Important Questions

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

Maharashtra State Board 11th Chemistry Important Questions Chapter 9 Elements of Group 13, 14 and 15

Question 1.
What are p-block elements?
Answer:

• Elements in which the differentiating electron (the last filling electron) enters the outermost p orbital are p-block elements.
• Since maximum six electrons can be accommodated in p-subshell i.e., three p-orbitals, the p-block contains six groups numbered from 13 to 18 in the modem periodic table.
• The p-block elements show greater variation in the properties than s-block elements.

Question 2.
Write the names of the elements present in groups 13, 14 and 15.
Answer:

Group	Name of family	Name of the elements
13	Boron family	Boron (5B), aluminium (13Al), gallium (31Ga), indium (49In), thallium (81Tl)
14	Carbon family	Carbon (6C), silicon (14Si), germanium (32Ge), tin (50Sn), lead (82Pb)
15	Nitrogen family	Nitrogen (7N), phosphorus (15P), arsenic (33AS), antimony (51Sb), bismuth (83Bi)

Question 3.
i. Write the general outer electronic configuration of the elements of group 13, group 14 and group 15.
ii. By how many electrons do their outer electronic configurations differ from their nearest inert gas?
Answer:
i.

Group	General outer electronic configuration
13	ns2 np1
14	ns2 np2
15	ns2 np3

ii. The outer electronic configurations of the elements group 13, group 14 and group 15 differ from their nearest inert gas by 5, 4 and 3 electrons, respectively.

Question 4.
In which form do the elements of groups 13,14 and 15 occur in nature?
Answer:

• The elements of groups 13, 14 and 15 do not occur in free monoatomic state and are found as compounds with other elements.
• They also occur in the form of polyatomic molecules (such as N₂, P₄, C₆₀) or polyatomic covalent arrays (such as graphite, diamond).

Question 5.
Write condensed electronic configurations of the following elements.
₁₃Al, ₄₉In, ₁₄Si, ₅₀Sn, ₁₅P, ₃₃As
Answer:
Condensed electronic configurations of
i. ₁₃Al: [Ne]3s² 3p¹
ii. ₄₉In: [Kr]4d¹⁰5s²5p¹
iii. ₁₄Si: [Ne]3s²3p²
iv. ₅₀Sn: [Kr]4d¹⁰5s²5p²
v. ₁₅P: [Ne]3s²3p³
vi. ₃₃As: [Ar]3d¹⁰4s²4p³

Note: Condensed electronic configurations of elements of groups 13, 14 and 15 are given in the table below.

Question 6.
Name the following.
i. A metalloid present in group 13.
ii. A group 13 element which is the third most abundant element in the earth’s crust.
Answer:
i. Boron
ii. Aluminium

Question 7.
Why boron is classified as a metalloid?
Answer:
Boron is glossy and hard solid like metals but a poor conductor of electricity like nonmetals. Since it exhibits properties of both metals and nonmetals, boron is classified as a metalloid.

Question 8.
Describe the variation in the electronegativity of group 13 elements.
Answer:

• In group 13, on moving down the group, the electronegativity decreases from B to Al.
• However, there is a marginal increase in the electronegativity from Al to Tl.
• This trend is a result of the irregularities observed in atomic size of elements.

Question 9.
Atomic numbers of the group 13 elements are in the order B < Al < Ga < In < Tl. Arrange these elements in increasing order of ionic radii of M³⁺.
Answer:

• The given elements are in an increasing order of their atomic number.
• The general outer electronic configuration of group 13 elements is ns²np¹.
• M³⁺ ion is formed by the removal of three electrons from the outermost shell ‘n’.
• In the M³⁺ ions, the ‘n-1’ shell becomes the outermost shell. Size of the ‘n-1’ shell increases down the group.

Therefore, the ionic radii of M³⁺ ion increases down the group in the following order:
B³⁺ < Al³⁺ < Ga³⁺ < In³⁺ < Tl³⁺

Question 10.
Why the atomic radius of Gallium is less than that of aluminium?
Answer:

• Atomic radius of the elements increases down the group due to addition of new shells.
• Electronic configuration of Al is [Ne]3s²3p¹ while that of Ga is [Ar]3d¹⁰4s²4p¹.
• As Al does not have d-electrons, it offers an exception to this trend.
• As we go from Al down to Ga the nuclear charge increases by 18 units. Out of the 18 electrons added, 10 electrons are in the inner 3d subshell of Ga. These d-electrons offer poor shielding effect.
• Therefore, the effect of attraction due to increased nuclear charge is experienced prominently by the outer electrons of Ga and thus, its atomic radius becomes smaller than that of Al.

Hence, the atomic radius of gallium is less than that of aluminium.

Question 11.
The values of the first ionization enthalpy of Al, Si and P are 577, 786 and 1012 kJ mol^-1 respectively. Explain the observed trend.
Answer:

• The trend shows increasing first ionization enthalpy from Al to Si to P.
• Al, Si and P belong to the third period in the periodic table and hence, they have same valence shell.
• As we move across a period from left to right, the nuclear charge increases. Due to this, electrons in the valence shell are held more tightly by the nucleus as we go from Al to Si to P.
• Therefore, more energy is required to remove an electron from its outermost shell.

Hence, the value of first ionization enthalpy increases from Al to Si to P.

Note: Atomic and physical properties of group 13 elements.

Question 12.
Name metal(s), nonmetal(s) and metalloid(s) of group 14.
Answer:
i. Metal: Tin, lead
ii. Nonmetal: Carbon
iii. Metalloid: Silicon, germanium

Question 13.
Explain the variation in the following properties of the group 14 elements,
i. Atomic radii
ii. Ionization enthalpy
iii. Electronegativity
Answer:
i. Atomic radii (Covalent radii):

• In the periodic table as we move down the group 14 from C to Pb, the atomic radii increases due to the addition of new shell at each succeeding element.
• However, the increase is comparatively less after silicon due to poor shielding by inner d- and f-electrons in the atoms.

ii. Ionization enthalpy:

• Due to increased effective nuclear charge, group 14 elements have higher value of ionization enthalpy than corresponding group 13 elements.
• In the periodic table, as we move down the group 14 from C to Sn, the ionization enthalpy decreases.
• From Si to Sn, the ionization enthalpy decreases slightly.
• However, from Sn to Pb, the ionization enthalpy increases slightly. It is due to the poor shielding effect of intervening d and f orbitals and increase in the size of the atoms.

iii. Electronegativity:

• Due to small atomic size, group 14 elements are slightly more electronegative than the corresponding group 13 elements.
• As we move down the group 14 from C to Si in the periodic table, the electronegativity decreases.
• The electronegativity values for elements from Si to Pb are almost the same.
• Among group 14 elements, carbon is the most electronegative with electronegativity of 2.5.

Question 14.
Explain why there is a phenomenal decrease in ionization enthalpy from carbon to silicon.
Answer:

• Carbon is the first element of group 14 and thus, it has the smallest atomic size.
• The ionization enthalpy of carbon (1086 kJ mol^-1) is very high due to its small atomic size (77 pm) and high electronegativity (2.5).
• However, the ionization enthalpy of silicon (786 kJ mol^-1) decreases phenomenally due to the increase in its atomic size (118 pm) and low electronegativity (1.8).

Note: Atomic and physical properties of group 14 elements.

Question 15.
Which type of elements are present in group 15? Mention their physical state.
Answer:

• Group 15 includes all the three traditional types of elements i.e., metals, nonmetals and metalloids.
• Nitrogen is a gas whereas the remaining group 15 elements are solids.
• The gaseous nitrogen and brittle phosphorus are nonmetals.
• Arsenic and antimony are metalloids while bismuth is moderately reactive metal.

Question 16.
Explain the trends in physical properties of group 15 elements.
i. Atomic and ionic radii
ii. Ionization enthalpy
iii. Electronegativity
Answer:
i. Atomic and ionic radii:

• Atomic size increases down the group with increasing atomic number.
• The effective nuclear charge in case of group 15 elements is larger than that of group 14 elements. Due to the increased effective nuclear charge, electrons are strongly attracted by the nucleus. Thus, the atomic and ionic radii of group 15 elements are smaller than the atomic and ionic radii of the corresponding group 14 elements.
• On moving down the group, number of shells increases which leads to increased shielding effect and as a result atomic radii and ionic radii increases.

ii. Ionization enthalpy:

• Due to extra stability of half-filled p-orbitals and relatively smaller size of group 15 elements, ionization enthalpy of group 15 elements is much greater than that of the group 14 elements in the corresponding periods.
• On moving down the group, increase in atomic size and screening effect overcome the effective nuclear charge and thus, ionization enthalpy decreases.

iii. Electronegativity:

• Due to smaller size and greater effective nuclear charge of atoms, group 15 elements have higher electronegativity values than group 14 elements.
• On moving down the group, electronegativity values decreases due to increase in the size of the atoms and shielding effect.
• Nitrogen is the most electronegative element among group 15 elements. However, there is not much of a difference between the electronegativity values of other elements of group 15.

Note: Atomic and physical properties of group 15 elements.

Question 17.
Write a note on the oxidation state of p-block elements with respect to groups 13, 14 and 15 elements.
Answer:

• Oxidation state is the primary chemical property of all elements.
• The highest oxidation state exhibited by the p-block elements is equal to the total number of valence electrons i.e., the sum of s- and p-electrons present in the valence shell. This is sometimes called the group oxidation state.
• In boron, carbon and nitrogen families, the group oxidation state is the most stable oxidation state for the lighter elements.
• Besides, the elements of groups 13, 14 and 15 exhibit other oxidation states which are lower than the group oxidation state by two units.
• The lower oxidation states become increasingly stable as we move down to heavier elements in the groups.

Note: Group oxidation states and common oxidation states with examples for groups 13, 14 and 15.

Question 18.
What are general oxidation states of group 13 elements? Explain.
Answer:

• The general oxidation states of group 13 are +1 and +3.
• The group 13 elements have the outermost electronic configuration ns² np¹.
• If only np¹ electron takes part in bonding, the oxidation state is +1 and if all the three electrons i.e., ns² np¹ take part in bonding, the oxidation state is +3. Hence, the expected oxidation states are +1 and +3.

Question 19.
Give reason: The increased stability of the oxidation state is lowered by 2 units than the group oxidation state in heavier p-block elements.
Answer:

• The increased stability of the oxidation state lowered by 2 units than the group oxidation state in heavier p-block elements is due to inert pair effect.
• In these elements, the two s-electrons are involved less readily in chemical reactions.
• This is because, in heavier p-block elements, the s-electrons of valence shell experience poor shielding than valence p-electrons due to ten inner d-electrons.

Hence, the increased stability of the oxidation state is lowered by 2 units than the group oxidation state in heavier p-block elements.

Question 20.
Why Tl¹⁺ ion is more stable than Tl³⁺?
Answer:

• Tl is a heavy element which belongs to group 13 of the p-block.
• The common oxidation state for this group is +3.
• In p-block, the lower oxidation state is more stable for heavier elements due to inert pair effect.

Hence, Tl¹⁺ ion is more stable than Tl³⁺ ion.

Question 21.
How can you explain the higher stability of BCl₃ as compared to TlCl₃?
Answer:

• Boron is a light element in group 13 and has outermost electronic configuration 2s² 2p¹ whereas thallium is a heavy element in group 13 and has outermost electronic configuration 6s² 6p¹.
• Because of its small ionic radius, boron forms stable compounds in +3 oxidation state.
• Thallium has a large atomic size and due to the inert pair effect forms more stable compounds with lower oxidation state +1 than compounds with +3 oxidation state.

Therefore, BCl₃ has higher stability than TlCl₃.

Question 22.
State the oxidation state for the following:
i. The group oxidation state of group 14 elements.
ii. The stable oxidation state for lead.
iii. Oxidation state of carbon in CH₄.
Answer:
i. +4
ii. +2
iii. -4

Question 23.
GeCl4 is more stable than GeCl₂ while PbCl₂ is more stable than PbCl₄. Explain.
Answer:

• Elements Ge and Pb belong to 4th and 6th period in the group 14.
• The group oxidation state of group 14 elements is +4.
• However, the stability of other oxidation state which is lower by 2 units i.e., +2, increases down the group due to inert pair effect.
• Therefore, the stability of the oxidation state +4 is more in Ge than in Pb while the stability of the oxidation state +2 is more in Pb than in Ge.

Hence, GeCl₄ is more stable than GeCl₂ while PbCl₂ is more stable than PbCl₄.

Question 24.
Name the elements of group 14 which are generally occur in +2 oxidation state.
Answer:
The elements of group 14 that are generally occur in +2 oxidation state are tin (Sn) and lead (Pb).

Question 25.
Discuss the nature of bonding in compounds of group 13, 14 and 15 elements.
Answer:

• The lighter elements in groups 13, 14 and 15 have small atomic radii and high ionization enthalpy values. They form covalent bonds with other atoms by overlapping of valence shell orbitals.
• As we move down the group, the value of ionization enthalpy decreases. The atomic radius increases since the valence shell orbitals are more diffused.
• The heavier elements in these groups tend to form ionic bonds. The first member of these groups belongs to second period and do not have d orbitals and hence, B, C and N cannot expand their octet.
• The subsequent elements in the group possess vacant d orbital in their valence shell, which can expand their octet forming a variety of compounds.

Question 26.
Explain the reactivity of groups 13, 14 and 15 elements towards air.
Answer:
i. Group 13 elements:
a. On heating with air or oxygen, group 13 elements form oxide of the type E₂O₃.
\(4 \mathrm{E}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{E}_{2} \mathrm{O}_{3(\mathrm{~s})}\) (where, E = B, Al, Ga, In, Tl)

b. At high temperature, group 13 elements also react with nitrogen present in the air to form corresponding nitrides.
\(2 \mathrm{E}_{(\mathrm{s})}+\mathrm{N}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{EN}_{(\mathrm{s})}\) (where, E = B, Al, Ga, In, Tl)

ii. Group 14 elements: The elements of group 14 on heating in air or oxygen form oxide of the type EO and EO₂ in accordance with the stable oxidation state and availability of oxygen.

iii. Group 15 elements: The elements of group 15 on heating in air or oxygen forms two types of oxide i.e., E₂O₃ and E₂O₅.

Due to increase in metallic character down the groups 13, 14 and 15, the nature of their oxides gradually varies from acidic through amphoteric to basic.
[Note: The temperature required for the reaction of nitrogen with oxygen is very high. This is produced by striking an electric arc.]

Question 27.
Classify the following oxides into acidic, basic or amphoteric.
B₂O₃, Ga₂O₃, Tl₂O₃, In₂O₃, Al₂O₃
Answer:

Acidic oxide	B2O3
Basic oxides	In2O3, Tl2O3
Amphoteric oxides	Al2O3, Ga2O3

Question 28.
Match the following.

	Column A		Column B
i.	N2O5	a.	Amphoteric
ii.	Bi2O3	b.	Acidic
iii.	Sb2O3	c.	Basic

Answer:
i – b,
ii – c,
iii – a

Note: Nature of stable oxides of groups 13, 14 and 15 elements

Question 29.
State TRUE or FALSE. Correct the false statement.
i. Sb is more stable in +3 oxidation state.
ii. Oxides of the type E₂O₅ are formed by group 15 elements.
iii. As₄O₆ is an acidic oxide.
Answer:
i. True
ii. True
iii. False
As₄O₆ is an amphoteric oxide.

Question 30.
What happens when the elements of groups 13, 14 and 15 react with water?
Answer:
i. Most of the elements of groups 13, 14 and 15 are unaffected by water.
ii. Aluminium reacts with water on heating and forms hydroxide while tin reacts with steam to form oxide.

iii. Lead is unaffected by water due to the formation of a protective film of oxide.

Question 31.
Why is phosphorus stored under water?
Answer:
Phosphorus is highly reactive and hence, it is stored under water to prevent its reaction with air as it catches fire on being exposed to air.

Question 32.
Explain the reactivity of group 13 elements towards halogens.
Answer:
i. All the elements of group 13 react directly with halogens to form trihalides (EX3).
2E_(S) + 3X_2(g) → 2EX_3(s) (where, E = B, Al, Ga, In and X = F, Cl, Br, I)
ii. Thallium is an exception as it forms monohalides (TlX).

Question 33.
Describe the reactivity of group 14 elements with halogens.
Answer:

• All the elements of group 14 (except carbon) react directly with halogens to form tetrahalides (EX₄).
• The heavy elements Ge and Pb form dihalides as well.
• Stability of dihalides increases down the group due to inert pair effect.
• The ionic character of halides also increases steadily down the group.

Question 34.
Discuss the reactivity of group 15 elements with halogens.
Answer:

• Elements of the group 15 reacts with halogens to form two series of halides i.e., trihalides (EX₃) and pentahalides (EX₅).
• The pentahalides possess more covalent character due to availability of vacant d orbitals of the valence shell for bonding.
• Nitrogen being second period element, does not have d orbitals in its valence shell, and therefore, does not form pentahalides.
• Trihalides of the group 15 elements are predominantly covalent except BiF₃. The only stable trihalide of nitrogen is NF₃.

Question 35.
Nitrogen does not form pentahalides. Give reason.
Answer:

• The electronic configuration of 7N is 1s² 2s² 2p³. It has 3 unpaired electrons which can form 3 covalent bonds, thus forming NX₃ molecule.
• Valence shell of nitrogen (n = 2) contains only s and p orbitals.
• Thus, due to the absence of d orbitals in the valence shell, nitrogen cannot expand its octet, therefore, it cannot form compounds like NCl₅ and NF₅.

Hence, nitrogen does not form pentahalides.

Question 36.
Nitrogen does not form NCl₅ or NF₅ but phosphorus can. Explain.
Answer:

• The electronic configuration of 7N is 1s2 2s2 2p3 while that of 15P is 1s² 2s² 2p⁶ 3s² 3p³.
• As phosphorus contains d orbitals, it can expand its octet to form MX₃ as well MX₅ compounds.
• However, due to absence of d orbitals, nitrogen cannot form MX₃ or MX₅.

Hence, Nitrogen does not form NCl₅ or NF₅ but phosphoms can form compounds like PCl₅ or PF₅.

Question 37.
Define catenation.
Answer:
The property of self-linking of atoms of an element by covalent bonds to form chains and rings is called catenation.

Question 38.
Explain catenation of group 14 elements.
Answer:
i. The property of self-linking of atoms of an element by covalent bonds to form chains and rings is called catenation.
ii. The strength of the element-element bond determines the tendency of an element to form a chain.
iii. Among the elements of group 14, the bond strength is maximum for C-C bond (348 kJ mol^-1). Hence, carbon has maximum tendency for catenation.

Bond	Bond strength (Bond enthalpy kJ mol-1)
C-C	348
Si-Si	297
Ge-Ge	260
Sn-Sn	240

iv. From the values of bond enthalpy, it can be concluded that the tendency to form chains is maximum for carbon and much lesser for silicon. Germanium has still lesser tendency and tin has hardly any tendency for catenation. Lead does not show catenation.
Therefore, the order of catenation of group 14 elements is C >> Si > Ge = Sn.

Question 39.
State TRUE or FALSE. Correct the false statement.
i. Among the group 14 elements, Ge does not show the property of catenation.
Answer:
i. False
Among the group 14 elements, Pb (lead) does not shows the property of catenation.

Question 40.
Define allotropy.
Answer:
When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc. the phenomenon is called allotropy.

Question 41.
i. What are allotropes?
ii. Name various allotropes of carbon.
Answer:
i. When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc. the phenomenon is called allotropy and the individual crystalline forms are called allotropes.
ii. Diamond, graphite, fiillerenes, graphene and carbon nanotubes are various allotropes of carbon.

Question 42.
Explain the structure of diamond.
Answer:
Structure of diamond:

• In diamond, each carbon atom undergoes sp3 hybridization and is linked to four other carbon atoms in tetrahedral manner.
• The C – C bond length is 154 pm.
• The tetrahedra are linked together forming a three-dimensional network structure involving strong C-C single bonds which makes diamond the hardest natural substance.

Question 43.
Write physical properties of diamond. Also, state its uses.
Answer:
i. Physical properties

• Diamond is the hardest natural substance.
• It has abnormally high melting point (3930 °C).
• It is a bad conductor of electricity.

ii. Uses: Diamond is used

• for cutting glass and in drilling tools.
• for making dies for drawing thin wire from metal.
• for making jewellery.

Question 44.
Describe the structure of graphite.
Answer:

• Graphite is composed of layers of two-dimensional sheets of carbon atoms.
• Each sheet is made up of hexagonal net of sp² carbons bonded to three neighbours forming three bonds.
• The fourth electron in the unhybrid p-orbital of each carbon is shared by all carbon atoms resulting in a π bond. These it electrons are delocalized over the whole layer.
• The C – C bond length in graphite is 141.5 pm.
• The individual layers are held by weak van der Waals forces and separated by 335 pm.
• Graphite is soft and slippery and is thermodynamically most stable allotrope of carbon.

Question 45.
Diamond is very hard whereas graphite is soft. Explain.
Answer:

• Diamond has three-dimensional network of sp³ hybridized carbon atoms joined by extended covalent bonds which are difficult to break. Therefore, diamond is hard.
• Graphite has two-dimensional sheet like structure, like layers of hexagonal rings formed from sp² hybridized carbon atoms. These layers are held by weak van der Waals forces, which can be broken easily. Therefore, graphite is soft and slippery.

Hence, diamond is very hard whereas graphite is soft.

Question 46.
i. What are fullerenes?
ii. How are they prepared?
Answer:
i. Fullerenes are allotropes of carbon in which carbon molecules are linked by a definite numbers of carbon atoms, for example as in C₆₀.
ii. Fullerenes are produced when an electric arc is struck between the graphite electrodes in an inert atmosphere of argon or helium. The soot formed contains significant amount of C₆₀ fullerene and smaller amounts of other fullerenes C₃₂, C₅₀, C₇₀ and C₈₄.

Question 47.
Discuss the structure and properties of fullerene (C6o).
Answer:

• C₆₀ has a shape like soccer ball and called Buckminsterfullerene or bucky ball.
• It contains 20 hexagonal and 12 pentagonal fused rings of carbon.
• The C₆₀ fullerene structure exhibit separations between the neighbouring carbons as 143.5 pm and 138.3 pm.
• Fullerenes are covalent and soluble in organic solvents.
• Fullerene C₆₀ reacts with group 1 metals forming solids such as K₃C₆₀.
• The compound K₃C₆₀ behaves as a superconductor below 18 K, which means that its carries electric current with zero resistance.

Question 48.
Explain the structure of carbon nanotubes.
Answer:

• Carbon nanotubes are cylindrical in shape consisting of rolled-up graphite sheet.
• Nanotubes can be single-walled (SWNTs) with a diameter of less than 1 nm or multi-walled (MWNTs) with diameter reaching more than 100 nm.
• Their lengths range from several micrometres to millimetres.

Question 49.
Describe the properties of carbon nanotubes.
Answer:

• Carbon nanotubes are robust. They can be bent, and when released, they will spring back to the original shape.
• Carbon nanotubes have high electrical or heat conductivities and highest strength-to-weight ratio for any known material to date.
• The researchers of NASA are combining carbon nanotubes with other materials into composites that can be used to build lightweight spacecraft.

Question 50.
What is graphene?
Answer:

• Isolated layer of graphite is called graphene.
• Graphene sheet is a two dimensional solid.
• It has unique electronic properties.

Question 51.
Explain the structure of various allotropes of phosphorus.
Answer:
Phosphorus is found in different allotropic forms. White and red phosphorus are important allotropes of phosphorus.
i. White (yellow) phosphorus:

1. White (yellow) phosphorus consists of discrete tetrahedral P₄ molecules.
2. The P – P – P bond angle is 60°.
3. White phosphorus is less stable and hence more reactive, because of angular strain in the P₄ molecule.

ii. Red phosphorus:

• Red phosphorus consists of chains of P₄ linked together by covalent bonds.
• Thus, it is polymeric in nature.

Question 52.
Enlist properties of
i. white phosphorus.
ii. red phosphorus.
Answer:
i. Properties of white phosphorus:

• It is translucent white waxy solid.
• It glows in the dark (chemiluminescence).
• It is insoluble in water but dissolves in boiling NaOH solution.
• It is poisonous.

ii. Properties of red phosphorus:

• It is stable and less reactive.
• It is odourless and possess iron grey lustre.
• It does not glow in the dark.
• It is insoluble in water.
• It is nonpoisonous.

Question 53.
How is red phosphorus prepared?
Answer:
Red phosphorus is prepared by heating white phosphorus at 573 K in an inert atmosphere.

Question 54.
State whether the following statement is TRUE or FALSE. Correct if false.
i. Covalent molecules have irregular shape described with the help of bond lengths and bond angles.
ii. It is difficult to understand the reactivity of covalent inorganic compounds from their structures.
iii. Inorganic molecules are often represented by molecular formulae indicating their elemental composition.
Answer:
i. False
Covalent molecules have definite shape described with the help of bond lengths and bond angles.
ii. False
The reactivity of covalent inorganic compounds is better understood from their structures.
iii. True

Question 55.
Describe structure of the following molecules.
i. Boron trichloride
ii. Aluminium chloride
iii. Orthoboric acid
Answer:
i. Structure of boron trichloride (BCl₃) molecule:

• Boron trichloride (BCl₃) is a covalent compound.
• In BCl₃ molecule, boron atom is sp² hybridized having one vacant unhybridized p orbital.
• B in BCl₃ has incomplete octet.
• BCl₃ is a nonpolar trigonal planar molecule.
• Each Cl – B – Cl bond angle is 120°.

ii. Structure of aluminium chloride (AlCl₃) molecule:

• Aluminium atom in aluminium chloride is sp² hybridized, with one vacant unhybrid p-orbital.
• Aluminium chloride exists as the dimer (Al₂Cl₆) formed by overlap of vacant 3d orbital of Al with a lone pair of electrons of Cl.

iii. Structure of orthoboric or boric acid (H₃BO₃) molecule:

• Orthoboric acid has central boron atom bound to three -OH groups.
• The solid orthoboric acid has layered crystal structure in which trigonal planar B(OH)₃ units are joined together by hydrogen bonds.

Question 56.
Which are the different crystalline forms of silica?
Answer:
Quartz, cristobalite and tridymite are the different crystalline forms of silica.
[Note: These crystalline forms are inter-convertible at a suitable temperature.]

Question 57.
Explain the structure of silicon dioxide.
Answer:

• Silicon dioxide (SiO₂), is also known as silica.
• It is a covalent three-dimensional network solid.
• In SiO₂, each silicon atom is covalently bound in tetrahedral manner to four oxygen atoms.
• The crystal contains eight membered rings having alternate silicon and oxygen atoms.

Question 58.
Discuss the nature and structure of the following compounds.
i. Nitric acid
ii. Phosphoric acid
Answer:
i. Nitric acid:

• Nitric acid (HNO₃) is a strong, oxidizing mineral acid.
• The central nitrogen atom is sp² hybridized.
• HNO₃ exhibits resonance phenomenon.
• Figure (a) represents resonating structures of HNO₃ while figure (b) represents resonance hybrid of HNO₃.

ii. Phosphoric acid (Orthophosphoric acid):

• Phosphorus forms number of oxyacids. Orthophosphoric acid (H₃PO₄) is a strong nontoxic mineral acid.
• It contains three ionizable acidic hydrogens.
• The central phosphorus atom is tetrahedral.

Question 59.
Give the molecular formula of crystalline borax.
Answer:
The crystalline borax has formula Na₂B₄O₇.10H₂O or Na₂[B₄O₅(OH)₄].8H₂O.

Question 60.
How is borax obtained from its mineral colemanite?
Answer:
Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

Question 61.
Why is the aqueous solution of borax alkaline?
Answer:
On hydrolysis, borax forms a strong base (NaOH) and a weak acid (H₃BO₃). The presence of the strong base makes borax solution alkaline.

Question 62.
What happens when borax is heated strongly?
Answer:
Borax is a white crystalline solid. On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.

Question 63.
Explain borax bead test.
Answer:
i. Borax bead test is used to detect coloured transition metal ions.
ii. On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.

iii. The borax bead consists of sodium metaborate and boric anhydride, which reacts with metals salts to form coloured bead.
e.g. When borax is heated in a Bunsen burner flame with CoO on a loop of platinum wire, a blue coloured Co(BO₂)₂ bead is formed.

Question 64.
Write the uses of borax.
Answer:
Borax is used

• to manufacture optical and hard borosilicate glasses.
• as a flux for soldering and welding.
• as a mild antiseptic in the preparation of medical soaps.
• in qualitative analysis for borax bead test.
• as a brightener in washing powder.

Question 65.
How are silicones prepared? Write their properties.
Answer:
i. Preparation of silicones:
a. Alkyl or aryl substituted silicon chlorides having general formula R_nSiCl_(4-n) (R = alkyl or aryl group) are used as the starting materials for manufacture of silicones.
b. When methyl chloride reacts with silicon in the presence of copper catalyst at a temperature 573 K, various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amounts of Me₄Si are formed.

c. Hydrolysis of dimethyldichlorosilane, (CH₃)₂SiCl₂ followed by condensation polymerisation yields straight chain silicone polymers.

d. The chain length of polymer can be controlled by adding (CH₃)₃SiCl at the end.

ii. Properties:

• Silicones are water repellent.
• They have high thermal stability.
• They are good electrical insulators.
• They are resistant to oxidation and chemicals.

Question 66.
Explain the preparation of ammonia from nitrogeneous organic matter.
Answer:
Ammonia is formed by the decomposition of nitrogeneous organic matter such as urea. It is therefore, present naturally in small quantities in air and soil.

Question 67.
Describe laboratory method for preparation of ammonia.
Answer:
Ammonia is prepared on laboratory scale by decomposition of the ammonium salts with calcium hydroxide or caustic soda.

Question 68.
How is ammonia manufactured by Haber process?
Answer:

• On the large scale, ammonia is prepared by direct combination of dinitrogen and dihydrogen by Haber process.
• In this process, dinitrogen reacts with dihydrogen under high pressure of 200 × 10⁵ Pa (200 atm) and temperature around 700 K to produce ammonia.
  N_2(g) + 2H_2(g) ⇌ 2NH_3(g); Δ_fH° = -46.1 kJ mol^-1
• Iron oxide with trace amounts of K₂O and Al₂O₃ is used as catalyst in Haber process.
• High pressure favours the formation of ammonia as equilibrium is attained rapidly under these conditions.

Question 69.
State the physical properties of ammonia.
Answer:

• Ammonia is a colourless gas with pungent odour.
• It has freezing point of 198.4 K and boiling point of 239.7 K.
• It is highly soluble in water.

Question 70.
What is liquor ammonia?
Answer:
The concentrated aqueous solution of ammonia (NH₃) is called liquor ammonia.

Question 71.
Give reason: Ammonia has higher melting and boiling points.
Answer:

• In solid and liquid state, NH₃ molecules get associated together through hydrogen bonding.
• As a result, extra amount of energy is required to break such intermolecular hydrogen bonds. Hence, ammonia has higher melting and boiling points.

Question 72.
Why is ammonia basic in aqueous solution?
Answer:
i. As ammonia is highly soluble in water, it readily forms OH^– ions in its aqueous solution.
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(a q)}^{-}\)
ii. Thus, due to the formation of OH^– ions, aqueous solution of ammonia is basic in nature.

Question 73.
How does the aqueous solution of ammonia react with the following salt solutions?
i. ZnSO₄
ii. FeCl₃
Answer:
Aqueous solution of ammonia precipitates out as hydroxides (or hydrated oxides) of metals solutions.

Question 74.
Write applications of ammonia.
Answer:
Ammonia is used in

• manufacture of fertilizers such as urea, diammonium phosphate, ammonium nitrate, ammonium sulphate etc.
• manufacture of some inorganic compounds like nitric acid.
• refrigerant (liq. ammonia).
• laboratory reagent in qualitative and quantitative analysis (aq. solution of ammonia).

Question 75.
Give reactions involved in the formation of Nessler’s reagent.
Answer:

Question 76.
How does ammonia react with Nessler’s reagent?
Answer:
Ammonia react with Nessler’s reagent (an alkaline solution of K₂HgI₄) to form a brown precipitate (Millon’s base).

Question 77.
Complete and write the balanced chemical equations for:
i. Ca₂B₆O₁₁ + Na₂CO₃ →
ii. CoO + B₂O₃ →
iii. AgCl + NH₃ →
iv. ZnSO₄ + 2NH₄OH →
v. a. 2KI + HgCl₂ →
b. 2KI + HgI₂ →
Answer:

Question 78.
Naina was preparing a compound in the laboratory. She added compound ‘A’ to (CaOH)₂ solution. As a result of this, a compound ‘B’ was obtained which had a pungent smell. On adding Nessler’s reagent to the compound ‘B’, a brown precipitate of compound ‘C’ was obtained.
Write the chemical reactions involved and identify ‘A’, ‘B’ and ‘C’.
Answer:
i. When ammonium chloride is mixed with (CaOH)₂ solution, ammonia is formed which has a pungent odour.

ii. Ammonia react with Nessler’ s reagent (an alkaline solution of K₂Hgl₄) to form a brown precipitate (Millon’ s base).

Multiple Choice Questions

1. The electronic configuration of boron family is ……………
(A) ns² np²
(B) ns² np⁵
(C) ns² np⁶
(D) ns² np¹
Answer:
(D) ns² np¹

2. ………… has noble gas core plus 14 f-electrons and 10 d-electrons.
(A) Gallium
(B) Indium
(C) Thallium
(D) Boron
Answer:
(C) Thallium

3. The group 15 element having inner electronic configuration as of argon is …………..
(A) Phosphorus (Z = 15)
(B) Antimony (Z = 51)
(C) Arsenic (Z = 33)
(D) Nitrogen (Z = 7)
Answer:
(C) Arsenic (Z = 33)

4. Which of the following is NOT a metalloid?
(A) B
(B) Sn
(C) Ge
(D) Sb
Answer:
(B) Sn

5. Among the group 13 elements, melting point is highest for …………..
(A) B
(B) Al
(C) Ga
(D) In
Answer:
(A) B

6. On moving down the group 14, the ionization enthalpy
(A) increases slightly from Si to Sn and decreases slightly from Sn to Pb
(B) increases throughout uniformly from Si to Pb
(C) decreases throughout uniformly from Si to Pb
(D) decreases slightly from Si to Sn and increases slightly from Sn to Pb
Answer:
(D) decreases slightly from Si to Sn and increases slightly from Sn to Pb

7. ………… is the most electronegative element of group 14.
(A) Carbon
(B) Silicon
(C) Germanium
(D) Tin
Answer:
(A) Carbon

8. The stability of +3 oxidation state in aqueous solution is in order ……………
(A) Al > Ga > In > Tl
(B) Tl > In > Ga > Al
(C) Al > Tl > Ga > In
(D) Tl > Al > Ga > In
Answer:
(A) Al > Ga > In > Tl

9. Group oxidation state of group 15 elements is ……………
(A) +4
(B) +1
(C) +3
(D) +5
Answer:
(D) +5

10. …………. cannot expand its octet due to absence of d orbital in its valence shell.
(A) Ga
(B) C
(C) As
(D) Ge
Answer:
(B) C

11. Which one of the following statements about boron and aluminium is INCORRECT?
(A) Both exhibit oxidation state of +3.
(B) Both form oxides of the formula M₂O₃.
(C) Both form trihalides, MX₃.
(D) Both form amphoteric oxides.
Answer:
(D) Both form amphoteric oxides.

12. Which of the following is basic oxide?
(A) Bi₂O₃
(B) CO₂
(C) B₂O₃
(D) SiO₂
Answer:
(A) Bi₂O₃

13. The reaction of Al with H₂O produces ……………
(A) Al₂O₃
(B) AlH₃
(C) Al(OH)₃
(D) Al₂H₆
Answer:
(C) Al(OH)₃

14. Which of the following is a stable halide of nitrogen?
(A) NF₃
(B) NCl₅
(C) NF₅
(D) NBr₅
Answer:
(A) NF₃

15. Catenation is the ability of …………..
(A) atoms to form strong bonds with similar atoms
(B) elements to form giant molecules
(C) an element to form multiple bonds
(D) an element to form long chains of identical atoms
Answer:
(D) an element to form long chains of identical atoms

16. Among the group 13 elements, the property of allotropy is shown by ………………
(A) indium
(B) aluminium
(C) thallium
(D) boron
Answer:
(D) boron

17. Thermodynamically stable allotrope of carbon is …………..
(A) diamond
(B) graphite
(C) buckyball
(D) all of these
Answer:
(B) graphite

18. White phosphorus contains discrete …………… molecules.
(A) P₅
(B) P₄
(C) P₆
(D) P₅₂
Answer:
(B) P₄

19. In white phosphorus, the P-P-P bond angle is ……………
(A) 60°
(B) 90°
(C) 109.5
(D) 120°
Answer:
(A) 60°

20. 3c-2e bonds are present in ………………
(A) NH₃
(B) B₂H₆
(C) H₃BO₃
(D) SiCl₄
Answer:
(B) B₂H₆

21. Which of the following is borax?
(A) Na₂B₄O₇.4H₂O
(B) Na₂B₄O₇.10H₂O
(C) H₃BO₃
(D) NaBO₂
Answer:
(B) Na₂B₄O₇.10H₂O

22. In Borax bead test, the coloured ions give characteristic coloured beads due to formation of …………….
(A) metal borates
(B) metal metaborates
(C) metal phosphates
(D) metal tetraborates
Answer:
(B) metal metaborates

23. The catalyst used in Haber process contains …………..
(A) nickel
(B) palladium
(C) iron
(D) platinum
Answer:
(C) iron

24. Which of the following is used as refrigerant?
(A) Nessler’s reagent
(B) Liq. ammonia
(C) Borax
(D) Diborane
Answer:
(B) Liq. ammonia

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

Maharashtra State Board 11th Chemistry Important Questions Chapter 8 Elements of Group 1 and 2

Question 1.
Why is hydrogen studied separately even though it appears at the top of group 1?
Answer:
Even though hydrogen appears at the top of the group 1 containing alkali metals, it is studied separately because many of its properties differ from that of the alkali metals.

Question 2.
Give reason: Hydrogen (H₂) molecule is also referred to as dihydrogen.
Answer:

• The nucleus of a hydrogen atom consists of one positively charged proton i.e., nuclear charge of +1 and one extranuclear electron.
• As this electron is in direct influence of nuclear attraction, hydrogen has a little tendency to lose this electron.
• However, it can easily pair with the other electron forming a covalent bond.
• Therefore, it exists in diatomic form as H₂ molecule and hence, it is also referred to as dihydrogen.

Question 3.
Why does hydrogen occur in a diatomic form rather than in monoatomic form under normal
conditions?
Answer:

• Hydrogen atom has only one electron in its valence shell having electronic configuration 1s¹.
• It can acquire stable configuration of helium by sharing this electron with another hydrogen atom.
• Therefore, it shares its single electron with electron of the other H-atom to achieve stable inert gas configuration of He.
• Thus, hydrogen readily forms diatomic molecule and exists as H₂ rather than in monoatomic form.

Question 4.
Write a note on occurrence of hydrogen.
Answer:

• In the free state hydrogen exists as dihydrogen gas.
• Hydrogen is most abundant element in the universe and constitutes 70% of the total mass of the universe.
• Hydrogen is also the principal element in the solar system.
• On the earth, hydrogen is the tenth most abundant element on mass basis and the third most abundant element on atom basis.

Question 5.
State whether the following statements are TRUE or FALSE. Correct the false statement.
1. Electronic configuration of hydrogen is 1s¹.
ii. H⁺ ion formed by loss of the electron from hydrogen atom exists freely.
iii. H⁺ is nothing but a proton.
iv. Metastable metallic hydrogen was discovered at Harvard university, USA, in January 2017.
Answer:
i. True
ii. False
Hydrogen atom does not exist freely and is always associated with other molecules i.e., H₃O⁺.
iii. True
iv. True

Question 6.
Explain the laboratory methods for preparation of dihydrogen.
Answer:
Laboratory methods for preparation of dihydrogen:
i. By action of dilute HCl on zinc granules: Zinc granules on reaction with dilute hydrochloric acid (HCl) liberates hydrogen gas.

ii. By action of aqueous NaOH on zinc: Zinc on reaction with aqueous sodium hydroxide (NaOH) forms soluble sodium zincate and liberates hydrogen gas.

Question 7.
Describe the industrial method of preparation of dihydrogen by electrolysis of pure water.
Answer:
i. Pure water is a poor conductor of electricity. Therefore, a dilute aqueous solution of acid or alkali is used to prepare dihydrogen by electrolysis.
ii. For example, electrolysis of dilute aqueous solution of sulphuric acid yields two volumes of hydrogen at cathode and one volume of oxygen at anode.

Question 8.
How is pure dihydrogen (> 99.5% purity) gas obtained from barium hydroxide?
Answer:
Electrolysis of warm aqueous solution of barium hydroxide using nickel electrodes yields pure dihydrogen (> 99.5% purity).

Question 9.
Explain the terms:
i. Syngas
ii. Water-gas shift reaction.
Answer:
i. Syngas:

• Syngas is the mixture of CO and H₂. It is also called ‘water-gas’.
• It is used for the synthesis of CH₃OH and many hydrocarbons, hence, the name syngas or ‘synthesis gas’.
• Production of syngas is also the first stage of gasification of coal.

ii. Water-gas shift reaction:
The carbon monoxide in the water-gas is transformed into carbon dioxide by reacting with steam in presence of iron chromate as catalyst.

This reaction is called water-gas shift reaction.

Question 10.
Complete the following chemical reactions:

Answer:

Question 11.
Enlist physical properties of dihydrogen.
Answer:
Physical properties of dihydrogen:

• Dihydrogen is a colourless, tasteless and odourless gas.
• It bums with a pale blue flame.
• It is a nonpolar and water-insoluble gas.
• It is lighter than air.

Question 12.
What is the action of dihydrogen on the following?
i. Metals
ii. Dioxygen
Answer:
i. Action of dihydrogen on metals:
a. Dihydrogen combines with all the reactive metals including alkali metals, calcium, strontium and barium at high temperature to form metal hydrides.
b. For example: Dihydrogen combines with metallic sodium at high temperature to yield sodium hydride.

ii. Action of dihydrogen on dioxygen: Dihydrogen reacts with dioxygen in the presence of catalyst or by heating to form water. This reaction is highly exothermic.

Question 13.
Explain the effect of high bond dissociation energy of H-H bond on chemical reactivity of dihydrogen?
Answer:

• The bond dissociation energy of H-H bond is very high i.e, 436 kJ mol^-1. and thus, it does not react easily under normal conditions.
• However, at high temperature or in the presence of catalysts, hydrogen combines with many metals and non-metals to form corresponding hydrides and halides respectively.

Question 14.
What happens when dihydrogen reacts with halogens?
Answer:
i. Dihydrogen reacts with halogens (X₂) to give the corresponding hydrogen halides (HX).

ii. Dihydrogen reacts with fluorine to form hydrogen fluoride even at very low temperature (-250°C) in dark.

iii. However, the reaction with iodine requires a catalyst as the vigour of reaction of dihydrogen decreases with increasing atomic number of halogen.

Question 15.
Explain the reducing nature of hydrogen with chemical reactions.
Answer:
Dihydrogen reduces oxides and ions of some metals that are less reactive than iron, to the corresponding number of halogen metals at moderate temperature.
e.g.
i. CuO_(s) + H_2(g) → Cu_(s) + H₂O_(l)
ii. Fe₃O_4(s) + 4H_2(g) → 3Fe_(s) + 4H₂O_(s)
iii. Pd²⁺_(aq) + H_2(g) → Pd_(s) + 2H⁺_(aq)

Question 16.
What is hydrogenation?
Answer:
Hydrogenation is the reaction in which hydrogen gas reacts with unsaturated organic compounds in the presence of a catalyst to form hydrogenated (saturated) compounds.

Question 17.
How does dihydrogen react with various organic compounds to give useful, commercially important products?
Answer:
i. Hydrogenation of unsaturated organic compounds:
e.g. Hydrogenation of unsaturated organic compounds such as vegetable oil using nickel catalyst gives saturated organic compounds such as solid edible fats like vanaspati ghee.

ii. Hydroformylation of olefins and subsequent reduction of aldehyde to form alcohol:

Question 18.
Explain hydroformylation reaction of olefins using a suitable example.
Answer:
Hydroformylation of olefins gives aldehydes which on further reduction gives alcohols.
e.g. i. Hydroformylation of propene gives butyraldehyde.

ii. Butyraldehyde further undergoes reduction to give n-butyl alcohol.

Question 19.
What are the uses of dihydrogen?
Answer:
Dihydrogen is used in

• the production of ammonia.
• the formation of vanaspati ghee by catalytic hydrogenation of oils.
• rocket fuel (mixture of liquid hydrogen and liquid oxygen).
• the preparation of important organic compounds like methanol in bulk quantity.
  \(2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{CO}_{(\mathrm{g})} \stackrel{\text { Cobaltcatalyst }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{OH}_{(l)}\)
• the preparation of hydrogen chloride (HCl) and metal hydrides.

Question 20.
Justify the placement of hydrogen in the group of alkali metals with the help of reaction with halogens.
Answer:
i. Hydrogen on reaction with halogens (X₂) give compounds with general formula HX.
e.g. H₂ + Cl₂ → 2HCl
ii. Similarly, alkali metals (M) on reaction with halogens (X₂) give compounds with general formula MX.
e.g. 2Na + Cl₂ → 2NaCl
iii. Thus, H₂ and alkali metals are monovalent elements and more electropositive than halogens. This similarity justifies the position of hydrogen in the group 1.

Question 21.
What do you mean by s-block elements? Where are they placed in the modern periodic table?
Answer:

• Elements of group 1 and group 2 in which the last electron enters into ‘ns’ subshell are s-block elements.
• The s-block elements are placed on the extreme left in the modem periodic table.

Question 22.
Name elements of group 1 and group 2.
Answer:

• Group 1 of the periodic table consists of the elements: hydrogen, lithium, sodium, potassium, rubidium, caesium and francium.
• Group 2 of the periodic table consists of elements: beryllium, magnesium, calcium, strontium, barium and radium.

Question 23.
What are alkali metals?
Answer:
The elements of group 1 except hydrogen are collectively called alkali metals.

Question 24.
What are alkaline earth metals?
Answer:
The elements of group 2 are collectively called alkaline earth metals because they occur as minerals in rocks.

Question 25.
Write a note on occurrence of group 1 and group 2 elements:
Answer:
i. Group 1 (alkali metals):

• Two elements of group 1 i.e., sodium and potassium are the sixth and seventh most abundant elements present in the earth’s crust.
• However, francium does not occur appreciably in nature because it is radioactive and has short half-life period.

ii. Group 2 (alkaline earth metals):

• The elements magnesium and calcium are found abundantly in earth’s crust.
• Radium is radioactive and is not easy to find.

Question 26.
Give reasons: s-block elements are never found in free state in nature.
Answer:

• s-Block elements contain group 1 and group 2 elements.
• The general outer electronic configuration of the group 1 elements is ns¹ and that of the group 2 elements is ns².
• The loosely held s-electrons in the valence shell of these elements can be easily removed to form metal ions.
• As a result, they are highly reactive in nature and always found in combined state.

Hence, s-block elements are never found in free state in nature.

Note: Electronic configurations of group 1 elements

Note: Electronic configurations of group 2 elements

Question 27.
Describe the physical properties of alkali and alkaline earth metals.
Answer:

• All the alkali and alkaline earth metals are silvery white in appearance.
• Due to their large atomic size they have low density.
• Both alkali and alkaline earth metals are soft, however, alkaline earth metals are harder than the alkali metals.
• Alkali metals are the most electropositive elements while alkaline earth metals are comparatively less electropositive than alkali metals.

Question 28.
Explain why do the group 1 and group 2 elements form diamagnetic and colourless compounds.
Answer:

• Unipositive ions of all the elements of group 1 have inert gas configuration and hence, they have no unpaired electron.
• Similarly, group 2 elements can lose their two valence shell electrons and form divalent ions that have inert gas configuration with no unpaired electrons.

Hence, due to the absence of unpaired electrons, compounds formed by group 1 and group 2 elements are diamagnetic and colourless.

Question 29.
Why do the properties of lithium and beryllium differ from the rest of the group 1 and group 2 elements?
Answer:
The properties of lithium and beryllium differ from the rest of the group 1 and group 2 elements due to their extremely small size and comparatively high electronegativity.

Question 30.
Complete the following table.

Group 1 elements	Group 2 elements
……………….	Alkaline earth metals
Outer electronic configuration: ……………….	Outer electronic configuration: ns2
Monovalent positive ions	……………….

Answer:

Group 1 elements	Group 2 elements
Alkali metals	Alkaline earth metals
Outer electronic configuration: ns1	Outer electronic configuration: ns2
Monovalent positive ions	Divalent positive ions

Question 31.
State the trends in the following properties of group 1 and group 2 elements down a group.
i. Atomic radii
ii. Ionic radii
iii. Ionization enthalpy
iv. Electronegativity
v. Standard reduction potential
Answer:

Sr. no.	Property	Down a group
i.	Atomic radii	Increases
ii.	Ionic radii	Increases
iii.	Ionization enthalpy	Decreases
iv.	Electronegativity	Decreases
V.	Standard reduction potential	Decreases

Question 32.
Give reasons: Potassium superoxide is used in breathing equipment used for mountaineers and in submarines and space.
Answer:
i. Potassium superoxide has ability to absorb carbon dioxide and give out oxygen at the same time.

ii. Due to this property of KO₂, it is used in breathing equipment used for mountaineers and in submarines and space.

Question 33.
What is the oxidation state of:
i. Na in Na₂O₂?
ii. K in KO₂?
Answer:
i. Oxidation state of Na in sodium peroxide (Na₂O₂):
Let x be the oxidation state of Na in Na₂O₂.
The net charge on peroxide ion \(\left(\mathrm{O}_{2}^{2-}\right)\) is -2.
Since any compound is electrically neutral, it has an overall charge as zero.
∴ 2x + (-2) = 0
∴ x = + 1
∴ Oxidation state of Na in Na₂O₂ is +1.

ii. Oxidation state of K in potassium dioxide/potassium superoxide (KO₂):
Let x be the oxidation state of K in KO₂
The net charge on superoxide ion \(\left(\mathrm{O}_{2}^{-}\right)\) is -1.
Since any compound is electrically neutral, it has an overall charge as zero.
∴ x + (-1) = 0
x = + 1
∴ Oxidation state of K in KO₂ is + 1.
[Note: Oxidation state of alkali metal is always +1.]

Question 34.
Magnesium strip slowly tarnishes on keeping in air but metallic calcium is readily attacked by air. Explain.
Answer:

• The reactivity of group 2 metals increases with increasing atomic radius and lowering of ionization enthalpy
  down the groups.
• Thus, calcium has lower ionization enthalpy. Therefore, calcium is more reactive than magnesium.
• Hence, Mg reacts slowly with air forming a thin film of oxide resulting into tarnishing, whereas Ca reacts readily at room temperature with oxygen and nitrogen in the air.

Question 35.
What happens when alkali metals react with hydrogen and halogens?
Answer:
i. Reaction with hydrogen: Alkali metals react with hydrogen at high temperature to form the Corresponding metal hydrides.

ii. Reaction with halogens: All the alkali metals react vigorously with halogens to produce their ionic halide salts.

[Note: As we move down the group, the reactivity of alkali metals towards hydrogen and halogens decreases in the following order: Li > Na > K > Rb > Cs.]

Question 36.
NaCl is an ionic compound but LiCl has some covalent character, explain.
Answer:

• Li⁺ ion has very small size and therefore, the charge density on Li⁺ is high.
• 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 LiCl bond.
• Na⁺ ion cannot distort the electron cloud of Cl^– due to the bigger size of Na⁺ compared to Li⁺.

Hence, NaCl is an ionic compound but LiCl has some covalent character.

Question 37.
Why is lithium iodide most covalent in nature among alkali halides?
Answer:

• Among the alkali metal ions, Li⁺ ion is the smallest cation while among halides, anion I^– has the largest size.
• Thus, electron cloud around I^– ion is easily distorted by Li⁺ ion leading to polarisation of anion and covalency.
• Also, the difference in electronegativities of Li and I is small.

Hence, lithium iodide is most covalent in nature among alkali halides.

Question 38.
Explain the reactivity of alkaline earth metals towards:
i. Water
ii. Hydrogen
iii. Halogens
Answer:
i. Reaction with water:
a. The elements of group 2 (alkaline earth metals) react with water to form metal hydroxide and evolve hydrogen gas.

b. Be does not react with water at all, Mg reacts with boiling water while Ca, Sr, Ba react vigorously even with cold water.

ii. Reaction with hydrogen: All alkaline earth metals except beryllium (Be), when heated with hydrogen form MH₂ type hydrides.

iii. Reaction with halogens: All the alkaline earth metals combine with halogens at high temperature to form their corresponding halides.

[Note: As we move down the group, the chemical reactivity of alkaline earth metals increases in the order Mg < Ca < Sr < Ba.]

Question 39.
Complete the following chemical equations.

Answer:

Question 40.
Describe the reducing nature of group 1 and group 2 elements.
Answer:
The reducing power of an element is measured in terms of standard electrode potential (E0) corresponding to the following transformation i.e, tendency to lose electron.
\(\mathrm{M}_{(\mathrm{s})} \longrightarrow \mathrm{M}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-}\)
i. Reducing nature of group 1 elements:

• All the alkali metals have high negative values of E⁰ which indicates that they have strong reducing nature and hence, they can be used as strong reducing agents.
• Lithium is the most powerful and sodium is the least powerful reducing agent in the group.

ii. Reducing nature of group 2 elements:

• All the alkaline earth metals have high negative values of stanard reduction potential (E⁰) and are strong reducing agents.
• However, reducing power of alkaline earth metals is less than that of alkali metals.

Question 41.
Explain the nature of the solution formed by group 1 and group 2 metals in liquid ammonia.
Answer:
i. The alkali metals are soluble in liquid ammonia and thus, they dissolve in it giving deep blue solutions which are conducting in nature.
M + (x + y)NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
ii. The blue colour of the solution is due to the ammoniated electron.
iii. These solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide.
\(\mathrm{M}_{(\mathrm{am})}^{+}+\mathrm{e}^{-}+\mathrm{NH}_{3(l)} \longrightarrow \mathrm{MNH}_{2(\mathrm{am})}+\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}\)
(where ‘am’ denotes solution in ammonia.)
iv. As a result, the blue colour of the solution changes to bronze and the solution becomes diamagnetic.
v. Similarly, the alkaline earth metals are also soluble in liquid ammonia which give deep blue-black coloured solutions.
M + (x + 2y) NH₃ → [M(NH₃)_x]²⁺ + 2[e(NH₃)_y]^–

Question 42.
Explain: Diagonal relationship in group 1 and group 2.
Answer:

• Elements belonging to the same group are expected to exhibit similarity and gradation in their properties.
• However, first alkali metal, lithium, and the first alkaline earth metal, beryllium, do not fulfil this expectation.
• Thus, lithium shows many differences when compared with the remaining alkali metals and shows similarity with magnesium, the second alkaline earth metal.
• Similarly, beryllium shows many differences with remaining alkaline earth metals and shows similarity with aluminium, the second element of the next main group i.e., group 13.
• The relative placement of these elements with similar properties in the periodic table is across a diagonal and thus, it is called diagonal relationship.

Question 43.
Explain diagonal relationship between lithium and magnesium with respect to:
i. Property of chlorides
ii. Thermal decomposition of their carbonates
Answer:
Both lithium and magnesium show similarities in various physical and chemical properties as follows:
i. Property of chlorides: Chlorides of lithium (LiCl) and magnesium (MgCl₂) are deliquescent as group 2 elements form deliquescent chlorides. These chlorides form corresponding hydrates (LiCl.2H₂O and MgCl₂.8H₂O) on crystallization from their aqueous solutions.

ii. Thermal decomposition of carbonates: Heating of lithium carbonate and magnesium carbonate results in their easy decomposition to form corresponding oxides and carbon dioxide (CO₂).

Question 44.
Mention the properties of lithium that differ from rest of the alkali group metals.
Answer:

• Reaction with nitrogen: Only lithium from alkali group metals reacts with nitrogen present in the air on heating, while rest of the members do not react with nitrogen.
• Thermal decomposition of carbonates: Alkali metal carbonates show no reaction on heating, while lithium carbonate decomposes on heating to form the corresponding oxide and liberate carbon dioxide gas.
• Property of chlorides: Lithium (LiCl) is deliquescent and forms corresponding hydrate (LiCl.2H₂O). Other alkali chlorides are not deliquescent and do not form hydrates.

Question 45.
Write a note on the diagonal relationship between Be and Al.
OR
What are the similarities between beryllium and aluminium?
Answer:
i. Beryllium is placed in the group 2 and period 2 of the modem periodic table. It resembles aluminium which is placed in group 13 and period 3.

ii. Due to nearly same charge to radius ratio of their ions, beryllium (\(\frac {2}{31}\) = 0.065) and aluminium (\(\frac {3}{53.55}\) = 0.056) exhibit diagonal relationship.
iii. Due to diagonal relationship, Be and Al show following similarities in their properties:
a. Nature of bonding: Both Be and Al have tendency to form covalent chlorides.

b. Lewis acids: BeCl₂ and AlCl₃ act as Lewis acids.
c. Solubility in organic solvents: BeCl₂ and AlCl₃ are soluble in organic solvents.
d. Nature of oxide: Both Be and Al form amphoteric oxides.

Question 46.
Explain the amphoteric nature of aluminium oxide with the help of reactions.
Answer:
Al₂O₃ (magnesium oxide) reacts with both acid (HCl) as well as base (NaOH) to form the corresponding products and therefore, it is amphoteric in nature.

Question 47.
Beryllium shows many differences with other alkaline earth metals. Discuss these differences with respect to chlorides and oxides.
Answer:
1. Properties of chlorides: Beryllium chloride is covalent whereas chlorides formed by other alkaline earth metals are ionic in nature. Beryllium chloride is a strong Lewis acid whereas chlorides formed by other alkaline earth metals are not Lewis acids. Beryllium chloride is soluble in organic solvents whereas chlorides formed by other alkaline earth metals are insoluble in organic solvents.

2. Properties of oxide: Beryllium oxide is amphoteric whereas oxides formed by other alkaline earth metals are basic in nature.

Question 48.
Complete the following chemical equations.

Answer:

Question 49.
Write the uses of
i. alkali metals
ii. alkaline earth metals
Answer:
i. Uses of alkali metals:

• Lithium metal is used in long-life batteries used in digital watches, calculators and computers.
• Liquid sodium is used for heat transfer in nuclear power stations.
• Potassium chloride is used as a fertilizer.
• Potassium is used in manufacturing potassium superoxide (KO₂) for oxygen generation. It is good absorbent of carbon dioxide.
• Caesium is used in photoelectric cells.

ii. Uses of alkaline earth metals:

• Beryllium is used as a moderator in nuclear reactors.
• Alloy of magnesium and aluminium is widely used as structural material and in aircrafts.
• Calcium ions are important ingredient in biological system, essential for healthy growth of bones and teeth.
• Barium sulphate is used in medicine as barium meal for intestinal X-ray.
• Radium is used in radiotherapy for cancer treatment.

Question 50.
State the importance of sodium and potassium in biological system.
Answer:

• Sodium ion is present as the largest supply in all extracellular fluids. These fluids provide medium for transporting nutrients to the cells.
• The concentration of sodium ion in extracellular fluid regulates the flow of water across the membrane.
• Sodium ions participate in the transmission of nerve signals.
• Potassium ions are the most abundant ions within the cells. They are required for maximum efficiency in the synthesis of proteins and also in oxidation of glucose.

Question 51.
How are the following ions of group 2 elements biologically important?
i. Mg²⁺
ii. Ca²⁺
Answer:
i. Magnesium ion (Mg²⁺)

• Mg²⁺ ions are important part of chlorophyll in green plants.
• They play an important role in the breakage of glucose and fat molecules, synthesis of proteins with enzymes and regulation of cholesterol level.

ii. Calcium ion (Ca²⁺)

• Ca²⁺ ions are important for bones and teeth in the form of apatite [Ca₃(PO₄)₂].
• They play an important role in blood clotting.
• Ca²⁺ ions are required for contraction and stretching of muscles.
• They are also required to maintain the regular beating of heart.

Question 52.
Explain Solvay process for manufacture of sodium carbonate.
Answer:
Sodium carbonate (Na₂CO₃) is commercially prepared by Solvay process. Preparation of sodium carbonate by Solvay process involves two stages.
i. In the first stage of Solvay process, carbon dioxide gas is bubbled through a concentrated solution of NaCl which is saturated with NH₃. This results in the formation of ammonium bicarbonate. Crystals of sodium bicarbonate separate as a result of the following reactions.


ii. Ammonium bicarbonate and sodium chloride undergoes double decomposition reaction to form sodium bicarbonate. As sodium bicarbonate has low solubility, it precipitates out in the form of crystals.
iii. In the second stage, the separated crystals of sodium bicarbonate are heated to obtain sodium carbonate (Na₂CO₃).

iv. NH₄Cl obtained in this process is treated with slaked lime, Ca(OH)₂, to recover NH₃ while CaCl₂ is obtained as a byproduct.

Question 53.
How is ammonia recovered in Solvay process? Name the important by-product obtained in the step?
Answer:
i. Ammonium chloride (NH₄Cl) is obtained during the Solvay process which is used for the preparation of Na₂CO₃. When NH₄Cl is treated with slaked lime, Ca(OH)₂, ammonia is recovered.

ii. Calcium chloride is obtained as an important by-product in this reaction.

Question 54.
Why potassium carbonate cannot be obtained by Solvay process?
Answer:
Potassium hydrogen carbonate (KHCO₃) is highly water soluble and cannot be precipitated out by reacting with potassium choride (KCl) and hence, potassium carbonate (K₂CO₃) cannot be obtained by Solvay process.

Question 55.
What is the action of heat on crystalline sodium carbonate (washing soda)?
Answer:
i. On heating washing soda (decahydrate of sodium carbonate) up to 373 K, it loses water molecules to form corresponding monohydrate.

ii. On heating above 373 K, monohydrate further loses water and changes into white anhydrous powder called soda ash.

Question 56.
Give reason: Aqueous solution of sodium carbonate is alkaline in nature.
Answer:
i. Sodium carbonate is hydrolysed by water as shown in the reaction given below.

ii. One of the products formed as a result of hydrolysis is NaOH which is a strong base.
Hence, aqueous solution of sodium carbonate is alkaline in nature due to formation of strong base (NaOH).

Question 57.
What are the uses of sodium carbonate?
Answer:
Uses of sodium carbonate:

• Due to its alkaline properties, sodium carbonate has an emulsifying effect on grease and dirt and hence, it is used as a cleaning material.
• It is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates.
  For example: Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaHCO_3(aq)
• It is used for commercial production of soap and caustic soda.
• Sodium carbonate is used as an important laboratory reagent.

Question 58.
Describe the preparation of sodium hydroxide by Castner-Kellner process.
OR
Explain the electrolysis method for preparation of sodium hydroxide.
Answer:
i. Sodium hydroxide (caustic soda) is commercially obtained by the electrolysis of aqueous sodium chloride solution (brine) in Castner-Kellner cell (mercury cathode cell).
ii. In Castner-Kellner cell, mercury is used as cathode, carbon rod as anode and brine solution is used as electrolyte which is subjected to electrolysis.

iii. During electrolysis, the following reactions take place:
a. At cathode: Sodium ions get reduced to metallic sodium, which combines with mercury to form sodium amalgam (Na-Hg).
\(\mathrm{Na}^{+}+\mathrm{e}^{-} \stackrel{\mathrm{Hg}}{\longrightarrow} \mathrm{Na} \text {-amalgam }\)
b. At anode: Chloride ions are oxidized and thus, chlorine gas is evolved.
\(\mathrm{Cl}^{-} \longrightarrow \frac{1}{2} \mathrm{Cl}_{2}+\mathrm{e}^{-}\)

iv. Sodium amalgam is then treated with water to obtain sodium hydroxide and hydrogen gas.

Question 59.
Enlist the physical properties of sodium hydroxide.
Answer:
Physical properties of sodium hydroxide:

• Sodium hydroxide (NaOH) is a white deliquescent solid.
• It has a melting point of 591 K.
• It is highly water soluble and gives a strongly alkaline solution.
• The surface of sodium hydroxide solution absorbs atmospheric CO₂ to form Na₂CO₃.

Question 60.
Explain how sodium hydroxide is commercially important.
Answer:
Commercial uses of sodium hydroxide:

• Sodium hydroxide is used in purification of bauxite (the aluminium ore).
• It is used in commercial production of soap, paper, artificial silk and many chemicals.
• It is used for mercerising cotton fabrics.
• It is used in petroleum refining.
• It is also used as an important laboratory reagent.

Question 61.
Calcium carbonate occurs naturally in which forms?
Answer:
Calcium carbonate (CaCO₃) occur naturally in the form of chalk, limestone and marble.

Question 62.
Describe the various methods used for preparation of calcium carbonate.
Answer:
i. a. Calcium carbonate is prepared by passing carbon dioxide through solution of calcium hydroxide (slaked lime). This results in the formation of water insoluble solid calcium carbonate.

b. However, excess carbon dioxide transforms the precipitate of CaCO₃ into water-soluble calcium bicarbonate and therefore, it has to be avoided.

ii. Calcium carbonate can also be prepared by adding solution of calcium chloride to a solution of sodium carbonate. This results in the formation of calcium carbonate as precipitate.

Question 63.
Why controlled addition of CO₂ is essential during preparation of calcium carbonate from slaked lime?
Answer:
When excess of CO₂ is present, it leads to the formation of water-soluble calcium bicarbonate.

Hence, while preparing calcium carbonate from slaked lime, controlled addition of CO₂ is essential.

Question 64.
Mention some physical properties of calcium carbonate.
Answer:

• Calcium carbonate is soft, light, white powder.
• It is practically insoluble in water.

Question 65.
What happens when:
i. calcium carbonate is thermally decomposed?
ii. calcium carbonate reacts with dilute mineral acids?
Answer:
i. When calcium carbonate is heated to 1200 K, it decomposes into calcium oxide along with evolution of carbon dioxide gas.

ii. Calcium carbonate reacts with dilute mineral acids such as HCl and H₂SO₄ to give the corresponding calcium salt and liberate carbon dioxide gas.

Question 66.
Give the important uses of calcium carbonate.
Answer:

• Calcium carbonate in the form of marble is used as building material.
• It is used in the manufacture of quicklime (CaO) which is the major ingredient of cement.
• A mixture of CaCO₃ and MgCO₃ is used as flux in the extraction of metals from their ores.
• It is required for the manufacture of high-quality paper.
• It is an important ingredient in toothpaste, chewing gum, dietary supplements of calcium and filler in cosmetics.

Question 67.
Match the pairs.

	Column A		Column B
i.	Castner-Kellner cell	a.	Na2CO
ii.	Slaked lime	b.	CaCO3
iii.	Solvay process	c.	NaOH
iv.	Limestone	d.	Ca(OH)2

Ans:
i – c,
ii – d,
iii – a,
iv – b

Question 68.
How is hydrogen peroxide prepared by the action of cold dilute H₂SO₄ on
i. Hydrated barium peroxide?
ii. sodium peroxide (Merck process)?
Answer:
Preparation of hydrogen peroxide by the action of cold dilute H₂SO₄ on
i. hydrated barium peroxide: When hydrated barium peroxide is treated with ice-cold dilute sulphuric acid, the precipitate of barium sulphate is obtained. This precipitate is then filtered off to get hydrogen peroxide solution.

ii. Sodium peroxide (Merck process): When small quantity of sodium peroxide is added to ice-cold solution of dilute sulphuric acid with stirring, it gives hydrogen peroxide.

Question 69.
Explain how hydrogen peroxide can be obtained by electrolysis method.
Answer:
i. H₂O₂ can be manufactured by electrolysis of 50% H₂SO₄. In this method, 50% solution of H₂SO₄ is subjected to an electrolytic oxidation to form peroxydisuiphuric acid at anode.

ii. On hydrolysis, peroxy sulphuric yields hydrogen peroxide.

iii. This method can be used for the laboratory preparation of D₂O₂.

Question 70.
Describe the industrial method for preparation of hydrogen peroxide.
OR
How is hydrogen peroxide obtained from 2-ethylanthraquinol?
Answer:
i. Industrially hydrogen peroxide is prepared by air-oxidation of 2-ethylanthraquinol.

ii. 2-Ethylanthraquinone is reduced back to 2-ethylanthraquinol by catalytic hydrogenation.

Question 71.
What are the physical properties of hydrogen peroxide?
Answer:

• Pure H₂O₂ is a very pale blue coloured liquid.
• Its boiling point is 272.4 K.
• H₂O₂ is miscible in water and forms a hydrate (H₂O₂. H₂O).

Question 72.
How is strength of H₂O₂ solution expressed?
Answer:

• Strength of aqueous solution of H₂O₂ is expressed in ‘volume’ units i.e., volume strength.
• The commercially marketed 30% (by mass) solution of H₂O₂ has volume strength of 100 volume.
• It means that 1 mL of 30% solution of H₂O₂ will give 100 mL oxygen at STP.

Thus, Volume strength refers to the volume of oxygen (O₂) in litres at STP obtained by decomposition of 1 litre of the sample.

Question 73.
Write reactions depicting oxidising and reducing action of hydrogen peroxide in acidic medium.
Answer:
H₂O₂ acts as a mild oxidising as well as reducing agent.
i. Oxidising action of H₂O₂ in acidic medium.
\(2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

ii. Reducing action of H₂O₂ in acidic medium.
\(2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{O}_{2}\)

Question 74.
Enlist uses of hydrogen peroxide.
Answer:

• Hydrogen peroxide is used as mouthwash, germicide and mild antiseptic.
• It is used as a preservative for milk and wine.
• It is used as a bleaching agent for soft materials, due to its mild oxidising property.
• Due to its reducing property, is used as an antichlor to remove excess chlorine from fabrics which have been bleached by chlorine.
• Nowadays it is used in environmental chemistry for pollution control and restoration of aerobic condition of sewage water.

Question 75.
Describe preparation and properties of lithium aluminium hydride (LAH).
Answer:
i. Preparation: Lithium hydride when treated with aluminium chloride, gives lithium aluminium hydride, (LiAlH₄).

ii. Properties:

• Lithium aluminium hydride is a colourless solid.
• It reacts violently with water and even with atmospheric moisture.

Question 76.
How is lithium aluminium hydride (LAH) useful in organic synthesis?
Answer:
i. LAH is a source of hydride (H^–) and therefore, it is used as a reducing agent in organic synthesis.
For example:

ii. It is useful in the preparation of PH₃ (phosphine).
4PCl₃ + 3LiAlH₄ → 4PH₂ + AlCl₃ + LiCl

Question 77.
Complete the following reactions by mentioning the reagent/reaction conditions under which these reactions are carried out.

Answer:

Question 78.
Calculate % (by mass) of a H₂O₂ solution which is 45.4 volume.
Answer:
Given: 45.4 volume H₂O₂ solution
To find: % (by mass) of H₂O₂
Formula: Percentage (%) by mass = \(\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100\)
calculation: 45.4 volume H₂O₂ solution means 1 L of this solution will liberate 45.4 L of O₂ at STP.
Hydrogen peroxide (H₂O₂) decomposes as:

Ans: % (by mass) of H₂O₂ in 45.4 volume H₂O₂ solution is 13.6%.

Question 79.
Calculate the strength (g/L) of 20 volume solution of hydrogen peroxide.
Solution:
Given: 20 volume H₂O₂ solution
To find: Strength of H₂O₂ (g/L)
Formula: 20 volume H₂O₂ solution means that 1 L of this solution will liberate 20 L of oxygen at S.T.P. Let us calculate the amount of H₂O₂ (in grams) which gives 20 L of oxygen at S.T.P. This amount will be present in 1 L of 20 volume solution of H₂O₂.
Hydrogen peroxide (H₂O₂) decomposes as:

22.7 L of O₂ at S.T.P. is produced from H₂O₂ = 68 g
20 L of O₂ at S.T.P. is produced from H₂O₂ = \(\frac {68}{22.7}\) × 20 = 59.912g = 59.912 g/litre
Ans: Strength of H₂O₂ in 20 volume H₂O₂ solution is 59.912 g/L.

Question 80.
Calculate the volume strength of a 5% solution of hydrogen peroxide.
Solution:
Given: 5% solution of H₂O₂
To find: Volume strength of H₂O₂ solution
Calculation: 100 mL of solution contains 5 g of H₂O₂
1000 mL of solution will contain \(\frac {5}{100}\) × 1000 = 50 g of H₂O₂

68 g of H₂O₂ will give O₂ at S.T.P. = 22.7 L
50 g of H₂O₂ is present in 1000 mL of H₂O₂ or 1 L of H₂O₂
50 g of H₂O₂ will give O₂ at S.T.P. = \(\frac {22.7}{68}\) × 50 = 16.691 L
∴ 1 L of H₂O₂ gives O₂ at S.T.P. = 16.691 L
∴ Strength of H₂O₂ = 16.691 volume
Ans: The given 5% H₂O₂ solution is equivalent to 16.691 volume solution of hydrogen peroxide.

Question 81.
Naina is a school going kid. Every morning her mother makes her drink a glass of milk. When she asked her mother that why she has to drink a glass of milk daily, her mother told her that it is beneficial in maintaining healthy bones and teeth. Regular consumption of milk is recommended because it is a rich source of calcium.
i. In which form is calcium important for bones and teeth?
ii. Calcium belongs to which family in the modern periodic table?
iii. Write its electronic configuration.
iv. Calcium contain how many valence electrons?
v. Give any two-biological importance of calcium.
Answer:
i. Calcium is important for bones and teeth in the form of apatite [Ca(PO₄)₂].
ii. It belongs to the family of alkaline earth metals in the modem periodic table.
iii. Electronic configuration of 20Ca is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².
iv. Calcium contains two valence electrons as it has two electrons in its outermost shell (4s).

v. Calcium ion (Ca²⁺)

• Ca²⁺ ions are important for bones and teeth in the fonn of apatite [Ca₃(PO₄)₂].
• They play an important role in blood clotting.
• Ca²⁺ ions are required for contraction and stretching of muscles.
• They are also required to maintain the regular beating of heart.

Multiple Choice Questions

1. Of all the elements present in the periodic table, ………… has the simplest atomic structure.
(A) lithium
(B) beryllium
(C) hydrogen
(D) helium
Answer:
(C) hydrogen

2. Electronic configuration of hydrogen is similar to that of ……………
(A) transition elements
(B) inert gases
(C) alkaline earth metals
(D) alkali metals
Answer:
(D) alkali metals

3. Isotopes are atoms of the same element having different ………….. number.
(A) neutron
(B) proton
(C) electron
(D) Both (A) and (B)
Answer:
(A) neutron

4. Tritium, \(\left({ }_{1}^{3} \mathrm{H}\right)\) ……………
(A) is an isotope of hydrogen
(B) contains one electron, one proton and two neutrons
(C) is a beta particle emitter
(D) all of these
Answer:
(D) all of these

5. In the electrolysis of acidified water using, ………….. is liberated at the anode.
(A) dihydrogen
(B) sulphate ions
(C) oxygen
(D) chloride ions
Answer:
(C) oxygen

6. Water gas is a mixture of ………….
(A) CO + H₂
(B) CO₂ + H₂
(C) O₂ + H₂
(D) CO + O₂
Answer:
(A) CO + H₂

7. During production of dihydrogen by water-gas shift reaction, which of the following is present as an impurity?
(A) Carbon monoxide
(B) Carbon dioxide
(C) Calcium carbonate
(D) Calcium oxide
Answer:
(B) Carbon dioxide

8. Solution is used to remove carbon dioxide present in the mixture along with dihydrogen.
(A) Sodium hydroxide
(B) Hydrochloric acid
(C) Magnesium chloride
(D) Sodium arsenite
Answer:
(D) Sodium arsenite

9. The reaction between dihydrogen and …………… is highly exothermic.
(A) halogens
(B) dioxygen
(C) dinitrogen
(D) metals
Answer:
(B) dioxygen

10. The elements of group 1 and group 2 belong to which block of the modem periodic table?
(A) d-block
(B) s-block
(C) p-block
(D) f-block
Answer:
(B) s-block

11. Which of the following is NOT an alkaline earth metal?
(A) Beryllium
(B) Barium
(C) Calcium
(D) Caesium
Answer:
(D) Caesium

12. The common oxidation state for alkali metals is …………….
(A) +2
(B) +1
(C) +3
(D) +4
Answer:
(B) +1

13. All alkaline earth metals have ………….. valence electrons in the outermost orbit.
(A) one
(B) two
(C) three
(D) four
Answer:
(B) two

14. Electronic configuration of potassium with respect to nearest noble gases is …………..
(A) [He]2s¹
(B) [Ne]3s¹
(C) [Ar]4s¹
(D) [Kr]5s¹
Answer:
(C) [Ar]4s¹

15. Which of the following is radioactive alkali metal?
(A) Rubidium
(B) Caesium
(C) Francium
(D) Beryllium
Answer:
(C) Francium

16. Which of the following element is rarest amongst s-block elements?
(A) Strontium
(B) Barium
(C) Radium
(D) Calcium
Answer:
(C) Radium

17. Which of the following is FALSE?
(A) Alkali metals readily loose electron to form monovalent M⁺ ions.
(B) In a group, from Li to Cs, atomic and ionic radii increase with atomic number.
(C) The monovalent ions of alkali metals are larger in size than the parent atoms.
(D) Ionization enthalpies decrease down the group from Li to Cs.
Answer:
(C) The monovalent ions of alkali metals are larger in size than the parent atoms.

18. The first ionization enthalpies of alkaline earth metals are …………. than those of the corresponding alkali metals.
(A) higher
(B) lower
(C) same
(D) none of these
Answer:
(A) higher

19. Which of the following alkaline earth metal does NOT react with water?
(A) Sr
(B) Mg
(C) Ca
(D) Be
Answer:
(D) Be

20. Oxides and hydroxides of alkaline earth metals except beryllium are ………….. in nature.
(A) acidic
(B) basic
(C) amphoteric
(D) neutral
Answer:
(B) basic

21. ………… is an excellent absorbent of carbon dioxide.
(A) KO₂
(B) KCl
(C) KOH
(D) KHCO₃
Answer:
(A) KO₂

22. Lithium shows diagonal relationship with ……………
(A) beryllium
(B) magnesium
(C) calcium
(D) boron
Answer:
(B) magnesium

23. The diagonal relationship between Li and Mg is due to the similarity in ……………
(A) ionic sizes
(B) electronegativity value
(C) polarizing power
(D) all of the above
Answer:
(D) all of the above

24. The alkali metal that reacts with nitrogen directly to form nitride is ……………
(A) Li
(B) Na
(C) K
(D) Rb
Answer:
(A) Li

25. In the Solvay process, the chief products are ……………
(A) CaCO₃ and Ca(HCO₃)₂
(B) Na₂CO₃ and NaHCO₃
(C) Na₂SO₄ and NaHSO₄
(D) CaCl₂ and Ca(NO₃)₂
Answer:
(B) Na₂CO₃ and NaHCO₃

26. In Castner-Kellner process for preparation of sodium hydroxide, ………….. is subjected to electrolysis.
(A) NaCl
(B) NaOH
(C) Na₂O
(D) Na2CO₃
Answer:
(A) NaCl

27. Which of the following method of preparation of H₂O₂ is known as Merck’s method?
(A) BaO₂.8H₂O(s) + H₂SO₄(aq) → BaSO₄(s) + H₂O₂(aq) + 8H₂O(l)
(B) Na₂O₂ + H₂SO₄ → Na₂SO₄ + H₂O₂
(C) BaO₂ + H₂O + CO₂ → BaCO₃↓ + H₂O₂
(D) 3BaO₂ + 2H₃PO₄ → Ba₃(PO₄)₂↓ + 3H₂O₂
Answer:
(B) Na₂O₂ + H₂SO₄ → Na₂SO₄ + H₂O₂

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 10 States of Matter Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 10 States of Matter

Question 1.
What are the three distinct physical forms of a substance?
Answer:
The three distinct physical forms of a substance are solid, liquid and gas.

Question 2.
What are the different forms (physical states) in which water exists?
Answer:
Water exists in the three different forms solid ice, liquid water and gaseous vapours.

Question 3.
Give the differences between the three states of matter.
OR
State the properties of three states of matter.
Answer:

Question 4.
With suitable diagram, explain how three states of matter are interconvertible by exchange of heat.
Answer:

• On heating, solid changes to liquid, which on further heating changes to gas.
• On cooling, gas condenses to liquid, which on further cooling change to solid.

Question 5.
What are intermolecular forces? Explain the role of these forces in different states of matter.
Answer:

• Intermolecular forces are the attractive forces as well as repulsive forces present between the neighbouring molecules.
• The attractive force decreases with the increase in distance between the molecules.
• The intermolecular forces are strong in solids, less strong in liquids and very weak in gases. Thus, the three physical states of matter can be determined as per the strength of intermolecular forces.
• The physical properties of matter such as melting point, boiling point, vapor pressure, viscosity, evaporation, surface tension and solubility can be studied with respect to the strength of attractive intermolecular forces between the molecules.
• During the melting process, intermolecular forces are partially overcome, whereas they are overcome completely during the vapourization process.

Question 6.
Name the different types of intermolecular forces.
Answer:
The types of intermolecular forces are as follows:

1. Dipole-dipole interactions
2. Ion-dipole interactions
3. Dipole-induced dipole interactions
4. Hydrogen bonding
5. London dispersion forces

Question 7.
Write a short note on dipole moment.
Answer:
Dipole moment:
i. Dipole moment (µ) is the product of the magnitude of the charge (Q) and the distance between the centres of positive and negative charge (r).
ii. It is designated by a Greek letter (µ) (mu) and its unit is Debye (D).
iii. Dipole moment is a vector quantity and is depicted by a small arrow with tail in the positive centre and head pointing towards the negative centre.
iv. For example, HCl is a polar molecule.

The crossed arrow above the structure represents an electron density shift. Thus, polar molecules have permanent dipole moments.

Question 8.
Explain dipole-dipole interactions.
Answer:
Dipole-dipole interactions:
i. When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of another polar molecule. Interaction between such molecules is termed as dipole-dipole interaction.
ii. These forces are generally weak, with energies of the order of 3-4 kJ mol^-1 and are significant only when molecules are in close contact, i.e., in a solid or a liquid state.
iii. For example, C₄H₉Cl, (butyl chloride), CH₃ – O – CH₃ (dimethyl ether), ICl (iodine chloride, B.P. 27 °C), are dipolar liquids.
iv. The molecular orientations due to dipole-dipole interaction in ICl liquid is shown in the following figure:

v. More polar the substance, greater the strength of its dipole-dipole interactions.

Question 9.
Explain the effect of dipole moment on boiling point with an example.
Answer:
Higher the dipole moment, stronger are the intermolecular forces. Thus, higher is the boiling point.
e.g. Dipole moment of dimethyl ether (CH₃ – O – CH₃) is 1.3 D while that of ethane (CH₃ – CH₂ – CH₃) is 0.1 D. Since, dipole moment of dimethyl ether is higher than that of ethane, the boiling point of dimethyl ether is higher than that of ethane.
Note: Dipole moments and boiling points of some compounds:

Question 10.
Explain ion-dipole interactions.
Answer:
Ion-dipole interactions:
i. An ion-dipole force is the result of electrostatic interactions between an ion (cation or anion) and the partial charges on a polar molecule.
ii. The strength of this interaction depends on the charge and size of an ion. It also depends on the magnitude of dipole moment and size of the molecule.
iii. Ion-dipole forces are particularly important in aqueous solutions of ionic substances. When an ionic compound is dissolved in water, the ions get separated and surrounded by water molecules. This process is called hydration of ions.
iv. For example, Na⁺ ion (cation) – H₂O interaction is shown in the following figure:

v. The charge density on Na+ is more concentrated than the charge density on Cl^– because Na⁺ is smaller in size than Cl^–. This makes the interaction between (Na⁺) and negative end of the polar H₂O molecule stronger than the corresponding interaction between (Cl^–) and positive end of the polar H₂O molecule.
vi. More the charge on cation, stronger is the ion-dipole interaction. For example, Mg²⁺ ion has higher charge and smaller ionic radius (78 pm) than Na⁺ ion (98 pm), hence Mg²⁺ ion is surrounded (hydrated) more strongly with water molecules and exerts strong ion-dipole interaction.
Thus, the strength of interaction increases with increase in charge on cation and with decrease in ionic size or radius.
Therefore, ion-dipole forces increase in the order: Na⁺ < Mg²⁺ < Al³⁺.

Question 11.
Write a short note on dipole-induced dipole interactions.
Answer:
Dipole-induced dipole interactions:
i. When polar molecules (like H₂O, NH₃) and nonpolar molecules (like benzene) approach each other, the polar molecules induce dipole in the nonpolar molecules. Hence, ‘Temporary dipoles’ are formed by shifting of electron clouds in nonpolar molecules.
ii. For example, ammonia (NH₃) is polar and has permanent dipole moment while Benzene (C₆H₆) is nonpolar and has zero dipole moment. The force of attraction developed between the polar and nonpolar molecules is of the type dipole-induced dipole interaction. This is shown in the following figure:

Question 12.
Explain briefly London dispersion forces.
Answer:
London dispersion forces:

• In nonpolar molecules and inert gases, only dispersion forces exist.
• Dispersion forces are also called as London forces or van der Waals forces.
• It is the weakest intermolecular force that develops due to interaction between two nonpolar molecules.
• In general, all atoms and molecules experience London dispersion forces, which result from the motion of electrons.
• At any given instant of time, the electron distribution in an atom may be asymmetrical, giving the atom a short-lived dipole moment. This momentary dipole on one atom can affect the electron distribution in the neighbouring atoms and induce momentary dipoles in them. As a result, weak attractive force develops.
• For example, substances composed of molecules such as O₂, CO₂, N₂, halogens, methane gas, helium and other noble gases show van der Waals force of attraction.
• The strength of London forces increases with increase in molecular size, molecular mass and number of electrons present in an atom or molecule.

Question 13.
Give reason: Benzene has zero dipole moment and has no dipole-dipole forces yet it exists in liquid state.
Answer:

• Benzene (C₆H₆) is nonpolar molecule and has zero dipole moment.
• In benzene, only London forces exist due to momentary dipoles.
• The strength of London forces increases with increase in molecular size, molecular mass and number of electrons present in an atom or molecule.
• Hence, due to the presence of London forces, benzene exists in liquid state.

Question 14.
Explain the term polarizability.
Answer:
Polarizability:

• When two nonpolar molecules approach each other, attractive and repulsive forces between their electrons and the nuclei will lead to distortions in the size of electron cloud, a property referred to as polarizability.
• Polarizability is a measure of how easily an electron cloud of an atom is distorted by an applied electric field.
• It is the property of atom.
• The ability to form momentary dipoles, that means, the ability of the molecule to become polar by redistributing its electrons is known as polarizability of the atom or molecule.

Question 15.
Describe how London dispersion forces affect the boiling points of isomeric compounds like n-pentane and neopentane.
Answer:

• More the spread out of shapes, higher the dispersion forces present between the molecules.
• London dispersion forces are stronger in a long chain of atoms where molecules are not compact.
• For example, n-Pentane boils at 309.4 K, whereas neopentane boils at 282.7 K.
• Both the substances have the same molecular formula, C₅H₁₂, but n-pentane is longer and somewhat spread out, whereas neopentane is more spherical and compact.

Question 16.
Write a short note on hydrogen bonding.
Answer:
Hydrogen bonding:

• The electrostatic force of attraction between positively polarised hydrogen atom of one molecule and a highly electronegative atom (which may be negatively charged) of other molecule is called as hydrogen bond.
• Strong electronegative atoms that form hydrogen bonds are nitrogen, oxygen and fluorine.
• A hydrogen bond is a special type of dipole-dipole attraction.
• Hydrogen bonds are generally stronger than usual dipole-dipole and dispersion forces, and weaker than true covalent or ionic bonds.
• Hydrogen bond is denoted by (….) dotted line.
  e.g. Water (H₂O) and ammonia (NH₃) show hydrogen bonding.

Question 17.
Explain intramolecular and intermolecular hydrogen bond with suitable examples.
Answer:
i. Hydrogen bond which occurs within one single molecule represents intramolecular hydrogen bond.
e.g. H-bonding in ethylene glycol:

ii. A hydrogen bond present between two like or unlike molecules represents intermolecular hydrogen bond.
e.g. H-bonding in H-F:

Question 18.
How does hydrogen bonding influence boiling points of compounds?
Answer:

• Due to the presence of hydrogen bonding in the compounds, more energy is required to break the bonds.
• Therefore, boiling point is more in case of liquid molecules containing hydrogen bond.
• Hydrogen bonds can be quite strong with energies up to 40 kJ/mol.
• The boiling point generally increases with increase in molecular mass, but the hydrides of nitrogen (NH3), oxygen (H₂O) and fluorine (HF) have abnormally high boiling points due to the presence of hydrogen bonding between the molecules.

[Note: Due to presence of H-bond, viscosity’ of liquid increases. Hydrogen bonds play vital role in determining structure and properties of proteins and nucleic acids present in all living organisms.]

Question 19.
Observe the following figure and answer the questions:

i. What do the dotted lines represent?
ii. A water molecule can form how many H-bonds?
iii. Is this an example of intramolecular H-bonding?
Answer:
i. The dotted lines represent hydrogen bonds.
ii. A water molecule can form four H-bonds.
iii. No, it is an example of intermolecular H-bonding.

Question 20.
Explain the relation between intermolecular forces and thermal energy.
Answer:

• Thermal energy is the measure of kinetic energy of the particles of matter that arises due to movement of particles.
• It is directly proportional to the temperature; that means, thermal energy increases with increase in temperature and vice versa.
• Three states of matter are the consequence of a balance between the intermolecular forces of attraction and the thermal energy of the molecules.
• If the intermolecular forces are very weak, molecules do not come together to make liquid or solid unless thermal energy is decreased by lowering the temperature.
• When a substance is to be converted from its gaseous state to solid state, its thermal energy (or temperature) has to be reduced. At this stage, the intermolecular forces become more important than thermal energy of the particles.

Note: Comparison of intermolecular forces:

Question 21.
State true or false. Correct the false statement.
i. London dispersion force is the weakest intermolecular force that develops due to interaction between two nonpolar molecules.
ii. More the charge on cation, stronger is the ion-dipole interaction.
iii. A hydrogen bond is a special type of dipole-induced dipole attraction.
iv. Thermal energy is directly proportional to the temperature.
Answer:
i. True
ii. True
iii. False,
A hydrogen bond is a special type of dipole-dipole attraction.
iv. True

Question 22.
Name the major intermolecular forces between:
i. Cl₂ and CBr₄
ii. SiH₄ molecules
iii.He atoms in liquid He
iv. HCl molecules in liquid HCl
v. He and a polar molecule
vi. Water molecules
Answer:
i. London dispersion forces
ii. London dispersion forces
iii. London dispersion forces
iv. Dipole-dipole interactions
v. Dipole-induced dipole
vi. Hydrogen bonding

Question 23.
Why is the chemistry of atmospheric gases an important subject of study?
Answer:
The chemistry of atmospheric gases is an important subject of study as it involves air pollution. O₂ in air is essential for survival of aerobic life.

Question 24.
What are the measurable properties of gases?
OR
Explain the following measurable properties of gases in detail: Mass, volume, pressure, temperature and diffusion.
Answer:
Measurable properties of gases are as follows:
i. Mass:

• The mass (m) of a gas sample is measure of the quantity of matter it contains.
• It can be measured experimentally.
• The SI unit of mass is kilogram (kg).
  1 kg = 10³ g.
• The mass of a gas is related to the number of moles (n) by the expression:
  n = \(\frac{\text { mass in grams }}{\text { molar mass in grams }}=\frac{\mathrm{m}}{\mathrm{M}}\)

ii. Volume:

• Volume (V) of a sample of gas is the amount of space it occupies.
• It is expressed in terms of different units like Litres (L), millilitres (mL), cubic centimetre (cm³), cubic metre (m³) or decimetre cube (dm³).
• The SI Unit of volume is cubic metre (m³).
• Most commonly used unit to measure the volume of the gas is decimetre cube or litre.

iii. Pressure:

• Pressure (P) is defined as force per unit area.
  Pressure = \(\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{f}}{\mathrm{a}}\)
• Pressure of gas is measured with ‘manometer’ and atmospheric pressure is measured by ‘barometer’.
• The SI unit of pressure is pascal (Pa) or Newton per metre square (N m^-2).

iv. Temperature:

• It is the property of an object that determines direction in which energy will flow when that object is in contact with another object.
• In scientific measurements, temperature (T) is measured either on the Celsius scale (°C) or the Kelvin scale (K).
• The SI unit of temperature of a gas is Kelvin (K).
• The Celsius and Kelvin scales are related by the expression: T(K) = t °C + 273.15

v. Density: Density (d) of a substance is the mass per unit volume.
d = \(\frac{\mathrm{m}}{\mathrm{V}}\)
∴ The SI unit of density is kg m^-3.
In the case of gases, relative density is measured with respect to hydrogen gas and is called vapour density.
∴ Vapour density = \(\frac{\text { Molar mass }}{2}\)

vi. Diffusion:
a. Diffusion is the process of mixing two or more gases to form a homogeneous mixture.
b. The volume of gas diffused per unit time is the rate of diffusion of that gas.

c. SI Unit for rate of diffusion is dm³ s^-1 or cm³ s^-1.

Question 25.
Convert:
i. 3.5 atm to mm Hg
ii. 1520 torr to atm
iii. 5 m³ to dm³
iv. 580 °c to Kelvin
Answer:
I. 3.5 atm to mm Hg:
1 atm = 760 mm Hg
∴ 3.5 atm = 3.5 × 760
= 2660 mm Hg

ii. 1520 torr to aim:
1 atm = 760 torr
∴ 1 torr = \(\frac {1}{760}\) atrn
∴ 1520 torr = \(\frac {1520}{760}\) = atm

iii. 5 m³ to dm³:
1 m³ = 10³ dm³
∴ 5m³ = 5 × dm³ = 5000 dm³

iv. 580 °C to Kelvin:
T(K) = t °C + 273.15
∴ T(K)= (580 °C) + 273.15 = 853.15 K

Question 26.
Name four measurable properties that are essential to study behaviour of gases.
Answer:

• Pressure
• Volume
• Temperature
• Number of moles

Question 27.
Explain Boyle’s law with the help of a diagram.
Answer:
Boyle’s law (Pressure-Volume relationship):
i. Statement: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
ii. Explanation:
The mathematical expression of Boyle’s law is:
P ∝ \(\frac{1}{\mathrm{~V}}\) (at constant T and n) k₁
∴ P = \(\frac{\mathrm{k}_{1}}{\mathrm{~V}}\) (where, k₁ is the proportionality constant)
On rearranging the above equation,
∴ PV = k₁ = constant
This implies that at constant temperature, product of pressure and volume of the fixed amount of a gas is constant.
Thus, when a fixed amount of a gas at constant temperature (T) occupying volume V₁ initially at pressure (P₁) undergoes expansion or compression, volume of the gas changes to V₂ and pressure to P₂.
According to Boyle’s law,
P₁V₁ = P₂V₂ = constant

iii. Schematic illustration of Boyle’s law:

Question 28.
Give the different graphical representations of Boyle’s law.
Answer:
i. Graph of pressure (P) versus volume (V) of a gas at constant temperature:
If the pressure (P) is plotted against volume (V) at constant temperature, a curve is obtained.

As the pressure increases, the volume decreases exponentially. The product of pressure and volume is always constant (PV = k). For a given mass of a gas, the value of k varies only with temperature.
[Note: Each curve is an isotherm as it is plotted at constant temperature, (iso = constant, therm = temperature).]

ii. Graph of PV versus pressure (P) of a gas constant temperature:
If the product of pressure and volume (PV) is plotted against pressure (P), a straight line is obtained parallel to x-axis (pressure axis).

iii. Graph of pressure (P) of a gas versus reciprocal of volume (1/V) at constant temperature:
If the pressure (P) of the gas is plotted against (1/V), a straight line is obtained passing through the origin.

[Note: At high pressure, deviation from Boyle’s law is observed in the behaviour of gases.]

Question 29.
Derive the relationship between density and pressure.
Answer:
Relationship between density and pressure:
With increase in pressure, gas molecules get closer and the density (d) of the gas increases. Hence, at constant temperature, pressure is directly proportional to the density of a fixed mass of gas.
From Boyle’s law,
PV = k₁ …….(1)
∴ V = \(\frac{\mathrm{k}_{1}}{\mathrm{P}}\)
But, d = \(\frac{\mathrm{m}}{\mathrm{V}}\)
On substituting, V from equation (2),
d = \(\frac{\mathrm{m}}{\mathrm{K}_{1}}\) × P
∴ d = k P …….(3)
where k = New constant
∴ d ∝ P
Above equation shows that at constant temperature, the pressure is directly proportional to the density of a fixed mass of the gas.

Question 30.
Write a short note on absolute temperature scale.
Answer:
Absolute temperature scale:

• Absolute temperature scale is related to Celsius temperature scale by the equation:
  T K = t °C + 273.15
• This also called thermodynamic scale of temperature.
• The units of this absolute temperature scale is called (K) in the honour of Lord Kelvin who determined the accurate value of absolute zero as -273.15 °C in the year 1854.

Question 31.
Explain Charles’ law with the help of a diagram.
Answer:
Charles’ law (Temperature-Volume relationship):
i. Statement: At constant pressure, the volume of a fixed mass of a gas is directly proportional to its temperature in Kelvin.

ii. Explanation:
For an increase of every degree of temperature, volume of the gas increases by \(\frac{1}{273.15}\) of its original value at 0 °C. This is expressed mathematically as follows:
V_t = V₀ + \(\frac{t}{273.15} V_{0}\) ………….(1)
Where V_t and V₀ are the volumes of the given mass of gas at the temperatures t °C and 0 °C. Rearranging the Eq. (1) gives

The equation (3) on rearrangement takes the following form:
\(\frac{\mathrm{V}_{\mathrm{t}}}{\mathrm{T}_{\mathrm{t}}}=\frac{\mathrm{V}_{0}}{\mathrm{~T}_{0}}\)
From this, a general equation can be written as follows:
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) …………(4)
\(\frac{\mathrm{V}}{\mathrm{T}}\) = K₂ = constant (at constant P and n)
∴ V = k₂T OR V ∝ T ……(5)
The equation (4) is the mathematical expression of Charles’ law.

iii. Schematic illustration of Charles’ law:

This shows that at constant pressure, gases expand on heating and contract on cooling.

Question 32.
Give the different graphical representation of Charles’ law.
Answer:
Graph of volume versus temperature at constant pressure:

• According to Charles’ law, the graph of volume of a gas (at given constant pressure, say P₁) versus its temperature in Celsius, is a straight line with a positive slope.
• On extending the line to zero volume, the line intercepts the temperature axis at -273.15 °C.
• At any other value of pressure, say P₂, a different straight line for the volume temperature plot is obtained, but we get the same zero-volume temperature intercept at -273.15 °C.
• The straight line of the volume versus temperature graph at constant pressure is called isobar.

[Note: Zero volume for a gas sample is a hypothetical state. In practice, all the gases get liquified at a temperature higher than -273.15 °C. This temperature is the lowest temperature that can be imagined but practically cannot be attained. It is the absolute zero temperature on the Kelvin scale (0 K).]

Question 33.
Write the statement for Gay-Lussac’s law.
Answer:
Statement for Gay-Lussac’s law:
At constant volume, pressure (P) of a fixed amount of a gas is directly proportional to its absolute temperature (T).

Question 34.
Give the mathematical expression for Gay-Lussac’s law.
Answer:
Gay-Lussac’s law (Pressure-Temperature relationship):
i. Statement: At constant volume, pressure (P) of a fixed amount of a gas is directly proportional to its absolute temperature (T).
ii. Explanation:
Gay-Lussac’s law can be mathematically expressed as:
P ∝ T
∴ P = k₃T
∴ \(\frac{\mathrm{P}}{\mathrm{T}}\) = Constant (at constant V and n)
Thus, according to Gay-Lussac’s law,
\(\frac{\mathrm{P}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2}}{\mathrm{~T}_{2}}\) = constant

Question 35.
Give the graphical representation of Gay-Lussac’s law.
Answer:
Graph of pressure versus temperature of a gas at constant volume:
When a graph is plotted between pressure (P) in atm and temperature (T) in kelvin, a straight line is obtained It is known as isochore.

Question 36.
Define the following terms:
i. Isotherm
ii. Isobar
iii. Isochore
Answer:
i. A graph of pressure (P) versus volume (V) at a constant temperature is known as isotherm.
ii. A graph of volume (V) versus absolute temperature (T) at a constant pressure is known as isobar.
iii. A graph of pressure (P) versus absolute temperature (T) at a constant volume is known as isochore.

Question 37.
State and explain Avogadro law.
Answer:
Avogadro law (Volume-Amount relationship):
i. Statement: Equal volumes of all gases at the same temperature and pressure contain equal number of molecules.
ii. Explanation:
V is directly proportional to n (number of moles) at constant ‘P’ and ‘T’.
V ∝ n
V = k₄ × n (where, k₄ is proportionality constant)
∴ \(\frac{\mathrm{V}}{\mathrm{n}}\) = constant (at constant T and P)
Note: Representation of Avogadro law

Question 38.
What is molar volume?
Answer:
The volume occupied by one mole of an ideal gas at STP is 22.414 L. This volume is known as molar volume.

Question 39.
Derive the relation between density of a gas and its molar mass.
Answer:
Relation between density of a gas and its molar mass:
According to Avogadro’s law, V ∝ n
Now, n = \(\frac{\mathrm{m}}{\mathrm{M}}\) (where, m is the mass of the gas and M is the molar mass of the gas)
∴ V ∝ \(\frac{\mathrm{m}}{\mathrm{M}}\)
M ∝ \(\frac{\mathrm{m}}{\mathrm{V}}\)
But, \(\frac{\mathrm{m}}{\mathrm{V}}\) = d (where, d is the density of the gas)
∴ d ∝ M
Thus, density of a gas is directly proportional to its molar mass.

Question 40.
The volume occupied by a given mass of a gas at 298 K is 25 mL at 1 atmosphere pressure.
Calculate the volume of the gas if pressure is increased to 1.25 atmosphere at constant temperature.
Solution:
Given: P₁ = Initial pressure = 1 atm, V₁ = Initial volume = 25 mL
P₂ = Final pressure = 1.25 atm
To find: V₂ = Final volume of the gas
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law, P₁V₁ = P₂V₂
Substituting the values of P₁, V₁ and P₂ in the above expression, we get
V₂ = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 25}{1.25}\) = 20 mL
Ans: The volume occupied by the gas is 20 mL.

Question 41.
The volume of a given mass of a gas is 0.6 dm3 at a pressure of 2 atm. Calculate the volume of the gas if its pressure is increased to 2.4 at the same temperature.
Solution:
Given: P₁ = Initial pressure = 2 atm
V₁ = Initial volume of given mass of the gas = 0.6 dm³
P₂ = Final pressure = 2.4 atm
To find: V₂ = Final volume of the gas
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
V₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2}}=\frac{2 \times 0.6}{2.4}=0.5 \mathrm{dm}^{3}\)
Ans: The volume of the given gas is 0.5 dm³.

Question 42.
What will be the minimum pressure required to compress 500 dm³ of air at 5 bar to 200 dm³ at 25 °C.
Solution:
Given: P₁ = Initial pressure = 5 bar
V₁ = Initial volume = 500 dm³; V₂ = Final volume = 200 dm³
To find: P₂ = Final pressure
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ P₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}}=\frac{5 \times 500}{200}\) = 12.5 bar
Ans: The minimum pressure required to compress 500 dm³ of air at 5 bar to 200 dm³ at 25 °C is 12.5 bar.

Question 43.
A balloon has certain volume at sea level. At what pressure (in kPa) will its volume be increased by 40% if the temperature is kept constant?
Solution:
Given: P₁ = Initial pressure = 101.325 kPa (∵ The pressure at sea level = 101.325 kPa)
V₁ = Initial volume at sea level = 100 dm³ (assumption)
V₂ = Final volume = (100 + 40) = 140 dm³
To find: P₂ = Final pressure
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}\)
∴ P₂ = \(\frac{101.325 \times 100}{140}\) = 72.375 kPa
Ans: The pressure at which volume of the given balloon will be increased by 40% at a given temperature is 72.375 kPa.

Question 44.
At 300 K, a certain mass of a gas occupies 1 × 10^-4 dm³ volume. Calculate its volume at 450 K and at the same pressure.
Solution:
Given: T₁ = Initial temperature = 300 K, V₁ = Initial volume = 1 × 10^-4 dm³,
T₂ = Final temperature = 450 K
To find: V₂ = Final volume
Formula: \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (at constant n and P)
Calculation: According to Charles’ law, at constant pressure.

Ans: The volume of given gas becomes 1.5 × 10^-4 dm³ at the temperature of 450 K and same pressure.

Question 45.
A certain mass of a gas occupies a volume of 0.2 dm³ at the temperature, x K. Calculate the volume of the gas if its absolute temperature is doubled at same pressure.
Solution:
Given: V₁ = Initial volume = 0.2 dm³, T₁ = Initial temperature = x K
T₂ = Final temperature = 2 × x = 2x K
To find: V₂ = Final volume of the gas
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The volume of given gas becomes 0.4 dm³ when the temperature is doubled.

Question 46.
The volume of a given mass of a gas at 0 °C is 0.2 dm³. Calculate its volume at 100 °C, if the pressure remains the same.
Solution:
Given: V₁ = Initial volume = 0.2 dm³, T₁ = Initial temperature = 0 °C = 273.15 K,
T₂ = Final temperature = 100 °C = 100 + 273.15 K = 373.15 K
To find: V₂ = Volume at 100 °C
Formula: \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The volume of gas at 100 °C is 0.273 dm³.

Question 47.
A glass container is sealed with a gas at 0.8 atm pressure and at 25 °C. The glass container sustains a pressure of 2 atm. Calculate the temperature to which the gas can be heated before bursting the container.
Solution:
Given: P₁ = Initial pressure = 0.8 atm, P₂ = Final pressure = 2 atm
T₁ = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
To find: T₂= Final temperature

Ans: The temperature to which the gas can be heated before bursting the container is 472 °C.

Question 48.
An 8.0 L of sample at 0 °C and 5.6 atm of pressure contains 2.0 moles of a gas. If more 1.0 mole of gas at the same temperature and pressure is added, calculate the final volume.
Solution:
Given: V₁ = Initial volume = 8.0 L
n₁ = Initial mol = 2.0 mol
n₂ = Final mol = (2.0 + 1.0) = 0.3 mol
To find: V₂ = Final volume

Ans: The final volume is 12 L.

Question 49.
What is an ideal gas equation?
Answer:
Ideal gas equation is obtained by combining three gas laws, namely, Boyle’s law, Charles’ law and Avogadro law. Mathematically, it is given as:
PV = nRT
where,
P = Pressure of gas, V = Volume of gas, n = number of moles of gas,
R = Gas constant, T = Absolute temperature of gas

Question 50.
Derive the ideal gas equation.
Answer:
According to Boyle’s law,
V ∝ \(\frac{1}{\mathrm{P}}\) (at constant T and n) ………..(1)
According to Charles’ law,
V ∝ T (at constant P and n) …….(2)
According to Avogadro’s law,
V ∝ n (at constant P and T) …….(3)
Combining relations (1), (2) and (3), we get
V ∝ \(\frac{\mathrm{nT}}{\mathrm{P}}\)
Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get
∴ V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\)
On rearranging the above equation, we get
PV = nRT
where,
P = Pressure of gas,
V = Volume of gas,
n = number of moles of gas,
R = Gas constant,
T = Absolute temperature of gas.
This is the ideal gas equation or equation of state.
[Note: In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

Question 51.
Deduce values of gas constant ‘R’ in different units.
Answer:
i. R in SI Unit (in Joules): Value of R can be calculated by using the SI units of P, V and T. Pressure P is measured in N m^-2 or Pa, volume V in meter cube (m³) and temperature T in Kelvin (K).

ii. R in litre atmosphere: If pressure (P) is expressed in atmosphere (atm) and volume in litre (L) or decimeter cube (dm³) and Temperature in kelvin (K), (that is, old STP conditions), then value of R is,
R = \(\frac{1 \mathrm{~atm} \times 22.414 \mathrm{~L}}{1 \mathrm{~mol} \times 273.15 \mathrm{~K}}\)
∴ R = 0.0821 L atm K^-1 mol^-1
OR
R = 0.0821 dm³ atm K^-1 mol^-1

iii. R in calories: We know, 1 calorie = 4.184 Joules
∴ R= \(\frac{8.314}{4.184}\) = 1.987 ≅ 2 cal K^-1 mol^-1

Question 52.
Derive the following expression:
M = \(\frac{\text { mRT }}{\text { PV }}\)
Answer:
According to ideal gas equation,
PV = nRT
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}\)
Now, for a known mass ‘m’ of gas having molar mass ‘M’, number of moles ‘n’ is given as:
n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
Therefore, \(\frac{m}{M}=\frac{P V}{R T}\)
On rearranging the equation, we get
M = \(\frac{\text { mRT }}{\text { PV }}\)

Question 53.
Derive the expression for combined gas law.
Answer:
The ideal gas equation is written as
PV = nRT …….(1)
On rearranging equation (1), we get,

The ideal gas equation used in this form is called combined gas law.

Question 54.
Derive the relation between density, molar mass and pressure.
Answer:
Relation between density, molar mass and pressure:
According to ideal gas equation,
PV = nRT …..(1)
On rearranging equation (1), we get
\(\frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\) ……….(2)
Now, n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
On substituting the value of n, equation (2) becomes
\(\frac{\mathrm{m}}{\mathrm{MV}}=\frac{\mathrm{P}}{\mathrm{RT}}\)
\(\frac{\mathrm{d}}{\mathrm{M}}=\frac{\mathrm{P}}{\mathrm{RT}}\) ……….(3)
where d = \(\frac{\mathrm{m}}{\mathrm{V}}\) = density of the gas
On rearranging the equation, we get
M = \(\frac{\mathrm{dRT}}{\mathrm{P}}\) ……(4)
This equation can be used to calculate molar mass of a gas in terms of its density.

Question 55.
State Boyle’s law in terms of density.
Answer:
Boyle’s law in terms of density is stated as ‘At constant temperature, pressure of a given mass of gas is directly proportional to its density’.

Question 56.
Derive the expression that relates partial pressure with mole fraction of a gas.
Answer:
The partial pressures of individual gases can be written in terms of ideal gas equation as follows:


Thus, partial pressure of a gas is obtained by multiplying the total pressure of mixture by mole fraction of that gas.

Question 57.
What is water vapour?
Answer:
The ‘gas’ above the surface of liquid water is described as water vapour.

Question 58.
Write a short note on aqueous tension.
Answer:
Aqueous tension:
i. When the liquid water is placed into a container and air above is pumped away and the container is sealed, then the liquid water evaporates and only water vapour remains in the above space. After sealing, the vapour pressure increases initially, then slows down as some water molecules condense back to form liquid water. After a few minutes, the vapour pressure reaches a maximum value, which is called the saturated vapour pressure. The pressure exerted by saturated water vapour is called aqueous tension (P_aq).
ii. Aqueous tension increases with increase in temperature.

Question 59.
Explain how pressure of a dry gas can be calculated using aqueous tension.
Answer:
i. When a gas is collected over water in a closed container, it gets mixed with the saturated water vapour in that space. Therefore, the measured pressure corresponds to the pressure of the mixture of that gas and the saturated water vapour in that space.
ii. Pressure of pure and dry gas can be calculated by using the aqueous tension. It is obtained by subtracting the aqueous tension from the total pressure of the moist gas.
∴ P_{Dry gas} = P_Total – P_aq
i.e., P_{Dry gas} = P_Total – Aqueous Tension
Note: Aqueous tension of water (vapour pressure) as a function of temperature.

Question 60.
A sample of N₂ gas was placed in a flexible 9.0 L container at 300 K at a pressure of 1.5 atm. The gas was compressed to a volume of 3.0 L and heat was added until the temperature reached 600 K. What is the new pressure inside the container?
Solution:
Given: V₁ = Initial volume = 9.0 L, V₂ = Final volume = 3.0 L,
P₁ = Initial pressure = 1.5 atm
T₁ = Initial temperature = 300 K, T₂ = Final temperature = 600 K
To find: P₂ = Final pressure
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law,

Ans: The new pressure inside the container is 9 atm.

Question 61.
A gas at 772 mm Hg and at 35 °C occupies a volume of 6.851 L. Calculate its volume at STP.
Solution:
Given: V₁ = Initial volume = 6.851 L
P₁ = Initial pressure = 772 mm Hg, P₂ = Final pressure = 760 mm Hg
T₁ = Initial temperature = 35 °C = 35 + 273.15 K = 308.15 K
T₂ = Final temperature = 273.15 K
To find: V₂ = Final volume
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law,

Ans: The volume of gas at STP is 6.169 L.

Question 62.
Find the temperature in °C at which volume and pressure of 1 mol of nitrogen gas becomes 10 dm³ and 2.46 atmosphere respectively.
Solution.
P = 2.46 atm, V = 10 dm³, n = 1 mol, R = 0.0821 dm³-atm K^-1 mol^-1
To find: Temperature (T)
Formula: PV = nRT
According to ideal gas equation,
PV = nRT
∴ T = \(\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{2.46 \times 10}{1 \times 0.0821}\)
T = 299.63 K
Temp, in °C = 299.63 – 273.15 = 26.48 °C
Ans: The temperature of the nitrogen gas under the given conditions is 26.48 °C.

Question 63.
Calculate the temperature of 5.0 mol of a gas occupying 5 dm³ at 3.32 bar.
(R = 0.083 bar dm³ K^-1 mol^-1)
Solution:
Given: n = number of moles = 5.0 mol, V = volume = 5 dm³
P = pressure = 3.32 bar, R = 0.083 bar dm³ K^-1 mol^-1
To find: Temperature (T)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
∴ T = \(\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{3.32 \times 5}{5.0 \times 0.083}\) = 40 K
Ans: The temperature of the gas is 40 K.

Question 64.
Calculate the number of moles of hydrogen gas present in a 0.5 dm³ sample of hydrogen gas at a pressure of 101.325 kPa at 27 °C.
Solution:
Given: V = 0.5 dm³ = 0.5 × 10^-3 m³, P = 101.325 kPa = 101.325 × 10³ Pa = 101.325 × 10³ Nm^-2
T = 27 °C = 27 + 273.15 K = 300.15 K, R = 8.314 J K^-1 mol^-1
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,

Ans: The number of moles of hydrogen gas present in the given volume is 0.020 moles.

Question 65.
A mixture of 28 g N₂, 8 g He and 40 g Ne has 20 bar pressure. What is the partial pressure of each of these gases?
Solution:
Given: m_N2 = 28 g, m_He = 8 g, m_Ne = 40 g,
P_Total = 20 bar
To find: Partial pressure of each gas
Formula: P₁ = x₁ × P_Total
Calculation: Determine the number of moles (n) of each gas using the formula: n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
Determine the mole fraction of each gas using the formula:

Ans: The partial pressure of nitrogen, helium and neon are 4 bar, 8 bar and 8 bar respectively.

Question 66.
What is an ideal gas?
Answer:
Ideal gas:

• The gases which obey’ ideal gas equation over a complete range of temperature and pressure are called ideal gases.
• For an ideal gas, the ratio of PV/RT = 1.
• In an ideal gas, there are no interactive forces between the molecules and the molecular volume is negligibly small compared to the volume occupied by the gas. The gas particles are considered as point particles.

Question 67.
What are real gases?
Answer:
Real gases:

• Gases, which do not obey ideal gas equation under all the conditions of temperature and pressure are called real gases.
• For real gases, the ratio of PV/RT will be either greater than 1 or less than 1.
• Real gases show deviation from ideal gas behaviour at higher pressures and lower temperatures.
• The intermolecular attractive forces are not negligible in real gases.
• In real gases, the actual volume of the molecules cannot be neglected as compared to the total volume of the container.

Question 68.
Explain the reason for deviations of gases from ideal behaviour.
Answer:
A deviation from the ideal behaviour is observed at high pressure and low temperature. It is due to two reasons.

• The intermolecular attractive forces are not negligible in real gases. These do not allow the molecules to collide the container wall with full impact. This results in decrease in the pressure.
• At high pressure, the molecules are very close to each other. The short range repulsive forces start operating and the molecules behave as small but hard spherical particles. The volume of the molecule is not negligible.
  Therefore, very less volume is available for molecular motion.
• At very low temperature, the molecular motion becomes slow and the molecules are attracted to each other due to the attractive force. Hence, the behaviour of the real gas deviates from the ideal gas behaviour.
• Deviation with respect to pressure can be studied by plotting pressure (P) vs volume (V) curve at a given temperature.
• From the graph, it is clear that at very high pressure, the measured volume is more than theoretically calculated volume assuming ideal behaviour. However, at low pressure, measured and theoretically calculated volumes approach each other.

Question 69.
What is compressibility factor (Z)?
Answer:
Compressibility factor (Z):
i. It is defined as the ratio of product PV and nRT.
Z = \(\frac{\mathrm{PV}}{\mathrm{nRT}}\)
ii. Deviation from ideal behaviour is measured in terms of compressibility factor.
iii. For ideal gases, Z = 1 under all conditions of temperature and pressure. Therefore, the graph of Z versus P will be a straight line parallel to pressure axis.
iv. For gases that deviate from ideal behaviour, value of Z deviates from unity.
Note: Variation of compressibility factor for some gases

Question 70.
Show that the compressibility factor can be represented as Z = \(\frac{V_{\text {real }}}{V_{\text {ideal }}}\)
Answer:
For real gas,

Thus, the compressibility factor (Z) is the ratio of actual molar volume of a gas to its molar volume if it behaved ideally, at that temperature and pressure.

Question 71.
Explain: Liquefaction of CO₂ with the help of pressure vs volume isotherm.
Answer:
Most gases behave like ideal gases at high temperature. For example, the PV curve of CO₂ gas at 50 °C is like the ideal Boyle’s law curve. As the temperature is lowered, the PV curve shows a deviation from the ideal Boyle’s law curve. At a particular value of low temperature, the gas gets liquified at certain increased value of pressure. For example, CO₂ gas liquifies at 38.98 °C and 73 atmosphere pressure. This is the highest temperature at which liquid CO₂ can exist. Above this temperature, liquid CO₂ cannot form even if very high pressure is applied. Other gases also show similar behaviour.

Question 72.
Define: Critical temperature, critical volume and critical pressure.
Answer:
i. The temperature above which a substance cannot be liquified by increasing pressure is called its critical temperature (T_c).
ii. The molar volume at critical temperature is called the critical volume (V_c).
iii. The pressure at the critical temperature is called the critical pressure (P_c).

Note: Critical constants for common gases

i. N₂ and O₂, have T_c values much below 0 °C and their P_c values are high. Consequently, liquefaction of O₂ and N₂ (and air) requires compression and cooling.
ii. The T_c value of CO₂ nearly equals the room temperature; however, its P_c value is very high. Therefore, CO₂ exists as gas under ordinary condition.

Question 73.
Water has T_c = 647.1 K and P_c = 220.6 bar. What do these values imply about the state of water under ordinary conditions?
Answer:
The T_c and P_c values of water are very high compared to the room temperature and common atmospheric pressure. As a result, water exists in liquid state under ordinary condition of temperature and pressure.

Question 74.
Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the gas particles. Critical temperatures of ammonia and carbon dioxide are 405.5 K and 304.10 K respectively. Which of these gases will liquefy first when you start cooling from 500 K to their critical temperature?
Answer:
When cooling of ammonia and carbon dioxide gas is started from 500 K, then ammonia reaches its critical temperature first (i.e., 405.5 K.) and hence, it is also the first to get liquefied.
When the cooling is continued further, carbon dioxide gas is liquefied as it reaches its critical temperature (i.e., 304.10 K).

Question 75.
CO₂ has T_c = 38.98 °C and P_c = 73 atm. How many phases of CO₂ coexist at
i. 50 °C and 73 atm
ii. 20 °C and 50 atm.
Answer:
i. 50 °C and 73 atm represent a condition for CO₂ above its T_c. Therefore, under this condition CO₂ exists only as single phase.
ii. 20° C and 50 atm represent a condition for CO₂ below its T_c. Therefore, under this condition two phases of CO₂, namely, liquid and gas can coexist.

Question 76.
In which of the following cases, water will have the highest and the lowest boiling point?
i. Water is boiled in an open vessel.
ii. Water is boiled in a pressure cooker.
iii. Water is boiled in an evacuated vessel.
Answer:
Higher the pressure to which a liquid is exposed, higher will be its boiling point. The pressure to which water is exposed is maximum in the pressure cooker and minimum in the evacuated vessel. Therefore, boiling point of water is highest in a pressure cooker and lowest in an evacuated vessel.

Question 77.
Define: Liquid state
Answer:
Liquid state is the intermediate state between solid state and gaseous state.

Question 78.
Give reason: Liquid possesses properties such as fluidity, definite volume and ability to take shape of the bottom of the container in which it is placed.
Answer:
Molecules of liquid are held together by moderately strong intermolecular forces and can move about within the boundary of the liquid. As a result, liquid possesses properties such as fluidity, definite volume and ability to take shape of the bottom of the container in which it is placed.

Question 79.
Name some measurable properties of liquid.
Answer:

• Density
• Boiling point
• Freezing point
• Vapour pressure
• Surface tension
• Viscosity.

Question 80.
What are the factors affecting vapour pressure?
Answer:
Factors affecting vapour pressure:

• Nature of liquid: Liquids having relatively weak intermolecular forces possess high vapour pressure. Such liquids are called volatile liquids.
  e. g. Petrol evaporates quickly than motor oil. Hence, petrol has higher vapour pressure than motor oil.
• Temperature: When the liquid is gradually heated, its temperature rises and its vapour pressure increases.

Question 81.
Explain how temperature affects surface tension.
Answer:
Surface tension is a temperature dependent property. When attractive forces are large, surface tension is large. Surface tension decreases as the temperature increases. With increase in temperature, kinetic energy of molecules increases. So, intermolecular forces of attraction decrease, and thereby surface tension decreases.

Question 82.
Mention some applications of surface tension.
Answer:
Applications of surface tension:

• Cleaning action of soap and detergent is due to the lowering of interfacial tension between water and oily substances. Due to lower surface tension, the soap solution penetrates into the fibre, surrounds the oily substance and washes it away.
• Efficacy of toothpastes, mouthwashes and nasal drops is partly due to presence of substances having lower surface tension. This increases the efficiency of their penetrating action.

Question 83.
Give reason: Liquid droplets acquire spherical shape.
Answer:
For a given volume of a liquid, spherical shape always imparts minimum surface area thereby reducing the surface tension. Hence, liquid droplets acquire spherical shape.

Question 84.
Define: Coefficient of viscosity
Answer:
Coefficient of viscosity is defined as the degree to which a fluid resists flow under an applied force, measured by the tangential frictional force per unit area per unit velocity gradient when the flow is laminar.

Question 85.
Describe various factors affecting viscosity of a liquid.
Answer:
Factors affecting viscosity of a liquid:
i. Temperature: Viscosity is a temperature dependent property.
Viscosity ∝ \(\frac{1}{\text { Temperature }}\)
ii. Nature of liquid: Viscosity also depends on molecular size and shape. Larger molecules have more viscosity and spherical molecules offer the least resistance to flow and therefore are less viscous. Greater the viscosity, slower is the liquid flow.

Question 86.
Describe three daily life instances where viscosity plays an important role.
Answer:

• Lubricating oils are viscous liquids. Gradation of lubricating oils is done on the basis of viscosity. A good quality lubricating oil does not change its viscosity with increase or decrease in temperature.
• Increase blood viscosity than the normal value is taken as an indication of cardiovascular disease.
• Glass panes of old buildings are found to become thicker with time near the bottom. This indicates that glass is not a solid but a supercooled viscous liquid.

Question 87.
For an experiment, a scientist fills different gases in four flasks as shown below:

i. What is the ratio of the number of molecules of the gases in flask A to flask B?
ii. Calculate the pressure exerted by nitrogen gas in flask B if the temperature is doubled.
iii. If the scientist transfers the gas in flask D to another flask of 2.5 L at 1 atm pressure, what will be the temperature of the gas in the new flask?
Answer:
i. 1:1
ii. P ∝ T (when V and n are constants)
∴ If temperature is doubled, pressure also doubles.
∴ P = 2 atm
iii. \(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (when P and n are constants)
∴ \(\frac{1}{298}=\frac{2.5}{T_{2}}\)
∴ T₂ = 745 K

Question 88.
A balloon containing 0.6 mol of helium gas has a volume of 1.5 L.
i. Assuming that the temperature and pressure remains constant, what happens to the volume of the balloon if an additional 0.6 mol of helium is added?
ii. Assuming that the temperature and pressure remains constant, what happens to the volume of the balloon if 0.3 mol of helium is removed?
Answer:
i. The volume of the balloon increases.
ii. The volume of the balloon decreases.

Question 89.
In an experiment conducted to study the diffusion of gases using same experimental conditions, following data were recorded.
Gas A: 50 cm³ of gas A takes 7 minutes to diffuse from one container to the adjacent container.
Gas B: 50 cm³ of gas B takes 10 minutes to diffuse from one container to the adjacent container.
i. What is the rate of diffusion of gas A?
ii. What is the rate of diffusion of gas B?
iii. Which gas has higher molecular mass?
Answer:
i. Volume of gas A diffused = 50 cm³
Time required for diffusion = 7 minutes = 7 × 60 seconds

∴ The rate of diffusion of gas A is 0.12 cm³ s^-1.
ii. The rate of diffusion of gas B is 0.083 cm³ s^-1.
iii. Gas B has higher molecular mass.

Multiple Choice Questions

1. Which of the following is CORRECT for both gases and liquids?
(A) Indefinite volume
(B) Definite shape
(C) Indefinite shape
(D) Definite volume
Answer:
(D) Definite volume

2. The composition of …………. in air is about 78% by volume.
(A) CO₂
(B) O₂
(C) N₂
(D) Ar
Answer:
(C) N₂

3. Which of the following expression at constant pressure represents Charles’s law?
(A) V ∝ \(\frac{1}{\mathrm{~T}}\)
(B) V ∝ \(\frac{1}{\mathrm{~T}^{2}}\)
(C) V ∝ T
(D) V ∝ d
Answer:
(C) V ∝ T

4. The pressure of 2 mole of ideal gas at 546 K having volume 44.8 L is …………….
(A) 2 atm
(B) 3 atm
(C) 7 atm
(D) 1 atm 10²³
Answer:
(A) 2 atm

5. At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen at 4 bar. The molar mass of gaseous molecule is …………….
(A) 28 g mol^-1
(B) 56 g mol^-1
(C) 112 g mol^-1
(D) 224 g mol^-1
Answer:
(C) 112 g mol^-1

6. As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
(A) Increases
(B) Decreases
(C) Remains same
(D) Becomes half
Answer:
(A) Increases

7. Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is ………….. (Atomic wt. of Cl = 35.5 u)
(A) 1.46
(B) 0.46
(C) 1.64
(D) 0.64
Answer:
(B) 0.46

8. The number of moles of H₂ in 0.224 L of hydrogen gas at STP (273 K, 1 atm) assuming ideal gas behaviour is …………..
(A) 1
(B) 0.1
(C) 0.01
(D) 0.001
Answer:
(C) 0.01

9. The volume occupied by 11.5 g of carbon dioxide at STP is approximately equal to:
(A) 5.9 L
(B) 22.5 L
(C) 86 L
(D) 259 L
Answer:
(A) 5.9 L

10. Which of the following is CORRECT regarding a fixed amount of ideal gas?
(A) Doubling the temperature, doubles the volume, provided the pressure remains the same.
(B) Doubling the temperature, halves the volume, provided the pressure remains the same.
(C) Doubling the pressure, doubles the volume, provided the temperature remains the same.
(D) Doubling the volume, doubles the pressure, provided the temperature remains the same.
Answer:
(A) Doubling the temperature, doubles the volume, provided the pressure remains the same.

11. When one mole of an ideal gas is heated from 300 K to 360 K at constant pressure of 1 atm, its volume …………….
(A) increases from V to 6.0V
(B) increases from V to 3.6V
(C) increases from V to 1.2V
(D) increases from V to 1.6V
Answer:
(C) increases from V to 1.2V

12. The partial pressure of a gas is obtained by multiplying the total pressure of mixture by …………… of that gas.
(A) molar mass
(B) moles
(C) mass
(D) mole fraction
Answer:
(D) mole fraction

13. The highest temperature at which liquid CO₂ can exist is ……………..
(A) 18.98 °C
(B) 38.98 °C
(C) 50.0 °C
(D) 73.9 °C
Answer:
(B) 38.98 °C

14. The SI unit of surface tension is ……………..
(A) Pascal
(B) N s m^-2
(C) km^-2 s
(D) N m^-1
Answer:
(D) N m^-1

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 7 Modern Periodic Table Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 1.
Mention features of Mendeleev’s periodic table.
Answer:
Features of Mendeleev’s periodic table:

• In Mendeleev’s periodic table, all 63 elements were arranged in an increasing order of their atomic masses. The serial or ordinal number of an element in the increasing order of atomic mass was referred to as its atomic number.
• Mendeleev’s periodic table consisted of vertical groups and horizontal series (now called periods).
• Elements belonging to the same group showed similar properties.
• Properties of elements in a series/period showed gradual variation from left to right.

Question 2.
Why was Mendeleev’s periodic table readily accepted by scientific community?
Answer:
Mendeleev’s periodic table was readily accepted by scientific community due to the following advantages:
i. Mendeleev had left some gaps corresponding to certain atomic numbers in the periodic table so as to maintain periodicity of the properties. When the elements corresponding to these atomic numbers were discovered, they fitted well into the gaps with their properties as predicted by Mendeleev’s periodic law.

ii. Mendeleev did not predicted presence of inert gases, however, they were discovered in later years. It was possible to accommodate inert gases in Mendeleev’s periodic table by creating an additional group without disturbing the position of other elements in his periodic table.

Question 3.
Give reason: Mendeleev’s periodic law was modified into modern periodic law.
Answer:

• Henry Moseley in 1913, studied X-ray spectra of large number of elements.
• He observed that the frequency of X-ray emitted from an element is related to atomic number (Z) of an element and not its atomic mass.
• Therefore, the atomic number, Z, was considered as more fundamental property of the atom than the atomic mass.
• As a result, Mendeleev’s periodic law was modified.

Question 4.
Define atomic number.
Answer:
Atomic number (Z) is the total number protons present in the atom of an element.

Question 5.
State the modern periodic law.
Answer:
Modern periodic law: “The physical and chemical properties of elements are a periodic function of their atomic numbers”.

Question 6.
Periods and groups present in the modern periodic table are numbered based on whose recommendation ?
Answer:
Numbering of the periods and groups in the modem periodic table is based on the recommendation provided by the International Union of Pure and Applied Chemistry (IUPAC).

Question 7.
Write a note on: Structure of the modern periodic table.
Answer:
Structure of the modern periodic table:
i. The modem periodic table also known as the Tong form of periodic table’ has number of boxes formed by the intersection of horizontal rows and vertical columns.
ii. The horizontal rows are called periods and the vertical columns are called groups.
iii. There are seven periods numbered from 1 to 7 and eighteen groups numbered from 1 to 18.
iv. There are total 118 boxes in the modem periodic table which are filled with 118 elements discovered till now including manmade elements.
v. The modem periodic table is divided into four blocks i.e., s-block, p-block, d-block and f-block.

• Two groups on the extreme left of the modem periodic table form the s-block.
• Six groups on the extreme right constitute the p-block.
• Ten groups in the centre form the d-block
• Two series at the bottom of the modem periodic table constitute the f-block. It contains fourteen elements in each series.

Question 8.
State the relationship between the modern periodic table and electronic configuration in periods.
Answer:

• The modem periodic table is based on the atomic numbers of the elements. When elements are arranged in an increasing order of atomic number (Z), periodicity is observed in their electronic configurations which reflects in the characteristic structure of the modem periodic table.
• The location of elements in the modem periodic table is correlated to quantum numbers of the last filled orbital.
• Along a period, the atomic number increases by one and one electron is added to the outermost shell which forms neutral atom of the next element.
• The period number is same as the principal quantum number ‘n’ of the valence shell of the elements.
• A period begins with filling of a particular shell and ends when the valence shell attains complete octet configuration (or duplet, in case of the first period).
• The next period begins with addition of electron to the next shell of higher energy compared to the previous period. e. g. First shell of the elements gets filled along the first period while second shell starts filling in the second period and addition of electrons continues till second shell (valence shell) attains stable electronic configuration.

Question 9.
Give reason: First period in the modern periodic table contains only two elements.
Answer:

• Elements present in the first period i.e., H and He contain only one shell which is also their valence shell and can accommodate maximum two electrons.
• As first shell can accommodate only two electrons, first period ends at He which has a complete duplet. Hence, first period in the modem periodic table contains only two elements.

Question 10.
Write names and electronic configurations of elements of first period in the modern periodic table. Identify which of them has the stable complete electronic configuration.
Answer:

• Hydrogen (H) : 1s², Helium (He) : 2s²
• Since helium has a complete duplet i.e., two electrons in its valence shell, it has the stable complete electronic configuration.

Question 11.
Explain how does the filling of electrons takes place in the modern periodic table across:
i. Second period
ii. Third period
Answer:
i. Filling of electrons in the second period:

• In the second period, electrons are filled in the second shell i.e., n = 2.
• This shell can accommodate a maximum of eight electrons and gets filled as the atomic number increases along the second period.
• The second period begins with Li (Z = 3): 1s² 2s¹ and ends up with Ne (Z = 10): 1s² 2s² 2p⁶.
• Neon has complete octet with 8 electrons in its valence shell. Therefore, the second period contains eight elements.

ii. Filling of electrons in the third period:

• The third period corresponds to the filling of the third shell i.e. n = 3.
• The third period also contains eight elements.
• It begins with the filling of electrons in the first element Na (Z = 11) : [Ne] 3s¹ and ends with the last element Ar (Z = 18) = [Ne] 3s² 3p⁶.
• The condensed electronic configurations for the elements of third period is [Ne] 3s^1-2 3p^1-6.

Question 12.
There are 18 elements in the fourth period of the modern periodic table. Explain.
Answer:

• The fourth period corresponds to the filling of fourth shell, n = 4.
• Therefore, it begins with filling of 4s subshell. The first two elements of the fourth period are K (Z = 19) : [Ar] 4s¹ and Ca (Z = 20) : [Ar] 4s².
• According to the aufbau principle, the next higher energy subshell is 3d, which can accommodate up to 10 electrons. Thus, filling of the 3d subshell results in the next 10 elements of the fourth period i.e., from Sc (Z = 21) : [Ar] 4s²3d¹ to Zn (Z = 30): [Ar] 4s²3d¹⁰.
• After filling of 3d subshell, the electrons enter the 4p subshell which can accommodate maximum 6 electrons. Hence, filling of 4p subshell results in the next 6 elements i.e., from Ga (Z = 31): [Ar] 4s²3d¹⁰ 4p¹ to Kr (Z = 36): [Ar] 4s² 3d¹⁰ 4p6.
• Thus, the elements in fourth period are: 2 elements (with 4s subshell), 10 elements (with 3d subshell) and 6 elements (with 4p subshell).
• Hence, there are 18 elements in the fourth period of the modem periodic table.

Question 13.
Why does the fifth period of the modern periodic table contain 18 electrons?
Answer:
The fifth period accommodates 18 elements as a result of successive filling of electrons in the 5s, 4d and 5p subshells.

Question 14.
What is the general trend followed while filling of electrons across a period in the modern periodic table.
Answer:

• A period begins by filling of one electron to the ‘s’ subshell of a new shell and ends when an element corresponding to the same shell attains complete octet (or duplet).
• Between these two ‘s’ and ‘p’ subshell of the valence shell, the inner subshells ‘d’ and ‘f’ are filled successively following the aufbau principle.

Question 15.
What is the subshell in which the last electron of the first element in the 6th period enters?
Answer:
The 6^th period begins by filling the last electron in the shell with n = 6. The lowest energy subshell of any shell is ‘s’. Therefore, the last electron of the first element in the 6th period enters the subshell ‘6s’.

Question 16.
How many elements are present in the 6th period? Explain.
Answer:

• The 6^th period begins by filling the last electron in the subshell ‘6s’ and ends by completing the subshell ‘6p’. Therefore, the sixth period has the subshells filled in increasing order of energy as 6s < 4f < 5d < 6p.
• The electron capacities of these subshells are 2, 14, 10 and 6, respectively. Therefore, the total number of elements in the 6^th period is 2 + 14 + 10 + 6 = 32.

Question 17.
How does electronic configuration vary down a group in the modern periodic table?
Answer:

• As we move from top to bottom in a group, a new shell gets added successively in the atom of an element. Therefore, the last electron enters in a new shell down the group.
• Hence, the general outer electronic configuration of the elements in a group remains the same. This holds true for groups 1, 2 and 3 elements.
• In the groups 13 to 18 the appropriate inner ‘d’ and ‘f’ subshells are completely filled and the general outer electronic configuration is the same down the groups 13 to 18.
• However, in the groups 4 to 12, the ‘d’ and ‘f subshells are introduced at a later stage (4th period for ‘d’ and 6th period for ‘f’) down the group. As a result, variation in the general outer configuration is introduced only at the

Note: General outer electronic configuration in groups 1 to 3 and 13 to 18.

Question 18.
On what basis is the modern periodic table divided into four blocks?
Answer:
The modem periodic table is divided into four blocks based on the subshell in which the last electron enters.

Question 19.
Why elements of group 1 and group 2 are known as s-block elements?
Answer:

• The subshell in which the last electron enters decides the block to which an element belongs.
• In group 1 and group 2 elements, the last electron is filled in the s subshell.

Therefore, the elements of group 1 and group 2 are known as s-block elements.

Question 20.
Elements belonging to which groups constitute the p-block and why?
Answer:

• Elements belonging to groups 13, 14, 15, 16, 17 and 18 constitute the p-block.
• The last electron in the p-block elements is filled in p subshell.
• As p subshell contains three degenerate p orbitals, it can accommodate up to 6 electrons.
• Therefore, the p-block elements belonging to six groups i.e., groups 13, 14, 15, 16, 17 and 18 in which last electron enters in p subshell constitute the p-block.

Question 21.
Give reason: Helium which is the first element of group 18 is placed in the p-block even though its last electron enters in s subshell.
Answer:

• Electronic configuration of helium is 1s² which indicates that it has a stable electronic configuration i.e., a complete duplet.
• The p-block ends with group 18 which is a family of inert gases having stable electronic configuration (complete octet except helium).
• Therefore, helium is placed with group 18 elements in p-block due to its stable electronic configuration even though its last electron enters in s subshell.

Question 22.
State the general outer electronic configuration of s-block and p-block elements.
Answer:
The general outer electronic configuration of s-block elements is ns^1-2.
The general electronic configuration for the p-block elements is ns²np^1-6.

Question 23.
There are total 10 groups in the d-block of the modern periodic table. Explain.
Answer:

• The d-block in the modem periodic table is formed as a result of filling the last electron in d orbital.
• As there are five orbitals in a d subshell, 10 electrons can successively be accommodated.

Hence, there are total 10 groups in the d-block of the modem periodic table i.e., group 3 to 12.

Question 24.
The last electron enters a (n-1)d orbital only after the ns subshell is completely filled. Explain.
Answer:
A d subshell is present in the shells with n ≥ 3 and according to the (n+1) rule, the energy of ns orbital is less than that of the (n-1)d orbital. As a result, the last electron enters a (n-1)d orbital only after the ns subshell is completely filled.

Question 25.
Chromium exhibit 4s¹ 3d⁵ electronic configuration instead of 4s² 3d⁴. Explain.
Answer:

• Completely filled or half-filled subshells are highly stable.
• In 4s¹ 3d⁵ configuration, both s and d subshells are half-filled.
• Thus, due to the extra stability associated with half-filled subshells, chromium exhibits 4s¹ 3d⁵ electronic configuration instead of 4s² 3d⁴.

Question 26.
What is the general outer electronic configuration of d-block and f-block elements?
Answer:
The general outer electronic configuration of the d-block elements is ns^0-2 (n-1)d^1-10 while the general outer electronic configuration of the f-block elements is ns² (n-1)d^0-1 (n-2)f^1-14.

Question 27.
Expected outer electronic configuration of europium (Eu) is 6s² 4f⁶ 5d¹. However, it exhibits different than expected outer electronic configuration.
i. Write the observed outer electronic configuration of Eu.
ii. What is the reason for this variation in electronic configuration?
Answer:
i. Observed outer electronic configuration of europium (Eu) is 6s² 4f⁷ 5d⁰.
ii. In the observed electronic configuration of Eu, 4f subshell is half-filled which is a highly stable configuration. Therefore, observed electronic configuration of Eu varies than expected.

Question 28.
Name the two series that constitute f-block.
Answer:
The f-block constitutes two series of 14 elements called the lanthanide and the actinide series which are placed one below the other.

Question 29.
State whether the following statements are true or false. Correct if false.
i. Position of the elements in the modern periodic table is related to the quantum number of their last filled orbital.
ii. Group number is same as the principal quantum number ‘n’ of the valence shell of the elements.
Answer:
i. True
ii. False
Period number is same as the principal quantum number ‘n’ of the valence shell of the elements.

Question 30.
Name the following.
i. Shortest period in the modern periodic table.
ii. Block which is placed separately at the bottom of the modern periodic table.
Answer:
i. First period
ii. f-Block

Question 31.
How can a period, group and block of the element be determined?
Answer:
The group, period and the block of the element can be determined on the basis of its electronic configuration.
i. Period: The principal quantum number of the valence shell corresponds to the period of the element.
e. g. The principal quantum number (n) of the valence shell (3s¹) of Na (1s² 2s² 2p⁶ 3s¹) is 3. This corresponds to third period.

ii. Block: The subshell in which the last electron enters, corresponds to the block of the elements (with exception being He).
e. g. The subshell 3d (in which the last electron enters) for Sc (1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²) corresponds to d block.

iii. Group: The group of the element is determined on the basis of number of electrons present in the outermost or penultimate [next to outermost, i.e. (n-1)] shell:

• For s-block elements, group number = number of valence electrons.
• For p-block elements, group number = 18 – number of electrons required to complete octet.
• For d-block elements, group number = 2 + number of (n-1)d electrons.

Question 32.
Outer electronic configurations of a few elements are given below. Explain them and identify the period, group and block in the periodic table to which they belong.
₂He: 1s², ₅₄Xe: 5s²5p⁶, ₁₆S: 3s²3p⁴, ₇₉Au: 6s¹5d¹⁰
Answer:
i. ₂He: 1s²
Here, n = 1. Therefore, ₂He belongs to the 1^st period.
The shell n = 1 has only one subshell, namely 1s. The outer electronic configuration 1s² of ‘He’ corresponds to the maximum capacity of 1s, the complete duplet. Therefore, He is placed at the end of the 1^st period in the group 18 of inert gases. So, ‘He’ belongs to p-block.

ii. ₅₄Xe: 5s²5p⁶
Here, n = 5. Therefore, ₅₄Xe belongs to the 5^th period.
The outer electronic configuration. 5s²5p⁶ corresponds to complete octet. Therefore, ₅₄Xe is placed in group 18 and belongs to p-block.

iii. ₁₆S: 3s²3p⁴
Here, n = 3. Therefore, ₁₆S belongs to the 3^rd period. The 3p subshell in ‘S’ is partially filled and short of completion of octet by two electrons. Therefore, ‘S’ belongs to (18 – 2) = 16^th group and p-block.

iv. ₇₉AU: 6s¹5d¹⁰
Here n = 6. Therefore, ‘Au’ belongs to the 6^th period.
The sixth period begins with filling of electron into 6s and then into 5d orbital.
The outer configuration of ‘Au’: 6s¹ 5d¹⁰ implies that (1 + 10) = 11 electrons are filled in the outer orbitals to give ‘Au’. Therefore ‘Au’ belongs to the group 11.
As the last electron has entered ‘d’ orbital ‘Au’ belongs to the d-block.

Question 33.
Predict the block, periods and groups to which the following elements belong.
i. Mg (Z = 12)
ii. V (Z = 23)
iii. Sb (Z = 51)
iv. Rn (Z = 86)
v. Na (Z = 11)
vi. Cl (Z = 17)
Answer:
i. Mg (Z = 12): Atomic number of Mg is 12. Electronic configuration is 1s² 2s² 2p⁶ 3s².
Block: Since the last electron enters s subshell (3 s), Mg belongs to s-block.
Period: n = 3. Therefore, it is present in the third period.
Group: For s-block element, group number = number of valence electrons = 2. Hence, it belongs to group 2.

ii. V (Z = 23): Atomic number of V is 23. Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s².
Block: Since the last electron enters d subshell (3d), V belongs to d-block.
Period: n = 4. Therefore, it is present in the fourth period.
Group: For d-block elements, group number = 2 + number of (n – 1)
d electrons = 2 + 3 = 5. Hence, it belongs to group 5.

iii. Sb (Z = 51): Atomic number of Sb is 51.
Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p³.
Block: Since the last electron enters p subshell (5p), Sb belongs to p-block.
Period: n = 5. Therefore, it is present in the fifth period.
Group: For p block elements, group number = 18 – number of electrons required to complete octet
= 18 – 3 = 15. Hence it belongs to group 15.

iv. Rn (Z = 86): Atomic number of Rn is 86.
Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6p⁶.
Block: Since the last electron enters p subshell (6p), Rn belongs to p-block.
Period: n = 6. Therefore, it is present in the sixth period.
Group: For p block elements, group number = 18 – number of electrons required to complete octet
= 18 – 0 = 18. Hence, it belongs to group 18.

v. Na (Z = 11): Atomic number of Na is 11. Electronic configuration is 1s² 2s² 2p⁶ 3s¹.
Block: Since the last electron enters s subshell (3s), Na belongs to s-block.
Period: n = 3. Therefore, it is present in the third period.
Group: For s-block element, number of the group = number of valence electrons = 1. Hence, it belongs to group 1.

vi. Cl (Z = 17): Atomic number of Cl is 17. Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
Block: Since the last electron enters p subshell (3p), Cl belongs to p-block.
Period: n = 3. Therefore, it is present in the third period.
Group: For p block elements, group number = 18 – number of electrons required to complete octet
= 18 – 1 = 17. Hence, it belongs to group 17.

Question 34.
State the characteristics of s-block elements.
Answer:

• The s-block contains the elements of group 1 (alkali metals) and group 2 (alkaline earth metals).
• They occur in nature only in combined state as they are reactive elements.
• Except Li and Be, compounds formed by all other s-block elements are predominantly ionic in nature.
• This is because they have only one or two valence electrons which they can lose readily forming M⁺ or M²⁺ ions.
• Since they can lose electrons easily, they have low ionization enthalpies, which decreases down the group resulting in increased reactivity.

Question 35.
What are main group elements?
Answer:
The p-block elements together with s-block elements are called main group elements or representative elements.

Question 36.
Give reason: Group 18 elements do not participate in chemical reactions readily.
Answer:

• Group 18 is the last group of p-block and include noble or inert gases.
• They have closed valence shells (complete duplet in the case of ‘He’ and complete octet in the case of the other noble gases).
• Therefore, they show very low chemical reactivity and thus, do not participate in chemical reactions readily.

Question 37.
Why nonmetals present in group 17 and 16 in the modern periodic table are highly reactive?
Answer:

• Nonmetals present in group 17 (halogen family) and group 16 (chalcogens) have highly negative electron gain enthalpies.
• As a result, they readily accept one or two electrons and form anions (X^– or X^2-) that have complete octet.
• Therefore, nonmetals present in group 17 and 16 are highly reactive.

Question 38.
Explain the composition of the p-block in the modern periodic table.
Answer:

• The p-block contains elements of groups 13 to 18.
• It contains all the three types of elements i.e., metals, nonmetals and metalloids.
• In the p-block, metals and nonmetals are separated from each other by a zig-zag line. The metals are present on the left and the nonmetals are present on the right side while the metalloids are present along the zig-zag line.

Question 39.
State whether the following statements are true or false. Correct if false.
i. Nonmetallic character increases from left to right across a period.
ii. Nonmetallic character increases down a group.
Answer:
i. True
ii. False
Nonmetallic character decreases down a group.

Question 40.
Differentiate between s-block and p-block elements.
Answer:
s-Block elements:

• s-Block contains group 1 and group 2 elements.
• It contains only metals.
• The last electron in the s-block elements enters in s orbital.
• General outer electronic configuration of s-block elements is ns^1-2.
• e.g. Na, K, Ca, Mg, etc.

p-Block elements:

• p-Block contains elements from groups 13 to 18.
• It contains metals, nonmetals as well as metalloids.
• The last electron in the p-block elements enters in p orbital.
• General outer electronic configuration of p-block elements is ns² np^1-6.
• e.g. C, N, O, F, etc.

Question 41.
Write a note on the characteristics of the d-block elements.
Answer:

• The d-block contains elements of the groups 3 to 12 which are all metals. They are also known as transition elements or transition metals.
• They form a bridge between chemically reactive s-block elements and less reactive elements of groups 13 and 14.
• Most of the d-block elements possess partially filled inner d orbitals. As a result, the d-block elements have properties such as variable oxidation state, paramagnetism, ability to form coloured ions. They can be used as catalysts.
• Zn, Cd, and Hg with configuration ns² (n-1)d¹⁰, (completely filled s and d subshells) do not show characteristic properties of transition metals as they are stable.

Question 42.
Explain in brief about the f-block elements.
Answer:

• The elements present in f-block are all metals and are placed in the two rows called lanthanide series (₅₈Ce to ₇₁Lu) and actinide series (₉₀Th to ₁₀₃Lr).
• The lanthanides are also known as rare earth elements while the actinide elements beyond ₉₂U are called transuranium elements.
• All the transuranium elements are manmade and radioactive.
• The last electron of the elements of these series is filled in the (n-2)f subshell, and therefore, these are called inner transition elements.
• These elements have very similar properties within each series.

Question 43.
What is lanthanide and actinide series?
Answer:
i. Lanthanide series: The fourteen elements after lanthanum (Z = 57) i.e., from cerium (Z = 58) to lutetium (Z = 71) are named after their preceding member (₅₇La) present in the third group and 6^th period and are called lanthanides. They are kept in separate series called lanthanide series at the bottom of the modem periodic table.

ii. Actinide series: The fourteen elements after actinium i.e., from thorium (Z = 90) to lawrencium (Z = 103) are named after ₈₉Ac present in third period and 7^th group. They are kept in separate series called actinide series at the bottom of the modem periodic table.

Question 44.
Differentiate between d-block and f-block elements.
Answer:
d-Block elements:

• d-Block contains elements from group 3 to group 12.
• It is present in the middle of the modern periodic table.
• They are also known as transition elements.
• The last electron in the d-block elements enters in d orbital.
• General outer electronic configuration of d-block elements is ns^0-2 (n-1)d^1-10 .
• e.g. Cu, Zn, Cr, Ti, V, etc.

f-Block elements:

• f-Block contains elements of lanthanide and actinide series.
• f-block elements are present below the modern periodic table as two separate rows.
• They are also known as inner transition elements.
• The last electron in the f-block elements enters in f orbital.
• General outer electronic configuration of f-block elements is ns² (n-1) d^0-1 (n-2) f¹¹⁴.
• e.g. Ce, Pr, Nd Th, U, Np, etc.

Question 45.
Chlorides of two metals are common laboratory chemicals and both are colourless. One of the metals reacts vigorously with water while the other reacts slowly. Place the two metals in the appropriate block in the periodic table. Justify your answer.
Answer:
i. Metals are present in all the four blocks of the periodic table.
ii. Salts of metals in the f-block and p-block (except AlCl₃) are not common laboratory chemicals. Therefore, the choice is between s- and d-block.
iii. From the given properties their placement is done as shown below:

• s-block: Metal that reacts vigorously with water.
• d-block: Metal that reacts slowly with water.

iv. The colourless nature of the less reactive metal in the d-block implies that the inner d subshell is completely filled.

Question 46.
What are periodic properties?
Answer:

• The elements in the modem periodic table (long form of periodic table) are arranged in such a way that on moving across a period or down the group, several properties of elements vary in regular fashion. These properties are called periodic properties.
• Atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valency and oxidation states are several properties of elements that show periodic variations.

Question 47.
What leads to the phenomena called effective nuclear charge and screening effect in an atom?
Answer:

• The periodic trends are explained in the terms of two fundamental factors, namely, attraction of extranuclear electrons towards the nucleus and repulsion between electrons belonging to the same atom.
• These attractive and repulsive forces operate simultaneously in an atom.
• This results in two interrelated phenomena called effective nuclear charge and screening effect in an atom.

Question 48.
Explain the concept of effective nuclear charge in detail.
Answer:
i. In a multi-electron atom, the positively charged nucleus attracts the negatively charged electrons around it, and there is mutual repulsion between the negatively charged extranuclear electrons.
ii. The repulsion by inner shell electrons results in pushing the outer shell electrons further away from the nucleus. Thus, the outer shell electrons are held less tightly by the nucleus.
iii. As a result, the attraction of the nucleus for an outer electron is partially cancelled and hence, an outer shell electron does not experience the actual positive charge present on the nucleus. This effect of the inner electrons on the outer electrons is called screening effect or shielding effect.
iv. The net nuclear charge actually experienced by an electron is called the effective nuclear charge (Z_eff).
The effective nuclear charge is lower than the actual nuclear charge (Z).
v. Effective nuclear charge (Z_eff) = Z – electron shielding
= Z – σ
Here σ (sigma) is called shielding constant or screening constant and the value of σ depends upon type of the orbital that the electron occupies.

Question 49.
Define:
i. Effective nuclear charge (Z_eff)
ii. Screening effect (or shielding effect)
Answer:
i. Effective nuclear charge (Z_eff): In multi-electron atom, the net nuclear charge actually experienced by an electron is called the effective nuclear charge (Z_eff).
ii. Screening effect (or shielding effect): In multi-electron atom, the effect of the inner electrons on the outer electrons is called screening effect or shielding effect of the inner/core electrons.

Question 50.
Explain the variations in effective nuclear charge
i. Across a period
ii. Down a group
Answer:
i. Across a period:

• As we move across a period, atomic number increases by one and thus, actual nuclear charge (Z) increases by +1 at a time.
• However, the valence shell remains the same and the newly added electron gets accommodated in the same shell. There is no addition of electrons to the core i.e., inner shells. Thus, shielding due to core electrons remains the same even though the actual nuclear charge increases.
• As a result, the effective nuclear charge (Z_eff) goes on increasing across a period.

ii. Down a group:

• As we move down a group, a new larger valence shell is added. As a result, there is an additional shell in the core.
• The shielding effect of the increased number of core electrons outweighs the effect of the increased nuclear charge. Thus, the effective nuclear charge experienced by the outer electrons decreases largely down a group.
• Hence, the effective nuclear charge (Z_eff) decreases down a group.

Question 51.
Define atomic radius.
Answer:
Atomic radius is one half of the internuclear distance between two adjacent atoms of a metal or two single bonded atoms of a nonmetal.

Question 52.
What is meant by covalent radius of the atom? Explain with suitable examples.
Answer:

• In the case of nonmetals (except noble gases), the atoms of an element are bonded to each other by covalent bonds.
• Bond length of a single bond is taken as sum of radii of the two single bonded atoms. This is called covalent radius of the atom.
• For example: Bond length of C-C bond in diamond is 154 pm. Therefore, atomic radius of carbon is estimated to be 77 pm which is half of bond length (\(\frac {1}{2}\) × 154 = 77).

Question 53.
How is atomic radius of a nonmetallic element estimated?
Answer:

• The atomic size of a nonmetallic element is estimated from the distance between the two atoms bound together by a single covalent bond. From this, the covalent radius of the element is estimated.
• The internuclear distance in a diatomic molecule of an element is its covalent bond length. Half the covalent bond length gives the covalent radius.
• Bond length of Cl-Cl bond in Cl₂ is measured as 198 pm. Therefore, the atomic radius of Cl is estimated to be 99 pm.

Question 54.
Define metallic radius.
Answer:
One half of the distance between the centres of nucleus of the two adjacent atoms of a metallic crystal is called as a metallic radius.

Question 55.
How is metallic radius determined in the case of metals? Give suitable example.
Answer:

• In the case of metals, distance between the adjacent atoms in metallic sample is measured. One half of this distance is taken as the metallic radius.
• For example: In beryllium, distance between the adjacent Be atoms is measured. One half of this distance is taken as the metallic radius of a Be atom.
• Distance between two adjacent Be atoms is 224 pm. Therefore, metallic radius of a Be atom is 112 pm.

Note: Atomic radii of some elements are given in the table below.

Question 56.
How is a cation and an anion formed?
Answer:
A cation (or positively charged ion) is formed by the removal of one or more electrons from the atom of an element whereas an anion (or negatively charged ion) is formed when the atom of an element gains one or more electrons.

Question 57.
Give reasons: Radius of a cation is smaller and that of an anion is larger as compared to that of their parent atoms.
Answer:

• A cation is formed by the loss of one or more electrons, therefore, it contains fewer electrons that the parent atom but has the same nuclear charge.
• As a result, the shielding effect is less and effective nuclear charge is larger within a cation. Thus, radius of a cation is smaller than the parent atom.
• However, an anion is formed by the gain of one or more electrons and therefore, it contains a greater number of electrons than the parent atom.
• Due to these additional electrons, anion experiences increased electronic repulsion and decreased effective nuclear charge. As a result, an anion has larger radius than its parent atom.

Hence, radius of a cation is smaller and that of an anion are larger as compared to that of their parent atoms.

Question 58.
Define: Isoelectronic species
Answer:
The atoms or ions which have the same number of electrons are called isoelectronic species.

Question 59.
Explain with example why the radii of isoelectronic species vary.
Answer:
i. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
ii. For example, F^– and Na⁺ both have 10 electrons but the nuclear charge on F^– is +9 which is smaller than that of Na⁺ which has the nuclear charge +11.
Hence, F has larger ionic radii (133 pm) than Na⁺ (98 pm).

Question 60.
What is the trend observed in the ionic size of the following isoelectronic species? Explain.
i. Na⁺, Mg²⁺, Al³⁺ and Si⁴⁺
ii. O^2-, F^–, Na⁺ and Mg²⁺
Answer:
i. Na⁺, Mg²⁺, Al³⁺ and Si⁴⁺

a. Among the given ions, the nuclear charge varies but the number of electrons remains the same and therefore, these are isoelectronic species.
b. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
c. The nuclear charge increases in the order Na⁺ < Mg²⁺ < Al³⁺ < Si⁴⁺ and thus, the ionic size decreases in the order Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺.

ii. O^2-, F^–, Na⁺ and Mg²⁺

a. Among the given ions, the nuclear charge varies but the number of electrons remains the same and therefore, these are isoelectronic species.
b. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
c. The nuclear charge increases in the order O^2- < F^– < Na⁺ < Mg²⁺ and thus, the ionic size decreases in the order O^2- > F^– > Na⁺ > Mg²⁺.

Question 61.
Identify the species having larger radius from the following pairs:
i. Na and Na⁺
ii. Na⁺ and Mg²⁺
Answer:
i. The nuclear charge is the same in Na and Na⁺. But Na⁺ has a smaller number of electrons and a smaller number of occupied shells (two shells in Na⁺, while three shells in Na). Therefore, radius of Na is larger.
ii. Na⁺ and Mg²⁺ are isoelectronic species. Mg²⁺ has a larger nuclear charge than that of Na⁺. Therefore, Na⁺ has larger radius.

Question 62.
Which of the following species will have the largest and the smallest size? Why?
Mg, Mg²⁺, Al, Al³⁺
Answer:

• Atomic radius decreases across the period. Hence, the atomic radius of Mg is larger than that of Al.
• Parent atoms have larger radius than their corresponding cations. Hence, the radius of Mg is larger than that of Mg²⁺ and the radius of Al is larger than that of Al³⁺.
• Mg²⁺ and Al³⁺ are isoelectronic. Among isoelectronic species, the one with larger nuclear charge will have smaller radius. Al³⁺ (Z = 13) has a larger nuclear charge than that of Mg²⁺ (Z = 12). Hence, the ionic radius of Al³⁺ is smaller than Mg²⁺.
• Therefore, the decreasing order of radius is Mg > Al > Mg²⁺ > Al³⁺.

Hence, species with the largest size is Mg and with the smallest size is Al³⁺.

Question 63.
Identify the element with more negative value of electron gain enthalpy from the following pairs. Justify.
i. Cl and Br
ii. F and O
Answer:
i. Cl and Br belong to the same group of halogens with Br having higher atomic number than CL As the atomic number increases down the group, the effective nuclear charge decreases. The increased shielding effect of core electrons can be noticed. The electron has to be added to a farther shell, which releases less energy and thus, electron gain enthalpy becomes less negative down the group. Therefore, Cl has more negative electron gain enthalpy than Br.

ii. F and O belong to the same second period with F having higher atomic number than O. As the atomic number increases across a period, atomic radius decreases, effective nuclear charge increases and electron can be added more easily. Therefore, more energy is released with gain of an electron as we move towards right in a period. Therefore, F has more negative electron gain enthalpy than O.

Question 64.
Explain the importance of electronegativity.
Answer:

• When two atoms of different elements form a covalent bond, the electron pair is shared unequally.
• Electronegativity represents attractive force exerted by the nucleus on shared electrons. Electron sharing between covalently bonded atoms takes place using the valence electron.
• It depends upon the effective nuclear charge experienced by electron involved in formation of the covalent bond.
• Electronegativity predicts the nature of the bond, or, how strong is the force of attraction that holds two atoms together.

Question 65.
Explain the trend in electronegativity
i. across a period
ii. down a group
Answer:
i. Across a period:
a. As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily.
b. Hence, due to the increase in effective nuclear charge, the tendency to attract shared electron pair in a covalent bond increases i.e., electronegativity increases from left to right across a period.
e. g. Li < Be < B < C < N < O <F.

ii. Down a group:
a. As we move down the group from top to bottom in the periodic table, the size of the valence shell goes on increasing.
b. However, the effective nuclear charge decreases as the shielding effect of the core electrons increases due to the increase in the size of the atoms.
c. Thus, the tendency to attract shared electron pair in a covalent bond decreases, decreasing the electronegativity down the group.
e.g. F > Cl > Br > I > At.

Question 66.
Explain the terms:
i. Valency of an element
ii. Oxidation state (or oxidation number)
iii. Chemical reactivity
Answer:
i. Valency of an element:

• Valency of an element indicates the number of chemical bonds that the atom can form giving a molecule.
• The most fundamental chemical property of an element is its combining power. This property is numerically expressed in terms of valency or valence.
• Valence does not have any sign associated with it.
• Valency of the main group elements is usually equal to the number of valence electrons (outer electrons) and/or equal to difference between 8 and the number of valence electrons.

ii. Oxidation state (or oxidation number):

• The oxidation state or oxidation number is a frequently used term related to valence.
• Oxidation number has a sign, + or – which is decided by the electronegativities of atoms that are bonded.

iii. Chemical reactivity:

• Chemical reactivity is related to the ease with which an element loses or gains the electrons.
• Chemical properties of elements depend on their electronic configuration.

Question 67.
What is the trend observed in the valency of main group elements?
Answer:
i. Valency of the main group elements is usually equal to the number of valence electrons (outer electrons) or it is equal to the difference between 8 and the number of valence electrons.
i.e., (8 – number of valence electrons).
ii. The valency remains the same down the group and shows a gradual variation across the period as atomic number increases from left to right.

Note: Periodic trends in valency of main group elements.

Question 68.
Give any two distinguishing points between metals and nonmetals.
Answer:
Metals:

1. Generally, metals exhibit good electrical conductivity.
2. They can form compounds by loss of valence electrons.

Nonmetals:

1. Generally, nonmetals exhibit poor electrical conductivity.
2. Nonmetals can form compounds by gain of valence electrons in valence shell.

Question 69.
Explain the variation of the following property of elements down a group and across a period.
i. Metallic character
ii. Nonmetallic character
Answer:
The variation observed in the metallic and nonmetallic character of elements can be explained in the terms of ionization enthalpy and electron gain enthalpy.
i. Metallic character:

• The ionization enthalpy decreases down the group. Thus, the tendency to lose valence electrons increases down the group and the metallic character increases down a group.
• However, the ionization enthalpy increases across the period and as a result metallic character decreases across a period.

ii. Nonmetallic character:

• Electron gain enthalpy becomes less negative as we move down the group and hence, nonmetallic character decreases down the group.
• However, electron gain enthalpy becomes more and more negative across the period and thus, nonmetallic character increases across the period.

Question 70.
Justify the position of most reactive and least reactive elements in the modern periodic table.
Answer:

• Chemical reactivity of elements depends on the ease with which it attains electronic configuration of the nearest inert gas by gaining or losing electrons.
• The elements preceding an inert gas react by gaining electrons in the outermost shell, whereas the elements which follow an inert gas in the periodic table react by loss of valence electrons. Thus, the chemical reactivity is decided by the electron gain enthalpy and ionization enthalpy values, which in turn, are decided by effective nuclear charge and finally by the atomic size.
• The ionization enthalpy is the smallest for the element on the extreme left in a period, whereas the electron gain enthalpy is the most negative for the second last element on the extreme right, (preceding to the inert gas which is the last element of a period).
• Thus, the most reactive elements lie on the extreme left and the extreme right (excluding inert gases) of the periodic table.

Question 71.
How can we predict chemical reactivity of elements based on their oxide formation reactions and the nature of oxides formed?
Answer:

• The chemical reactivity can be illustrated by comparing the reaction of elements with oxygen to form oxides and the nature of the oxides.
• Alkali metals present on the extreme left of the modem periodic table are highly reactive and thus, they react vigorously with oxygen to form oxides such as Na₂O which reacts with water to form strong bases like NaOH.
• The reactive elements on the right i.e., halogens react with oxygen to form oxides such as Cl₂O₇ which on reaction with water form strong acids like HClO₄.
• The oxides of the elements in the centre of the main group elements are either amphoteric (Al₂O₃) neutral (CO, NO) or weakly acidic (CO₂).

Question 72.
Write the chemical equations for reaction, if any, of (i) Na₂O and (ii) Al₂O₃ with HCl and NaOH both. Correlate this with the position of Na and Al in the periodic table, and infer whether the oxides are basic, acidic or amphoteric.
Answer:
i. Na₂O + 2HCl → 2NaCl + H₂O
Na₂O + NaOH → No reaction
As Na₂O reacts with an acid to form salt and water it is a basic oxide. This is because Na is a reactive metal lying on the extreme left of the periodic table.

ii. Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
As Al₂O₃ reacts with an acid as well as base to form a salt and water. It is an amphoteric oxide. Al is a moderately reactive element lying in the centre of main group elements in the periodic table.

Question 73.
Comment on the chemical reactivity of d-block and f-block elements.
Answer:

• d-block (transition) elements and f-block (inner transition) elements exhibit very small change in atomic radii.
• Therefore, the transition and inner transition elements belonging to the individual series have similar chemical properties.
• Their ionization enthalpies are intermediate between those of s-block and p-block elements. Thus, d-block and f-block elements generally show moderate reactivity.

Question 74.
Ge, S and Br belong to the groups 14, 16 and 17, respectively. Predict the empirical formulae of the compounds those can be formed by (i) Ge and S, (ii) Ge and Br.
Answer:
From the group number we understand that the general outer electronic configuration and number of valence electrons and valencies of the three elements are:

i. S is more electronegative than Ge. Therefore, the empirical formula of the compound formed by these two elements is predicted by the method of cross multiplication of the valencies:

ii. Br is more electronegative than Ge. The empirical formula of the compound formed by these two elements is predicted by the method of cross multiplication of valencies:

[Note: More electronegative element is written on right hand side in cross multiplication method.]

Question 75.
The first ionization enthalpies of 5 elements of second period are given below:

Element	1st IE values (kJ mol-1)
I	520
II	1681
III	1086
IV	2080
V	899

Based on the above data, answer the following questions:
i. Identify the element having highest atomic number.
ii. If element I is lithium, how will you explain its low value of first ionization enthalpy?
iii. Explain why ionization enthalpies are always positive.
Answer:
i. Element IV. The first ionization enthalpy increases with increase in atomic number along a period. Hence, the element IV having highest IE will have highest atomic number among the given elements.
ii. Alkali metals have only one electron in their valence shell which can be easily lost resulting in the stable noble gas configuration. Therefore, lithium shows low value of first ionization enthalpy.
iii. Energy is always required to remove electrons from an atom. Hence, ionization enthalpies have positive value.

Question 76.
From the elements Mg, Ar, Cl, Sr, P and S, choose one that fits each of the below given descriptions:
i. An element having two valence electrons.
ii. An element having properties similar to that of O.
iii. A noble gas.
iv. An alkaline earth metal,
v. An element having electronic configuration 1s²2s²2p⁶3s²3p³.
Answer:
i. Magnesium (Mg)
ii. Sulphur (S)
iii. Argon (Ar)
iv. Strontium (Sr)
v. Phosphorus (P)

Multiple Choice Questions

1. Mendeleev’s periodic table had …………… elements.
(A) 75
(B) 83
(C) 63
(D) 118
Answer:
(C) 63

2. The serial or ordinal number of an element in Mendeleev’s periodic table was recognized as ………….
(A) neutron number
(B) valency
(C) principal quantum number
(D) proton number
Answer:
(D) proton number

3. Mendeleev predicted the existence of …………..
(A) aluminium
(B) silicon
(C) tellurium
(D) germanium
Answer:
(D) germanium

4. According to Mendeleev’s periodic law, the physical and chemical properties of elements are the periodic function of their …………..
(A) atomic weights
(B) atomic numbers
(C) molecular formulas
(D) molecular weights
Answer:
(A) atomic weights

5. Moseley showed that the fundamental property of an element is ……………
(A) atomic number
(B) atomic mass
(C) both A and B
(D) none of these
Answer:
(A) atomic number

6. According to periodic law of elements, the variation in properties of elements is related to their ……………
(A) densities
(B) atomic masses
(C) atomic sizes
(D) atomic numbers
Answer:
(D) atomic numbers

7. At present, how many elements are known?
(A) 118
(B) 110
(C) 114
(D) 120
Answer:
(A) 118

8. The long form of the periodic table consists of how many periods?
(A) 5
(B) 8
(C) 10
(D) 7
Answer:
(D) 7

9. According to quantum mechanical model of the atom, the properties of elements can be correlated to their …………….
(A) atomic number
(B) atomic mass
(C) valency
(D) electronic configuration
Answer:
(D) electronic configuration

10. The fourth, fifth and sixth periods are long periods and contain ……………
(A) 18, 18 and 36
(B) 18, 28 and 32
(C) 18, 15 and 31
(D) 18, 18 and 32
Answer:
(D) 18, 18 and 32

11. f-block elements are also known as ……………
(A) transition elements
(B) inert gas elements
(C) normal elements
(D) inner transition elements
Answer:
(D) inner transition elements

12. Which of the following forms a bridge between reactive s-block elements and less reactive group 13 and 14 elements?
(A) Inert gases
(B) Transition metals
(C) Halogens
(D) Inner transition metals
Answer:
(B) Transition metals

13. ………… elements are known as chalcogens.
(A) Group 17
(B) Group 18
(C) Group 16
(D) Group 1
Answer:
(C) Group 16

14. The name ‘rare earth elements’ is used for …………..
(A) lanthanides only
(B) actinides only
(C) both lanthanides and actinides
(D) alkaline earth metals
Answer:
(C) both lanthanides and actinides

15. Atomic number of V is 23 and its electronic configuration is …………….
(A) 1s² 2s² 2p⁶ 3p⁶ 3d³ 4s²
(B) 1s² 2s² 2d³ 3p⁶ 2p⁶ 4s²
(C) 2s² 1s² 2p⁶ 3s² 3d³ 4s²
(D) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²
Answer:
(D) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

16. Aluminium belongs to …………. elements.
(A) s-block
(B) p-block
(C) d-block
(D) f-block
Answer:
(B) p-block

17. In P^3-, S^2- and Cl^– ions, the increasing order of size is ………….
(A) Cl^– < S^2- < P^3-
(B) P^3- < S^2- < Cl^–
(C) S^2- < Cl^– < P^3-
(D) S^2- < P^3- < Cl^–
Answer:
(A) Cl- < S^2- < P^3-

18. The CORRECT order of radii is ……………
(A) N < Be < B
(B) F^–< O^2- <N^3-
(C) Na < Li < K
(D) Fe³⁺ < Fe²⁺ < Fe⁴⁺
Answer:
(B) F^– < O^2- <N^3-

19. Which of the following species will have the largest size Mg, Mg²⁺, Fe, Fe³⁺?
(A) Mg
(B) Mg²⁺
(C) Fe
(D) Fe³⁺
Answer:
(C) Fe

20. Which one of the following is CORRECT order of the size?
(A) I > I^– >I⁺
(B) I > I⁺ > I^–
(C) I⁺ > I^– > I
(D) I^– > I > I⁺
Answer:
(D) I^– > I > I⁺

21. The CORRECT order of increasing radii of the elements Na, Si, Al and P is ……………
(A) Si < Al < P < Na
(B) Al < Si < P < Na
(C) P < Si < Al < Na
(D) Al < P < Si < Na
Answer:
(C) P < Si < Al < Na

22. The metallic and nonmetallic properties of elements can be judged by their ……………
(A) electron gain enthalpy
(B) ionization enthalpy
(C) electronegativity
(D) valence
Answer:
(C) electronegativity

23. Which element has the most negative electron gain enthalpy?
(A) Sulphur
(B) Fluorine
(C) Chlorine
(D) Hydrogen
Answer:
(C) Chlorine

24. Which of the properties remain unchanged on descending a group in the periodic table?
(A) Atomic size
(B) Density
(C) Valency electrons
(D) Metallic character
Answer:
(C) Valency electrons

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound

Question 1.
State the different types of waves.
Answer:

1. Waves which require a material medium for their propagation are called mechanical waves. Example: Sound waves, string waves, seismic waves, etc.
2. Waves which do not require material medium for their propagation are called electromagnetic waves. Example: Light waves, radio waves, y-rays, etc.
3. The wave associated with any object when it is in motion is called as matter wave.
4. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
5. Waves are also of stationary type.

Question 2.
Define the following terms. Give their SI units.
i. Period
ii. Frequency
iii. Velocity
Answer:
i. Period (T):
The time taken by the particle of a medium to complete one vibration is called period of the wave.
SI unit: second (s)

ii. Frequency (n):
The number of vibrations performed by a particle per second is called frequency of a wave.
SI unit: hertz (Hz)

iii. Velocity (v):
The distance covered by a wave per unit time is called the velocity of the wave.
SI unit: m/s

Question 3.
State the properties that should be possessed by a medium for a mechanical wave to propagate through it.
Answer:

1. The medium must be continuous and elastic so that it can regain its original state as soon as the deforming forces are removed.
2. The medium should possess inertia. It must be capable of storing energy and transferring it in the form of waves.
3. The frictional resistance of the medium should be negligible to avoid damped oscillations.

Question 4.
What are two types of progressive waves? State two characteristics of progressive waves.
Answer:
Progressive waves are classified into two types:
a. Transverse progressive waves
b. Longitudinal progressive waves.

Characteristics of progressive waves:
1. All the vibrating particles of medium have same amplitude, period and frequency.
2.. State of oscillation i.e., phase changes from particle to particle.

Question 5.
A violin string emits sound of frequency 510 Hz. How far will the sound waves reach when string completes 250 vibrations? The velocity of sound is 340 m/s.
Answer:
Given: n = 510 Hz, v = 340 m/s,
number of vibrations = 250
To find: Distance
Formula: v = nλ
Calculation:
From formula,
λ = \(\frac {v}{n}\) = \(\frac {340}{510}\) = \(\frac {2}{3}\) m
Distance covered in 1 vibration = \(\frac {2}{3}\) m
∴ Distance covered in 250 vibration
= \(\frac {2}{3}\) × 250 = 166.7 m 3
Answer: The distance covered by sound waves is 166.7 m

Question 6.
The speed of sound in air is 330 m/s and that in glass is 4500 m/s. What is the ratio of the wavelength of sound of a given frequency in the two media?
Answer:
Given: v_air = 330 m/s, v_glass = 4500 m/s
To find: \(\frac {λ_{air}}{λ_{glass}}\)
Formula: v = nλ
Calculation: From formula,
v_air = n λ_air
v_glass = n λ_glass
∴ \(\frac {λ_{air}}{λ_{glass}}\) = \(\frac {v_{air}}{v_{glass}}\) = \(\frac {330}{4500}\) = 7.33 × 10^-2

Question 7.
The velocity of sound in gas is 498 m/s and in air is 332 m/s. What is the ratio of wavelength of sound waves in gas to air?
Answer:
v_g = 498 m/s, v_a = 332 m/s
To find: Ratio of wavelengths (\(\frac {λ_g}{λ_a}\))
Formula: v = nλ
Calculation:
Frequency of sound wave remains same.
From formula,
For gas λ_g = \(\frac {v_g}{n}\) and for air λ_ag = \(\frac {v_a}{n}\)
∴ \(\frac {v_g}{v_a}\) = \(\frac {v_g}{v_a}\) = \(\frac {498}{332}\) = \(\frac {3}{2}\)
∴ \(\frac {v_g}{v_a}\) = 3 : 2

Question 8.
A human ear responds to sound waves of frequencies in the range of 20 Hz to 20 kHz. What are the corresponding wavelengths, if the speed of sound in air is 330 m/s? Answer:
Given: v₁ = v_g = 330 m/s, n₁ = 20 Hz,
n₂ = 20 kHz = 20 × 10³ Hz
To find: Wavelength (λ₁ and λ₂)
Formula: v = nλ
Calculation:
From formula,
λ₁ = \(\frac {v_1}{n_1}\) = \(\frac {330}{20}\) = 16.5 m
λ₂ = \(\frac {v_2}{n_2}\) = \(\frac {330}{20×10^3}\) = 16.5 × 10^-3 = 0.0165 m

Question 9.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (i) the reflected sound, (ii) the transmitted sound? Speed of sound in air is 340 m s^-1 and in water is 1486 m s^-1
Answer:
Given: n = 1000 kHz = 10⁶ Hz,
v_a = 340 m/s,
v_w = 1486 m/s
To find: Reflected wavelength (λ_R),
Transmitted wavelength (λ_T),
Formula: v = nλ
Calculation:
In different medium, frequency of sound wave remains same.
From formula,
Sound is reflected in air,
i. ∴ λ_R = \(\frac {v_a}{n}\) = \(\frac {330}{10^6}\) = 3.4 × 10^-4 m
Sound is transmitted in water,
ii. ∴ λ_T = \(\frac {v_w}{n}\) = \(\frac {1486}{10^6}\) = 1.486 × 10^-3 m

Question 10.
The wavelength of a sound note is 1 m in air and 2.5 m in a liquid. Find the speed of sound in the liquid, if the speed of the sound in air is 330 m/s.
Answer:
Given: λ_a = 1 m, λ_l = 2.5 m, v_a = 330 m/s,
To find: Speed of sound (v_l)
Formula: v = nλ
Calculation:
From formula,
Since the frequency n remains the same,
v_a = nλ_a and v_l = nλ_l
∴ \(\frac {v_l}{v_a}\) = \(\frac {λ_l}{λ_a}\)
∴ v_l = v_a \(\frac {λ_l}{λ_a}\) = 330 × \(\frac {2.5}{1}\) = 825 m/s

Question 11.
Define a polarised wave.
Answer:
A wave in which the vibrations of all the particles along the path of a wave are constrained to a single plane is called a polarised wave.

Question 12.
Write down the main characteristics of longitudinal waves.
Answer:
Characteristics of longitudinal waves:

1. All the particles of medium in the path of wave vibrate in a direction parallel to the direction of propagation of wave with same period and amplitude.
2. When longitudinal wave passes through a medium, the medium is divided into alternate compressions (high pressure zone) and rarefactions (low pressure zone).
3. A compression and adjacent rarefaction form one cycle of a longitudinal wave. The distance between any two consecutive points having same phase (successive compression or rarefactions) is called wavelength of the wave.
4. For propagation of longitudinal waves, the medium should possess the property of elasticity of volume (bulk modulus). Thus, longitudinal waves can travel through solids, liquids and gases. Longitudinal wave cannot travel through vacuum or empty space.
5. The compressions and rarefactions advance in the medium and are responsible for transfer of energy.
6. When longitudinal wave advances through medium, there is periodic variations in pressure and density along the path of wave and with time.
7. Since the direction of vibration of particles and direction of propagation of wave are same or parallel, longitudinal waves cannot be polarised.

Question 13.
State the mathematical expression for a transverse progressive wave travelling along the positive and negative x-axis.
Answer:
i. Consider a transverse progressive wave whose particle position is described by x and displacement from equilibrium position is described by y.
Such a sinusoidal wave can be written as follows:
∴ y (x, t) = a sin (kx – ωt + ø) ……… (1)
where a, k, ω and ø are constants,
y (x, t) = displacement as a function of position (x) and time (t)
a = amplitude of the wave,
ω = angular frequency of the wave
(kx₀ – ωt + ø) = argument of the sinusoidal wave and is the phase of the particle at x at time t.

ii. At a particular instant, t = t₀,
y (x, t₀) = a sin (kx – ωt₀ + ø)
= a sin (kx + constant)
Thus at t = t₀, shape of wave as a function of x is a sine wave.

iii. At a fixed location x = x₀
y(x₀, t) = a sin (kx₀ – ωt + ø)
= a sin (constant – ωt)
Hence the displacement y, at x = x₀ varies as a sine function.

iv. This means that the particles of the medium, through which the wave travels, execute simple harmonic motion around their equilibrium position.

v. For (kx – ωt + ø) to remain constant, x must increase in the positive direction as time t increases. Thus, the equation (1) represents a wave travelling along the positive x axis.

vi. Similarly, a wave travelling in the direction of the negative x axis is represented by,
y(x, t) = a sin (kx + ωt +ø) …….(2)

Question 14.
Explain the Laplace’s correction to the Newton’s formula for the velocity of sound in air.
Answer:
Laplace’s correction:
Laplace suggested that the compression or rarefaction takes place too rapidly. Heat produced during compression and lost during rarefaction does not get sufficient time for dissipation. Due to this, the whole heat content remains same. Thus, it is an adiabatic process.

According to Laplace, E is the adiabatic modulus of elasticity of air medium which is given by,
E = γP ….(1)
where P = pressure of the air medium γ = ratio of specific heat of air at constant pressure (c_p) and specific heat of air at constant volume (c_v). i.e., γ = c_p/c_v.

iii. Using equation, v = \(\sqrt{\frac {E}{ρ}}\), we have velocity of sound in air,
v = \(\sqrt{\frac {γP}{ρ}}\), …. [From equation (1)]
For air, γ = 1.41
At NTP, P = 0.76 × 13600 × 9.8 N/m²
ρ = 1.293 kg/m³.
∴ v = \(\sqrt{\frac {1.41×0.76×13600×9.8}{1.293}}\) = 332.35 m/s
This value is in close agreement with the experimental value.

Question 15.
What is the effect of temperature on the velocity of sound in air?
Answer:
Effect of temperature on velocity of sound:
i. Let v₀ and v be the velocity of sound in air at T₀ and T Kelvin respectively. Let ρ₀ and p be the densities of gas at temperature T0 and T respectively.

ii. Considering the number of moles n = 1 for the gas, we have,

iii. From above formula, we can conclude that velocity of sound in air increases with increase in temperature.

Question 16.
Show that for 1 °C rise in temperature, the velocity of sound in air increases by 0.61 m/s.
Answer:
Let v₀ = velocity of sound at 0 °C or 273 K
v = velocity of sound at t °C or (273 + 1) K
we have, v ∝ √T

Hence, velocity of sound increases by 0.61 m/s when temperature increases by 1 °C.

Question 17.
Suppose you are listening to an out-door live concert sitting at a distance of 150 m from the speakers. Your friend is listening to the live broadcast of the concert in another country and the radio signal has to travel 3000 km to reach him. Who will hear the music first and what will be the time difference between the two? Velocity of light = 3 × 10⁸ m/s and that of sound is 330 m/s.
Answer:
Distance between me and the speakers
(s₁) = 150 m, distance radio signal has to travel (S₂) = 3000 km, v_sound 330 m/s, v_light = 3 × 10⁸ m/s
Time taken by sound to reach me,
= \(\frac {s_1}{v_sound}\) = \(\frac {150}{330}\) = 0.4546 s
Time taken by the broadcasted sound (done by
EM waves = \(\frac {s_2}{v_light}\) = \(\frac {3000km}{30×10^5km/s}\) = \(\frac {3×1^30}{3×10^5}\) = 10^-2 s
∴My friend will hear the sound first.
The time difference will be = 0.4546 – 0.01
= 0.4446 s.

Question 18.
Consider a closed box of rigid walls so that the density’ of the air inside it is constant. On heating, the pressure of this enclosed air is increased from P₀ to P. It is now observed that sound travels 1.5 times faster than at pressure P₀. Calculate P/P₀.
Answer:
Given: v_P = 1.5 v_P0
To find: Ratio of pressure (\(\frac {p}{p_0}\))

Question 19.
The densities of nitrogen and oxygen at NTP are 1.25 kg/m³ and 1.43 kg/m³ respectively. If the speed of sound in oxygen at NTP is 320 m/s, calculate the speed in nitrogen, under the same conditions of temperature and pressure, (γ for both the gases is 1.4)
Answer:
Given: ρ_N = 1.25 kg/m³, ρ = 1.43 kg/m³,
v₀ = 320 m/s,
To find: Speed in nitrogen (v_N)
Formula: v = (\(\sqrt{\frac {γP}{ρ}}\) )
Calculation: From formula,

Question 20.
Find the temperature at which the velocity of sound in air will be 1.5 times its velocity at 0 °C
Answer:
Given: \(\frac {p}{p_0}\) = 1.5, T₀ = 0 °C = 273 K
To find: Temperature (T)
Formula: \(\frac {v}{v_0}\) = \(\sqrt{\frac {T}{T_0}}\)
Calculation:
From formula,
\(\frac {T}{T_0}\) = (\(\frac {v}{v_0}\))²
∴ T = T₀ (\(\frac {v}{v_0}\))²
∴ T = 273 (1.5)² = 614.25 K = 341.25 °C

Question 21.
The velocity of sound in air at 27 °C is 340 m/s. Calculate the velocity of sound in air at 127 °C.
Answer:
Given: T₁ = 27 °C = 27 + 273 = 300 K,
v₁ = 340 m/s,
T₂ = 127 °C = 127 + 273 = 400 K
To find: Velocity (v₂)
Formula: \(\frac {v_1}{v_2}\) = \(\sqrt{\frac {T_1}{T_2}}\)
Calculation: From formula,
v₂ = v₁ \(\sqrt{\frac {T_2}{T_1}}\) = 340, \(\sqrt{\frac {400}{300}}\)
= 340 × 1.1547
∴ v₂ = 392.6 m/s

Question 22.
The wavelength of a note is 27 m in air when the temperature is 27 °C. What is the wavelength when the temperature is increased to 37 °C?
Answer:
Given: λ₁ = 27 m,
T₁ = 27 °C = 273 + 27 = 300 K,
T₂ = 37 °C = 273 + 37 = 310 K
To find: Wavelength (λ₂)

Question 23.
We cannot hear an echo at every place. Give reason.
Answer:

1. Echo of sound depends upon the temperature of the surrounding and distance between source and reflecting surface.
2. To hear a distinct echo at 22 °C, the minimum distance required between the source of sound and reflecting surface should be 17.2 metre.
3. The velocity of sound depends on the temperature of air. Thus, the minimum distance will change with temperature. Hence, we cannot hear an echo at every place.

Question 24.
Write a short note on reverberation.
Answer:

1. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
2. Sound wave gets reflected multiple times if the distance between reflecting surface and source of sound is less than 15 m.
3. During reverberation, the time interval between the successive reflections of a sound is small.
4. As a result, the reflected sound waves overlap and produce a continuously increasing loud sound which is at times difficult to understand. Measures to decrease reverberation:
5. Reverberation can be decreased by making the walls and roofs rough and by using curtains in the hall to avoid reflection of sound.
6. Chairs and wall surfaces should be covered with sound absorbing materials.
7. Porous cardboard sheets, perforated acoustic tiles, gypsum boards, thick curtains etc. should be used on the ceilings and walls.

Question 25.
Define acoustics.
Answer:
The branch of physics which deals with the study of production, transmission and reception of sound is called acoustics.

Question 26.
State the conditions that must be satisfied for proper acoustics in an auditorium along-with their remedies.
Answer:
i. Acoustics of an auditorium should be such that the sound is heard sufficiently loudly at all the points in the auditorium. The surface behind the speaker should be parabolic with the speaker at its focus for uniform distribution of sound in the auditorium. Reflection of sound helps to maintain good loudness through the entire auditorium.

ii. Echoes and reverberations should be reduced. More absorptive reflecting surfaces and full auditoriums help in reducing echoes.

iii. Unnecessary focusing of sound, poor audibility zone or region of silence should be avoided. Curved surface of the wall or ceiling should be avoided for this purpose.

iv. Echelon effect which arises due to the mixing of sound produced in the hall by the echoes of sound produced in front of regular structure like stairs should be reduced. Stair type construction in the hall must be avoided for this purpose.

v. To avoid outside stray sound from entering, the auditorium should be sound-proof when closed.

vi. Inside fittings, seats, etc. should not produce any sound for proper acoustics. Air conditioners instead of fans and soft action door closers should be used.

Question 27.
State the applications of acoustics observed in nature.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Question 28.
State the medical applications of acoustics.
Answer:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Question 29.
State the underwater applications of acoustics.
Answer:

1. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
2. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
3. Motion and position of submerged objects like submarine can be measured with the help of this system.

Question 30.
State the applications of acoustics in environmental and geological studies.
Answer:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

ii. Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics.

iii. This is useful in studying the properties of the Earth. Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 31.
A man shouts loudly close to a high wall. He hears an echo. If the man is at 40 m from the wall, how long after the shout will the echo be heard? (speed of sound in air = 330 m/s)
Answer:
Given s = 40m, v = 330 m/s
To Find: time (t)
Formula: Time = distance \(\frac {distence}{speed}\)
Calculation:
The distance travelled by the sound wave
= 2 × distance from man to wall.
= 2 × 40 = 80 m.
From formula,
∴ Time taken to travel the distance
\(\frac {distence}{speed}\) = \(\frac {80}{30}\) = 0.24 s

Question 32.
Write a short note on pitch of sound note.
Answer:

1. Pitch refers to the sharpness or shrillness of sound.
2. Increase in frequency of sound results in increase in the pitch and the sound is said to be sharper.
3. Tone refers to a single frequency of a wave.
4. A note may contain single or multiple tones.
5. High frequency is generally referred as high pitch or high tone.
6. Generally, speech of the men is of low pitch (shrill) and that of the women is of high pitch (sharp). Tones of an acoustic guitar are sharper than that of a base guitar. Sound of table is sharper than that of a dagga.

Question 33.
Write a short note on quality (timbre) of sound note.
Answer:
i. Timbre of a sound refers to the quality of the sound which depends upon the mixture of tones and overtones in the sound. Same sound played on different musical instruments feels significantly different and the musical instrument from which the sound generated can be easily identified.

Question 34.
Write a short note on loudness of sound.
OR
Explain how loudness affects the characteristics of sound.
Answer:
Loudness:
i. Loudness depends upon the intensity of vibration.

ii. Intensity of a wave is proportional to square of the amplitude (I ∝ A²) and is measured in the (SI) unit ofW/m²

iii. The human response to intensity is not linear, i.e., a sound of double intensity is louder but not doubly loud.

iv. Under ideal conditions, for a perfectly healthy human ear, the least audible intensity is I₀ = 10^-12 W/m².

v. Loudness of a sound of intensity I (measured in unit bel) is given by,
L₂ = log₁₀ (\(\frac {I}{I_0}\)) ………….. (1)

vi. Decibel is the commonly used unit for loudness.

vii. As, 1 decibel or 1 dB = 0.1 bel.
∴ 1 bel = 10 dB. Thus, loudness in dB is 10 times loudness in bel.
∴ L_dB = 10L_bel = 10 log₁₀ (\(\frac {I}{I_0}\))
For sound of least audible intensity I₀
L_dB = 10 log₁₀ (\(\frac {I_0}{I_0}\)) = 10 log₁₀ (1) = 0 ………… (2)
This corresponds to threshold of hearing.

viii. For sound of 10 dB,
10 = 10 log₁₀ (\(\frac {I}{I_0}\))
∴ (\(\frac {I}{I_0}\)) = 10 ¹ or I = 10 I₀
For sound of 20 dB,
20 = 10 log₁₀ (\(\frac {I}{I_0}\))
= (\(\frac {I}{I_0}\)) = 10² or I = 100 I₀ and so on.

ix. This implies, loudness of 20 dB sound is felt double that of 10 dB, but its intensity is 10 times that of the 10 dB sound. Similarly, 40 dB sound is left twice as loud as 20 dB sound but its intensity is 100 times as that of 20 dB sound and 10000 times that of 10 dB sound. This is the power of logarithmic or exponential scale. If we move away from a (practically) point source, the intensity of its sound varies inversely with square of the distance, i.e., I ∝ \(\frac {1}{r^2}\).

Question 35.
When heard independently, two sound waves produce sensations of 60 dB and 55 dB respectively. How much will the sensation be if those are sounded together, perfectly in phase?
Answer:
L₁ = 60 dB = 10 log₁₀ \(\frac {I_1}{I_0}\)
∴ \(\frac {I_1}{I_0}\) = 10⁶
∴ I₁ = 10⁶I₀
Similarly, I₂ = 10^5.5 I₀
As the waves combine perfectly in phase, the vector addition of their amplitudes will be given by A² = (A₁ + A₂)² = A\(_1^2\) + A\(_2^2\) + 2A₁, A₂ As intensity is proportional to square of the amplitude.
∴ I = I₁ + I₂ + 2\(\sqrt {I_1I_2}\) = 10⁵ I₀ (10¹ +10^0.5 + 2\(\sqrt {10^{1.5}}\))
= 10⁵I₀(10 + 3.1623 +2 × 10^0.75)
= 24.41 × 10⁵I₀ = 2.441 × 10⁶I₀
∴ L = 10 log₁₀ (\(\frac {I}{I_0}\)) = 10 log₁₀ (2.441 × 10⁶)
= 10[log₁₀ (2.441) + log₁₀(10⁶)]
= 10(0.3876 + 6)
L = 63.876 dB ~ 64 dB

Question 36.
The noise level in a class-room in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increases to 100?
Answer:
Loudness of sound is given as,

∴ L_B – L_A = 0.301 × 10 = 3.01
∴ L_B = L_A + 3.01 = 53.01 dB

Question 37.
Calculate the decibel increase if there is a two-fold increase in the intensity of a wave. (Given: log₁₀ 2 = 0.3010)
Answer:
L = 10 log₁₀ \(\frac {I}{I_0}\) decibel
L’ = 10 log₁₀ \(\frac {2I}{I_0}\) decibel
L’ – L = 10 (log₁₀ \(\frac {2I}{I_0}\) – log₁₀ (\(\frac {I}{I_0}\))
= 10 log₁₀ 2
= 10 × 0.3010
∴ L’ – L = 3.01 dB

Question 38.
Derive the expression for apparent frequency when listener is stationary and source is moving away from the listener.
Answer:

i. Consider a source of sound S moving away from a stationary listener L with velocity v_s. Let the speed of sound with respect to the medium be v (always positive). The listener uses a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point Si which is at a distance d from the listener, emit a crest. This crest reaches the listener at time t₁, given as, t₁ = d/v. …………(1)

iii. Let T₀ be the time period at which the waves are emitted.
At t = T₀, distance travelled by the source away from the stationary listener to reach point S₂ = v_sT₀.
∴ Distance of point S₂ from the listener = d + v_sT₀.
At S₂, The source emits second crest. This crest reaches the listener at t₂, given as,
t₂ = T₀ + (\(\frac {d+v_sT_0}{v}\)) …………. (2)

iv. Similarly, the time taken by the (p+1)^th crest (where, p is an integer, p = 1, 2, 3,…), emitted by the source at time pT₀, to reach the listener is given as,
t_p+1 = pT₀ + (\(\frac {d+pv_sT_0}{v}\)) …………. (3)
∴ the listener’s detector counts p crests in the time interval,
t_p+1 – t₁ = pT₀ + (\(\frac {d+pv_sT_0}{v}\)) – \(\frac {d}{v}\)
The period of wave as recorded by the listener is,

Where, n = frequency recorded by the listener (apparent frequency)
n₀ = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving away from the listener.

Question 39.
Derive an expression for apparent frequency when listener is stationary and source is moving towards the listener.
Answer:

i. Consider a source of sound S moving towards a stationary listener L with velocity v_s. Let the speed of sound with respect to the medium be v (always positive). The listener use a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point S₁ which is at a distance d from the listener, emit a Crest. This crest reaches the listener at time t₁, given as,
∴ t₁ = d/v. ……….(1)

iii. Let T₀ be the time period at which the waves are emitted.
At t = T₀, distance travelled by the source away from the stationary listener to reach point S₂ = v_sT₀.
Distance of S₂ from the listener = d – v_sT₀.
At S₂, The source emits second crest. This crest reaches the listener at
t₂ = T₀ + (\(\frac {d-v_sT_0}{v}\)) ………….. (2)

iv. Similarly, the time taken by the (p+1)^th crest (where, p = 1,2,3,…), emitted by the source at time pT₀, to reach the listener is given as,
t_p+1 = pT₀ + (\(\frac {d-pv_sT_0}{v}\)) ……………. (3)
∴ the listener’s detector counts p crests in the time interval,
t_p+1 – t₁ = pT₀ + (\(\frac {d-pv_sT_0}{v}\)) – \(\frac {d}{v}\)
∴ the period of wave as recorded by the listener is,

Where, n = frequency recorded by the listener (apparent frequency)
n₀ = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving towards the listener.

Question 40.
Derive the expression for apparent frequency when the source is stationary and the listener is moving towards the source.
Answer:

i. Consider a listener approaching a stationary source S with velocity v_L as shown in figure. Let the speed of sound with respect to the medium be v (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L₁ is at an initial distance d from the source.
At time t = t₁ the listener receives the first wave at the position L₂.
Distance travelled by the listener towards the stationary source during time t₁ = v_Lt₁.
Distance travelled by the sound wave during time t₁ = d – v_Lt₁
∴ time taken by the sound wave to travel this distance, t₁ = \(\frac {d-v_Lt_1}{v}\)
∴ t₁ = \(\frac {d}{v+v_L}\) ………….. (1)

iii. Let at time t = T₀ (time period of the waves emitted by the source), the source emits a second wave.
At t = t₂, the listener receives the second wave. Distance travelled by the listener towards the stationary source during time t₂ = v_Lt₂.
Distance travelled by the sound wave during time t₂ = d – v_Lt₂
∴ time taken by the sound wave to travel this distance = \(\frac {d-v_Lt_2}{v}\)
However, this time should be counted after T₀, as the second wave was emitted at t = T₀.
∴ t₂ = T₀ + \(\frac {d-v_Lt_2}{v}\)
∴ t₂ = \(\frac {vT_0+d}{v+v_L}\) …………. (2)

iv. Similarly, for the third wave, we get,
t₃ = 2T₀ + \(\frac {d-v_Lt_3}{v}\)
∴ t₃ = \(\frac {2vT_0+d}{v+v_L}\) …………. (3)

v. Extending this argument to the (p + 1)^th wave, we get,
t_p+1 = pT₀ + \(\frac {d-v_Lt_{p+1}}{v}\)
∴ t_p+1 = \(\frac {pvT_0+d}{v+v_L}\) …………. (4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.

This is the expression for apparent frequency when the source is stationary and the listener is moving towards the source.

Question 41.
Derive the expression for apparent frequency when the source is stationary and the listener is moving away from the source.
Answer:

i. Consider a listener moving away a stationary source S with velocity V_L. Let the speed of sound with respect to the medium be y (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L₁ is at an initial distance d from the source.
At time t = t₁ the listener receives the first wave at the position L₂.
Distance travelled by the listener away from the stationary source during time t₁ = v_Lt₁.
Distance travelled by the sound wave during time t₁ = d + v_Lt₁
∴ time taken by the sound wave to travel this distance,
t₁ = \(\frac {d+v_Lt_1}{v}\)
∴ t₁ = \(\frac {d}{v-v_L}\) ………….. (1)

iii. Let at time t = T₀ (time period of the waves emitted by the source), the source emits a second wave.
At t = t₂, the listener receives the second wave. Distance travelled by the listener away from the stationary source during time t₂ = v_Lt₂.
∴ Distance travelled by the sound wave during time t₂ = d + v_Lt₂.
∴ time taken by the sound wave to travel this distance = \(\frac {d+v_Lt_2}{v}\)
However, this time should be counted after T₀, as the second wave was emitted at t = T₀.
∴ t₂ = T₀ \(\frac {d+v_Lt_2}{v}\) ………….. (2)
∴ t₂ = \(\frac {vT_0+d}{v-v_L}\)

iv. Similarly for the third wave, we get
t₃ = 2T₀ \(\frac {d+v_Lt_3}{v}\)
∴ t₃ = \(\frac {2vT_0+d}{v-v_L}\) …………..(3)

v. Extending this argument to the (p + 1)^th wave, we get,
t_p+1 = pT₀ + \(\frac {d+v_Lt_{p+1}}{v}\)
∴ t_p+1 = \(\frac {pvT_0+d}{v-v_L}\) …………..(4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.

This is the expression for apparent frequency when the source is stationary and the listener is moving away from the source.

Question 42.
State the common properties between Doppler effect of sound and light.
Answer:
i. The recorded frequency is different than the emitted frequency in case of relative motion between listener (or observer) and source (of sound or light waves).

ii. In case of relative approach, recorded frequency > emitted frequency.

iii. In case of relative recede, recorded frequency < emitted frequency.

iv. For values of listener velocity (v_L) or source velocity (v_s) much smaller then wave speed (speed of sound or light).
n = n₀ (1±\(\frac {v_r}{v}\))
Where, vr = relative velocity
n = actual frequency of the source
n₀ = apparent frequency of the source
v = velocity of sound in air.
(upper sign is used during relative approach and lower sign is during relative recede.)

v. If velocities of source and observer (listener) are along different lines, their respective components along the line joining them should be chosen for longitudinal Doppler effect and the same mathematical treatment is applicable.

Question 43.
State the major difference between Doppler effects of sound and light.
Answer:

1. Speed of light being absolute, only relative velocity between the observer and the source matter irrespective of who is in motion. However, for obtaining exact Doppler shift for sound waves, it is absolutely important to know who is in motion.
2. In case of light, classical and relativistic Doppler effects are different while sound only has classical doppler effect.
3. Presence of wind affects the velocity of sound which affects the Doppler shift in sound.

Question 44.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s^-1 (b) recedes from the platform with a speed of 10 m s^-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given: v_s = 10 m/s, v = 340 m/s, n₀ = 400 Hz
Apparent frequency (n), velocity of sound (v_s) in each case
Formulae:
i. n = n₀ (\(\frac {v}{v-v_s}\))
ii. n = n₀ \(\frac {v}{v+v_s}\)
Calculation:
a. As the train approaches the platform, using formula (i),
n = 400 (\(\frac {340}{340-10}\)) = 421.12 Hz

b. As the train recedes from the platform, using formula (ii),
n = 400 (\(\frac {340}{340+10}\)) = 388.57 Hz

ii. The relative motion of source and observer results in the apparent change in the frequency but has no effect on the speed of sound. Hence, the speed of sound remains unchanged in both the cases.

Question 45.
A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the observer with the speed of 72 km/hour. (Speed of sound in air = 340 m/s)
Answer:
Given: v_s = 72 km/ hr = 20 m/s, n₀ = 640 Hz,
v = 340 m/s
To find: Difference in apparent frequencies
(n_A – n’_A)
Formulae:
i. When the train moves towards the stationary observer then,
n_A = n₀ (\(\frac {v}{v-v_s}\))
ii. When the train moves away the stationary observer then,
n’_A = n₀ (\(\frac {v}{v+v_s}\))
Calculation: From formula (i),
n_A = 640 (\(\frac {340}{340-20}\))
∴ n_A = 680 Hz
From formula (ii),
n’_A = 640 (\(\frac {340}{340+20}\))
∴ n’_A = 604.4 Hz
Difference between n_A and n’_A
= n_A – n’_A = 75.6 Hz

Question 46.
The speed limit for a vehicle on road is 120 km/hr. A policeman detects a drop of 10% in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? (Given: Velocity of sound=340 m/s).
Answer:
Given: Speed limit, v_L = 120 km/hr
n’_A = n_A – \(\frac {10}{100}\) n_A = 0.9 n_A
Velocity of sound, v = 340 m/s
To Find: Velocity of source (v_s)
i. n_A = (\(\frac {v}{v-v_s}\))n
ii. n’_A = (\(\frac {v}{v+v_s}\))n
Calculation:
Dividing formula (i) by (ii),

Question 47.
A stationary source produces a note of frequency 350 Hz. An observer in a car moving towards the source measures the frequency of sound as 370 Hz. Find the speed of the car. What will be the frequency of sound as measured by the observer in the car if the car moves away from the source at the same speed? (Assume speed of sound = 340 m/s)
Answer:
Given: n₀ = 350 Hz, v = 340 m/s,
n_A = 370 Hz
To find: Speed (v_L), Frequency (n_A)
Formulae:
i. When the car moves towards the stationary source then,
n_A = n₀ (\(\frac {v+v_s}{v}\))

ii. When the car moves away from the stationary source then,
n_A = n₀ (\(\frac {v-v_L}{v}\))
Calculation: From formula (i),
370 = 35o (\(\frac {340+v_L}{340}\))
∴ 359.43 =340 +v_L
∴ v_L = 19.43 m/s
From formula (ii),
∴ n_A = 35o (\(\frac {340-20}{340}\)) = \(\frac {35}{34}\) × 320
∴ n_A = 329.41 Hz

Question 48.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Blowing of wind changes the velocity of sound. As the wind is blowing in the direction of sound, effective speed of sound v_e = v + v_w = 340 + 10 = 350 m/s
As the source and listener both are at rest, frequency is unchanged, i.e., n = 400 Hz.
∴ wavelength, λ = \(\frac {v_e}{n}\) = \(\frac {350}{400}\) = 0.875 m
For still air, v_w = 0 and v_e = v = 340 m/s
Also, as observer runs towards the stationary train v_L = 10 m/s and v_s = 0
The frequency now heard by the observer,
n = n₀ (\(\frac {v+v_L}{v}\)) = 400 (\(\frac {340+10}{340}\))
= 411.76 Hz
As the source is at rest, wavelength does not change i.e., λ’ = λ = 0.875 m
Comparing the answers, it can be stated that, the situations in two cases are different.

Question 49.
A SONAR system fixed in a submarine operates at a frequency 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Frequency of SONAR (source)
n = 40 kHz = 40 × 10³ Hz
Speed of sound waves, v = 1450 m s^-1
Speed of the listener, v_L = 360 km h^-1
= 360 × \(\frac {5}{18}\) ms^-1
= 100 m s^-1
Since, the source is at rest and the observer moves towards the source (SONAR).
We have,
n = n₀ (\(\frac {v+v_L}{v}\)) = 40 × 10³ × (\(\frac {1450+100}{14540}\))
∴ n = 4.276 × 10⁴ Hz
This frequency n’ is reflected by the enemy ship and is observed by the SONAR (which now acts as observer). Therefore, in this case v_s = 100 m s^-1
Apparent frequency,
n = n₀ (\(\frac {v}{v-v_s}\))
= 4.276 × 10⁴ × (\(\frac {1450}{1450-100}\)) = 4.59 × 10⁴ Hz
∴ n = 45.9 kHz

Question 50.
A rocket is moving at a speed of 220 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1200 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (i) the frequency of the sound as detected by the target and (ii) the frequency of the echo as detected by the rocket (velocity of sound = 330 m/s)
Answer:
Given: v_s = 220 m/s, v_L = 0 m/s, n = 1200 Hz
To find: Apparent frequency (n)
i. n = n₀ (\(\frac {v}{v-v_s}\))
ii. n = n₀ \(\frac {v+v_L}{v}\)
Calculation: For first case, observer is stationary and source i.e., rocket is moving towards the target.
Hence, using formula (i),
frequency of sound as detected by the target,
n = 1200 (\(\frac {330}{330-220}\)) = 3600 Hz
For second case, target acts as a source with frequency 3600 Hz as it is the source of echo. While rocket detector acts as an observer. This means, v_s = 0 and V_L = 220 m/s
Using formula (ii),
frequency of echo as detected by the rocket,
n = 3600 (\(\frac {330+220}{330}\)) = 600 Hz

Question 51.
A bat is flying about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, frequency of sound emitted by bat,
n = 40 kHz
Velocity of bat, v_s = 0.03 v
where v is velocity of sound.
The bat is moving towards the flat wall. This is the case of source in motion and the observer at rest.
Therefore, the frequency of sound reflected at the wall is,
n = n_s (\(\frac {v}{v-v_s}\)) = n × (\(\frac {v}{v-0.03v}\))
= n × \(\frac {1}{0.97}\) = \(\frac {n}{0.97}\)
The frequency n’ is reflected by the wall and is again received by the bat moving towards the wall. This is the case of an observer moving towards the source with velocity v_L = 0.03 v.
The frequency observed by bat,

Question 52.
A bat, flying at velocity V_B = 12.5 m/s, is followed by a car running at velocity V_c = 50 m/s. Actual directions of the velocities of the car and the bat are as shown in the figure below, both being in the same horizontal plane (the plane of the figure). To detect the car, the bat radiates ultrasonic waves of frequency 36 kHz. Speed of sound at surrounding temperature is 350 m/s.

There is an ultrasonic frequency detector fitted in the car. Calculate the frequency recorded by this detector. The ultrasonic waves radiated by the bat are reflected by the car. The bat detects these waves and from the detected frequency, it knows about the speed of the car. Calculate the frequency of the reflected waves as detected by the bat. (sin 37° = cos 53° ~ 0.6, sin 53° = cos 37° ~ 0.8)
Answer:
The components of velocities of the bat and the car, along the line joining them, are
v_c cos 53° ~ 50 × 0.6 = 30 m s^-1 and
v_B cos 37° ~ 12.5 × 0.8 = 10 m s^-1

Doppler shifted frequency n = n₀ (\(\frac {v±v_1}{v±v_s}\))
upper signs to be used during approach, lower signs during recede.
Case I: Frequency radiated by the bat
n₀ = 36 × 10³ Hz,
The source (bat) is receding, while the listener (car) is approaching
v_S = v_B cos 37° = 10 m/s and
V_L = v_C cos 53° = 30 m/s
∴ Frequency detected by the detector in the car,
n = n₀ (\(\frac {v+v_L}{v+v_s}\))
∴ n = 36 × 10³ (\(\frac {350+30}{350+10}\)) = 36 × 10³ × \(\frac {38}{36}\)
∴ n = 38 × 10³ Hz = 38 kHz

Case II: The car is the source.
Emitted frequency by the car, is given as,
n₀ = 38 × 10³ Hz,
Car, the source, is approaching the listener (bat).
Now bat-the listener is receding while car the source is approaching,
∴ v_s = v_c cos 53° = 30 m/s
∴ v_L = v_B cos 37° = 10 m/s
∴ n = n₀ (\(\frac {v-v_L}{v-v_s}\))
∴ n = 38 × 10³ (\(\frac {350-10}{350-30}\)) = 38 × 10³ × \(\frac {34}{33}\)
= 39.15 × 10³ Hz
∴ n = 39.15 kHz

Question 53.
Source of sound is placed at one end of a copper bar of length 1 km. Two sounds are heard at the other end at an interval of 2.75 seconds, (speed of sound in air = 330 m/s)
i. Why do we hear two sounds?
ii. Find the velocity of sound in copper.
Answer:
i. Two sounds are heard because sound travels through air as well as through copper.

ii. In air, t₁ = \(\frac {distence}{time}\) = \(\frac {1000}{330}\) = 3.03 s
As the time interval is 2.75 seconds and sound travels faster in copper.
∴ In copper, t₂ = 3.03 – 2.75 = 0.28 s
∴ velocity of sound in copper = \(\frac {1000}{0.28}\) = 3571 m/s

Question 54.
If all the persons mentioned in the table below are listening to a match commentary on the same channel at their respective locations positioning at same distance from television, then will they hear the same line of the commentary at same instant of time? Justify your answer.

Name of a person	Location	Humidity
Aijun	Bangalore	65 %
Virendra	Hyderabad	56%
Vikas	Delhi	54%
Nilesh	Mumbai	75%

Answer:
As the order of humidity for the above locations is Mumbai > Bangalore > Hyderabad > Delhi.
As velocity of sound increases with increase in humidity, the order of velocity of sound at their respective locations is Mumbai > Bangalore > Hyderabad > Delhi.
Hence, the order of persons who would listen the line of commentary first to last is Nilesh, Arjun, Virendra, Vikas.

Question 55.
Speed of sound is greater during day than at night. True or False? Justify your answer.
Answer:
True. At night, the amount of CO₂ in atmosphere increases the density of atmosphere. Since, Speed of sound is inversely proportional to the square root of density. Hence, speed of sound is greater during day than in night.

Question 56.
Case I: During summer (33 °C), Prakash was waiting for a train at the platform, train arrived tt seconds after he heard train’s whistle.
Case II: During winter (19 °C), train arrived t₂ seconds after Prakash heard the sound of train’s whistle.
i. Will t₂ be equal to t₁? Justify your answer.
ii. Calculate the velocity of sound in both the cases.
(velocity of sound in air at 0 °C = 330 m/s)
Answer:
i. Velocity of sound is directly proportional to square root of absolute temperature.
Hence, whistle’s sound will be first heard by Prakash in summer than in winter.
Therefore, the time interval between sound and train reaching Prakash in summer will be more than in winter.
i.e.,t₁ > t₂

ii. When t = 33 °C
∴ v₁ = v₀ + 0.61t
= 330 + 0.61 × 33
∴ v₁ = 350.13 m/s
When t = 19 °C
v₂ = v₀ + 0.61t
= 330 + 0.61 × 19
v₂ = 341.59 m/s

Question 57.
You are at a large outdoor concert, seated 300 m from speaker system. The concert is also being broadcast live. Consider a listener 5000 km away who receives the broadcast. Who will hear the music first, you or listener and by what time difference? (Speed of light = 3 × 10⁸ m/s and speed of sound in air = 343 m/s)
Answer:
s₁ = 300 m,
v₁ = 343 m/s,
∴ t₁ = \(\frac {s_1}{v_1}\) = \(\frac {300}{343}\) = 0.8746 s
Now,
s₂ = 5000 km = 5 × 10⁶ m,
v₂ = c = 3 × 10⁸ m/s
∴ t₂ = \(\frac {s_2}{c}\) = \(\frac {5×10^6}{3×10^8}\) = 0.0167 s
∴ t₂ < t₁
∴ Listener will hear the music first.
Time difference = t₁ – t₂
= 0.8746 – 0.0167
= 0.8579 s
The listener will hear the music first, about 0.8579 s before the person present at the concert.

Question 58.
When a source of sound moves towards a stationary observer, then the pitch increases. Give reason.
Answer:
When a source of sound moves towards a stationary observer, then the increase in pitch is due to actual or apparent change in wavelength. When the source of sound moves towards an observer at rest, waves get compressed and the effective velocity of sound waves relative to source becomes less than the actual velocity. Hence the wavelength of sound waves an decreases which results into increase in pitch.

Question 59.
In the examples given below, state if the wave motion is transverse, longitudinal or a combination of both?

1. Light waves travelling from Sun to Earth.
2. ultrasonic waves in air produced by a vibrating quartz crystal.
3. Waves produced by a motor boat sailing in water.

Answer:

1. Light waves (from Sun to Earth) are electromagnetic waves which are transverse in nature.
2. Ultrasonic waves in air are basically sound waves of frequency greater than the audible frequencies. Therefore, these waves are longitudinal.
3. The water surface is cut laterally and pushed backwards by the propeller of motor boat. Therefore, the waves produced are a mixture of longitudinal and transverse waves.

Question 60.
Internet my friend
Answer:
https://hyperphysics.phys-astr.gsu.edu/ hbase/hframe.html
[Students are expected to visit the above mentioned website and collect more information about sound.]

Multiple Choice Questions

Question 1.
Water waves are …………….
(A) longitudinal
(B) transverse
(C) both longitudinal and transverse
(D) neither longitudinal nor transverse
Answer:
(C) both longitudinal and transverse

Question 2.
Sound travels fastest in ……………..
(A) water
(B) air
(C) steel
(D) kerosene oil
Answer:
(C) steel

Question 3.
At room temperature, velocity of sound in air at 10 atmospheric pressure and at 1 atmospheric pressure will be in the ratio ……………..
(A) 10 : 1
(B) 1 : 10
(C) 1 : 1
(D) cannot say
Answer:
(C) 1 : 1

Question 4.
In a gas, velocity of sound varies directly as ………………
(A) square root of isothermal elasticity.
(B) square of isothermal elasticity.
(C) square root of adiabatic elasticity.
(D) adiabatic elasticity.
Answer:
(C) square root of adiabatic elasticity.

Question 5.
At a given temperature, velocity of sound in oxygen and in hydrogen has the ratio …………………
(A) 4 : 1
(B) 1 : 4
(C) 1 : 1
(D) 2 : 1
Answer:
(B) 1 : 4

Question 6.
With decrease in water vapour content in air, velocity of sound …………………..
(A) increases
(B) decreases
(C) remains constant
(D) cannot say
Answer:
(B) decreases

Question 7.
The temperature at which speed of sound in air becomes double its value at 0 °C is ……………….
(A) 546 °C
(B) 819 °C
(C) 273 °C
(D) 1092 °C
Answer:
(B) 819 °C

Question 8.
The velocity of sound in air at NTP is 330 m/s. What will be its value when temperature is doubled and pressure is halved?
(A) 330 m/s
(B) 165 m/s
(C) 330 √2 m/s
(D) \(\frac {330}{√2}\) m/s
Answer:
(D) \(\frac {330}{√2}\) m/s

Question 9.
A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second have wave velocity
(A) 2.5 m/s
(B) 5.0 m/s
(C) 8.0 m/s
(D) 10.0 m/s
Answer:
(D) 10.0 m/s

Question 10.
A radio station broadcasts at 760 kHz. What is the wavelength of the station?
(A) 395 m
(B) 790 m
(C) 760 m
(D) 197.5 m
Answer:
(A) 395 m

Question 11.
If the bulk modulus of water is 2100 MPa, what is the speed of sound in water?
(A) 1450 m/s
(B) 2100 m/s
(C) 0.21 m/s
(D) 21 m/s
Answer:
(A) 1450 m/s

Question 12.
If speed of sound in air at 0°C is 331 m/s. What will be its value at 35° C?
(A) 331 m/s
(B) 366 m/s
(C) 351.6 m/s
(D) 332 m/s.
Answer:
(C) 351.6 m/s

Question 13.
For a progressive wave, in the usual notation
(A) v = λT
(B) n = \(\frac {v}{λ}\)
(C) T = λv
(D) λ = \(\frac {1}{n}\)
Answer:
(B) n = \(\frac {v}{λ}\)

Question 14.
At normal temperature, for an echo to be heard the reflecting surface should be at a minimum distance of ………………. m.
(A) 34.4
(B) 17.2
(C) 10
(D) 20
Answer:
(B) 17.2

Question 15.
In a transverse wave, are regions of negative displacement.
(A) rarefactions
(B) compressions
(C) crests
(D) troughs
Answer:
(D) troughs

Question 16.
If pressure of air gets doubled at constant temperature then velocity of sound in air ……………….
(A) gets doubled
(B) remains unchanged
(C) √2 times initial velocity
(D) becomes half
Answer:
(B) remains unchanged

Question 17.
Wave motion has ……………………
(A) single periodicity.
(B) double periodicity.
(C) only periodicity in space.
(D) only periodicity in time.
Answer:
(B) double periodicity.

Question 18.
The speed of the mechanical wave depends upon ………………
(A) elastic properties of the medium only.
(B) density of the medium only.
(C) elastic properties and density of the medium
(D) initial speed.
Answer:
(C) elastic properties and density of the medium

Question 19.
Longitudinal waves CANNOT be …………………
(A) reflected
(B) refracted
(C) scattered
(D) polarised
Answer:
(D) polarised

Question 20.
Wavelength of the transverse wave is 30 cm. If the particle at some instant has displacement 2 cm, find the displacement of the particle 15 cm away at the same instant.
(A) 2 cm
(B) 17 cm
(C) -2 cm
(D) -17 cm
Answer:
(C) -2 cm

Question 21.
The wavelength of sound in air is 1.5 m and that in liquid is 2 m. If the velocity of sound in air is 330 m/s, the velocity of sound in liquid is
(A) 330 m/s
(B) 440 m/’s
(C) 495 m/s
(D) 660 m/s
Answer:
(B) 440 m/’s

Question 22.
The velocity of sound in a gas is 340 m/s at the pressure P, what will be the velocity of the gas when only pressure is doubled and temperature same?
(A) 170 m/s
(B) 243 m/s
(C) 340 m/s
(D) 680 m/s
Answer:
(C) 340 m/s

Question 23.
Choose the correct statement.
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
(B) For 1 °C rise in temperature, velocity of sound decreases by 0.61 m/s.
(C) For 1 °C rise in temperature, velocity of sound decreases by \(\frac {1}{273}\) m/s.
(D) For 1 °C rise in temperature, velocity of sound increases by \(\frac {1}{273}\) m/s.
Answer:
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.

Question 24.
A sound note emitted from a certain source has a velocity of 300 m/s in air and 1050 m/s in water. If the wavelength of sound note in air is 2 m, the wavelength in water is …………
(A) 2 m
(B) 6 m
(C) 7 m
(D) 12 m
Answer:
(C) 7 m

Question 25.
A thunder clap was heard 6 seconds after a lightening flash was seen. If the speed of sound in air is 340 m/s at the time of observation, the distance of the listener from the thunder clap is ………………
(A) 56.6 m
(B) 346 m
(C) 1020 m
(D) 2040 m
Answer:
(D) 2040 m

Question 26.
The speed of sound in air at NTP is 330 m/s. The period of sound wave of wavelength 66 cm is …………………
(A) 0.2 s
(B) 0.1
(C) 0.1 × 10^-2 s
(D) 0.2 × 10^-2 s
Answer:
(D) 0.2 × 10^-2 s

Question 27.
If the velocity of sound in hydrogen is 1248 m/s, the velocity of sound in oxygen is [Given: M_O = 32 and M_H = 2]
(A) 1248 m/s
(B) 624 m/s
(C) 312 m/s
(D) 300 m/s
Answer:
(C) 312 m/s

Question 28.
If the source is moving away from the observer, then the apparent frequency …………..
(A) will increase
(B) will remain the same
(C) will be zero
(D) will decrease
Answer:
(D) will decrease

Question 29.
The working of SONAR is based on …………………
(A) resonance
(B) speed of a star
(C) Doppler effect
(D) speed of rotation of sun
Answer:
(C) Doppler effect

Question 30.
The formula for speed of a transverse wave on a stretched spring is ……………… (m = linear mass density, T = tension in Spring)
(A) v = \(\sqrt{\frac {m}{T}}\)
(B) v = \(\sqrt{\frac {T}{m}}\)
(C) v = (\(\frac {m}{T}\))^{\(\frac {3}{2}\)}
(D) v = (\(\frac {T}{m}\))^{\(\frac {3}{2}\)}
Answer:
(B) v = \(\sqrt{\frac {T}{m}}\)

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 13 Respiration and Energy Transfer Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 1.
What is Phosphorylation?
Answer:
It is the formation of ATP, by addition inorganic phosphate to ADP.
ADP + Pi → ATP

Question 2.
What are the different ways of Phosphorylation?
Answer:
Phosphorylation occurs in three different ways as – photophosphorylation, substrate-level phosphorylation and oxidative phosphorylation.

Question 3.
What is Substrate-level phosphorylation? Where does it occur?
Answer:
Substrate-level phosphorylation is a direct phosphorylation of ADP by transfer of a phosphate group from any suitable substrate. It occurs in cytoplasm of the cells and matrix of mitochondria.

Question 4.
What is oxidative phosphorylation? Mention the site of this reaction.
Answer:
In oxidative phosphorylation ATP is synthesize by using the energy released during the oxidation of substrates like NADH + H⁺ and FADH₂. This occurs on the inner mitochondrial membrane only.

Question 5.
When is ATP hydrolysed?
Answer:
ATP is hydrolysed whenever energy is required for any metabolic reactions.

Question 6.
What is respiration?
Answer:
Respiration is a catabolic process wherein complex organic substrate is oxidized to simple components to generate biological energy, i.e. ATP.

Question 7.
Give an example for anabolic and catabolic process.
Answer:
Anabolic process- Photosynthesis (Biosynthetic process).
Caiabolic process- Respiration (Breakdown process).

Question 8.
Name the following.

1. The substrate that undergoes oxidation during oxidative phosphorylation
2. The phosphorylation that occurs only in inner mitochondria! membrane
3. Two ways of cellular respiration

Answer:

1. NADH+H⁺ and FADH₂
2. Oxidative phosphorylation
3. Anaerobic and aerobic respiration

Question 9.
What is anaerobic respiration?
Answer:
1. Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation.
2. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol.

Question 10.
What is glycolysis? Describe various steps involved in glycolysis.
OR
Describe the process involved in formation of pyruvic acid from glucose molecule.
OR
What is glycolysis? Describe the process of glycolysis with the help of schematic representation.
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 11.
Write down the overall chemical reaction of a process in which glucose in broken down to two molecules of pyruvic acid.
Answer:
Overall reaction of glycolysis:
Glucose + 2 ATP + 2iP + 4 ADP + 2 NAD⁺ → 2 Pyruvate + 2 ADP + 4 ATP + 2 NADH + 2 Ha⁺ + 2 H₂O

Question 12.
Write explanatorv notes on: Glycolysis
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 13.
How glycolysis is regulated?
Answer:
Glycolysis is strongly regulated by the complex interplay between ATP consumption, NADH₂ regeneration and regulation of various glycolytic enzymes like hexokinase, PFK-1, pyruvate kinase, etc. Besides, it is also controlled by hormones like glucagon, epinephrine and insulin.

Question 14.
Where does glycolysis take place in a cell?
Answer:
Glycolysis takes place in the cytoplasm of a cell.

Question 15.
What are the products of cleavage in glycolysis?
Answer:
Dihydroxyacetone phosphate (DHAP) and 3-phosphoglyceraldehyde (3-PGAL) are the products of cleavage in glycolysis.

Question 16.
Where does dehydration occur in glycolysis?
Answer:
In glycolysis, dehydration occurs when 2-Phosphoglyceric acid loses a water molecule (dehydration) to form phosphoenol pyruvic acid in presence of the enzyme enolase.

Question 17.
Name the process which is common to both aerobic and anaerobic respiration.
Answer:
Glycolysis is common to both aerobic and anaerobic respiration.

Question 18.
Name the enzymes that catalyse the irreversible reactions.
Answer:
Hexokinase, Phosphoffuctokinase, Phosphoglycerate kinase and Pyruvate kinase are the enzymes that catalyse the irreversible reactions.

Question 19.
Where glycolysis is the only source of energy production?
Answer:
Glycolysis is only source of energy production in erythrocytes, renal medulla, brain and sperm.

Question 20.
Which type of muscle fibres are rich in myoglobin?
Answer:
Red muscles are rich in myoglobin.

Question 21.
Which type of muscle fibre mainly performs anaerobic respiration?
Answer:
White muscle fibres mainly performs anaerobic respiration.

Question 22.
What is lactic acid fermentation?
Answer:
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H⁺ to form lactic acid as the product and NAD⁺ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process. It is represented as follows:

Question 23.
What is the fate of pyruvate in yeast?
Answer:
Yeast shows both aerobic and anaerobic respiration depending upon the presence or absence of oxygen.
1. In absence of oxygen, the pyruvate undergoes anaerobic respiration where it is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H to ethanol and carbon dioxide. This type of anaerobic respiration is termed alcoholic fermentation.
2. In the presence of oxygen however, it can respire aerobically to produce C02 and H20.

Question 24.
What is alcoholic fermentation?
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H+ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation. It is represented as follows:

Question 25.
Name the process of respiration which does not involve intake of oxygen (0₂) and release of carbon dioxide (C0₂).
Answer:
Lactic acid fermentation is the process of anaerobic respiration which does not involve intake of oxygen (0₂) and release of carbon dioxide (C0₂).

Question 26.
Why yeast stops multiplying in the culture after alcoholic fermentation?
Answer:
After alcoholic fermentation the multiplication of yeast stops because the alcohol formed during the process kills the yeast cells.

Question 27.
What is aerobic respiration?
Answer:

1. Aerobic respiration occurs in the presence of free molecular oxygen during oxidation of glucose.
2. In this type of respiration, the glucose is completely oxidized to C0₂ and H₂0 with release of large amount of energy.
3. It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Question 28.
Where does aerobic respiration occur in eukaryotic cell?
Answer:
Aerobic respiration occurs in the mitochondria in eukaryotes.

Question 29.
Explain the conversion of pyruvic acid to acetyl CoA.
OR
Describe the connecting link between glycolysis and Krebs cycle.
Answer:

1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 30.
Name the following.
1. Which enzymes converts pyruvic acid to acetyl CoA?
2. Name the coenzyme required by pyruvate dehydrogenase.
Answer:
1. Pyruvate dehydrogenase
2. Thiamin (vitamin Bi)

Question 31.
Why is acetyl Co-A called connecting link between glycolysis and Krebs cycle?
Answer:

1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 32.
Why vitamin Bi is important for maintaining good health?
Answer:
1. Pyruvate dehydrogenase (PDH) requires thiamin (vitamin Bi) as a co-enzyme. It cannot function in absence of vitamin B₁.
2. Thiamin deficiency causes many disorders such as pyruvic acidosis and lactic acidosis, which are life threatening conditions. Hence, it is required to maintain good health.

Question 33.
Describe Citric acid cycle.
OR
With the help of schematic representation explain Krebs cycle.
Answer:

1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
5. This reaction is an oxidative decarboxylation process and produces H⁺ ions and electrons along with carbon dioxide.
6. During the process NAD⁺ is reduced to NADH+H⁺.
7. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
8. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
9. Regeneration of oxaloacetic acid occurs to complete the cycle.
10. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD⁺ or FAD⁺ as the coenzyme.
11. The coenzymes are consequently reduced to NADH⁺H⁺ and FADH₂ respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
12. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Question 34.
What is the site of Krebs cycle in mitochondria?
Answer:
Krebs cycle takes place in the mitochondrial matrix.

Question 35.
Match the following:

Column I	Column II	Column III
1. Krebs cycle	(a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate	(p) Pyruvate dehydrogenase
2. Glycolysis	(b) Succinyl CoA → Succinate	(q) NADH dehydrogenase
3. ETS	(c) Pyruvic acid → Acetyl CoA	(r) Two ATP are produced
4. Link reaction	(d) Transfer of electrons from Complex to Ubiquinone	(s) Succinyl CoA – synthetase

Answer:

Column I	Column II	Column III
1. Krebs cycle	(b) Succinyl CoA → Succinate	(s) Succinyl CoA – synthetase
2. Glycolysis	(a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate	(r) Two ATP are produced
3. ETS	(d) Transfer of electrons from Complex I to Ubiquinone	(q) NADH dehydrogenase
4. Link reaction	(c) Pyruvic acid → Acetyl CoA	(p) Pyruvate dehydrogenase

Question 36.
Explain ETS.
OR
Illustrate the mechanism of electron transport system.
OR
Give an account of ATP generation steps during ETS.
OR
Explain the mechanism of electron transportation system (ETS).
OR
Explain terminal oxidation.
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube (Tube B).
6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Question 37.
Explain the significance of electron transport system.
Answer:
Significance of ETS:

1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H⁺ and FADH₂) for recycling.
4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
5. ETS produces water molecules.

Question 38.
What is oxidative phosphorylation?
Answer:
Oxidative phosphorylation: It is a metabolic pathway that uses energy released by the oxidation of substrates to produce ATP.

1. Oxidative phosphorylation takes place in the mitochondrial membrane.
2. Many intermediate products during respiration are oxidised and release 2H⁺.
3. The released hydrogen is trapped by NAD⁺ or FAD⁺.
4. Electrons pass through electron transport system to produce ATP and metabolic water.

Question 39.
Give the balance sheet of ATP formed in aerobic respiration.
Answer:

Step of Respiration	Consumption	Production				Total	Net benefit
		Substrate level phosphorylation	1 Oxidative Phosphorylation
			NADH+ H+	FADH2	Total
Glycolysis	2	4	2 × 3 = 6	___________	6	10	8
Pyruvate → AcetylCoA	_______________	___________	2 × 3 = 6	___________	6	6	6
Krebs cycle	_______________	1 × 2 = 2	6 × 3 = 18	2×2 = 4	22	24	24
Total	2	[6]	30 + 4 = [34]			m	38

Question 40.
With the help of an experiment explain how yeast respires anaerobically.
Answer:
Respiration in yeast can be demonstrated with the help of an experiment.
Anaerobic respiration in yeast:

1. A pinch of dry baker’s yeast suspended in water containing 10ml of 10% glucose in a test tube (test tube A).
2. The surface of the liquid is covered with oil to prevent entry of air and the test tube is closed tightly with rubber stopper to prevent leakage.
3. One end of a short-bent glass tube is inserted through it to reach the air inside the tube.
4. Other end of the glass tube is connected by a polyethylene or rubber tubing to another bent glass tube fitted into a stopper.
5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube
   (Tube B).
6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Question 41.
With the help of an experiment explain how germinating seeds respire.
Answer:

1. Seed coats of a few germinating seeds (peas, beans or gram) are removed and are then put in a test tube filled with mercury.
2. After closing the test tube with the thumb, it is vertically inverted in a trough of mercury and the thumb is carefully removed.
3. Being lighter than mercury, the seeds rise to the closed upper end of the test Uibe. No gas is seen at first in the test tube.
4. As germination proceeds, a gas begins to collect at the top of the mercury in the test tube.
5. On introducing a pellet of potassium hydroxide into the tube, it rises to the top and absorbs the gas. The mercury again fills the tube.
6. The potassium hydroxide reacts with carbon dioxide gas to produce potassium carbonate and water.
7. The gas therefore disappears. Evidently germinating seeds produce carbon dioxide by anaerobic respiration in the absence of oxygen in the mercury column.

[Note: Mercury is toxic chemical. Proper precautions must be taken while performing this experiment ]
[Note: Students can scan the adjacent QR code to get conceptual clarity with the aid of a relevant example.]

Question 42.
Give the significance of respiration.
OR
Write a short note on significance of respiration.
Answer:

1. Respiration provides energy for biosynthesis of biomolecules.
2. It is also a source of energy for cell division, growth, repairs and replacement of worn out parts, movements,
   locomotion etc.
3. Various intermediates of Krebs cycle are used as building blocks for synthesis of other complex compounds.
4. Coupled with photosynthesis, it helps to maintain the balance between C0₂ and O₂ in the atmosphere.
5. Anaerobic respiration (fermentation) is used in various industries such as dairies, bakeries, distilleries, leather industries, paper industries etc. It is used in the commercial production of alcohol, organic acids, vitamins, antibiotics etc.

Question 43.
Match the following:

Column I	Column II	Column III
1. Krebs cycle	(a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate	(p) Pyruvate dehydrogenase
2. Glycolysis	(b) Succinyl CoA → Succinate	(q) NADH dehydrogenase
3. ETS	(c) Pyruvic acid → Acetyl CoA	(r) Two ATP are produced
4. Link reaction	(d) Transfer of electrons from Complex I to Ubiquinone	(s) Succinyl CoA – synthetase

Answer:

Column I	Column II	Column III
1. Krebs cycle	(b) Succinyl CoA → Succinate	(s) Succinyl CoA – synthetase
2. Glycolysis	(a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate	(r) Two ATP are produced
3. ETS	(d) Transfer of electrons from Complex I to Ubiquinone	(q) NADH dehydrogenase
4. Link reaction	(c) Pyruvic acid → Acetyl CoA	(p) Pyruvate dehydrogenase

Question 44.
Apply Your Knowledge

Question 1.
While teaching respiration professor said that oxygen is important for respiration in humans, Rakesh kept thinking, where exactly oxygen is required during cellular respiration? What would be the correct explanation for his doubt?
Answer:
Oxygen is very important for life, we humans cannot survive without oxygen. Glycolysis, link reaction and Krebs cycle do not involve oxygen. In cellular reaction, oxygen is required only during ETS where it acts as an electron acceptor hence leading to terminal oxidation where it gets converted into water called metabolic water.

The process of respiration is very fast and occurs continuously as cell require continuous supply of energy to carry out metabolic activities, thus, we require oxygen for cellular respiration even when we are sleeping.

Question 2.
Sonal while studying ETS had a doubt, why FADH2 yields only 2 ATP’s whereas NADH₂ yields three ATP’s? What would be the correct explanation for her doubt?
Answer:

1. During ETS, NADH₂ and FADH₂ undergoes oxidation where they are converted to NAD⁺ and FAD⁺ respectively.
2. During this conversion, hydrogen is removed, which latter forms proton and electron. This electron so formed is passed on to different electron carriers. During this transfer of electron, energy is released which is used in ATP synthesis.
3. In case of NADH₂, energy is released at three places hence results into formation of 3 ATPs whereas, in case of FADH₂ energy is released at only two places, hence results into formation of only 2 ATPs.
4. Here, the ATP formation occurs by Chemiosmotic hypothesis.

Question 45.
Quick Review

Various steps involved in glycolysis:

Step(s)	Substrate	Enzyme	End product(s)
Phosphorylation	Glucose + ATP	Hexokinase	Glucose – 6 – Phosphate + ADP
Isomerisation	Glucose – 6-Phosphate	Phosphohexose isomerase	Fructose – 6 – Phosphate
Phosphorylation	Fructose – 6-Phosphate + ATP (Phosphate donor)	Phosphofructokinase	Fructose 1,6-Diphosphate + ADP
Cleavage	Fructose -1, 6-Diphosphate	Aldolase	3 – Phosphoglyceraldehyde + Dihydroxyacetone phosphate
Phosphorylation and Dehydrogenation	3-PGAL + H3PO4 + NAD+	Triosephosphate dehydrogenase	1, 3 Diphosphoglyceric acid + NADH + H+
Dephosphorylation	1, 3-DPGA + ADP	Diphosphoglycerate kinase (Mg2)	ATP + 3-Phosphoglyceric acid
Rearrangement	3-PGA	Phosphoglycerate mutase	2-Phosphoglyceric acid
Dehydration	2-PGA	Enolase	Phosphoenol Pyruvic acid + H20
Dephosphorylation	Phosphoenol Pyruvic acid +ADP	Pyruvate kinase	Pyruvic acid + ATP

Various steps involved in Krebs cycle:

Step(s)	Substrate	Enzyme	End product(s)
Condensation	Acetyl-CoA+Oxalo- acetic acid + H20	Citrate synthase	Citric acid (6C) + Coenzyme- A
Dehydration Hydration	It is again completed in two steps: Citric acid Cis-Aconitic acid + H20	Aconitase Aconitase	Cis-Aconitic acid + H20 Iso-citric acid(6C)
Oxidative decarboxylation	a. Isocitric acid + NAD+	Isocitrate dehydrogenase	Oxalo succinic acid(6C) + NADH + H+
	b. Oxalosuccinic acid	Isocitrate dehydrogenase	a – ketoglutaric acid(5C) + CO
Oxidative Decarboxylation. (-2H) (-C02)	a-Ketoglutaric acid + H20 + NAD + Coenzyme A	a-Ketoglutarate dehydrogenase complex	Succinyl-CoA(4C) + NADH + H++co2
Substrate level phosphorylation	Succinyl – Co A + GDP + iP	Succinyl CoA synthetase	Succinic acid (4C) + Coenzyme-A + GTP
Dehydrogenation (-2H)	Succinic acid + FAD (H-acceptor)	Succinate dehydrogenase	Fumaric acid (4C) + FADH2
Hydration(+H20)	Fumaric acid + H20	Hydration(+H20)	Fumaric acid + H20
Dehydrogenation (-2H)	Malic acid + NAD+ (H-acceptor)	Dehydrogenation (-2H)	Malic acid + NAD+ (H-acceptor)

Question 46.
Exercise

Question 1.
Define phosphorylation.
Answer:
It is the formation of ATP, by addition inorganic phosphate to ADP.
ADP + Pi → ATP

Question 2.
Mention the different ways of phosphorylation.
Answer:
Phosphorylation occurs in three different ways as – photophosphorylation, substrate-level phosphorylation and oxidative phosphorylation.

Question 3.
Define substrate-level phosphorylation.
Answer:
Substrate-level phosphorylation is a direct phosphorylation of ADP by transfer of a phosphate group from any suitable substrate. It occurs in cytoplasm of the cells and matrix of mitochondria.

Question 4.
What is fermentation?
Answer:
1. Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation.
2. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol.

Question 5.
Where does glycolysis occurs?
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 6.
What is glycolysis? Explain with the help of schematic representation.
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 7.
Define fermentation. What are the different types of fermentation?
Answer:
1. Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation.
2. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol.
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H⁺ to form lactic acid as the product and NAD⁺ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process.
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H⁺ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Question 8.
Name the products of lactic acid fermentation.
Answer:
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H⁺ to form lactic acid as the product and NAD⁺ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process.

Question 9.
Write the chemical reaction of lactic acid fermentation.
Answer:
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H⁺ to form lactic acid as the product and NAD⁺ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process.

Question 10.
Explain alcoholic fermentation.
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H⁺ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Question 11.
Less energy is produced during anaerobic respiration that in aerobic respiration. Justify.
Answer:
Anaerobic respiration produces less energy because:

1. Incomplete breakdown of respiratory substrate takes place.
2. Some of the products of anaerobic respiration can be oxidised further to release energy which shows that anaerobic respiration does not liberate the whole energy contained in the respiratory substrate.
3. NADH₂ does not produce ATP, as electron transport is absent.
4. Only 2 ATP molecules are generated from one molecule of glucose during anaerobic respiration.

Question 12.
Enlist steps involved in aerobic respiration.
Answer:
It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Question 13.
Write the reaction of connecting link between glycolysis and Krebs cycle.
Answer:
This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 14.
How glycolysis is regulated?
Answer:
Glycolysis is strongly regulated by the complex interplay between ATP consumption, NADH₂ regeneration and regulation of various glycolytic enzymes like hexokinase, PFK-1, pyruvate kinase, etc. Besides, it is also controlled by hormones like glucagon, epinephrine and insulin.

Question 15.
Write the significance of ETS.
Answer:
Significance of ETS:

1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H⁺ and FADH₂) for recycling.
4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
5. ETS produces water molecules.

Question 16.
Give the schematic representation of glycolysis.
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 17.
Describe the formation of Acetyl Co-A in respiration.
Answer:

1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 18.
Give the schematic representation of the overall view of Krebs cycle.
Answer:

1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
5. This reaction is an oxidative decarboxylation process and produces H⁺ ions and electrons along with carbon dioxide. During the process NAD⁺ is reduced to NADH+H⁺.
6. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
7. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
8. Regeneration of oxaloacetic acid occurs to complete the cycle.
9. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD⁺ or FAD⁺ as the coenzyme.
10. The coenzymes are consequently reduced to NADH+H⁺ and FADH₂ respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
11. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Question 19.
Explain why the respiratory pathway is an amphibolic pathway?
Answer:

1. Respiration is considered as a catabolic process; however, it is not entirely correct in case of Krebs cycle.
2. Many reactions of Krebs cycle involve oxidation of acetyl CoA to release energy and C0₂.
3. However, the breakdown of respiratory substrates provides intermediates like a-ketoglutarate, oxaloacetate are used as precursors for synthesis of fatty acids, glutamic acid and aspartic acid respectively.
4. Thus, as the same respiratory process acts as catabolic as well as anabolic pathway for synthesis of various intermediate metabolic products, it is called amphibolic pathway.

Question 20.
Write a detailed note on the connecting link between glycolysis and Krebs cycle.
Answer:

1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 21.
Give the diagrammatic representation of ETS.
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH⁺H⁺ and FADH₂, electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C.
7. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
8. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
9. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
10. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
11. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
12. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question 22.
Explain the process of terminal oxidation. Give its significance (any two points).
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH⁺H⁺ and FADH₂, electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C.
7. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
8. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
9. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
10. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
11. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
12. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question 23.
1. Enlist the main steps involved in aerobic respiration and their place of occurrence.
2. Differentiate between photosynthesis and aerobic respiration.
Answer:
1. Glycolysis takes place in the cytoplasm of a cell.
2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.,
3. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria. and These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.

Question 24.
1. Give the schematic representation of ETS.
2. Differentiate between respiration and combustion.
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH⁺H⁺ and FADH₂, electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C.
7. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
8. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
9. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
10. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
11. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
12. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question 25.
Enlist the different respiratory substrate.
Answer:
Respiratory substrates are the molecules that are oxidized during respiration to release energy which can be used for ATP synthesis. Carbohydrates, fats and proteins are the common respiratory substrate. Glucose is the most common respiratory substrate.

Question 26.
Define R.Q. What is its value for fats?
Answer:
1. Respiratory quotient (R.Q.) or respiratory ratio is the ratio of volume of CO₂ released to the volume of O₂ consumed in respiration.
2. R.Q. = Volume of CO₂ released/Volume of O₂ consumed

Question 27.
Write the significance of respiration.
Answer:

1. Respiration provides energy for biosynthesis of biomolecules.
2. It is also a source of energy for cell division, growth, repairs and replacement of worn out parts, movements,
   locomotion etc.
3. Various intermediates of Krebs cycle are used as building blocks for synthesis of other complex compounds.
4. Coupled with photosynthesis, it helps to maintain the balance between C0₂ and O₂ in the atmosphere.
5. Anaerobic respiration (fermentation) is used in various industries such as dairies, bakeries, distilleries, leather industries, paper industries etc. It is used in the commercial production of alcohol, organic acids, vitamins, antibiotics etc.

Question 47.
Multiple Choice Questions:

Question 1.
Respiration is regarded as a ________ process.
(A) catabolic
(B) anabolic
(C) reduction
(D) synthetic
Answer:
(A) catabolic

Question 2.
Anaerobic respiration is also called as ________ .
(A) Glycolysis
(B) fermentation
(C) phosphoryaltion
(D) decarboxylation
Answer:
(B) fermentation

Question 3.
The first compound formed in glycolysis is
(A) Glucose-6-phosphate
(B) Glucose-1,6-biphosphate
(C) Fructose-6-phosphate
(D) Pyruvic acid
Answer:
(A) Glucose-6-phosphate

Question 4.
Which of the following compounds in last step of glycolysis gives pyruvic acid?
(A) 3 – phosphoglyceraldehyde
(B) dihydroxy acetone phosphate
(C) phosphoenolpyruvate
(D) 2-phosphoglycerate
Answer:
(C) phosphoenolpyruvate

Question 5.
In glycolysis, dehydration occurs during formation of
(A) 3 – phosphoglyceraldehyde
(B) 2 – phosphoglycerate
(C) phosphoenolpyruvate
(D) dihydroxyacetone phosphate
Answer:
(B) 2 – phosphoglycerate

Question 6.
In which of the following steps dehydrogenation occurs?
(A) Glucose → Glucose 6-phosphate
(B) 3-phophoglcerate → 2-phophoglccratc
(C) phosphoenolpyruvate → pyruvate
(D) 3 – phosphoglyceraldehyde → 1, 3-bisphosphoglycerate
Answer:
(D) 3 – phosphoglyceraldehyde → 1, 3-bisphosphoglycerate

Question 7.
The compound common to both aerobic and anaerobic respiration is
(A) C0₂
(B) pyruvic acid
(C) acetyl CoA
(D) free oxygen
Answer:
(B) pyruvic acid

Question 8.
Which compound is found both in respiration and photosynthesis?
(A) Phosphoglycerate
(B) Phosphoglyceraldehyde
(C) Both (A) and (B)
(D) Succinic acid
Answer:
(C) Both (A) and (B)

Question 9.
Which type of respiration does not release C0₂?
(A) Aerobic respiration
(B) Alcoholic fermentation
(C) Lactic acid fermentation
(D) Krebs cycle
Answer:
(C) Lactic acid fermentation

Question 10.
What is the overall goal of glycolysis, Krebs cycle and electron transport system?
(A) Synthesis of ATP in fermentation reaction
(B) Carbohydrates
(C) Nucleic acids
(D) ATP in small stepwise units
Answer:
(D) ATP in small stepwise units

Question 11.
The intermediate between glycolysis and TCA cycle is:
(A) Lactic acid
(B) Acetaldehyde
(C) Fructose-6-phosphate
(D) Acetyl Co-A
Answer:
(D) Acetyl Co-A

Question 12.
During alcoholic fermentation, decarboxylation of pyruvate gives
(A) acetaldehyde
(B) lactic acid
(C) ethyl alcohol
(D) methyl alcohol
Answer:
(A) acetaldehyde

Question 13.
Where the link reaction occurs in prokaryotes?
(A) cytoplasm
(B) mitochondrial matrix
(C) cell membrane
(D) mitochondrial membrane
Answer:
(A) cytoplasm

Question 14.
In Krebs cycle, dehydration of substrate occurs
(A) once
(B) twice
(C) thrice
(D) four times
Answer:
(A) once

Question 15.
Which of the following steps generate ATP without ETS?
(A) Pyruvic acid → Acetyl Co-A
(B) ∝-ketoglutarate → Succinic acid
(C) Iso-citric acid → Oxalosuccinic acid
(D) Succinyl Co-A → Succinic acid
Answer:
(D) Succinyl Co-A → Succinic acid

Question 16.
In Krebs cycle, the acid which undergoes oxidative decarboxylation is
(A) citric acid
(B) succinic acid
(C) malic acid
(D) ∝-ketoglutaric acid
Answer:
(D) ∝-ketoglutaric acid

Question 17.
During Krebs cycle, fumaric acid gets converted into malic acid by
(A) decarboxylation
(B) dehydrogenation
(C) dehydration
(D) hydration
Answer:
(D) hydration

Question 18.
Krebs cycle is also called TCA cycle because
(A) the first compound formed is citric acid.
(B) it was discovered by Sir Hans Krebs.
(C) organic acids formed have 3 carboxylic acid groups.
(D) acetyl Co-A is formed
Answer:
(C) organic acids formed have 3 carboxylic acid groups.

Question 19.
Which of the following compound is the acceptor of Acetyl Co-A in Krebs cycle?
(A) Oxaloacetic acid
(B) Fumaric acid
(C) Malic acid
(D) Oxalo succinic acid
Answer:
(A) Oxaloacetic acid

Question 20.
Which of the following compounds is formed in Krebs cycle from fumaric acid?
(A) Oxalo acetic acid
(B) Malic acid
(C) ∝-KGA
(D) Citric acid
Answer:
(B) Malic acid

Question 21.
Which of the following step of aerobic respiration would be omitted when fatty acids are used as respiratory substrate?
(A) Glycolysis
(B) Krebs cycle
(C) Electron transfer chain reaction
(D) Terminal oxidation.
Answer:
(A) Glycolysis

Question 22.
During Krebs cycle, decarboxylation occurs _______ time/s.
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(B) 2

Question 23.
The conversion of malic acid to oxalo acetic acid is catalyzed by
(A) malate reductase
(B) malate thiokinase
(C) fumarase
(D) malate dehydrogenase
Answer:
(D) malate dehydrogenase

Question 24.
Electron carriers of oxidative phosphorylation are present on
(A) outer membrane of mitochondria.
(B) inner membrane of mitochondria.
(C) thylakoid membrane of chloroplast.
(D) matrix of mitochondria.
Answer:
(B) inner membrane of mitochondria.

Question 25.
Which of the following derives maximum energy per molecule of glucose?
(A) Alcoholic fermentation.
(B) Lactic acid fermentation.
(C) Aerobic respiration in unicellular organisms.
(D) Glycolysis in liver cells.
Answer:
(C) Aerobic respiration in unicellular organisms.

Question 26.
The cytochrome which donates de-energised electron to oxygen is
(A) cytochrome-a
(B) cytochrome-b
(C) cytochrome-a₃
(D) cytochrome-c
Answer:
(C) cytochrome-a₃

Question 27.
In terminal oxidation FADH₂ is oxidized by
(A) complex I
(B) complex II
(C) complex III
(D) complex IV
Answer:
(B) complex II

Question 28.
The net gain of energy from a molecule of glucose in the aerobic respiration is
(A) 38
(B) 35
(C) 70
(D) 76
Answer:
(A) 38

Question 29.
Each molecule of NADH₂ through ETS yields
(A) 1 ATP
(B) 2 ATPs
(C) 3 ATPs
(D) 4 ATPs
Answer:
(C) 3 ATPs

Question 30.
One glucose molecule, through ETS yields
(A) 2 ATP molecules
(B) 3 ATP molecules
(C) 34 ATP molecules
(D) 38 ATP molecules
Answer:
(C) 34 ATP molecules

Question 31.
The adenosine triphosphate (ATP) gain during glycolysis, connecting link and Krebs cycle respectively are _______ .
(A) 8, 6,24
(B) 8,24, 6
(C) 24, 8, 6
(D) 6, 8, 24
Answer:
(A) 8, 6,24

Question 32.
The respiratory quotient (R.Q.) of carbohydrate is ________ .
(A) 0.7
(B) 1
(C) 0.9
(D) 0.1
Answer:
(B) 1

Question 33.
R. Q. for proteins is about _______ .
(A) 0.7
(B) 0.8
(C) 0.9
(D) 1.0
Answer:
(C) 0.9

Question 48.
Competitive Corner:

Question 1.
Which of the following statements regarding mitochondria is INCORRECT?
(A) Inner membrane is convoluted with infoldings.
(B) Mitochondrial matrix contains single circular DNA molecule and ribosomes.
(C) Outer membrane is permeable to monomers of carbohydrates, fats and proteins.
(D) Enzymes of electron transport are embedded in outer membrane.
Hint: Enzymes of electron transport are embedded in inner membrane.
Answer:
(D) Enzymes of electron transport are embedded in outer membrane.

Question 2.
Where is the respiratory electron transport system (ETS) located in plants?
(A) Intermembrane space
(B) Mitochondrial matrix
(C) Outer mitochondrial membrane
(D) Inner mitochondrial membrane
Answer:
(D) Inner mitochondrial membrane

Question 3.
In case of anaerobic respiration, the R.Q is _____ .
(A) always less than one
(B) always more than one
(C) always infinity
(D) Variable on the basis of substrate.
Answer:
(C) always infinity

Question 4.
The net gain of ATP molecules during aerobic breakdown of one glucose molecule is _______ .
(A) 40
(B) 38
(C) 36
(D) 30
Answer:
(B) 38

Question 5.
During glycolysis the compounds PGAL and DHAP are formed from fructose 1,6- diphosphate by _______ .
(A) cleavage
(B) isomerisation
(C) phosphorylation
(D) condensation
Answer:
(A) cleavage

Question 6.
Number of oxygen molecules utilized in glycolysis is ________ .
(A) 0
(B) 2
(C) 4
(D) 6
Answer:
(A) 0