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Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

January 7, 2025January 6, 2025 by Bhagya

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

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

1. Choose the most correct option.

Question A.
A sample of pure water, whatever the source always contains …………. by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
Answer:
a. 88.9

Question B.
Which of the following compounds can NOT demonstrate the law of multiple proportions ?
a. NO, NO₂
b. CO, CO₂
c. H₂O, H₂O₂
d. Na₂S, NaF
Answer:
d. Na₂S, NaF

Question C.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
a. – 40°
b. + 40°
c. – 80°
d. – 20°
Answer:
a. – 40°

Question D.
SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
Answer:
b. Ampere

Question E.
In the reaction N₂ + 3H₂ → 2NH₃, the ratio by volume of N₂, H₂ and NH₃ is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
Answer:
d. gaseous volumes

Question F.
Which of the following has maximum number of molecules ?
a. 7 g N₂
b. 2 g H₂
c. 8 g O₂
d. 20 g NO₂
Answer:
b. 2 g H₂

Question G.
How many g of H₂O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
Answer:
a. 4.5

Question H.
The number of molecules in 22.4 cm³ of nitrogen 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

Question I.
Which of the following has the largest number of atoms ?
a. 1g Au_(s)
b. 1g Na_(s)
c. 1g Li_(s)
d. 1g Cl_2(g)
Answer:
c. 1g Li_(s)

2. Answer the following questions.

Question A.
State and explain Avogadro’s law.
Answer:
i. In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.

ii. Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:

According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:

Question B.
Point out the difference between 12 g of carbon and 12 u of carbon.
Answer:
12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom.

Question C.
How many grams does an atom of hydrogen weigh ?
Answer:
The mass of a hydrogen atom is 1.6736 × 10^-24 g.

Question D.
Calculate the molecular mass of the following in u.
a. NH₃
b. CH₃COOH
c. C₂H₅OH
Answer:
i. Molecular mass of NH₃ = (1 × Average atomic mass of N) + (3 × Average atomic mass of H)
= (1 × 14.0 u) +(3 × 1.0 u)
= 17 u

ii. Molecular mass of CH₃COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u)
= 60 u

iii. Molecular mass of C₂H₅OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u)
= 46 u
Ans: i. The molecular mass of NH₃ = 17 u
ii. The molecular mass of CH₃COOH = 60 u
iii. The molecular mass of C₂H₅OH = 46 u

Question E.
How many particles are present in 1 mole of a substance ?
Answer:
The number of particles in one mole is 6.0221367 × 10²³.

Question F.
What is the SI unit of amount of a substance ?
Answer:
The SI unit for the amount of a substance is mole (mol).

Question G.
What is meant by molar volume of a gas ?
Answer:
The volume occupied by one mole of a gas at standard temperature (0 °C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm³.

Question H.
State and explain the law of conservation of mass.
Answer:
Law of conservation of mass:

The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
Antoine Lavoisier who is often referred to as the father of modem chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
Both his experiments resulted in increased weight of products.
After several experiments, in burning of phosphorus, he found that the weight gained by the phosphoms was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.

Question I.
State the law of multiple proportions.
Answer:
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”.

3. Give one example of each

Question A.
Homogeneous mixture
Answer:
Homogeneous mixture: Solution (An aqueous solution of sugar)

Question B.
Heterogeneous mixture
Answer:
Heterogeneous mixture: Suspension (of sand in water)

Question C.
Element
Answer:
Element: Gold

Question D.
Compound
Answer:
Compound: Distilled water.

4. Solve problems :

Question A.
What is the ratio of molecules in 1 mole of NH₃ and 1 mole of HNO₃.
Answer:
One mole of any substance contains particles equal to 6.022 × 10²³.
1 mole of NH₃ = 6.022 × 10²³ molecules of NH₃
I mole of HNO₃ = 6.022 × 10²³ molecules of HNO₃
∴ Ratio = \(\frac{6.022 \times 10^{23}}{6.022 \times 10^{23}}\) = 1 : 1
Ans: The ratio of molecules is = 1 : 1.

Question B.
Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen

Molar volume of a gas = 22.4 dm³ mol^-1 = 22.4 L at STP

Ans: Number of moles of hydrogen = 0.02 mol

Question C.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
∴ Mass of 18 atoms of hydrogen = 18 × 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u

Question D.
Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ x = \(\frac{254 \mathrm{u}}{127 \mathrm{u}}\) = 2 atoms

b. 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol^-1
Now,

Now,
Number of atoms = Number of moles × Avogadro’s constant
= 2 mol × 6.022 × 10²³ atoms/mol
= 12.044 × 10²³ atoms
= 1.2044 × 10²⁴ atoms
Ans. i.Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = 1.2044 × 10²⁴ atoms

Question E.
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. 5 mg carbon = 5 × 10^-3 g carbon
Atomic mass of carbon = 12 u
∴ Molar mass of carbon 12 g mol^-1

b. 12 mg carbon = 12 × 10^-3 g carbon

Number of atoms = Number of moles × Avogadro’s constant
Number of atoms of carbon = 1 × 10^-3 mol × 6.022 × 10²³ atoms/mol
= 6.022 × 10²⁰ atoms
Ans: Number of moles of carbon in his homework writing = 4.167 × 10^-4mol
Number of atoms of carbon in 12 mg homework writing = 6.022 × 10²⁰ atoms

Question F.
Arjun purchased 250 g of glucose (C₆H₁₂O₆) for Rs 40. Find the cost of glucose per mole.
Answer:
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C₆H₁₂O₆
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C₆H₁₂O₆).
Molecular mass of glucose
= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)
=180 u
∴ Molar mass of glucose = 180 g mol^-1

Now,
\(\frac {250}{180}\) mol of glucose cost = Rs 40
1 mol glucose cost = x
∴ x = \(\frac{40 \times 180}{250}\) = Rs 28.8/mol of glucose
Ans. The cost of glucose per mole is Rs 28.8.

[ Calculation using log table:
\(\frac{40 \times 180}{250}\)
= Antilog₁₀ [log₁₀(40) + log₁₀(180) + log₁₀(250)]
= Antilog₁₀[1.6021 + 2.2553 – 2.3979]
= Antilog₁₀[1.4595] = 28.80 ]

Question G.
The natural isotopic abundance of ¹⁰B is 19.60% and ¹¹B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron(B)

Ans. Average atomic mass of boron = 10.84 u

Question H.
Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 °C
b. 30 °C
Answer:
a. Given: Temperature in degree Celsius =40°C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 40 °C in the formula,
°F = \(\frac {9}{5}\) (°C)+32
= \(\frac {9}{5}\) (40) + 32
= 72 + 32
= 104 °F

b. Given: Temperature in degree Celsius = 30 °C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 30 °C in the formula,
°F = \(\frac {9}{5}\)(°C) + 32
= \(\frac {9}{5}\)(30) + 32
= 54 + 32
= 86 °F
Ans: i. The temperature 40 °C corresponds to 104 °F.
ii. The temperature 30 °C corresponds to 86 °F.

Question I.
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: 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
Calculator: Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH₃COOH
= (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12 u) + (4 × 1 u) + (2 × 16 u) = 60 u
∴ Molar mass of acetic acid = 60 g mol^-1

Now,
Number of molecules of acetic acid = Number of moles × Avogadro’s constant
= 0.367 mol × 6.022 × 10²³ molecules/mol
= 2.210 × 10²³ molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 × 10²³ molecules

Question J.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 32 = 64 g) of oxygen to give (2 × 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g

Question K.
Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles × Avogadro’s constant
= 0.4 mol × 6.022 × 10²³ atoms/mol
= 2.4088 × 10²³ atoms of N

b. 1.6 g of Sulphur (S)
Molar mass of sulphur = 32 g mol^-1

Number of atoms of S = Number of moles × Avogadro’s constant
= 0.05 mol × 6.022 × 10²³ atoms/mol
= 0.3011 × 10²³ atoms
= 3.011 × 10²² atoms of S
Ans: a. Number of nitrogen atoms in 0.4 mole = 2.4088 × 10²³ atoms of N
b. Number of sulphur atoms in 1.6 g = 3.011 × 10²² atoms of S

Question L.
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Answer:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
% of oxygen = \(\frac{1.2}{3.2}\) × 100 = 37.5%
% of metal = \(\frac{2.0}{3.2}\) × 100 = 62.5%
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g
% of oxygen = \(\frac{0.85}{2.27}\) × 100 = 37.44 ≈ 37.5%
% of metal = \(\frac{1.42}{2.27}\) × 100 = 62.56 ≈ 62.5%
Therefore, irrespective of the source, the given compound contains same elements in the same proportion. The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight”. Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.

Question M.
In two moles of acetaldehyde (CH₃CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer:
Molecular formula of acetaldehyde: C₂H₄O
Moles of acetaldehyde = 2 mol
a. Number of moles of carbon atoms = Moles of acetaldehyde × Number of carbon atoms
= 2 × 2
= 4 moles of carbon atoms

b. Number of moles of hydrogen atoms = Moles of acetaldehyde × Number of hydrogen atoms
= 2 × 4
= 8 moles of hydrogen atoms

c. Number of moles of oxygen atoms = Moles of acetaldehyde × Number of oxygen atoms
= 2 × 1
= 2 moles of oxygen atoms

d. Number of molecules of acetaldehyde = Moles of acetaldehyde × Avogadro number
= 2 mol × 6.022 × 10²³ molecules/mol
= 12.044 × 10²³ molecules of acetaldehyde
Ans: i. Number of moles of carbon, hydrogen and oxygen are 4, 8, 2 respectively,
ii. Number of molecules of acetaldehyde = 12.044 × 10²³

Question N.
Calculate the number of moles of magnesium oxide, MgO in
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of Mg = 24 and O = 16)
Answer:
Given: i. Mass of MgO = 80 g
ii. Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae: Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
Calculation: i. Molecular mass of MgO = (1 × Average atomic mass of Mg) + (1 × Average atomic mass of O)
= (1 × 24u) + (1 × 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol^-1
Mass of MgO = 80 g
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{80 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 2 mol

ii. Mass of MgO = 10 g, Molar mass of MgO = 40 g mol^-1
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{10 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.25 mol
Ans: i. The number of moles in 80 g of magnesium oxide, MgO = 2 mol
ii. The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol

Question O.
What is volume of carbon dioxide, CO₂ occupying by i. 5 moles and ii. 0.5 mole of CO₂ gas measured at STP.
Answer:
Given: i. Number of moles of CO₂ = 5 mol
ii. Number of moles of CO₂ = 0.5 mol
To find: Volume at STP
Formula: Number of moies of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Calculation: Molar volume of a gas 22.4 dm³ mol^-1 at STP.
Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
∴ i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5mol × 22.4 dm³ mol^-1 = 112 dm³
ii. Volume of the gas at STP Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm³ mol^-1 = 11.2 dm³
Ans: i. Volume of 5 mol of CO₂ = 112 dm³
ii. Volume of 0.5 mol of CO₂ = 11.2 dm³

Question P.
Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1.
Answer:
The molecular formula of potassium chlorate is KClO₃.
Required chemical equation:

2 moles of KClO₃ = 2 × 122.5 = 245 g
3 moles of O₂ at STP occupy = (3 × 22.4 dm³) = 67.2 dm³
Thus, 245 g of potassium chlorate will liberate 67.2 dm³ of oxygen gas.
Let ‘x’ gram of KClO₃ liberate 6.72 dm³ of oxygen gas at S.T.P.
∴ x = \(\frac{245 \times 6.72}{67.2}\) = 24.5 g
Ans: Mass of potassium chlorate required = 24.5 g

Question Q.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH₂)₂CO. Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH₂)₂CO
Molar mass of urea = 60 g mol^-1

∴ Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 2.247 × 10²³ atoms of hydrogen
∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 1.124 × 10²³ atoms of nitrogen
∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of carbon
∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 10²³ atoms of H, 1.124 × 10²³ atoms of N, 0.562 × 10²³ atoms of C and 0.562 × 10²³ atoms of O.

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

Hence, (0.5 × 32 = 16 g) of sulphur will combine with (0.5 × 32 = 16 g) of oxygen to give (0.5 × 64 = 32 g) sulphur dioxide.
Ans: Mass of sulphur dioxide produced = 32 g

5. Explain

Question A.
The need of the term average atomic mass.
Answer:

Several naturally occurring elements exist as a mixture of two or more isotopes.
Isotopes have different atomic masses.
The atomic mass of such an element is the average of atomic masses of its isotopes.
For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.

Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.

Question B.
Molar mass.
Answer:
i. The mass of one mole of a substance (element/compound) in grams is called its molar mass.
ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
e.g.

Element	Atomic mass (u)	Molar mass (g mol^-1)
H	1.0	1 0
C	12.0	12.0
O	16.0	16.0

iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
e.g.

Polyatomic substance	Molecular/formula mass (u)	Molar mass (g mol^-1)
O₂	32.0	32.0
H₂O	18.0	18.0
NaCl	58.5	58.5

Question C.
Mole concept.
Answer:

Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon -12 isotope.
One mole is the amount of substance which contains 6.0221367 × 10²³ particles/entities.

Question D.
Formula mass with an example.
Answer:

The formula mass of a substance is the sum of atomic masses of the atoms present in the formula.
In substances such as sodium chloride, positive (sodium), and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na⁺) ion is surrounded by six chlorides (Cl^–) ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units.
Therefore, NaCl is just the formula that is used to represent sodium chloride though it is not a molecule.
In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass.

e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

Question E.
Molar volume of gas.
Answer:
i. It is more convenient to measure the volume rather than mass of the gas.
ii. It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm³ at standard temperature (0 °C) and pressure (1 atm) (STP).
iii. The volume of 22.4 dm³ at STP is known as molar volume of a gas.
iv. The relationship between number of moles and molar volume can be expressed as follows:

[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol^-1]

Question F.
Types of matter (on the basis of chemical composition).
Answer:
Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.

They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
1. Metals:

They have a lustre (a shiny appearance).
They conduct heat and electricity.
They can be drawn into wire (ductile).
They can be hammered into thin sheets (malleable).
e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.

2. Nonmetals:

They have no lustre, (except diamond, iodine)
They are poor conductors of heat and electricity, (except graphite)
They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine

3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.

ii. Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.

Mixtures are of two types:

Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
e.g. Suspension, which contains insoluble solid in a liquid.

11th Chemistry Digest Chapter 1 Some Basic Concepts of Chemistry Intext Questions and Answers

Can you tell? (Textbook Page No. 1)

Question 1.
Which are mixtures and pure substances from the following?
i. Sea water
ii. Gasoline
iii. Skin
iv. A rusty nail
v. A page of textbook
vi. Diamond
Answer:

No.	Material	Pure substance or mixture
i.	Seawater	Mixture
ii.	Gasoline	Mixture
iii.	Skin	Mixture
iv.	A rusty nail	Mixture
V.	A page of textbook	Mixture
vi.	Diamond	Pure substance

Can you tell? (Textbook Page No. 2)

Question 1.
Classify the following as element and compound.
i. Mercuric oxide
ii. Helium gas
iii. Water
iv. Table salt
v. Iodine
vi. Mercury
vii. Oxygen
viii. Nitrogen
Answer:

No.	Material	Element or compound
i.	Mercuric oxide	Compound
ii.	Helium gas	Element
iii.	Water	Compound
iv.	Table salt	Compound
V.	Iodine	Element
vi.	Mercury	Element
vii.	Oxygen	Element
viii.	Nitrogen	Element

Can you tell? (Textbook Page No. 6)

Question 1.
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:

If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced.

Can you recall? (Textbook Page No. 6)

Question 1.
What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:

The smallest indivisible particle of an element is called an atom.
A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
Every atom of an element has definite mass. The order of magnitude of mass of one atom is 10^-27 kg.
Isotopes are the atoms of the same element having same atomic number but different mass number.

Try this (Textbook Page No. 8)

Question 1.
Find the formula mass of CaSO₄, if atomic mass of Ca = 40.1 u, S =32.1 u and O = 16.0 u.
Solution:
Formula mass of CaSO₄
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= (40.1) + 32.1 + (4 × 16.0) = 136.2 u
Ans: Formula mass of CaSO₄ = 136.2 u

Can you recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items.

Try this (Textbook Page No. 10)

Question 1.
Calculate the volume in dm³ occupied by 60.0 g of ethane at STP.
Solution:
Given: Mass of ethane at STP = 60.0 g
To find: Volume of ethane
Formulae:

Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP
Molecular mass of ethane = 30 g mol^-1

∴ Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 2 mol × 22.4 dm³ mol^-1 = 44.8 dm³
Ans: Volume of ethane = 44.8 dm³

Activity :

Activity 1.
Collect information of various scientists and prepare charts of their contribution in chemistry.
Answer:

Scientists		Contributions
Joseph Louis Gay-Lussac (1778 – 1850) (French chemist and physicist)	i.	Formulated the gas law.
	ii.	Collected samples of air at different heights and recorded temperatures and moisture contents.
	iii.	Discovered that the composition of atmosphere does not change with increasing altitude.
Amedeo Avogadro (1776 – 1856) (Italian scholar)	i.	Published article in French journal on determining the relative masses of elementary particles of bodies and proportions by which they enter combinations.
Amedeo Avogadro (1776 – 1856) (Italian scholar)	ii.	Published a research paper titled “New considerations on the theory of proportions and on determination of the masses of atoms.”

11th Std Chemistry Questions And Answers:

11th Chemistry Chapter 3 Exercise Basic Analytical Techniques Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 3 Basic Analytical Techniques Textbook Exercise Questions and Answers.

Basic Analytical Techniques Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 3 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 3 Exercise Solutions

1. Choose the correct option

Question A.
Which of the following methods can be used to seperate two compounds with different solubilities in the same solvent?
a. Fractional crystallization
b. Crystallization
c. Distillation
d. Solvent extraction
Answer:
a. Fractional crystallization

Question B.
Which of the following techniques is used for seperation of glycerol from soap in soap industry ?
a. Distillation under reduced pressure
b. Fractional distillation
c. Filtration
d. Crystallization
Answer:
a. Distillation under reduced pressure

Question C.
Which technique is widely used in industry to seperate components of mixture and also to purify them ?
a. Steam distillation
b. Chromatography
c. Solvent extraction
d. Filtration
Answer:
b. Chromatography

Question D.
A mixture of acetone and benzene can be seperated by the following method :
a. Simple distillation
b. Fractional distillation
c. Distillation under reduced pressure
d. Sublimation
Answer:
b. Fractional distillation

Question E.
Colourless components on chromatogram can not be observed by the following :
a. Using UV light
b. Using iodine chamber
c. Using the spraying reagent
d. Using infrared light
Answer:
d. Using infrared light

2. Answer the following

Question A.
Which of the following techniques is used for purification of solid organic compounds?
a. Crystallisation
b. Distillation
Answer:
Solid (crude/impure) organic compounds can be purified by crystallization.

Question B.
What do you understand by the terms
a. residue
b. filtrate.
Answer:
a. Residue: In the process of filtration, the insoluble (undissolved) impurities which remain on the filter paper are called residue.

b. Filtrate: In the process of filtration, the liquid which pass through the filter paper and collected in the beaker is called filtrate.

Question C.
Why is a condenser used in distillation process?
Answer:
In the process of distillation, a liquid is converted into its vapour and the vapour is then condensed back to liquid on cooling. The condenser has a jacket with two outlets through which water is circulated. Hence, to provide efficient cooling, a condenser is used.

Question D.
Why is paper moistened before filtration?
Answer:
Before filtration, filter paper is moistened with appropriate solvent to ensure that it sticks to the funnel and does not let the air to pass through the leaks.

Question E.
What is the stationary phase in Paper Chromatography?
Answer:
Paper chromatography is a type of partition chromatography in which a special quality paper, namely Whatman paper 1 is used. The water trapped in the fibres of the paper acts as stationary phase.

Question F.
What will happen if the upper outlet of the condenser is connected to the tap instead of the lower outlet?
Answer:

If water enters through upper outlet of condenser, the water will quickly flow down under the influence of gravity. This allows only a small section of the condenser to be cooled enough.
If water enters through lower outlet of condenser, the entire condenser will be filled with water before it leaves out providing maximum cooling to the condenser. This results in maximum recovery of purified liquid.

Hence, water must be allowed to enter through lower outlet of condenser during distillation process.

Question G.
Give names of two materials used as stationary phase in chromatography.
Answer:

Alumina
Silica gel

Question H.
Which properties of solvents are useful for solvent extraction?
Answer:

Organic compound must be more soluble in the organic solvent, than in water.
Solvent should be immiscible with water and be able to form two distinct layers.

Question I.
Why should spotting of mixture be done above the level of mobile phase ?
Answer:

If spotting of a mixture is done at the level of mobile phase, then solvent will come in contact with the sample spot.
Sample spot will dissolve in the mobile phase and its components will move all over the plate resulting in no distinct separation.

Hence, spotting of mixture should be done above the level of mobile phase.

Question J.
Define : a. Stationary phase b. Saturated solution
Answer:
a. Stationary phase:
Stationary phase is a solid or a liquid supported on a solid which remains fixed in a place and on which different solutes are adsorbed to a different extent.

b. Saturated solution:
A saturated solution is a solution which cannot dissolve additional quantity of a solute.

Question K.
What is the difference between simple distillation and fractional distillation?
Answer:

No.	Simple distillation	Fractional distillation
i.	If in a mixture the difference in boiling points of two liquids is appreciable/large, they are separated from each other using the simple distillation.	If in a mixture the difference in boiling points of two liquids is not appreciable/large, they are separated from each other using the fractional distillation.
ii.	Simple distillation assembly is used.	fractionating column is fitted in distillation assembly.
e.g.	Mixture of acetone (b.p. 329 K) and water (b.p. 373 K) can be separated by this method.	Mixture of acetone (b.p. 329 K) and methanol (b.p. 337.7 K) can be separated by this method.

Question L.
Define a. Solvent extraction
b. Distillation.
Answer:
a. Solvent extraction:
Solvent extraction is a method used to separate an organic compound present in an aqueous solution, by shaking it with a suitable organic solvent in which the compound is more soluble than water.

b. Distillation:
The process in which liquid is converted into its vapour phase at its boiling point and the vapour is then condensed back to liquid on cooling is known as distillation.

Question M.
List the properties of solvents which make them suitable for crystallization.
Answer:
The solvent to be used for crystallization should have following properties:

The compound to be crystallized should be least or sparingly soluble in the solvent at room temperature but highly soluble at high temperature.
Solvent should not react chemically with the compound to be purified.
Solvent should be volatile so that it can be removed easily.

Question N.
Name the different types of Chromatography and explain the principles underlying them.
Answer:
Depending on the nature of the stationary phase i.e., whether it is a solid or a liquid, chromatography is classified into adsorption chromatography and partition chromatography.
i. Adsorption chromatography: This technique is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.

Adsorption chromatography is further classified into two types:

Column chromatography
Thin-layer chromatography

ii. Partition chromatography: This technique is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. For example, paper chromatography

Question O.
Why do we see bands separating in column chromatography?
Answer:

In column chromatography, the solutes get adsorbed on the stationary phase and depending on the degree to which they are adsorbed, they get separated from each other.
The component which is readily adsorbed are retained on the column and others move down the column to various distances forming distinct bands.

Hence, we see bands separating in column chromatography.

Question P.
How do you visualize colourless compounds after separation in TLC and Paper Chromatography?
Answer:
i. Thin-layer chromatography (TLC): If components are colourless but have the property of fluorescence then they can be visualized under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The iodine vapours are adsorbed by the components and the spots appear brown. Also, spraying agent like ninhydrin can also be used (for amino acids).

ii. Paper Chromatography: The spots of the separated colourless components may be observed either under ultra-violet light or by the use of an appropriate spraying agent.

Question Q.
Compare TLC and Paper Chromatography techniques.
Answer:

Chromatography technique	TLC	Paper chromatography
Principle	It is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.	It is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.
Stationary phase	Solid (adsorbent like silica gel or alumina over a glass plate)	Liquid (water trapped in the fibres of a Paper)
Mobile phase	Liquid (single solvent/mixture of solvents)	Liquid (single solvent/mixture of solvents)
Visualization of components of a mixture	Similar to TLC the coloured components are visible as coloured spots and the colourless components are observed under UV light or using a spraying agent.

3. Label the diagram and explain the process in your words.

Answer:
When filtration is carried out using a vacuum pump it is called filtration under suction. It is a faster and more efficient technique than simple filtration. The diagram is as follows:

ii. Procedure:

The assembly for filtration under suction consists of a thick wall conical flask with a sidearm (Buchner flask).
The flask is connected to a safety bottle by rubber tube through the side arm.
Buchner funnel (a special porcelain funnel with a porous circular bottom) is fitted on the conical flask with the help of a rubber cork.
A circular filter paper of correct size is placed on the circular porous bottom of the Buchner funnel and the funnel is placed on the flask.
Filter paper is moistened with a few drops of water or solvent.
Suction is created by starting the pump and filtration is carried out.

iii. Crystals are collected on the filter paper and filtrate in the flask.

11th Std Chemistry Questions And Answers:

11th Biology Chapter 4 Exercise Kingdom Animalia Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 4 Kingdom Animalia Textbook Exercise Questions and Answers.

Kingdom Animalia Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 4 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 4 Exercise Solutions

1. Choose correct option

Question (A)
Which of the following belongs to a minor phylum?
(a) Comb jelly
(b) Jelly fish
(c) Herdmania
(d) Salpa
Answer:
(a) Comb jelly

Question (B)
Select the animal having venous heart.
(a) Crocodile
(b) Salamander
(c) Rohu
(d) Toad
Answer:
(c) Rohu

Question (C)
In Ascaris, _______ .
(a) mesoglea is present
(b) endoderm is a discontinuous layer
(c) mesoderm is present in patches
(d) body cavity is absent
Answer:
(c) mesoderm is present in patches

Question (D)
Which of the following is INCORRECT in case of birds?
(a) Presence of teeth
(b) Presence of scales
(c) Nucleated RBCs
(d) Hollow bones
Answer:
(a) Presence of teeth

Question (E)
Chitinous exoskeleton is a characteristic of ________ .
(a) Dentalium
(b) Antedon
(c) Millipede
(d) Sea urchin
Answer:
(c) Millipede

2. Answer the following questions.

Question (A)
Reptiles are known for having three chambered heart. Which animal shows a near four chambered condition in reptiles?
Answer:
Crocodiles have a four chambered heart.

Question (B)
The circulatory system has evolved from open to closed type in Animal Kingdom. Which Phylum can be called first to represent closed circulation?
Answer:
Phylum Annelida is the first phylum to represent closed circulation.

Question (C)
Pinna is part of external ear and it is found in mammals. Do Aves and Reptiles show external ear in any form?
Answer:
No, Aves and Reptiles do not show external ear in any form. They possess tympanum which represents the ear.

Question (D)
Fish and frog can respire in water. Can they respire through their skin? If yes, why do they have gills?
Answer:
1. Yes, fishes and frogs can respire through their skin.
2. The larval stage of frog i. e. tadpole respires through gills. During metamorphosis, tadpoles lose their gills and develop lungs.
3. Frogs do not have scales and breathe through their skin underwater.
4. Fishes respire primarily via gills. The body of fishes is covered with scales which limits cutaneous respiration in them.

Question (E)
Birds need to keep their body light to help in flying. Hence, they show presence of some organs only on one side. How their skeleton helps in reducing their weight?
Answer:

In birds, the forelimbs are modified into wings for flying.
They possess stream-lined body to reduce resistance during flight.
Bones are hollow or pneumatic to reduce body weight.
In order to reduce body weight, urinary bladder is absent. Also, females possess only left ovary and oviduct.
Body is covered by feathers to facilitate flying.

Question (F)
Cnidarians and Ctenophorans are both diploblastic. Which other character do they have in common, which is not found in other phyla?
Answer:
Cnidarians and ctenophorans show tissue level of body organization. They have blind sac body plan and radially symmetrical body.

Question (G)
Crab and Snail both have a protective covering. Is it made up of the same material?
Answer:
No, the protective covering is not made up of same material in crab and snail. The protective covering of crabs is made up of chitin and that of snails is made up of calcium carbonate.

Question (H)
Sponge and sea star show calcareous protective material. Do they belong to the same Phylum?
Answer:
No, they do not belong to same phylum. Sponges belong to phylum Porifera and sea star belongs to phylum Echinodermata.
1. Adult echinoderms are radially symmetrical but larval forms are bilaterally symmetrical.
2. Larvae of echinoderms are free-swimming.

Question (I)
Fish and snake both have scales. How do these scales differ from each other?
Answer:
Fishes have dermal scales covering the body surface whereas snakes have epidermal scales or scutes.

Question (J)
Lower Phyla like Arthropods and Cnidarians show metamorphosis. Is it also found in any class of Phylum Chordata?
Answer:
Yes, it is also found in class Amphibia of phylum Chordata.

Question 3.
Draw neat labelled diagram.
A. Sycon
B. Aurelia
C. Amphioxus
D. Catla
E. Balanoglossus
F. Scolidon
Answer:
A. Sycon

B. Aurelia

C. Amphioxus

D. Catla

E. Balanoglossus

F. Scolidon

Question 4.
Match the following.

Phylum	Characters
1. Annelida	(a) Tube feet
2. Mollusca	(b) Ostia
3. Ctenophora	(c) Radula
4. Porifera	(d) Parapodia
5. Echinodermata	(e) Comb plates

Answer:

Phylum	Characters
1. Annelida	(d) Parapodia
2. Mollusca	(c) Radula
3. Ctenophora	(e) Comb plates
4. Porifera	(b) Ostia
5. Echinodermata	(a) Tube feet

5. Identify the animals given in pictures and write features of its phylum/class.

Question 1.

Answer:
The organism in the given picture is Comb jelly (Red midwater Comb jelly) and it belongs to phylum Ctenophora.

Question 2.

Answer:
The organism in the given picture is Eel and it belongs to phylum Chordata.

Question 3.

Answer:
The given organism in the given picture is Dolphin and it belongs to class Mammalia.

Question 4.

Answer:
The given organism is Snake and it belongs to class Reptilia

Question 5.

Answer:
The given organism is Sea urchin and belongs to phylum Echinodermata.

Question 6.

Answer:
The given organism is flying lizard and belongs to class Reptilia.

Question 7.

Answer:
The organism is Herdmania and belongs to Phylum Chordata (Subphylum Urochordata).

Question 8.

Answer:
The organism in the given picture is Nautilus and it belongs to phylum Mollusca.

Question 9.

Answer:
The organism in the given picture is Amphioxus and it belongs to Phylum Chordata (Subphylum Cephalochordata).

6. Observe and identify body symmetry of given animals.

Question 1.
Observe and identify body symmetry of given animals.

Answer:
Fig i. represents asymmetry
Fig ii. represents radial symmetry
Fig iii. represents bilateral symmetry

Practical/Project:

Question 1.
Study different animals in kingdom Animalia and prepare the chart with detail scientific information.
Answer:
Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

Habitat: They are aquatic, mostly marine but few species are found in fresh water.
Forms: They are sedentary animals (attached to substratum or rock).
Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
division of labour among cells. Hence, their body is considered as a colony of different types of cells.
Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.
Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
Sponges have great power of regeneration.
e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Characteristics of members belonging to phylum Cnidaria:

Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
Forms: They are sessile or free swimming.
Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
Body Symmetry: They have radially symmetrical body.
Germ layer: They are diploblastic.
Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
Digestion: They have extracellular and intracellular digestion.
Reproduction: Cnidarians reproduce asexually and sexually.

Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

Habitat: They are exclusively marine.
Forms: They are free swimming animals.
Germ layers: Members of this phylum are diploblastic.
Body Symmetry: They are radially symmetrical.
Body plan: The animals of this phylum show blind-sac body plan.
Body organization: They show tissue level organization.
Locomotion: It is earned out by eight rows of ciliated comb plates.
Bioluminescence: It is the characteristic feature of the members of this phylum.
Digestion: It is extracellular and intracellular.
Reproduction: Reproduction is sexual with indirect development.
Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplana

11th Biology Digest Chapter 4 Kingdom Animalia Intext Questions and Answers

Can you recall? (Textbook Page No. 29)

(i) What is the basis for classification?
Answer:
Grades of organization, body symmetry, body cavity, germ layers and segmentation form the basis for classification.

(ii) Who proposed Five kingdom classification system?
Answer:
Robert Whittaker proposed the five kingdom system of classification.

(iii) What is the need and importance of classification?
Answer:
Need and importance for classification:
a. Classification facilitates the identification of animals with great accuracy.
b. The study of animals becomes convenient.
c. It helps in understanding the relationship of animals with other living organisms.
d. It helps to understand the habitat of each animal along with its role in nature.
e. By studying few animals from a group, we can gain a better understanding about the entire group.
f. It helps in understanding different adaptations shown by animals.
g. It gives an idea about evolution of animals.

Observe and discuss. (Textbook Page No. 29)

Discuss the criteria of classification.
Answer:
1. The given diagrams represents the number of germ layers and body symmetry used as criteria for animal classification.
2. Number of germ layers:
(a) When an organism shows only two germ layers, they are called diploblastic animals. In this case, the outer ectoderm is separated from the inner endoderm by a non-living substance called mesoglea.
(b) When an organism shows three germinal layers, they are called triploblastic animals. The three layers are namely – outer ectoderm, middle mesoderm and inner endoderm.
3. Body symmetry:
Body symmetry implies to the similarity in shape, size and number of parts on the opposite sides of a median line when body is divided into two halves by an imaginary line along different plane. Animals may be asymmetrical, radially symmetrical or bilaterally symmetrical.
(a) Asymmetrical animals:
An animal is said to be asymmetrical when its body cannot be divided into two identical halves in any plane.
(b) Radially symmetrical animals:
In certain animals, body can be cut or divided into two similar halves in a number of planes wherein, all the cuts (planes) pass through the centre. This type of symmetry is called radial symmetry.
(c) Bilaterally symmetrical animals:
In this type, the body of the animal can be bisected or divided in two equal or identical halves by a single median or vertical plane.

Internet my friend. (Textbook Page No. 30)

Which are the larval stages of Porifera.
Answer:
Larval stages of Porifera:
Parenchymula – Flagellate larvae of calcinean sponges
Amphiblastula – Free swimming larval stage of Sycon and many other calcareous sponges. Rhagon— Larval stage which give rise to the leuconoid condition in demospongiae.
[Students are expected to find more information about the larval stages of Porifera on internet.]

Find out. (Textbook Page No. 31)

Information about coral reefs and sea fan.
Answer:
Coral reefs:

A coral reef is an underwater ecosystem characterized by reef building corals.
Coral reefs constitute 25% of all marine species on the planet.
They belong to phylum Cnidaria.
There are three main types of coral reefs – fringing, barrier and atoll. Coral reefs provide ecosystem services for tourism, fisheries and shoreline protection.
They cannot survive in high temperatures, thus due to climate change there is a sharp decline in their population.

Sea fan or Gorgonia:

1. It is a soft coral composed of numerous polyps – cylindrical, sessile (attached) forms that grow together in a flat, fan-like pattern.
2. It belongs to phylum Cnidaria.
3. It does not produce calcium carbonate skeletons.
[Students can find out more information about coral reefs and sea fan using internet ]

Can you tell? (Textbook Page No. 32)

(i) State the parasitic adaptations in Liver fluke and Ascaris.
Answer:
Parasitic adaptations in Liver fluke:
a. Presence of hooks and suckers
b. Body covered with cuticle
c. Lacks digestive system
d. They are hermaphrodites

Parasitic adaptations in Ascaris:
a. Presence of muscular pharynx for sucking the food.
b. Body covered by tough, thick and resistant cuticle.
c. Secretes enzymes against the enzymes secreted by the host.
d. Respiration is anaerobic.
e. Reproductive system is highly developed.

(ii) Give example of free living platyhelminth.
Answer:
Planaria

Find out. (Textbook Page No. 33)

What are the merits and demerits of hermaphroditism?
Answer:
Hermaphroditism is the condition in which an organism possesses reproductive organs of both the sexes.

Merits of hermaphroditism:
a. Assured fertilization which reduces the risk of a species to become extinct due to unavailability of mating partner.
b. Energy required for searching out mating partner is conserved.
c. Frequency of mating is maximized.

Demerits of hermaphroditism:
a. More energy is required to maintain both the reproductive systems.
b. Limited gene diversity.
[Source: http://floydbiology. blogspot. com/2012/06/httpmattc-thinks. html]
[Students are expected to find more information using the internet.]

Why are leeches used in Ayurveda?
Answer:
a. Leeches are used in blood purification therapy to treat many diseases as they suck impure blood from the affected site of the patient’s body.
b. The anticoagulant – hirudin present in saliva of leech, inhibits the coagulation of blood and makes blood thinner. This dissolves the clots found in vessels and facilitates the blood supply.

What is the role of earthworms in agriculture? What is vermicompost?
Answer:
Role of earthworms in agriculture:
a. Earthworms loosen the soil by burrowing deep into it, thus they help to aerate the soil.
b. This continuous digging of soil also helps the water to reach the roots quickly.
c. Earthworms can decompose the organic matter from the soil and convert it into rich manure.
d. This helps in increasing the fertility of soil which ultimately increases the crop production.
e. Earthworm castings are rich in nutrients which act as natural fertilizer.
Vermicompost:
Vermicompost is the product of vermicomposting. It is organic manure produced as vermicast by earthworm feeding on biological waste material and plant residues.

Can you tell? (Textbook Page No. 34)

(i) Explain the term metameric segmentation.
Answer:
In some animals, body consists of many segments arranged along the length of the body. When the external segmentation coincides with the internal segmentation, it is called as metameric segmentation and the phenomenon is called metamerism.

(ii) Give characteristics of Arthropoda.
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs:Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

(iii) Enlist the harmful Arthropods.
Answer:
Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silkworms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

Find out. (Textbook Page No. 34)

(i) Why is phylum Arthropoda considered as most successful phylum?
Answer:
Phylum Arthropoda is considered as most successful phylum because of the following reasons:
a. Phylum Arthropoda is the largest phylum of kingdom Animalia. It includes various forms like lobsters, prawns, crabs, insects, millipedes, locust, honeybees, etc.
b. They are omnipresent (present everywhere). Arthropods show great variety of adaptations as their habitat varies from terrestrial to aquatic habitat.
c. Several others factors also contribute to the success of the phylum which includes:
1. The exoskeleton of arthropods is made up of tough chitinous exoskeleton. This enables them to survive on lands in almost all environment and is a great defense against predators.
2. They possess jointed appendages which allow complex movements.
3. They exhibit moulting or eedysis.
4. They have metamerically segmented body helping in movement around diverse environments.

(ii) What do we mean by parthenogenesis?
Answer:
Development of an egg into a complete individual without fertilization is known parthenogenesis. It is found in many non-vertebrates such as bees, rotifers and even some lizards and birds (turkey).

(iii) What do we mean by living fossil?
Answer:
A member of a living animal or plant species that is almost identical to species known from the fossil record (not the recent fossil record), i.e. they have changed very little over a long period.
[Source:https://www. encyclopedia, com/earth-and-environment/ecology-and- environmentalism/environmental-studies/living-fossil]

(iv) How the bees produce honey?
Answer:
a. Bees produce honey using the nectar of flowering plants. A bee sucks the nectar and stores it in a honey sac until it returns to the hive.
b. The nectar is then transferred to worker bees in the hive who suck the nectar from the honey sac through their proboscis. This nectar contains 70% water and 20% honey. Honeybees get rid of excess water by swallowing and regurgitating the nectar again and again. They also fan their wings over filled cells of honeycomb.

When most of the water has evaporated from the honeycomb, the bee seals the comb with a secretion of liquid from its abdomen which eventually hardens into beeswax. This is how the honey bees use nectar to produce a thick, sticky and sweet honey.

(v) What will happen if arthropods do not moult?
Answer:
a. Moulting or eedysis is a periodic shedding of the outer cuticle layer of body in arthropods.
b. The outer layer of body of arthropods is formed of tough, non-living chitinous substance.
c. If arthropods do not moult, they cannot grow and mature into adult forms

Can you tell? (Textbook Page No. 34)

Why do Molluscs have shell?
Answer:
Molluscs are soft-bodied animals. Thus, the calcareous shell provides supports and protects the organisms from predators.

Can you tell? (Textbook Page No. 36)

Give salient features of phylum Echinodermata.
Answer:
Salient features of phylum Echinodermata (Echinus – spines, derma – skin)

Habitat: These are exclusively marine.
Forms: Members of this phylum are solitary, sedentary or free-living and gregarious, benthic.
Body symmetry: These animals are radially symmetrical with pentamerous symmetry.
Shape: Members of Echinodermata are spherical, elongated or star-shaped.
Body: The endoskeleton is made up of calcareous ossicles. Spines are formed on the body. Hence, they are known as echinoderms. The body has two sides oral and aboral and lacks definite divisions. Mouth is ventrally present on oral surface and anus on aboral surface.
Water vascular system: Presence of water vascular system is the peculiar character of echinoderms. The madrepOrite is the opening of water vascular system through which water enters. Water vascular system is useful in locomotion, food capturing, respiration.
Digestion: Digestive system is complete.
Respiration: Peristomial gills, papillae, respiratory tree, etc. are used for respiration.
Circulatory and excretory systems: Absent in echinoderms.
Nervous system: Nervous system is simple with a nerve ring around the mouth and radial nerves in arms.
Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external.
Development is indirect, i.e. through larval stages. They show high power of regeneration.

e.g. Sea lily (Antedon), Sea star (Asterias), Sea cucumber (Cucumaria), Brittle star (Ophiothrix), Sea urchin (Echinus).

Can you tell? (Textbook Page No. 36)

Can you tell? (Textbook Page No. 36)
Answer:
1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered as notochord. It is now placed as a separate phylum under Non-Chordata.
2. It possesses certain characteristics of both Chordates and Non-chordates.
3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Find out. (Textbook Page No. 36)

Why Balanoglossus is considered as connecting link between Non-chordates and chordates?
Answer:
Balanoglossus belongs to phylum Hemiehordata. For Explanation:

Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered as notochord. It is now placed as a separate phylum under Non-Chordata.
It possesses certain characteristics of both Chordates and Non-chordates.
Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Observe and discuss. (Textbook Page No. 36)

Compare and contrast between Non-Chordates and Chordates.

Answer:

Non-chordates	enoraates
1. Notochord is absent.	Notochord present at least in the early embryonic life.
2. Nerve cord is ventral, paired and ganglionated.	Nerve cord is single, dorsal and non-ganglionated.
3. The heart, if present is dorsal.	The heart is ventral in position.
4. Pharyngeal gill slits are absent.	Pharyngeal gill slits are present at least in embryonic stage.
5. Post-anal tail is absent.	Post-anal tail is present at least in embryonic stage.

Can you tell? (Textbook Page No. 37)

Herdmania is called a Chordate. Explain.
Answer:
1. Herdmania belongs to phylum Urochordata.
2. It is called a chordate as it shows the following features:
a. Presence of notochord at least in early embryonic life. (In Herdmania, notochord is present in the tail of the larval forms).
b. Presence of hollow, dorsal nerve chord, running throughout the length of the body.
c. Presence of pharyngeal gill slits.
d. Presence of post-anal tail.

Can you tell? (Textbook Page No. 37)

Give characteristics of Petromyzon. Comment on its mode of nutrition.
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

Members of class Cyclostomata are jaw-less and eel like organisms.
Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
Median fms are present but paired fins are absent.
They are ectoparasites on fishes.
They have sucking circular mouth, without jaws.
Cranium and vertebral column are made up of cartilage.
Their digestive system lacks stomach.
Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
Heart is two chambered with one auricle and one ventricle.
Gonad is single, large and without gonoduct.
Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.
e.g Petromyzon (Lamprey), Myxine (Hagfish).

Can you tell? (Textbook Page No. 38)

(i) What is the lateral line system?
Answer:
a. Lateral line system is the system with mechanoreceptors called neuromasts, for the detection of watei current.
b. These neuromasts are arranged in an interconnected network along the head and body.
c. Lateral line system also known as lateralis system.

(ii) Why Piscian heart is called a venous heart?
Answer:
a. Pisces have two-chambered heart. They have single and closed circulation.
b. Heart of Pisces receives blood only from veins and thus always shows presence of deoxygenated blood which it pumps directly to the gills for oxygenation.
Thus, the heart of Pisces is called a venous heart.

Can you tell? (Textbook Page No. 40)

Amphibians do not have exoskeleton. Give reason.
Answer:
1. Amphibians live in both water and on land.
2. They perform cutaneous respiration (i. e. gaseous exchange across the skin or outer integument.) under water and when on land, they respire through lungs.
Thus, to facilitate cutaneous respiration, amphibians do not have exoskeleton.

Can you tell? (Textbook Page No. 40)

Why are amphibians and reptilians called poikilotherms?
Answer:
Amphibians and reptilians are called poikilotherms as they cannot maintain a constant body temperature. Their body temperature changes according to the change in surrounding temperature.

Can you tell? (Textbook Page No. 41)

Give adaptations in Aves for flying.
Answer:

In birds, the forelimbs are modified into wings for flying.
They possess stream-lined body to reduce resistance during flight.
Bones are hollow or pneumatic to reduce body weight.
In order to reduce body weight, urinary bladder is absent. Also, females possess only left ovary and oviduct.
Body is covered by feathers to facilitate flying.

Can you tell? (Textbook Page No. 41)

(i) Aves and mammals are homeotherms. Give reason.
Answer:
a. Aves and mammals can generate heat to maintain their body temperature.
b. They keep their body temperature constant, irrespective of fluctuations in environmental temperature. Thus, Aves and mammals are homeotherms.

(ii) How mammals differ from other groups of animals?
Answer:
Features of class Mammalia (mammae: breasts, nipple):

Special feature: Presence of mammary glands (milk-producing glands) for the nourishment of young ones. Mammary glands are modified sweat glands.
Habitat: Mammals are omnipresent (present everywhere). These are mostly terrestrial, some are aquatic and few are aerial and arboreal (living on trees).
Locomotion: Limbs are the organs of locomotion and are modified for walking, climbing, burrowing, swimming, etc.
Body division: Body is differentiated into head, neck, trunk and tail. They have external ear (pinna).
Body temperature: Mammals are homeotherms or warm-blooded animals.
Exoskeleton: It is in the form of hair, fur, nails, hooves, horns, etc.
Skin: Skin is glandular and has sweat glands and sebaceous (oil) glands.
Mouth cavity: Mammals show heterodont dentition (various types of teeth like incisors, canines, premolars and molars).
Circulation: Heart is ventral in position, four chambered with two auricles and two ventricles. RBCs are biconcave and enucleated (except camel). Blood is red in colour.
Respiration: Respiration takes place by lungs.
Nervous system: Brain is highly developed. Cerebrum shows a transverse band called corpus callosum.
Reproduction and development: Only few mammals are oviparous, e.g. Duck billed platypus. Some have pouches for development of immature young ones. These are called marsupials, e.g. Kangaroo. Most of the mammals are placental and viviparous.

Do yourself. (Textbook Page No. 41)

Observe different animals in your surrounding, write detailed classification and write down the characteristics of animals in following format.
Answer:

11th Std Biology Questions And Answers:

11th Biology Chapter 7 Exercise Cell Division Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 7 Cell Division Textbook Exercise Questions and Answers.

Cell Division Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 7 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 7 Exercise Solutions

1. Choose correct option

Question (A)
The connecting link between Meiosis – I and Meiosis – II is ……….. .
(a) interphase – I
(b) interphase – II
(c) interkinesis – III
(d) anaphase – IV
Answer:
(c) interkinesis – III

Question (B)
Synapsis is pairing of ………………. .
(a) any two chromosomes
(b) non – homologous chromosomes
(c) sister chromatids
(d) homologous chromosomes
Answer:
(d) homologous chromosomes

Question (C)
Spindle apparatus is formed during which stage of mitosis?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) S-phase

Question (D)
Chromosome number of a cell is almost doubled up during _______ .
(a) G₁ – phase
(b) S – phase
(c) G₂-phase
(d) G₀-phase
[Note: Due to DNA replication the DNA content of cell doubles during S-phase. But the number of chromosomes remain the same.]
Answer:
(b) S – phase

Question (E)
How many meiotic divisions are necessary for formation of 80 sperms?
(a) 80
(b) 40
(c) 20
(d) 10
Answer:
(c) 20

Question (F)
How many chromatids are present in anaphase – I of meiosis – I of a diploid cell having 20 chromosomes?
(a) 4
(b) 6
(c) 20
(d) 40
Answer:
(d) 40

Question (G)
In which of the following phase of mitosis chromosomes are arranged at equatorial plane?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) Metaphase

Question (H)
Find incorrect statement.
(a) Condensation of chromatin material occurs in prophase.
(b) Daughter chromatids are formed in anaphase.
(c) Daughter nuclei are formed at metaphase.
(d) Nuclear membrane reappears in telophase.
Answer:
(c) Daughter nuclei are formed at metaphase.

Question (I)
Histone proteins are synthesized during
(a) G₁ phase
(b) S – phase
(c) G₂ – phase
(d) Interphase
Answer:
(b) S – phase

2. Answer the following questions

Question (A)
While observing a slide, student observed many cells with nuclei. But some of the nuclei were bigger as compared to others but their nuclear membrane was not so clear. Teacher inferred it as one of the phase in the cell division. Which phase may be inferred by teacher?
Answer:
Prophase.

Question (B)
Students prepared a slide of onion root tip. There were many cells seen under microscope. There was a cell seen under microscope. There was a cell with two groups of chromosomes at opposite ends of the cell. This cell is in which phase of mitosis?
Answer:
Anaphase.

Question (C)
Students were shown some slides of cancerous cells. Teacher made a comment as if there would have been a control at one of its cell cycle phase, there wouldn’t have been a condition like this. Which phase the teacher was referring to?
Answer:
The phase teacher was referring would be G_i phase.

Question (D)
Some Mendelian crossing experimental results were shown to the students. Teacher informed that there are two genes located on the same chromosome. He enquired if they will be ever separated from each other?
Answer:

Genes are located on chromosomes at specific distance and position.
The greater this distance, the greater the chance that a crossover can occur between the genes and the greater the chances of recombination.
The chances of recombination are less between the genes that are placed closed to each other on the chromosome.
Therefore, due to recombination the two genes located on the same chromosome have possibility of separating from each other.

Question (E)
Students were observing a film on Paramoecium. It underwent a process of reproduction. Teacher said it is due to cell division. But students objected and said that there was no disappearance of nuclear membrane and no spindle formation, how can it be cell division? Can you clarify?
Answer:

Paramoecium is a unicellular organism. The division in Paramoecium occurs by amitosis.
It is the simplest mode of cell division.
In amitosis, nucleus elongates and a constriction appears. This constriction deepens and divides the nucleus in two daughter nuclei followed by the division of cytoplasm.

Question (F)
Is the meiosis responsible for evolution? Justify your answer.
Answer:

Meiosis ensures that organisms produced by sexual reproduction contain correct number of chromosomes.
Meiosis exhibits genetic variation by the process of recombination.
Variations increase further after union of gametes during fertilization creating offspring with unique characteristics. Thus, it creates diversity of life and is responsible for evolution.

Question (G)
Why mitosis and meiosis – II are called as homotypic division?
Answer:
1. In mitosis, the chromosome number and genetic material of daughter cells remain same as that of the parent cell.
2. In meiosis – II, two haploid cells formed during first meiotic division divide further into four haploid cells. This division is identical to mitosis. The daughter cells formed in second meiotic division are similar to their parent cells with respect to the chromosome number formed in meiosis -1. Hence mitosis and meiosis – II are called homotypic division.

Question (H)
Write the significance of mitosis.
Answer:

As mitosis is equational division, the chromosome number is maintained constant.
It ensures equal distribution of the nuclear and the cytoplasmic content between the daughter cells, both quantitatively and qualitatively. Therefore, the process of mitosis also maintains the nucleo-cytoplasmic ratio.
The DNA is also equally distributed.
It helps in growth and development of organisms.
Old and worn-out cells are replaced through mitosis.
It helps in the asexual reproduction of organisms and vegetative propagation in plants.

Question (I)
Enlist the different stages of prophase – I.
Answer:
1. Prophase -I:
It is the most complicated and longest phase of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:
The volume of the nucleus increases.
The chromosomes become long distinct and coiled.
They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies. j Lep
The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during Gj phase of interphase.]

b. Zygotene:
Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:
Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points. These points are called chiasmata (Appear like a cross-X).
Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

d. Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

e. Diakinesis:
The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
The terminal chiasmata exist till the metaphase.
The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

3. Draw labelled diagrams and write explanation

Question (A)
With the help of suitable diagram, describe the cell cycle.
Answer:
1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.
2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.
The interphase is subdivided into three sub-phases as G₁ – phase, S-phase and G₂-phase.
a. G₁ – phase (First gap period/First Gap Phase):
It begins immediately after cell division.
RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.
b. S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
c. G₂ – phase (Second growth phase/Second Gap Phase):
Metabolic activities essential for cell division occur during this phase.
Various proteins which are necessary for the cell division are also synthesized in this phase.
Apart from this, RNA synthesis also occurs during this phase.
In animal cells, a daughter pair of centrioles appears near the pre-existing pair.

Question (B)
Distinguish between mitosis and meiosis.
Answer:

Mitosis	Meiosis
(a) It occurs in somatic cells and stem cells.	It occurs in germ cells.
(b) In this nucleus divides only once.	In this nucleus divides twice (Meiosis I and Meiosis II)
(c) In these two daughter cells are formed.	In these four daughter cells are formed.
(d) Daughter cells formed by mitotic division are diploid (2n).	Daughter cells formed by meiotic division are haploid (n)•
(e) In mitosis, crossing over does not take place.	In meiosis, crossing over takes place.
(f) Mitosis plays an important role in growth, repair, healing and development.	Meiosis is important for formation of haploid gametes and spores.

Question (C)
Draw labelled diagrams and write explanation Draw the diagram of metaphase.
Answer:
Metaphase:
a. Chromosomes are completely condensed and appear short.
b. Centromere and sister chromatids become very prominent.
c. All the chromosomes are arranged at equatorial plane of cell. This is called metaphase plate.
d. Mitotic spindle is fully formed in this phase.
e. Centromere of each chromosome divides horizontally into two, each being associated with a chromatid. [Note: The centromeres divide at the beginning of anaphase so that the two chromatids of each chromosome become separated from each other.
Source: Cell Division, Donald B. McMillan, Richard J. Harris, in An Atlas of Comparative Vertebrate Histology, 2018.]

Question 4.
Match the following column – A with column – B

Column I (Phases)	Column II (Their events)
1. Leptotene	(a) Crossing over
2. Zygotene	(b) Desynapsis
3. Pachytene	(c) Synapsis
4. Diplotene	(d) Bouquet stage

Answer:

Column I (Phases)	Column II (Their events)
1. Leptotene	(d) Bouquet stage
2. Zygotene	(c) Synapsis
3. Pachytene	(a) Crossing over
4. Diplotene	(b) Desynapsis

Question 5.
Is the given figure correct? Why?

Answer:
1. The given figure is incorrect as the spindle fibres are not attached to centromere of the chromosomes.
2. During metaphase, chromosomes are attached to spindle fibres with the help of centromeres.

Question 6.
If an onion has 16 chromosomes in its leaf cell, how many chromosomes will be there in its root cell and pollen grain.
Answer:
1. The chromosomes in root cell will be 16 as root cell is a diploid cell.
2. The chromosomes in pollen grain will be 8 as pollen grain is a haploid cell.

7. Identify the following phases of mitosis and label the ‘A’ and ‘B’ given in diagrams.

Question (i)

Answer:
The diagram shown is of Metaphase.
A: Chromosomes arranged on metaphase plate

Question (ii)

Answer:
The diagram shown is of Anaphase.
B: Chromatids moving to opposite poles.

Practical / Project:

Question 1.
Fix the onion root tips at different durations of the day starting from 6am up to 9am at the intervals of half an hour. Prepare the slide of each fixed root tip and analyse the relation between time and phase of mitosis.
Answer:
Mitotic division is an equational division in which one parent cell give rise to two daughter cells with equal number of chromosomes in daughter cells and mother cell. It has four sub phases: prophase, metaphase, anaphase, telophase.

Mitosis is affected by temperature and time. Mitotic index is high in morning so the mitosis is observed clearly in the morning. (Mitotic index is defined as the ratio between the number of cells in a population undergoing mitosis to the total number of cells in a population. )
[Note: Students catt use above information for reference and perform this activity on their own.]

11th Biology Digest Chapter 7 Cell Division Intext Questions and Answers

Can you recall? (Textbook Page No. 76)

How do your wounds heal?
Answer:
a. A wound is an injury to living tissue.
b. Healing of wound take place by mitosis.
c. Repetitive mitotic divisions near the site of injury results in healing of wound.

Can you tell? (Textbook Page No. 79)

What is cell cycle?
Answer:

Sequential events occurring in the life of a cell is called cell cycle.
Interphase and M – phase are the two phases of cell cycle.
Cell undergoes growth or rest during interphase and divides during M – phase.

Discuss with teacher (Textbook Page No. 76)

Some cells do not have gap phase in their cell cycle whereas some cells spend maximum part of their life in gap phase. Search for such cells. Some cells are said to be in G₀ phase. What is this G₀ phase?
Answer:

G₀ is the phase of the cell cycle in eukaryotes in which many cell types stop dividing. It is also called a quiescent stage.
If cells are deprived of appropriate growth factors, they stop at the Gi checkpoint of the cell cycle. Their growth and division are arrested and they remain in G₀ phase.
Mature neurons and muscle cells remain in G₀ phase.

Question 5.
Can you tell? (Textbook Page No. 79)
Answer:
1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.
2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.
The interphase is subdivided into three sub-phases as G₁ – phase, S-phase and G₂-phase.
a. G₁ – phase (First gap period/First Gap Phase):
It begins immediately after cell division.
RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.
b. S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
c. G₂ – phase (Second growth phase/Second Gap Phase):

Metabolic activities essential for cell division occur during this phase.
Various proteins which are necessary for the cell division are also synthesized in this phase.
Apart from this, RNA synthesis also occurs during this phase.
In animal cells, a daughter pair of centrioles appears near the pre-existing pair.

Internet my friend (Textbook Page No. 77)

What is Karyogram or Karyotype?
Answer:
1. A karyotype is a representation of condensed chromosomes arranged in pairs.
2. Analysis of the karyotype of a particular individual indicates whether the individual has a normal set of chromosomes or whether there are abnormalities in number or appearance of individual chromosomes.

Can you tell? (Textbook Page No. 79)

Which are the steps of mitosis?
Answer:
Steps in mitosis are Karyokinesis and Cytokinesis. Karyokinesis includes four stages – Prophase, Metaphase, Anaphase and Telophase.

Internet my friend (Textbook Page No. 79)

How the life span of a cell is decided?
Answer:

Life span of different cells vary greatly.
Life span of a cell is decided by its growth rate, metabolic activities and cell size.
The life span of a cell can be analysed in laboratory by applying carbon-14 technique to DNA.
This method is commonly used in archaeology and paleontology to find the age of fossils. Same can be applied to determine the life span of a cell.

Do yourself (Textbook Page No. 80)

Write down the explanation of prophase I in your own words.
Answer:
1. Prophase -I:
It is the most complicated and longest phas0e of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:

The volume of the nucleus increases.
The chromosomes become long distinct and coiled.
They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies.
The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during G_j phase of interphase.]

b. Zygotene:

Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:

Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points.
These points are called chiasmata (Appear like a cross-X).
Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

e. Diakinesis:

The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
The terminal chiasmata exist till the metaphase.
The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

Curiosity Box: (Textbook Page No. 81)

(i) What is exact structure of synaptonemal complex?
Answer:
Synaptonemal complexes are zipper like structures assembled between homologous chromosomes during the prophase of first meiotic division.
[Source: ncbi.nlm. nih.gov/pubmed/8743892]

(ii) What is structure of chiasma?
Answer:
Chiasma is a X-shaped point of attachment between two non-sister chromatids of a homologous chromosomes.

(iii) Which type of proteins are involved in formation of spindle fibres?
Answer:
Spindle fibres are formed from microtubules with many accessory proteins.

(iv) Why and how spindle fibres elongate and some contract?
Answer:
a. Spindle fibres elongate for assembly of chromosomes at equatorial plane of the cell during metaphase and spindle fibres contract for pulling chromosomes towards opposite poles during anaphase.
b. The spindle fibres elongate (polymerize) by incorporating subunits of the protein tubulin and contract

(v) What is the role of centrioles in formation of spindle apparatus?
Answer:
Centriole plays an important role in cell division. Centrioles help organize microtubule assembly and forms spindle apparatus that separate the chromosomes during cell division.

Curiosity box (Textbook Page No. 81)

What would have happened in absence of meiosis?
Answer:

Gametes are produced by the process of meiosis which are essential for sexual reproduction.
Diploid organisms have two set of chromosomes (one paternal and one maternal).
For a diploid organism to undergo sexual reproduction it needs to create gametes that contain only one set of chromosomes so the number of chromosomes remains same in the next generation.
In absence of meiosis, the chromosome number of parents and their offsprings will differ in every generation; hence no species will hold its characters.
Also, there will be no crossing over of homologous chromosomes. Thus, there will be no variations with respect to the changing environment in progeny to maintain their existence, which may lead to extinction of species.

Can you tell? (Textbook Page No. 82)

(i) What is the difference between mitosis and meiosis?
Answer:

Mitosis	Meiosis
(a) It occurs in somatic cells and stem cells.	It occurs in germ cells.
(b) In this nucleus divides only once.	In this nucleus divides twice (Meiosis I and Meiosis II)
(c) In these two daughter cells are formed.	In these four daughter cells are formed.
(d) Daughter cells formed by mitotic division are diploid (2n).	Daughter cells formed by meiotic division are haploid (n) •
(e) In mitosis, crossing over does not take place.	In meiosis, crossing over takes place.
(f) Mitosis plays an important role in growth, repair, healing and development.	Meiosis is important for formation of haploid gametes and spores.

(ii) What is difference between meiosis – I and meiosis – II?
Answer:

Meiosis I	Meiosis II
(a) Diploid cell is divided into two haploid cells.	Two haploid cells formed in meiosis I divides further into four haploid cells.
(b) This division is called heterotypic division.	This division is called homotypic (equational) division.
(c) It consists of prophase – I, metaphase – I, anaphase -1, telophase -1 and cytokinesis.	It consists of prophase – II, metaphase – II, anaphase – II, telophase – II and cytokinesis.
(d) Number of chromosomes is reduced to half, i.e. from diploid to haploid state.	In meiosis II number of chromosomes remain the same.
(e) It is complicated and long duration division.	It is simple and short duration division.
(f) Telophase I results into 2 daughter cells.	Telophase II results in 4 daughter cells.

(iii) Elaborate the process of recombination.
Answer:
a. Recombination is exchange of genetic material between paternal and maternal chromosomes during gamete formation.
b. The points where crossing over takes place is known as chiasmata.
c. Chromatids acquire new combinations of alleles by physically exchanging segments in crossing-over.
d. The exchange of genetic material between homologous chromosomes involves accurate breakage and joining of DNA molecules through a complex mechanism.
e. It is catalyzed by enzymes.

Do Yourself (Textbook Page No. 82)

Prepare a concept map on cell division in following box.
Answer:
Refer Quick Review

Internet My Friend (Textbook Page No. 82)

Different types of proteins like cyclins, maturation promoting factor (MPF), cyclosomes, enzymes like cyclin dependent kinases (CDK) play important role in control of cell cycle. Collect more information about these proteins and enzymes from internet, prepare a power-point presentation and present it in the class.
Answer:

The regulation of the cell cycle involves an internal control system consisting of proteins called cyclins and enzymes called cyclin-dependent kinases.
A Cdk is a protein kinase. When the kinase of the Cdk is activated upon binding to a cyclin, it phosphorylates target proteins in the cell, regulating their activities.
Those proteins play important roles in initiating or regulating significant events of the cell cycle, such as DNA replication, mitosis, and cytokinesis.
Maturation Promoting Factor (MPF) triggers the cell’s passage into the mitotic phase.
[Note: Students are expected to perform the above activity by their own with the help of information provided in the answer.]

11th Std Biology Questions And Answers:

11th Biology Chapter 2 Exercise Systematics of Living Organisms Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 2 Systematics of Living Organisms Textbook Exercise Questions and Answers.

Systematics of Living Organisms Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 2 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 2 Exercise Solutions

1. Choose correct option.

Question (A)
Which of the following shows single stranded RNA and lacks protein coat?
(a) Bacteriophage
(b) Plant virus
(c) Viroid
(d) Animal virus
Answer:
(c) Viroid

Question (B)
Causative agent of red tide is ________ .
(a) Dinoflagellate
(b) Euglenoid
(c) Chrysophyte
(d) Lichen
Answer:
(A) Dinoflagellate

Question (C)
Select odd one out for Heterotrophic bacteria.
(a) Nitrogen fixing bacteria
(b) Lactobacilli
(c) Methanogens
(d) Cyanobacteria
Answer:
(c) Methanogens or (d) Cyanobacteria

Question (D)
Paramoecium: Ciliated Protist :: Plasmodium: _______ .
(a) Amoeboid protozoan
(b) Ciliophora
(c) Flagellate protozoan
(d) Sporozoan
Answer:
(d) Sporozoan

2. Answer the following

Question (A)
What are the salient features of Monera?
Answer:
Salient features of Kingdom Monera:

Size: The organisms included in this kingdom are microscopic, unicellular and prokaryotic.
Occurrence: These are omnipresent. They are found in all types of environment which are not generally inhabited by other living beings.
Nucleus: These organisms do not have well-defined nucleus. DNA exists as a simple double-stranded circular single chromosome called as nucleoid. Apart from the nucleoid they often show presence of extrachromosomal DNA which is small circular called plasmids.
Cell wall: Cell wall is made up of peptidoglycan (also called murein) which is a polymer of sugars and amino acids.
Membrane-bound cell organelles: Membrane-bound cell organelles like mitochondria, chloroplast, endoplasmic reticulum are absent. Ribosomes are present, which are smaller in size (70S) than in eukaryotic cells.
Nutrition: Majority are heterotrophic, parasitic or saprophytic in nutrition. Few are autotrophic that can be either photoautotrophs or chemoautotrophs.
Reproduction: The mode of reproduction is asexual or with the help of binary fission or budding. Very rarely, sexual reproduction occurs by conjugation method.
Examples:
Archaebacteria: e.g. Methanobacillus, Thiobacillus, etc.
Eubacteria: e.g. Chlorobium, Chromatium, and Cyanobacteria e.g. Nostoc, Azotobacter, etc.

Question (B)
What will be the shape of a bacillus and coccus type of bacteria?
Answer:
The shape of bacillus type of bacteria is rod-shaped and coccus is spherical.

Question (C)
Why is binomial nomenclature important?
Answer:
Binomial nomenclature is important because:

The binomials are simple, meaningful and precise.
They are standard since they do not change from place to place.
These names avoid confusion and uncertainty created by local or vernacular names. The organisms are known by the same name throughout the world.
The binomials are easy to understand and remember.
It indicates phylogeny (evolutionary history) of organisms.
It helps to understand inter-relationship between organisms.

3. Write short notes

Question (A)
Write a note on useful and harmful bacteria.
Answer:
(i) Useful bacteria:
Most of the bacteria act as a decomposer. They breakdown large molecules in simple molecules or minerals. Examples of some useful bacteria:
Lactobacillus’. It helps in curdling of milk.
Azotobacter. It helps to fix nitrogen for plants.
Streptomyces: It is used in antibiotic production such as streptomycin.
Methanogens: These are used for production of methane (biogas) gas from dung.
Pseudomonas spp. and Alcanovorax borkumensis: These bacteria have the ability to destroy the pyridines and other chemicals. Hence, used to clear the oil spills.

(ii)Harmful bacteria:
This includes disease causing bacteria. They cause various diseases like typhoid, cholera, tuberculosis, tetanus, etc. Examples of some harmful bacteria:
Salmonella typhi: It is a causative organism of typhoid.
Vibrio cholerae: It causes cholera.
Mycobacterium tuberculosis’. It causes tuberculosis.
Clostridium tetani: It causes tetanus.
Clostridium spp.: It causes food poisoning.
Many forms of mycoplasma are pathogenic.
Agrobacterium , Erwinia, etc are the pathogenic bacteria causing plant diseases.
Animals and pets also suffer from bacterial infections caused by Brucella, Pastrurella, etc.

Question (B)
Write short note on five kingdom system.
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

Kingdom Monera
Kingdom Protista
Kingdom Plantae
Kingdom Fungi
Kingdom Animalia

Question (C)
Write short note on useful fungi.
Answer:
Economic importances of fungi are as follows:
1. Role of fungi in medicine:
(a) Antibiotic penicillin is obtained from Penicillium.
(b) Drugs like cyclosporine, immunosuppressant drugs, precursors of steroid hormones, etc are isolated from fungi.

2. Role of fungi in industries:
(a) Yeast is used in bread making. It causes dough to rise and make the bread light and spongy. It is also used in breweries or wine making industries. Sugars present in grapes are fermented by using yeast. This results in production of alcohol which is used for making wine.
(b) Lichen is a symbiotic association of algae and fungi are used in preparation of litmus paper which is used as acid-base indicator.

3. Role of fungi in food:
(a) Fungi like mushrooms are consumed as a food. These are rich source of protein.
(b) Fungi genus Penicillium helps in ripening of cheese.

4. Role of fungi as biocontrol agents:
(a) Fungi help to control growth of weeds.
(b) Pathogenic fungi like Fusarium sp., Phytophthorapalmivora, Alternaria crassa, etc act as mycoherbicides.

Question 4.
Complete tree diagram in detail.

Answer:

5. Draw neat labelled diagrams

Question (A)
Draw neat and labelled diagram of Paramoecium.
Answer:
Characteristics:

It belongs to kingdom Protista. It is further classified as animal like protist.
It lacks cell wall.
It shows heterotrophic and holozoic nutrition.
It is a ciliated protozoan where locomotion is due to cilia.
It has gullet (a cavity) which opens on the cell surface.

Quesiton (B)
Draw neat and labelled diagram of Euglena.
Answer:
Characteristics:
It belongs to kingdom Protista. It is further classified into euglenoids.

1. Dinoflagellates:

They are aquatic (mostly marine) and autotrophic (photosynthetic).
They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
The cell wall is made up of cellulosic stiff plates.
A pair of flagella is present, hence they are motile.
They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

2. Euglenoids:

They lack cell wall but have a tough covering of proteinaceous pellicle.
Pellicle covering provides flexibility and contractibility to Euglena.
They possess two flagella, one short and other long.
They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

Question (C)
Draw a neat labelled diagram of TMV.
Answer:

Question 6.
Complete chart and explain in your word.

Answer:

Depending upon the host, viruses are classified into three types as:
1. Plant virus
2. Animal virus
3. Bacterial virus (Bacteriophage)

1. Plant virus:
(a) Generally, they are rod shaped or cylindrical with helical symmetry.
(b) Majority of plant viruses have RNA as their genetic material. (Exception: Cauliflower Mosaic Virus has double stranded DNA as genetic material)
(c) Plant viruses cause disease in plants, e.g. Tobacco Mosaic Virus (TMV).

2. Animal virus:
(a) Generally, they are polyhedral in shape with radial symmetry.
(b) They have either DNA or RNA as genetic material.
(c) It causes disease to majority of animals including human beings, e.g. Influenza virus.

3. Bacteriophage:
(a) They have tadpole-like shape.
(b) They infect bacteria and hence are called as bacteriophage.
(c) Bacteriophages were discovered by Twort.
(d) Bacteriophages have double stranded DNA as the genetic material.
(e) Its body consists of head, collar and tail.

Question 7.
Identify the following diagram, label it and write detail information in your words.

Answer:
The given figure represents Bacteriophage.

Question 8.
The scientific name of sunflower is given below. Identify the correctly written name.
(A) Helianthus annus
(B) Helianthus Annus
(C) Helianthus annuus L.
(D) Helianthus annuus l.
Answer:
The correctly written scientific name of sunflower is Helianthus annuus L.

Question 9.
Match the following.

Kingdom	Examples
1. Monera	a. Riccia
2. Protista	b. Cyanobacteria
3. Plantae	c. Rhizopus
4. Fungi	d. Diatoms

Answer:

Kingdom	Examples
1. Monera	b. Cyanobacteria
2. Protista	d. Diatoms
3. Plantae	a. Riccia
4. Fungi	c. Rhizopus

Question 10.
Complete the following.
1. Plant-like Protista – [ ]
2. [ ] – Entamoeba

Practical /Project:

Question 1.
Make a group of students. Observe living organisms in your school/college campus and try to write their characters with respect to habit, habitat, mode of nutrition, growth- determinate or indeterminate, type of reproduction – vegetative reproduction, asexual reproduction, sexual reproduction. With the help of similarity and dissimilarity, try to classify organisms into different categories. Similar work should implement for animal group.
Answer:
The common living organisms observed near school/college are:
1. Plants
Habit: Herb, shrub, tree, etc.
Habitat: Terrestrial or aquatic
Mode of nutrition: Autotrophic
Growth: Indeterminate
Types of reproduction: Vegetative, asexual and sexual reproduction.

2. Animal e.g. dog, cats, cow, etc.
Habitat: Terrestrial
Mode of nutrition: Heterotrophs
Growth: Determinate
Types of reproduction: Only sexual reproduction

3. Birds e.g. Crow, sparrow, etc.
Habitat: Aviary (shows diverse habitat)
Mode of nutrition: Heterotrophs Growth: Determinate
Types of reproduction: Only sexual reproduction
[Note: Students are expected to collect more information about characteristics of living organisms and classify them into different categories]

Question 2.
Find out types of lichens and its economic importance.
Answer:
Types of lichens are:
1. Based on fungal components:
(a) Ascolichens:
In this category, the fungal partner belongs to Ascomycetes group of fungi.
(b) Basidiolichens:
Here, the fungal partner belongs to Basidiomycetes group of fungi.
(c) Deuterolichens:
In this category, the fungal partner belongs to Deuteromycetes group of fungi.

2. Based on their forms:
(a) Crustose lichen:
These lichens show crust-like growth.
These lichens grow on rocks and bark of the trees,
e. g. Graphis, Lecanora, Haematomma, etc.
(b) Foliose lichen:
These lichens grow on trees in the hilly regions.
The thallus is like a dry forked leaf,
e. g. Parmelia, Collema, Peltigera
(c) Fruticose lichen:
These lichens are seen on the branches of trees hanging down.
They are cylindrical, well branched and pendulous, with hair-like outgrowths,
e. g. Usnea, Cladonia, Alectoria, etc.

3. Economic importance of lichens:

(a) Lichen as food and fodder:
Many species of lichens are used as food by animals including man. Lichens contain a substance lichenin which is similar to carbohydrate making them edible. Parmelia is used in curry powder in India. Lichens like Cladonia, Citraria, Evernia, Parmelia are used as fodder as they form a favourite food for reindeers and cattles.

(b) Lichens in medicine:
Lichens contain usnic acid due to which they are used in medicines. Usnea and Cladonia species are used as an antibiotic against Gram positive bacteria.
Species like Lobaria, Citraria are useful in respiratory disease like T.B., Peltigera is useful in hydrophobia, Parmelia is used in treatment of epilepsy, whereas Usnea is used in urinary disease. Some lichens are also used in medicine due to their anticarcinogenic property.

(c) Industrial use of lichens:
1. Lichens are used in various dyes for colouring fabrics.
2. Species like Rocella and Lasallia are used in preparation of litmus paper which is acid-base indicator.
3. In Sweden and Russia, lichens are used for production of alcohol.
4. Orcein is a biological stain obtained from Orchrolechia androgyna and O. tortaria.
5. Some lichens are also used in tanning process in leather industry.
6. Evernia and Ramalina are the sources of essential oils which are used in preparation of soaps and other cosmetics.

(d) Other uses of lichens:
1. Lichens are used in cosmetics.
2. Some lichens like Everniaprunastri also known as oakmoss is used in making perfumes.
3. Lichen is also used as a preservative for beer.

11th Biology Digest Chapter 2 Systematics of Living Organisms Intext Questions and Answers

Can you tell? (Textbook Page No. 7)

Enlist uses of taxonomy?
Answer:
Uses of taxonomy are as follows:

It is used to assign each organism an appropriate place in a systematic framework of classification.
It is used to group animals and plants by their characteristics and relationships.
It is used to classify organisms based upon their similarities and differences.
It is used for nomenclature of an organism. Assigning a name to an organism is essential for its identification without confusion throughout the scientific world.
It is used to serve as an instrument for identification of an organisms. A newly isolated organism can be placed to its nearest relative or can be identified as a new organism with unknown characteristics.
It becomes easier to understand the evolutionary trends in different groups of organisms.

Can you tell? (Textbook Page No. 7)

Which characters of organisms are visible characters?
Answer:
The visible characters of organisms include habit, colour, mode of respiration, growth, reproduction, etc.

Can you tell? (Textbook Page No. 7)

What is evolution?
Answer:

It is believed that the life originated on earth in its very simple form.
Constant struggle of the early living beings gave rise to more and more perfect forms of life.
This struggle and progress are evolution which led to formation of diverse life forms.

Can you tell? (Textbook Page No. 7)

What is DNA barcoding?
Answer:
DNA barcoding is a new method for identification of any species based on its DNA sequence, which is obtained from a tiny tissue sample of the organism under study.

Can you tell? (Textbook Page No. 7)

Name the recent approaches in taxonomy.
Answer:
The recent approaches in taxonomy includes:

Morphological Approach
Embryological Approach
Ecological Approach
Behavioral Approach / Ethological Approach
Genetical Approach / Cytological Approach
Biochemical Approch
Numerical Taxonomy

Can you tell? (Textbook Page No. 9)

Make a flow chart showing taxonomic hierarchy.
Answer:
Kingdom → Sub-kingdom → Division / Phylum → Class → Cohort / Order → Family Genus → Species

Do Yourself (Textbook Page No. 16)

Complete the table (given on textbook Page No.16) through collecting information about sunflower, tiger with characteristic features.
(i) Sunflower:

Category	Taxon	Characteristics
Kingdom	Plantae	Autotrophic, photosynthetic, cell wall present.
Sub-kingdom	Phanerogamae	Seed producing plants, reproductive structures are visible.
Division	Angiospermae	Seeds are enclosed within the fruit.
Class	Dicotyledonae	Two cotyledons, tap root system, reticulate venation, pentamerous symmetry of flower, vascular bundle open.
Order	Asterales	Capitulum inflorescence, showing ray florets and disc florets.
Family	Asteraceae	Aster family
Genus	Helianthus	–
Species	annuus	–

(ii) Tiger:

Category	Taxon	Characteristics
Kingdom	Animalia	Multicellular eukaryotes, cell wall absent, heterotrophic nutrition.
Phylum	Chordata	Notochord present
Class	Mammalia	Presence of mammary gland
Order	Carnivora	Carnivorous in nature
Family	Felidae	Cat-like mammals
Genus	Panthera	Large cats
Species	tigris	–

Can you tell? (Textbook Page No. 9)

Why horse and ass are considered to be two different species or animals?
Answer:
1. Species is a group of organisms that can interbreed under natural conditions to produce fertile offsprings.
2. Horse and ass (donkey) are considered to be two different species or animals, because they cannot interbreed under natural condition to produce fertile offspring.

Internet my friend: (Textbook Page No. 9)

(i) Collect the information about most recent system of classification of living organisms and Kingdom System of Classification, e.g. Search for APG system of classification for Plants.
Answer:
[Note: Students are expected to collect more information about most recent system of classification of living organisms and Kingdom System of Classification from internet on their own.]

(ii) Collect the information about classification systems for all types of organisms.
Answer:
[Note: Students are expected to collect more information about classification systems for all types of organisms from internet on their own.]

Can you recall? (Textbook Page No. 6)

What is Five Kingdom system of classification?
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

Kingdom Monera
Kingdom Protista
Kingdom Plantae
Kingdom Fungi
Kingdom Animalia

Can you tell (Textbook Page No. 14)

Classify fungi into their types.
Answer:
Fungi are classified into four types on the basis of their structure, mode of spore formation and fruiting bodies as follows:
1. Phycomycetes:
Members of this class are commonly called as algal fungi.
These are consisting of aseptate coenocytic hyphae.
They grow well in moist and damp places on decaying organic matter as well as in aquatic habitats or as parasites on plants.
e.g. Mucor, Rhizopus (bread mold), Albugo (parasitic fungus on mustard).

2. Ascomycetes:
These are commonly called as sac fungi.
These are multicellular. Rarely they are unicellular (e.g. Yeast).
Hyphae are branched and septate.
They can be decomposers, parasites or coprophilous (grow on dung).
Some varieties of this class are consumed as delicacies such as morels and truffles.
Neurospora is useful in genetic and biochemical assays.
e.g. Aspergillus, Penicillium, Neurospora, Claviceps, Saccharomyces (unicellular ascomycetes).

3. Basidiomycetes:
These are commonly called as club fungi.
They have branched septate hyphae.
e.g. Agaricus (mushrooms), Ganoderma (bracket fungi), Ustilago (smuts), Puccinia (rusts), etc.

4. Deuteromycetes:
It is a group of fungi which are known to reproduce only asexually.
They are commonly called imperfect fungi.
They are mainly decomposers, while few are parasitic, e.g. Alternaria.

Can you tell? (Textbook Page No. 14)

Write a note on economic importance of fungi.
Answer:
Economic importances of fungi are as follows:
1. Role of fungi in medicine:
(a) Antibiotic penicillin is obtained from Penicillium.
(b) Drugs like cyclosporine, immunosuppressant drugs, precursors of steroid hormones, etc are isolated from fungi.

3. Role of fungi in food:
(a) Fungi like mushrooms are consumed as a food. These are rich source of protein.
(b) Fungi genus Penicillium helps in ripening of cheese.

4. Role of fungi as biocontrol agents:
(a) Fungi help to control growth of weeds.
(b) Pathogenic fungi like Fusarium sp., Phytophthorapalmivora, Alternaria crassa, etc act as mycoherbicides.

Can you tell? (Textbook Page No. 14)

Why are fungi considered as heterotrophic organisms?
Answer:
In fungi, chloroplast is absent, thus they cannot synthesize their own food by photosynthesis. Fungi decompose the organic matter by breaking down with the help of enzymes from which they absorb nutrients. Thus, exhibiting heterotrophic mode of nutrition.

Can you tell? (Textbook Page No. 14)

What are coenocytic hyphae?
Answer:
1. In filamentous fungi, body consists of mycelium which is formed by a network of hyphae.
2. When these hyphae are non-septate, multinucleated, they are known as coenocytic hyphae.

Can you tell? (Textbook Page No. 14)

(i) How are fungi different from plants?
Answer:
Fungi are different from plants because:
(a) They lack chloroplast hence, do not perform photosynthesis and are heterotrophic in nutrition. Whereas plants are autotrophic and prepare their own food by photosynthesis.
(b) They are separated from Plantae based on their saprophytic mode of nutrition.
(c) Fungi are decomposers of ecosystem whereas plants are producers of ecosystem.
(d) In fungi, cell wall is made up of fungal cellulose or chitin. Whereas in plants, cell wall is made up of cellulose and pectic compounds.

(ii) Have you seen any diseased plant in your farm?
Answer:
Yes, I have seen some diseased plants in our farm.
There are different pathogens like fungi, bacteria, viruses that cause diseases in plants.
The common plant diseases are:
(a) Leaf rust disease: It is caused by fungus Puccinia triticina. It is the most common rust disease of wheat.
(b) Blight disease in rice: It is caused by harmful bacteria Xanthomonas oryzae. It causes wilting of seedlings and yellowing and drying of leaves.
(c) Early blight of potato: It is caused by fungi Alternaria solani. It causes ‘bulls eye’ patterned leaf spots and tuber blight on potato.
(d) Crown gall disease: It is caused by Agrobacterium tumefaciens. This pathogen infects the plant and forms rough surfaced galls on stem and roots.
[Students are expected to write their observations about diseased plants found informs]

Can you tell? (Textbook Page No. 14)

Complete the following table:
Answer:

Plantae	Animalia
1. Autotrophic mode of nutrition.	Heterotrophic mode of nutrition.
2. They do not show locomotion.	They show locomotion.
3. Cell wall is present.	Cell wall is absent.
4. Chloroplast present.	Chloroplast absent.
5. They do not possess nervous system.	They possess well developed nervous system, i
6. Reproduction can be both sexual and asexual.	Mainly shows sexual reproduction.

Can you tell? (Textbook Page No. 15)

Why are viruses called infectious nucleoproteins?
Answer:
1. Viruses are acellular, highly infectious and ultramicroscopic.
2. Viruses possess their own genetic material in the form of either DNA or RNA, but never both. The genetic material in viruses is covered by a protein coat (capsid), hence called nucleoprotein.
3. They do not show any activity outside the body of host but once they enter their specific host cell, they start multiplying within the living host cells.
4. Viruses lack their own metabolic machinery, they make use of the cellular machinery of the host i.e. ribosome for the synthesis of protein during their reproduction and therefore, they cause severe infection. Thus, they are called infectious nucleoproteins.

Can you tell? (Textbook Page No. 15)

Describe genetic material in plant and animal viruses as well as in bacteriophages.
Answer:
The genetic material in different viruses is as given below:
1. Plant virus: (b) Majority of plant viruses have RNA as their genetic material. (Exception: Cauliflower Mosaic Virus has double-stranded DNA as genetic material)
2. Animal virus: (b) They have either DNA or RNA as genetic material.
3. Bacteriophage: (d) Bacteriophages have double-stranded DNA as the genetic material.

Can you tell? (Textbook Page No. 15)

Differentiate between viruses and viroids.
Answer:

Viruses	Viroids
1. They have high molecular weight.	They have low molecular weight.
2. They are larger in size.	They are smaller in size.
3. They can infect plant, animals and bacteria.	They mainly infect plants.
4. The genetic material can be ss-RNA, ds-RNA or DNA.	The genetic material is single stranded circular RNA.
5. Protein coat is present.	Protein coat is absent.
6. mosaic disease is a plant disease caused by viruses.	Tomato chloric dwarf is a plant disease caused by viroids.

Internet my friend. (Textbook Page No. 15)

In modern medicine, certain infectious neurological diseases were found to be transmitted by abnormally folded proteins. These proteins are called prions. The word prion comes from ‘proteinaceous infectious particle’, e.g. mad cow disease in cattle, Jacob’s disease in human.
Find more information about prions.
Answer:
Prions:
1. A prion is a misfolded form of a protein generally present in brain cells.
2. When the prion gets into a cell containing the normal form of the protein, the prion somehow converts normal protein molecules to the misfolded prion versions.
3. Several prions then aggregate into a complex that can convert other normal proteins to prions.
4. Prions can be transmitted through blood, surgical instruments and contaminated food.
5. Diseases caused by prions are Bovine Spongiform Encephalopathy in cattles, Kuru and Creutzfeldt – Jakob disease in humans.
[Note: Students are expected to search for more information about Prions on internet]

11th Std Biology Questions And Answers:

11th Biology Chapter 11 Exercise Study of Animal Type: Cockroach Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 11 Study of Animal Type: Cockroach Textbook Exercise Questions and Answers.

Study of Animal Type: Cockroach Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 11 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 11 Exercise Solutions

Question (A)
Chemical nature of chitin is ____________ .
(A) protein
(B) carbohydrate
(C) lipid
(D) glycoprotein
Answer:
(B) carbohydrate & (D) glycoprotein

Question (B)
Cockroach has ___________ type of mouthparts.
(A) sponging
(B) chewing and biting
(C) piercing and sucking
(D) lapping
Answer:
(B) chewing and biting

Question (C)
Spiracle is a part of ________ system of cockroach.
(A) circulatory
(B) respiration
(C) reproductive
(D) nervous
Answer:
(B) respiration

Question (D)
________ is a part of digestive system.
(A) Trachea
(B) Hypopharynx
(C) Haemocyte
(D) Seminal vesicle
Answer:
(B) Hypopharynx

Question (E)
_________ is also called as brain of cockroach.
(A) Supra-oesophageal ganglion
(B) Sub-oesophageal ganglion
(C) Hypo-cerebral ganglion
(D) Thoracic ganglion
Answer:
(A) Supra-oesophageal ganglion

2. Answer the following questions

Question (A)
Describe the digestive system of cockroach.
OR
With the help of neat and labelled diagram, describe the digestive system of cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.

(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.

(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.

(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.

(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is divisible into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (B)
Give an account on tracheal system of cockroach.
Answer:
1. Cockroach has an internal respiratory system of air tubes called tracheal system by which the air is brought into the body and is in contact with every part of the body. It allows the exchange of gases directly between the air and tissues without the need of blood.
These air tubes of internal respiratory system begin at the opening on body surface called spiracles.

2. Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called trachea. The spiracles let the air into and out of trachea.

3. Trachea: The trachea form a definite pattern of branching tubes arranged transversely as well as longitudinally. They are about 1mm thick and have spiral or annular thickening of chitin. The inner lining of chitin prevents the trachea from collapsing. Each trachea further branches into smaller tubes called tracheoles.

4. Tracheoles: These are fine intracellular tubes that penetrate deep into tissues. They are thin and not lined by chitin. They end blindly in the cells. Each tracheole at the blind end is filled with a watery fluid through which exchange of gases takes place. The content of this fluid keeps changing. At high muscular activity, part of fluid part is drawn into the tissues to enable more and rapid oxygen intake.

Question (C)
Describe the nervous system of cockroach.
Answer:
Nervous system in cockroach:
Nervous system of cockroach is ventral, solid and ganglionated. It consists of central nervous system (CNS), peripheral nervous system (PNS) and autonomous nervous system (ANS).
Central nervous system (CNS): Central nerv ous system consists of nerve ring and ventral nerve cord.
1. Nerve ring consists of:
a. a pair of supra-oesophageal ganglia
b. a pair of circum-oesophageal connectives
c. a pair of sub-oesophageal ganglia
(a) Supra-oesophageal ganglia or cerebral ganglia: A pair of supra-oesophageal ganglia is collectively known as the brain. Brain is present in head, above the oesophagus and between antennal bases. Each supra-oesophageal ganglion is formed by the fusion of three small ganglia – protocerebram, deutocerebrum and tritocerebrum.
(b) Circum-oesophageal connectives: Supra-oesophageal ganglia are connected to sub-oesophageal ganglion by a pair of lateral nerves called as circum-oesophageal connectives. Connectives arise from supra-oesophagial ganglia.
(c) Sub-oesophageal ganglia: It is a bilobed and present below the oesophagus, in head. It is also formed by the fusion of three pairs of ganglia.

2. Ventral nerve cord:
a. It arises from the sub-oesophageal ganglion. It is present along mid-ventral position, in perineural sinus.
b. It is double ventral nerve cord and consists of nine segmental, paired ganglia.
c. First three pairs of segmental ganglia are large and known as thoracic ganglia. The other six pairs of segmental ganglia are in abdomen (abdominal ganglia).
d. 6th abdominal ganglion is the largest and it is present in 7th abdominal segment.
e. There is no ganglion in 6th segment.

Peripheral nervous system (PNS):

The peripheral nervous system comprises of nerves that arise from various ganglia of CNS.
Six pairs of nerves arise from the supra-oesophageal ganglia.They supply to the eyes, antenna and labrum.
Nerves arising from the sub-oesophageal ganglion supply to the mandibles, maxillae and labium.
Nerves arising from the thoracic ganglia supply to the wings, legs and internal thoracic organs.
Nerves from abdominal ganglia go to the abdominal organs of respective abdominal segments.
Autonomic nervous system (ANS): It consists of four ganglia and a retrocerebral complex.

The ganglia are as follows:

Frontal ganglion: It is present above the pharynx and in front of brain.
Hypocerebral ganglion: It is present on the anterior region of oesophagus.
Ingluvial ganglion: It is present on crop. It is also called as visceral ganglion.
Ventricular ganglion: It is present on gizzard.

Question (D)
With the help of neat labelled diagram, describe female reproductive system of cockroach.
Answer:

Female reproductive system of cockroach consists a pair of ovaries, a pair of oviducts, vagina, spermatheca and accessory glands.
Ovaries are primary reproductive organs. They are paired and lie lateral in position in 2nd – 6lh abdominal segments. Each ovary is formed of a group of 8 ovarian tubules or ovarioles, containing a chain of developing ova. All ovarioles of an ovary open in lateral oviduct of respective side.
The lateral oviducts unite to form a common oviduct or vagina.
Common oviduct or vagina opens into the Bursa copulatrix (genital chamber), the female organ of copulation.
Spermatheca, is a sperm storing structure present in the 6th segment opens into genital chamber. It receives the sperms during copulation and store them for fertilization.
Collaterial glands are accessory paired glands that open in genital chamber.
Female gonapophyses consists of six chitinous plates surrounding the genital pore.
In males, genital pouch or genital chamber lies at the hind end of abdomen which is bounded dorsally by 9th and 10th terga and ventrally b; male genital pore and gonapophysis.

Question (E)
Draw a labelled diagram of digestive system of a cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.
(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.
(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.
(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.
(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food,

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

Question (F)
A student observed that the cockroaches are killed for dissection by simply putting them in soap water. He inquired whether soap is so poisonous. Teacher said it is due to its peculiar respiratory system. How?
Answer:
Cockroaches when put in soap solution, the solution enters into their body through the small respiratory openings called spiracles. The spiracles lead to trachea which further branches into smaller tubes called tracheoles. Each of these tracheoles has body fluid which acts as a stationary medium for diffusion. The soap solution rapidly diffuses through the entire respiratory system which may result in suffocation and eventually lead to the death of cockroach.

Question (G)
Describe the circulatory system of cockroach.
Answer:

1. Haemolymph: Haemolymph is colourless as it is without any pigment. It consists of plasma and seven types of blood cells/haemocytes. Plasma consists of water with some dissolved organic and inorganic solutes. It is rich in nutrients and nitrogenous wastes like uric acid.Cockroach has open circulatory system. It consists of colourless blood (haemolymph), a dorsal blood vessel (heart and dorsal aorta) and haemocoel.

2. Haemocoel: The body cavity of cockroach (haemocoel) can be divided into three sinuses due to two diaphragms i.e. dorsal and ventral diaphragm. These diaphragms are thin, fibromuscular septa (sing.septum)
which remain attached to terga along lateral sides at intermittent points.
(a) Dorsal diaphragm: It has 12 pairs (10 abdominal and 2 thoracic) of fan-like alary muscles. Alary muscles are triangular with pointed end attached to terga at lateral side and broad end lies between the heart and dorsal diaphragm.
(b) Ventral diaphragm: It is flat and present just above the ventral nerve cord. Laterally, it is attached to sterna at intermittent points.
(e) Sinuses: The coelom of cockroach is divided into three sinuses – pericardial sinus, perivisceral sinus and perineural sinus.

1. Pericardial sinus: It is dorsal, very small and contains dorsal vessel.
2. Perivisceral sinus: It is middle and largest sinus. It contains fat bodies and almost all major visceral organs of alimentary canal and reproductive system.
3. Perineural sinus: It is ventral, small and contains ventral nerve cord. It is continuous into legs. All the three sinuses communicate with each other through the pores present between two successive points of attachments of diaphragms.
4. Dorsal blood vessel: This is present in pericardial sinus, just below the tergum. It is divisible into posterior heart and anterior aorta (dorsal aorta/cephalic vessel).
(a) Heart: It is about 2.5 cm long, narrow, muscular tube that is open anteriorly and closed posteriorly. It starts from 9th abdominal segment and extends anteriorly upto 1st thoracic segment. Heart of cockroach is 13 chambered, out of which 10 chambers are in abdominal region and 3 chambers are in thoracic region. Each chamber has a pair of vertical slit-like incurrent aperture or opening called ostium (plural: ostia). Ostia are present along lateral side in the posterior region of first 12 chambers. Each ostium has lip-like valves that allow the flow of blood from sinus to heart only.
(b) Anterior aorta: Heart is continued by a short, thin-walled vessel called dorsal aorta. It lies in head region and opens in haemocoel.

3. Answer the following questions

Question (A)
How will you identify male or female cockroach?
Answer:
Male and female cockroach can be identified with the help of following differences:

Male cockroach	Female cockroach
1. Abdomen is relatively long and narrow.	Abdomen is short and broad.
2. 7th tergum covers 8,h tergum.	7th tergum covers 8th and 9th terga.
3. Antennae are longer in size.	Antennae are shorter in size.
4. Anal styles are present.	Anal styles are absent.
5. Brood pouch is absent.	Brood pouch is present.
6. All 9 sterna visible.	Only 7 sterna visible.

Question (B)
Write a note on: Gizzard of cockroach.
Answer:
Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

Question (C)
Give the systematic position of cockroach.
Answer:
Systematic position of cockroach:

Classi	fication	Reasons
Kingdom	Animalia	Cell wall absent, heterotrophic nutrition.
Phylum	Arthropoda	They have jointed appendages. Body is chitinous and segmented.
Class	Insecta	They possess two pairs of wings and three pairs of walking legs.
Genus	Periplaneta	Straight wings and nocturnal.
Species	americana	Originated in the continent of America.

Question (D)
What would have happened if cockroach did not have gizzard?
Answer:
1. The gizzard in cockroach is a spherical organ which has chitinous teeth and bristles.
2. The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.
3. If the cockroach did not have gizzard, the food will not be crushed into small particles and unfiltered food will enter the hindgut. Thus, digestion will be affected in the absence of gizzard.

Question (E)
What is the functional difference between eyes of cockroach and human being?
Answer:
1. Cockroaches have compound eyes whereas humans have simple eyes.
2. Eyes of cockroach possess several ommatidia that collectively form an image and help them to detect even the slightest movement of its predator. They provide mosaic or hazy vision.
3. Human eyes contain single lens and a clear image is formed on the retina. Humans have binocular vision which provides an improved perception of depth and gives a three-dimensional image of their surroundings.

Question (F)
What is the functional difference between respiratory systems of cockroach and human being?
Answer:
The functional difference between the respiratory systems of cockroach and human being is that in respiratory system of cockroach transport of gases does not occur via. blood whereas in human respiratory system transport of gases takes place via blood. In cockroach, the circulatory system has no role in respiratory process whereas in humans, circulatory system plays an important in respiratory process.

4. Explain the following in short.

Question (A)
What are anal cerci?
Answer:
1. Anal cerci are a pair of appendages at the end of the abdomen that arise from the 10th segment of the body of both male and female cockroach.
2. They are sensitive to wind movements and detect vibrations.

Question (B)
What is ganglion?
Answer:
1. Ganglion is a group of nerve cell bodies.
2. It represents the brain in advanced invertebrates.

Question (C)
Write a short note on hypopharynx.
Answer:
Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

Question (D)
What is mesentron?
Answer:
Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

Question (E)
Location of tergum.
Answer:
1. Tergum is a chitinous plate located in the abdomen of cockroach.
2. The abdomen is elongated and made up of ten segments. Each segment has a dorsal tergum and ventral sternum. Tergum is jointed to the sternum laterally by a soft cuticle called pleura.

Question (F)
What is ootheca?
Answer:
1. The secretion of collaterial glands forms a capsule around them is called as ootheca or egg case.
2. It is about 8 mm long and ranges from dark reddish to blackish brown.
3. Ootheca contains 14 to 16 fertilized eggs in two rows.
4. They are dropped or glued to a suitable surface, like a crack or crevice with good humidity near a food source.
5. A female cockroach on an average, produces 9 to 10 oothecae during its lifespan.

Question (G)
How many chambers are present in heart of a cockroach?
Answer:
13 chambers are present in heart of a cockroach, out of which 10 chambers are in abdominal region and 3 are in thoracic region.

Practical/Project:

Question 1.
Visit to nearest sericulture farm and study the life cycle of silk worm.
Answer:

The life cycle of the silk moth consists of four stages namely, egg, larva, pupa and adult.
Thousands of eggs deposited by female moths are incubated artificially to reduce the incubation period.
Larvae hatching out of eggs are released on mulberry plants to obtain nourishment from mulberry leaves.
After feeding for 3 – 4 weeks, larvae move to branches of mulberry plant.
The silk thread is formed from the secretion of salivary glands of larvae.
Larvae spin this thread around themselves forming a cocoon, which may be spherical in shape.
Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
These fibres are then unwound, processed and reeled.
Different kinds of fabric are woven from silk threads.

[The life cycle of silkworm is given for reference. Students are expected to visit the nearest sericulture farm and attempt this activity on their own.]

11th Biology Digest Chapter 11 Study of Animal Type: Cockroach Intext Questions and Answers

Can you recall? (Textbook Page No. 127)

How many different types of animals are present around us?
Answer:
Animals on earth show great diversity. The different types of animals present around us are;
a. Unicellular and multicellular
b. Prokaryotic and eukaryotic
c. Vertebrates and invertebrates
d. Unisexual and hermaphrodite
e. Aquatic, terrestrial, amphibian, reptilian, aerial, etc.

Can a person do a complete detailed study of each of those animals?
Answer:
Yes, a person can do a complete detailed study of each of those animals. Classification of animals based on characteristics into various groups has made it easier to study them.

Which phylum is most diverse and populous?
Answer:
Phylum Arthropoda is most diverse and populous.

Curiosity box: (Textbook Page No. 127)

Why do insects need moulting?
Answer:
a. Insects undergo metamorphosis (change of form or structure in an individual after hatching or birth). Each time an insect enters the next growth stage it has to molt.
b. Moulting is the process in which formation of new chitinous exoskeleton and subsequent shedding of the old one occurs.
c. The insects need moulting as their exoskeleton is rigid unlike the skin and does not allow the body to grow.

What is the difference between simple and compound eyes?
Answer:

Simple eyes	Compound eyes
1. Simple eyes contain single lens and several sensory cells.	Compound eyes contain several lenses (around 2000) called ommatidia (sing. Ommatidium).
2. Single lens collect light and focuses onto retina to form a single image	Each ommatidium forms an image of an object thereby forming several images of an ob ject.
3. Simple eye does not form a complex image but can detect movement of the object.	Compound eye forms a complex image of an object 1 and detects even a slightest movement of the object.

Use your brainpower. (Textbook Page No. 131)

Why do body cavity of cockroach is called as haemocoel?
Answer:
The body cavity of cockroach is known as haemocoel as it is filled with haemolymph (blood). Cockroaches have open type of circulation thus; the body cavity is filled with haemolymph.

Internet my friend. (Textbook Page No. 136)

Collect the information about techniques and objectives of rearing the cockroaches in countries like China and make a Powerpoint presentation including video clips.
Answer:
1. Cockroach rearing industry is a booming industry in China. Cockroaches are reared in more than hundred farms in China. A giant farm in China produces around 6 billion cockroaches.
2. It is believed that cockroaches can be used to prepare a medicine that can prevent stomach cancer. They are also used to treat compost waste.
[Students can search on internet for more information about the techniques and objectives of rearing the cockroaches]

11th Std Biology Questions And Answers:

11th Chemistry Chapter 5 Exercise Chemical Bonding Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

Chemical Bonding Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 5 Exercise Solutions

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO₃
b. CO₂
c. H₂S
d. Cl₂O
Answer:
b. CO₂

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order ?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. Energy levels in an atom
b. the shapes of molecules and ions.
c. the electrone getivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO₂?

	C=O bond	CO₂ molecule
A	polar	non-polar
B	non-polar	polar
C	polar	polar
D	non-polar	non-polar

Answer:

	C=O bond	CO₂ molecule
A	polar	non-polar

Question E.
Which O₂ molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH₄
b. C₂H₂
c. NH₃
d. H₂O
Answer:
d. H₂O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H₂)
b. Water (H₂O)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H₂ and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp² Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C₂H₆
c. C₂H₄
d. CF₃Cl
e. SO₂
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

σ (sigma) bond	π (pi) bond
1. It is formed when atomic orbitals overlap along internuclear axis.	1. It is formed when atomic orbitals overlap side-ways (laterally).
2. Electron density is high along the axis of the molecule (i.e., internuclear axis).	2. Electron density is zero along the axis of the molecule (i.e., internuclear axis).
3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released.	3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released.
4. It is a strong bond.	4. It is a weak bond.
5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals.	5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
b. Boron: The electronic configuration of boron is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\). The valency is expected to be 1 but it is 3 as in BF₃.
c. Carbon: The electronic configuration of carbon is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) . The valency is expected to be 2, but observed valency is 4 as in CH₄.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH₃ (107°18′) and H₂O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH₄) molecule on the basis of sp³ hybridization:
i. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\);
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH₄ molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

One lone pair and three bond pairs are present in ammonia molecule.
The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

Two lone pairs and two bond pairs are present in water molecule.
The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s² 2s² 2p_x¹ 2p_y¹
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH₃) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H₂
b. F₂
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\). During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF₂ molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H₂O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF₃ molecule is planar but NH₃ pyramidal. Explain.
Answer:
i. In the BF₃ molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH₃ molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH₃ molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH₃ is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF₄
b. NF₃
c. HCN
d. H₂S
Answer:
a. CF₄: There are four bond pairs on the central atom. Hence, shape of CF₄ is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF₃: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF₃ is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H₂S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C₂H₂).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF₂, AlCl₃, LiCl, CaCl₂, NaCl.
Answer:
AlCl₃ > BeF₂ > CaCl₂ > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF₆ compound.
Answer:
The total number of electrons around sulphur (S) in SF₆ is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

formation of the excited state and
mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
where, N_b is the number of electrons present in bonding MOs and N_a is the number of electrons present in antibonding MOs.

Question I.
Why is O₂ molecule paramagnetic?
Answer:
The electronic configuration of O₂ molecule is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

by loss and gain of electrons
by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl₂, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K⁺ + e^–
Cl + e^– → Cl^–
K⁺ + Cl^– → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in \(\mathrm{NH}_{4}^{+}\) will have formal charge +1?
Answer:
In \(\mathrm{NH}_{4}^{+}\), nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF₇?
Answer:
Lewis structure of IF₇ is:

In IF₇, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H₂ stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s¹. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H₂ molecule attains stability due to duplet formation. Hence, H₂ is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br₂, N₂, I₂, NH₃
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br₂: Nonpolar
v. N₂: Nonpolar
vi. I₂: Nonpolar
vii. NH₃: Polar

11th Std Chemistry Questions And Answers:

11th Biology Chapter 9 Exercise Morphology of Flowering Plants Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 9 Morphology of Flowering Plants Textbook Exercise Questions and Answers.

Morphology of Flowering Plants Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 9 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 9 Exercise Solutions

1. Choose correct option

Question (A)
Which one of the following will grow better in moist and shady region?
(a) Opuntia
(b) Orchid
(c) Mangrove
(d) Lotus
Answer:
(b) Orchid

Question (B)
A particular plant had a pair of leaves at each node arranged in one plane. What is the arrangement called?
(a) Alternate phyllotaxy
(b) Decussate phyllotaxy
(c) Superposed phyllotaxy
(d) Whorled phyllotaxy
Answer:
(c) Superposed phyllotaxy

Question (C)
In a particular flower the insertion of floral whorls was in such a manner, so the ovary was below other three whorls, but its stigma was taller than other three whorls. What will you call such flower?
(a) Hypogynous
(b) Perigynous
(c) Inferior ovary
(d) Half superior – half inferior
Answer:
(c) Inferior ovary

Question (D)
Beet and Arum both store food for perennation.
Are the examples for two different types?
(a) Beet is a stem but Arum is a root
(b) Beet is a root but Arum is a stem
(c) Beet is a stem but Arum is a leaf
(d) Beet is a stem but Arum is an inflorescence
Answer:
(b) Beet is a root but Arum is a stem

2. Answer the following questions

Question (A)
Two of the vegetables we consume are nothing but leaf bases. Which are they?
Answer:
Onion, Garlic

Question (B)
Opuntia has spines but Carissa has thorns. What is the difference?
Answer:

In Opuntia, stem is modified into leaf like photosynthetic organ known as phylloclade.
Spines growing on phylloclade of Opuntia are leaves, modified to reduce the loss of water through transpiration.
Thoms in Carissa are modified apical buds. They provide protection against browsing animals.
Thus, spines in Opuntia and thorns in Carissa have different origin and function.

Question (C)
Teacher described Hibiscus as solitary Cyme. What it means?
Answer:
1. In Cymose inflorescence, growth of peduncle is finite and it terminates into flower.
2. In Hibiscus, flower is borne singly at the tip of peduncle. Hence, teacher described Hibiscus as solitary cyme.

3. Write notes on

Question (A)
Fusiform root.
Answer:
Fusiform root:
1. Fusiform root is the modification of tap root for food storage.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)

Question (B)
Racemose inflorescence.
Answer:
Racemose inflorescence

Question (C)
Fasciculated tuberous root.
Answer:
Fasciculated tuberous root:
1. Fasciculated tuberous roots are modification of adventitious roots for storage of food.
2. Fasciculated tuberous roots do not develop any definite shape like modified tap roots.
3. a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

Question (D)
Region of cell maturation.
Answer:
Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question (E)
Rhizome.
Answer:
Rhizome:

Rhizome is a modification of underground stem for storage of food.
It is prostrate, dorsiventrally thickened and brownish in colour.
It grows either horizontally or obliquely beneath the soil.
Rhizome shows nodes and intemodes. It bears terminal and axillary buds at nodes.
Terminal bud under favourable conditions produces aerial shoot which degenerates at the end of favourable condition.
Growth of rhizome takes place with lateral buds, such growth is known as sympodial growth, e.g. Ginger (Zingiber officinale), Turmeric {Curcuma domestica), Canna etc.
In plants where rhizomes grow obliquely, terminal bud brings about growth of rhizomes. This is known as monopodial growth, e.g. Nymphea, Nelumbo (Lotus), Pteris (Fern) etc.
Rhizomes perform functions like storage of food, vegetative propagation and perennation.

Question (F)
Stolon.
Answer:
Stolons:
1. The slender lateral branch arising from the base of main axis is known as stolon.
2. In some plants it is above ground (wild strawberry).
3. Primarily stolon shows upward growth in the form of ordinary branch, but when it bends and touches the ground terminal bud grows into new shoot and develops adventitious roots.
e.g. Wild Strawberry, Jasmine, Mentha, etc. [Any one example]

Question (G)
Leaf venation.
Answer:
Leaf venation:

Arrangement of veins and veinlets in leaf lamina is known as venation.
Veins are responsible for conduction of water and minerals as well as food.
The structural framework of the lamina is developed by veins.
There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question (H)
Cymose inflorescence.
Answer:
Cymose inflorescence.

Question (I)
Perianth.
Answer:
Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

Question (J)
Write a short note on vexillary aestivation.
Answer:
Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

Question (K)
Write a short note on axile placentation.
Answer:
Axile placentation: Placentation: The mode of arrangement of ovules on the placenta within the ovary is called placentation.
Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.

Question 4.
Identify the following figures and write down the types of leaves arrangement.

Answer:
1. The given figures represent phyllotaxy. It is the arrangement of leaves on the stem and branches in a specific manner.
2. Figure ‘a’ and ‘b’ represents, alternate phyllotaxy. In this type of phyllotaxy, single leaf arises from each node of a stem. e.g. Mango
3. Figure ‘c’ represents opposite decussate phyllotaxy. In this type of phyllotaxy, a pair of leaf arise from each node and the consecutive pair at right angle to the previous one. e.g. Calotropis.

5. Students were on the excursion to a botanical garden. They noted following observation. Will you be able to help them in understanding those conditions?

Question (A)
A wiry outgrowth was seen on a plant arising from in between the leaf and stem.
Answer:
A wiry outgrowth on a plant arising from in between the leaf and stem can be an axillary stem tendril. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.

Question (B)
There was a green plant with flat stem, but no leaves. The entire plant was covered by soft spines.
Answer:
Student must have observed phylloclade, which is a modification of stem.
Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

Question (C)
Many oblique roots were given out from the lower nodes, apparently for extra support.
Answer:
a. Students must have observed adventitious roots in monocotyledonous plants like maize, sugarcane, wheat, etc.
b. Adventitious roots develop from any part of a plant other than radicle.
c. In such plants, adventitious roots arise from the lower node of a stem and provide extra support to the plant. These roots are also called as stilt roots.

Question (D)
Many plants in the marshy region had upwardly growing roots. They could be better seen during low tide.
Answer:
a. Plants growing in marshy region (halophytes) produce upwardly growing roots called as
pneumatophores or respiratory roots.
b. The main root system of these plants does not get sufficient air for respiration as soil is water logged.
c. Due to this, mineral absorption of plant also gets affected.
d. To overcome this problem underground roots, develop special roots which are negatively geotropic; growing vertically upward.
e. These roots are conical projections present around main trunk of plant.
f. Respiratory roots show presence of lenticels which helps in gaseous exchange.

Question (E)
A plant had leaves with long leaf apex, which was curling around a support.
Answer:
a. Students must have observed leaf tip tendril.
b. In some weak stems, leaf apex modifies into thin, green, wiry, coiled structure called as leaf tendril.
c. Such leaf tendrils, help in climbing by curling around a support.

Question (F)
A plant was found growing on other plant. Teacher said it is not a parasite. It exhibited two types of roots.
Answer:
a. Student must have observed an epiphytic plants like Dendrobium, Vanda growing on other plant.
b. The two types of roots exhibited by this plant must be clinging roots and epiphytic roots.
c. Clinging roots:
1. Clinging roots are tiny roots develop along intemodes, show disc at tips.
2. It exudes sticky substance which enables plant to get attached to the substratum without damaging it.

d. Epiphytic roots:
1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question (G)
While having lunch onion slices were served to them. Teacher asked which part of the plant are you eating?
Answer:
a. The edible part of an onion is fleshy leaves.
b. Onion is a bulb, in which stem is highly reduced, discoid and possesses adventitious roots at the base.
c. This stem bears a whorl of fleshy leaves which store food material.
d. The scale leaves or fleshy leaves are arranged in concentric manner over the stem. Some outer scale leaves become thin and dry. Thus, it is also called as tunicated or layered bulb.

Question (H)
Students observed large leaves of coconut and small leaves of Mimosa. Teacher asked it what way they are similar?
Answer:
a. Both large leaves of coconut and small leaves of Mimosa show pinnately compound leaves.
b. In both plants, leaf lamina is divided into number of leaflets.
c. Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf.

Question (I)
Teacher showed them Marigold flower and said it is not one flower. What the teacher meant?
Answer:
a. Marigold flower is an inflorescence in which flowers are produced in a definite manner on a peduncle.
b. In Marigold, racemose type of inflorescence can be observed.
c. In this, peduncle condenses to form a flat rounded structure called receptacle.
d. Opening of flower centripetal i.e. younger flowers are towards the centre and open later, while older flowers towards the periphery and open first.

Question (J)
Students cut open a Papaya fruit and found all the seeds attached to the sides. Teacher inquired about the possible placentation of Papaya ovary.
Answer:
a. In Papaya, seeds are attached to the sides of a fruit. Thus, parietal placentation is possible in papaya ovary,
b. In parietal placentation, ovules are placed on the inner wall of unil unilocular ovary of multicarpellary, syncarpus gynoecium.

Question 6.
Match the following.

Answer:
(i-c-1), (ii-e-3), (iii-a-4), (iv-b-5), (v-d-2)
[Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 7.
Observe the following figures and label the different parts.

Answer:

8. Differentiate with diagrammatic representation.

Question (A)
Differentiate with diagrammatic representation: Racemose and cymose inflorescence.
Answer:

Question (B)
Differentiate with diagrammatic representation: Reticulate and parallel venation
Answer:

Question (C)
Differentiate with diagrammatic representation: Taproot and Adventitious roots
Answer:

Practical / Project:

Question 1.
Collect different leaves from nearby region and observe variation in margin, leaf base, apex etc.
[Note: Students can scan the given Q.R code to study the different le
af margin, leaf base and apex.]

Question 2.
Find out and make a note of economically important plant from family Fabaceae, Solanaceae and Liliaceae.
Answer:
1. Economically important plant from family Fabaceae:
Family Fabaceae includes many pulses like gram, arhar, moong, soybean; edible oil seeds like soybean, groundnut; dye (lndigofera); fibres which can be obtained from Sun hemp, Sesbania trifolium which can be used as fodder; some plants are ornamental like lupin, sweet pea; some medicinal plants like muliathi.

2. Economically important plant from family Solanaceae:
Family Solanaceae includes many plants which are good source of food e.g. tomato, brinjal, potato; Spice e.g. chilli; Medicine e.g. belladonna, ashwagandha; Ornamental plants like Petunia.

3. Economically important plant from family Liliaceae:
Family Liliaceae includes many ornamental plants like tulip, Gloriosa, Medicinal plants like Aloe vera. Asparagus and source of colchicine, e.g. Colchicum autumnale.

Question 3.
Collect different leaves from garden and observe their veins and classify it.

11th Biology Digest Chapter 9 Morphology of Flowering Plants Intext Questions and Answers

Use your brainpower. (Textbook Page No. 102)

Why underground stem is different from roots?
Answer:
Underground stems are modified to perform different functions like storage of food, perennation and vegetative propagation. However, they differ from root in having nodes and intemodes.

Use your brainpower. (Textbook Page No. 104)

Why the stem has to perform photosynthesis in xerophytes?
Answer:
1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert.
2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration.
3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

Internet My Friend. (Textbook Page No. 106)

Collect information of types of leaf venation.
Answer:
1. Figure ‘R’ shows types of reticulate venation. WTien the veins and veinlets form a network, it is called
reticulate venation.
On the basis of number of mid-veins, reticulate venation is of two types:
a. Pinnate or unicostate: It is with single midrib e.g. Peepal, Mango.
b. Palmate or multicostate: It is with two or more prominent veins. It is further divided into convergent or divergent.
1. Multicostate convergent reticulate: Many prominent veins appear from the base of leaf lamina and converged in a curved manner towards the leaf apex. e.g. Zizyphus
2. Multicostate divergent reticulate: Prominent veins arise from the single point at the base of leaf lamina
and then diverge from one another towards the leaf margin, e.g. Cucurbita

2. Figure ‘P’ shows types of parallel venation. When veins run almost parallel to one another it is called parallel venation. It is of two types:
a. Unicostate: In this, lamina has single prominent mid vein from which many lateral parallel veins arise at regular intervals, e.g. Banana
b. Multicostate: In this, two or more mid veins run parallel to each other. It is further divided into convergent or divergent.

1. Multicostate convergent parallel:
Many prominent veins arise from the leaf base and then converge at leaf apex. e.g. Grasses
2. Multicostate divergent parallel:
Many prominent veins arise from the leaf base and then diverge towards margin, e.g. Borassus flabellifer (Toddy palm)

Observe and Discuss. (Textbook Page No. 112)

Observe and Discuss.
Answer:
1. Figure ‘a’ shows fruit of tomato.

It is a simple fruit as it develops from a single flower with bicarpellary syncarpous gynoecium.
It is a berry, because it has fleshy endocarp and many seeds.

2. Figure ‘b’ shows fruit of Custard apple.

It is an aggregate fruit, because it develops from a single flower with polycarpellary, apocarpous gynoecium.
Here, the ovary of each carpel gives rise to a part of the fruit called fruitlet. Hence, it is called an aggregation of fruitlets.
Custard apple can be further described as Etaerio of berries.

3. Figure ‘c’ shows fruit of pineapple.

It is a composite fruit, because it develops from a complete inflorescence.
Pineapple can be further described as Sorosis, as it develops from catkin type of inflorescence.

4. Figure ‘d’ shows fruit of milkweed.

It is a simple dehiscent dry fruit.
It has many seeds. When pericarp becomes dry and thin, it breaks open by one ventral suture.

Activity. (Textbook Page No. 113)

Study the family Liliaceae, prepare a table of following characteristics.
Answer:

Symmetry of flower	Actinomorphic
Bisexual/ Unisexual	Bisexual
Calyx	Absent
Corolla	Absent
Androecium	Stamens six, arranged in two whorls of 3 each, epiphyllous
Gynoecium	Tricarpellary, syncarpous, trilocular ovary with many ovules
Aestivation	Valvate
a. Calyx	Absent
b. Corolla	Absent
Placentation	Axile
Position of ovary	Superior ovary
Types of fruit	Capsule, rarely berry

11th Std Biology Questions And Answers:

11th Chemistry Chapter 8 Exercise Elements of Group 1 and 2 Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 8 Elements of Group 1 and 2 Textbook Exercise Questions and Answers.

Elements of Group 1 and 2 Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 8 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 8 Exercise Solutions

1. Explain the following

Question A.
Hydrogen shows similarity with alkali metals as well as halogens.
Answer:

Electronic configuration of hydrogen is 1s¹ which is similar to the outer electronic configuration of alkali metals of group 1 i.e., ns¹.
However, 1s¹ also resembles the outer electronic configuration of group 17 elements i.e., ns² np⁵.
By adding one electron to H, it will attain electronic configuration of the inert gas He which is 1s² and by adding one electron to ns² np⁵ we get ns²np⁶ which is the outer electronic configuration of the remaining inert gases.
Therefore, some chemical properties of hydrogen are similar to those of alkali metals while some resemble halogens.

Hence, hydrogen shows similarity with alkali metals as well as halogens.

Question B.
Standard reduction potential of alkali metals have high negative values.
Answer:

The general outer electronic configuration of alkali metals is ns¹.
They readily lose one valence shell electron to achieve stable noble gas configuration and hence, they are highly electropositive and are good reducing agents.

Hence, standard reduction potentials of alkali metals have high negative values.

Question C.
Alkaline earth metals have low values of electronegativity; which decrease down the group.
Answer:

Electronegativity represents attractive force exerted by the nucleus on shared electrons.
The general outer electronic configuration of alkaline earth metals is ns². They readily lose their two valence shell electrons to achieve stable noble gas configuration. They are electropositive and hence, they have low values of electronegativity.

Question D.
Sodium dissolves in liquid ammonia to form a solution which shows electrical conductivity.
Answer:
i. Sodium dissolves in liquid ammonia giving deep blue coloured solutions which is electrically conducting in nature.
Na + (x + y) NH₃ → [Na(NH₃)_x]+ + [e(NH₃)_y]^–
ii. Due to formation of ions, the solution shows electrical conductivity.

Question E.
BeCl₂ is covalent while MgCl₂ is ionic.
Answer:

Be²⁺ ion has very small ionic size and therefore, it has very high charge density.
Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl^–) which is larger in size.
This results in partial covalent character of the bond in BeCl₂.
Mg²⁺ ion has very less tendency to distort the electron cloud of Cl^– due to the bigger size of Mg²⁺ as compared to Be²⁺.

Hence, BeCl₂ is covalent while MgCl₂ is ionic.

Question F.
Lithium floats an water while sodium floats and catches fire when put in water.
Answer:

When lithium and sodium react with water, hydrogen gas is released. Due to these hydrogen gas bubbles, lithium and sodium floats on water.
eg. 2Na + 2H₂O → 2Na⁺ + 2OH^– + H₂↑
The reactivity of group 1 metals increases with increasing atomic radius and lowering of ionization enthalpy down the group.
Thus, sodium having lower ionization enthalpy, is more reactive than lithium.
Hence, lithium reacts slowly while sodium reacts vigorously with water.
Since the reaction of sodium with water is highly exothermic, it catches fire when put in water.

2. Write balanced chemical equations for the following.

Question A.
CO₂ is passed into concentrated solution of NaCl, which is saturated with NH₃.
Answer:

Question B.
A 50% solution of sulphuric acid is subjected to electrolyte oxidation and the product is hydrolysed.
Answer:

Question C.
Magnesium is heated in air.
Answer:

Question D.
Beryllium oxide is treated separately with aqueous HCl and aqueous NaOH solutions.
Answer:
Beryllium oxide (BeO) is an amphoteric oxide and thus, it reacts with both acid (HCl) as well as base (NaOH) to give the corresponding products.
i. \(\mathrm{BeO}+\underset{(\text { Acid })}{2 \mathrm{HCl}} \longrightarrow \mathrm{BeCl}_{2}+\mathrm{H}_{2} \mathrm{O}\)
ii. \(\mathrm{BeO}+\underset{(\text { Base })}{2 \mathrm{NaOH}} \longrightarrow \mathrm{Na}_{2} \mathrm{BeO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

3. Answer the following questions

Question A.
Describe the diagonal relationship between Li and Mg with the help of two illustrative properties.
Answer:
a. The relative placement of these elements with similar properties in the periodic table is across a diagonal and is called diagonal relationship.
b. Lithium is placed in the group 1 and period 2 of the modem periodic table. It resembles with magnesium which is placed in the group 2 and period 3.

ii. Li and Mg show similarities in many of their properties.
e. g.
a. Reaction with oxygen:
1. Group 1 elements except lithium, react with oxygen present in the air to form oxides (M₂O) as well as peroxides (M₂O₂) and superoxides (MO₂) on further reaction with excess of oxygen.
2. This anomalous behaviour of lithium is due to its resemblance with magnesium as a result of diagonal relationship.
3. As group 2 elements form monoxides i.e., oxides, lithium also form monoxides.

b. Reaction with nitrogen:
1. All the group 1 elements react only with oxygen present in the air to form oxides while group 2 elements react with both nitrogen and oxygen present in the air forming corresponding oxides and nitrides.
2. However, lithium reacts with oxygen as well as nitrogen present in the air due to its resemblance with magnesium.

Question B.
Describe the industrial production of dihydrogen from steam. Also write the chemical reaction involved.
Answer:
Three stages are involved in the industrial production of dihydrogen from steam.
i. Stage 1:
a. Reaction of steam on hydrocarbon or coke (C) at 1270 K temperature in presence of nickel catalyst gives water-gas which is a mixture of carbon monoxide and hydrogen.
1. Reaction of steam with hydrocarbon:

2. Reaction of steam with coke or carbon (C):

b. Sawdust, scrapwood, etc. can also be used in place of carbon.

ii. Stage 2:
Water-gas shift reaction: When carbon monoxide in the water-gas reacts with steam in the presence of iron chromate (FeCrO₄) as catalyst, it gets transformed into carbon dioxide. This is called water-gas shift reaction.

iii. Stage 3: In the last stage, carbon dioxide is removed by scrubbing with sodium arsenite solution.

Question C.
A water sample, which did not give lather with soap, was found to contain Ca(HCO₃)₂ and Mg(HCO₃)₂. Which chemical will make this water give lather with soap? Explain with the help of chemical reactions.
Answer:

Soap does not lather in hard water due to presence of the soluble salts of calcium and magnesium in it. So, the given water sample is hard water.
Hardness of hard water can be removed by removal of these calcium and magnesium salts.
Sodium carbonate is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates. Thus, it will make water give lather with soap.
e.g. Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaHCO_3(aq)

Question D.
Name the isotopes of hydrogen. Write their atomic composition schematically and explain which of these is radioactive ?
Answer:
i. Hydrogen has three isotopes i.e., hydrogen \(\left({ }_{1}^{1} \mathrm{H}\right)\), deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) and tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) with mass numbers 1, 2 and 3 respectively.
ii. They all contain one proton and one electron but different number of neutrons in the nucleus.
iii. Atomic composition of isotopes of hydrogen:

iv. Tritium is a radioactive nuclide with half-life period 12.4 years and emits low energy β^– particles.
v. Schematic representation of isotopes of hydrogen is as follows:

4. Name the following

Question A.
Alkali metal with smallest atom.
Answer:
Lithium (Li)

Question B.
The most abundant element in the universe.
Answer:
Hydrogen (H)

Question C.
Radioactive alkali metal.
Answer:
Francium (Fr)

Question D.
Ions having high concentration in cell sap.
Answer:
Potassium ions (K⁺)

Question E.
A compound having hydrogen, aluminium and lithium as its constituent elements.
Answer:
Lithium aluminium hydride (LiAlH₄)

5. Choose the correct option.

Question A.
The unstable isotope of hydrogen is …..
a. H-1
b. H-2
c. H-3
d. H-4
Answer:
c. H-3

Question B.
Identify the odd one.
a. Rb
b. Ra
c. Sr
d. Be
Answer:
a. Rb

Question C.
Which of the following is Lewis acid ?
a. BaCl₂
b. KCl
c. BeCl₂
d. LiCl
Answer:
c. BeCl₂

Question D.
What happens when crystalline Na₂CO₃ is heated ?
a. releases CO₂
b. loses H₂O
c. decomposes into NaHCO₃
d. colour changes.
Answer:
b. loses H₂O

Activity :

1. Collect the information of preparation of dihydrogen and make a chart.
2. Find out the s block elements compounds importance/uses.
Answer:
1.

2. Uses of s-block elements:
Group 1 elements (alkali metals):
a. Lithium: Lithium is widely used in batteries.
b. Sodium:

Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
Sodium is also used as an important reagent in the Wurtz reaction.
It is used in the manufacture of sodium vapour lamp.

c. Potassium:

Potassium has a vital role in biological system.
Potassium chloride (KCl) is used as a fertilizer.
Potassium hydroxide (KOH) is used in the manufacture of soft soaps and also as an excellent absorbent of carbon dioxide.
Potassium superoxide (KO₂) is used as a source of oxygen.

d. Caesium: Caesium is used in devising photoelectric cells.

Group 2 elements (alkaline earth metals):
a. Magnesium: Magnesium hydroxide [Mg(OH)₂] in its suspension form is used as an antacid.
b. Calcium: Compounds of calcium such as limestone and gypsum are used as constituents of cement and mortar.
c. Barium: BaSO₄ being insoluble in H₂O and opaque to X-rays is used as ‘barium meal’ to scan the X-ray of human digestive system.
[Note: Students are expected to collect additional information about preparation of dihydrogen and uses of s-block elements on their own.]

11th Chemistry Digest Chapter 8 Elements of Group 1 and 2 Intext Questions and Answers

Can you recall? (Textbook Page No. 110)

Question 1.
Which is the first element in the periodic table?
Answer:
Hydrogen is the first element in the periodic table.

Question 2.
What are isotopes?
Answer:
Many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element. Isotopes of an element have the same atomic number (number of protons) but different atomic mass numbers due to different number of neutrons in their nuclei.

Question 3.
Write the formulae of the compounds of hydrogen formed with sodium and chlorine.
Answer:
Hydrogen combines with sodium to form sodium hydride (NaH) while it reacts with chlorine to form hydrogen chloride (HCl).

Can you tell? (Textbook Page No. 110)

Question 1.
In which group should hydrogen be placed? In group 1 or group 17? Why?
Answer:

Hydrogen contains one valence electron in its valence shell and thus, its valency is one. Therefore, hydrogen resembles alkali metals (group 1 elements) as they also contain one electron in their valence shell (alkali metals tend to lose their valence electron).
However, hydrogen also shows similarity with halogens (group 17 elements) as their valency is also one because halogens tend to accept one electron in their valence shell.
Due to this unique behaviour, it is difficult to assign any definite position to hydrogen in the modem periodic table.

Just think! (Textbook Page No. 112)

Question 1.
\(2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaH}_{(\mathrm{s})}\)
In the above chemical reaction which element does undergo oxidation and which does undergo reduction?
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + H_2(g) → 2Na⁺ + 2H^–
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and each hydrogen atom gains one electron to form H. This can be represented as follows:

iii. Na is oxidised to NaH by loss of electrons while the elemental hydrogen is reduced to NaH by gain of electrons.

Can you recall? (Textbook Page No. 113)

Question i.
What is the name of the family of reactive metals having valency one?
Answer:
The family of reactive metals having valency one is known as alkali metals (group 1).

Question ii.
What is the name of the family of reactive metals having valency two?
Answer:
The family of reactive metals having valency two is known as alkaline earth metals (group 2).

11th Std Chemistry Questions And Answers:

11th Chemistry Chapter 9 Exercise Elements of Group 13, 14 and 15 Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 9 Elements of Group 13, 14 and 15 Textbook Exercise Questions and Answers.

Elements of Group 13, 14 and 15 Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 9 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 9 Exercise Solutions

1. Choose correct option.

Question A.
Which of the following is not an allotrope of carbon ?
a. buckyball
b. diamond
c. graphite
d. emerald
Answer:
d. emerald

Question B.
………… is inorganic graphite.
a. borax
b. diborane
c. boron nitride
d. colemanite
Answer:
c. boron nitride

Question C.
Haber’s process is used for preparation of ………….
a. HNO3
b. NH3
c. NH2CONH2
d. NH4OH
Answer:
b. NH3

Question D.
Thallium shows different oxidation state because ……………
a. of inert pair effect
b. it is inner transition element
c. it is metal
d. of its high electronegativity
Answer:
a. of inert pair effect

Question E.
Which of the following shows most prominent inert pair effect ?
a. C
b. Si
c. Ge
d. Pb
Answer:
d. Pb

2. Identify the group 14 element that best fits each of the following description.

A. Non-metallic element
B. Form the most acidic oxide
C. They prefer +2 oxidation state.
D. Forms strong π bonds.
Answer:
i. Carbon (C)
ii. Carbon
iii. Tin (Sn) and lead (Pb)
iv. Carbon

3. Give reasons.

A. Ga³⁺ salts are better reducing agent while Tl³⁺ salts are better oxidising agent.
B. PbCl₄ is less stable than PbCl₂
Answer:
A. i. Both gallium (Ga) and thallium (Tl) belong to group 13.
ii. Ga is lighter element compared to thallium Tl. Therefore, its +3 oxidation state is stable. Thus, Ga+ loses two electrons and get oxidized to Ga³⁺. Hence, Ga+ salts are better reducing agent.
iii. Thallium is a heavy element. Therefore, due to the inert pair effect, Tl forms stable compounds in +1 oxidation state. Thus, Tl³⁺ salts get easily reduced to Tl¹⁺ by accepting two electrons. Hence, Tl³⁺ salts are better oxidizing agent.
[Note: This question is modified so as to apply the appropriate textual concept.]

B. i. Pb has electronic configuration [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p².
ii. Due to poor shielding of 6s² electrons by inner d and f electrons, it is difficult to remove 6s² electrons (inert pair).
iii. Thus, due to inert pair effect, +2 oxidation state is more stable than +4 oxidation state.
Hence, PbCl₄ is less stable than PbCl₂.

4. Give the formula of a compound in which carbon exhibit an oxidation state of

A. +4
B. +2
C. -4
Answer:
A. CCl₄
B. CO
C. CH₄

5. Explain the trend of the following in group 13 elements :

A. atomic radii
B. ionization enthalpy
C. electron affinity
Answer:
A. Atomic radii:

In group 13, on moving down the group, the atomic radii increases from B to Al.
However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.
However, the atomic radii again increases from Ga to Tl.
Therefore, the atomic radii of the group 13 elements varies in the following order:
B < Al > Ga < In < Tl

B. Ionization enthalpy:

Ionization enthalpies show irregular trend in the group 13 elements.
As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.
However, there is a marginal difference in the ionization enthalpy from Al to Tl.
The ionization enthalpy increases slightly for Ga but decreases from Ga to In.
In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly.
Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.
The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.

C. Electron affinity:
a. Electron affinity shows irregular trend. It first increases from B to A1 and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.

b. From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.
Note: Electron affinity of group 13 elements:

6. Answer the following

Question A.
What is hybridization of Al in AlCl₃?
Answer:
Al is sp² hybridized in AlCl₃.

Question B.
Name a molecule having banana bond.
Answer:
Diborane (B₂H₆)

7. Draw the structure of the following

Question A.
Orthophosphoric acid
Answer:

Question B.
Resonance structure of nitric acid
Answer:

8. Find out the difference between

Question A.
Diamond and Graphite
Answer:
Diamond:

It has a three-dimensional network structure.
In diamond, each carbon atom is sp³ hybridized.
Each carbon atom in diamond is linked to four other carbon atoms.
Diamond is poor conductor of electricity due to absence of free electrons.
Diamond is the hardest known natural substance.

Graphite:

It has a two-dimensional hexagonal layered structure.
In graphite, each carbon atom is sp² hybridized.
Each carbon atom in graphite is linked to three other carbon atoms.
Graphite is good conductor of electricity due to presence of free electrons in its structure.
Graphite is soft and slippery.

Question B.
White phosphorus and Red phosphorus
Answer:
White phosphorus:

It consists of discrete tetrahedral P₄ molecules.
It is less stable and more reactive.
It exhibits chemiluminescence.
It is poisonous.

Red phosphorus:

It consists chains of P₄ molecules linked together by covalent bonds.
It is stable and less reactive.
It does not exhibit chemiluminescence.
It is nonpoisonous.

9. What are silicones? Where are they used?
Answer:
i. a. Silicones are organosilicon polymers having R₂SiO (where, R = CH₃ or C₆H₅ group) as a repeating unit held together by

b. Since the empirical formula R₂SiO (where R = CH₃ or C₆H₅ group) is similar to that of ketones (R₂CO), these compounds are named as silicones.

ii. Applications: They are used as

insulating material for electrical appliances.
water proofing of fabrics.
sealant.
high temperature lubricants.
for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

10. Explain the trend in oxidation state of elements from nitrogen to bismuth.
Answer:

Group 15 elements have five valence electrons (ns² np³). Common oxidation states are -3, +3 and +5. The range of oxidation state is from -3 to +5.
Group 15 elements exhibit positive oxidation states such as +3 and +5. Due to inert pair effect, the stability of +5 oxidation state decreases and +3 oxidation state increases on moving down the group.
Group 15 elements show tendency to donate electron pairs in -3 oxidation state. This tendency is maximum for nitrogen.
The group 15 elements achieve +5 oxidation state only through covalent bonding.
e. g. NH₃, PH₃, ASH₃, SbH₃, and BiH₃ contain 3 covalent bonds. PCl₅ and PF₅ contain 5 covalent bonds.

11. Give the test that is used to detect borate radical is qualitative analysis.
Answer:
i. Borax when heated with ethyl alcohol and concentrated H₂SO₄, produces volatile vapours of triethyl borate, which bum with green edged flame.

ii. The above reaction is Used as a test for the detection and removal of borate radical \(\left(\mathrm{BO}_{3}^{3-}\right)\) in qualitative analysis.

12. Explain structure and bonding of diborane.
Answer:

Electronic configuration of boron is 1s² 2s² 2p¹. Thus, it has only three valence electrons.
In diborane, each boron atom is sp³ hybridized. Three of such hybrid orbitals are half filled while the fourth sp3 hybrid orbital remains vacant.
The two half-filled sp³ hybrid orbitals of each B atom overlap with 1s orbitals of two terminal H atoms and form four B – H covalent bonds. These bonds are also known as two-centred-two-electron (2c-2e) bonds.
When ‘1s’ orbital of each of the remaining two H atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three-centred-two- electron bonds (3c-2e) or banana bonds.
Hydrogen atoms involved in (3c-2e) bonds are the bridging H atoms i.e., H atoms in two B – H – B bonds.
In diborane, two B atoms and four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.

13. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is white crystalline solid and used for inorganic qualitative analysis.

a. Name the compound produced.
b. Write the reaction that explains its formation.
Answer:
a. Borax
b. Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

14. Ammonia is a good complexing agent. Explain.
Answer:
i. The lone pair of electrons on nitrogen atom facilitates complexation of ammonia with transition metal ions. Thus, ammonia is a good complexing agent as it forms complex by donating its lone pair of electrons.

ii. This reaction is used for the detection of metal ions such as Cu²⁺ and Ag⁺.

15. State true or false. Correct the false statement.

A. The acidic nature of oxides of group 13 increases down the graph.
B. The tendency for catenation is much higher for C than for Si.
Answer:
A. False
The acidic nature of oxides of group 13 decreases down the group. It changes from acidic through amphoteric to basic.
B. True

16. Match the pairs from column A and B.

	Column A		Column B
i.	BCl₃	a.	Angular molecule
ii.	SiO₂	b.	Linear covalent molecule
iii.	CO₂	c.	Tetrahedral molecule
		d.	Planar trigonal molecule

Answer:
i – d,
ii – c,
iii – b

17. Give the reactions supporting basic nature of ammonia.
Answer:
In the following reactions ammonia reacts with acids to form the corresponding ammonium salts which indicates basic nature of ammonia.

18. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
Answer:
i. When silver nitrate (AgNO₃) is added to an aqueous solution of salt sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed.

ii. On adding ammonium hydroxide (NH₄OH) to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.

11th Chemistry Digest Chapter 9 Elements of Group 13, 14 and 15 Intext Questions and Answers

Can you recall? (Textbook Page No. 123)

Question 1.
If the valence shell electronic configuration of an element is 3s²3p¹, in which block of the periodic table is it placed?
Answer:
The element having valence shell electronic configuration 3s² 3p¹ must be placed in the p-block of the periodic table as its last electron enters in p-subshell (3p).

Can you recall? (Textbook Page No. 127)

Question 1.
What is common between diamond and graphite?
Answer:
Both diamond and graphite are made up of carbon atoms as they are two allotropes of carbon.

Can you recall? (Textbook Page No. 129)

Question i.
Which element from the following pairs has higher ionization enthalpy?
B and TI, N and Bi
Answer:
Among B and Tl, boron has higher ionization enthalpy while, among N and Bi, nitrogen has higher ionization enthalpy.

Question ii.
Does boron form covalent compound or ionic?
Answer:
Yes, boron forms covalent compound.

Try this. (Textbook Page No. 131)

Question 1.
Find out the structural formulae of various oxyacids of phosphorus.
Answer: