain the need to design degradable chemicals.
Answer:

• Environment protection is the prime concern which has lead to the need for designing chemicals that degrade and can be discarded easily. These chemicals and their degradation products should be non-toxic, non-bioaccumulative or should not be environmentally persistent.
• This principle aims at waste product being automatically degradable to clean the environment. Thus the preference for biodegradable polymers and pesticides.
• To make the separation and segregation easier for the consumer an international plastic recycle mark is printed on larger items.

Question 9.
Define the role of real time analysis in pollution prevention.
Answer:

• There is a dire need to develop improvised analytical methods to allow for real time, in process monitoring and control prior to the formation of hazardous substances.
• It is very much important for the chemical industries and nuclear reactors to develop or modify analytical
  methodologies so that continuous monitoring of the manufacturing and processing unit is possible.

Question 10.
Define the role of safer chemistry in accident prevention.
Answer:
(1) It is needed to develop chemical processes that are safer and minimize the risk of accidents. It is important to select chemical substances used in a chemical reaction in such a way that they can minimize the occurrence of chemical accidents, explosions, fire and emissions.

(2) For example : Chemical process that works with the gaseous substances can lead to relatively higher possibilities of accidents including explosion as compared to the system working with nonvolatile liquid and solid substances.

Question 11.
Green chemistry plays an important role in sustainable development. Explain.
Answer:
Sustainable development is a development that protects the environment and the world’s resources. We can achieve sustainable development by adapting the twelve principles of green chemistry.

Green chemistry designs safer chemicals which are less toxic. It normally leads to low cost, use of less energy, environmentally friendly solvents and less production of waste. Green chemistry works on the principle of atom economy and minimum or no waste production. It encourages the use of renewable feed stocks and reduces the use of toxic and hazardous chemicals. It eliminates majorly stoichiometry reactions and prefers to use catalysis. It preserves the environment and safety requirements with added benefit of cost reduction.

Question 12.
How are nanomaterials classified ?
Answer:
Nanoparticles, nanowires and nanotubes can be classified according to dimensions. The nano structured materials may be large organic molecules, inorganic cluster compounds and metallic or semiconductor particles.

Question 13.
What are zero, one and two dimensional nanoscale system ?
Answer:

• Zero-Dimensional Nanostructures : A zero-dimensional structure is one in which all three dimensions are in the nanoscale.
  For example : Nanoparticles.
• One-Dimensional Nanostructures : A one-dimensional nanostructure is one in which two dimensions are in the nanoscale. For example : Nanowires and Nano rods.
• Two-Dimensional Nanostructures : A two-dimensional nanostructure is one in which one dimension is in the nanoscale. For example : Thin films.

Question 14.
State the different characteristic features of nanoparticles.
Answer:
The nanoparticle science is special as at such a small scale, different laws dominate than what we experience in our everyday life.

The characteristic features like optical properties, catalytical activities, have huge surface area and good thermal properties mechanical strength electrical conductivity vary than that of bulk material.

(1) Colour : At nanoscale this optical property behaves differently. Elemental gold has nice shining yellow colour, but nanoparticles of gold show red colour.

(2) Catalytic activity : Since the surface area of nanoparticles is large they show increased catalytic activity. They are usually heterogenous catalyst that means catalysts are solid form and the reactions occur on the surface of the catalyst. These catalysts can be easily separated and recycled. For example : Pd, Pt metal nanoparticles used in hydrogenation reactions. Ti02, ZnO are used in photocatalysis. Gold in bulk is unreactive but the nanoparticles of gold behave as very good catalyst for organic reactions.

(3) Surface area : High surface-to-volume ratio is a very important characteristic of nanoparticles. Bulk material if subdivided into a group of individual nanoparticles, the total volume remains the same, but the collective surface area is largely increased. With large surface area for the same volume, these small particles react much faster because more surface area provides more number of reaction sites, leading to more chemical reactivity. Explanation of increase in surface area with decrease in particle size.

(4) Thermal strength : The melting point of nanomaterial changes drastically with size.

For example : Sodium clusters (Nan) of 1000 atoms melts at 288 K, 10000 atoms melt at 303 K and bulk sodium melts at 371 K.

(5) Mechanical strength : The mechanic al strength of nano clusters increase the hardness of the metal.

For example : nanoparticles of copper and palladium clusters with diameter in the range of 5-7 nm have hardness up to 500 r. greater than the bulk metal.

(6) Electrical conductivity : At nanoscale level the electrical conductivity changes. For example : Carbon nanotubes behave as a conductor or semiconductor whereas carbon is nonconductor.

Question 15.
Describe the two methods of synthesising nanomaterials (nanoparticles).
Answer:
The two methods of synthesising nanomaterials :
(1) Bottom-up and
(2) Top down methods :

(1) Bottom-up method : Synthesis of nanoparticles in the bottom-up approach molecular components arrange themselves into more complex assemblies atom by atom, molecule by molecule and cluster by cluster from the bottom. Example : synthesis of nanoparticles by colloidal dispersion.

(2) Top-down method : In the top-down approach, involves nanomaterials being synthesised from bulk material by breaking the material. The bulk solids are disassembled into finer pieces until they are constituted of only few atoms. Example : Nanoparticles are synthesised by colloidal dispersion.

Question 16.
Discuss the various analytical tools used for characterization of nanoparticles.
Answer:
The analytical tools used for characterization of nanoparticles are

• U.V visible spectroscopy – It gives the preliminary confirmation of formation of nanoparticles.
• X-ray Diffraction (XRD) – The information given by this tool is about particle size, crystal structure and geometry.
• Scanning electron microscopy (SEM) : This is used to study the structure of surface of material that is the morphology of the material.
• Transmission electron microscopy (TEM) gives information about the particles size.
• (FTIR) Fourier transform infrared spectroscopy gives information about absorption of functional groups and binding nature of the nanomaterial.

Question 17.
Give evidence of use of nanoparticles by humans in ancient times with appropriate examples.
Answer:
There is enough evidence that nanomaterials have been produced and used by humans in ancient times. For example :

• Gold and silver nanoparticles trapped in the glass matrix gives ruby red colour in some ancient glass paintings.
• The decorative glaze or metallic film known as lustre found on some medieval pottery is due to certain spherical metallic nanoparticles.
• Carbon black is a nanostructured material that is used in tyres of car to increase the life of tyre. (Discovery in 1900). Carbon nanotubes are made up of graphite sheets with nanosized diameter. They have highest strength.
• Fumed silica, a component of silicon rubber, coatings, sealants and adhesives is also a nanostructured material.

Question 18.
Explain the different applications of nanoparticles.
Answer:
The contribution of nanochemistry in number of innovative products in various disciplines due to their unique physical, chemical, optical, structural, catalytic properties. Few applications are as follows :

• Nanoparticles contribute to stronger, lighter, cleaner and smarter surfaces and systems. They are used in the manufacture of scratchproof eyeglasses, transport, sunscreen, crack resistant paints, etc.
• Used in electronic devices like Magnetoresistive Random Access Memory (MRAM).
• Nanotechnology plays an important role in water purification techniques. Silver nanoparticles are used in water purification system to get safe drinking water.
• Self cleansing materials : Lotus is an example of self cleansing. Nanostructures on lotus leaves repel water which carries dirt as it rolls off. Lotus effect is the basis of self cleaning windows.

Question 19.
State the advantages of nanoparticles and nanotechnology.
Answer:

• Nanotechnology has revolutionalized electronics and computing.
• Nanotechnology has benefited the energy sector by making solar power more economical and energy storage more efficient.
• Nanotechnology has transformed the medical field with the manufacture of smart dmgs which help cure the life threatening diseases like cancer and diabetes faster and without side effects.

Question 20.
State the disadvantages of nanoparticles and nanotechnology.
Answer:
Despite the benefits that nanotechnology offers to the world, it is accompanied by certain disadvantages and potential risks.

The standard of living has been raised by nanotechnology but at the same time it has increased the environmental pollution. The kind of pollution caused by nanotechnology is very dangerous for living organism, it is called nano pollution.

Nanoparticles can be potential health hazard depending on the size, chemical composition and shape. They can be inhaled and can be deposited in the human respiratory tract and in the lungs, causing lung damage.

Question 21.
Name the development that meets the needs of the present, without compromising the ability of future generation to meet their own need.
Answer:
Sustainable development

Question 22.
Give name of father of green chemistry.
Answer:
Paul T. Anastas

Question 23.
Environmentally safe chemistry is known as.
Answer:
Green chemistry

Question 24.
How many principles does green chemistry have ?
Answer:
Twelve

Question 25.
Which principle of green chemistry has its perspective largely towards petrochemicals?
Answer:
Use of renewable feedstocks.

Question 26.
Name the chemical which leachs out of plastic packaging materials.
Answer:
Phthalate

Question 27.
Name the materials having structural components with at least one dimension in the nanometer scale.
Answer:
Nanomaterials.

Question 28.
Name the class of nanomaterial i.e. nanotubes, fibres, nanowires belong to.
Answer:
Two dimensions are in the nanoscale.

Question 29.
Name the nanoparticles used in sunscreen.
Answer:
Zinc oxide (ZnO) and Titanium dioxide (TiO₂).

Question 30.
What is the colour of gold nanoparticles ?
Answer:
Red

Question 31.
Name the nanoparticles used as catalyst in hydrogenation reaction.
Answer:
Palladium and Platinum.

Question 32.
Name the two approaches used to synthesize nanomaterials.
Answer:
Bottom up and Top down.

Question 33.
Give the name of the wet chemical synthetic process for nanomaterials.
Answer:
Sol-gel process.

Question 34.
Give the steps involved in preparation of nanoparticle using sol-gel process.
Answer:
Hydrolysis, polycondensation, drying, thermal decomposition.

Question 35.
Name the analytical techniques used for characterisation of nanomaterials.
Answer:
u.v-visible spectroscopy, x-ray diffraction (XRD), scanning electron microscopy (SEM), Transmission electron microscopy (TEM), Fourier transform infrared spectroscopy (FTIR).

Question 36.
Name the technique used to analyse particle size, crystal structure and geometry of a nanoparticle.
Answer:
x-ray diffraction (XRD)

Question 37.
Name the analytical technique used to study the morphology (structure of surface) of a material.
Answer:
Scanning electron microscopy (SEM)

Question 38.
Which innovative material has been developed using the lotus effect ?
Answer:
Self cleansing material

Question 39.
Which are the sectors that are revolutionalized by nanoparticles ?
Answer:
Electronics, energy sector and medical fields.

Question 40.
What are the disadvantages of nanotechnology ?
Answer:
Nano pollution and lung damage.

Question 41.
Name the scientist who coined the word nanotechnology.
Answer:
Nario Taniguchi (University professor at Tokyo in 1974).

Question 42.
Select and write the most appropriate answer from the given alternatives for each subquestion:

1. The measure of the amount of atoms from the starting materials that are present in the useful product at the end of chemical process is known as
(a) catalyst
(b) atom economy
(c) design of safer chemicals
(d) design for efficient energy
Answer:
(b) atom economy

2. The atom economy of the following reaction is CH₃ – CH₂ – CH₂ – CH₂ – OH + NaBr + H₂SO₄ → CH₃ – CH₂ – CH₂ – CH₂ – Br + NaHSO₄ + H₂O
(a) 49.81%
(b) 49%
(c) 50%
(d) 100%
Answer:
(a) 49.81%

3. Green chemistry reduces risk by
(a) developing the process for reuse and recycle of solvents and chemicals
(b) inventing technologies to clean the environ-ment
(c) minimize the use of chemicals
(d) reducing or eliminating the use or generation of hazardous chemicals in chemical products and process
Answer:
(d) reducing or eliminating the use or generation of hazardous chemicals in chemical products and process

4. Chemical synthesis should be designed to mini-mizes the use of
(a) liquid fuels
(b) solid fuels
(c) gaseous fuels
(d) energy
Answer:
(d) energy

5. The chemistry that applies across the life cycle of a chemical product like design, manufacture and use is called
(a) eco-friendly chemistry
(b) green chemistry
(c) environmental chemistry
(d) inorganic chemistry
Answer:
(b) green chemistry

6. According to the principles of green chemistry the chemicals involved in the production must be
(a) non-hazardous
(b) toxic
(c) polluting
(d) highly toxic
Answer:
(a) non-hazardous

7. Which of the following is not one of the twelve principles of green chemistry ?
(a) using renewable feedstocks
(b) designing safer chemicals and products
(c) maximizing atom economy
(d) avoiding the use of catalysts
Answer:
(d) avoiding the use of catalysts

8. Chemical synthesis based on principle of green chemistry encourages the use of
(a) hazardous chemicals
(b) reactions with low atom efficiency
(c) catalyst
(d) high energy requirements
Answer:
(c) catalyst

9. The plastic bottles made of HDPE are used to store household cleaner and shampoo can be recycled to make
(a) carpets, furniture, new containers
(b) detergent bottles, fencing, floor tiles, pens
(c) custom-made products
(d) cables, mudflaps, panelling, roadway gutters
Answer:
(b) detergent bottles, fencing, floor tiles, pens

10. The plastic ketch-up bottles and syrup bottles made from polypropylene (pp) can be recycled to make
(a) battery cables, brooms, ice scrapers, rakes
(b) envelopes, floor tiles, lumber
(c) custom-made products
(d) carpet, furniture, new containers
Answer:
(a) battery cables, brooms, ice scrapers, rakes

11. The role of green chemistry aims to
(a) design chemical processes and products that maximize profits
(b) design safer chemicals and products by reduc¬ing or eliminating the use and generation of hazardous substances
(c) design processes and products that work efficiently
(d) utilize non-renewable feedstocks
Answer:
(b) design safer chemicals and products by reducing or eliminating the use and generation of hazardous substances

12. The study of phenomena and manipulation of materials of atomic, molecular and macromolecular scales where properties differ significantly from those at a large scale is called
(a) nanoscience
(b) nanochemistry
(c) nanotechnology
(d) nanomaterial
Answer:
(a) nanoscience

13. The term nanotechnology was first used by whom and when ?
(a) Richard Feynman, 1959
(b) Nario Taniguchi, 1974
(c) Eric Drexter, 1986
(d) Sumia Lijima, 1991
Answer:
(b) Nario Taniguchi, 1974

14. Which one of these statements is NOT true ?
(a) Gold at the nanoscale is red.
(b) A very highly useful application of nanochem¬istry is medicine.
(c) Sunscreen contains nanoparticles of zinc oxide (ZnO) and (SiO₂) silicon oxide.
(d) Silicon at nanoscale is not an insulator
Answer:
(c) Sunscreen contains nanoparticles of zinc oxide (ZnO) and (SiO₂) silicon oxide.

15. Which of the historical works mentioned below contain nanotechnology?
(a) Lycurgus cup
(b) Medieval stained glass windows in churches
(c) Damascus steel swords
(d) All of the above
Answer:
(d) All of the above

16. The nanometer scale is conventionally defined as
(a) 10 – 100nm
(b) 1 – 100nm
(c) 1 – 1000 nm
(d) 1 – 10000 nm
Answer:
(b) 1-100 nm

17. The material synthesized on the nanometer scale possess
(a) same bulk properties
(b) different bulk properties
(c) unique optical, magnetic, electrical properties
(d) no change in properties
Answer:
(c) unique optical, magnetic, electrical properties

18. Nanomaterials of zero dimension is
(a) one in which all three dimensions are in the nanoscale
(b) one in which two dimensions are in the nanoscale
(c) one in which one dimension is in the nanoscale
(d) None of the above
Answer:
(a) one in which all three dimensions are in the nanoscale

19. The science which deals with the design and synthesis of material on nanoscale with different size and shape is called
(a) nanoscience
(b) nanochemistry
(c) nanophysics
(d) nanotechnology
Answer:
(b) nanochemistry

20. Elemental has a shining yellow colour, but the colour of nanoparticles of gold is
(a) green
(b) yellow
(c) red
(d) blue
Answer:
(c) red

21. The surface area of nanoparticles
(a) is the same as in bulk
(b) increases with the same volume of the bulk
(c) decreases with the same volume of the bulk
(d) does not change with particle size
Answer:
(b) increases with the same volume of the bulk

22. The nanomaterial based catalyst are usually
(a) homogeneous catalyst
(b) heterogeneous catalyst
(c) good catalyst
(d) bad catalyst
Answer:
(b) heterogeneous catalyst

23. The catalyst used in photocatalysis is
(a) gold
(b) Raney Ni
(c) TiO₂
(d) AI₂O₃
Answer:
(c) TiO₂

24. Nanosized copper clusters have the mechanical strength of
(a) less than the bulk copper wire
(b) 100% more than the bulk metal
(c) 200% more than the bulk metal
(d) 500% more than the bulk metal
Answer:
(d) 500% more than the bulk metal

25. The most common method used for synthesis of nanomaterials is
(a) sol-gels method
(b) only sol method
(c) only gel method
(d) colloidal dispersion method
Answer:
(a) sol-gels method

26. What is the information obtained from uv-visible spectroscopy when used for nanomaterials ?
(a) morphology of structure
(b) preliminary conformation of formation of nanoparticle
(c) particle size
(d) functional group present
Answer:
(b) preliminary conformation of formation of nanoparticle

27. What information of the nanoparticles is obtained from transmission electron microscopy technique ?
(a) structure
(b) functional group
(c) particle size
(d) geometry
Answer:
(c) particle size

28. The analytical tool used to study the structure of surface of nanoparticle i.e. morphology is
(a) Absorption spectroscopy
(b) Scanning electron microscopy
(c) Emission spectroscopy
(d) Nuclear magnetic resonance spectroscopy
Answer:
(b) Scanning electron microscopy

29. The constituents of carbon nanotubes are
(a) nanosized graphite sheets
(b) nanosized carbon black
(c) nanosized coal black
(d) None of the above
Answer:
(a) nanosized graphite sheets

30. Self cleansing windows are example of the
(a) Nanoparticle effect
(b) Crompton effect
(c) Lotus effect
(d) Tyndal effect
Answer:
(c) Lotus effect

31. Which highly effective and cost effective nano-particles are used for water purification ?
(a) gold nanoparticle
(b) copper nanoparticle
(c) silver nanoparticle
(d) silica nanoparticle
Answer:
(c) silver nanoparticle

32. The sectors revolutionalized by nanotechnology are
(a) electronics and computing
(b) energy
(c) medicine
(d) All of the above
Answer:
(d) All of the above

33. Name the body part that gets affected by the hazardous nano pollution.
(a) heart
(b) brain
(c) lungs
(d) eyes
Answer:
(c) lungs

34. The pollution caused by nanotechnology is known as
(a) air pollution
(b) nano pollution
(c) ground pollution
(d) environmental pollution
Answer:
(b) nano pollution.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives

Question 1.
What are halogen derivatives of hydrocarbons?
Answer:
The replacement of hydrogen atom/s in aliphatic or aromatic hydrocarbons by halogen atom/s results in the formation of halogen derivatives of hydrocarbons.

Question 2.
How are halogen derivatives of hydrocarbons classified?
Answer:
Halogen derivatives of alkane are classified as :

1. Monohalogen derivative (or alkyl halide) : It is a halogen derivative of an alkane in which one hydrogen atom is replaced by one halogen atom and it is also called alkyl halide. E.g. C₂H₅Br.
2. Poh halogen derivatives : These are halogen derivatives in which more than one hydrogen atoms of an alkane are substituted by corresponding number of halogen atoms.

They are classified as follows :
(i) Dihalogen derivatives : The compounds formed by the substitution of two hydrogen atoms of an alkane by two halogen atoms are called dihalogen derivatives.

They are further classified as :

• Vicinal dihalides (Two halogen atoms on vicinal or adjacent carbon atoms)
• Geminal dihalides : CH₃ – CHCI₂ (Two halogen atoms on the same carbon atom)

(ii) Trihalogen derivatives : The compounds formed by the substitution of three hydrogen atoms of an alkane by three halogen atoms are called trihalogen derivatives.

(iii) Tetrahalogen derivatives : The compounds formed by the substitution of four hydrogen atoms of an alkane by four halogen atoms are called tetrahalogen derivatives. E.g. CCI₄.

Question 3.
What are alkyl halides? How are they classified?
Answer:
The compound formed by the replacement of one hydrogen atom in an alkane by a halogen atom is called an alkyl halide. The halogen atom is bonded to sp³ hybridised carbon. Alkyl halides are classified into the following three classes depending on the type of the carbon to which halogen atom is bonded.

(1) Primary (1°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a primary carbon atom is called primary alkyl halide.

[Primary (1°) carbon atom i.e., the carbon atom which is attached to only one carbon atom.]

They are represented by the general formula R – ^1°CH₂ – X.

(2) Secondary (2°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a secondary carbon atom is called secondary alkyl halide. [Secondary (2°) carbon i.e., the carbon atom which is attached to two other carbon atoms.]

They are represented by the general formula (R and R’ can be same or different)

(3) Tertiary (3°) alkyl halide : Alkyl halide in which halogen atom is bonded to a tertiary carbon atom is called tertiary alkyl halide. [Tertiary (3°) carbon i.e., the carbon atom which is attached to three other carbon atoms.]

They are represented by the general formula  (R, R’ and R” may be same or different)

Question 4.
Explain the following :
(1) Alkyl halide or haloalkanes
(2) Allylic halides
(3) Benzylic halide
(4) Vinylic halide
(5) Haloalkyne
(6) Aryl halide or haloarenes.
Answer:
(1) Alkyl halide or haloalkanes : In alkyl halides or haloalkanes the halogen atom is bonded to sp3 hybridized carbon which is a part of saturated carbon chain.
Example :

(2) Allylic halides : In allylic halides, halogen atom is bonded to a sp³ hybridized carbon atom next to a carbon-carbon double bond.

Example :

(3) Benzylic halide : In benzylic halides, halogen atom is bonded to a sp³ hybridized carbon atom which is further bonded to an aromatic ring.
Example :

(4) Vinylic halides : In vinylic halides, halogen atom is bonded to a sp² hybridized carbon atom of aliphatic chain. Vinylic halide is a haloalkene.
Example :

(5) Haloalkyne : In haloalkynes, halogen atom is bonded to a sp hybridized carbon atom.
Example :

(6) Aryl halides or haloarenes : In aryl halides, halogen atom is directly bonded to the sp² hybridized carbon atom of aromatic ring.
Example :

Question 5.
Give the IUPAC names of the following :
Answer:

Question 6.
Draw the structures of the following compounds:
Answer:

Question 7.
Write the structure of-
(a) 3-chloro-3-ethylhex-l-ene
(b) 1-Iodo-2, 3-dimethylbutane
(c) 1, 3, 5-tribromobenzene
Answer:

Question 8.
Write structures of
(a) 2-iodo-3-methyl pentane
(b) 3-chiorolleNane
(c) 1-chloro-2, 2-dimethyl propane
(d) 1-chloro-4-ethyl cyclohexane.
Answer:

Question 9.
Write the possible isomers of monochloro derivatives of 2,3-Dimethylbutane and write their IUPAC names.
Answer:
The given parent hydrocarbon has molecular formula, C₆H₁₄. The monochloro derivative of this compound has molecular formula C₆H₁₃CI.
The parent hydrocarbon is,

Hence the structures of isomers of monochioroderivative are.

Question 10.
Write structures and IUPAC names of all possible isomers of C₅H₁₁Br and classify them as l°/2°/3°.
Answer:
C₅H₁₁Br is a monohalogen derivative.

Question 11.
How are following compounds obtained from alcohols :
(1) ethyl chloride C₂H₅CI
(2) isopropyl chloride (CH₃ CHCI – CH₃)
(3) tert-butyl chloride (CH₃)₃ – CI?
Answer:
Alcohols in the presence of Lucas reagent which is a solution of concentrated HCI and ZnCI₂ form alkyl halides. Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols.

(3) Tertiary alcohols don’t need ZnCI₂ to react with HCI.

The order of reactivity of alcohols with a given halo acid is 3° >2°> 1°.

Question 12.
How are following compounds prepared from alcohols:
(1) ethyl bromide (C₂H₅Br)
(2) isopropyl bromide (CH₃ – CHBr – CH₃)
(3) tert-butyl bromide (CH₃)₃ C – Br?
Answer:
(1) Ethyl alcohol on heating with conc. hydrobromic acid (48%) forms ethyl bromide.
OR
When ethyl alcohol is treated with a mixture of NaBr and H₂SO₄, ethyl bromide is formed. Here HBr is generated in situ.

(2) Isopropyl alcohol, on reaction with NaBr and dil. H₂SO₄ forms isopropyl bromide.

(3) Tertiary alcohol on reaction with sodium bromide and dil. H₂SO₄ forms tert-butyl bromide.

Question 13.
How is ethyl iodide obtained from ethyl alcohol?
Answer:
When ethyl alcohol is treated with sodium or potassium iodide in 95 % phosphoric acid, ethyl iodide is formed. Here HI is generated in situ.

Question 14.
How will you prepare the following :

(1) Ethyl chloride (chloroethane) from ethyl alcohol using
(i) PCI₃
(ii) PCI₅ and
(iii) SOCI₂.
Answer:
(i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.

(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.

(2) Isopropyl chloride (2-chloropropane) from isopropyl alcohol using
(i) PCI₃
(ii) PCI₅
(iii) SOCI₂.
Answer:
When isopropyl alcohol is refluxed with phosphorus trichloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with phosphorus pentachloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, isopropyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for the preparation of alkyl chloride.

(3) Ethyl bromide (bromoethane) from ethyl alcohol.
Answer:
When ethyl alcohol is treated with a mixture of red phosphorus and bromine or hydrobromic acid (phosphorus tribromide is generated in situ), ethyl bromide is formed.

(4) Ethyl iodide (do ethane) from ethyl alcohol.
Answer:
When ethyl alcohol is heated with a mixture of red phosphorus and iodine, (phosphorus triiodide is generated in situ), ethyl iodide is formed.

Question 15.
Explain halogenation of methane.
Answer:
Halogenation : A reaction of alkanes with halogens (CI₂, Br₂, I₂) in the presence of appropriate conditions forming a mixture of alkyl halides.
(1) Chlorination:

(2) When excess of chlorine is used, tetrachioro methane, a major product is obtained. When excess of methane is used, chioromethane, a major product is obtained. The order of reactivity of halogens towards alkane is

(3) lodination:

However, iodination reaction is a reversible reaction. HI being a strongest reducing agent reduces methyl iodide back to methane.

(4) Fluorination: A reaction of alkane with fluorine is explosive and also hydrofluoric acid is poisonous and corrosive. Hence, alkyl fluorides are not prepared by halogenation of alkane.

Question 16.
Predict the possible products of the following reaction :
(1) Bromination of propane
(2) Bromination of n-butane
(3) Bromination of 2-Methyl propane.
Answer:
(1) Bromination of propane

(2) Bromination of n-butane

(3) Bromination of 2-Methyl propane

Question 17.
How are following compounds prepared by halogenation of ethane :
(1) Chloroethane
(2) Bromoethane
(3) Iodoethane?
Answer:
(1) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorine in the presence of diffused sunlight or U.V. light or at high temperature, chloroethane is obtained.

(2) Bromination of ethane : When ethane is heated with Br₂ in the presence of anhydrous AlBr₃, bromoethane is obtained.

(3) Iodination : When ethane is reacted with I₂ in the presence of suitable oxidising agents like-HgO or HIO₃ or dilute HNO₃ iodoethane is obtained.

Question 18.
Direct iodination of alkanes is not possible.
Answer:
(1) Direct iodination of alkanes using iodine is highly reversible.
\(\mathrm{RH}+\mathrm{I}_{2} \rightleftharpoons \mathrm{RI}+\mathrm{HI}\)
(2) Hydroiodic acid HI being strong reducing agent, it reduces RI to alkane RH.
(3) The reaction takes place only in the presence of a suitable oxidizing agent like HgO, HIO₃ or dilute HNO₃ which decomposes HI. Hence, direct iodination of alkanes is not possible.

Question 20.
How are following compounds obtained from alkenes :

Answer:
(1) Ethene on reaction with hydrogen chloride forms

(2) Ethene on reaction with hydrogen bromide forms

(3) Propene on reaction with hydrogen iodide forms

(4) but-2-ene on reaction with hydrogen iodide forms 2-iodobutane.

Question 19.
State and explain Markovnikov’s rule.
Answer:
Markovnikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent gets attached to that carbon atom of the double bond which carries less number of hydrogen atoms.

Example : Addition of HBr’to unsymmetrical alkene like propene gives two products.

Isopropyl bromide is the major product, since the negative part (Br^–) of HBr is attached to carbon atom of a double bond with less number of hydrogen atoms.

Question 20.
Explain peroxide effect.
OR
Write a note on the Kharasch-Mayo effect.
OR
Explain the addition of HBr to (unsymmetrical alkene) propane in the presence of benzoyl peroxide.
Answer:
The addition of HBr to an unsymmetrical alkene (propane) in the presence of benzoyl peroxide takes place in the opposite orientation to that of Markovnikov’s rule and this is known as Kharasch-Mayo effect or peroxide effect or Anti-Markovnikov addition.

Question 21.
Write the structure of alkyl halide obtained by the action hydrogen bromide on 2-Methyiprop-1-ene in the presence of peroxide.
Answer:
In the presence of peroxide. HBr to 2-Methyl prop-I-cne forms l-Bromo-2-methylpropane.

Question 22.
How are alkyl iodides prepared from alkyl chlorides/bromides?
Answer:
Alkyl iodide is prepared by treating alkyl chloride or alkyl bromide with sodium iodide, in the presence of dry acetone, sodium chloride or sodium bromide precipitates from the solution and can be separated by filtration. This reaction is known as Finkelstein reaction.

Question 23.
How are alkyl fluorides prepared with alkyl chlorides/alkyl bromides?
Answer:
When alkyl chloride or alkyl bromide is heated with metallic fluorides like AgF, CaF₂, CoF₂ or Hg₂F₂, alkyl fluoride is formed.

This reaction is known as Swarts reaction.

Question 24.
Explain the preparation of haloarenes using electrophilic substitution.
Answer:
When arene is treated with chlorine or bromine in dark at ordinary temperature in the presence of lewis acid as a catalyst like Fe, FeCI₃ or anhydrous AlCI₃, aryl chloride or aryl bromide is formed.

When toluene is brominated in dark at ordinary temperature in the presence of iron, a mixture of ortho and para bromo tolerene is obtained.

Ortho and para isomers can be easily separated as there is large difference in melting points of ortho and para isomers.

Question 25.
Write a note on Sandmeyer’s reaction.
Answer:
Aryl halides are most commonly prepared by replacement of nitrogen of diazonium salt. The replacement of diazonium group by -Cl or -Br using cuprous salt is called Sandmeyer’s reaction. When a primary aromatic amine (like aniline) suspended in cold F1C1, is treated with sodium nitrite, a diazonium salt (benzene diazonium chloride) is formed. When diazonium salt is treated with cuprous chloride or cuprous bromide, aryl halide (chlorobenzene or bromobenzene) is formed.

When benzene diazonium salt is mixed with potassium iodide, iodobenzene is formed.

Question 26.
Define the following :
Answer:
(1) Monochromatic light : It consists of rays of single wavelength vibrating in different planes perpendicular to the direction of propagation of the light.

(2) Plane polarized light : A light having oscillations only in one plane perpendicular to direction of propagation of light is known as plane polarized light.

(3) Optical isomerism : The steroisomerism in which the isomers have different spatial arrangements of groups/atoms around a chiral atom is called optical isomerism.

(4) Optical activity : The property of a substance by which it rotates plane of polarization of incident plane polarized light is known as optical activity.

(5) Optically active compound : The compound which rotate the plane of plane polarized light is called optically active compound.

(6) Enantiomers : The optical isomers which are non-superimposable mirror images of each other are called enantiomers or enantiomorphs or optical antipodes.
Example : 2-chlorobutane, lactic acid

(7) Chiral carbon atom : Carbon atom in a molecule which carries four different groups/atoms is called chiral carbon atom.

Chiral atom in a molecule is marked with asterisk (*)

For example : C in lactic acid

(8) Chiral molecule : When a molecule contains one chiral atom, it acquires a unique property i.e. it is non- superimposable with its mirror image is said to be chiral molecule.

(9) Chirality : The relationship between a chiral molecule and its mirror image is similar to the relationship between left and right hands. Therefore it is called handedness or chirality.

(10) Dextrorotatory substance or r/-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the right hand side (RHS) is called dextrorotatory substance (or isomer) and denoted by d or (+) sign.

(11) Laevorotatory substance or /-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the left hand side (LHS) is called laevorotatory substance (or isomer) and denoted by / or (-) sign.

(12) Racemic mixture or Racemate : A mixture containing equimolar quantities of dextro (d) and laevo (/) optical
isomers which is optically inactive due to molecules of one enantiomer is cancelled by equal and opposite optical rotation due to molecules of the other enantiomer is called a racemic mixture or racemate. It is represented as (dl) or (+).

Question 27.
Calculate the number of isomers for 2-chlorobutane.
Answer:
The number of optical isomers possible for a compound is 2ⁿ where n = number of asymetric carbon atoms.
As n = 1 for 2-ehlorobutane, 2ⁿ = 2¹ = 2. Hence, it has two optical isomers.

Question 28.
How many optical isomers are possible for C₅H₁₁ CI?
Answer:
The number of optical isomers : 3.

Question 29.
How many optical isomers are possible for glucose?
Answer:
The number of optical isomers : 16.

Question 30.
Draw the structures and indicate the chiral carbon atoms in
(1) Lactic acid
(2) 2-Chlorobutane.
Answer:
(1) In lactic acid structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, COOH, OH, CH₃ and H.
(2) In 2-chlorobutane structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, -CH₂ – CH₃ (ethyl), CH₃ (methyl), Cl and H.

Question 31.
Identify chira! and achiral molecules.

Question 32.
Complete the following reactions and explain optical activity of the products formed:
(i) Pent-1-ene with HBr
Answer:

(ii) Pent-2-ene with HBr
Answer:

Question 33.
C₆H₁₂ (A) on treatment with HCI produced a compound Y. Which is optically active, what is structure A?
Answer:

Question 34.
A racemic mixture is optically inactive. Explain.
Answer:

• A racemic mixture contains equimolar (or equimolecular) quantities of the dextrorotatory (d-) and laevorotatory (l-) isomers (enantiomers) of a compound.
• The d-enantiomer rotates the plane of plane-polarized light to the right, while the l-enantiomer rotates the same to the left to the same extent.
• The quantities of the d- and l-enantiomers being the same, both the rotations are of the same magnitude, but of opposite directions. Hence, they cancel each other. Hence, a racemic mixture is optically inactive.
• It is represented as dl or ( + ). Example : ( ± ) lactic acid

Question 35.
Explain Fischer projection formula with illustration.
OR
Write a note on Fischer projection formula.
Answer:
Fischer projection formula or cross formula : The three dimensional (3-D) view of a molecule is presented on plane of paper. A Fischer projection formula can be drawn by visualizing the main carbon chain verical in the molecule. Each carbon on the vertical chain is represented by a cross.

Conventionally the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines represent the bonds going below the carbon.

Question 36.
Explain Wedge formula with illustration
OR
Write a note on Wedge formula.
Answer:
Wedge formula : When a tetrahedral carbon is imagined to be present in the plane of paper all the four bonds at this carbon cannot lie in the same plane. The bonds in the plane of paper are represented by normal lines, the bonds projecting above the plane of paper are represented by solid wedges (or simply by bold lines) while bonds going below the plane of paper are represented by broken wedges (or simply by broken lines).

Question 37.
Give a laboratory test to confirm the presence of halogen in the original organic compound.
Answer:
Haloalkanes are of neutral type in aqueous medium. On warming with aqueous sodium or potassium hydroxide the covalently bonded halogen in haloalkane is converted to halide ion.

When this reaction mixture is acidified by adding dilute nitric acid and silver nitrate solution is added a precipitate of silver halide is formed which confirms presence of halogen in the original organic compound.

Question 38.
Define the following :
Answer:
(1) Mechanism of a reaction : It is a step by step description of exactly how the reactants are transformed into products in as much details as possible.
(2) Substitution reaction : When a group bonded to a carbon in a substrate is replaced by another group to get a product with no change in state of hybridization of that carbon, the reaction is called substitution reaction.

Question 39.
Describe the action of aqueous KOH (or NaOH) on :
(1) ethyl bromide
(2) isopropyl bromide
(3) tert-butyl chloride
(4) methyl bromide
(5) 2-chlorobutane.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 40.
Describe the action of sodium ethoxide on
(1) ethyl bromide
(2) methyl bromide :
OR
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from alkyl halides?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained. In this reaction halide (-X) of alkyl halide is replaced by an alkoxy group (-OR). This reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (or unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

(1) Simple (symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.

e.g., When ethyl bromide is heated with sodium ethoxide, diethyl ether is formed.

(2) Mixed (unsymmetrical) ether : When an alkyl halide and sodium alkoxide having different alkyl groups are heated, unsymmetrical ether is obtained.

When methyl bromide is heated sodium ethoxide, ethyl methyl ether is formed.
When ethyl bromide is heated with sodium meihoxide, ethyl methyl ether is formed.

Question 41.
What is the action of silver salt of carboxylic acid on alkyl halide?
Answer:
When an alkyl halide (R – X) is heated with silver salt of carboxylic acid (R -COOAg). an ester is obtained.

Question 42.
Describe the action of alcoholic silver acetate on
(1) methyl bromide
(2) ethyl bromide.
Answer:
(1) Methyl bromide : When methyl bromide is heated with an alcoholic silver acetate, methyl acetate is formed.

(2) Ethyl bromide : When ethyl bromide is heated with an alcoholic silver acetate, ethyl acetate is formed.

Question 43.
What is the action of alcoholic silver propionate on ethyl bromide?
Answer:
When ethyl bromide is heated with an alcoholic silver propionate. ethyl propionate is formed.

Question 44.
Describe the action of excess of ammonia on (I) ethyl bromide (2) n.propyl bromide.
Answer:
(1) Ethyl bromide : When ethyl bromide is boiled under pressure with an excess of alcoholic ammonia, ethylamine (ethanamine) is formed. This is known as ‘ammonolysis of ethyl bromide.

(2) n-propyl bromide : When n-propy1 bromide is boiled under pressure with an excess of ammonia, n-propyl amine (propanamine) is formed.

Question 45.
What is ammonolysis? Give a suitable example for the reaction.
Answer:
When an alkyl halide is boiled under pressure with an excess of alcoholic solution of ammonia (NH₃), corresponding (primary amine) alkyl amine is formed. This reaction is known as ammonolysis of alkyl halide.

(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 46.
Describe the action of aqueous alcoholic potassium cyanide on
(1) ethyl bromide
(2) methyl iodide.
Answer:
Ethyl bromide : When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.

(2) Methyl iodide : When methyl iodide is boiled with alcoholic solution of potassium cyanide, methyl cyanide is formed.

Question 47.
Describe the action of alcoholic silver cyanide on
(1) ethyl bromide
(2) methyl chloride.
OR
Explain isocyanide reaction of
(1) ethyl bromide
(2) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl isocyanide is formed.

(2) Methyl chloride : When .methyl chloride is heated with alcoholic silver cyanide, methyl isocyanide is formed.

The above reactions (1) and (2) are called isocyanide reaction.

Question 48.
Describe the action of potassium nitrite on
(i) ethyl bromide,
(ii) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is treated with potassium nitrite, ethyl nitrite is formed.

(2) Methyl chloride : When methyl chloride is treated with potassium nitrite, methyl nitrite is formed.

Question 49.
Describe the action of silver nitrite on (1) ethyl chloride (2) n-propyl bromide.
Answer:
(1) Ethy chloride : When ethyl chloride is treated with silver nitrite, nitroethane is obtained.

(2) n-Propyt bromide: When n-propyl bromide is treated with silver nitrate, nitropropane is obtained.

Question 50.
How will you convert (the following:

(1) Ethyl bromide to ethanol.
Answer:

(2) Ethyl bromide to propane nitrile.
Answer:

(3) Ethyl bromide to ethyl amine.
Answer:

(4) Ethyl bromide to ethyl acetate.
Answer:

(5) Ethyl bromide to ethyl isocyanide.
Answer:

(6) Ethyl bromide to ethyl methyl ether.
Answer:

(7) Ethyl bromide to n-butane.
Answer:

(8) Ethyl bromide to Ethyl magnesium bromide.
Answer:

Question 51.
Define the following :
Answer:
(1) Nucleophilic bimolecular reaction (SN²) : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends on the concentration of the substrate and the nucleophile is called a nucleophilic bimolecular reaction.

Example :

(2) SN¹ reaction : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends only on the concentration of the substrate is called nucleophilic unimolecular or first order reaction or SN¹ reaction.

Question 52.
Explain, the mechanism of alkaline hydrolysis (reaction with aqueous KOH) of tert-butyl bromide (2-Bromo-2-methylpropane) with energy profile diagram.
OR
Explain only reaction mechanism for alkaline hydrolysis of tert-butyl bromide.
Answer:
(i) Consider alkaline hydrolysis of tert-butyl bromide (2-Bromo-2 methylpropane) with aqueous NaOH or KOH.

(ii) Kinetics of the reaction : Due to steric hindrance of voluminous three methyl groups around carbon, nucleophile OH- cannot attack carbon atom directly. Hence, the reaction takes place in two steps.

Step I : This involves heterolytic fission of C – Br covalent bond in the substrate forming carbocation and Br^–. This is a slow process.

Step II : This step involves attack of nucleophile OH- or carbocation forming C – OH bond and product tert-butyl alcohol. Since it involves ionic charge neutralisation, it is a fast step.

Rate Determining Step (R.D.S.) : Since the first step is a slow step, it is R.D.S., and therefore the rate of the reaction depends on the concentration of only one reactant, (CH₃) C – Br.
Rate = R = k [(CH₃)₃ C – Br] where k is a rate constant of the reaction.

SN¹ reaction : The reaction between tert.butyl bromide and hydroxide ion to form tert.butyl alcohol follows a first-order kinetics. The rate of this reaction depends only on the concentration of one substance (tert-butyl bromide) and is independent of the concentration of alkali added. It is an unimolecular first (1st) order Nucleophilic Substitution reaction denoted as SN reaction.

Stereochemistry and mechanism of the reaction : The reaction takes place in two steps and both the steps involve formation of transition states (T.S.).

T.S. -1 for first step :

In this transition state, C – Br bond is partially broken, so that carbon atom carries partial positive charge (+δ) and Br carries partial negative charge (-δ) which further breaks forming carbocation and Br . Tert-butyl cation (carbocation) has a planar structure and the CH₃ – C – CH₃ bond angle is 120°. It is the intermediate of the reaction. It is unstable. In this step, hybridisation of carbon atom changes from sp³ (tetrahedral geometry) to sp (planar geometry).

T.S. – II for second step :

In this transition state, C – OH bond is partially fonned so that carbon atom carries partial positive charge (+ δ) and OH carries partial negative charge ( -δ) which further forms tert-butyl alcohol.

Formation of a racemic mixture : Since OH^– has equal probability of the attack on carbocation from frontside and from backside, the products obtained are equal. In case of optical active alkyl halide, a racemic mixture is obtained.

Question 53.
Discuss SN² mechanism of methyl bromide using aqueous KOH. Draw energy profile diagram.
OR
Discuss the mechanism of alkaline hydrolysis of methyl bromide or Bromomethane.
Answer:
(1) Consider alkaline hydrolysis of methyl bromide (Bromomethane). CH₃Br with aqueous NaOH or KOH.

(2) Stereochemistry and Kinetics of the reaction iR.D.S.) : This hydrolysis reaction takes place only in one step which is a rate determining step i.e. R.D.S. The rate of hydrolysis reaction depends on the Concentration of CH3Br
and 0H which are present in the R.D.S. of the reaction.
Rate = R = k [CH₃Br] (OH]
where k is rate constant of the reaction.

SN² reaction : The reaction between methyl bromide and hydroxide ion to form methanol follows a second order kinetics, since the rate of the reaction depends on the concentrations of two reacting species, namely methyl bromide and hydroxide ion it is bimolecular second order (2nd) Nucleophilic Substitution reaction denoted by SN².

(3) Mechanism of the reaction :
(i) It is a single step mechanism. The reaction takes place in the following steps :

(ii) Backside attack of the nucleophile : Nucleophile, OH^– attacks carbon atom of CH₃Br from back side i. e. from opposite side to that of the leaving group i.e. Br^– to experience minimum steric repulsion and electrostatic repulsion between the incoming nucleophile (OH^–) and leaving Br^–.

(iii) Transition state : When a nucleophile, OH^– approaches carbon atom of CH₃Br, the potential energy of the system increases until a transition state (T.S.) of maximum potential energy is formed in which C – Br bond is partially broken and C – OH bond is partially formed. The negative charge is equally shared by both incoming nucleophile- OH^– and outgoing, leaving group-Br^–. (Thus, the total negative charge is diffused.)

(iv) In CH₃Br, carbon atom is sp³ -hybridized and CH₃Br molecule is tetrahedral. The hybridisation of carbon atom changes to sp² hybridisation. The transition state contains pentacoordinate carbon having three δ (sigma) bonds in one plane making bond angles of 120° with each other i.e., H₁; H₂ and H₃ atoms lie in one plane while two partial covalent bonds containing Br and OH lie collinear and on opposite sides perpendicular to the plane.

(v) Inversion of configuration : The transition state decomposes fast by the complete breaking of the C-Br bond and the new C-OH bond is formed on the other side. The breaking of C-Br bond and the formation of C-OH bond take place simultaneously. The energy required to break the C-Br bond is partly obtained from the energy released when C-OH bond is formed. The formation of product CH₃OH is accompanied by complete or 100% inversion of configuration forming again sp³-hybridized carbon atom giving tetrahedral CH₃OH molecule. But in this structure the positions of H₂ and H₃ atoms in the reactant (CH₃Br) and in product are on the opposite side. This inversion of configuration is called Walden inversion.

Question 54.
Discuss the factors influencing S_N1 and S_N2 mechanism.
Answer:
(1) Nature of substrate : S_N2 : The transition state (T.S.) of S_N2 mechanism + is pentacoordinate, it is crowded. As a result S_N2 mechanism is favoured in primary halides and least favoured in tertiary halides.

S_N1 : A planar carbocation intermediate is formed in S_N1 reaction. Bulky alkyl groups can be easily accommodated in planar carbocation and it has no steric crowding. As a result SN1 mechanism is favoured in tertiary halides and least favoured in primary halides.

The carbocation intermediate is stabilized by + effect of alkyl substituents and also by hyperconjugation y effect of alkyl substituents containing a-hydrogens. As a result, S_N1 mechanism is favoured in tertiary halides and least favoured in primary halides.

Thus, tertiary alkyl halides undergo nucleophilic substitution by S_N1 mechanism while primary halides follow S_N2 mechanism.

(2) Nucleophilicity of the reagent : A strong nucleophile attacks the substrate faster and favours S_N2 mechanism. The rate of S_N1 mechanism is independent of the nature of nucleophile. Nucleophile does not react in the 1st step (slow step) of S_N1. Nucleophile reacts fast after the carbocation intermediate is formed.

(3) Solvent polarity : (1) S_N1 reaction proceeds more rapidly in polar protic solvents than in aprotic solvent. Polar protic solvent decreases the rate of S_N2 reaction. (2) In S_N2 mechanism, rate depends on substrate as well as nucleophile. A polar solvent stabilizes nucleophile by solvation. Thus solvent deactivates the nucleophile by stabilizing it. Hence, aprotic solvents or solvent of low polarity will favour S_N2 mechanism.

Question 55.
How does relative reactivity for alkaline hydrolysis with respect to S_N2 and S_N1 vary in the following alkyl halides :
(1) Bromomethane
(2) Bromoethane
(3) 2-Bromopropane
(4) 2-Bromo-2-methylpropane ?
Answer:
(A) Relative reactivity for S_N2 mechanism decreases in the order of :

(B) Relative reactivity for S_N1 mechanism decreases in the order of :
2-Bromo-2-methylpropane > 2-Bromopropane > Bromoethane > Bromomethane

Question 56.
Explain with reason the relative order of reactivity of l°/2°/3° alkyl halides by S_N1 mechanism.
Answer:
In alkaline hydrolysis of an alkyl halide by S_N1 mechanism, the formation of carbocation as an intermediate product is involved.

The increasing order of a stability of carbocation is,

The stability order for carbocation is 3° > 2° > 1°.
Therefore the increasing order of reactivity by S_N1 mechanism of alkyl halides is
(1°) primary < (2°) secondary < (3°) tertiary

Question 57.
Which one of the following is more easily hydrolysed in S_N1 and S_N2 reaction by aqueous KOH, C₆H₅ CHCIC₆H₅ and C₆H₅CH₂CI?
Answer:
In S_N1 reaction C₆H₅CHCI C₆H₅ will be more easily hydrolysed than C₆H₅CH₂CI
In S_N2 reaction C₆H₅ CH₂CI will be more easily hydrolysed than C₆H₅CHCIC₆H₅.

Question 58.
Choose the member that will react faster than the following pairs by S_N1 mechanism.
(1) l-bromo-2, 2-dimethyl propane or 2-bromopropane.

Answer:
The reactivity of S_N1 reaction depends on the steric hindrance, in 2-bromopropane, a-carbon atom is attached to two methyl groups suffers greater steric hindrance to nucleophilic attack than l-bromo-2, 2-dimethyl propane. Hence, 2-bromopropane react faster by S_N1 mechanism.

(2) 2-Iodo-2-methyl butane or 2-iodio-3-methyl butane.

Answer:
Since, 2-Iodo-2-methyl butane is a tertiary alkyl halide, it undergoes SN^-1 reaction faster than 2-iodo-3-methyl butane.

(3) 1-Chloro propane or 2-chloropropane.

Answer:
Since, 2-chloropropane is a secondary alkyl halide, it undergoes SN^-1 reaction faster than 1-chloropropane.

(4) 2-Iodo-2-methyl butane or tert-butyl chloride.

Answer:
Since, iodine is a better leaving group than chloride 2-iodo-2-methyl butane undergo SN^-1 reaction faster than tert-butyl chloride.

Question 59.
Write a note on elimination reaction.
OR
Explain dehydrohalogenation reaction.
Answer:
When alkyl halide having at least one β-hydrogen is boiled with alcoholic solution of potassium hydroxide (KOH), an alkene is formed due to elimination of hydrogen atom from β-carbon and halogen atom from α-carbon, is called dehydrohalogenation.

Tertiary butyl bromide when heated with alcoholic solution of potassium hydroxide forms isobutylene.

This reaction is called β-elimination (or 1,2-elimination) reaction as it involves elimination of halogen and a β-hydrogen atom.

As hydrogen and halogen is removed in this reaction it is also known as dehydrohalogenation reaction.

Question 60.
Describe the action of alcoholic potassium hydroxide (aic. KOH) on
(1) ethyl bromide
(2) n-propyl bromide
(3) isopropyl bromide
(4) tert-butyl chlorIde.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 61.
Describe the action of alc.KOH on 2-bromobutane.
When 2-bromobutane is boiled with alc.KOH on 2-bromobutane, a mixture of but-l-ene and but-2-ene is formed.
Answer:

Question 62.
Explain Saytzelf’s rule with suitable example.
Answer:
Saytzcff’s rule : In dehydrohalogenation reaction the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Hence the number of alkyl substituents on doubly bonded Carbon atoms increases, the stability of the alkene giving its major products.

Hence the increasing stability of alkenes is.

There are two types of fi hydrogens (β₁ and β₂) therefore two alkenes are expected.

Question 63.
What is a Grignard reagent ?
Answer:
Grignard reagent : An organometallic compound in which the divalent magnesium is directly linked to an alkyl group (R -) and a halogen atom (X), and has general formula R – Mg – X is called Grignard reagent. OR When alkyl halide is treated with magnesium in dry ether as solvent, it gives alkyl magnesium halide. It is known as Grignard reagent.

The carbon-magnesium bond is highly polar and magnesium-halogen bond is in ionic in nature. Grignard reagent is highly reactive. It is an important reagent and used in the preparation of a large number of organic compounds.

Question 64.
How is Grignard reagent prepared ?
Answer:
Grignard reagent is an alkyl magnesium halide, R – Mg – X obtained by the reaction of alkyl halide R – X with magnesium (Mg) in dry ether.

When an alkyl halide like CH₃I is added from a dropping funnel to a flask containing pieces of pure Mg in pure and dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH₃ – Mg – I is formed.

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesium iodide.

Question 65.
Write a note on Grignard reagent.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 66.
Describe the action of water on
(1) methyl magnesium iodide
(2) ethyl magnesium iodide.
Answer:
(1) Methyl magnesium iodide : When methyl magnesium iodide is treated with water, methane is obtained

(2) Ethyl magnesium iodide : When ethyl magnesium iodide is treated with water, ethane is obtained.

Question 67.
Describe the action of ammonia on
(1) ethyl magnesium bromide
(2) n-propyl magnesium chloride.
Answer:
(1) Ethyl magnesium bromide : When ethyl magnesium bromide is treated with ammonia, ethane is formed.

(2) n-Propyl magnesium chloride : When n-propyl magnesium chloride is treated with ammonia, propane is formed.

Question 68.
Explain Wurtz reaction. OR Explain the action of sodium with alkyl halides.
Answer:
(1) When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher alkane is formed. This is called Wurtz reaction or Wurtz coupling reaction.

(2) In this reaction the alkyl radicals from two molecules of the reacting alkyl halide combine or couple to form the higher alkane.

(3) Thus, methyl bromide reacts with sodium in ether to form ethane (C₂H₆), while ethyl bromide under the same conditions forms n-butane (C₄H₁₀).

(4) If a mixture of two different alkyl halides is treated with Na in dry ether, then a mixture of alkanes is obtained called self coupling products. For example, a mixture of CH₃Br and C₂H₅Br gives propane along with C₂H₆ and C₄H₁₀.

Question 69.
Explain the reaction of haloarene with alkyl halide and sodium metal.
Write a note on Wurtz-Fittig reaction.
Answer:
When an alkyl halide and an aryl halide is treated with sodium metal in dry ether the corresponding alkylarene (alkyl benzene) is formed. The reaction is known as Wurtz-Fittig reaction. This reaction allows alkylation of alkyl halides.

Question 70.
Describe the action of aryl halide on sodium metal.
Answer:
Aryl halide reacts with sodium metal in dry ether, biphenyl is formed. This reaction is known as Fittig reaction.

Question 71.
Identify the product A of following reaction.

Answer:

Question 72.
Explain the following substitution reactions of chlorobenzene :
(1) Halogenation
(2) Nitration
(3) Sulphonation.
Answer:
(1) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrous ferric chloride, a mixture of ortho and para-dichlorobenzene (major product) is formed.

(2) Nitration : When chlorobenzene is heated with nitrating mixture (cone, nitric acid -I- cone, sulphuric acid) a mixture of l-chloro-4-nitro benzene (major product) and l-chloro-2-nitrobenzene is formed.

(3) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixture of 4-chlorobenzene sulphonic acid (major product) and 2-chlorobenzene sulphonic acid is formed.

Question 73.
Describe the action of the following on chlorobenzene :
(1) Methyl chloride in the presence of anhydrous AICI₃
(2) Acetyl chloride in the presence of anhydrous AICI₃.
Answer:
(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 74.
Write a note on Friedel Craft’s reaction.
Answer:

(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 75.
Convert 1-chlorobutane into the following compounds :

Answer:
(1) 1-Chlorobutane to butan-l-ol :

(2) 1-Chlorobutane to 1-iodobutane :

(3) 1-Chlorobutane to n-butyl cyanide (CH₃ – CH₂ – CH₂ – CH₂ – CN) :

(4) l-Chlorobutane to

Question 76.
Predict the expected product of substitution reactions :
(1) Isobutyl chloride + sodium ethoxide.
Answer:

(2) n-butyl chloride + sodium.
Answer:

(3) 1-chloropropane + aq. potassium hydroxide.
Answer:

(4) Aniline + NaNO₂/HCl.
Answer:

Question 77.
Write the products:

Question 78.
Identify A and B in the following :

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Question 79.
State the uses of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or Chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freons
(6) DDT (p, p’-Dichlorodiphenyl trichloroethane).
Answer:
(1) Dichloromethane (CH₂CI₂) :

• Dichloromethane dissolves wide range of organic compounds, hence it is used as solvent for many chemical reactions.
• It is used as a solvent as a paint remover and degreaser.
• It is used as propellant in aerosols and as a fumigant pesticide for grains and strawberries.
• It is used to decaffinate tea or coffee.

(2) Trichloromethane or Chloroform (CHCI₃) :

• Chloroform in the production of chlorofluoromethane, freon refrigerant R-22.
• It is used as solvent in pharmaceuticals, pesticides, gums, fats, resins and dye industry.
• It is a good source of dichlorocarbene species.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Carbon tetrachloride is used in the manufacture of refrigerants.
• It is used as a dry cleaning agent and as a pesticide for stored grains.
• It is very useful solvent for oils, fats and resins. It serves as a source of chlorine.

(4) Iodoform (CHI₃) :

• Iodoform is used as antiseptic, dressing of wounds and sores.
• On small scale it is used as disinfectant.

(5) Freons :

• Freons are widely used as propellants in aerosol, products of food, cosmetics and pharmaceutical industries.
• Freons containing bromine in their molecules are used as fire extinguishers.
• They are used in aerosol insecticides, solvent for cleaning clothes and metallic surfaces.
• It is used as foaming agents in the preparation of foamed plastics and in production of certain fluorocarbons.
• It is used as refrigerants and air conditioning purposes.

Question 80.
State the environmental effects of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freon.s (CCI₂F₂, CCI₃F, CHCIF₂)

Answer:
(1) Dichloromethane (CH₂CI₂) :

• Higher levels of dichloromethane in air causes nausea, numbness in fingers and toes, dizziness.
• Lower levels of dichloromethane causes impaired vision and hearing.
• Direct contact with eyes can damage cornea.

(2) Trichloromethane or chloroform (CHCI₃) :

• When chloroform is exposed to air in the presence of sunlight, it slowly oxidised to phosgene, a poisonous compound, therefore it is stored in dark, amber coloured bottles.
• Chloroform vapour when inhaled for a short time causes dizziness, headache and fatigue and if inhaled for a long time affects central nervous system.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Exposure to carbon tetrachloride causes eye irritation, damages nerve cells, vomiting sensation, dizziness, unconciousness or death. Long exposure to chloroform may affect liver.
• When mixed with air it causes depletion of the ozone layer, which affects human skin leading to cancer.

(4) Iodoform (CHI₃) : Iodoform has a strong smell. It causes irritation to skin and eyes. It may cause respiratory irritation or breathing difficulty, dizziness, nausea, depression of central nervous system, visual disturbance.

(5) Freons (CCI₂F₂, CCI₃F, CHCIF₂) :

• Freon as refrigerant causes ozone depletion.
• Freons have low toxicity and low biological activity.
• Freons from propane group are more toxic in nature.
• Regular large inhalation of freon results in breathing problems, organ damage, loss of consciousness.

(6) DDT :

• DDT is not readily metabolised by animals.
• It is deposited and stored in fatty tissues.
• Exposure to high doses of DDT may cause vomiting, tremors or shakiness.
• Laboratory animal studies showed adverse effect of DDT on liver and reproduction.
• DDT is a pressistent organic pollutant, readily absorbed in soils and tends to accumulate in the ecosystem.
• When dissolved in oil or other lipid, it is readily absorbed by the skin. It is resistant to metabolism.
• There is a ban on use of DDT due to all these adverse effects.

Question 81.
What is the chemical name of freon?
Answer:
The chemical name of freon is Dichlorodifluoromethane.

Question 82.
What is the chemical name of DDT ?
Answer: The chemical name of DDT is p, p’-Dichlorodiphenyltrichloroethane.

Activity :
(1) Collect detailed information about Freons and their uses.
(2) Collect information about DDT as a persistent pesticide.
Reference books :
(1) Organic chemistry by Morrison, Boyd, Bhattacharjee, 7th edition, Pearson.
(2) Organic chemistry by Finar, Vol 1, 6th edition, Pearson

Multiple Choise Questions

Question 83.
Select and write the most appropriate answer from the given alternatives for each sub-question :

Question 1.

Answer:
(b) CH₃ – CH₂ – CH₂ – I

Question 2.
The rate of S_N2 reaction depends on the concentra¬tion of
(a) only the substrate
(b) only the reagent
(c) both the substrate and the reagent
(d) neither the substrate nor the reagent
Answer:
(c) both the substrate and the reagent

Question 3.
In S_N2 reaction, the hydrolysis of alkyl halide shows
(a) the retention of configuration
(b) the inversion of configuration
(c) both retention and inversion of configuration
(d) no change in the configuration
Answer:
(b) the inversion of configuration

Question 4.
The one step exothermic reaction is
(a) S_N1
(b) S_N2
(C) S_N
(d) S2_N
Answer:
(b) S_N2

Question 5.
Which of the following is correct about S_N2 mechanism?
(a) Two step reaction
(b) Complete inversion of configuration
(c) Formation of carbonium ion
(d) Favoured by polar solvent
Answer:
(b) Complete inversion of configuration

Question 6.
Which of the following is not a nucleophile?
(a) Ammonia
(b) Ammonium ion
(c) Primary amine
(d) Secondary amine
Answer:
(d) Secondary amine

Question 7.
Which of the following undergoes nucleophilic substitution exclusively by S_N2 mechanism ?
(a) ethyl chloride
(b) isopropyl chloride
(c) chlorobenzene
(d) benzyl chloride
Answer:
(d) benzyl chloride

Question 8.
Which of the following is most reactive towards nucleophilic substitution reaction ?
(a) CH₂ = CH – CI
(b) CH₃CH = CHCI
(c) C₆H₅CI
(d) CICH₂ – CH = CH₂
Answer:
(d) CICH₂ – CH = CH₂

Question 9.
The stability order of carbocation is
(a) 2° > 3° > 1°
(b) 3° > 2° > 1°
(c) 3° > 1° > 2°
(d) 1° > 3° > 2°
Answer:
(b) 3° > 2° > 1°

Question 10.

(a) ethane
(b) propane
(c) n-butane
(d) n-pentane
Answer:
(c) n-butane

Question 11.
Which of the following characteristic properties of the enantiomers is correct?
(a) The enantiomers possess same physical and chemical properties
(b) The enantiomers are optically active compounds
(c) The enantiomers have different optical rotations
(d) All of these
Answer:
(d) All of these

Question 12.
The optically inactive compound is
(a) glucose
(b) lactic acid
(c) isopropyl alcohol
(d) 2-bromo butane
Answer:
(c) isopropyl alcohol

Question 13.
A compound with the molecular formula CH₂OH(CHOH)₃CH₂OH has optically active forms
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 14.
A racemic mixture consists of
(a) equal amount of d and l isomers
(b) unequal amounts of d and / isomers
(c) unknown amounts of d and / isomers
(d) only d isomers
Answer:
(a) equal amount of d and l isomers

Question 15.
Which of the following compounds is not optically active ?
(a) Lactic acid
(b) Secondary butyl chloride
(c) n-propyl iodide
(d) Glucose
Answer:
(c) n-propyl iodide

Question 16.
Which of the following compounds shows optical activity ?
(a) n-butyl chloride
(b) isobutyl chloride
(c) sec-butyl chloride
(d) t-butyl chloride
Answer:
(c) sec-butyl chloride

Question 17.
The major product of the following reaction is

Answer:
(c)

Question 18.
The above reaction is known as

(a) Wurtz-Fittig reaction
(b) Friedel Craft’s reaction
(c) Sandmeyer’s reaction
(d) Swarts reaction
Answer:
(b) Friedel Craft’s reaction

Question 19.
Iodoform is used as
(a) an anaesthetic
(b) an antiseptic
(c) an analgesic
(d) an antibiotic
Answer:
(b) an antiseptic

Question 20.
p, p’-dichlorodiphenyl trichloroethane is used as
(a) insecticide
(b) anaesthetic
(c) antiseptic
(d) refrigerant
Answer:
(a) insecticide

Question 21.
The order of reactivity in nucleophilic substitution reaction is
(a) CH₃F < CH₃C1 < CH₃I < CH₃Br
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I
(c) CH₃F < CH₃Br < CH₃C1 < CH₃I
(d) CH₃I < CH₃Br < CH₃C1 < CH₃F
Answer:
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I

Question 22.
Racemate is
(a) optically active
(b) optically dextro rotatory
(c) optically inactive
(d) optically laevorotatory
Answer:
(c) optically inactive

Question 23.
The number of asymmetric carbon atoms in glucose are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 24.
The geometry of carbon lum ion is
(a) Tetrahedral
(b) planar
(c) linear
(d) pyramidal
Answer:
(b) planar

Question 25.
In its nucleophilic substitution reaction, aryl halide resembles
(a) Vinyl chloride
(b) allyl chloride
(e) Benzyl chloride
(d) ethyl chloride
Answer:
(a) Vinyl chloride

Question 26.
The weakest C-Cl bond is present in

Answer:
(d)

Question 27.
Which alkyl halide among the following com¬pounds has the highest boiling point ?
(a) (CH₃)₃CCI
(b) CH₃CH₂CH₂CH₂CI
(c) CH₃CH₂CH₂C1
(d) CH₃CH(CH₃)CH₂CI
Answer:
(b) CH₃CH₂CH₂CH₂CI

Question 28.
It is difficult to break C-Cl bond in CH₂ = CH – CI due to
(a) Hyper conjugation
(b) Resonance
(c) Electromeric effect
(d) Inductive effect
Answer:
(b) Resonance

Question 29.
Which one of the following when heated with metallic sodium will not give the corresponding alkane ?

Answer:
(c)

Question 30.
The most reactive alkyl halide towards SN2 reac¬tion is
(a) CH₃X
(b) R₃CX
(C) R₂CHX
(d) RCH₂X
Answer:
(a) CH₃X

Question 31.
The number of electrons surrounding the carbon- ium ion is
(a) 6
(b) 8
(c) 10
(d) 7
Answer:
(a) 6

Question 32.
The lowest stability of carbocation among the compounds

Answer:
(a)

Question 33.
Carbon atom in methyl carbocation contains how many pairs of electrons?
(a) 8
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 34.
The optically inactive compound is
(a) Glucose
(b) Lactic acid
(c) 2-Chlorobutane
(d) 2-Chloropropane
Answer:
(d) 2-Chloropropane

Question 35.
The hydrogen halide which does not obey Markownikv rule in presence of peroxide is
(a) HC1
(b) HBr
(c) HF
(d) HI
Answer:
(b) HBr

Question 36.
Which one of the following is NOT used to prepare alkyl halide from an alcohol ?
(a) SOCl₂
(b) PC1₃
(c) HC1 + ZnCl₂
(d) NaCl
Answer:
(d) NaCl

Question 37.
The total number of electrons present in the central carbon atom of a free radical is
(a) 7
(b) 8
(c) 9
(d) 6
Answer:
(a) 7

Question 38.
In which of the following pairs both are nucleophiles ?
(a) BF₃, AICI₃
(b) NO⁺₂, Cl^–
(c) CN^–, NH₃
(d) Br⁺, BC1₃
Answer:
(c) CN^–, NH₃

Question 39.
Which one of the following alkane is NOT formed in Wurtz reaction ?
(a) Methane
(b) Ethane
(c) Propane
(d) Butane
Answer:
(a) Methane

Question 40.
Which of the following groups has highest priority according to R, S convention?
(a) CH₂OH
(b) COOH
(c) COCH₃
(d) COOCH₃
Answer:
(d) COOCH₃

Question 41.
The halogen atom in aryl halides is
(a) o- and p-di reefing
(b) m-directing
(c) o, m and p-di reefing
(d) only m-directing
Answer:
(a) o- and p-di reefing

Question 42.
Chlorobenzene can be obtained by benzene diazonium chloride by
(a) Friedel Craft’s reaction
(b) Wurtz reaction
(c) Gatterman’s reaction
(d) Fittig reaction
Answer:
(c) Gatterman’s reaction

Question 43.
Which of the following carbocations is least stable ?


Answer:
(c)

Question 44.
But-l-ene on reaction with HCI in the presence of sodium peroxide yields
(a) n-butyl chloride
(b) isobutyl chloride
(c) secondary butyl chloride
(d) tertiary butyl chloride
Answer:
(c) secondary butyl chloride

Question 45.
Carbon tetrachloride is used as
(a) anaesthetic
(b) antiseptic
(c) dry cleaning agent
(d) fire extinguisher
Answer:
(c) dry cleaning agent

Question 46.
Identify the product D in the following sequence of reactions :

(a) 2, 2-dimethyl butane
(b) 2, 3-dimethyl butane
(C) hexane
(d) 2, 4-dimethylpentane
Answer:
(b) 2, 3-dimethyl butane

Question 47.
The preparation of alkyl fluoride from alkyl chlor ide, in presence of metallic fluorides is known as
(a) Williamson’s reaction
(b) Finkeistein reaction
(c) Swarts reaction
(d) Wurlz reaction
Answer:
(c) Swarts reaction

Question 48.
UPAC name of the following compound is

(a) 3-Bromo-3, 4-dimetbyiheptane
(b) 3,4-dimethyl-3-bromoheptane
(c) 5-Bromo-4,5-dimethylheptane
(d) 4,5-dimethyl-5-bromoheptane
Answer:
(a) 3-Bromo-3, 4-dimetbyiheptane

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 9 Coordination Compounds Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 9 Coordination Compounds

Question 1.
What are double salts?
Answer:
Double salts are crystalline molecular or addition compounds containing more than one salt in simple molecular proportions soluble in water and in solution they ionise and exhibit all the properties of the constituent ions.

For example, K₂SO_4⁺ A1₂(SO₄)_3^– 24H₂O
\(\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{~K}_{(\mathrm{aq})}^{+}+2 \mathrm{Al}_{(\mathrm{aq})}^{3+}+4 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+24 \mathrm{H}_{2} \mathrm{O}_{0}\)

Question 2.
Define coordination compound.
Answer:
Coordination compound : It consists of a central metal ion or atom surrounded by atoms, molecules or anions called ligands by coordinate bonds, e.g. cisplatin Pt(NH₃)₂Cl₂, [Cu(NH₃)₄]SO₄.

Question 3.
Define Lewis bases and Lewis acids with respect to a coordination compound.
Answer:

• Lewis bases : In a coordination compound the ligands being electron pair donors they are Lewis bases.
• Lewis acids : The central metal atom or ion being electron acceptor behaves as a Lewis acid.
• For example, in the coordination compound, [Cu(NH₃)₄]²⁺, NH₃ is a Lewis base and Cu²⁺ is a Lewis acid.

Question 4.
Define coordination sphere. Give example.
Answer:
Coordination sphere : A coordination entity consisting of a central metal atom or ion and the coordinating groups like neutral molecules or anions (ligands) written inside a square bracket is together called coordination sphere. This is a discrete structural unit. The ionisable groups (generally ions) called counter ions are written outside the bracket.

For example, in the coordination compound K₄[Fe(CN)₆], the coordination sphere is [Fe(CN)₆]^4- while K⁺ represents counter ion.

Question 5.
Define and explain charge number of a complexion.
Answer:
Charge number of a complexion : The net charge carried by a complexion or a coordination entity is called its charge number.

Explanation :
(i) Charge number is equal to the algebraic sum of the charges carried by central metal atom or ion and all the ligands attached to it.
(ii) E.g. consider anionic complex, [Fe(CN)₆]^4-.
Charge number of [Fe(CN)₆]^4- = Charge on Fe²⁺ ions + 6 x charge on CN^– = ( + 2) + 6( -1) = – 4 Hence charge number of [Fe(CN)₆]^4- is – 4.

Question 6.
Explain the oxidation state of a metal in a complex.
Answer:

• The oxidation state of a metal atom or ion in the complex is the apparent charge carried by it in the complex.
• It depends upon the atomic number and electronic configuration of the metal atom or ion.
• The coordination number, the formula and geometry of a complex depend upon the oxidation state of the metal
  atom or ion.

Question 7.
What is the charge on a monodentate ligand X in the complex, [NiX₄]^2-?
Answer:
The charge number of the complex ion is – 2. Nickel being divalent, its oxidation state is + 2. If the charge on monodentate ligand X is y, then Charge number = charge on Ni²⁺ charge on 6X – 2 = + 2 + 4 xy
∴ y = – 1
Hence the charge on ligand X is – 1.

Question 8.
Calculate the oxidation state of a metal in the following complexes :
(a) [Fe(NH₃)₆](NO₃)₃
(b) Ni(CO)₅.
Answer:
(a) [Fe(NH₃)₆](NO₃)₃ ⇌ [Fe(NH₃)₆]³⁺ + 3NO₃^–

NH₃ is a neutral ligand, and the charge number of complex ion is + 3.
If the oxidation state of Fe is x then,
+ 3 = x + 6(0)
∴ x = + 3
∴ The oxidation state of Fe is +3.

(b) Ni(CO)₅ is a neutral complex and CO is a neutral ligand. If the oxidation state of Ni is x, then zero = x + 5 x (zero)
∴ x = zero.
The oxidation state of Ni is zero.

Question 9.
Define and explain the term coordination number (C.N.) of a metal in the complex.
Answer:
Coordination number or legancy (C.N.) : The number of (monodentate) ligands which are directly bonded by coordinate bonds to central metal atom or ion in a coordination compound is called coordination number (C.N.) of the metal atom or ion.

Explanation :

• The coordination number (C.N.) is a characteristic property of the metal and its electronic configuration.
• C.N. takes the values from 2 to 10, of which 4 and 6 are very common.
• The light transition metals show C.N. 4 and 6 while the heavier transition metals show C.N. 8.
• The geometry and shape of a complex compound depends upon C.N. of the metal.

Question 10.
Mention primary valence, secondary valence and coordination number in the following complexes :
(a) [Cu(NH₃)JCI₂
(b) [Co(NH₃)₃CI₃]
(C) K4[Fe(CN)₆]
(d) [CoF₆]₃
(e) [Pt(NH₃)₂Cl₂]
(f) [Pt(NH₃)₂(P_y)₃CI₂]
(g) Cr(CO)₆
(h) [Ni(CN)₄]^2-
Answer:

Question 11.
Classify the following complexes as homoleptic and heteroleptic complex :
(a) [Cu(NH₃)₄]SO₄;
(b) [Cu(en)₂(H₂O)CI]²⁺
(c) [Fe(H₂O)₅(NCS)]²⁺
(d) Tetraaminezinc(II) nitrate.

Question 12.
Summarise the rules of IUPAC nomenclature of coordination compounds.
Answer:
Following rules are followed for naming coordination compounds recommended by IUPAC :

1. In case of a complexion or a neutral molecule, name the ligand first and then the metal.
2. The names of anionic ligands are obtained by changing the ending -ide to -o and -ate to -ato.
3. The name of a complex is one single word. There must not be any space between different ligand names as well as between ligand name and the name of the metal.
4. After the name of the metal, write its oxidation state in Roman number which appears in parentheses without any space between metal name and parentheses.
5. If complex has more than one ligand of the same type, the number is indicated with prefixes, di-, tri-, tetra-, penta-, hexa- and so on.
6. For the complex having more than one type of ligands, they are written in an alphabetical order. Suppose two ligands with prefixes are tetraaqua and dichloro. While naming in alphabetical order, tetraaqua is written first and then dichloro.
7. If the ligand itself contains numerical prefix in its name, then display number by prefixes bis for 2, tris for 3, tetrakis for 4 and so forth. Put the ligand name in parentheses. For example, (ethylenediamine)3 or (en)3 would appear as tris (ethylenediamine) or tris(ethane-l, 2-diamine).
8. The metal in cationic or neutral complex is specified by its usual name while in the anionic complex the name of metal ends with ‘ate’.

Question 13.
State effective atomic number (EAN).
OR
State and explain effective atomic number (EAN). How is it calculated?
Answer:
Effective atomic number (EAN) : It is the total number of electrons present around the central metal atom or ion and calculated as the sum of electrons of metal atom or ion and the number of electrons donated by ligands.

It is calculated by the formula : EAN = Z – X + Y where.
Z = Atomic number of metal atom
X = Number of electrons lost by a metal atom forming a metal ion
Y = Total number of electrons donated by all ligands in the complex.

Generally the value of EAN is equal to the atomic number of the nearest inert element.

Explanation : Consider a complex ion [Co(NH₃)₆]³⁺
Oxidation state of cobalt is + 3 hence X = 3.
There are six ligands, hence Y = 2 x 6 = 12
Atomic number of cobalt. Z = 27
∴ EAN = Z – X + Y = 27 – 3 + 12 = 36.

Question 14.
Find effective atomic number (EAN) in the following complexes :
(1) [Ni(CO)₄]
(2) [Fe(CN)₆]^4-
(3) [Co(NH₃)₆]³⁺
(4) [Zn(NH₃)J²⁺
(5) [Pt(NH₃)₆]⁴⁺
Answer:

Question 15.
What is effective atomic number (EAN) in the following complexes ?
(1) [Fe(CN)₆]^3-
(2) [CU(NH₃)₄]²⁺
(3) [Pt(NH₃)₄]²⁺
Answer:

Question 16.
Calculate EAN in the following complexes :
(1) [Cr(H₂O)₂(NH₃)₂(en)]CI₃;
(2) [Ni(en)₂]SO₄;
(3) Na3[Cr(C₂O₄)₃].
Answer:

Question 17.
Define in coordination compounds :
(1) Isomerism
(2) Isomers.
Answer:

1. Isomerism : It is the phenomenon in coordination compounds having same molecular formula but different physical and chemical properties due to different arrangements of the ligands around the central metal atom or ion in the space.
2. Isomers : The isomers are the coordination compounds having same molecular formula but different physical and chemical properties due to the difference in arrangements of the ligands in the space.

Question 18.
Mention the types of isomerisms in coordination compounds.
Answer:
There are two principal types of isomerisms in coordination compounds as follows :
(A) Stereoisomerism
(B) Structural isomerism (OR Constitutional isomerism)

(A) Stereoisomerism is further classified as :

• Geometrical isomerism
• Optical isomerism

(B) Structural isomerism is further classified as :

• Ionisation isomerism
• Linkage isomerism
• Coordination isomerism
• Solvate (or hydrate) isomerism

Question 19.
Why does stereoisomerism arise in the coordination compounds?
Answer:
In the coordination compounds (complexes) the ligands are linked to the central metal atom or ion by coordinate bonds which are directional in nature and hence give rise to the phenomenon of stereoisomerism.

In this isomerism, the different stereoisomers have different arrangements of ligands (atoms, molecules or ions) in space around the central metal atom or ion. Hence they have different physical and chemical properties and give rise to the phenomenon of stereoisomerism.

Question 20.
Define, in coordination compounds : (1) Stereoisomerism (2) Stereoisomers.
Answer:
(1) Stereoisomerism The phenomenon of isomerism in the coordination compounds arising due to different spatial positions of the ligands in the space around the central metal atom or ion is called stereoisomerism.

(2) Stereoisomers : The coordination compounds having same molecular formula but different stereoisomerism due to different spatial arrangements of the ligand groups in the space around the central metal atom or ion are called stereoisomers.

Question 21.
Define :
(1) Geometrical isomerism and
(2) Geometrical isomers.
Answer:
(1) Geometrical isomerism : The phenomenon of isomerism in the heteroleptic coordination compounds with the same molecular formula but different spatial arrangement of the ligands in the space around the central metal atom or ion is called geometrical isomerism.

(2) Geometrical isomers : The heteroleptic coordination compounds having same molecular formula but different geometrical isomerism due to different spatial arrangements of the ligands in the space around the central metal atom or ion are called geometrical isomers.

Question 22.
Define cis and trails isomers in the coordination compounds.
Answer:
(1) Cis-isomer : A heteroleptic coordination compound in which two similar ligands are arranged adjacent to each other is called cis-isomer. For example,

Gis-Diamminedichloroplatinum(II)

(2) Trans-isomer : A heteroleptic coordination compound in which two similar ligands are arranged diagonally opposite to each other is called trans-isomer. For example,

Trans-Diamminedichloroplatinum(II)

Question 23.
Write structures for geometrical isomers of Diamminebromochloroplatinum(II).
Answer:

Question 24.
Explain the geometrical isomerism of the octahedral complex of the type [MA₄B₂] with a suitable example.
Answer:

• Consider an octahedral complex of a metal M with coordination number six and monodentate ligands a and b having formula [MA₄B₂],
• CA-isomer is obtained when both the B ligands occupy adjacent (1,2) positions.
• Trans-isomer is obtained when the ligands B occupy the opposite (1,6) positions.
• For example, consider a complex [CO(NH₃)₄CI₂]⁺. The structures of cis and trans isomers are
  

Question 25.
Explain the geometrical isomerism of the octahedral complex of the type [M(AA)₂B₂]^n± with a suitable example.
Answer:

• Consider an octahedral complex of metal M with coordination number six and a bidentate ligand AA and monodentate ligand B having molecular formula [M(AA)₂B₂]^n± .
• Bidentate ligand AA has two identical coordinating atoms.
• Cis- isomer is obtained when two bidentate AA ligands as well as two ‘B’ ligands are at adjacent positions.
• Trans-isomer is obtained when two AA ligands and two B ligands are at opposite positions.
• For example, consider a complex [Co(en)₂CI₂]⁺.
  

Question 26.
Explain the geometrical isomerism of the octahedral complex of the type [MA4BC] with suitable example.
Answer:

• Consider an octahedral complex of metal M with coordination number six and monodentate ligands A, B andC.
• Cis-isomer is obtained when both the ligands B and C occupy adjacent (1,2) positions.
• Trans-isomer is obtained when the ligands B and C occupy opposite positions.
• For example, consider a complex [Pt(NH₃)₄BrCI] of the type [MA₄BC],
  

Question 27.
Define : (1) Optical isomerism (2) Optical isomers.
Answer:
(1) Optical isomerism : The phenomenon of isomerism in which different coordination compounds having same molecular formula have different optical activity is called optical isomerism.

(2) Optical isomers : Different coordination compounds having same molecular formula but different optical activity
are called optical isomers.

Question 28.
Explain : (1) Plane polarised light (2) Optical activity.
Answer:
(1) Plane polarised light : A monochromatic light having vibrations only in one plane is called a plane polarised light. This light is obtained by passing monochromatic light through NICOL prism.

(2) Optical activity : A phenomenon of rotating a plane of a plane polarised light by an optically active substance is
called optical activity. This substance is said to be optically active.

Question 29.
Explain : (1) Dextrorotatory substance (2) Laevorotatory substance. (1 mark each)
Answer:

1. Dextrorotatory substance : An optically active substance which rotates the plane of a plane polarised light to right hand side is called dextrorotatory or d isomer denoted by d.
2. Laevorotatory substance : An optically, active substance which rotates the plane of a plane polarised light to the left hand side is called laevorotatory or l isomer and denoted by l.

Question 30.
What are the conditions for the optical isomerism in coordination compounds?
Answer:

• Optical isomerism is exhibited by those coordination compounds which possess chirality.
• There should not be the presence of element of symmetry which makes the complex optically inactive.
• The mirror images of the complex molecule or ion must be non-superimposable with the molecule or ion. B

Question 31.
What are enantiomers?
Answer:
Enantiomers : The two forms of the optical active complex molecule which are mirror images of each other are called enantiomers.

There are two forms of enantiomers, d form and l form.

Question 32.
Draw diagrams for the optical isomers of a complex, [Co(en)₃]³⁺.
Answer:
The complex [Co(en)₃]³⁺ has two optical isomers.

Question 33.
Explain the optical isomerism in the octahedral complex with two symmetrical bidentate chelating ligands.
Answer:
The octahedral complexes of the type [M(AA)2Q2]”±, in which two symmetrical bidentate chelating ligands like AA and two monodentate ligands like a are coordinated to the central metal atom or ion exhibit optical isomerism and two optical isomers d and l can be resolved. For example, [Pt Cl2(en)2]2.

The cis-form is unsymmetrical and optically active while the trans-form is symmetrical and hence optically inactive. The optical isomers of cis-form (d and ) of this complex along with trans-form are shown below,

Question 34.
When are optical isomers called chiral?
Answer:
When the mirror images of optical isomers of the complex are nonsuperimposable they are said to be chiral. For example, [Co(en)₂(NH₃)₂]³⁺.

Question 35.
Define and explain ionisation isomerism.
Answer:
Ionisation isomerism : The phenomenon of isomerism in the metal complexes in which there is an exchange of ions between coordination (or inner) sphere and outer sphere is known as ionisation isomerism.

Explanation :

• Ionisation isomers have same molecular formula but different arrangement of ions in the inner sphere and outer sphere in the complex,
• Hence on ionisation, these ionisation isomers produce different ions in the solution. This ionisation isomerism is also called ion-ion exchange isomerism.

Examples:
(A) [CO(NH₃)₄CI₂]Br and (B) [Co(NH₃)₄CIBr] Cl

Ionisation :
(A) [CO(NH₃)₄CI₂] Br ⇌ [CO(NH₃)₄CI₂]⁺ + Br-
(B) [Co(NH₃)₄CIBr]Cl ⇌ [CO(NH₃)₄ClBr]⁺ + CP
In the isomers (A) and (B), there is an exchange of ions namely Br^– and CI^–.

Question 36.
Define : (1) Linkage isomerism (2) Linkage isomers.
OR
What is linkage isomerism ? Explain with an example.
Answer:
(1) Linkage isomerism : The phenomenon of isomerism in which the coordination compounds have same metal atom or ion and same ligand but bonded through different donor atoms or linkages is known as linkage isomerism.

(2) Linkage isomers : The coordination compounds having same metal atom or ion and ligand but bonded through different donor atoms or linkages are called linkage isomers.
For example : Nitro complex [CO(NH₃)₅NO₂]CI₂ = (Yellow) and nitrito complex [CO(NH₃)₅ONO]CI₂ (Red)

Question 37.
Explain linkage isomers with NO₂^– group as a ligand.
Answer:
(1) Nitro group (_–NO₂^–) is an ambidentate ligand. NO₂^– group may link to central metal atom, through N or O.
(2) The two linkage isomers are, [CI: → Ag ← : NO₂]^– and [CI: → Ag ← O-NO]^–
Choloronitroargentate(I) ion and Chloronitritoargentate(I) ion

Question 38.
Write linkage isomers of a complex having constituents Co³⁺, 5NH3 and NO^–₂.
Answer:
(i) NO^–₂ is an ambidentate ligand which can be linked through N or O.
(ii) The linkage isomers are as follows :
(a) [CO(NH₃)₅(NO₂)]²⁺ Pentaamminenitrocobalt(III) ion
(b) [CO(NH₃)₅(ONO)]²⁺ Pentaamminenitritocobalt(III) ion

Question 39.
Define: (1) Coordination isomerism (2) Coordination isomers.
Answer:
(1) Coordination isomerism : The phenomenon of isomerism in the ionic coordination compounds having the same molecular formula but different complex ions involving the interchange of ligands between cationic and anionic spheres of different metal. ions is called coordination isomerism.

(2) Coordination isomers : The ionic coordination compounds having same molecular formula but different complexions duc to interchange of ligands between cationic and anionic spheres of different metal ions are called coordination isomers.

For example,

Question 40.
Give three examples of coordination isomers. (I mark each)
Answer:

• [Cu(NH₃)₄] [PICI₄] and I Pt(NH₃)₄] [ICuCl4₄]
• [Cr(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₄(CN)₂] [Cr(NH₃)₂(CN)₄]
• [Cr(NH₃)₆] [Cr(SCN)₆] and [Cr(NH₃)₄(SCN)₂] [Cr(SCN)₄(NH₃)₂]

Question 41.
Define Solvate or Hydrate isomerism.
Answer:
Solvate ate or Hydrate isomerism : The phenomenon of isomerism in the coordination compounds arising due to the exchange of solvent or H₂O molecules inside the coordination sphere and outer sphere of the complex is known as solvate or hydrate isomerism.

Question 42.
Define solvate or hydrate isomers.
OR
What are hydrate isomers? Explain with examples.
Answer:
Solvate or Hydrate isomers : The coordination compounds having the same molecular formula but differ in the number of solvent or H₂O molecules inside the coordination sphere and outer sphere of the complexes are called solvate or hydrate isomers.

For example : [Cr(H₂O)₆] CI₃; [Cr(H₂O)₅CI]CI₂ H₂O; and [Cr(H₂O)₄CI₂] CI 2H₂O.

Question 43.
A coordination compound has the formula COCI₃ 6H₂O. Write the hydrate isomers of the complex.
Answer:
The possible hydrate isomers of the coordination compounds having molecular formula COCI₃ 6H₂O are as follows :
(1) [CO(H₂O)₆]CI₃;
(2) [CO(H₂O)₅CI]CI₂ H₂O
(3) [CO(H₂O)₄CI₂] Cl – 2H₂O
(4) [Co(H₂O)₃ CI₃] 3H₂O.

Question 44.
Explain the steps involved in describing the bonding in coordination compounds using valence bond theory.
Answer:

• Vacant d-orbitais of metal ion form coordination bonds with ligands.
• s, p orbitais along with vacant d-orbitais of metal ion take part in hybridisation.
• The number of vacant hybrid orbitais formed is equal to number of hybridising orbitais which is equal to the number of ligand donor atoms or coordination number of the metal.
• The metal-ligand coordination bonds are formed by the overlap between the vacant hybrid orbitais of metal and the filled orbitais of the ligands.
• The hybrid orbitais used by the metal ion point in the direction of the ligands.
• When inner (n – 1)d orbitais of metal ion are used in the hybridisation then the complex is called (a) inner orbital complex while when outer nd orbitais are used, complexes are called (b) outer orbital complexes. .

Question 45.
Explain the steps involved ¡n the metal-ligand bonding.
Answer:

• Find the oxidation state of central metal ion in the complex.
• Write the valence shell electronic configuration of metal ion.
• From the formula of the complex determine the number of ligands and find the number of metal ion orbitais required for bonding.
• Find the orbitais of metal ion available for hybridisation and the type of hybridisation involved.
• Represent the electronic configuration of metal ion after hybridisation.
• Exhibit filling of hybrid orbitais after complex formation.
• Determine the nunther of unpaired electrons and predict magnetic property of the complex.
• Find whether the complex is low spin or high spin (applicable for octahedral complexes with d⁴ or d⁸ electronic configuration.)

Question 46.
What are the salient features of valence bond theory (VBT)?
Answer:
The salient features of valence bond theory (VBT) are as follows :

1. According to this theory, a central metal atom or ion present in a complex provides a definite number of vacant orbitals (s, p, d and) to accommodate the electrons from the ligands for the formation coordinate bonds with the metal ion atom.
2. The number of vacant orbitals provided by the central metal atom or ion is the same as the coordination number of the metal. For example : Cu²⁺ provides 4 vacant orbitals in the complex. [Cu(NH₃)₄]²⁺.
3. The vacant orbitals of metal atom or ion undergo hybridisation forming the same number of hybridised orbitals, since the bonding with the hybrid orbitals is stronger.
4. Each ligand has one or more orbitals containing one or more lone pairs of electrons.
5. The shape or geometry of the complex depends upon the type of hybridisation of the metal atom.
6. When inner orbitals namely (n – 1) d orbitals in transition metal atom or ion hybridise, the complex is called inner complex and when outer orbitals i.e., nd orbitals hybridise then the complex is called outer complex.
7. When the central metal atom or ion in the complex contains one or more unpaired electrons the complex is
   paramagnetic while if all the electrons are paired, the complex is diamagnetic.

Question 47.
What is the spin pairing process in the coordination compound?
Answer:
When the ligands approach the metal atom or ion for the formation of a complex, they influence the valence electrons of metal atom or ion. Accordingly the ligands are classified as (A) strong ligands and (B) weak ligands.

(A) Strong ligands :

• They cause the pairing of unpaired electrons present in the metal atom or ion.
• Spin pairing process :
  • The process of pairing of unpaired electrons in metal atom or ion due to the presence of ligands in the complex is called spin pairing process.
  • This spin pairing process decreases the number of unpaired electrons and hence decreases the paramagnetic character of the complex.
  • The strong ligands also promote the outer ns electrons to the vacant inner (n – 1)d orbitals.

(B) Weak ligands : The weak ligands have no effect on the electrons in the valence shell of a metal atom or ion.
Strong ligands : CO, CN, ethylenediammine (en), NH₃, EDTA, etc.
Weak ligands : CI^–, I^–, OH^–, etc.

[Note : If a complex has n number of unpaired electrons then the magnetic moment, μ is given by ‘spin only’ formula μ = n(n + 2) B.M. where B.M. (Bohr Magneton) is the unit of magnetic moment. Hence from the magnitude of p, the number of unpaired electrons in the complex and its structure can be evaluated.]

Question 48.
Explain the structure of octahedral complex, [CO(NH₃)₆]³⁺ on the basis of valence bond theory.
Answer:
(1) Hexaamminecobalt(III) ion, [CO(NH₃)₆]³⁺ is a cationic complex, the oxidation state of cobalt is + 3 and the coordination number is 6.

(2) Electronic configuration : ₂₇CO [Ar]¹⁸ 3d⁷ 4s²
Electronic configuration : Co³⁺ [Ar]¹⁸ 3d⁶ 4s° 4p°

(3) Since NH₃ is a strong ligand, due to spin pairing effect, all the four unpaired electrons in 3d orbital are paired giving two vacant 3d orbitals.

(4) Since the coordination number is Co³⁺ ion gets six vacant orbitais by hybridisation of two 3d vacant orbitais, One 4s and three 4p orbitais forming six d²sp³ hybrid orbitais giving octahedral geometly. It is an inner complex.

(5) 6 lone pairs of electrons from 6NH₃ ligands are accommodated in the six vacant d²sp³ hybrid orbitals. Thus six hybrid orbitals of Co³⁺ overlap with filled orbitals of NH³ forming 6 coordinate bonds giving octahedral geometry to the complex.

Since the complex has all electrons paired, it is diamagnetic.

Question 49.
Explain the geometry of [CoF₆]^3- on the basis of valence bond theory.
Answer:
(1) Hexafluorocobaltate(III) ion, [CoF]^3- is an anionic complex, the oxidation state of cobalt is +3 and the coordination number is 6.
(2) Electronic configuration : ₂₇Co [Ar]¹⁸ 3d⁷ 4s² 4p° 4d°
Electronic configuration : Co³⁺ [Ar]18 3d⁶ 4s° 4p° 4d°

(3) Since F is a weak ligand, there is no spin pairing effect and Co³⁺ possesses 4 unpaired electrons.
(4) Since the coordination number is 6, the Co³⁺ ion gets six vacant orbitals by hybridisation of one 45 orbital, three 4p orbitals and two 4d orbitals forming six sp³d² hybrid orbitals giving octahedral geometry.

(5) 6 lone pairs of electrons from 6F^– ligands are accommodated in the six vacant sp³d² hybrid orbitals. Thus six hybrid orbitals of Co³⁺ overlap with filled orbitals of F^– forming 6 coordinate bonds giving octahedral geometry to the complex. It is an outer complex.

As the complex has 4 unpaired electrons it is paramagnetic.
Magnetic movement μ is,

Question 50.
Explain the structure of tetrachloronickelate(II) [NiCI]^2- on the basis of valence bond theory.
Answer:
(1) Tetrachloronickelate(II) ion is an anionic complex, oxidation state of Ni is +2 and the coordination number is 4.
(2) Electronic configuration : ₂₈Ni [Ar]¹⁸ 3d⁸ 4s² 4p°
Electronic configuration : Ni²⁺ [Ar]¹⁸ 3d⁸ 4s° 4p°

(3) Since the coordination number is 4, it gets 4 vacant hybrid orbitals by sp³ -hybridisation of one 4s and three 4p orbitals giving tetrahedral geometry to the complex.

(4) As Cl is a weak ligand, 2 unpaired electrons in Ni²⁺ remain undisturbed.
(5) 4 lone pairs of electrons from 40 ligands are accommodated in the vacant four sp³ hybrid orbitals. Thus four sp³ hybrid orbitals of Ni²⁺ overlap with filled orbitals of Cl^– forming 4 coordination bonds, giving tetrahedral geometry to the complex.
Since the complex has 2 unpaired electrons, it is paramagnetic.

Magnetic moment \(\mu \text { is, } \mu=\sqrt{n(n+2)}=\sqrt{2(2+2)}=2.83 \text { B.M. }\)

Question 51.
Explain the structure of [Ni(CN)₄]^2- on the basis of valence bond theory.
Answer:
(1) Tetracyanonickelate (II) ion, INi(CN)₂]^2- is an anionic complex, oxidation state of Ni is + 2 and the coordination number is 4.
(2) Electronic configuration : ₂₂Ni [Ar]¹⁸ 3d⁸ 4s² 4p°
Electronic configuration: Ni²⁺ [Ai]¹⁸ 3d⁸ 4s° 4p°

(3) Since CN^– is a sa-ong ligand, one of the unpaired electrons in 3d orbital is promoted giving two paired electrons and one vacant 3d orbital.

(4) Since the coordination number is 4, Ni²⁺ gets 4 vacant hybrid orbitais by hybridisation of one 3d, one 4s and two 4p orbitais forming four dsp² hybrid orbitaIs. This has square planar geometry.

(5) 4 lone pairs from 4CN^– ligands are accommodated in the vacant four dsp² hybrid orbitais. Thus four dps² hybrid orbitais of Ni²⁺ overlap with filled orbitais of CN forming 4 coordinate bonds giving square planar geometry to the complex. It is an inner complex.

Since the complex ion has all electrons paired, it is diamagnetic.

Question 52.
What are the limitations of valence bond theory?
Answer:
In case of the coordination compounds, the valence bond theory has the following limitations :
(1) It cannot explain the spectral properties (colours) of the complex compounds.
(2) Even if the magnetic moments can be calculated from the number of unpaired electrons, it cannot explain the magnetic moment arising due to orbital motion of electrons.
(3) It cannot explain why the metal ions with the same oxidation state give inner complexes and outer complexes with different ligands.
(4) In every complex, it cannot explain magnetic properties based on geometry of the complex.
(5) Quantitative interpretations of thermodynamic and kinetic stabilities of the coordination compounds cannot be accounted.
(6) The complexes with weak field ligands and strong field ligands cannot be distinguished.
(7) It cannot predict the tetrahedral and square planar geometry of complexes with coordination number 4.
(8) The order of reactivity of inner complexes of d³, d⁴, d⁵ and d⁶ metal ions cannot be explained.
(9) It cannot explain the rates and mechanisms of reactions of the coordination compounds.

Question 53.
What are the assumptions of Crystal Field Theory (CFT)?
Answer:
Bethe and van Vleck developed Crystal Field Theory (CFT) to explain various properties of coordination compounds. The salient features of CFT are as follows :

1. In a complex, the central metal atom or ion is surrounded by various ligands which are either negatively charged ions (F^–, CI^–, CN^–, etc.) or neutral molecules (H₂O, NH₃, en, etc.) and the most electronegative atom in them points towards central metal ion.
2. The ligands are treated as point charges involving purely electrostatic attraction between them and metal ion.
   • The central metal ion has five, (n – 1)d degenerate orbitals namely d_xy, d_yz, d_zx, d_{(x² – y²)} and d_z².
   • When the ligands approach the metal ion, due to repulsive forces, the degeneracy of <i-orbitals is destroyed and they split into two groups of different energy, t_2g and e_g orbitals. This effect is called crystal field splitting which depends upon the geometry of the complex.
   • The T-orbitals lying in the direction of ligands are affected to a greater extent while those lying in between the ligands are affected to a less extent.
   • Due to repulsion, the orbitals along the axes of ligands acquire higher energy while those lying in between the ligands acquire less energy.
   • Hence repulsion by ligands give two sets of split up orbitals of metal ion with different energies.
   • The energy difference between two sets of d-orbitais after splitting by ligands is called crystal field splitting energy (CFSE) and represented by Δ₀ or by arbitrary term 10D_q. The value of Δ or 10D_q depends upon the geometry of the complex.
     
3. The electrons of metal ion occupy the split d-orbitais according to Hund’s rule. aufbau principle and those orbitais with minimum repulsion and the farthest away from the ligands.
4. CFI’ does not account for overlapping of orbitais of central metal ion and ligands, hence does not consider covalent nature of the complex.
5. From the crystal field stability energy, the stability of the complexes can he known.
   

Question 54.
What is crystal field splitting?
Answer:
The splitting of five degenerate d-orbitals of the transition metal ion into different sets of orbitals (to_2g and e_g) having different energies in the presence of ligands in the complex is called crystal field splitting.

Question 55.
What is crystal field stabilisation energy?
Answer:
Crystal field stabilisation energy (CFSE) It is the change in energy achieved by preferential filling up of the orbitals by electrons in the complex of metal atom or ion.

CFSE is expressed as a negative .quantity i.e., CFSE < 0. Higher the negative value more is the stability of the complex. m

Question 56.
Explain the factors affecting Crystal Field Splitting parameter (Δ₀).
Answer:
Crystal Field Splitting parameter (Δ₀) depends on. (a) Strength of the ligands and (b) Oxidation state of the metal.

(a) Strength of the ligands : Since strong field ligands like CN^–, en, etc. approach closer to the central metal ion, it results in a large crystal field splitting and hence Δ₀ has higher values.
(b) Oxidation state of the metal A metal ion with the higher positive charge draws the ligands closer to it which results in large separation of t_2g and e_g set of orbitals. The complexes involving metal ions with low oxidation state have low values of Δ₀. For example [Fe(NH₃)₆]³⁺ has higher Δ₀ than [Fe(NH₃)₆]²⁺. H

Question 57.
Explain the octahedral geometry of complexes using crystal field theory.
Answer:
(1) In an octahedral complex [MX₆]ⁿ⁺, the metal atom or ion is placed at the centre of regular octahedron while six ligands occupy the positions at six vertices of the octahedron.

(2) Among five degenerate d-orbitals. two orbitals namely d_{x² – y²} and d_z² are axial and have maximum electron density along the axes, while remaining three c-orbitals namely d_xy, d_yz and d_yz are planar and have maximum electron density in the planes and in-between the axes.

(3) Hence, when the ligands approach a metal ion, the orbitals d_{x² – y²} and d_z² experience greater repulsion and the orbitals d_zy, d_yz and d_zx experience less repulsion.

(4) Therefore the energy of d_{x² – y²} and d_z² increases while the energy of d_xy, d_yz and d_zx decreases and five d-orbital lose degeneracy and split into two point groups. The orbitals d_xy, d_yz and d_yz form t_2g group of lower energy while d_{x² – y²} and d_z² form e group of higher energy.

Thus t_2g has three degenerate orbitals while e_g has two degenerate orbitals.

(5) Experimental calculations show that the energy of t_2g orbitals is lowered by 0.4Δ₀ or 4D_q and energy of e_g. orbital is increased by 0.6 Δ₀ or 6D_q Thus energy difference between t_2g and e_g orbitals is Δ₀ or 10D_g which is crystal field splitting energy.

(6) CFSF increases with the increasing strength of ligands and oxidation stale of central metal ion.

Question 58.
Explain the tetrahedral geometry of complexes using crystal field theory.
Answer:
(1) In the tetrahedral complex, [MX₄]^n±, the metal atom or ion is placed at the centre of the regular tetrahedron and the four ligands, are placed at four corners of the tetrahedron.

(2) The ligands approach the central metal atom or ion in-between the three coordinates x, y and The orbitals d_xy, d_yz and d are pointed towards ligands and experience greater repulsion while the axial orbitals d_x² – _y² and d_z² lie in-between metal-ligand bond axes and experience comparatively less experience.

Fig. 9.12 (a) and (b) : Tetrahedral geometry hasing central metal atoll) (M) at the centre and four ligands (L) al the four corners

(3) Therefore energy of d_xy, d_yz and dd_zx orbitals is increased while that of d_x²y² and d_z² is lowered. Hence 5d-orbitals lose their degeneracy and split into two point groups, namely f2? of higher energy (d_xy, d_yz and d_zx) and e_g of lower energy (d_{x² – y²} and d_z²).

(4) The experimental calculations show that the energy of t_2g orbitals is increased by 0.4 Δ₀ or 4D_q and energy of e_g is lowered by 0.6 Δ₀ or 6D_g. Thus energy difference between t_2g and e_g orbitals is Δ₀ or 10D_g which is crystal field splitting energy (CFSE).

(5) This explains that the entry of each electron in e_g orbitals, stabilises the complex by 0.6 Δ₀ or 6D_q While the entry of each electron in t_2g orbitals destabilises the tetrahedral complex by 0.4 Δ₀ or 4D_g.

(6) In case of strong field ligands, the electrons prefer to pair up in e_g orbitals giving low spin (LS) complexes while in case of weak held ligands, the electrons prefer to enter higher energy t_2g orbitals giving more unpaired electrons and hence form high spin (HS) complexes.

Table 9.4 : Properties of complexes

Question 59.
Give valence bond description for the bonding in the complex [VCI₄]^–. Draw box diagrams for free metal ion. Which hybrid prbitals are used by the metal? State the number of unpaired electrons.
Answer:

Since CI^– is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = 2
Type of hybridisation = sp³

Geometry of complex ion = Tetrahedral
The complex ion is paramagnetic.

Question 60.
Write a note on colour in coordination compounds.
Answer:

• A large number of coordination compounds show wide range of colours due to d – d transition of electron and this can be explained by crystal field theory (CFT).
• The complex absorbs the light in one visible region (400 nm to 700 nm) and transmits the light in different visible region giving complementary colour.
• Consider an octahedral purple coloured complex of [Ti(H₂O)₆]³⁺ which absorbs green light and transmits purple colour. Similarly [Cu(H₂O)₆]²⁺ absorbs the light in the red region of radiation spectrum and transmits in the blue region, hence the complex appears blue.
• The absorption of light arises due to d-d transition of electron from lower energy level (t_2g) to higher energy level (e_g) in octahedral complex.
• The energy required for transition depends upon crystal field splitting energy Δ₀. If Δ₀ = ΔE, then the energy of an absorbed photon (hv) is \(\Delta E=h v=\frac{h c}{\lambda}\) where λ, v and c are wavelength, frequency and velocity of the absorbed light.
• Higher the magnitude of Δ₀ or ΔE, higher is the frequency or lower is the wavelength of the absorbed radiation.
• Since Δ₀ depends upon nature of metal atom or ion, its oxidation state, nature of ligands and the geometry of the complex, different coordination compounds have different colours.

Question 61.
Explain the purple colour of the complex, [Ti(H₂O)₆]³⁺ with the help of crystal field theory.
Answer:
(1) [Ti(H₂O)₆]³⁺ is an octahedral complex, oxidation state of titanium is +3 (Ti³⁺) and the coordination number is 6.

(2) Electronic configuration = ₂₂Ti [Ar]¹⁸ 3d² 4s²
Electronic configuration = Ti³⁺ + [Ar]¹⁸ 3d¹
OR

(3) According to crystal field theory, 3d orbitals undergo crystal field splitting giving higher energy e_g, two orbitals and lower energy t_2g, three orbitals.

(4) The crystal field splitting energy (CFSE), Δ₀ is found to be 3.99 x 10^-19 J/ion from the spectrochemical studies.

(5) The absorption of radiation of wavelength λ or frequency v results in the transition of one unpaired electron from Photon energy t_2g orbital to e_g orbital.

(6) The wavelength of the absorbed radiation will be.

(7) Hence the complex [Ti(H₂O)₆]³⁺ absorbs the green radiation of wavelength 498 nm in the visible region and transmits the complementary purple light. Therefore the complex is purple coloured.

Question 62.
An octahedral complex absorbs the radiation of wavelength 620 nm. Find the crystal field splitting energy.
Answer:
Crystal field splitting energy Δ₀ is given by,

Question 63.
What are the applications of coordination compounds ?
Answer:
(1) In biology : Several biologically important natural compounds are metal complexes which play an important role in number of processes occurring in plants and animals.

For example, chlorophyll in plants is a complex of Mg²⁺ ions, haemoglobin in blood is a complex of iron, vitamin B₁₂ is a complex of cobalt.

(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more stable, more effective and can be assimilated easily.

For example, platinum complex [Pt(NH₃)₂CI₂] known as cisplatin is effectively used in cancer treatment. EDTA is used to treat poisoning by heavy metals like lead.

(3) To estimate hardness of water :

• The hardness of water is due to the presence Mg²⁺ and Ca²⁺ ion in water.
• The strong field ligand EDTA forms stable complexes with Mg²⁺ and Ca²⁺. Hence these ions can be removed by adding EDTA to hard water.

Similarly these ions can be selectively estimated due to the difference in their stability constants.

(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to supply continuously the metal ions in small amounts.

For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)₂] and K[Au(CN)₂] are used.

Multiple Choice Questions

Select and write the most appropriate answer from the given alternatives for each subquestion :

Question 1.
The coordination number of cobalt in the complex [Co(en)₂Br₂]CI₂ is
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(c) 6

Question 2.
EDTA combines with cations to form
(a) chelates
(b) polymers
(c) clathrates
(d) non-stoichiometric compounds
Answer:
(a) chelates

Question 3.
Which one of the following compounds can exhibit coordination isomerism?
(a) [Co(en)₂CI₂]Br
(b) [CO(NH₃)₆] [Cr(CN)₆]
(c) [Co(en)₃]CI₃
(d) [CO(NH₃)₅NO₂]CI₂
Answer:
(b) [CO(NH₃)₆] [Cr(CN)₆]

Question 4.
Which of the following compounds can exhibit linkage isomerism?
(a) [Co(en)₃]CI₃
(b) [Co(en)₂CI₂]CI
(c) [Co(en)₂NO₂Br]CI
(d) [Co(NH₃)₅CI]Br₂
Answer:
(c) [Co(en)₂NO₂Br]CI

Question 5.
Oxidation number of cobalt in K[COCI₄] is
(a) +1
(b) -1
(c) +3
(d) -3
Answer:
(b) -1

Question 6.
The correct structure of [Cr(H₂O)₆]³⁺ is ……………………
(a) octahedral
(b) tetrahedral
(c) square pyramidal
(d) trigonal bipyramidal
Answer:
(b) tetrahedral

Question 7.
Amongst the following ions which one has the highest paramagnetism?
(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [CU(H₂O)₆]²⁺
(d) [Zn(H₂O)₆]²⁺
Answer:
(b) [Fe(H₂O)₆]²⁺

Question 8.
The geometry of [Ni(CN)₄]^3- and [NiCI₄]^-2 are
(a) both tetrahedral
(b) both square planar
(c) tetrahedral and square planar respectively
(d) square planar and tetrahedral respectively
Answer:
(d) square planar and tetrahedral respectively

Question 9.
The complex cis-[Pt(NH₃)₂CI₂] is used in treatment of cancer under the name.
(a) Aspirin
(b) Eqanil
(c) cisplatin
(d) transplatin
Answer:
(c) cisplatin

Question 10.
[CO(NH₃)₆]³⁺ is an orbital complex and is in nature.
(a) inner, paramagnetic
(b) inner, dimagnetic
(c) outer, paramagnetic
(d) outer, dimagnetic
Answer:
(b) inner, dimagnetic

Question 11.
The IUPAC name of [Ni(Co)₄] is
(a) tetra carbonyl nickel (O)
(b) tetra carbonyl nickel (II)
(c) tetra carbonyl nickelate (O)
(d) tetra carbonyl nickelate (II)
Answer:
(a) tetra carbonyl nickel (O)

Question 12.
The number of ions produced by the complex [CO(NH₃)₄CI₂] Cl is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 13.
The dimagnetic species is
(a) [Ni(CN)₄]^2-
(b) [NiCl₄]^2-
(c) [CoCI₄]^2-
(d) [CoF₆]^2-
Answer:
(a) [Ni(CN)₄]^2-

Question 14.
Which one of the following is an inner orbital complex as well as diamagnetic in behaviour (Atomic no. Zn = 30, Cr = 24, Co = 27, Ni = 28)
(a) [Zn(NH₃)₆]²⁺
(b) [Cr(NH₃)₆]³⁺
(c) [CO(NH₃)₆]³⁺
(d) [Ni(NH₃)₆]²⁺
Answer:
(c) [CO(NH₃)₆]³⁺

Question 15.
Among [Ni(Co)J, [Ni(CN)4]2A [NiClJ2- Species, the hybridisation states at the Nickel atom are respectively
(a) sp³, dsp², sp³
(b) sp³, dsp², dsp²
(c) dsp², sp³, sp³
(d) sp³, sp³, dsp²
Answer:
(a) sp³, dsp², sp³

Question 16.
The strongest ligand in the following is
(a) CN^–
(b) Br^–
(c) HO^–
(d) F^–
Answer:
(a) CN^–

Question 17.
Magnetic moment of (NH4)2 (MnBr4) is BM
(a) 5.91
(b) 4.91
(c) 3.91
(d) 2.91
Answer:
(a) 5.91

Question 18.
The complex which violates EAN rule is
(a) Fe(CO)₅
(b) [Fe(CN)₆]^3-
(c) Ni(CO)₄
(d) [Zn(NH₃)₄]CI₂
Answer:
(b) [Fe(CN)₆]^3-

Question 19.
EDTA is a ligand of the type
(a) bidentate
(b) tridentate
(c) tetradentate
(d) hexadentate
Answer:
(d) hexadentate

Question 20.
The cationic complex among the following is
(a) K₃[Fe(CN)₆]
(b) Ni(CO)₄
(c) K₂HgI₄
(d) [CO(NH₃)₆]CI₂
Answer:
(d) [CO(NH₃)₆]CI₂

Question 21.
If Z is the atomic number of a metal, X is number of electrons lost forming metal ion and Y is the number of electrons from the ligands then EAN is
(a) Z + X + Y
(b) X – Z + Y
(c) Z – X + Y
(d) X + Z – Y
Answer:
(c) Z – X + Y

Question 22.
Octahedral complex has hybridisation,
(a) dsp²
(b) d³sp³
(c) dsp³
(d) d²sp³
Answer:
(d) d²sp³

Question 23.
Inner complex has hybridisation,
(a) d²sp³
(b) sp³d²
(c) sp³d
(d) sp³d³
Answer:
(a) d²sp³

Question 24.
The number of unpaired electrons in [CO(NH3)6]3+ is
(a) 0
(b) 1
(c) 2
(d) 4
Answer:
(a) 0

Question 25.
The number of unpaired electrons in [NiClJ2- and [Ni(CN)4]2_ are respectively,
(a) 2, 2
(b) 2, 0
(c) 0, 0
(d) 1, 2
Answer:
(b) 2, 0

Question 26.
Among the following complexes, the highest magnitude of crystal field stabilisation energy will be for [Co(H₂O)₆]³⁺, [CO(CN)₆]^3-, [Co(NH₃)₆]³⁺, [CoF₆]^3-
(a) [Co(H₂O)₆]³⁺
(b) [CO(CN)₆]^3-
(c) [Co(NH₃)₆]³⁺
(d) [CoF₆]^3-
Answer:
(b) [CO(CN)₆]^3-

Question 27.
The number of unpaired electrons in a low spin octahedral complex ion of d1 is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 28.
The number of unpaired electrons in a high spin octahedral complex ion of d⁷ is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(d) 3

Question 29.
Ligand used in the estimation of hardness of water is
(a) EDTA
(b) DBG
(c) chloride
(d) bromo
Answer:
(a) EDTA

Question 30.
Which of the following complexes will give a white precipitate on treatment with a solution of barium nitrate?
(a) [Cr(NH₃)₄SO₄] CI
(b) [CO(NH₃)₄CI₂] NO₂
(c) [Cr(NH₃)₄CI₂] SO₄
(d) [CrCI₃(H₂O)₄]CI
Answer:
(c) [Cr(NH₃)₄CI₂] SO₄

Question 31.
What is effective atomic number of Fe (z = 26) in [Fe(CN)₆]^4-?
(a) 12
(b) 30
(c) 26
(d) 36
Answer:
(d) 36

Question 32.
Cisplatin compound is used in the treatment of
(a) malaria
(b) cancer
(c) AIDS
(d) yellow fever
Answer:
(b) cancer

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 5 Electrochemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 5 Electrochemistry

Question 1.
What is electrochemistry ?
Answer:
Electrochemistry : It is the branch of physical chemistry which involves the study of the inter-relation between chemical changes and electrical energy and also concerned with the electrical properties of electrolytic solutions such as resistance and conductance.

Question 2.
What is electric conduction?
Answer:
The transfer of electric charge or electrons from one point to another is called electric conduction which results in an electric current.

Question 3.
What are the electric conductors?
Answer:
The substances that allow the flow of electricity or electric charge transfer through them are called the electric conductors.

Question 4.
What is a flow of electricity or a transfer of electric charge?
Answer:
The flow of electricity or a transfer of electric charge through a conductor involves the transfer of electrons from one point to the other point. This takes place under the influence of applied electric potential.

Question 5.
What are the types of electric conductors? On what basis are they classified ?
Answer:
The electric conductors are classified according to the mechanism of the transfer of electrons or charge. There are two types of conductors as follows :

(i) Electrons (or metallic) conductors : The electric conductors through which the conduction of electricity takes place by a direct flow of electrons under the influence of applied potential are called electronic conductors.

In this case, there is no transfer of matter like atoms or ions. For example, solid and molten metals such as Al, Cu, etc.

(ii) Electrolytic conductors : The conductors in which the conduction of electricity takes place by the migration of positive ions (cations) and negative ions (anions) of the electrolyte are called electrolytic conductors. In this, the conduction involves the transfer of matter and it is accompanied with chemical changes. For example, solutions of electrolytes (strong and weak), molten salts.

Question 6.
Distinguish between electronic and electrolytic conductors.
Answer:
Electronic conductors:

1. The flow of electricity takes place by direct flow of electrons through the conductor.
2. The conduction does not involve the transfer of a matter.
3. No chemical change is involved during conduction.
4. The resistance of the conductor increases and conductivity decreases with the increase in temperature.
5. The conductance of metallic conductors is very high.
6. Examples are solid or molten metals, such as Al, Cu, etc.

Electrolytic conductors:

1. The electron transfer takes place by the migration of ions (cations and anions) of the electrolyte.
2. The conduction involves the transfer of a matter.
3. Chemical changes are always involved during the passage of an electric current.
4. The resistance decreases and the conductivity increases with the increase in temperature.
5. The conductance of the electrolytes is comparatively low.
6. Examples are aqueous solutions of acids, bases or salts.

Question 7.
What information is provided by measurement of conductivities of solutions?
Answer:

• The conducting and nonconducting properties of solutions can be identified by the measurement of their conductivities.
• The substances like sucrose and urea which do not dissociate in aqueous solutions have same conductivity as that of water. Hence they are nonelectrolytes.
• The substances like KCl, CH₃COOH, NaOH, etc. dissociate in their aqueous solutions and their conductivities are higher than water. Hence they are electrolytes.
• On the basis of high or low electrical conductivity, the electrolytes can be classified as strong and weak electrolytes. The solutions of strong electrolytes have high conductivities while solutions of weak electrolytes have lower conductivities.

Question 8.
What is Ohm’s law?
Answer:
Ohm’s law : According to Ohm’s law, the electrical resistance R of a conductor is equal to the electric potential difference, V divided by the electric current, I.
R = \(\frac{V}{I}\) ohm

Question 9.
What are SI units of
(a) electrical resistance
(b) potential and
(c) electric current?
Answer:
(a) The SI unit of electrical resistance is Ohm denoted by Ω (omega).
(b) The SI unit of potential is volt denoted by V.
(c) The SI unit of electric current is ampere denoted by A.

Question 10.
How is electrical conductance of a solution denoted ? What are its units ?
Answer:
The electrical conductance of a solution is denoted by G and it is the reciprocal of resistance, R.
G = \(\frac{1}{R}\)
The unit of G is siemens denoted by S or Ω^-1.
Hence we can write, S = Ω^-1 = AV^-1 = CV^-1S^-1 where A is ampere and C is coulomb.

Question 11.
What is electrical conductance? What are its units ?
Answer:
The reciprocal of the electrical resistance of a solution is called the conductance. It is represented by G.
∴ Conductance (G) = \(\frac{1}{\text { Resistance }}=\frac{1}{\mathrm{R}}\)
The conductance has units of reciprocal of ohm (Ω^-1, ohm^-1 or mho). In SI units, conductance has units as Siemens, (S). (1 S = 1 Ω^-1 = 1 ohm^-1 = 1 mho = AV^-1 = CV^-1 S, where C represents electric charge in coulomb, and A represents current strength)

Question 12.
What is specific conductance or conductivity?
Answer:
The reciprocal of specific resistance or resistivity is called specific conductance or conductivity.
If ρ is the resistivity then,
conductivity = \(\frac{1}{\text { resistivity }}=\frac{1}{\rho}\)
Conductivity is denoted by κ (kappa), where κ = \(\frac{1}{\rho}\)
It is the conductance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) It is the conductance of a conductor of volume 1 m³ (or in C.G.S. units, the volume of 1 cm³).

Question 13.
What are the units of specific conductance or conductivity?
Answer:
If ρ is a resistivity and κ is conductivity or specific conductance, then

(where S is Siemens)
(In C.G.S. system, the units of κ are Ω^-1 cm^-1 or S cm^-1 which are commonly used.)

Question 14.
Define molar conductivity. What is the significance of it ?
Answer:
Molar conductivity: It is defined as a conductance of a volume of the solution containing ions from one mole of an electrolyte when placed between two parallel plate electrodes 1 cm apart and of large area, sufficient to accommodate the whole solution between them, at constant temperature. It is denoted by ∧_m.

Thus, the significance of molar conductivity is the conductance due to ions from one mole of an electrolyte.

Question 15.
Obtain a relation between conductivity (κ) and molar conductivity (∧_m).
Answer:
Conductivity or specific conductance (κ) is the conductance of 1 cm³ of the solution in C.G.S. units, while molar conductivity is the conductance of a solution containing one mole of an electrolyte. Consider C molar solution, i.e., C moles of an electrolyte present in 1 litre or 1000 cm³ of the solution.
∴ C moles of an electrolyte are present in 1000 cm³ solution.
∴ 1 mole of an electrolyte is present in \(\frac{1000}{\mathrm{C}}\) cm
solution.
Now,
∴ Conductance of 1 cm³ of this solution is κ,
∴ Conductance of \(\frac{1000}{\mathrm{C}}\) cm³ of the solution is \(\frac{\kappa \times 1000}{C}\)
This represents molar conductivity, ∧_m.
∴ ∧_m = \(\frac{\kappa \times 1000}{C}\) cm² mol^-1 (in C.G.S units)
[In case of SI units :
Consider a solution in which C moles of an electrolyte are present in 1 m³ of solution.
Conductivity κ is the conductance of 1 m³ of solution.
∵ C moles of an electrolyte are present in 1 m³ solution.
∴ 1 mol of an electrolyte is present in \(\frac{1}{C}\) solution.
∵ Conductance of 1 m³ of this solution is κ.
∴ Conductance of \(\frac{1}{C}\) m³ of the solution is \(\frac{\kappa}{\mathrm{C}}\)
This represents molar conductivity, ∧_m.
∴ ∧_m = \(\frac{\kappa}{\mathrm{C}}\)Ω^-1 m² mol^-1 (In SI units).]

Question 16.
What are the units of molar conductivity, ∧_m?
Answer:
In SI units: Conductivity κ is expressed in Ω^-1m^-1 (or S m^-1) and concentration of the solution is expressed in mol m^-3.

In C.G.S. units : Conductivity is expressed in Ω^-1 cm^-1 (or S cm^-1) and concentration of the solution is expressed in mol L^-1 or moles in 1000 cm³ of the solution.

Question 17.
Explain the variation of molar conductivity with concentration for strong and weak electrolytes.
OR
How is the molar conductivity of strong electrolytes at zero concentration determined by graphical method? Why is this method not useful for weak electrolytes?
Answer:
(i) As the dilution of an electrolytic solution increases, the dissociation of the electrolyte increases, hence the total number of ions increases, therefore, the molar conductivity increases.

Fig. 5.5 : Variation of molar conductivity with \(\sqrt{\mathbf{c}}\)
(ii) The increase in molar conductivity with increase in dilution or decrease in concentration is different for strong and weak electrolytes.
(iii) On dilution, the molar conductivity of strong electrolytes increases rapidly and approaches to a maximum limiting value at infinite dilution or zero concentration and represented as ∧ ∞ or ∧₀ or ∧⁰_m. In case of weak electrolytes which dissociate less as compared to strong electrolytes, the molar conductivity is low and increases slowly in high concentration region, but increases rapidly at low concentration or high dilution. This is because the extent of dissociation increases with dilution rapidly.
(v) ∧₀ values for strong electrolytes can be obtained by extrapolating the linear graph to zero concentration (or infinite dilution). However ∧₀ for the weak electrolytes cannot be obtained by this method, since the graph increases exponentially at very high dilution and does not intersect ∧_m axis at zero concentration.

Question 18.
Why has the molar conductance of an electrolyte the maximum value at infinite dilution ?
Answer:

• As the dilution of an electrolytic solution increases or concentration decreases, the dissociation of an electrolyte increases.
• At infinite dilution, the dissociation of an electrolyte is complete (100% dissociation). Hence all the ions from one mole of an electrolyte are available to carry electricity.

Therefore the molar conductance at infinite dilute (∧₀) for a given electrolyte has the highest or limiting value. It is always constant for the given electrolyte at constant temperature.

Question 19.
State Kohlrausch’s law.
OR
State and explain Kohlrausch’s law of independent migration of ions.
Answer:
(A) Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

(B) Explanation : Both the ions, cation and anion of the electrolyte make a definite contribution to the molar conductivity of the electrolyte at infinite dilution or zero concentration (∧₀).
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the molar ionic conductivities of cation and anion respectively at infinite dilution, then
∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\)
This is known as Kohlrausch’s law of independent migration of ions.
For an electrolyte, B_x A_y giving x number of cations and y number of anions,
∧₀ = x\(\lambda_{+}^{0}\) + y\(\lambda_{-}^{0}\)

(C) Applications of Kohlrausch’s law :
(1) With this law, the molar conductivity of a strong electrolyte at zero concentration can be determined. For example,
\(\wedge_{0(\mathrm{KCl})}=\lambda_{\mathrm{K}^{+}}^{0}-\lambda_{\mathrm{Cl}^{-}}^{0}\)
(2) ∧₀ values of weak electrolyte with those of strong electrolytes can be obtained. For example,

Question 20.
State Kohlrausch’s law and write mathematical expression of molar conductivity of the given solution at infinite dilution.
Answer:
Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

This law of independent migration of ions is represented as
∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).
where ∧₀ is the molar conductivity of the electrolyte at infinite dilution or zero concentration while \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the molar ionic conductivities of cation and anion respectively at infinite dilution.

Question 21.
Explain the determination of molar conductivity of a weak electrolyte at infinite dilution or zero concentration using Kohlrausch’s law.
Answer:
Molar conductivity of a weak electrolyte at infinite dilution or zero concentration cannot be measured experimentally.
Consider the molar conductivity (∧₀) of a weak acid, CH₃COOH at zero concentration. By Kohlrausch s law, ∧_0CH3COOH = λ⁰_CH3COOH^– + λ⁰ _H⁺ where λ⁰ _CH3COO^– and λ⁰ _H⁺ are the molar ionic conductivities of CH₃COO^– and H⁺ ions respectively.

If ∧_0CH3COONa, ∧_0HCl and ∧_0NaCl are the molar conductivities of CH₃COONa, HCl and NaCl respectively at zero concentration, then by
Kohlrausch’s law,

Hence, from ∧₀ values of strong electrolytes, ∧₀ of a weak electrolyte CH₃COOH, at infinite dilution can be calculated.

Question 22.
How is the degree of dissociation related to the molar conductance of the electrolytic solution ?
Answer:
(i) At zero concentration or at infinite dilution, the molar conductivity has a maximum value denoted by ∧₀.
(ii) This is due to complete dissociation of the weak electrolyte making all the ions available from one mole of the electrolyte to carry electricity at zero concentration.
(iii) If α is the degree of dissociation, then

This suggests that at zero concentration or infinite dilution, the electrolyte is completely (100%) dissociated.

Question 23.
Write the relation between molar conductivity and molar ionic conductivities for the following electrolytes :
(a) KBr, (b) Na₂SO₄, (c) AlCl₃.
Answer:
(a) If ∧₀ is molar conductivity of an electrolyte at infinite dilution and \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are molar ionic conductivities then,

Question 24.
How is molar conductivity of an electrolytic solution measured ?
Answer:
The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone bridge.

Fig. 5.6 : Measurement of conductance

The measurement of molar conductivity of a solution involves two steps as follows :
Step I : Determination of cell constant of the conductivity cell :
KCl solution (0.01 M) whose conductivity is accurately known (κ = 0.00141 Ω^-1 cm^-1) is taken in a beaker and the conductivity cell is dipped. The two electrodes of the cell are connected to one arm while the variable known resistance (R) is placed in another arm of Wheatstone bridge.

A current detector D’ which is a head phone or a magic eye is used. J is the sliding jockey (contact) that slides on the arm AB which is a wire of uniform cross section. A source of A.C. power (alternating power) is used to avoid electrolysis of the solution.

By sliding the jockey on wire AB, a balance point (null point) is obtained at C. Let AC and BC be the lengths of wire.
If R_solution is the resistance of KCl solution and R_x is the known resistance then by Wheatstone’s bridge principle,
\(\frac{R_{\text {solution }}}{\mathrm{BC}}=\frac{R_{x}}{\mathrm{AC}}\)
∴ R_solution = \(\mathrm{BC} \times \frac{R_{x}}{\mathrm{AC}}\)
Then the cell constant ‘b’ of the conductivity cell is obtained by, b = κ_KCl × R_solution.

Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (R_s) is measured by Wheatstone bridge method by similar manner by obtaining a null point at D.
The conductivity (κ) of the given solution is, cell constant b
κ = \(\frac{\text { cell constant }}{R_{\mathrm{s}}}=\frac{b}{R_{\mathrm{s}}}\)

Step III: Calculation of molar conductivity :
The molar conductivity (∧_m) is given by,

Since the concentration of the solution is known, ∧_m can be calculated.

Solved Examples 5.3

Question 25.
Solve the following :

(1) The resistance of a solution is 2.5 × 10³ ohm. Find the conductance of the solution.
Solution :
Given : Resistance of solution = R = 2.5 × 10³ Ω
Conductance of solution = G = ?
G = \(\frac{1}{R}\)
= \(\frac{1}{2.5 \times 10^{3}}\) ohm^-1 (Ω^-1 or S)
= 4 × 10^-3 Ω^-1 (or S)
Ans. Conductance = G = 4 × 10^-3 Ω^-1

(2) A conductivity cell has two electrodes 20 mm apart and of cross section area 1.8 cm². Find the cell constant.
Solution :
Given: Distance between two electrodes = l
= 20 mm
= 2 cm
Cross section area = a = 1.8 cm
Cell constant = b = ?
b = \(\frac{l}{a}=\frac{2}{1.8}\) = 1.111 cm^-1
Ans. Cell constant = 1.111 cm^-1

(3) The conductivity of 0.02 M AgNO₃ at 25 °C is 2.428 × 10^-3 Ω^-1 cm^-1. What is its molar conductivity ?
Solution :
Given : Concentration of solution = C = 0.02 M AgNO₃
Temperature = T = 273 + 25 = 298 K
Conductivity = κ = 2.428 × 10^-3 Ω^-1 cm^-1 (or S cm^-1)
Molar conductivity = ∧_m = ?
∧_m = \(\frac{\kappa \times 1000}{C}\)
= \(\frac{2.428 \times 10^{-3} \times 1000}{0.02}\)
= 121.4 Ω^-1 cm² mol^-1 (or 121.4 S cm² mol^-1)
Ans. Molar conductivity = ∧_m
= 121.4 Ω^-1 cm² mol^-1

(4) 0.05 M NaOH solution offered a resistance of 31.6 in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm^-1, calculate the molar conductivity of NaOH solution.
Solution :
Given : Concentration = C = 0.05 M NaOH
Resistance = R = 31.6 Ω
Cell constant = b = 0.367 cm^-1

Ans. Molar conductivity = ∧_m = 232.2 Ω^-1 cm² mol^-1

(5) A conductivity cell filled with 0.1 M KCl gives at 25 °C a resistance of 85.5 ohms. The conductivity of 0.1 M KCl at 25° is 0.01286 ohm^-1 cm^-1. The same cell filled with 0.005 M HCl gives a resistance of 529 ohms. What is the molar conductivity of HCl solution at 25 °C ?
Solution :
Given : Resistance of KCl solution = R_KCl = 85.5 Ω
Conductivity of KCl solution = κ_KCl
= 0.01286 ohm^-1 cm^-1
Concentration = C = 0.005 M HCl
Resistance of HCl solution = R_soln = 529 ohms
Molar conductivity of HCl = ∧_m(HCl) = ?

Ans. Molar conductivity of HCl solution = ∧_m(HCl)
= 416 ohm^-1 cm² mol^-1

(6) The molar conductivity of 0.05 M BaCl₂ solution at 25 °C is 223 Ω^-1 cm² mol^-1. What is its conductivity?
Solution :
Given : Molar conductivity = ∧_m
= 223 Ω^-1 cm² mol^-1
Concentration = C = 0.05 M BaCl₂
Conductivity = κ = ?

Ans. Conductivity = κ = 0.01115 Ω^-1 cm^-1

(7) Conductivity of a solution is 6.23 × 10^-5 Ω^-1 cm^-1 and its resistance is 13710 Ω. If the electrodes are 0.7 cm apart, calculate the cross-sectional area of electrode.
Solution :
Given : κ = 6.23 × 10^-5 Ω^-1 cm^-1
R = 13710 Ω
l = 0.7 cm
a = ?

Ans. Cross sectional area of electrode = 0.8195 cm²

(8) A conductivity cell filled with 0.01 M KCl gives at 25 °C the resistance of 604 ohms. The conductivity of KCl at 25 °C is 0.00141 Ω^-1 cm^-1. The same cell filled with 0.001 M AgNO₃ gives a resistance of 6529 ohms. Calculate the molar conductivity of 0.001 M AgNO₃ solution at 25 °C.
Solution :
Given : Resistance of KCl solution = R_KCl
= 604 ohm (Ω)
Conductivity of KCl solution = κ_KCl
= 0.00141 Ω^-1 cm^-1
Concentration = C = 0.001 M AgNO₃
Resistance of solution = R_sol = 6529 ohm (Ω)
Molar conductivity = ∧_m = ?

cell constant b
= 130.4 Ω^-1 cm² mol^-1
Ans. Molar conductivity of AgNO₃ solution = ∧_m
= 130.4 Ω^-1 cm² mol^-1

(9) Resistance and conductivity of a cell containing 0.001 M KCl solution at 298 K are 1500 Ω and 1.46 × 10^-4 S.cm^-1 respectively. What is cell constant.
Solution :
Given : Resistance of KCl solution = 1500 Ω, conductivity of KCl solution = κ = 1.46 × 10^-4 S.cm^-1, Cell constant = b = ?
Cell constant = Conductivity (k) × Resistance
= 1.46 × 10^-4 × 1500
= 0.219 cm^-1
Ans. Cell constant = 0.219 cm^-1

(10) A conductivity cell filled with 0.02 M H₂SO₄ gives at 25 °C resistance of 122 ohms. If the molar conductivity of 0.02 H₂SO₄ is 618 Ω^-1 cm² mol^-1, what is the cell constant?
Solution :
Given : Concentration = C = 0.02 M H₂SO₄
Resistance of H₂SO₄ solution = R_soln = 122 Ω
Molar conductivity = ∧_m = 618 Ω^-1 cm² mol^-1
Cell constant = b = ?

Ans. Cell constant = b = 1.51 cm^-1

(11) A conductivity cell filled with 0.02 M AgNO₃ gives at 25 °C resistance of 947 ohms. If the cell constant is 2.3 cm^-1, what is the molar conductivity of 0.02 M AgNO₃ at 25 °C?
Solution :
Given : Concentration = C = 0.02 M AgNO₃
Resistance of solution = R_soln = 947 Ω
Cell constant = b = 2.3 cm^-1
Molar conductivity = ∧_m = ?
Conductivity of soln = κ

Ans. Molar conductivity = ∧_m
= 121.5 Ω^-1 m² mol^-1

(12) Resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohms. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. [Given : Conductivity of 0.1 M KCl solution is 1.29 Sm^-1.]
Solution:
Given : Resistance of 0.1 M KCl solution = R₁ = 100 Ω
Resistance of 0.02 M KCl solution = R₂ = 520 Ω
Conductivity of 0.02 M KCl solution = κ₂ = ?
Molar conductivity of 0.02 M KCl solution = ∧_m = ?
Conductivity of 0.1 M KCl solution = κ₁
= 1.29 S m^-1
Cell constant = b = κ₁ × R₁ = 1.29 × 100
= 129 m^-1
= 1.29 cm^-1

(13) The molar conductivities at zero concentration (or at infinite dilution) of CH₃COONa, HCl and NaCl in Ω^-1 cm² mol^-1 are 90.8,426.2 and 126.4 respectively. Calculate the molar conductivity of CH₃COOH at infinite dilution.
Solution :

(14) The molar conductivities at zero concentrations of NH₄Cl, NaOH and NaCl are respectively 149.7Ω^-1 cm² mol^-1, 248.1 Ω^-1 cm² mol^-1 and 126.5 Ω^-1 cm² mol^-1. What is the molar conductivity of NH₄OH at zero concentration ?
Solution :

(15) What is the molar conductivity of AgI at zero concentration if the ∧₀ values of NaI, AgNO₃ and NaNO₃ are respectively 126.9 Ω^-1 cm² mol^-1, 133.4 Ω^-1 cm² mol^-1 and 121.5 Ω^-1 cm² mol^-1 ?
Solution :

Adding equations (i) and (ii) and subtracting equation (iii) we get equation I.

(16) Molar conductivity of KCl at infinite dilution is 150.3 S cm² mol^-1. If the molar conductivity of K⁺ is 73.4, calculate that of Cl^–.
Solution :
Given : Molar conductivity at infinite dilution
= ∧_(KCl) = 150.3 S cm² mol^-1
Molar conductivity of K⁺
= \(\lambda_{\mathrm{K}^{+}}^{0}\) = 73.4 S cm² mol^-1
Molar conductivity of Cl^– = \(\lambda_{\mathrm{Cl}^{-}}^{0}\) = ?
By Kohlrausch’s law,

(17) Molar conductivities at infinite dilution of Mg²⁺ and Br^– are 105.8 Ω^-1 cm² mol^-1 and 78.2 Ω^-1 cm² mol^-1 respectively. Calculate molar conductivity at zero concentration of MgBr₂.
Solution :
Given : \(\lambda_{\mathrm{Mg}^{2+}}^{0}\) = 105.8 Ω^-1 cm² mol^-1
\(\lambda_{\mathrm{Br}^{-}}^{0}\) = 78.2 Ω^-1 cm² mol^-1
\(\wedge_{0\left(\mathrm{MgBr}_{2}\right)}\) = ?
By Kohlrausch’s law,
\(\wedge_{0\left(\mathrm{MgBr}_{2}\right)}\) = \(\lambda_{\mathrm{Mg}^{2+}}^{0}\) + 2\(\lambda_{\mathrm{Br}^{-}}^{0}\)
= 105.8 + 2 × 78.2 = 105.8 + 156.4
= 262.2 Ω^-1 cm² mol^-1
Ans. Molar conductivity of MgBr₂ at zero concentration = \(\wedge_{0\left(\mathrm{MgBr}_{2}\right)}\) = 262.2 Ω^-1 cm² mol^-1

(18) The molar conductivity of 0.1 M CH₃COOH at 25 °C is 15.9 Ω^-1 cm² mol^-1. If the molar conductivities of CH3COO^– and H⁺ ions in Ω^-1 cm² mol^-1 at zero concentration are 40.8 and 349.6 respectively, calculate degree of dissociation of 0.1 M CH₃COOH.
Solution :
Given : Concentration = C = 0.1 M CH₃COOH
Molar conductivity = ∧_m = 15.9 Ω^-1 cm² mol^-1
\(\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0}\) = 40.8 Ω-1 cm² mol^-1;
\(\lambda_{\mathrm{H}^{+}}^{0}\) = 349.6 Ω^-1 cm² mol^-1
Degree of dissociation = α = ?
By Kohlrausch’s law,
\(\wedge_{0\left(\mathrm{CH}_{3} \mathrm{COOH}\right)}=\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0}+\lambda_{\mathrm{H}^{+}}^{0}\)
= 40.8 + 349.6
= 390.4 Ω^-1 cm² mol^-1
α = ∧_m/∧0
= \(\frac{15.9}{390.4}\) = 0.0407
Ans. The degree of dissociation of CH₃COOH = 0.0407

(19) The dissociation constant of a weak monoacidic base is 1.2 × 10^-5 at 25 °C. The molar conductivity of the base at zero concentration is 354.8 Ω^-1 cm² mol^-1 at 25°C. Calculate the percentage dissociation and molar conductivity of the weak base at 0.1 M concentration.
Solution :
Given : Dissociation constant of the base = K_b = 1.2 × 10^-5
Concentration = C = 0.1 M
∧₀ = 354.8 Ω^-1 cm² mol^-1
Percentage dissociation = ?
∧_m = ?
K_a = \(\frac{\mathrm{C} \alpha^{2}}{1-\alpha}\); For a week electrolyte, α is small,
∴ K_a = cα²;
∴ α = \(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}=\left(\frac{1.2 \times 10^{-5}}{0.1}\right)^{\frac{1}{2}}\) = 1.0954 × 10^-2
∴ Percentage dissociation = α × 100
= 1.0954 × 10-2 × 100 = 1.0954%
Now, α = \(\frac{\wedge_{\mathrm{m}}}{\wedge_{0}}\)
∴ ∧_m = α × ∧₀ = 1.954 × 10^-2 × 354.8
= 6.932 Ω^-1 cm² mol^-1
Ans. Percentage dissociation = 1.0954
Molar conductivity = ∧_m
= 6.932 Ω^-1 cm² mol^-1

Question 26.
What is an electrochemical cell? What does it consist of?
Answer:
Electrochemical cell : It consists of two electronic conductors such as metal plates dipping into an electrolytic or ionic conductor which is an aqueous electrolytic solution or a pure liquid of a molten electrolyte.

Question 27.
What are electrochemical reactions ?
Answer:

1. Electrochemical reactions : The chemical reactions occurring in electrochemical cells which involve transfer of electrons from one species to other are called electrochemical reactions. They are redox reactions.
2. These reactions are made of two half reactions namely oxidation at one electrode (anode) and reduction at another electrode (cathode) of the electrochemical cell.
3. The net reaction is the sum of the above two half reactions.

Question 28.
Define electrode.
Answer:
Electrode : The arrangement consisting of a metal rod dipping in an aqueous solution or molten electrolyte containing ions and conduct electric current due to oxidation or reduction half reactions occurring on its surface is called an electrode.

The electrodes which take part in the reactions are called active electrodes while those which do not take part in the reactions are called inert electrodes.

Question 29.
Define : (a) Anode (b) Cathode.
Answer:
(a) Anode : An electrode of an electrochemical cell, at which oxidation half reaction occurs due to the loss of electrons from some species is called an anode.
(b) Cathode : An electrode of an electrochemical cell at which reduction half reaction occurs due to gain of electrons by some species is called a cathode.

Question 30.
What are the types of electrochemical cells ?
Answer:
There are two types of electrochemical cells as follows :

1. Electrolytic cells
2. Voltaic or galvanic cells.

Question 31.
Define : (1) Electrolytic cell (2) Voltaic or galvanic cell.
Answer:
(1) Electrolytic cell : An electrochemical cell in which a non-spontaneous chemical reaction is forced to occur by passing direct electric current into the solution from the external source and where electrical energy is converted into chemical energy is called an electrolytic cell. E.g. voltameter, electrolytic cell for deposition of a metal.

(2) Voltaic or galvanic cell : An electrochemical cell in which a spontaneous chemical reaction occurs producing electricity and where a chemical energy is converted into an electrical energy is called voltaic cell or galvanic cell. E.g. Daniell cell, dry cell, lead storage battery, fuel cells, etc.

Question 32.
Define electrolysis.
Answer:
Electrolysis : The process of a non-spontaneous chemical decomposition of an electrolyte by the passage of an electric current through its aqueous solution or fused mass and in which electrical energy is converted into chemical energy is called electrolysis. E.g. Electrolysis of fused NaCl.

Question 33.
Describe electrolysis of aqueous NaCl.
Answer:
(1) Construction of an electrolytic cell : It consists of a vessel containing aqueous solution of NaCl. Two inert electrodes (graphite electrodes) are dipped in it and connected to an external source of electricity like battery. The electrode connected to the negative terminal is a cathode and that connected to a positive terminal is an anode.

(2) Working of the cell :
(A) NaCl_(aq) and H₂O_(l) dissociate as follows :

(3) Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : There are Na⁺ and H⁺ ions but since H⁺ are more reducible than Na⁺, they undergo reduction liberating hydrogen and Na⁺ are left in the solution.
2H₂O_(l) + 2e^– → H_2(g) + \(2 \mathrm{OH}_{(\mathrm{aq})}^{-}\) (reduction)
E⁰ = -0.83 V
(ii) Oxidation half reaction at anode : At anode there are Cl^– and OH^–. But Cl^– ions are preferably oxidised due to less decomposition potential.

Net cell reaction : Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write,

Since Na⁺ and OH^– are left in the solution, they form NaOH_(aq).

(4) Results of electrolysis :

• H₂ gas is liberated at cathode.
• Cl₂ gas is liberated at anode.
• NaOH is formed in the solution and it reacts basic.

Question 34.
Define and explain the following electrical units : (1) Coulomb (2) Ampere (3) Volt (4) Joule (5) Ohm.
Answer:
(1) Coulomb : It is a quantity of electricity obtained when one ampere current flows for one second.
It is the unit of quantity of electricity.
Q = I × t Coulomb (C)
where Q is the charge or quantity of electricity in coulombs.

(2) Ampere : It is a strength of an electric current obtained when one coulomb of electricity is passed through a circuit for one second.
∴ I = Q/t

(3) Volt : It is the potential difference between two points of an electric conductor required to send a current of one amphere through a resistance of one ohm.
∴ V = I × R
where V is the potential difference in volts and R is the resistance of a conductor in ohms.

(4) Joule : It is the electrical work or energy produced when one coulomb of electricity is passed through a
potential difference of one volt.
∴ Electrical work = Q × V J
where Q is electrical charge in coulombs and V is the potential difference.

(5) Ohm : It is the resistance of an electrical conductor across which when potential difference of 1 volt is applied, a current of one ampere is obtained. It has units, Ω or per siemens.

Question 35.
Explain quantitative aspects of electrolysis.
Answer:
(1) Calculation of quantity of electricity : If an electric current of strength I A is passed through the cell for t seconds, then quantity of electricity (Q) obtained is given by,
Q = I × t C (Coulomb)

(2) Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed = \(\frac{Q(\mathrm{C})}{F\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)}\)

(3) Calculation of moles of product formed : Consider one mole of ions, \(\mathbf{M}_{(\mathrm{aq})}^{n^{+}}\) which will require n moles of electrons for reduction.
\(\mathbf{M}_{(\mathrm{aq})}^{n^{+}}\) + ne^– → M (Reduction half reaction)

(4) Calculation of mass of product : Mass, W of product formed is given by,
W = moles of product × molar mass of product (M)

When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses W₁ and W₂ of products produced are given by,

Question 36.
Define Faraday.
Answer:
Faraday : It is defined as the quantity of the electric charge carried by one mole of electrons.
It has value, 1F = 96500 C/mol

Question 37.
Obtain a charge on one electron from Faraday’s value.
Answer:

• One Faraday is the electric charge on one mole of electrons (6.022 × 10²³ electrons).
• 1 Faraday = 96500 (per mol of electrons).
• Hence the charge on one electron is, change on one electron = \(\frac{96500}{6.022 \times 10^{23}}\)
  = 1.602 × 10^-9 C.

Solved Examples 5.4-5.5

Question 38.
Solve the following :

(1) An electric current of 100 mA is passed through an electrolyte for 2 hours, 20 minutes and 20 seconds. Find the quantity of electricity passed.
Solution :
Given : Electric current = I = 100 mA
= 100 × 10^-3 A
= 0.1 A
Time = t = 2 hrs + 20 min + 20 s
= 2 × 60 × 60 + 20 × 60 + 20
= 8420 s
The quantity of electricity = Q = ?
Q = I × t
= 0.1 × 8420
= 842 C
Ans. Quantity of electricity passed, Q = 842 C

(2) An electric current of 500 mA is passed for 1 hour and 30 minutes. Calculate the
(i) Quantity of electricity (or charge)
(ii) Number of Faradays of electricity
(iii) Number of electrons passed (Charge on 1 electron = 1.602 × 10^-19 C)
Solution :
Given : Electric current = I = 500 mA
= 500 × 10^-3 A = 0.5 A
Time = t = 1 hr + 30 min
= 1 × 60 × 60 + 30 × 60
= 5400 s
(i) The quantity of electricity = Q = ?
(ii) Number of Faradays of electricity = ?
(iii) Number of electrons passed = ?

(i) Q = I × t = 0.5(A) × 5400(s) = 2700 C

(iii) 1F is the electric charge on 6.022 × 10²³ electrons.
∴ 0.028F is the charge on,
0.028 × 6.022 × 10²³ = 1.686 × 10²² electrons
∴ Number of electrons passed = 1.686 × 10²²
Ans. (i) The quantity of electricity = Q = 2700 C
(ii) Number of Faradays of electricity = 0.028 F
(iii) Number of electrons passed = 1.686 × 10²²

(3) How much electricity in terms of Faraday is required to produce :
(a) 20 g of Ca from molten CaCl₂
(b) 40 g of Al from molten Al₂O₃
(Given : Molar mass of Calcium and Aluminium are 40 g mol^-1 and 27 g mol^-1 respectively.)
Solution :

(4) For the following conversions,
calculate
(i) number of moles of electrons
(ii) number of Faradays
(iii) Amount of electricity :
(A) 0.1 mol conversion of Zn²⁺ to Zn.
(B) 0.08 mol conversion of \(\mathbf{M n O}_{4}^{2-}\) to Mn²⁺
(C) 1.1 mol conversion of \(\mathrm{Cr}_{2} \mathbf{O}_{7}^{2-}\) to Cr³⁺.
Solution :
(i) Number of moles of electrons = ?
(ii) Number of Faradays = ?
(iii) Amount of electricity = Q = ?
(A) Number of moles of Zn²⁺ =0.1 mol
Zn²⁺ + 2e^– → Zn
(i) ∵ 1 mol Zn²⁺ requires 2 mol electrons
0.1 mol Zn²⁺ will require
∴ 0.1 × 2 = 0.2 mol electrons
(ii) ∵ 1 mol electrons = 1 Faraday
∴ 0.2 mol electrons = 0.2 × 1
= 0.2 Faradays
(iii) ∵ 1 Faraday = 96500 C
∴ 0.2 Faraday = 96500 × 0.2 = 48250 C
Amount of electricity required =48250C

(B) Number of moles of \(\mathrm{MnO}_{4}^{-}\) = 0.08 mol
\(\mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\)
(i) ∵ 1 mol \(\mathrm{MnO}_{4}^{-}\) requires 5 mol electrons
∴ 0.08 mol \(\mathrm{MnO}_{4}^{-}\) will require
5 × 0.08 = 0.4 mol electrons
(ii) Number of Faradays = 0.4 × 1 = 0.4
(iii) Amount of electricity = Q = 0.4 × 96500
= 38600 C

(C) Number of moles of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) = 1.1 mol

(i) ∵ 1 mol \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) requires 6 mol electrons
∴ 1.1 mol \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) will require
6 × 1.1 = 6.6 mol electrons
(ii) Number of Faradays = 1 × 6.6 = 6.6
(iii) Amount of electricity = 6.6 × 96500
= 6.369 × 10⁵ C
Ans.
(A) (i) Number of moles of electrons = 0.2 mol
(ii) Number of Faradays = 0.2
(iii) Amount of electricity = 48250 C

(B) (i) Number of moles of electrons = 0.4 mol
(ii) Number of Faradays = 0.4
(iii) Amount of electricity = 38600 C

(C) (i) Number of moles of electrons = 6.6 mol
(ii) Number of Faradays = 6.6
(iii) Amount of electricity = 6.369 × 10⁵ C

(5) What mass of aluminium is produced at the cathode during the passage of 4 ampere current through Al₂(SO₄)₃ solution for 100 minutes? Molar mass of aluminium is 27 g mol^-1.
Solution :
Given : I = 4 A; t = 100 × 60 = 600 s
F = 96500 C mol^-1, M = 27 g mol^-1, W_Al = ?

(6) How long will it take to produce 2.415 g Ag metal from its salt solution by passing a current of 3 amperes? How many moles of electrons are required ? Molar mass of Ag is 107.9 gmol^-1.
Solution :
Given : Electric current = I = 3A
Mass of Ag produced = 2.415 g
Molar mass of Ag = Atomic mass of Ag
= 107.9 gmol^-1
Time = t = ? Number of moles of electrons = ?
Reduction half reaction at cathode :

From the reaction,
∵ 1 mole of Ag requires 1 mole of electrons
∴ 0.02238 mole of Ag will require,
0.02238 mol electrons
∵ 1 mole of electrons carries a charge of 96500 C,
∴ 0.02238 mole of electrons will carry a charge, 0.02238 × 96500 = 2160 C
∴ Quantity of electricity passed = Q = 2160 C
Let I be the current strength and t be time of electrolysis. Then,
∵ Q = I × t
∴ t = \(\frac{Q}{I}=\frac{2160}{3}\) = 720 s = \(\frac{720}{60}\) min = 12 min.
Ans. Time of electrolysis = 12 min
Moles of electrons = 0.02238 mol

(7) What current strength in ampere will be required to produce 2.369 × 10^-3 kg of Cu from CuSO4 solution in one hour? How many moles of electrons are required? Molar mass of copper is 63.5 gmol^-1.
Solution :
Given : Mass of Cu produced = 2.369 × 10^-3 kg
= 2.369 g
Time = t = 1 hr = 1 × 60 × 60 = 3600 s
Molar mass of Cu = 63.5 g mol^-1
Strength of current = I = ?
1 Faraday = 96500 C = 1 mol electrons
1 mol Cu = Molar mass of Cu = 63.5 g
Reduction half reaction :

Moles of Cu deposited = \(\frac{2.369}{63.5}\) = 0.0373 mol Cu
From the reaction,
∵ 1 mol of Cu requires 2 mol electrons
∴ 0.0373 mol Cu will require 2 × 0.0373
= 0.0746 mol electrons
Now,
∵ 1 mol electrons = 96500 C
∴ 0.0746 mol electrons = 96500 × 0.0746 = 7199 C
∴ Quantity of electricity required = Q = 7199 C
∴ Q = I × t
∴ Current, I = \(\frac{Q}{t}=\frac{7199}{3600}\) = 2A
Ans. Current strength = I = 2A
Moles of electrons required = 0.0746 mol

(8) A current of 6 amperes is passed through AlCl₃ solution for 15 minutes using Pt electrodes, when 0.504 g Al is produced. What is the molar mass of Al ?
Solution :
Given : Electric current = I = 6 A
Time = t = 15 min = 15 × 60 s = 900 s
Mass of Al produced = 0.504 g
Molar mass of Al = ?
Reduction half reaction,
\(\mathrm{Al}_{(\mathrm{aq})}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{(\mathrm{aq})}\)
Quantity of electricity passed = Q = I × t
= 6 × 900 = 5400 C
Number of moles of electrons = \(\frac{Q}{F}=\frac{5400}{96500}\)
= 0.05596 mol
From half reaction,
∵ 3 moles of electrons deposit 1 mole Al
∴ 0.05596 moles of electrons will deposit,
\(\frac{0.05596}{3}\) = 0.01865 mol Al
Now,
∵ 0.01865 mole Al weighs 0.504 g
∴ 1 mole Al will weigh, \(\frac{0.504}{0.01865}\) = 27 g
Hence molar mass of Al is 27 g mol^-1
Ans. Molar mass of Al = 27 g mo^-1

(9) How many moles of electrons are required for the reduction of (i) 3 moles of Zn²⁺ to Zn,
(ii) 1 mol of Cr³⁺ to Cr ?
How many Faradays of electricity will be required in each case ?
Solution :
(i) Given : For reduction of 3 mol Zn²⁺ to Zn;
Number of moles of electrons required = ?
Reduction half reaction,
Zn²⁺ + 2e^– → Zn
∵ 1 mole of Zn²⁺ requires 2 moles of electrons
∴ 3 moles of Zn²⁺ will require,
∵ 3 × 2 = 6 moles of electrons
∴ 1 mole of electrons = 1 F 6 moles of electrons = 6 F

(ii) Given : Reduction of 1 mol of Cr³⁺ to Cr :
Reduction half reaction,
Cr³⁺ + 3e^– → Cr
Hence 1 mole of Cr³⁺ will require 3 moles of electrons
∵ 1 mole of electrons = 1
∴ 3 moles of electrons = 3 F
Ans. (i) 6 mol electrons and 6 Faradays.
(ii) 3 mol electrons and 3 Faradays.

(10) In an electrolysis of AgNO₃ solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol^-1.)
Solution :
Given : Mass of Ag deposited = 0.7 g
Molar mass of Ag = 107.9 g mol^-1
Quantity of electricity = Q = ?
Reduction half reaction is,
Ag⁺ + e^– → Ag
1 mole of Ag = 107.9 g Ag requires 1 mole of electrons
∴ 0.7 g Ag will require, \(\frac{0.7}{107.9}\) = 6.49 × 10^-3 mole of electrons
∵ 1 mole of electrons carry 96500 C charge
∴ 6.49 × 10^-3 mole of electrons will carry, 96500 × 6.49 × 10^-3 = 626 C
Ans. Quantity of electricity required = 626 C,

(11) Calculate the amounts of Na and Chlorine gas produced during the electrolysis of fused NaCl by the passage of 1 ampere current for 25 minutes. Molar masses of Na and Chlorine gas are 23 g mol^-1 and 71 g mol^-1 respectively.
Solution :
Given : Electric current = I = 1 ampere
Time = t = 25 minutes = 25 × 60 s = 1500 s
Molar mass of Na = 23 g mol^-1
Molar mass of Cl₂ = 71 g mol^-1
Mass of Na produced = ?, Mass of Cl₂ produced = ?
Reactions during electrolysis :

Quantity of electricity = Q = I × t = 1 × 1500
= 1500 C
Number of moles of electrons passed
= \(\frac{Q}{F}=\frac{1500}{96500}\) = 0.01554
From half reaction (i),
∵ 2 moles of electrons deposit 2 moles of Na
∴ 0.01554 moles of electrons will deposit, \(\frac{0.01554 \times 2}{2}\) = 0.01554 mol Na
Mass of Na = Moles of Na × Molar mass of Na
= 0.01554 × 23 = 0.3572 g Na
From half reaction (ii)
∵ 2 moles of electrons produce 1 mole Cl₂
∴ 0.01554 moles of electrons will produce,
\(\frac{0.01554 \times 1}{2}\) = 7.77 × 10^-3 × 71
∴ Mass of Cl₂ gas = Moles of Cl₂ × Molar mass
= 7.77 × 10^-3 × 71
= 0.5518 g
Ans. Mass of Na deposited = 0.3572 g
Mass of Cl2 liberated = 0.5518 g

(12) Calculate the mass of Mg and the volume of Chlorine gas at NTP produced during the electrolysis of molten MgCl₂ by the passage of 2 amperes of current for 1 hour. Molar masses of Mg and Cl2 are 24 g mol-1 and 71 g mol-1 respectively.
Solution :
Given : Electric current = I = 2A
Time = t = 1 hr = 1 × 60 × 60 s = 3600 s
Molar mass of Mg = 23 g mol-1
Molar mass of Cl₂ = 71 g mol-1
Mass of Mg produced = ?
Volume of Cl₂ at NTP produced = ?
Reactions during electrolysis :
(i) Mg²⁺ + 2e^– → Mg (Reduction half reaction)
(ii) 2Cl^– → Cl_2(g) + 2e^– (Oxidation half reaction)
Quantity of electricity passed = Q = I × t
= 2 × 3600 = 7200 C
∵ 1 Faraday = 1 mol electrons
∴ Number of moles of electrons passed
= \(\frac{Q}{F}=\frac{7200}{96500}\) = 0.07461 mol
From half reaction (i),
∵ 2 moles of electrons deposit 1 mole of Mg
∴ 0.07461 moles of electrons will deposit, \(\frac{0.07461 \times 1}{2}\) = 0.037305 mol Mg
Mass of Mg = Moles of Mg × Molar mass of Mg
= 0.037305 × 24 = 0.8953 g Mg
From half reaction (ii),
∵ 2 moles of electrons produce 1 mol Cl₂ gas
∴ 0.07461 moles of electrons will produce,
\(\frac{0.07461}{2}\) = 0.037305 mol Cl₂
∵ 1 mole of Cl₂ occupies 22.4 dm at NTP
∴ 0.037305 mole of Cl₂ will occupy,
22.4 × 0.037305 = 0.8356 dm3
∴ Volume of Cl₂ gas produced
= 0.8356 dm³
= 0.8356 × 10³ cm³
= 835.6 cm³
Ans. Mass of Mg produced = 0.8953 g
Volume of Cl_2(g) at NTP produced = 835.6 cm³

(13) How many Faradays would be required to plate out one mole of free metal from the following cations?
(a) Mg²⁺ (b) Cr³⁺ (c) Pb²⁺ (d) Cu⁺
Solution :
(a) Reduction half reaction :

∵ 1 mol electrons = 1 Faraday
Since to deposite 1 mol Mg, two moles of electrons are required,
∴ To plate one mole Mg, 2 Faradays of electricity will be required.

(b) Reduction half reaction :
\(\mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}_{(\mathrm{s})}\)
∴ 1 mol Cr will require 3 mol electrons, hence 3 Faradays of electricity are required.

(c) Reduction half reaction :
\(\mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}_{(\mathrm{s})}\)
∴ 1 mol Pb will require 2 mol electrons, hence 2 Faradays are required.

(d) Reduction half reaction :
\(\mathrm{Cu}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}\)
∴ 1 mol Cu will require 1 mol electrons hence one Faraday of electricity is required.

(14) In a certain electrolysis experiment, 0.561 g of Zn is deposited in one cell containing ZnSO₄ solution. Calculate the mass of Cu deposited in another cell containing CuSO₄ solution in series with ZnSO₄ cell. Molar masses of Zn and Cu are 65.4 g mol^-1 and 63.5 g mol^-1 respectively.
Solution :
Given : Mass of Zn deposited = W_Zn = 0.561 g
Molar mass of Zn = 65.4 g mol^-1
Molar mass of Cu = 63.5 g mol^-1
Mass of Cu deposited = ?
Number of moles of Zn deposited
= \(\frac{\text { Mass of Zn deposited }}{\text { Molar mass of } \mathrm{Zn}}=\frac{0.561}{65.4}\)
= 8.578 × 10^-3 mol Zn
Reactions of electronics:
(i) Zn⁺⁺ + 2e^– → Zn (Half reaction in ZnSO₄ cell)
(ii) Cu⁺⁺ + 2e^– → Cu (Half reaction in CuSO₄ cell)
Mole ratio of Zn

∴ Mass of Cu produced
= moles of Cu × molar mass of Cu
= 8.578 × 10^-3 × 63.5
= 0.5447 g Cu
Ans. Mass of Cu deposited = 0.5447 g

(15) Two electrolytic cells, one containing AlCl₃ solution and the other containing ZnSO₄ solution are connected in series. The same quantity of electricity is passed through the cells. Calculate the amount of Zn deposited in ZnSO₄ cell if 1.2 g of Al are deposited in AlCl₃ cell. The molar masses of Al and Zn are 27 g mol^-1 and 65.4 g mol^-1 respectively.
Solution :
Given : Mass of Al deposited = 1.2 g
Molar mass of Al = 27 g mol^-1
Molar mass of Zn = 65.4 g mol^-1
Mass of zinc deposited = ω_Zn = ?
Reduction reactions in electrolysis :

Number of moles of Al deposited = \(\frac{1.2}{27}\)
= 0.04444 mol.
From reaction (i),
∵ 1 mol Al requires 3 mol electrons
∴ 0.04444 mol Al requires 3 × 0.04444
= 0.1333 mol electrons
Hence 0.1333 moles of electrons are passed through both the cells in the series.
From reaction (ii),
∵ 2 moles of electrons deposit 1 mol Zn
∴ 0.1333 moles of electrons will deposit, \(\frac{0.1333}{2}\) = 0.06665 mol Zn
Mass of Zn deposited = 0.06665 × 65.4 = 4.36 g
Ans. Mass of Zn deposited = 4.36 g

(16) How much quantity of electricity in coulomb is required to deposit 1.346 × 10^-3 kg of Ag in 3.5 minutes from AgNO₃ solution ?
(Given : Molar mass of Ag is 108 × 10^-3 kg mol^-1)
Solution :
Given : Mass of Ag deposited = 1.346 × 10^-3 kg
Molar mass of Ag = 108 × 10^-3 kg mol^-1
Time = t = 3.5 × 60 s

∵ 108 × 10^-3 kg Ag requires 1 Faraday
1.346 × 10^-3 kg Ag will require,
\(\frac{1.346 \times 10^{-3}}{108 \times 10^{-3}}\) = 0.01246 F
∵ If F = 96500 C
∴ 0.01246 F = 96500 × 0.01246 = 1202 C
Ans. Amount of electricity required = 1202 C

(17) How many electrons will have a total charge of 1 Coulomb ?
Solution :
Given : Charge = 1 Coulomb
Number of electrons = ?
1 Faraday = 96500 C per mol electrons
∵ 96500 C electric charge is present on 1 mol electrons
∴ 1C charge is present on \(\frac{1}{96500}\) mol electrons
∴ Number of electrons = \(\frac{1}{96500}\) × 6.022 × 10²³
= 6.24 × 10¹⁸ electrons
Ans. 1 Coulomb charge is present on 6.24 × 10¹⁸ electrons.

(18) A constant electric current flows for 4 hours through two electrolytic cells connected in series. One contains AgNO₃solution and second contains CuCl₂ solution. During this time, 4 grams of Ag are deposited in the first cell.
(a) How many grams of Cu are deposited in the second cell?
(b) What is the current flowing in amperes? (Atomic mass : Cu = 63.5 gmol^-1; Ag = 107.9 gmol^-1)
Solution :
Given : Mass of Ag deposited = 4 g
Molar mass of Cu = 63.5 g mol^-1
Molar mass of Ag = 107.9 g mol^-1
Time = t = 4 hrs = 4 × 60 × 60 = 14400 s
Mass of Cu deposited = W_Cu = ?
Current = I = ?
(a) Number of moles of Ag deposited
\(=\frac{\text { Mass of } \mathrm{Ag}}{\text { Molar mass of } \mathrm{Ag}}=\frac{4}{107.9}\)
= 0.03707 mol of Ag
Reactions of electrolysis :
(i) Ag⁺ + e^– → Ag (Half reaction in AgNO₃ cell)
(ii) Cu²⁺ + 2e^– → Cu (Half reaction in CuCl₂ cell)
Mole ratio of Ag

∴ Mass of Cu produced = 0.01854 × 63.5 = 1.177 g

(b) From the reaction,
∵ 1 mol Ag⁺ requires 1 mol electrons
∴ 0.03707 mol Ag will require 0.03707 mol electrons
∵ 1 mol electrons = 1 Faraday
∴ 0.03707 mol electrons = 0.03707 Faraday
∵ 1 Faraday = 96500 C
∴ 0.03707 Faraday
= 0.03707 × 96500 = 3577 C
∴ Quantity of electricity = Q = 3577 C.
Q = I × t
∴ I = \(\frac{\mathrm{Q}}{t}=\frac{3577}{14400}\) = 0.25 A
Ans. (a) Mass of Cu deposited = 1.177 g
(b) Current passed = 0.25 A

(19) The passage of 0.95 A current for 40 minutes deposited 0.7493 g Cu from CuSO₄ solution. Calculate the molar mass of Cu.
Solution :
Given : Electric current = I = 0.95 A
Time = f = 40 min = 40 × 60 = 2400 s
Mass of Cu deposited = 0.7493 g
Molar mass of Cu = ?
Reduction half reaction,
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}\)
Quantity of electricity = Q = I × t
= 0.95 × 2400
= 2280 C
Number of moles of electrons = \(\frac{2280}{96500}\)
= 0.02362 mol
∵ 2 mol electrons deposit 1 mol Cu
∴ 0.02362 mol electrons will deposit,
\(\frac{0.02362}{2}\) = 0.01181 mol Cu
Now,
0.01181 mol Cu weighs 0.7493 g
∴ 1 mol of Cu weigh, \(\frac{0.7493 \times 1}{0.01181}\) = 63.44 g
Hence molar mass of Cu 63.44 g mol^-1
Ans. Molar mass of Cu = 63.44 g mol^-1

(20) A quantity of 0.3 g of Cu was deposited from CuSO₄ solution by passing 4A through the solution for 3.8 min. Calculate the value of Faraday constant. (Atomic mass of Cu = 63.5 g mol^-1)
Solution :
Given : Mass of Cu deposited = 0.3 g
Electric current = I = 4A
Time = t = 3.8 min = 3.8 × 60 = 228 s
Value of Faraday = ?
Quantity of electricity passed = Q = I × t
= 4 × 228 = 912 C
Reduction half reaction,
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}\)
Number of moles of Cu deposited 0.3
= \(\frac{0.3}{63.5}\) = 0.004724 mol
From reduction half reaction,
1 mol Cu ≡ 2 mol electrons
∴ 0.004724 mol Cu = 2 × 0.004724
= 0.009448 mol electrons
Now
∵ 0.009448 mol electrons = 912 C
∴ 1 mol electrons = \(\frac{912}{0.009448}\) = 96528 C
∵ 1 Faraday charge is equal to charge on 1 mol electrons
∴ 1 Faraday = 96528 C
Ans. 1 Faraday = 96528 C

(21) In the electrolysis of water, one of the half reactions is
2H⁺_(aq) + 2e^– → H_2(g)
Calculate the volume of H₂ gas collected at 25 °C and 1 atm pressure by passing 2A for 1h through the solution. R = 0.08205 L atm K^-1 mol^-1.
Solution :
Given : Reduction half reaction :

Temperature = T = 273 + 25 = 298 K
Pressure = P = 1 atm
Electric current = I = 2A
Time = t = 1 hr = 1 × 60 × 60 = 3600 s
R = 0.08205 L atm K-1 mol-1
Volume of H₂ = V_H2 = ?
Quantity of electricity passed = Q = I × t
= 2 × 3600 = 7200 C
Number of moles of electrons = \(\frac{Q}{F}\)
\(\frac{7200}{96500}\) = 0.0746 mol
From the reaction,
∵ 2 mol electrons produces 1 mol H₂ gas
∴ 0.0746 mol electrons will produce \(\frac{0.0746}{2}\)
= 0.0373 mol H₂.
pV_H2 = nRT

= 0.912 L
Ans. Volume H₂ gas = 0.912 L

(22) Calculate the current strength and number of moles of electrons required to produce 2.369 × 10^-3 kg of Cu from CuSO₄ solution in one hour. (Molar mass of Cu is 63.5 g/mol)
Solution :
Given : Mass of Cu deposited = 2.369 × 10^-3 kg;
t = 1 hr = 3600 s
Molar mass of Cu = 63.5 g mol^-1
I = ?; Number of moles of electrons = ?

∵ For 63.5 × 10^-3 kg Cu Q = 2 × 96500 C
∴ For 2.369 × 10^-3 kg Cu

Ans. I = 2A; Number of moles of electrons = 0.07461

Question 39.
Define : Galvanic cell or voltaic cell.
Answer:
Galvanic or voltaic cell : An electrochemical cell which is used to produce electrical energy by a spontaneous chemical reaction inside it is called an electrochemical cell. In this chemical energy is converted into electrical energy.
Example : Daniell cell.

Question 40.
Define : Half cell or Electrode.
Answer:
Half cell or Electrode : It is a metal electrode dipped in the electrolytic solution and capable of establishing oxidation reduction equilibrium with one of the ions of electrolyte solution and develop electrode potential. E.g. Zn in ZnSO₄ solution.

Question 41.
What are the functions of a salt bridge ?
Answer:
The functions of a salt bridge are :

1. It maintains the electrical contact between the two electrode solutions of the half cells.
2. It prevents the mixing of electrode solutions.
3. It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
4. It eliminates the liquid junction potential.

Question 42.
What are the conventions used to write galvanic cell or cell diagram (cell formula) ?
Answer:
A galvanic cell or voltaic cell is represented by a short notation or diagram which includes electrodes, aqueous solutions of ions and other species that may or may not involve in the cell reaction.
The following conventions are used to represent the cell or write the cell notation :
(1) The metal electrodes or the inert electrodes like platinum are placed at the ends of the cell formula.
(2) The galvanic cell consists of two half cells or electrodes. The electrode on the extreme left hand side is anode where oxidation takes place and it carries negative (-) charge while extreme right hand electrode is cathode where reduction takes place and it carries positive (+) charge.
(3) The gases or insoluble substances are placed in the interior positions adjacent to the metal electrode.
(4) A single vertical line is written between two phases like solid electrode and aqueous solution containing ions.
(5) A double vertical line is drawn between two solutions of two electrodes which indicates a salt bridge connecting them electrically.
(6) The concentration of solutions or ions or pressures of gases are written in brackets along with the substances in the cell.
(7) Different ions in the same solution are separated by a comma.

(8) Examples of electrochemical cells :
(i) Daniel cell is represented as,

Question 43.
How to write cell reaction for a galvanic cell ?
Answer:
(1) A galvanic cell consists of two half cells or electrodes.
(2) Write oxidation half reaction for left hand electrode which is an anode and reduction half reaction for right hand electrode which is a cathode.
(3) Balance the number of electrons in the oxidation and reduction reactions.
(4) By adding both the reactions, overall cell reaction is obtained.
(5) For example, consider following cell :

Question 44.
Why is anode in a galvanic cell considered to be negative?
Answer:

1. According to IUPAC conventions, the electrode of a galvanic cell where de-electronation or oxidation takes place releasing electrons is called anode. Zn_(s) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^–
2. The electrons released due to oxidation reaction are accumulated on the metal electrode surface charging it negatively.

Hence anode in the galvanic cell is considered to be negative.

Question 45.
Why is cathode in a galvanic cell considered to be positive electrode?
Answer:
(1) According to IUPAC conventions, the electrode of the galvanic cell where electronation or reduction takes place is called cathode. In this, the electrons from the metal electrode are removed by cations required for their reduction.
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2e^– → Cu_(s)

(2) Since the electrons are lost, the metal electrode acquires a positive charge.
Hence cathode in the galvanic cell is considered to be positive.

Question 46.
Give the cell reactions in the case of the following cells :
(1)

Answer:

(2)

Answer:

(3) Pt, H_2(g) | H⁺_(aq) || Cl^–_(aq) | Cl_2(g), Pt
Answer:

(4) Ni_(s)|Ni²⁺ (1 M) || Al³⁺ (1 M) | Al_(s)
Answer:

Question 47.
Represent the half cells or electrodes for the following reactions :

Answer:

Question 48.
Formulate a cell from the following electrode reactions :
(a) Cl_2(g) + 2e^– → 2Cl^–_(aq)
(b) 2I^–_(aq) → I_2(s) + 2e^–
Answer:
(a) Cl_2(g) + 2e^– → 2Cl^–_(aq) (Reduction half reaction)
(b) 2I^–_(aq) → I_2(s) + 2e^– (Oxidation half reaction)
The galvanic cell is,
Pt |I_2(s)|I^–_(aq) (1 M) | Cl^–_(aq) (1 M) | Cl₂(g, P_Cl2)|Pt

Question 49.
Formulate a cell for each of the following reactions :
(a) \(\mathrm{Sn}_{\text {(aq) }}^{2+}\) + 2AgCl_(s) → \(\mathrm{Sn}_{(\mathrm{aq})}^{4+}\) + 2Ag(s) + \(2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
(b) Mg_(s) + Br_2(l) → \(\mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{Br}_{(\mathrm{aq})}^{-}\)
Answer:
(a) \(\mathrm{Sn}_{\text {(aq) }}^{2+}\) + 2AgCl_(s) → \(\mathrm{Sn}_{(\mathrm{aq})}^{4+}\) + 2Ag(s) + \(2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
The overall reaction takes place into two steps :
(i) \(\mathrm{Sn}_{\text {(aq) }}^{2+}\) → Sn⁴⁺ + 2e^– (Oxidation half reaction)
(ii) 2AgCl_(s) + 2e^– → 2Ag_(s) + \(2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\) (Reduction half reaction)
Hence the cell is,

(b) Mg_(s) + Br_2(l) → \(\mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{Br}_{(\mathrm{aq})}^{-}\)
The overall reaction takes place into two steps :

Question 50.
What is electrode potential?
Answer:
(1) Electrode potential : It is defined as the difference of electrical potential established due to electrode half reaction between metal electrode and the solution around it at equilibrium at constant temperature.

(2) Explanation : When a metal is immersed into a solution containing its ions there arises oxidation (or reduction) reaction involving a release of electrons (or gain of electrons). This gives rise to the formation of an electrical double layer, consisting of a charged metal surface and an ionic layer. The potential across this double layer i.e., between metal and the solution is an electrode potential.

Question 51.
Define :
(1) Oxidation potential,
(2) Reduction potential.
Answer:
(1) Oxidation potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium developed due to oxidation reaction at anode and at constant temperature.

(2) Reduction potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium developed due to reduction reaction at cathode and at constant temperature.

Question 52.
What is a standard state of a substance ?
Answer:
The standard state of a substance is that state in which the substance has unit activity or concentration at 25 °C. i.e., For solution having concentration 1 molar, gas at 1 atm, pure liquids or solids are said to be in their standard states.

Question 53.
Define the following terms :
(1) Standard electrode potential
(2) Standard oxidation potential
(3) Standard reduction potential.
Answer:
(1) Standard electrode potential : It is defined as the difference of electrical potential between metal electrode and the solution around it equilibrium when all the substances involved in the electrode reaction are in their standard states of unit activity or concentration at constant temperature.

(2) Standard oxidation potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium due to oxidation reaction, when all the substances involved in the oxidation reaction are in their standard states of unit activity or concentration at constant temperature.

(3) Standard reduction potential : It is defined as the difference of electrical potential between metal electrode and the solution around it at equilibrium due to reduction reaction, when all the substances involved in the reduction reaction are in their standard states of unit activity or concentration at constant temperature.

Question 54.
What is the standard potential of an electrode according to IUPAC convention?
Answer:
Standard reduction potential : According to IUPAC convention, the standard potential of an electrode due to reduction reaction at 298 K is taken as the standard reduction potential. In this active mass of the substance has unit value.

Question 55.
What is cell potential or emf of a cell ?
Answer:
Cell potential or emf of a cell : It is defined as the potential difference between two electrodes, responsible for an external flow of electrons from the left hand electrode at higher potential (anode), to the right hand electrode at lower potential (cathode), when connected to form an electrochemical or galvanic cell.

Since there is oxidation reaction at left hand electrode (LHE) or anode and reduction reaction at right hand electrode (RHE) or cathode, emf of the galvanic, Ecell, is given by
E_cell = (E_oxi)_anode + (E_red)_cathode
Since by IUPAC conventions, generally reduction potentials are used, hence, for the given cell,
(∵ E_oxi = -E_red)
∴ E_cell = (E_red)_cathode – (E_red)_anode
Similarly, standard emf of the cell, E⁰_cell is given by
E⁰_cell = (E⁰_red)_cathode – (E⁰_red)_anode

Question 56.
Explain dependence of cell potential on concentration.
OR
Explain Nernst equation for cell potential.
Answer:
Consider following general reaction taking place in the galvanic cell.
aA + bB → cC + dD
The cell voltage is given by,

where,
T → temperature
R → Gas constant
F → Faraday
n → Number of electrons in the redox cell reaction.

This is Nernst equation for cell potential. It is used to calculate cell potential and electrode potentials.

Question 57.
State (or write) Nernst equation for the electrode potential and explain the terms involved.
Answer:
The Nernst equation for the single electrode reduction potential for a given ionic concentration in the solution in the case, \(M_{(a q)}^{n+}\) + ne^– → M_(s) is given by

\(\mathrm{E}_{\mathrm{M}^{\mathrm{n}+} / \mathrm{M}}\) is the single electrode potential,
\(E_{\mathrm{M}^{n+} / \mathrm{M}}^{0}\) is the standard reduction electrode potential,
R is the gas constant = 8.314 JK^-1 mol^-1
T is the absolute temperature,
n is the number of electrons involved in the reaction,
F is Faraday (96500 C)
[Mn⁺] is the molar concentration of ions.

Question 58.
Obtain Nernst equation for the following cell :

Answer:
Electrode reactions and a cell reaction for the given cell are,

Here, n = 2
By Nernst equation, the cell potential is given by,

Question 59.
Obtain Nernst equation for the electrode potential for the electrode, \(\mathrm{Zn}_{(\mathrm{aq})}^{2+} \mid \mathrm{Zn}_{(\mathrm{s})}\).
Answer:
For the electrode, \(\mathrm{Zn}_{(\mathrm{aq})}^{2+} \mid \mathrm{Zn}_{(\mathrm{s})}\),
the reduction reaction is,
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}_{(\mathrm{s})}\) ∴ n = 2
By Nernst equation, the reduction electrode potential is given by,

where E⁰_zn²⁺/zn is the standard electrode potential of zinc electrode.

Question 60.
Obtain a relation between cell potential and Gibbs energy for the cell reaction.
Answer:
Consider a galvanic cell which involves n number of electrons in the overall cell reaction. Since one mole of electrons involve the electric charge equal to one Faraday (F) which is equal to 96500 C, the total charge involved in the reaction is,
Electric charge = n × F
If Ecell is the cell potential, then Electrical work = n × F × E_cell
According to thermodynamics, electric work is equal to decrease in Gibbs energy, -ΔG, we can write,
Electric work = n × F × E_cell = -ΔG
∴ ΔG = -nFE_cell
Under standard conditions, we can write
∴ ΔG⁰ = -nF\(E_{\text {cell }}^{0}\)
where \(E_{\text {cell }}^{0}\) is the standard cell potential and ΔG⁰ is the standard Gibbs free energy change.

Question 61.
Write Nernst Equation for the following reactions :
(a) Cr_(s) + 3Fe³⁺_(aq) → Cr³⁺_(aq) + 3Fe²⁺_(aq)
(b) Al³⁺_(aq) + 3e^– → Al_(s)
Answer:
(a) Cr_(s) + 3Fe³⁺_(aq) → Cr³⁺_(aq) + 3Fe²⁺_(aq)
The cell formulation is,
Cr_(s)|Cr³⁺_(aq) || Fe³⁺_(aq), Fe²⁺_(aq)| Pt
Hence cell potential is,

Question 62.
A single electrode potential can’t be measured but the cell potential can be measured. Explain.
Answer:
(1)


According to Nemst theory, electrode potential is the potential difference between the metal and ionic layer around it at equilibrium, i.e. the potential across the electric double layer.

(2) For measuring the single electrode potential, one part of the double layer, that is metallic layer can be connected to the potentiometer but not the ionic layer. Hence, single electrode potential can’t be measured experimentally.

(3) When an electrochemical cell is developed by combining two half cells or electrodes, they can be connected to the potentiometer and the potential difference or cell potential can be measured.
E_cell = E₂ – E₁
where E₁ and E₂ are reduction potentials of two electrodes.

(4) If one of the electrode potentials is known or arbitrarily assumed and E_cell is measured by potentiometer, then potential of another electrode can be obtained. Therefore it is necessary to choose a reference electrode with arbitrarily fixed potential and measure the potentials of other electrodes.

(5) Therefore Standard Hydrogen Electrode (SHE) is selected assuming arbitrary potential 0.0 volt. Hence potentials of all other electrodes are referred to as hydrogen scale potentials.

Question 63.
Describe the construction and working of the standard hydrogen electrode (S.H.E.). Give its advantages and disadvantages.
OR
What is the standard hydrogen electrode
OR
Primary reference electrode? Write the construction and working of it.
Answer:
A single electrode potential cannot be measured, but the cell potential can be measured experimentally. Hence, it is necessary to have a reference electrode. S.H.E. is a primary reference electrode.
(1) Construction :
(1) The standard hydrogen electrode (S.H.E.) consists of a glass tube at the end of which a piece of platinised platinum foil is attached as shown in Fig. 5.14. Around this plate there is an outer jacket of glass which has a side inlet through which pure and dry hydrogen gas is bubbled at one atmosphere pressure. The inner tube is filled with a little mercury and a copper wire is dipped into it. This provides an electrical contact with the platinum foil. The outer jacket ends into a broad opening.

(2) The whole assembly is kept immersed in a solution containing hydrogen ions (H⁺) of unit activity.
(3) This electrode is arbitrarily assigned zero potential.
(4) The platinised platinum foil is used to provide an electrical contact for the electrode. This permits rapid establishment of the equilibrium between the hydrogen gas adsorbed by the metal and the hydrogen ions in solution.

(2) Representation of S.H.E. :
H⁺ (1 M) | H₂ (g, 1 atm) | Pt

(3) Working :
Reduction : H⁺_(aq) + e^– ⇌ \(\frac {1}{2}\)H_2(g) E⁰ = 0.00 V
H₂ gas in contact with H⁺_(aq) ions attains an equilibrium establishing a potential.

(4) Applications of SHE : A reversible galvanic cell with the experimental (indicator) electrode, Zn²⁺ (1M) | Zn_(s) and SHE can be developed as follows :

Thus the potential can be directly obtained.

(5) Disadvantages (Drawbacks or Difficulties) :

• It is difficult to construct and handle SHE.
• Pure and dry H₂ gas cannot be obtained.
• Pressure of H₂ gas cannot be maintained exactly at 1 atmosphere.
• The active mass or concentration of H⁺ from HCl cannot be maintained exactly unity.

Question 64.
How is the potential of hydrogen electrode obtained?
Answer:
Hydrogen gas electrode is represented as,
H⁺_(aq) | H₂ (g, P_H2) | Pt
Electrode reduction reaction is,
2H⁺_(aq) + 2e^– → H_2(g)
By Nernst equation, the reduction potential is,

If H₂ gas is passed at 1 atm, then P_H2 = 1 atm

Question 65.
Draw the diagram for the determination of standard electrode potential with SHE.
Answer:
Consider the following cell :
Zn | Zn²⁺_(aq) || HCl | H₂(g, 1atm) | Pt

Question 66.
A voltaic cell consisting of Fe²⁺_(aq)|Fe_(s) and Bi³⁺_(aq) | Bi_(s) electrodes is constructed. When the circuit is closed, mass of Fe electrode decreases and that of Bi electrode increases.
(a) Write cell formula, (b) Which electrode is cathode and which electrode is anode ? (c) Write electrode reactions and overall cell reaction.
Answer:
(a) Since the mass of Fe electrode decreases, it undergoes oxidation and it is an anode or an oxidation electrode while as the mass of Bi electrode increases, there is a reduction of Bi³⁺ to Bi and it is cathode or a reduction electrode. Hence the cell formula is,
\(\mathrm{Fe}_{(\mathrm{s})}\left|\mathrm{Fe}_{\mathrm{(aq})}^{2+}(1 \mathrm{M}) \| \mathrm{Bi}_{(\mathrm{aq})}^{3+}(1 \mathrm{M})\right| \mathrm{Bi}\)

(b) The left hand electrode is an anode and right hand electrode is a cathode.

(c) Reactions :

Solved Examples 5.7 – 5.9

Question 67.
Solve the following :

(1) Write the reaction and calculate the potential of the half cell,
\(\mathbf{Z n}_{(\mathbf{a q})}^{2+}\) (0.2M) | Zn. (E⁰_Zn²⁺/Zn = – 0.76 V).
Solution :
Given : E⁰_Zn²⁺/Zn = -0.76 V
Concentration of Zn²⁺ = [Zn²⁺] = 0.2 M
E_Zn²⁺/Zn = ?
Reduction reaction for the half cell,

= – 0.76 + 0.0296 (-0.6990)
= -0.76 – 0.02069
= -0.78069 V
Ans. E⁰_Zn²⁺/Zn = -0.78069 V

(2) Write a reaction and calculate the potential of the electrode, \(\mathrm{Cl}_{(\mathrm{aq})}^{-}\) (0.05 M) | Cl₂ (g, 1 atm) | Pt E⁰_Cl2/Cl^– = 1.36 V.
Solution :
Given : Reduction reaction :

= 1.36 – 0.0592 (- 2 + 0.6990)
= 1.36 – 0.0592 (-1.3010)
= 1.36 + 0.077
= 1.437 V
Ans. Potential of the electrode = 1.437 V

(3) Calculate the potential of the electrode,
pH = 4.5 | H₂ (g, 1 atm) |Pt.
Solution :
Given : pH = 4.5

∴ E_H⁺/H2 = -0.0592 pH
= -0.0592 × 4.5
= -0.2664 V
Ans. E_H⁺/H2 = -0.2664 V

(4) If the standard cell potential of Daniell cell is 1.1 V, calculate standard free energy change for the cell reaction.
Solution :
Given : Daniell cell :

= – 2 × 96500 × 1.1
= -212300 J
= -212.3 kJ
Ans. Standard free energy change = ΔG⁰
= -212.3 kJ

(5) Write balanced equations for the half reactions and calculate the reduction potentials at 25 °C for the following half cells :
(a) Cl^– (1.2 M) | Cl₂(g, 3.6 atm) E⁰ = 1.36 V
(b) Fe²⁺ (2 M) | Fe_(s) E⁰ = – 0.44 V
Solution :
(a) Given : Half cell,
\(\mathrm{Cl}_{(\mathrm{aq})}^{-}\) (1.2 M) | Cl₂(g, 3.6 atm)|Pt
E⁰_Cl2/Cl^– = 1.36 V
The reduction reaction:

= 1.36 – 0.0296 (-0.3979)
= 1.36 + 0.01178
= 1.37178
≅ 1.372 V

(b) Given: Half cell, \(\mathrm{Fe}_{(\mathrm{aq})}^{2+}\) (2M) |Fe_(s)
E⁰ _Fe²⁺/Fe = -0.44 V
The reduction reaction:

= – 0.44 – 0.0296 × (- 0.3010)
= -0.44 + 0.00891
= -0.43109 V
Ans. (a) Half reaction : Cl_2(g) + 2e^– → \(2 \mathrm{Cl}_{\text {(aq) }}^{-}\)
E_Cell = 1.372 V
(b) \(\mathrm{Fe}_{(\mathrm{aq})}^{2+}\) + 2e^– → Fe_(s)
E_Cell = -0.43109 V.

(6) Using Nernst equation, calculate the potentials for the following half reactions :

Solution :
(a) Given :

= 0.535 – 0.0296 [ – 4 + 0.9542]
= 0.535 – 0.0296 [-3.0458]
= 0.535 + 0.0902
= 0.6252 V

Ans. (a) Potential of the half cell = 0.6252 V
(b) Potential of the half cell = 0.7118 V.

(7) Write the cell reaction and calculate the standard potential of the cell,
Ni_(s) | Ni²⁺(1 M) || Cl^–(1M) | Cl₂ (g, 1 atm) | Pt
E⁰_Cl2 = 1.36 V and E⁰Ni = – 0.25 V.
Solution :

= 1.36 – (-0.25)
= 1.36+ 0.25 = 1.61V
Ans. Cell reaction : Ni_(s) + Cl_2(g) → \(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
E⁰_Cell = 1.61 V

(8) Write the cell reaction and calculate cell potential and standard free energy change for a cell reaction in the following cell :
\(\mathbf{A l}_{(\mathrm{s})}\left|\mathbf{A l}_{(\mathbf{a q})}^{3+}(1 \mathbf{M}) \| \mathbf{C d}_{(\mathbf{a q})}^{2+}(1 \mathrm{M})\right| \mathbf{C} d\)
E⁰_Al³⁺/Al = -1-66 V and E⁰_cd²⁺/cd = -0.403 V
Solution :

Since concentrations of ions are 1 M each, it is a standard cell, hence the cell potential is E⁰_Cell.

Standard free energy change ΔG⁰ is given by,
ΔG⁰= – nFE⁰_Cell
= -6 × 96500 × 1.257
= – 727800 J
= -727.8 kJ
Ans. Cell reaction :

(9) Write the cell reaction and calculate cell potential and the standard free energy change for the cell reaction in the following cell :
Pt | H₂ (g, 1 atm) | \(\mathbf{H}_{(\mathrm{aq})}^{+}\)(1M) || \(\mathrm{Cu}_{\text {(aq) }}^{2+}\) (1M) | Cu_(s).
Mention anode and cathode and direction of flow of electrons in the external circuit. (E⁰_Cu²⁺/Cu = 0.337 V)
Solution :
Given : E⁰_Cu²⁺/Cu = 0.337 V;
E⁰_H⁺/H2 = E⁰_SHE = 0.0V
Pt | H₂(g, 1 atm) | \(\mathbf{H}_{(\mathrm{aq})}^{+}\) (1M) || \(\mathrm{Cu}_{\text {(aq) }}^{2+}\) (1M) | Cu_(s)

Anode : Hydrogen gas electrode (LHE)
Cathode : Copper electrode (RHE)
E⁰_Cell = E⁰_Cu²⁺/Cu – E⁰_H⁺/H2
= 0.337 – (0.0)
= 0.337 V
ΔG⁰ = – nFE⁰ = -2 × 96500 × 0.337
= – 65040J
= – 65.04 kJ
Electrons in the external circuit will flow from (LHE) hydrogen gas electrode to (RHE) copper electrode.
Ans. Cell reaction : H_2(g) + \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) → 2 \(\mathrm{H}_{(\mathrm{aq})}^{+}\) + Cu_(s)
Cell potential = E⁰_Cell = 0.337 V
ΔG⁰ = -65.04 kJ

(10) Calculate the reduction potential of the electrode, Zn²⁺ (0.02 M) | Zn_(s). E⁰_Zn⁺⁺/Zn = – 0.76 V.
Solution :
Given :E⁰_red = E⁰_Zn⁺⁺/Zn = -0.76 V;
Concentration of \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) = [Zn²⁺] = 0.02 M
The reduction reaction for the electrode,
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) +2e^– → Zn_(s); ∴ n = 2
The reduction potential is given by,

= – 0.76 + 0.0296 (- 1.6990)
= -0.76 – 0.0296 × 1.6990
= -0.76 – 0.0503
= -0.8103 V
Ans. E_red = E_Zn²⁺/Zn = -0.8103 V

(11) Calculate the potential of the following cell at 25 °C :
Zn | Zn²⁺(0.6 M) ||H⁺(1.2 M) | H₂ (g, 1 atm) | Pt
E°Zn2+/Zn = -0.763 V
Solution :
Given : E⁰_Zn²⁺/Zn = -0.763 V;
Concentrations : [Zn²⁺] = 0.6 M; [H⁺] = 1.2 M
[H₂]_g = 1 atm
Cell potential = E_cell = ?

= 0.763 – 0.0296 × (- 0.3801)
= 0.763 + 0.01125
= 0.77425 V
Ans. Cell potential = E⁰_cell = 0.77425 V

(12) The following redox reaction occurs in a galvanic cell.
2Al_(s) + 3Fe²⁺(1 M) → 2Al³⁺(1 M)+ 3Fe_(s)
(a) Write the cell notation.
(b) Identify anode and cathode
(c) Calculate E⁰_cell if E⁰_anode = – 1.66 V and E⁰_cathode = – 0 44 V
(d) Calculate ΔG⁰ for the reaction.
Solution :

(a) In the cell reaction, Al is oxidised from zero to 3+ while Fe³⁺ is reduced from 3+ to zero. Hence the cell notation is,

(b) Anode : Al electrode at LHE
Cathode : Fe electrode at RHE

(d) The standard free energy change ΔG⁰ is given by,
ΔG⁰ = – nFE⁰_cell
= – 6 × 96500 × 1.22
= – 70640 J
= – 706.4 kJ
Ans. (a) Cell notation :
\(\mathrm{Al}_{(\mathrm{s})}\left|\mathrm{Al}_{(\mathrm{aq})}^{3+}(1 \mathrm{M}) \| \mathrm{Fe}_{(\mathrm{aq})}^{2+}(1 \mathrm{M})\right| \mathrm{Fe}_{(\mathrm{s})}\)
(b) Anode : Al; Cathode : Fe
(c) E⁰_cell = 1.22 V
(d) ΔG⁰ = – 706.4 kJ

(13) Construct a cell consisting of \(\mathbf{N i}_{(\mathrm{aq})}^{2+}\) | Ni(s) half cell and H⁺ | H_2(g) | Pt half cell.
(a) Write the cell reaction
(b) Calculate emf of the cell if [Ni²⁺] = 0.1M,
P_H2 = 1 atm [H⁺] = 0.05 M and
E⁰_Ni = – 0.257 V.
Solution :


= 0.257 – 0.0296 × 1.6020
= 0.257 – 0.04742
= 0.20958
≅ 0.2096 V
Ans.

(14) Calculate the cell potential of the following cell at 25°C,

Standard reduction potentials (SRP) of Zn and Cu are -0.76 V and 0.334 V respectively.
Solution:
Given:

(15) Set up the cell consisting of \(\mathbf{H}_{\text {(aq) }}^{+} \mid \mathbf{H}_{2(\mathrm{~g})}\) and \(\mathbf{P b}_{(\mathbf{a q})}^{2+}\) | Pb_(s) electrodes. Calculate the emf at 25 °C of the cell if [Pb²⁺] = 0.1 M,
[H⁺] = 0.5 M and hydrogen gas is at 2 atm pressure. E⁰_pb²⁺/pb = – 0.126 V.
Solution :
Given : Half cells :

Concentrations : [H⁺] = 0.5 M; [Pb²⁺] = 0.1M;
[H₂]_g = P_H2 = 2 atm; E⁰_H⁺/H2 = E_SHE = 0.0 V;
E⁰_pb²⁺/pb = -0.126 V
Since E⁰_pb²⁺/pb (reduction) < E⁰_H⁺/H2 the Pb electrode is anode and hydrogen gas electrode is cathode.
The cell formulation :


= 0.126 – 0.0296 (- 0.0969)
= 0.126 + 0.002868
= 0.128868
≅ 0.1289 V
Ans. E_cell = 0.1289 V

(16) Consider a galvanic cell that uses the half reactions,
2H⁺_(aq) + 2e^– → H_(g)
Mg²⁺_(aq) + 2e^– → Mg_(s)
Write balanced equation for the cell reaction. Calculate E⁰_cell, E_cell and ΔG⁰ if concentrations are 1M each and P_H2 = 10 atm
E⁰_Mg²⁺/Mg = -2.37 V.
Solution :



The standard free energy change ΔG⁰ is given by
ΔG⁰= – nFE⁰_cell
= – 2 × 96500 × 2.37
= -457400 J
= – 457.4 kJ
Ans. E⁰_cell = 2.37 V; E_cell = 2.3404 V;
ΔG⁰ = -457.4 kJ

(17) Calculate E_cell and ΔG for the following at 28 °C : Mg_(s) + Sn²⁺ (0.04M) → Mg²⁺ (0.06M) + Sn_(s)
E⁰_cell = 2.23 V
Is the reaction spontaneous ?
Solution:
Given:
Mg_(s) + Sn²⁺ (0.04 M) → Mg²⁺ (0.06 M) + Sn
[Sn²⁺] = 0.4 M
[Mg²⁺] = 0.06 M
E⁰_cell = 2.23V
E_cell = ?
ΔG = ?

2.23 – 0.0296 × 0.1761
= 2.23 – 0.005213
= 2.224V
ΔG = – nFE
= – 2 × 96500 × 2.224
= – 4.292 × 105 J
= -429.2 kJ
Since ΔG is negative, the electrochemical reaction is spontaneous.

(18) The standard potentials for Sn²⁺/Sn and Fe²⁺/Fe half reactions are -0.136 V and -0.440 V respectively. At what relative concentrations of Sn²⁺ and Fe²⁺ will these have the same reduction potentials?
Solution :



Hence when relative concentrations of Sn²⁺ and Fe²⁺ i.e., [Sn²⁺]/[Fe²⁺] = 5.37 × 10^-11, both the electrodes will have same potential.

(19) Write the cell reaction and calculate the emf of the cell at 25 °C.
Cr_(s) | Cr³⁺(0.0065 M) || Co²⁺(0.012 M) | Co_(s)
E⁰_Co = – 0.280 V, E⁰_Cr = – 0.74 V
What is ΔG for the cell reaction ?
Solution :
Given :

By Nernst equation,

Ecell = 0.4463 V
ΔG = -nFE_Cell
= – 6 × 96500 × 0.4463
= – 258407 J
= – 258.4 kJ
Ans. Cell reaction :
\(2 \mathrm{Cr}_{(\mathrm{s})}+3 \mathrm{Co}_{(\mathrm{aq})}^{2+} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{Co}_{(\mathrm{s})}\)
E_Cell = 0.4463 V; ΔG = – 258.4 kJ.

(20) Calculate E⁰_Cell, ΔG⁰ and equilibrium constant for the reaction 2Cu⁺ → Cu²⁺ + Cu.
E⁰_Cu⁺/Cu = 0.52 V and E⁰_Cu²⁺,Cu+ = 0.16 V.
Solution :
Given : Cell reaction : 2Cu⁺_(aq) → Cu²⁺_(aq) +Cu_(s)
E⁰_Cu⁺/Cu = 0.52V; E⁰_Cu²⁺,Cu+ = 0.16 V
1F = 96500 C
E⁰_Cell = ? ΔG⁰ = ? K=?
(i) The formulation of the cell :

(ii) ΔG⁰ = – nFE⁰_Cell = – 1 × 96500 × 0.36
= – 34740 J
= – 34.74 kJ
(iii) Electrochemical redox reactions are considered as reversible reactions. If K is the equilibrium constant for the electrochemical redox reaction, then

Ans. E⁰_Cell = 0.36 V; ΔG⁰ = – 34.74 kJ; Equilibrium constant = K= 1.2 × 10⁶ mol^-1 dm³.

(21) Calculate the equilibrium constant for the redox reaction at 25 °C.
Sr_(s) + Mg²⁺ → Sr²⁺_(aq) + Mg_(s),
that occurs in a galvanic cell. Write the cell formula.
E⁰_Mg = – 2.37 V and E⁰_Sr = – 2.89 V.
Solution :
Given :
Cell reaction : Sr_(s) + Mg²⁺ → Sr²⁺_(aq) + Mg_(s)
E⁰ _Mg²⁺/Mg = -2.37 V; E⁰ _Sr²⁺/Sr = -2.89 V
Equilibrium constant K = ?
The formation of the cell :

Ans. Equilibrium constant = K = 3.698 × 10¹⁷

(22) The equilibrium constant for the following reaction at 25 °C is 2.9 × 10⁹. Calculate standard voltage of the cell.
Cl_2(g) + 2Br^–_(aq) ⇌ Br_2(l) + 2Cl^–_(aq)
Solution :
Given : Cell reaction : Cl_2(g) + 2Br^–_(aq) ⇌ Br_2(l) + 2Cl^–_(aq)
Equilibrium constant = K = 2.9 × 10⁹ atm^-1
Standard voltage of the cell = E⁰_Cell = ?
The formulation of the cell :

(23) Write the cell representation and calculate equilibrium constant for the following redox reaction :
Ni_(s) + 2 Ag⁺_(aq) (1M) → Ni²⁺_(aq) (1 M) + 2Ag_(s)
at 25 °C
E⁰_Ni = – 0.25 V and E⁰_Ag = 0.799 V
Solution :
Given : E⁰_Ni²⁺/Ni = – 0.25 V; E⁰_Ag⁺/Ag = 0.799 V
Equilibrium constant = K = ?

(24) Calculate the cell potential of the following galvanic cell :
Pt|H₂ (g, 1 atm)|\(\mathbf{H}_{\text {(aq) }}^{+} \mathbf{p H}\) = 3.51||Calomel electrode
E_cal = 0.242 V at 25 °C.
Solution :

= -0.0592 × pH
= -0.0592 × 3.5
= -0.2072 V
∴ E_cell = E_cal – E_H⁺/H2
= 0.242 – (-0.2072)
= 0.242 + 0.2072
= 0.4492 V
Ans. E_cell = 0.4492 V

Question 68.
How are the voltaic cells classified ?
Answer:
The voltaic cells are classified as primary and secondary voltaic cells.
(1) Primary voltaic cells : These are the voltaic cells in which the electrical energy or cell potentials are developed within the cells due to oxidation and reduction reactions at the reversible electrodes.

The chemicals and electrode materials consumed during the discharging can be regenerated by passing the current in opposite direction from the external source of electricity i.e., these cells can be recharged. For example, Daniell cell. There are the examples where the primary cells can’t be recharged. E.g. Dry cell.

(2) Secondary voltaic cells :
(i) These are the voltaic cells in which the electrical energy or cell potentials are not developed within the cell but electrical energy can be stored or cell potentials can be regenerated by passing electricity from the external source of electricity. Since the electrical energy obtained is second hand, these cells are called secondary cells or accumulators or storage cells.

(ii) These cells can be recharged by passing electric current in opposite direction from the external source of higher emf. Therefore the secondary cells are reversible cells. For example, lead accumulator (lead storage battery).

Question 69.
Explain the construction and working of a dry cell (or Leclanche’s cell).
OR
Write a note on dry cell.
Answer:
(A) Principle :

• Leclanche’s cell is a primary voltaic cell.
• It doesn’t contain mobile liquid electrolyte but contains moist viscose aqeuous paste of the electrolytes.
• It is an irreversible voltaic cell which can’t be recharged.

(B) Construction :
(i) It consists of a small zinc vessel which serves as an anode (negative electrode).

(ii) The zinc vessel contains a porous paper bag containing an inert graphite (C) electrode which serves as cathode, immersed in a paste of MnO₂ and carbon black. This paper bag divides the dry cell into two compartments, namely anode and cathode compartments.
(iii) The rest of the cell is filled with a moist paste of NH₄Cl and ZnCl₂ which acts as an electrolyte for zinc anode.
(iv) The graphite rod is fitted with a metal cap and the cell is sealed to prevent the drying of moist paste by evaporation.

(C) The dry cell can be represented as,
^– Zn|ZnCl_2(aq), NH₂Cl_(aq), MnO_2(s)|C⁺.

(D) Reactions in the dry cell :
(i) Oxidation at zinc anode :
Zn_(s) → \(\mathrm{Zn}_{\text {(aq) }}^{2+}\) + 2e^– (oxidation half reaction)
(ii) Reduction at graphite (C) cathode :
The electrons released in the oxidation reaction at anode, flow to cathode through external circuit.
Hydrogen in NH₄ ion is reduced to molecular hydrogen which reduces MnO₂ to Mn₂O₃.

(iii) Zn²⁺ react with NH₃ and form a complex.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+4 \mathrm{NH}_{3(\text { aq) }} \longrightarrow\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}{ }_{(\mathrm{aq})}\)
Since Zn²⁺ ions are removed, the overall cell reaction can’t be reversed.

(E) Uses of dry cell :

• Dry cell is used as a source of electric power in radios, flashlights, torches, clocks, etc.
• Since they are available in small size and portable, they can be used conveniently.

Question 70.
Describe the construction and working of lead accumulator (lead storage cell).
OR
Draw a neat labelled diagram of the lead accumulator. Explain the reactions involved in discharging and charging this cell. Represent this cell using cell conventions.
Answer:
(A) Principle :
(1) The lead accumulator is a secondary electrochemical cell since electrical energy and emf are not developed within the cell but it is previously stored by passing an electric current. Hence it is also called lead accumulator or lead storage battery.
(2) It is reversible since the electrochemical reaction can be reversed by passing an electric current in opposite direction and consumed reactants can be regenerated.
(3) Hence battery can be charged after it is discharged.

(B) Construction : In a lead accumulator, the negative terminal (anode) is made up of lead sheets packed with spongy lead, while the positive terminal (cathode) is made up of lead grids packed with PbO₂.

Sulphuric acid of about 38% strength (%w/w) or specific gravity 1.28 or 4.963 molar is the electrolyte in which the lead sheets and lead grids are dipped. The positive terminal and negative terminal are alternatively arranged in the electrolyte and are separately interconnected.

(C) Representation of lead accumulator :

(D) Working of a lead accumulator :
(1) Discharging : When the electric current is withdrawn from lead accumulator, the following reactions take place :
Oxidation at the – ve electrode or anode :

(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is

OR
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)
The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During the operation, the acid is consumed and its concentration decreases and specific gravity decreases from 1.28 to 1.17. As a result, the emf of the cell decreases. The emf of a fully charged cell is about 2.0 V.

(ii) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the – ve electrode or cathode :
PbSO_4(s) + 2e^– → Pb(s) + \(\mathrm{SO}_{4(a q)}^{2-}\)
Oxidation at the + ve electrode or anode :
PbSO_4(s) + 2H_2(l)O → PbO_2(s) + 4H⁺_(aq) + \(\mathrm{SO}_{4(a q)}^{2-}\) + 2e^–
The net reaction during charging is
2PbSO_4(s) + 2H₂O_(l) → Pb_(s) + PbO_2(s) + 4H⁺_(aq) + 2\(\mathrm{SO}_{4(a q)}^{2-}\)
OR
2PbSO_4(s) + 2H₂O → Pb_(s) + PbO_2(s) + 2H₂SO_4(aq)
The emf of the accumulator depends only on the concentration of H₂SO₄.

(E) Applications :

1. It is used as a source of d.c. electric supply.
2. It is used in automobile in ignition circuits and lighting the head lights by connecting 6 batteries giving 12V potential.
3. It is also used in invertors.

Question 71.
In lead accumulator which electrode is coated with PbO₂ ? Anode or cathode ?
Answer:
In lead accumulator, cathode is coated with PbO₂.

Question 72.
Write net charging and discharging reactions for lead storage battery.
Answer:
For lead storage battery :
Net charging reaction :
2PbSO_4(s) + 2H₂O_(l) → Pb_(s) + PbO_2(s) + 2H₂SO_4(aq)
Net discharging reaction :
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)

Question 73.
Write a note on Nickel-Cadmium (NICAD) cell.
Answer:
(1) Nickel-Cadminum (NICAD) cell is a secondary dry cell.
(2) It is rechargable, hence it is a reversible cell.
(3) It consists of a cadmium electrode in contact with an alkali and acts as anode while nickel (IV) oxide, NiO₂ in contact with an alkali acts as cathode. The alkali used is moist paste of KOH.
(4) Reactions in the cell :
(i) Oxidation at cadmium anode :
Cd_(s) + 2OH^–_(aq) → Cd(OH)_2(s) + 2e^–
(ii) Reduction at Ni02(s) cathode :
NiO_2(s) + 2H₂O_(l) + 2e^– → Ni(OH)_2(s) + 2OH^–_(aq)
The overall cell reaction is the combination of above two reactions.
Cd_(s) + NiO_2(s) + 2H₂O_(l) → Cd(OH)_2(s) + Ni(OH)_2(s)
(5) Since the net cell reaction doesn’t involve any electrolytes but solids, the voltage is independent of the concentration of alkali electrolyte.
(6) The cell potential is about 1.4 V.
(7) This cell has longer life than other dry cells.

Question 74.
Write a note on mercury battery.
Answer:
(1) Mercury battery is a rechargeable secondary dry cell.
(2) It consists of zinc anode amalgamated with mercury.
(3) The cathode consists of a paste of Hg and carbon.
(4) The electrolyte is a paste of KOH and ZnO in a strong alkaline medium.
(5) Reactions:

(6) The overall reaction involves only solid substances and electrolytic composition remains unchanged.
(7) Therefore mercury battery provides constant voltage (1.35 V) over a long period.
(8) It is superior to Leclanche’s cell in durability.
(9) Uses : It is used in hearing aids, electric watches, pacemakers, etc.

Question 75.
Describe the construction and working of hydrogen-oxygen (H₂-O₂) fuel cell.
Answer:
(A) Principle :
(i) The functioning of the fuel cell is based on the combustion reaction like,
2H_2(g) + O_2(g) → 2H₂O_(g) is exothermic redox reaction and hence it can be used to produce electricity.
(ii) The reactants of this fuel cell can be continuously supplied from outside, hence this can be used to supply electrical energy for a very long period.

(B) Construction :
(i) In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.

(ii) The electrolyte used is hot aqueous KOH solution in which porous anode and cathode carbon rods are immersed.
(iii) H₂ is continuously bubbled through anode while O₂ gas is bubbled through cathode.

(C) Working (cell reactions) :
(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H₂O.
2H_2(g) + 4OH^–_(aq) → 4H₂O_(l) + 4e^– (oxidation half reaction)
(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH-.
O_2(g) + 2H₂O_(l) + 4e^– → 4OH^–_(aq) (reduction half reaction)
(iii) Net cell reaction: Addition of both the above reactions at anode and cathode gives a net cell reaction.
2H_2(g) + O_2(g) → 2H₂O_(l) (overall cell reaction)

(D) Representation of the cell :

The overall cell reaction is an exothermic combustion reaction. However in this, H₂ and O₂ gases do not react directly but react through electrode reactions. Hence the chemical energy released in the formation of O-H bonds in H₂O, is directly converted into electrical energy.

(E) Advantages :

1. The fuel cell operates continuously as long as H₂ and O₂ gases are supplied to the electrodes.
2. The cell reactions do not cause any pollution.
3. The efficiency of this galvanic cell is the highest about 70% as compared to ordinary galvanic cells.

(F) Drawbacks of H₂-O₂ fuel cell :

1. The cell requires expensive electrodes like Pt, Pd.
2. In practice, voltage is less than 1.23 volt due to spontaneous reactions at the electrodes.
3. H₂ gas is expensive and hazardous.

(G) Applications :

1. It was successfully used in spacecraft.
2. It has potential applications in automobiles, power generators for domestic and industrial uses.

Question 76.
What are the applications of the fuel cells?
Answer:

1. Fuel cells have been used in the space programme providing electrical energy for a long duration.
2. The fuel cells have been used in automobiles on experimental basis.
3. In case of H₂-O₂ fuel cell, used in spacecraft, the water produced is used for drinking for astronauts.
4. The fuel cells using methanol as a fuel for combustion are used in electronic products such as cell phones and laptop computers.
5. The fuel cells have many potential applications as power generators for domestic and industrial uses.

Question 77.
In what way fuel cell differs from ordinary galvanic cells ?
Answer:

1. Fuel cell is a modified galvanic cell in which the thermal energy of combustion reactions is directly converted into electrical energy.
2. In the fuel cell, the reactants are not placed within the cell like ordinary galvanic cells, but they are continuously supplied to the electrodes from outside reservoir.
3. They cannot be recharged unlike ordinary galvanic cell.

Question 78.
Define electrochemical series or electromotive series.
Answer:
Electrochemical series (Electromotive series) : It is defined as the arrangement in a series of electrodes of elements (metal or non-metal in contact with their ions) with the electrode half reactions in the decreasing order of their standard reduction potentials.

Question 79.
Explain electrochemical series or electromotive series.
Answer:
The conventions used in the construction of electrochemical series (or electromotive series) are as follows :

• The (reduction) electrodes or half cells of the elements are written on the left hand side of the series and they are arranged in the decreasing order of their standard reduction potentials (E°red).
• Reduction half reactions are written for each half cell in such a way that the species with higher oxidation state and electrons are on left hand side while reduced species with lower oxidation state are on right hand side.
• The standard reduction potential of standard hydrogen electrode is 0.00 V, i.e., E⁰_H⁺/H2 = 0.0 V. The electrodes and half cell reactions with positive E⁰_red values are located above hydrogen and those with negative E⁰_red values below hydrogen. Above hydrogen, positive E⁰_red values increase, while below hydrogen negative E⁰ values increase.
• The positive E⁰_red values indicate the tendency for reduction and the negative E⁰_red values indicate the tendency for oxidation.
• The elements, whose electrodes are at the top of the series having high positive values for E⁰_red are good oxidising agents.
• The elements, whose electrodes are at the bottom of the series having high negative values for E⁰_red are good reducing agents.

Question 80.
What are the applications of electrochemical series (or electromotive series) ?
Answer:
The applications of electrochemical series (or electromotive series) are as follows :
(1) Relative strength of oxidising agents in terms of E⁰_red values : The E⁰_red value is a measure of the tendency of the species to be reduced i.e., to accept electrons and act as an oxidising agent. The species mentioned on left hand side of the half reactions are oxidising agents.

The substances in the upper positions in the series and hence in the upper left side of the half reactions have large positive E⁰_red values hence are stronger oxidising agents. For example, F₂, Ce⁴⁺, Au³⁺, etc. As we move down the series, the oxidising power decreases. Hence from the position of the elements in the electrochemical series, oxidising agents can be selected.

(2) Relative strength of reducing agents in terms of E⁰_red values : The lower E⁰_red value means lower tendency to accept electrons but higher tendency to lose electrons. The tendency for reverse reaction or oxidation increases as E⁰_red becomes more negative and we move towards the lower side of the series. For example, Li, K, Al, etc. are good reducing agents.

(3) Identifying the spontaneous direction of reaction : From the standard reduction potentials, E⁰_red, the spontaneity of a redox reaction can be determined. The difference between E⁰_red values for any two electrodes represents cell potential E⁰_cell, constituted by them.

If E°cell is positive then the reaction is spontaneous while if E⁰_cell is negative the reaction is non-spontaneous. For example, E⁰_Mg²⁺/Mg and E⁰_Ag⁺/Ag have values -2.37 V and 0.8 V respectively. Then Mg will be a better reducing agent than Ag. Therefore Mg can reduce Ag⁺ to Ag.

The corresponding reactions will be:

Therefore above reaction in the forward direction will be spontaneous while in the reverse direction will be non-spontaneous since for it E⁰_cell = -3.17V.

(4) Calculation of standard cell potential E⁰_cell : From the electrochemical series, the standard cell potential, E⁰_cell from the E⁰_red values for the half reactions given can be calculated.
For example,

Question 81.
Write any four applications of electrochemical series.
Answer:
The applications of electrochemical series are as follows :

1. Predicting relative strength of oxidising agents.
2. Predicting relative strength of reducting agents.
3. Identifying the spontaneous direction of a reaction.
4. To calculate the standard cell potential E°cell.

Multiple Choice Questions

Question 82.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. The cell constant of a conductivity cell is given by
(a) l × a
(b) \(\frac{a}{l}\)
(c) \(\frac{1}{l \times a}\)
(d) \(\frac{l}{a}\)
Answer:
(d) \(\frac{l}{a}\)

2. A conductivity cell has two platinum electrodes of area 1.2 cm² and 0.92 cm apart. Hence the cell constant is
(a) 1.104 cm^-1
(b) 1.304 cm^-1
(c) 0.906 cm^-1
(d) 0.767 cm^-1
Answer:
(d) 0.767 cm^-1

3. The conductivity of 0.02 M KI solution is 4.37 × 10^-4 Ω^-1 cm^-1. Hence its molar conductivity is
(a) 8.74 × 10^-6 Ω^-1 cm² mol^-1
(b) 21.85 Ω^-1 cm² mol^-1
(c) 4.58 × 10^-4 Ω^-1 cm² mol^-1
(d) 136.5 Ω^-1 cm² mol^-1
Answer:
(b) 21.85 Ω^-1 cm2 mol^-1

4. The specific conductance of 0.02 M HCl is 8.2 × 10^-3 Ω^-1 cm^-1. Hence its molar conductivity is
(a) 164 Ω^-1 cm² mol^-1
(b) 6.1 × 10³Ω^-1 cm² mol^-1
(c) 239.6 S cm² mol^-1
(d) 410 S cm² mol^-1
Answer:
(d) 410 S cm² mol^-1

5. Molar conductivity of an electrolyte is given by,

Answer:
(c) \(\wedge_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathrm{C}}\)

6. The units of molar conductivity are
(a) Ω cm^-2 mol^-1
(b) Ω^-1 cm² mol^-1
(c) Ω^-1 cm^-1 mol^-1
(d) Ω^– cm^-1 mol^-2
Answer:
(b) Ω^-1 cm² mol^-1

7. If conductivity is expressed in Ω^-1 m^-1 and concentration of the electrolytic solution in mol m^-3 then, the molar conductance is given by

Answer:
(b) \(\wedge_{\mathrm{m}}=\frac{\kappa}{C}\)

8. Kohlrausch’s law is represented as

Answer:
(a) \(\wedge_{0}=\lambda_{+}^{0}+\lambda_{-}^{0}\)

9. The degree of dissociation of a weak electrolyte is given by

Answer:
(c) α = \(\frac{\wedge_{\mathrm{m}}}{\wedge_{0}}\)

10. ∧₀ for CH₃COOH is 390.7 Ω^-1 cm² mol^-1. If ∧₀ for CH₃COOK, and HBr in Ω^-1 cm² mol^-1 are 115 and 430.4 respectively, then ∧0 for KBr is
(a) 74.6 Ω^-1 cm² mol^-1
(b) 180.6 Ω^-1 cm² mol^-1
(c) 154.7 Ω^-1 cm² mol^-1
(d) 706.1 Ω^-1 cm² mol^-1
Answer:
(c) 154.7 Ω^-1 cm² mol^-1

11. The molar conductivity of cation and anion of salt BA are 180 and 220 mhos respectively. The molar conductivity of salt BA at infinite dilution is
(a) 90 mhos · cm² · mol^-1
(b) 110 mhos · cm² · mol^-1
(c) 200 mhos · cm² · mol^-1
(d) 400 mhos · cm² · mol^-1
Answer:
(d) 400 mhos · cm² · mol^-1

12. If ∧_m and ∧₀ are the molar conductivities of a weak electrolyte at concentration C and at zero concentration, then the dissociation constant Ka is given by

Answer:
(b) K_a = \(\frac{\wedge_{\mathrm{m}}^{2} \times \mathrm{C}}{\Lambda_{0}\left(\wedge_{0}-\wedge_{\mathrm{m}}\right)}\)

13. What is the ratio of volumes of H₂ and O₂ liberated during electrolysis of acidified water ?
(a) 1 : 2
(b) 2 : 1
(c) 1 : 8
(d) 8 : 1
Answer:
(b) 2 : 1

14. What weight of copper will be deposited by passing 2 Faradays of electricity through a cupric salt? (atomic mass = 63.5)
(a) 63.5 g
(b) 31.75 g
(c) 127 g
(d) 12.7 g
Answer:
(a) 63.5 g

15. The S.I. unit of cell constant for conductivity cell is
(a) m^-1
(b) S·m^-2
(c) cm^-2
(d) S·dm²·mol^-1
Answer:
(a) m^-1

16. The charge of how many coulomb is required to deposit 1.0 g of sodium metal (molar mass 23.0 g mol^-1) from sodium ions is
(a) 2098
(b) 96500
(c) 193000
(d) 4196
Answer:
(d) 4196

17. The amount of electricity equal to 0.05 F is
(a) 48250 C
(b) 3776 C
(c) 4825 C
(d) 4285 C
Answer:
(c) 4825 C

18. The number of electrons that have a total charge of 965 coulombs is
(a) 6.022 × 10²³
(b) 6.022 × 10²²
(c) 6.022 × 10²¹
(d) 3.011 × 10²³
Answer:
(c) 6.022 × 10²¹

19. When 0.2 Faraday of electricity is passed through an electrolytic solution, the number of electrons involved are
(a) 96500
(b) 1.603 × 10^-19
(c) 1.2046 × 10²³
(d) 12 × 10⁶
Answer:
(c) 1.2046 × 10²³

20. When a charge of 0.5 Faraday is passed through AlCl₃ solution, the amount of aluminium deposited at the cathode is (Atomic weight of Al = 27)
(a) 4.5
(b) 18
(c) 27
(d) 2.7
Answer:
(a) 4.5

21. The quantity of electricity required to deposit 54 g of silver from silver nitrate solution is
(a) 0.5 Coulomb
(b) 0.5 Ampere
(c) 0.5 Faraday
(d) 0.5 Volt
Answer:
(c) 0.5 Faraday

22. Passage of 5400 C of electricity through an electrolyte deposited 5.954 × 10^-3 kg of the metal with atomic mass 106.4. The charge on the metal ion is
(a) + 1
(b) + 2
(c) + 3
(d) + 4
Answer:
(a) + 1

23. On calculating the strength of current in amperes if a charge of 840 C (coulomb) passes through an electrolyte in 7 minutes, it will be
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

24. On passing 1.5 F charge, the number of moles of aluminium deposited at cathode are [Molar mass of Al = 27 gram mol^-3]
(a) 1.0
(b) 13.5
(c) 0.50
(d) 0.75
Answer:
(c) 0.50

25. Number of faradays of electricity required to liberate 12 g of hydrogen is
(a) 1
(b) 8
(c) 12
(d) 16
Answer:
(c) 12

26. Daniell cell is
(a) Secondary cell
(b) Irreversible cell
(c) primary irreversible cell
(d) primary reversible cell
Answer:
(d) primary reversible cell

27. In the representation of galvanic cell, the ions in the same phase are separated by a
(a) single vertical line
(b) comma
(c) double vertical lines
(d) semicolon
Answer:
(b) comma

28. In the Daniell cell, reduction occurs at the
(a) anode
(b) zinc rod
(c) negative electrode
(d) positive electrode
Answer:
(d) positive electrode

29. The standard hydrogen electrode is represented as
(a) \(\mathrm{H}_{(\mathrm{aq})}^{+}\)|H₂(g, 1 atm) | Pt
(b) \(\mathrm{H}_{(\mathrm{aq})}^{+}\) 1M | H₂(g, 1 atm) | Pt
(c) \(\mathrm{H}_{(\mathrm{aq})}^{+}\) 1M|H_2(g)|Pt
(d) \(\mathrm{H}_{(\mathrm{aq})}^{+}\) 0.1M|H₂(g, 1 atm) | Pt
Answer:
(b) \(\mathrm{H}_{(\mathrm{aq})}^{+}\) 1M | H₂(g, 1 atm) | Pt

30. The essential condition to set a standard hydrogen electrode is
(a) 298 K
(b) pure and dry H₂ gas at 1 atm
(c) solution containing H⁺ at unit activity
(d) all of these
Answer:
(d) all of these

31. In hydrogen-oxygen fuel cell, the carbon rods are immersed in hot aqueous solution of
(a) KCl
(b) KOH
(c) H₂SO₄
(d) NH₄Cl
Answer:
(b) KOH

32. The emf of cell is 1.3 volt. The positive electrode has potential of 0.5 volt. The potential of negative electrode is
(a) 0.8 V
(b) -0.8 V
(c) 1.8 V
(d) – 1.8 V
Answer:
(b) -0.8 V

33. The electrode potential of a silver electrode dipped in 0.1 M AgNO₃ solution at 298 K is (E⁰_red of Ag = 0.80 volt)
(a) 0.0741 V
(b) 0.0591 V
(c) 0.741 V
(d) 0.859 V
Answer:
(c) 0.741 V

34. Which of the following species gains electrons more easily ?
(a) Na⁺
(b) H⁺
(c) Mg⁺
(d) Hg⁺
Answer:
(b) H⁺

35. In Nernst equation the constant 0.0592 at 298 K represents the value of

Answer:
(d) \(\frac{2.303 R T}{F}\)

36. The concept of electrode potential is explained on the basis of
(a) Arrhenius’ theory
(b) Ostwald’s theory
(c) Nemst’s theory
(d) Faraday’s law
Answer:
(c) Nemst’s theory

37. The standard reduction potentials of metals A and B are x and y respectively. If x > y, the standard emf of the cell containing these electrodes would be
(a) 2x – y
(b) y – x
(c) x – y
(d) x + y
Answer:
(c) x – y

38. The emf of the cell,

(E⁰_red = 0.34 V)
(a) -1.34
(b) 0.34 V
(c) -0.34 V
(d) 1.34
Answer:
(b) 0.34 V

39. The Electromotive Force of the following Cell Cu|Cu⁺⁺ (1 M)||A⁺g (1 M)|Ag is …………….. if E⁰_cu⁺⁺ = 0.33 V and E⁰ _Ag⁺⁺/Ag = 0.79 V
(a) 0.46 V
(b) – 0.46 V
(c) 1.12 V
(d) – 112 V
Answer:
(a) 0.46 V

40. The standard cell potential of the following cell is 0.463 V Cu|Cu⁺⁺ (1 M)||Ag⁺ (1 M)|Ag. If E⁰_Ag = 0.8 V, what is the standard potential of Cu electrode ?
(a) 1.137 V
(b) 0.337 V
(c) 0.463 V
(d) – 0.463 V
Answer:
(b) 0.337 V

41. The metal which cannot displace hydrogen from dil. H₂SO₄ solution is
(a) Zn
(b) Al
(c) Fe
(d) Ag
Answer:
(d) Ag

42. In the Lead storage battery during discharging
(a) pH of the electrolyte increases
(b) pH decreases
(c) pH remain unchanged
(d) pH increases or decreases depends on the extent of discharging
Answer:
(a) pH of the electrolyte increases

43. During the discharging of a lead storage battery,
(a) H₂SO₄ is consumed
(b) PbSO₄ is consumed
(c) Pb²⁺ ions are formed
(d) Pb is formed
Answer:
(a) H₂SO₄ is consumed

44. In lead accumulator, anode and cathode are
(a) (Pb + PbO₂), Pb
(b) Pb, PbO₂
(b) PbO₂, Pb
(d) Pb, (Pb + PbO₂)
Answer:
(d) Pb, (Pb + PbO₂)

45. The efficiency of the hydrogen-oxygen fuel cell is about
(a) 20%
(b) 40%
(c) 70%
(d) 90%
Answer:
(c) 70%

46. The strongest oxidizing agent among the species In³⁺ (E⁰ = – 1.34 V), Au³⁺ (E⁰ = 1.4 V), Hg²⁺ (E⁰ = 0.86 V), Cr³⁺ (E⁰ = – 0.74 V) is
(a) Cr³⁺
(b) Au³⁺
(c) Hg²⁺
(d) In³⁺
Answer:
(b) Au³⁺

47. The reaction, \(2 \mathrm{Br}_{(\mathrm{aq})}^{-}+\mathrm{Sn}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Br}_{2(\mathrm{l})}+\mathrm{Sn}_{(\mathrm{s})}\)
with the standard potentials, E⁰_Sn = -0.114 V, E⁰_Br2 = + 1.09 V, is
(a) spontaneous in reverse direction
(b) spontaneous in forward direction
(c) at equilibrium
(d) non-spontaneous in reverse direction
Answer:
(a) spontaneous in reverse direction

48. The cell potential of the following cell is
(E⁰_Al³⁺/Al = – 1.66 V)

(a) 1.66 V
(b) -1.66 V
(c) 0.5533 V
(d) 2.14 V
Answer:
(a) 1.66 V

49. The standard reduction potentials of Sn, Hg and Cr are – 1.36 V, 0.854 V and – 0.746 V respectively. The increasing order of oxidising power of the given elements is
(a) Sn < Hg < Cr
(b) Hg < Cr < Sn
(c) Sn < Cr < Hg
(d) Cr < Hg < Sn
Answer:
(c) Sn < Cr < Hg

50. If standard reduction potentials for Pb, K, Zn and Cu are -0.126 V, -2.925 V, -0.763 V and 0.337 V, the decreasing order of reducing power is
(a) Zn > Pb > K > Cu
(b) Cu > Pb > Zn > K
(c) K > Zn > Pb > Cu
(d) K > Pb > Cu > Zn
Answer:
(c) K > Zn > Pb > Cu+