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The p-Block Elements — NCERT Class 11 Chemistry

Complete NCERT Solutions for CBSE Class 11 Chemistry Chapter 11 — all intext and chapter-end exercise questions on Group 13 and Group 14 elements, boron compounds, carbon allotropes, silicones, silicates and zeolites answered clearly.

Updated 11 June 202616 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

Read each NCERT textbook question first, attempt it on your own, then compare your answer with the detailed solutions below. Focus on understanding the reasoning — not just the final answer — so you can tackle similar questions in exams confidently.


Key Concepts (Quick Recap)

  • Group 13 elements: B, Al, Ga, In, Tl — general electronic configuration ns²np¹
  • Anomalous properties of boron: small size, high ionization enthalpy, high electronegativity, non-metallic character, absence of d-orbitals, forms covalent compounds
  • Borax: Na₂B₄O₇·10H₂O — used as a flux and in borax bead test
  • Boric acid: H₃BO₃ — monobasic Lewis acid, accepts OH⁻ from water
  • Diborane: B₂H₆ — electron-deficient, 3-centre 2-electron (3c-2e) bonds
  • Aluminium: extraction by Hall-Héroult process, amphoteric, forms protective oxide layer
  • Group 14 elements: C, Si, Ge, Sn, Pb — general electronic configuration ns²np²
  • Catenation: ability of an element to form chains of bonds with itself; carbon shows maximum catenation
  • Allotropes of carbon: diamond (sp³), graphite (sp²), fullerene (C₆₀)
  • Oxides of carbon: CO (neutral, poisonous), CO₂ (acidic, greenhouse gas)
  • Silicones: (R₂SiO)ₙ — organosilicon polymers, water-repellant, heat-resistant
  • Silicates: SiO₄⁴⁻ tetrahedral units forming various structures (chains, sheets, 3D frameworks)
  • Zeolites: aluminosilicates used as catalysts and ion-exchange materials

Intext Questions

Page 270–271 (Group 13 Elements, Boron Properties)

Q1. Discuss the pattern of variation in the oxidation states of Group 13 elements.
Answer

The common oxidation state of Group 13 elements is +3 due to the loss of all three valence electrons (ns²np¹). However, as we move down the group, the +1 oxidation state becomes more stable due to the inert pair effect. In Tl, the +1 state is more stable than +3 because the two s-electrons (6s²) tend to remain paired and do not participate in bonding. Thus: B and Al predominantly show +3; Ga shows both +1 and +3; In shows both; Tl prefers +1.

Q2. How do atomic and ionic radii vary in Group 13?
Answer

Atomic radii increase down the group from B to Al as expected. However, Ga has a smaller atomic radius than Al due to poor shielding by d-electrons (the d-electrons in Ga do not shield the nuclear charge effectively). From Ga to In to Tl, radii increase normally. Similar anomalies are observed in ionic radii.

Q3. Why does boron show anomalous behaviour?
Answer

Boron differs from the rest of Group 13 because of: (i) extremely small size, (ii) very high ionization enthalpy, (iii) high electronegativity (closer to carbon than to aluminium), (iv) absence of d-orbitals, and (v) only four valence orbitals (one 2s and three 2p) available for bonding. As a result, boron is a non-metal, forms predominantly covalent compounds, and has a maximum covalency of 4 (using sp³ hybrids).

Q4. Discuss the trends in ionization enthalpy in Group 13.
Answer

Ionization enthalpy decreases from B to Al (as expected with increased size), but then the trend is irregular. Ga has higher IE₁ than Al because the d-electrons in Ga shield the nuclear charge poorly, making the outer electrons more tightly held. IE₁ then decreases from Ga to In, and Tl has a slightly higher IE₁ than In due to the presence of filled 4f and 5d orbitals which provide poor shielding (similar to lanthanide contraction effect).

Page 277–278 (Borax, Boric Acid, Diborane)

Q5. What is borax? What is its molecular formula?
Answer

Borax is sodium tetraborate decahydrate with the molecular formula Na₂B₄O₇·10H₂O. On heating, it loses water and swells to form a white porous mass. On further heating, it melts to form a glassy bead of sodium metaborate and boric anhydride: Na₂B₄O₇ → 2NaBO₂ + B₂O₃. The borax bead test is used to identify certain metal ions.

Q6. Explain the action of heat on boric acid.
Answer

On heating, boric acid (H₃BO₃) first loses water at around 370 K to form metaboric acid (HBO₂), which on further heating at around 430 K gives boric oxide (B₂O₃): H₃BO₃ →⁽³⁷⁰ ᴷ⁾ HBO₂ + H₂O HBO₂ →⁽⁴³⁰ ᴷ⁾ B₂O₃ + H₂O

Q7. Describe the structure of diborane.
Answer

In diborane (B₂H₆), each boron atom is sp³ hybridised. The four terminal B–H bonds are normal covalent bonds (2-centre 2-electron bonds). The two bridging B–H–B bonds are 3-centre 2-electron bonds (banana bonds), where a hydrogen atom bridges two boron atoms. The terminal H atoms and the two B atoms lie in the same plane, while the bridging H atoms lie above and below this plane.

Q8. Why is boric acid considered a Lewis acid?
Answer

Boric acid does not donate a proton (H⁺). Instead, it accepts a pair of electrons from a hydroxide ion (OH⁻) of water, acting as a Lewis acid: H₃BO₃ + H₂O → [B(OH)₄]⁻ + H⁺ The boron atom in H₃BO₃ has only six electrons (incomplete octet) and can accept an electron pair from OH⁻, forming [B(OH)₄]⁻. This releases H⁺, making the solution weakly acidic.

Page 286–287 (Group 14 Elements, Carbon Allotropes)

Q9. Discuss the general trends in Group 14 elements.
Answer

Group 14 elements (C, Si, Ge, Sn, Pb) show the following trends: (i) Metallic character increases down the group — C and Si are non-metals, Ge is a metalloid, Sn and Pb are metals. (ii) Ionization enthalpy decreases down the group. (iii) Atomic and ionic radii increase. (iv) The +4 oxidation state becomes less stable down the group due to the inert pair effect, while +2 becomes more stable. (v) Catenation tendency decreases from C to Pb.

Q10. Why does carbon show maximum catenation?
Answer

Carbon shows maximum catenation because: (i) the C–C bond is very strong (bond enthalpy ≈ 355 kJ mol⁻¹), (ii) carbon atoms can form single, double and triple bonds with each other, and (iii) the small size of carbon allows effective overlap of orbitals. The bond strength decreases down the group (Si–Si ≈ 222 kJ mol⁻¹, Ge–Ge ≈ 167 kJ mol⁻¹), so the catenation tendency also decreases.

Q11. What are the allotropes of carbon?
Answer

The three main allotropes of carbon are: 1. Diamond: Each carbon is sp³ hybridised, bonded to four other carbon atoms in a tetrahedral arrangement. It is the hardest known natural substance, a poor conductor, and has a very high melting point. 2. Graphite: Each carbon is sp² hybridised, forming hexagonal layers. It is soft, a good conductor of electricity (due to delocalised π-electrons), and used as a lubricant. 3. Fullerene (C₆₀): A cage-like structure of 60 carbon atoms arranged in 20 hexagons and 12 pentagons (like a football). Each carbon is sp² hybridised.

Page 293–295 (Silicones, Silicates, Zeolites)

Q12. What are silicones? How are they prepared?
Answer

Silicones are organosilicon polymers with the general formula (R₂SiO)ₙ, where R is an alkyl or aryl group. They are prepared by hydrolysis of chlorosilanes followed by condensation polymerisation: R₂SiCl₂ + 2H₂O → R₂Si(OH)₂ + 2HCl R₂Si(OH)₂ units undergo condensation to form linear, cyclic, or cross-linked polymers with Si–O–Si linkages. Silicones are water-repellant, heat-resistant, and chemically inert.

Q13. What are silicates? Explain their structure.
Answer

Silicates are minerals containing SiO₄⁴⁻ tetrahedral units. Each SiO₄⁴⁻ unit shares oxygen atoms with neighbouring units, forming various structures:

  • Orthosilicates: isolated SiO₄⁴⁻ units (e.g., olivine)
  • Pyrosilicates: two tetrahedra sharing one O (Si₂O₇⁶⁻)
  • Chain silicates: linear chains sharing two O per tetrahedron (pyroxenes) or double chains (amphiboles)
  • Sheet silicates: three O shared per tetrahedron (e.g., talc, mica)
  • 3D framework silicates: all four O shared (e.g., quartz, feldspar)
Q14. What are zeolites? State their uses.
Answer

Zeolites are aluminosilicates with a three-dimensional framework structure in which some Si⁴⁺ atoms are replaced by Al³⁺, creating a negative charge that is balanced by cations (Na⁺, K⁺, Ca²⁺). They have a porous structure with channels and cavities. Uses: (i) as catalysts in petrochemical cracking (shape-selective catalysis), (ii) as ion-exchange agents in water softening, and (iii) in removing toxic ions from waste water.


Chapter-End Exercise

Q1. Discuss the general characteristics of Group 14 elements with reference to their electronic configuration, oxidation states and trends in chemical reactivity.
Answer

Group 14 elements have the general electronic configuration ns²np². They show two common oxidation states: +4 and +2. The +4 state is more stable for lighter elements (C, Si), while +2 becomes more stable for heavier elements (Sn, Pb) due to the inert pair effect. Reactivity increases down the group: carbon is relatively inert, silicon reacts with halogens and alkalis, germanium is more reactive, and tin and lead are quite reactive. All form oxides (MO₂ and MO), hydrides (MH₄), and halides (MX₄ and MX₂).

Q2. Why does boron form only covalent compounds?
Answer

Due to its extremely small size and very high ionization enthalpy (801 kJ mol⁻¹ for B), boron cannot lose its three valence electrons to form B³⁺ ions. The sum of the first three ionization enthalpies is too large to be compensated by lattice enthalpy. Therefore, boron forms only covalent compounds by sharing its electrons.

Q3. Explain why CO is poisonous.
Answer

CO binds strongly to haemoglobin in blood, forming carboxyhaemoglobin, which is about 200 times more stable than oxyhaemoglobin. This prevents oxygen from binding to haemoglobin, reducing the oxygen-carrying capacity of blood and leading to oxygen deprivation in tissues, which can be fatal.

Q4. What is the state of hybridization of carbon in (a) CO₃²⁻ (b) diamond (c) graphite?
Answer

(a) CO₃²⁻: sp² hybridised (trigonal planar, three equivalent resonance structures) (b) Diamond: sp³ hybridised (each carbon bonded to four others in a tetrahedral arrangement) (c) Graphite: sp² hybridised (each carbon bonded to three others in hexagonal layers; one unhybridised p-orbital per carbon forms delocalised π-bonds)

Q5. Write the reactions involved in the preparation of aluminium by Hall-Héroult process.
Answer

Aluminium is extracted by the electrolysis of molten alumina (Al₂O₃) dissolved in cryolite (Na₃AlF₆): At cathode: Al³⁺ (melt) + 3e⁻ → Al(l) At anode: C(s) + O²⁻ (melt) → CO(g) + 2e⁻ Overall: 2Al₂O₃ + 3C → 4Al + 3CO₂ Cryolite lowers the melting point of alumina from 2323 K to about 1140 K and improves conductivity.

Q6. Explain the difference in the behaviour of CO and CO₂.
Answer

CO is a neutral oxide and a reducing agent. It is poisonous and forms carboxyhaemoglobin with haemoglobin. CO₂ is an acidic oxide and dissolves in water to form carbonic acid (H₂CO₃). CO has a triple bond (C≡O), making it quite stable but reactive as a ligand. CO₂ has a linear structure (O=C=O), is non-poisonous, and is a greenhouse gas.

Q7. Discuss the shape of B₂H₆ using VSEPR theory.
Answer

B₂H₆ cannot be fully explained by VSEPR theory because it involves 3-centre 2-electron (3c-2e) bonds. However, around each boron atom, there are four bonds (two terminal B–H bonds and two bridging B–H–B bonds), giving an approximate tetrahedral arrangement of bonds around each boron. The molecule has D₂ₕ symmetry with the terminal H atoms and B atoms in one plane and the bridging H atoms perpendicular to this plane.

Q8. Give reasons: (i) Diamond is used as an abrasive. (ii) Graphite is used as a lubricant.
Answer

(i) Diamond is the hardest known natural substance because each carbon atom is sp³ hybridised and bonded to four other carbon atoms in a rigid 3D tetrahedral network. This makes it ideal for cutting, grinding and polishing hard materials. (ii) In graphite, each carbon is sp² hybridised, forming hexagonal layers held together by weak van der Waals forces. These layers can slide over each other easily, making graphite an excellent dry lubricant.

Q9. Explain why boron trifluoride is a Lewis acid but aluminium trichloride acts as a Lewis acid only in the presence of a suitable donor.
Answer

BF₃ acts as a Lewis acid because boron has an incomplete octet (only 6 electrons) and readily accepts an electron pair. In AlCl₃, aluminium also has an incomplete octet, but AlCl₃ exists as a dimer (Al₂Cl₆) in which each Al achieves an octet through chlorine bridges (3c-2e bonds). Thus, AlCl₃ acts as a Lewis acid only when a strong donor molecule is present that can break the dimer.

Q10. Explain the difference between the structures of borax and boric acid.
Answer

Borax (Na₂B₄O₇·10H₂O) contains two BO₃ triangles and two BO₄ tetrahedra linked together, forming the [B₄O₅(OH)₄]²⁻ anion with Na⁺ ions and water of crystallisation. Boric acid (H₃BO₃) has a layered structure in which BO₃ units are connected through hydrogen bonds. Each boron is sp² hybridised and forms three B–O bonds with OH groups in a trigonal planar arrangement.

Q11. Write the balanced equations for the reactions of (a) Borax with HCl (b) Aluminium with NaOH (c) Aluminium with HCl.
Answer

(a) Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl (b) 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂↑ (c) 2Al + 6HCl → 2AlCl₃ + 3H₂↑

Q12. What happens when aluminium is treated with (a) dilute HCl (b) dilute NaOH (c) hot concentrated NaOH?
Answer

(a) 2Al + 6HCl → 2AlCl₃ + 3H₂↑ (aluminium dissolves, hydrogen evolved) (b) 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂↑ (sodium aluminate and hydrogen formed) (c) 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂↑ (sodium tetrahydroxoaluminate formed)

Q13. Explain the reaction of CO₂ with (a) lime water (b) sodium hydroxide.
Answer

(a) CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O (turns lime water milky; excess CO₂ dissolves the precipitate: CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂) (b) CO₂ + 2NaOH → Na₂CO₃ + H₂O (sodium carbonate formed; with excess CO₂: Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃)

Q14. Explain the amphoteric nature of aluminium with suitable reactions.
Answer

Aluminium reacts with both acids and bases, demonstrating its amphoteric nature: With acid: 2Al + 6HCl → 2AlCl₃ + 3H₂↑ With base: 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂↑ Aluminium oxide also shows amphoteric character: With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O With base: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Q15. What are silicones? Give their uses.
Answer

Silicones are organosilicon polymers with the general formula (R₂SiO)ₙ containing Si–O–Si linkages. Uses: (i) as water-repellant coatings for fabrics and leather, (ii) as heat-resistant lubricants, (iii) in surgical and cosmetic implants (biocompatible), (iv) as insulating material in electrical equipment, (v) in making silicone rubber seals and gaskets, (vi) as polish and wax additives.

Q16. What is the action of heat on (a) Borax (b) Boric acid (c) Aluminium hydroxide?
Answer

(a) Borax: Na₂B₄O₇·10H₂O →⁽ˣᵗᵃˡ⁾ Na₂B₄O₇ → 2NaBO₂ + B₂O₃ (borax bead) (b) Boric acid: H₃BO₃ →⁽³⁷⁰ ᴷ⁾ HBO₂ →⁽⁴³⁰ ᴷ⁾ B₂O₃ (c) Al(OH)₃: 2Al(OH)₃ →⁽ᴴᵉᵃᵗ⁾ Al₂O₃ + 3H₂O

Q17. Explain the structure of diborane. Discuss the nature of bonds in it.
Answer

In B₂H₆, each boron is sp³ hybridised. The molecule has: (i) Four terminal B–H bonds — normal covalent 2-centre 2-electron bonds. (ii) Two bridging B–H–B bonds — 3-centre 2-electron bonds (banana bonds). The four terminal hydrogen atoms and the two boron atoms lie in one plane. The two bridging hydrogen atoms lie above and below this plane. The terminal B–H bond length (119 pm) is shorter than the bridging B–H bond length (133 pm).

Q18. Why is CO₂ a gas while SiO₂ is a solid?
Answer

CO₂ exists as discrete linear molecules (O=C=O) with weak van der Waals forces between molecules, so it is a gas at room temperature. SiO₂ forms a giant 3D network covalent structure where each silicon is bonded to four oxygen atoms (sp³) and each oxygen is shared between two silicon atoms, creating a rigid macromolecular structure that is a solid with a very high melting point.

Q19. Explain why boron halides are Lewis acids. How does the Lewis acidity vary among BX₃?
Answer

Boron halides (BX₃) are Lewis acids because boron has only six electrons (incomplete octet) and can accept an electron pair. The Lewis acidity order is: BBr₃ > BCl₃ > BF₃. Although BF₃ would be expected to be the strongest acid based on electronegativity, the pπ-pπ back-bonding from fluorine to boron in BF₃ reduces the electron deficiency of boron, making BF₃ the weakest Lewis acid among the boron trihalides.

Q20. Give reasons for the following: (i) C and Si are non-metals but Ge is a metalloid. (ii) Pb shows +2 as the more stable oxidation state.
Answer

(i) Going down Group 14, the atomic size increases and ionization enthalpy decreases. C and Si have small sizes and high ionization enthalpies, making electron loss difficult (non-metals). Ge has intermediate properties — its ionization enthalpy is low enough to show some metallic character, but it also forms covalent bonds (metalloid). (ii) In Pb, the 6s² electrons are tightly held due to poor shielding by 4f and 5d electrons (inert pair effect). These two electrons do not participate easily in bonding, making the +2 state more stable than +4. Pb⁴⁺ acts as an oxidising agent.

Q21. Write balanced equations for the following: (a) B₂H₆ + H₂O → (b) B₂H₆ + NH₃ → (c) Al₂O₃ + C + Cl₂ →
Answer

(a) B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂↑ (diborane hydrolysed to boric acid) (b) 3B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂ (borazine or inorganic benzene formed) (c) Al₂O₃ + 3C + 3Cl₂ → 2AlCl₃ + 3CO (aluminium chloride formed)

Q22. Describe the manufacturing process of aluminium.
Answer

Aluminium is manufactured by the Hall-Héroult process: 1. Purification of bauxite (Al₂O₃·2H₂O) by Bayer's process to obtain pure alumina. 2. Electrolysis of alumina dissolved in molten cryolite (Na₃AlF₆) at about 1173 K in a steel tank lined with carbon (cathode) and carbon anodes. 3. Reactions: Cathode: Al³⁺ (melt) + 3e⁻ → Al(l) Anode: C + O²⁻ → CO + 2e⁻ (or C + 2O²⁻ → CO₂ + 4e⁻) 4. The molten aluminium collects at the bottom and is tapped off. 5. Cryolite lowers the melting point and increases conductivity.

Q23. Why does boron not form B³⁺ ions?
Answer

The sum of the first three ionization enthalpies of boron is very high (about 6887 kJ mol⁻¹). The energy required to remove three electrons is not compensated by hydration enthalpy or lattice enthalpy. Therefore, boron does not form B³⁺ ions and instead forms covalent compounds.

Q24. Why does the +2 oxidation state become more stable down Group 14?
Answer

As we go down Group 14, the inert pair effect becomes more pronounced. The ns² electrons experience increasing effective nuclear charge due to poor shielding by inner d and f electrons. In Pb, the 6s² pair is particularly inert, making the +2 oxidation state more stable than +4. PbCl₂ is stable while PbCl₄ is unstable and acts as an oxidizing agent.


Internal Links

  • [Chapter 11 Guide — The p-Block Elements](/school/cbse/class/11/chemistry/cbse-p-block-elements)
  • [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
  • [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)

Frequently Asked Questions

Boron is the first element of Group 13 and shows anomalous behaviour due to its extremely small atomic size, high ionization enthalpy, high electronegativity, and absence of d-orbitals. These factors make boron predominantly non-metallic, forming covalent compounds rather than ionic ones, unlike the other Group 13 elements.