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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