11.1. Discuss the pattern of variation in the oxidation states of
(i) B to Tl (ii) C to Pb.

Answer:  

(i) B to Tl
Common oxidation states are +1 and +3. The stability of +3 oxidation state decreases from B to Tl. +1 oxidation state increases from B to Tl.
(ii)C to Pb
The common oxidation states are +4 and +2. Stability of +4 oxidation state decreases from C to Pb.
Details can be seen from the text part.

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

Answer:

BCl₃ is quite stable. Because there is absence of d- and f-electrons in boron three valence electrons (2s² 2p_x1) are there for bonding with chlorine atom. In Tl the valence s-electron (6s²) are experiencing maximum inert pair effect. Thus, only 6p¹ electron is available for bonding. Therefore, BCl₃ is stable but TlCl₃ is comparatively unstable.

11.3. Why does borontrifluori.de behave as a Lewis acid?

Answer:

In BF₃, central atom has only six electrons after sharing with the electrons of the F
atoms. It is an electron deficient compound and thus behaves as a Lewis acid.

11.4. Consider the compounds, BCl₃ and CCl₄. How will they’behave with water justify?

Answer: 

In BCl₃, there is only six electrons in the valence shell of B atom. Thus, the octet is incomplete and it can accept a pair of electrons from water and hence BCl₃ undergoes hydrolysis. Whereas, in CCl₄, C atom has 8 electrons and its octet is complete. That’s why it has no tendency to react with water.
CCl₄ + H₂0 —————> No reaction

11.5. Is boric acid a protonic acid? Explain.

Answer: 

Boric acid is a Lewis acid, it is not a protonic acid.
Boric acid accepts electrons from hydroxyl ion of H₂0 molecule.
B (OH)₃ + 2HOH ————> [B (OH)₄]^– + H₃O⁺

11.6. Explain what happens when boric acid is heated.

Answer: 

On heating boric acid above 370 K, it forms metaboric acid, HB02 which on further heating yields boric oxide B₂O₃.

11.7. Describe the shapes of BF₃ and BH₄^–. Assign the hybridisation of boron in these species.

Answer: 

In BF₃, boron is SP² hybridized.
∴ shape of BF₃ = planar.
In [BH₄]^–, boron is sp3 hybridized, thus the shape is tetrahedral.

11.8. Write reactions to justify amphoteric nature of aluminium.

Answer: 

Aluminium reacts with acid as well as base. This shows amphoteric nature of aluminium.
2Al(s) + 6HCl(dil.) ——> 2AlCl₃(aq) + 3H₂(g)
2Al(s) + 2NaOH(aq) + 6H₂O(l) ———-> 2Na+ [Al(OH)₄]^–(aq) + 3H₂(g)

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

Answer: 

Electron deficient species are those in which the central atom in their molecule has the tendency to accept one or more electron pairs. They are also known as Lewis acid. BCl₃ and  SiCl₄ both are electron deficient species.
Since, in BCl₃, B atom has only six electrons. Therefore, it is an electron deficient compound.
In SiCl₄ the central atom has 8 electrons but it can expand its covalency beyond 4 due to the presence of d-orbitals.
Thus, SiCl₄ should also be considered as electron deficient species.

11.10. Write the resonance structure of CO₃^2- and HCO₃^– .

Answer:

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

Answer:

(a) CO₃^2- (sp²) (b) Diamond (sp³) (c) Graphite (sp²)

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

Answer:

• Since diamond exists as a three dimensional network solid, it is the hardest substance known with high density and high melting point.
  Whereas in graphite, any two successive layers are held together by weak forces of attraction. This makes graphite soft.
• In graphite, carbon atom is sp² hybridized whereas in diamond, carbon atom is sp³ hybridized.
• Unlike diamond, graphite is good conductor of heat and electricity.

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 Pbl₄.

Answer:

• PbCl₂  + Cl₂ ———> PbCl₄.
  This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect.
  heat
• PbCl₄ ———> PbCl₂ + Cl₂
  Because Pb2+ is more stable than Pb⁴⁺ due to inert pair effect.
• Pbl₄ does not exist because I- ion being a powerful reducing agent reduces Pb⁴⁺ ion to Pb²⁺ ion in solution.
• Pb4+ + 2I^– ——-> Pb²⁺ + l₂
  Pb(IV)                       Pb(II)

11.14. Suggest reason why the B-F bond lengths in BF₃ (130 pm) and BF^– (143 pm) differ.

Answer:

In BF₃ ‘B’ is sp² hybridised and in BF₄^– ‘B’ is sp³ hybridised. Thus, the difference in bond length is due to the state of hybridisation.

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 since the molecule is symmetrical (planar). Thus the polarities cancel out.

11.16. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF₃ is bubbled through. Give reason.

Answer:

Since, anhydrous HF is covalent compound and weak acid due to high bond dissociation energy. AlF₃ does not dissolve in HF.
Whereas NaF is ionic compound.
3NaF + AlF₃ ———> Na₃[AlF₆]
Na₃[AlF₆] + 3BF₃(g) ——-> AlF₃ + 3Na+ [BF]^–

11.17. Suggest a reason as to why CO is poisonous.

Answer:

CO reacts with haemoglobin to form carboxyhaemoglobin which can destroy the oxygen carrying capacity of haemoglobin and the man dies of suffocation.

11.18. How is excessive content of C0₂ responsible for global warming?

Answer:

Excess of C0₂ absorbs heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is radiated back to the earth. As a result, temperature of the earth increases. This is the cause of global warming.

11.19. Explain structures ofdiborane and boric acid.

Answer:

Boric acid contains planar BO₃^3- ions which are linked together through hydrogen bonding shown in the fig.

11.20. What happens when
(a) Borax is heated strongly
(b) Boric acid is added to water
(c) Aluminium is treated with dilute NaOH
(d) BF₃ is reacted with ammonia?

Answer:

11.21. Explain the following reactions.
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper.
(b) Silicon dioxide is treated with hydrogen fluoride.
(c) CO is heated with ZnO.
(d) Hydrated alumina is treated with aqueous NaOH solution.

Answer:

11.22. Give reasons:
(i) Cone. HNO₃ can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv) Diamond is used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.

Answer:

(i) Al reacts with cone. HNO₃ to form a very thin film of aluminium oxide on its surface which protects it from further reaction.
2Al(s) + 6HNO₃(conc.)———> Al₂O₃(s) + 6NO₂(g) + 3H₂O(l)
(ii) NaOH reacts with Al to evolve H₂ gas. Thus the pressure of the gas produced can be used for clogged drains.
2Al(s) + 2NaOH(aq) + 2H₂O(l) ——-> 2NaAlO₂(aq) + 3H₂(g)
(iii) Graphite has layered structure which are held by weak van der Waals forces. Thus, graphite cleaves easily between the layers, therefore it is very soft and slippery. That’s why it is used as lubricant.
(iv) Diamond is used as an-abrasive because it is an extremely hard substance.
(v) Alloys of aluminium, like duralumin, is used to make aircraft body due to some of its property like toughness, lightness and resistant to corrosion.
(vi) Generally, aluminium metal does not react with water quickly but, when it is kept overnight, it reacts slowly with water in presence of air.
2Al(s) + O₂(g) + H₂O(l) ——–> Al2O₃(S) + H₂(g)
a very small amount of (in ppm) Al³⁺ produced in the solution is injurious to health if the water is used for drinking purposes.
(vii) Aluminium is generally unaffected by air and moisture and it is also good conductor of electricity. That’s why it is used in transmission cables.

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

Answer:

Because there is increase in atomic size on moving from carbon to silicon, the screening effect increases. Thus the force of attraction of nucleus for the valence electron decreases as compared to carbon. Thus the ionization enthapy decreases from carbon to silicon.

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

Answer:

Due to poor shielding effect of d-electrons in Ga, the electrons in gallium experience great force of attraction by nucleus as compared to Al.

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


In diamond, carbon is SP³ -hybridized. Since, diamond is three dimensional network solid, it is hardest substance with high density whereas graphite has a layered structure. The various layers are formed by van der Waals forces of attraction that’s why graphite is soft and slippery.

11.26. (a) Classify following oxides as neutral, acidic, basic or amphoteric
CO, B₂O₂, Si0₂, C0₂, Al₂O₃, PbO₂, Tl₂O₃
(b) Write suitable equations to show their nature.

Answer: 

(a) Neutral — CO
Acidic — B₂O₂, Si0₂, C0₂ Basic — Tl₂O₃ Amphoteric — Tl₂O₃, PbO₂
(b)-CO does not react with acid as well as base at room temperature.
Being acidic B₂O₃, Si0₂ and C0₂ reacts with alkalis to form salts.

11.27. In some of the reactions thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences.

Answer: 

Tl shows both the oxidation state +1 and +3 due to inert pair effect. Tl forms basic oxide like group I elements. TlO₂ is strongly basic.

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:

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

Answer:  

(a) Inert pair effect: The pair of electron in the valence shell does not take part in bond formation is called inert pair effect.
(b)Allotropy: It is the property of the 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.
(c)Catenation: The property to form chains or rings not only with single bonds but also with multiple bonds with itself is called catenation.
For example, carbon forms chains with (C-C) single bonds and also with multiple bonds (C = C or C = C).

11.30. A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When cone.H₂SO₄is added to a hot solution of X, white crystal of an acid Z separates out.

Answer:

11.31. Write balanced equations for:

Answer:

11.32. Give one method for industrial preparation and one for laboratory preparation of CO and C0₂ each.

Answer:

11.33. An aqueous solution of borax is
(a) neutral (b) amphoteric (c) basic (d) acidic

Answer: 

Borax is a salt of a strong base (NaOH) and a weak acid (H₃BO₃), therefore, it is basic in nature, i.e., option (c) is correct.

11.34. Boric acid is polymeric due to
(a) its acidic nature (b) the presence of hydrogen bonds
(c) its monobasic nature (d) its geometry

Answer: 

Boric acid is polymeric due to the presence of H-bonds. Therefore, option (b) is correct.

11.35. The type of hybridisation of boron in diborane is
(a) sp (b) sp²(c) sp³(d) dsp²

Answer:

In B₂H₆, B is sp³-hybridized. Therefore, option (c) is correct.

11.36. Thermodynamically the most stable form of carbon is
(a)diamond (b) graphite (c) fullerenes (d) coal 

Answer: 

Thermodynamically the most stable form of carbon is graphite, i.e., option (b) is correct.

11.37. Elements of group 14
(a) exhibit oxidation state of +4 only (b) exhibit oxidation state of +2 and +4
(c) form M^2-and M⁴⁺ ion (d) form M²⁺ and M⁴⁺ ions.

Answer:

Due to inert pair effect, elements of group 14 exhibit oxidation states of +2 and +4. Thus, option (b) is correct.

11.38. If the starting material for the manufacture of silicons is RSiCl₃ write the structure of the product formed.

Answer:

Hydrolysis of aikyltrichlorosilanes gives cross-linked silicons.