NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements

NCERT Class 11 Chemistry Textbook Solved Questions

Question 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.

Question 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.

Question 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.

Question 4. Consider the compounds, BCl_3, 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

Question 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 H₂O molecule.
B (OH)₃ + 2HOH ————> [B (OH)₄]^– + H₃O⁺

Question 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)

Question 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.

Question 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)

Question 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.

Question 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)

Question 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.

Question 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.

Question 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]^–

Question 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.

Question 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.

Question 19. Explain structures of diborane and boric acid.
Answer: Boric acid contains planar BO₃^3- ions which are linked together through hydrogen bonding shown in the fig.

Question 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:

Question 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.

Question 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.

Question 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.

Question 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.

n 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.

Question 26. (a) Classify the 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.

Question 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.

Question 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:

Question 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).

Question 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:

Question 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.

Question 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.

Question 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.

Question 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.

Question 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.

Question 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.

NCERT Solutions for Class 11 Chemistry Chapter 11 Very Short Answer Type Questions

Question 1. Why is boron used in nuclear reactions?
Answer:  Because Boron can absorb neutrons.

Question 2. Why does boron form stable electron deficient compounds?
Answer:  Boron has three electrons in its valence shell that boron show three electrons with other elements and form an electron-deficient compound.

Question 3. By giving a balanced equation show how B(OH)₃ behaves as an acid in water.
Answer: B(OH)₃ + 2HOH ———> [B(OH)₄]^– + H₃O⁺.

Question 4. What is dry ice? Why is it so called? 
Answer: Carbon dioxide can be obtained as a solid in the form of dry ice by allowing the liquified CO₂ to expand rapidly.
Unlike ordinary ice it does not melt and hence does not wet the surface on which it is kept. Thus it is called dry ice.

Question 5. Name the element of group 14 which exhibits maximum tendency for catenation.
Answer: Carbon.

Question 6. What is the basic building unit of all silicates?
Answer: SiO₄^4- is the basic unit of all silicates.

Question 7. Why do boron halides form addition compounds with NH₃?
Answer: Boron halides are lewis acids and can accept a pair of electrons from amines to form addition product.

Question 8. What happens when NaBH₄ reacts with iodine?
Answer: 2NaBH₄ + I₂ ———–> B₂H₆ + 2NaI + H₂.

Question 9. Out of CCl₄ and SiCl₄ which one react with water and why?
Answer: Due to the presence of d-orbitals in Si, SiCl₄ reacts with water. CCl₄ does not react with water due to the absence of d-orbitals in C atom.

Question 10. Which oxide of carbon is regarded as anhydride of carbonic acid?
Answer: CO₂ is regarded as anhydride of carbonic acid.
H₂CO₃ ————> H₂O + CO₂

Question 11. What happens when boric acid is heated?
Answer:

NCERT Solutions for Class 11 Chemistry Chapter 11 Short Answer Type Questions ( The p Block Elements )

Question 1. What is meant by catenation? Why does ‘C show the property of catenation to maximum extent?
Answer:  It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to maximum extent because it is small in size and can form  Pπ-Pπmultiple bonds to itself.

Question 2. Give the chemical reactions as evidence for each of the following observations.
(i) Tin (II) is a reducing agent whereas lead (II) is not.
(ii) Gallium (I) undergoes disproportionation reaction.
Answer: (i) Due to inert pair effect Pb²⁺ is more stable than Pb⁴⁺. Whereas Sn⁴⁺ is more stable than Sn²⁺.
Thus Sn²⁺ is a good reducing agent and Pb²⁺ is not.
(ii) 3Ga+ ——–> 2Ga + Ga³⁺
This is because Ga³⁺ is more stable than Ga⁺.

Question 3. Describe two similarities and two dissimilarities between B and Al.
Answer: Similarities:

• Both have same number of valence electrons.
• Both have similar electronic configuration.

Dissimilarities:

• B is a non-metal where Al is a metal.
• B forms acidic oxide whereas Al forms amphoteric oxides.

Question 4. (a) What is general formula of silicons?
(b) How are linear silicons obtained?
Answer: (a) R₂SiO
(b) Linear silicons are obtained by the hydrolysis of R₂SiCl₂ (chlorosilanes).

Question 7.Give reason.
(i) C and Si are always tetravalent but Ge, Sn, Pb show divalency.
(ii) Gallium has higher ionization enthalpy than Al. Explain.
Answer: (i) Ge, Sn and Pb show divalency due to inert pair effect, Pb²⁺ is more stable than Pb⁴⁺ .
(ii) Due to poor shielding effect of d-electrons in Ga, effective nuclear charge increases as compared to Al. Thus the I.E of Ga is higher than Al.

Question 8. Give reason why boron and aluminium tend to form covalent compounds.
Answer: Sum of the three ionization enthalpies of both the elements are very high. Thus they have no tendency to lose electrons to form ioriic compound. Instead they form covalent compounds.

NCERT Solutions for Class 11 Chemistry Chapter 11 Long Answer Type Questions ( The p Block Elements )

Question 1. Explain the differences in properties of diamond and graphite based upon their structures.
Answer:

Question 2. Give reasons:
(a) Why do Boron halides form an additional compounds with NH₃?
(b) The tendency for catenation decreases down the group in Group 14.
(c) PbO₂ is a stronger oxidizing agent than SnO₂.
Answer:  (a) It is because BX₃ is an electron-deficient compound and NH₃ is an electron-rich compound.
(b) As we move down group 14, the atomic size increases and thus the strength of the element decreases down the group thus the bond dissociation enthalpy decreases steadily consequently the tendency for catenation decreases down the group.
(c) PbO₂ and SnO₂ both are present in +4 oxidation state. But due to stronger inert pair effect Pb²⁺ ion is more stable than Sn²⁺ ion.
In other way Pb⁴⁺ ions is more easily reduced to Pb²⁺ions. Thus PbO₂ acts as a stronger oxidising agent than SnO₂.

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