NCERT Solutions Class 11 Chemistry - The p-Block Elements

NCERT QUESTIONS: The p - Block Elements

Q 11.1. Discuss the pattern of variation in the oxidation states of

(i) B to Tl and                                   (ii) C to Pb.

Ans: (i) B to Tl

The electric configuration of group 13 elements is ns² np¹. Therefore, the most common oxidation state exhibited by them should be +3. However, it is only boron and aluminium which practically show the +3 oxidation state.

The remaining elements, i.e., Ga, In, Tl, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, Tl ( +1) is more stable than Tl ( +3). This is because of the inert pair effect.

The stability of the 3 oxidation state decreases on moving down the group.

(ii) C to Pb

The electronic configuration of group 14 elements is ns² np². Therefore, the most common oxidation state exhibited by them should be 4. However, the 2 oxidation state becomes more and more common on moving down the group. C and Si mostly show the 4 state. 

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

Ans: 

• Boron and thallium belong to group 13 of the periodic table.
• In this group, the +1 oxidation state becomes more stable on moving down the group.
• BCl3 is more stable than TlCl3 because the +3 oxidation state of B is more stable than the +3 oxidation state of Tl.
• In Tl, the +3 state is highly oxidising and it reverts back to the more stable +1 state.

Q 11.3. Why does boron trifluoride behave as a Lewis acid?

Ans: 

• The electric configuration of boron is ns² np¹.
• Boron has three electrons in its valence shell.
• It can form only three covalent bonds.
• This means that there are only six electrons around boron.
• Boron's octet remains incomplete.
• When one atom of boron combines with three fluorine atoms, its octet remains incomplete.
• Hence, boron trifluoride remains electron-deficient.
• Boron trifluoride acts as a Lewis acid.

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

Ans: Being a Lewis acid, BCl₃ readily undergoes hydrolysis. Boric acid is formed as a result.

CCl₄ completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When CCl₄ and water are mixed, they form separate layers.

Q 11.5. Is boric acid a protic acid? Explain. 

Ans: Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.

It behaves as an acid by accepting a pair of electrons from ^–OH ion.

Ques 11.6: Explain what happens when boric acid is heated.

Ans: On heating orthoboric acid (H₃BO₃) at 370 K or above, it changes to metaboric acid (HBO₂). On further heating, this yields boric oxide B₂O₃.

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

Ans: (i) BF₃

As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three sp² hybridised orbitals of boron with the sp orbitals of three halogen atoms. Boron is sp² hybridised in BF₃.

(ii) BH₄^–

Boron-hydride ion (BH₄^–) is formed by the sp³ hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.Q 11.8. Write reactions to justify amphoteric nature of aluminium. 

Ans: A substance is called amphoteric if it displays characteristics of both acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour.

(i) 

(ii) 

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

Ans: In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet.

(i) BCl₃
BCl₃ is an appropriate example of an electron-deficient compound. B has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6. However, it is still short of two electrons to complete its octet.

(ii) SiCl₄
The electronic configuration of silicon is ns² np². This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, SiCl₄ is not an electron-deficient compound.

Q 11.10. Write the resonance structures of  and .

Ans: (a)

(b)

There are only two resonating structures for the bicarbonate ion.

Q 11.11. What is the state of hybridisation of carbon in (a) CO₃²⁻  (b) diamond (c) graphite?

Ans: The state of hybridisation of carbon in:

(a)  CO₃²⁻

C in CO₃²⁻ is sp² hybridised and is bonded to three oxygen atoms.

(b) Diamond

Each carbon in diamond is sp³ hybridised and is bound to four other carbon atoms.

(c) Graphite

Each carbon atom in graphite is sp² hybridised and is bound to three other carbon atoms.

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

Ans: 
 

Q 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.

Ans: (a)

• Lead belongs to group 14 of the periodic table.
• The two oxidation states displayed by this group are 2 and 4.
• On moving down the group, the 2 oxidation state becomes more stable.
• The 4 oxidation state becomes less stable due to the inert pair effect.
• Hence, PbCl4 is much less stable than PbCl2.
• However, the formation of PbCl4 takes place when chlorine gas is bubbled through a saturated solution of PbCl2.

(c) Lead is known not to form PbI₄. Pb ( +4) is oxidising in nature and I^- is reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises I^– to I² and itself gets reduced to Pb(II).

Q 11.14. Suggest reasons why the B–F bond lengths in BF₃ (130 pm) and BF₄^– (143 pm) differ.

Ans: The B–F bond length in BF₃ is shorter than the B–F bond length in . BF₃ is an electron-deficient species. With a vacant p-orbital on boron, the fluorine and boron atoms undergo pπ–pπ back-bonding to remove this deficiency. This imparts a double-bond character to the B–F bond.

This double-bond character causes the bond length to shorten in BF₃ (130 pm). However, when BF₃ coordinates with the fluoride ion, a change in hybridisation from sp² (in BF₃) to sp³ (in  ) occurs. Boron now forms 4σ bonds and the double-bond character is lost. This accounts for a B–F bond length of 143 pm in  ion.

 Q 11.15. If B–Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.

Ans: 

• Electronegativity difference between B and Cl results in a polar B–Cl bond.
• The BCl3 molecule is non-polar.
• BCl3 has a trigonal planar shape.
• It is a symmetrical molecule.
• The dipole moments of the B–Cl bonds cancel each other out.
• This cancellation leads to a zero-dipole moment.

 Q 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 reasons.

Ans:  

• Hydrogen fluoride (HF) is a covalent compound.
• It has a very strong intermolecular hydrogen-bonding.
• HF does not provide ions.
• Aluminium fluoride (AlF) does not dissolve in HF.
• Sodium fluoride (NaF) is an ionic compound.
• When NaF is added to the mixture, AlF dissolves.
• This is due to the availability of free F–.
• The reaction involved in the process is:

• Boron trifluoride (BF3) is added to the solution.
• Aluminium fluoride precipitates out of the solution.
• The tendency of boron to form complexes is much more than that of aluminium.
• When BF3 is added to the solution, B replaces Al from the complexes.
• This replacement occurs according to the following reaction:

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

Ans: Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The CO–Hb complex is more stable than the O₂–Hb complex. The former prevents Hb from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the CO–Hb complex is about 300 times more stable than the O₂–Hb complex.

Q 11.18. How is excessive content of CO₂ responsible for global warming? 

Ans: 

• Carbon dioxide is a very essential gas for our survival.
• However, an increased content of CO2 in the atmosphere poses a serious threat.
• An increment in the combustion of fossil fuels has led to greater levels of carbon dioxide.
• The decomposition of limestone contributes to the increase in CO2 levels.
• A decrease in the number of trees has also resulted in higher carbon dioxide levels.
• Carbon dioxide has the property of trapping the heat provided by sunrays.
• Higher the level of carbon dioxide, higher is the amount of heat trapped.
• This results in an increase in the atmospheric temperature.
• Consequently, this causes global warming.

Q 11.19. Explain structures of diborane and boric acid. 

Ans: (a) Diborane

• B2H6 is an electron-deficient compound.
• B2H6 has only 12 electrons – 6 e– from 6 H atoms and 3 e– each from 2 B atoms.
• Thus, after combining with 3 H atoms, none of the boron atoms has any electrons left.
• X-ray diffraction studies have shown the structure of diborane as:
• 2 boron and 4 terminal hydrogen atoms (Ht) lie in one plane, 
• while the other two bridging hydrogen atoms (Hb) lie in a plane perpendicular to the plane of boron atoms.
• Of the two bridging hydrogen atoms, one H atom lies above the plane and the other lies below the plane.
• The terminal bonds are regular two-centre two-electron (2c – 2e–) bonds,
•  while the two bridging (B–H–B) bonds are three-centre two-electron (3c – 2e–) bonds. 

(b) Boric acid

Boric acid has a layered structure. Each planar BO₃ unit is linked to one another through H atoms. The H atoms form a covalent bond with a BO₃ unit, while a hydrogen bond is formed with another BO₃ unit. In the given figure, the dotted lines represent hydrogen bonds.

 Q 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?

Ans: (a) When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

(b) When boric acid is added to water, it accepts electrons from ^–OH ion.(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

(d) BF₃ (a Lewis acid) reacts with NH₃ (a Lewis base) to form an adduct. This results in a complete octet around B in BF₃.

Q 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.

Ans: (a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicone polymers called methyl-substituted chlorosilanes (MeSiCl₃, Me₂SiCl₂, Me₃SiCl, and Me₄Si) are formed.

(b) When silicon dioxide (SiO₂) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF₄). Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.

The SiF₄ formed in this reaction can further react with HF to form hydrofluorosilicic acid.

(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.

Q 11.22. Give reasons:

(i) Conc. 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.

Ans: (i) Concentrated HNO₃ can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.

(ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.

(iii) Graphite has a layered structure and different layers of graphite are bonded to each other by weak Van der Waals’ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.

(iv) In diamond, carbon is sp³ hybridised. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason, diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.

(v) Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as Cu, Mn, Mg, Si, and Zn. It is very malleable and ductile. Therefore, it is used in making aircraft bodies.

(vi) The oxygen present in water reacts with aluminium to form a thin layer of aluminium oxide. This layer prevents aluminium from further reaction. However, when water is kept in an aluminium vessel for long periods of time, some amount of aluminium oxide may dissolve in water. As aluminium ions are harmful, water should not be stored in aluminium vessels overnight.

(vii) Silver, copper, and aluminium are among the best conductors of electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminium is a very ductile metal. Thus, aluminium is used in making wires for electrical conduction.

Q 11.23. Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? 

Ans: Ionisation enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy (786 kJ). This is because of an appreciable increase in the atomic sizes of elements on moving down the group.

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

Ans:

• Ga has one shell more than Al.
• The size of Ga is lesser than Al.
• This is due to the poor shielding effect of the 3d-electrons.
• The shielding effect of d-electrons is very poor.
• The effective nuclear charge experienced by the valence electrons in gallium is much more than in the case of Al.

Q 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?

Ans: Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes.

Diamond:

The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest naturally-occurring substances. It is used as an abrasive and for cutting tools.

Graphite:

It has sp² hybridised carbon, arranged in the form of layers. These layers are held together by weak Van der Waals’ forces. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.

Q 11.26. (a) Classify following oxides as neutral, acidic, basic or amphoteric:

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

(b) Write suitable chemical equations to show their nature.

Ans: (1) CO = Neutral

(2) B₂O₃ = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium metaborate.

(3) SiO₂ = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium silicate.

(4) CO₂ = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium carbonate.

(5) Al₂O₃ = Amphoteric

Amphoteric substances react with both acids and bases. Al₂O₃ reacts with both NaOH and H₂SO₄.

(6) PbO₂ = Amphoteric

Amphoteric substances react with both acids and bases. PbO₂ reacts with both NaOH and H₂SO₄.

(7) Tl₂O₃ = Basic

Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium chloride.

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

Ans:

• Thallium belongs to group 13 of the periodic table.
• The most common oxidation state for this group is 3.
• Heavier members of this group also display the 1 oxidation state.
• This happens because of the inert pair effect.
• Aluminium displays the 3 oxidation state.
• Alkali metals display the 1 oxidation state.
• Thallium displays both the oxidation states.
• Therefore, it resembles both aluminium and alkali metals.
• Thallium, like aluminium, forms compounds such as TlCl3 and Tl2O3.
• It resembles alkali metals in compounds Tl2O and TlCl.

.Q 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.

Ans: The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.

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

Ans: (a) Inert pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group 13 elements, the electronic configuration is ns² np¹ and their group valency is +3. However, on moving down the group, the +1 oxidation state becomes more stable. This happens because of the poor shielding of the ns² electrons by the d- and f- electrons. As a result of the poor shielding, the ns² electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.

(b) Allotropy

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes.

(c) Catenation

The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in Si and S.

Q 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 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.

Ans: The given salt is alkaline to litmus. Therefore, X is a salt of a strong base and a weak acid. Also, when X is strongly heated, it swells to form substance Y. Therefore, X must be borax.

When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material Y. Hence, Y must be a mixture of sodium metaborate and boric anhydride.

When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are formed.

Q 11.31. Write balanced equations for:

(i) BF₃ LiH →  
(ii) B₂H₆  H₂O →
(iii) NaH B₂H₆ → 
(iv) H₃BO₃ →
(v) Al NaOH →
(vi) B₂H₆  NH₃ →

Ans:

(iv)   

Q 11.32. Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.

Ans: Carbon dioxide

In the laboratory, CO₂ can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:

CO₂ is commercially prepared by heating limestone. The reaction involved is as follows:

Carbon monoxide

In the laboratory, CO is prepared by the dehydration of formic acid with conc. H₂SO₄, at 373 K. The reaction involved is as follows:

CO is commercially prepared by passing steam over hot coke. The reaction involved is as follows:

Q 11.33. An aqueous solution of borax is

(a) neutral                                                      (b) amphoteric

(c) basic                                                         (d) acidic

Ans: (c) Borax is a salt of a strong base (NaOH) and a weak acid (H₃BO₃). It is, therefore, basic in nature.

Q 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

Ans: (b) Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.

Q 11.35. The type of hybridisation of boron in diborane is

(a) sp
(b) sp² 
(c) sp³  
(d) dsp²

Ans: (c) Boron in diborane is sp³ hybridised.

Q 11.36. Thermodynamically the most stable form of carbon is

(a) diamond 
 (b) graphite
 (c) fullerenes 
 (d) coal

Ans: (b) Graphite is thermodynamically the most stable form of carbon.

Q 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 M^{4 +}   ion                                          

(d) form M²⁺  and M⁴⁺  ions

Ans: (b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4. However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.

Q 11.38. If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.

Ans: 

(i)

(ii)