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

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

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NCERT Class 11 Chemistry Texbook 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 2px1) 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.

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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, 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

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 H20 molecule.
B (OH)3 + 2HOH ————> [B (OH)4]– + H3O+

Question 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.

Question 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.

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 10. Write the resonance structure of CO32- and HCO3– .
Answer:

Question 11. What is the state of hybridisation of carbon in
(a) CO32- (b) diamond (c) graphite?
Answer: (a) CO32- (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 ofdiborane and boric acid.
Answer: Boric acid contains planar BO33- ions which are linked together through hydrogen bonding shown in the fig.

Question 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:

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.


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.

Question 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.

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 31. Write balanced equations for:

Answer:

Question 32. Give one method for industrial preparation and one for laboratory preparation of CO and C02 each.
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 M2-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.

MORE QUESTIONS SOLVED

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: SiO44- 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:

Question 12. CO2 is gas while SiO2 is solid at room temperature. Give reason.
Answer:  The molecules of CO2 are held together by weak van der Waals forces of attraction which can be easily overcome by collisions of the molecules at room temperature. Consequently CO2 is a gas.
On the other hand silicon atom forms four single covalent bonds with O-atom which are tetrahedrally arranged and form a three-dimensional structure. Thus SiO2 is a high melting solid.

Question 13. What is producer gas ?
Answer: Producer gas is a mixture of CO and N2 in the ratio of 2 :1

Question 14. Write the state of hybridisation of’B’ in BF3.
Answer:  Hybridisation of ‘B’ in BF3 is sp2 .

Question 15. Mention the state of hybridisation of B in BH4.
Answer: sp2.

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

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 -Pπ multiple bonds to itself.

Question 2. Give the chemical reactions as an 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 5. What happens when
(i) Quick lime is heated with coke?
(ii) Carbon monoxide reacts with Cl2?
Answer: (i)CaO + 3C ————> CaC2 + CO
(ii)CO + Cl2 ————> COCl2

Question 6. What are Fullerenes ? How are they prepared ?
Answer: Fullerenes are the allotropes of carbon. Its structure is like a soccer ball.
They are prepared by heating graphite in electric arc in presence of inert gases such as helium or argon.

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

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 addition compound with NH3?
(b) The tendency for catenation decreases down the group in Group 14.
(c) PbO2 is a stronger oxidising 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 the 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.

Question 3. How is boron obtained from borax? Give chemical equations with reaction conditions.
Answer:

NCERT Solutions for Class 11 Chemistry Chapter 11 Multiple Choice Questions

Question 1. Boron has an extremely high melting point because of
(a) its ionic crystal structure
(b) the strong binding forces in the covalent polymer
(c) atomic size (d) allotropy
Question 2. Which of the following compound is an important catalyst as well as a Lewis acid?
(a) AI2S2 (b) BF3 (C) S4N4 (d) N2H4
Question 3. Carbon-60 contains
(a) 20 pentagons and 12 hexagons (b) 12 pentagons and 20 hexagons
(c) 30 pentagons and 30 hexagons (d) 24 pentagons and 36 hexagons
Question 4. The diamond molecule contains
(a) sp2-hybridized carbon atoms connected by single bonds
(b) sp2-hybridized carbon atoms connected by double bonds
(c) sp3-hybridized carbon atoms connected by single bonds
(d) sp3-and sp2-hybridized carbon atoms connected by single bonds
Question 5. Which of the following is the most ionic?
(a) CCl4 (b) PbCl2 (c) PbCl4 (d) SiCl4
Question 6. Silicon carbide (SiC) is known as
(a) quartz (b) tridynite (c) corundum (d) carborundum
Question 7. Which of the following is a purely acidic oxide?
(a) SiO2 (b) SnO2 (c) PbO (d) MnO2
Question 8. Silicones are a group of organosilicon polymers containing
(a) Si – O – Si linkages (b) O – Si – O linkages
(c) Si – C – Si linkages (d) Si – Si – O linkages
Question 9. Which of the following molecules have zero dipole moment?
(a) CS2 (b) CO2 (c) CCl2 (d) CH2Cl2
Question 10. Silicon shows a diagonal relation with
(a) magnesium (b) phosphorous (c) carbon (d) boron
Answer:  1.(b) 2.(d) 3.(b) 4.(c) 5.(b)
6.(d) 7.(a) 8.(a) 9.(a) and(b) 10.(d)

NCERT Solutions for Class 11 Chemistry Chapter 11 Hots Questions

Question 1. Give reason for the following observations:-
(a) The tendency for catenation decrease down the group in Group 14.13.
(b) The decreasing stability of+3 oxidations state with increasing atomic number in group (c)Molten aluminium bromide is a poor conductor of electricity.
Answer: (a) It is due to decrease in bond dissociation energy which is due to increase in atomic size.
C-C > Si-Si > Ge-Ge > Sn-Sn > Pb-Pb
(b) It is due to inert pair effect.
(c) Molten AlBr3 is poor conductor of electricity because it is covalent compound.

Question 2. (a) Why do Boron halides form addition compound with NH3?
(b) Assign appropriate reason for each of the following observations:-
(i) Anhydrous AlCl3 is used as a catalyst in many organic reactions.
(ii) No form of elemental silicon is comparable to graphite.
Answer:  (a) It is because BX3 is electron deficient whereas NH3 is electron rich.
(b) (i) It is Lewis acid.
(ii) It cannot form p π- pπ bond due to large size.

Question 3. Give the chemical reaction as an 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+.
(ii) 3Ga+ —–> 2Ga + Ga3+
This is because Ga3+ is more stable than Ga+.