Question 11.1

Discuss the pattern of variation in the oxidation states of

• B to Tl
• C to Pb

Answer.

• Common oxidation states are +1 and +3. Stability of +3 oxidation state decreases from B to Tl and +1 oxidation state increases from B to Tl.
• Common oxidation states are +4 and +2. Stability of +4 oxidation state decreases from C to Pb.

 

 

Question 11.2

How can you explain higher stability of BCl₃ as compared to TlCl₃?

Answer.

Inert pair effect becomes more stable as we move down the group.

Tl is more stable in its +1 oxidation state and TlCl₃ is not stable as Tl shows an oxidation state of +3 in TlCl₃.

 

 

Question 11.3

Why does boron triflouride behave as a Lewis acid?

Answer.

.In BF₃, central atom has only 6 electrons after sharing with the electrons of the fluorine atoms.

It is an electron deficient compound and acts as a Lewis acid.

 

Question 11.4

Consider the compounds, BCl₃ and CCl₄. How will they behave with water? Justify.

Answer.

In BCl₃, there are six electrons in the valence shell of boron atom.

The octet is incomplete and it can accept a pair of electrons from H₂O. Hence, BCl₃ undergoes hydrolysis.

In CCl₄, carbon atom has 8 electrons and its octet is complete. So, it has no tendency to react with water.

CCl₄ + H₂O --> No reaction.

 

Question 11.5

Is boric acid a protic acid? Explain.

Answer.

Boric acid is a Lewis acid. Boric acid accepts electrons from hydroxyl ion of water molecule.

 B(OH)₃ + 2HOH ^-  --> [B(OH)₄]^- + H₃O⁺

 Question 11.6

Explain what happens when boric acid is heated.

Answer.

On heating boric acid (above 370 K), it forms metaboric acid, HBO₂ which on further heating yields boric acid B₂O₃.

H₃BO₃ ---> HBO₂ ---> B₂O₃  (By heating)

 

Question 11.7

Describe the shapes of BF₃ and BH₄^-. Assign the hybridization of boron in these species.

Answer.

In BF₃, boron has sp² hydridisation.

So, shape of BF₃ is plannar.

In [BH₄]^-, boron is sp³ hybridized. Thus, its shape is tetrahedral.

 

 

Question 11.8

Write reaction to justify amphoteric nature of aluminum.

Answer.

Aluminum reacts with acid as well as base. This shows the amphoteric nature of aluminum.

2Al(s) + 6HCl(dil) ---> 2AlCl₃(aq) + 3H₂(g)

2Al(s) + 2NaOH(aq) + 6H₂O(l) ---> 2 Na⁺[Al(OH)₄^]- (aq) + 3H₂(g)

 

Question 11.9

What are electron deficient compounds? Are BCl₃ and SiCl₄ electron deficient species? Explain.

Answer.

Electron deficient species are those species which have central atom in their molecule which have tendency

to accept one or more electron pairs.

These are also called as Lewis acid.

BCl₃ and SiCl₄ are electron deficient species.

In BCl₃, B atoms has only six electrons, thus it is electron deficient.

In SiCl₄, central atom has 8 electrons, it expands its covalency beyond four due to the presence of d-orbitals.

So, it is considered as electron deficient species.

 

 

Question 11.10

Write the resonance structures of CO₃^2- and HCO₃^-.

Answer.

CO₃^2-

HCO₃^-

 

 

Question 11.11

What is the state of hybridization of carbon in (a) CO₃^2- (b) diamond (c) graphite?

Answer.

• Sp²
• Sp³
• Sp²

 

 

Question 11.12

Explain the difference in properties of diamond and graphite on the basis of their structures.

Answer.

DIAMOND	GRAPHITE
It has crystalline lattice.	It has layered structure.
It is made up of tetrahedral units.	It has a planar geometry.
C-C bond in diamond is 154 pm.	C-C bond in graphite is 141.5 pm.
Rigid covalent bond network is difficult to break.	Its layers can be separated easily as it is soft.

Question 11.13

Rationalise the given statements and give chemical reactions:

• Lead (II) chloride reacts with Cl₂ to give PbCl₄.
• Lead (IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, PbI₄.

Answer.

 

 

Question 11.14

Suggest reasons why the B-F bond lengths in BF₃ (1130 pm) and BF₄^- (143 pm) differ.

Answer.

In BF₃, boron is sp² hybridised.

In BF₄^-, boron is sp³ hybridized.

The difference in bond length is due to the state of hydridisation.

 

 

Question 11.15

If B-Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.

Answer.

B-Cl bond has dipole moment because of polarity.

In BCl₃, molecule is symmetrical (planar). Thus, polarities cancel out each other.

 

 

Question 11.16

Aluminum triflouride is insoluble in anhydrous HF but dissolves on addition of NaF.

Aluminum triflouride precipitates out of the resulting solution when gaseous BF₃ is bubbled through. Give reasons.

Answer.

Anhydrous HF is a covalent compound and weak acid. It us due to high bond dissociation energy.

AlF₃ does not dissolve in HF.

NaF is an ionic compound.

 

3NaF + AlF₃ ---> Na₃[AlF₆]

Na₃[AlF₆] + 3BF₃(g) ---> AlF₃ + 3Na⁺ [BF]^-

 

Question 11.17

Suggest a reason as to why CO is poisonous.

Answer.

Carbon monoxide reacts with haemoglobin to form carboxyhaemoglobin which can destroy oxygen carrying capacity

of Hb and lead to death of a person due too suffocation.

 

 

Question 11.18

How is excessive content of CO₂ responsible for global warming?

Answer.

Excess of carbon dioxide absorbs heat radiated by the earth.

Some of this heat is dissipated into the atmosphere while remaining heat is radiated back to the Earth.

As a result, temperature of the Earth increases. This causes global warming.

 

 

Question 11.19

Explain structures of diborane and boric acid.

Answer.

Boric acid contains BO₃^3- which are linked together through hydrogen bonding.

STRUCTURE OF DIBORANE MOLECULE

 

 

Question 11.20

What happens when

• Borax is heated strongly
• Boric acid is added to water
• Aluminum is treated with dilute NaOH
• BF₃ is reacted with ammonia

Answer.

 

 

Question 11.21

Explain the following reactions

• Silicon is heated with methyl chloride at high temperature in the presence of copper.
• Silicon dioxide is treated with hydrogen fluoride.
• CO is heated with ZnO
• Hydrated alumina is treated with aqueous NaOH solution.

Answer.

• Mixture of monomethyl chlorosilianes, dimethyl chlorosiliances and trimethyl chlorosilianes along with a
• small amount of tetramethyl silanes is formed.

        

 

 

Question 11.22

Give reasons.

• HNO₃ can be transported in aluminum container.
• A mixture of dilute NaOH and aluminum pieces is used to open drain.
• Graphite is used as lubricant.
• Diamond is used as an abrasive.
• Aluminum alloys are used to make aircraft body.
• Aluminum utensils should not be kept in water overnight.
• Aluminum wire is used to make transmission cables.

Answer.

• Aluminum reacts with concentrated HNO₃ to form a very thin film of aluminum oxide on its surface which protects it from further reaction.

         2Al(s) + 6HNO₃(conc) ---> Al₂O₃ (s) + 6NO₂(g) + 3H₂O(l)

• NOH react with aluminum to evolve hydrogen gas, so the pressure of the gas produced can be used for clogged drains.

          2Al(s) + 2NaOH(aq) + 2H₂O(l) ---> 2NaAlO₂(aq) + 3H₂(g)

• Graphite has layered structure which is held by weak Van der Waals forces.
• Thus, graphite is very soft and slippery, due to which it is used as a lubricant.
• Diamond is used as an abrasive because it is an extremely hard substance.
• Alloy of aluminum is used to make aircraft body due to its properties like toughness, lightness and corrosion resistance.
• Aluminum metal does not react with water quickly. But when it is kept overnight, it reacts slowly with water in the presence of air.

         2Al(s) + O₂(g) + H₂O(l) --->₃(s) + H₂(g)

• Aluminum is unaffected by air and moisture. It is good conductor of electricity. So it is used in transmission cables.

 

Question 11.23

Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?

Answer.

Ionization enthalpy of carbon is very high that is 1086 kJ/mol.

Increase in atomic size increases on moving from carbon to silicon, screening effect also increases.

The force of attraction of nucleus for the valence electron decreases as compared to carbon.

Hence, ionization enthalpy decreases from carbon to silicon.

 

 

Question 11.24

How would you explain the lower atomic radius of Ga as compared to Al?

Answer.

Because of the poor shielding effect of d-orbitals in Ga,

electrons in gallium experience greater force of attraction by nucleus as compared to aluminum.

 

 

Question 11.25

What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite.

What is the impact of structure on physical properties of two allotropes?

Answer.

Allotropes are the different forms of an element which have same chemical properties but different physical properties due to their structures.

Diamond

Carbon is sp³-hybridized, in diamond. Diamond is 3-D network solid. It is the hardest substance with high density.

Graphite

Graphite has a layered structure formed by Van der waals forces of attraction. Graphite is soft and slippery.

 

 

 

Question 11.26

• Classify following oxides as neutral, acidic, basic and amphoteric.

CO, B₂O₃, SiO₂, CO₂, Al₂O₃, PbO₂, Tl₂O₃

• Write suitable chemical equations to show their nature.

Answer.

• Neutral oxides – CO

Acidic oxides – B₂O₂, SiO₂, CO₂

Basic oxides – Tl₂O₃

Amphoteric oxides – Tl₂O₃, PbO₂

• CO does not react with acid as well as base at room temperature.

B₂O₃ reacts with basic oxides forming meta-borates.

 

 

 

Question 11.27

In some of the reactions thallium resembles aluminum, whereas in others it resembles with group I metals.

Support this statement by giving some evidences.

Answer.

Tl shows both oxidation state +1 and +3 due to inert pair effect.

Thallium forms basic oxide like group I elements. TlO₂ is strongly basic.    

 

 

Question 11.28

When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B).

Compound (A) is soluble in dilute HCl to form compound (C).

The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D).

Write suitable equations to support their identities.

Answer.

Given metal X gives white ppts with sodium hydroxide and precipitate dissolves in excess of sodium hydroxide.

See the reactions below:

 

 

Question 11.29

What do you understand by (a) inert pair effect (b) allotropy and (c) catenation/

Answer.

• Inert pair effect – The pair of electron in the valence shell does not take part in bond formation is called inert pair effect.
• Allotropy is the property of an element by which an element can exist in two or more forms which have same chemical
• properties but different physical properties due to their structures.
• Catenation is the property to form chains or rings not only with single bonds but also with multiple bonds with itself.

 

 

Question 11.30

A certain salt X, gives the following results.

• Its aqueous solution is alkaline to litmus.
• It swells up to a glassy material Y on strong heating.
• When conc. H₂SO₄ is added to a hot solution of X, white crystal of an acid Z separates out.

Write equations for all the above reactions and identify X, Y and Z.

Answer.

 

 

Question 11.31

Write balanced equations for

• BF₃ + LiH -->
• B₂H₆ + H₂O -->
• NaH + B₂H₆ -->
• H₃BO₃ -->
• Al + NaOH -->
• B₂H₆ + NH₃ -->

Answer.

 

 

 

Question 11.32

Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.

Answer.

 

 

Question 11.33

An aqueous solution of borax is

• Neutral
• Amphoteric
• Basic
• Acidic

Answer.

Correct option is (c)

Borax is a salt of a strong base (NaOH) and a weak acid (H₃BO₃). Hence, it is basic in nature.

 

Question 11.34

Boric acid is polymeric due to

• Its acidic nature
• The presence of hydrogen bonds
• Its monobasic nature
• Its geometry

Answer.

Correct option is (b)

Boric acid is polymeric due to the presence of hydrogen bonds.

Question 11.35

The type of hybridization of boron in diborane is

(a) sp

(b) sp²

(c) sp³

(d) dsp²

Answer.

Correct option is (c)

In B₂H₆, boron is sp³-hybridized.

 

 

Question 11.36

Thermodynamically the most stable form of carbon is

• Diamond
• Graphite
• Fullerenes
• Coal

Answer.

Correct option is (b)

Graphite is thermodynamically most stable form of carbon.

 

 

Question 11.37

Elements of group 14

• Exhibit oxidation state of +4 only
• Exhibit oxidation state of +2 and +4.
• Form M^2- and M⁴⁺ ions
• Form M²⁺ and M⁴⁺ ions

Answer.

Correct option is (b)

Most stable form of carbon (thermodynamically) is graphite.

 

 

Question 11.38

If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.

Answer.

Hydrolysis of alkyltrichlorosilanes gives cross-linked silicons.