Chapter 2 — p-Block Elements I Exercise Solutions
This chapter is best revised with comparison tables. Focus on group 13 and group 14 trends.
Key Trend Points
| Property | General idea |
|---|
| Atomic size | Increases down the group |
| Ionisation energy | Generally decreases down the group |
| Metallic character | Increases down the group |
| Oxidation states | Inert pair effect affects heavier elements |
Short Answer Points
- Boron is non-metallic and electron deficient.
- Aluminium is amphoteric.
- Carbon shows catenation due to strong C-C bond.
- Silicon forms silicates and silicones.
Reaction Answer Pattern
1. Write the reactants. 2. Add condition if required. 3. Balance the equation. 4. Mention product property or use.
Worked Examples
Example 1 — Combustion of diborane
B₂H₆ + 3O₂ → B₂O₃ + 3H₂O (highly exothermic)
Example 2 — Inert pair effect (model answer)
Q: Why is the +1 state more stable than +3 for thallium (Tl)?
Down group 13 the two ns electrons stay paired and become reluctant to take part in bonding (the inert pair effect). So heavier elements like Tl show the lower +1 oxidation state as more stable than +3.
Example 3 — Amphoteric oxide
Show that aluminium oxide is amphoteric.
With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
With base: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
Reacting with both acid and base → amphoteric.
Common Mistakes
- Confusing group 13 and group 14 oxidation states.
- Missing inert pair effect explanation.
- Writing unbalanced equations.
- Giving vague uses of compounds.
Exam Checklist
- Trends are compared clearly.
- Anomalous behaviour is explained.
- Equations are balanced.
- Uses are specific.
How to Use This Page
- Start with the Key Concepts table to recall definitions quickly.
- Practice MCQs to test whether you remember group trends and anomalous behaviour.
- Use Fill in the Blanks and True or False for quick factual recall.
- Answer Briefly and Answer in Detail sections cover the most-asked exam questions.
- Review Common Mistakes to Avoid before writing your exam.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Inert pair effect | Reluctance of ns² electrons to participate in bonding, making lower oxidation states more stable in heavier p-block elements |
| Catenation | Ability of an element to form chains of self-linked atoms (e.g., C-C bonds) |
| Borax | Sodium tetraborate (Na₂B₄O₇·10H₂O), used as a flux and in borax bead test |
| Diborane (B₂H₆) | An electron-deficient compound with two 3-centre 2-electron (banana) bonds |
| Orthoboric acid | H₃BO₃, a weak monobasic Lewis acid that accepts OH⁻ from water |
| Aluminium extraction | Bayer's process for concentration + Hall-Heroult for electrolytic reduction |
| Alum | Double salt: M₂SO₄·Al₂(SO₄)₃·24H₂O (M = K, NH₄, etc.), used in water purification |
| Allotropy | Existence of an element in different structural forms (e.g., diamond and graphite for carbon) |
| Silicones | Polymers with alternating Si-O backbone and organic side groups (R₂SiO)ₙ |
| Silicates | Minerals containing SiO₄⁴⁻ tetrahedra linked in various patterns |
| Zeolites | Microporous aluminosilicates used as molecular sieves and catalysts |
| Anomalous behaviour of boron | Small size, high ionisation energy, non-metallic character — differs from rest of group 13 |
| Anomalous behaviour of carbon | High catenation, multiple bond formation, allotropic forms — differs from rest of group 14 |
I. Choose the Best Answer (MCQs)
1. The element that shows the maximum catenation in group 14 is:
Answer
(a) Silicon (b) Germanium (c) Carbon (d) Tin
Answer(c) Carbon — Carbon has the strongest C-C bond and shows maximum catenation among group 14 elements.
2. Borax bead test is used to identify:
Answer
(a) Alkali metals (b) Transition metal cations (c) Halide ions (d) Sulphate ions
Answer(b) Transition metal cations — Borax forms metal metaborate beads with characteristic colours when heated with transition metal salts.
3. Which of the following is an electron-deficient compound?
Answer
(a) CH₄ (b) C₂H₆ (c) B₂H₆ (d) SiH₄
Answer(c) B₂H₆ — Diborane has only 12 valence electrons for 12 bonds, making it electron-deficient with 3-centre 2-electron bonds.
4. Orthoboric acid acts as a Lewis acid because it:
Answer
(a) Donates a proton (b) Accepts an electron pair from water (c) Donates an electron pair (d) Releases H⁺ directly
Answer(b) Accepts an electron pair from water — H₃BO₃ does not donate a proton directly; it accepts OH⁻ from water, acting as a Lewis acid.
H₃BO₃ + H₂O → [B(OH)₄]⁻ + H⁺
5. The inert pair effect is most prominent in:
Answer
(a) Boron (b) Aluminium (c) Gallium (d) Thallium
Answer(d) Thallium — The inert pair effect increases down the group; thallium (Tl) shows +1 as more stable than +3.
6. Silicones are:
Answer
(a) Inorganic polymers with Si-Si backbone (b) Organic polymers with C-C backbone (c) Polymers with Si-O backbone and organic side groups (d) Ionic compounds of silicon
Answer(c) Polymers with Si-O backbone and organic side groups — Silicones have the general formula (R₂SiO)ₙ with alternating silicon and oxygen atoms and organic groups attached to silicon.
7. In the extraction of aluminium by Hall-Heroult process, the electrolyte is:
Answer
(a) Pure molten alumina (b) Alumina dissolved in molten cryolite (c) Alumina dissolved in water (d) Aluminium chloride solution
Answer(b) Alumina dissolved in molten cryolite — Cryolite lowers the melting point and increases conductivity.
8. Which allotrope of carbon is the hardest natural substance?
Answer
(a) Graphite (b) Fullerene (c) Diamond (d) Carbon black
Answer(c) Diamond — Diamond has a 3-dimensional tetrahedral network of sp³ hybridised carbon atoms, making it extremely hard.
9. Borax on heating gives:
Answer
(a) Boron and oxygen (b) Sodium metaborate and boric anhydride (c) Boric acid and sodium oxide (d) Diborane and water
Answer(b) Sodium metaborate and boric anhydride — Borax decomposes on heating:
Na₂B₄O₇·10H₂O →(heat) Na₂B₄O₇ →(more heat) 2NaBO₂ + B₂O₃
10. Which of the following is a beryllium anomaly analogue in group 13?
Answer
(a) Boron shows properties different from the rest of group 13 (b) Aluminium is more metallic than boron (c) Gallium has a low melting point (d) Indium shows +3 oxidation state
Answer(a) Boron shows properties different from the rest of group 13 — Boron is a non-metal, electron-deficient, and forms covalent compounds, unlike the metallic rest of group 13.
11. Zeolites are:
Answer
(a) Silicates with chain structure (b) Aluminosilicates with a framework of pores (c) Simple silicon oxides (d) Borosilicate glass
Answer(b) Aluminosilicates with a framework of pores — Zeolites are microporous aluminosilicates used as molecular sieves and ion exchangers.
12. Potash alum has the formula:
Answer
(a) K₂SO₄·Al₂(SO₄)₃·24H₂O (b) Na₂SO₄·Al₂(SO₄)₃·24H₂O (c) KAl(SO₄)₂·12H₂O (d) Both (a) and (c) are equivalent formulae
Answer(d) Both (a) and (c) are equivalent formulae — The two formulae represent the same compound; K₂SO₄·Al₂(SO₄)₃·24H₂O = 2[KAl(SO₄)₂·12H₂O].
II. Fill in the Blanks
1. The element that shows maximum catenation in the periodic table is ________. Carbon (C)
2. Borax bead test is based on the formation of ________ when borax is heated with metal salts. Metal metaborates
3. Orthoboric acid is a ________ acid (Lewis / Bronsted). Lewis
4. The hybridisation of boron in diborane (B₂H₆) for terminal B-H bonds is ________. sp²
5. In group 13, the element that shows the most pronounced inert pair effect is ________. Thallium (Tl)
6. The general formula of silicones is ________. (R₂SiO)ₙ
7. The process used for the concentration of bauxite ore is ________. Bayer's process (leaching with NaOH)
8. Potash alum is used in ________ purification. Water
9. The allotrope of carbon that is a good conductor of electricity is ________. Graphite
10. Zeolites are microporous ________ used as molecular sieves. Aluminosilicates
III. True or False
1. Boron is a metal. — False. Boron is a non-metal (metalloid) and is the only non-metal in group 13.
2. The inert pair effect becomes more pronounced down group 13. — True. The ns² electrons become more reluctant to participate in bonding as the atomic size increases.
3. Diborane has 3-centre 2-electron bonds. — True. The two bridging B-H-B bonds in diborane are 3-centre 2-electron (banana) bonds.
4. Aluminium oxide reacts only with acids. — False. Aluminium oxide is amphoteric; it reacts with both acids and bases.
5. Graphite is harder than diamond. — False. Diamond is the hardest natural substance; graphite is soft and slippery due to its layered structure.
6. All group 14 elements show catenation. — True. All group 14 elements show catenation, but the tendency decreases down the group: C >> Si > Ge > Sn > Pb.
7. Orthoboric acid is a tribasic acid. — False. Orthoboric acid (H₃BO₃) is a monobasic Lewis acid; it accepts only one OH⁻ from water.
8. Silicones are thermally stable and water-repellent. — True. Silicones have high thermal stability and are hydrophobic due to the organic side groups.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Borax | (e) Na₂B₄O₇·10H₂O, used as flux |
| 2. Diborane | (a) B₂H₆, electron-deficient compound |
| 3. Orthoboric acid | (f) H₃BO₃, Lewis acid |
| 4. Inert pair effect | (c) ns² electrons stay paired, +1 state stable |
| 5. Diamond | (h) sp³ hybridised carbon, hardest natural substance |
| 6. Graphite | (g) sp² hybridised carbon, good conductor |
| 7. Silicones | (b) Polymers with Si-O backbone |
| 8. Zeolites | (d) Microporous aluminosilicates |
Answer1-e, 2-a, 3-f, 4-c, 5-h, 6-g, 7-b, 8-d
V. Answer Briefly (2-Mark Questions)
1. What is the inert pair effect? Explain with reference to group 13 elements.
Answer
The inert pair effect is the tendency of the ns² electrons in heavier p-block elements to remain paired and not participate in bonding, due to poor shielding by d and f electrons. In group 13, boron shows +3, but as we go down, the +1 oxidation state becomes more stable. Thallium (Tl) prefers +1 because the 6s² electrons act as an "inert pair," making Tl⁺ more stable than Tl³⁺.
2. Explain the anomalous behaviour of boron in group 13.
Answer
Boron differs from other group 13 elements because: (1) It is a non-metal while others are metals. (2) It has a very small atomic size and high ionisation energy. (3) It forms only covalent compounds, unlike the ionic compounds of aluminium. (4) It shows a maximum coordination number of 4 (not 6 like Al) due to its small size. (5) It forms electron-deficient compounds like diborane with 3-centre 2-electron bonds.
3. Show that aluminium oxide is amphoteric with chemical equations.
Answer
Aluminium oxide reacts with both acids and bases, proving its amphoteric nature:
With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
With base: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄] (sodium tetrahydroxoaluminate)
or simply: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O (sodium aluminate)
4. What are silicones? Give their general formula and two uses.
Answer
Silicones are synthetic polymers with a backbone of alternating silicon and oxygen atoms, with organic groups (methyl or phenyl) attached to silicon. Their general formula is (R₂SiO)ₙ. They are thermally stable, water-repellent, and chemically inert. Uses: (1) As sealants and water-repellent coatings, and (2) As insulating materials in electrical equipment.
5. Differentiate between diamond and graphite.
Answer
Diamond has sp³ hybridised carbon atoms in a 3-dimensional tetrahedral network, making it the hardest natural substance, a poor conductor of electricity, and transparent. Graphite has sp² hybridised carbon atoms arranged in hexagonal layers held by weak van der Waals forces, making it soft, slippery, a good conductor of electricity (due to delocalised π electrons), and opaque.
6. What is the borax bead test? How does it work?
Answer
The borax bead test is a qualitative test for identifying transition metal cations. When borax (Na₂B₄O₇·10H₂O) is heated on a platinum wire loop, it forms a colourless glassy bead of sodium metaborate and boric anhydride (B₂O₃). When this bead is heated with a metal salt, a coloured metaborate is formed. For example, cobalt gives a blue bead (Co(BO₂)₂), and chromium gives a green bead.
Na₂B₄O₇ → 2NaBO₂ + B₂O₃
CuO + B₂O₃ → Cu(BO₂)₂ (blue-green bead)
VI. Answer in Detail (5-Mark Questions)
1. Describe the extraction of aluminium from bauxite with equations.
Answer
Aluminium is extracted from bauxite (Al₂O₃·2H₂O) in two main stages:
Stage 1 — Concentration (Bayer's Process): Bauxite ore is crushed and digested with concentrated NaOH solution under pressure at 473–523 K. Alumina dissolves as sodium aluminate, while impurities (Fe₂O₃, SiO₂, TiO₂) remain as red mud. The solution is filtered, cooled, and seeded with Al(OH)₃ crystals.
Al₂O₃·2H₂O + 2NaOH →(473-523 K) 2NaAlO₂ + 3H₂O
2NaAlO₂ + 4H₂O → 2NaOH + 2Al(OH)₃ (precipitation by seeding)
2Al(OH)₃ →(heat) Al₂O₃ + 3H₂O (calcination at 1473 K)
Stage 2 — Hall-Heroult Process (Electrolytic Reduction): Pure alumina is dissolved in molten cryolite (Na₃AlF₆) at about 950 °C in an iron cell lined with carbon (cathode). Carbon rods serve as the anode. On passing current, Al³⁺ ions are reduced at the cathode.
At cathode: Al³⁺ + 3e⁻ → Al
At anode: C(graphite) + O²⁻ → CO + 2e⁻ and C + 2O²⁻ → CO₂ + 4e⁻
Overall: 2Al₂O₃ → 4Al + 3O₂
The oxygen liberated reacts with the carbon anode to form CO and CO₂, so anodes must be replaced periodically. The aluminium obtained is about 99.5% pure and is further refined by Hoopes process if needed.
2. Discuss the important compounds of boron — borax, boric acid, and diborane.
Answer
(a) Borax (Na₂B₄O₇·10H₂O): Also called sodium tetraborate decahydrate, borax is a white crystalline solid. On heating, it first loses water molecules, then decomposes to sodium metaborate and boric anhydride, forming a glassy bead used in the borax bead test.
Na₂B₄O₇·10H₂O →(heat) Na₂B₄O₇ →(red heat) 2NaBO₂ + B₂O₃
Borax is used as a flux in welding, in the borax bead test, in making optical and borosilicate glass, and as a mild antiseptic.
(b) Orthoboric Acid (H₃BO₃): Orthoboric acid is obtained by treating borax with sulphuric acid:
Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + Na₂SO₄
It is a weak monobasic Lewis acid — it does not donate a proton but accepts OH⁻ from water:
H₃BO₃ + H₂O → [B(OH)₄]⁻ + H⁺
On heating, it forms metaboric acid and then boric anhydride:
H₃BO₃ →(100 °C) HBO₂ →(160 °C) H₂B₄O₇ →(red heat) B₂O₃
(c) Diborane (B₂H₆): Diborane is the simplest boron hydride. It is an electron-deficient compound with 12 valence electrons. It has two types of B-H bonds: four terminal B-H bonds (normal 2-centre 2-electron bonds) and two bridging B-H-B bonds (3-centre 2-electron or "banana" bonds). Diborane burns in air to give boron trioxide and water:
B₂H₆ + 3O₂ → B₂O₃ + 3H₂O
It hydrolyses with water:
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂
3. Explain the trends in group 14 elements (carbon family).
Answer
Group 14 elements (C, Si, Ge, Sn, Pb) show the following trends:
(a) Atomic and Ionic Radii: Atomic radii increase down the group from carbon to lead as additional shells are added. However, the increase is not uniform due to d- and f-orbital filling (lanthanide contraction effect on Pb).
(b) Ionisation Energy: Ionisation energy decreases down the group as atomic size increases. Carbon has the highest first ionisation energy. There is a slight irregularity at Ge-Sn due to poor shielding by d-electrons.
(c) Metallic Character: Increases down the group. Carbon and silicon are non-metals (silicon is a metalloid), germanium is a metalloid, while tin and lead are metals.
(d) Oxidation States: The common oxidation states are +4 and +2. The +4 state is more stable for lighter elements (C, Si), while the +2 state becomes more stable for heavier elements (Sn, Pb) due to the inert pair effect. Carbon also shows -4 in carbides.
(e) Catenation: The ability to form self-linked chains decreases down the group. Carbon shows maximum catenation (forms millions of organic compounds). Silicon shows limited catenation, and tin and lead show almost none.
Catenation tendency: C >> Si > Ge ≈ Sn > Pb
(f) Allotropy: All group 14 elements show allotropy. Carbon has diamond and graphite (and fullerene). Silicon has amorphous and crystalline forms. Tin has grey (α) and white (β) forms.
4. What are silicates and zeolites? Explain their structures and uses.
Answer
(a) Silicates: Silicates are minerals composed of SiO₄⁴⁻ tetrahedra linked in various patterns through shared oxygen atoms. The classification depends on how the tetrahedra are linked:
1. Orthosilicates (Nesosilicates): Independent SiO₄⁴⁻ units. Example: Olivine (Mg₂SiO₄). 2. Pyrosilicates (Sorosilicates): Two tetrahedra share one oxygen → Si₂O₇⁶⁻. Example: Thortveitite. 3. Chain silicates (Inosilicates): Tetrahedra share two oxygens forming chains — single chain (SiO₃²⁻)ₙ or double chain (Si₄O₁₁⁶⁻)ₙ. Example: Asbestos. 4. Sheet silicates (Phyllosilicates): Each tetrahedron shares three oxygens → (Si₂O₅²⁻)ₙ. Example: Talc, mica. 5. 3D framework silicates (Tectosilicates): All four oxygens are shared → SiO₂ (quartz). When Al replaces some Si, aluminosilicates form.
(b) Zeolites: Zeolites are microporous aluminosilicates with a 3D framework of SiO₄ and AlO₄ tetrahedra. They have regularly spaced pores of molecular dimensions, making them excellent molecular sieves. Uses: (1) As molecular sieves to separate molecules by size. (2) As ion exchangers in water softening. (3) As catalysts in petrochemical cracking (e.g., ZSM-5). (4) In removing hardness of water by exchanging Ca²⁺/Mg²⁺ with Na⁺.
5. Discuss the anomalous behaviour of carbon in group 14.
Answer
Carbon shows anomalous behaviour compared to other group 14 elements due to its small size, high electronegativity, and high ionisation energy:
1. Maximum Catenation: Carbon forms millions of compounds by linking with itself (C-C bonds) and with other elements. This is because the C-C bond energy (348 kJ/mol) is very high, comparable to C-H and C-O bonds. Silicon, in contrast, has a much lower Si-Si bond energy (222 kJ/mol).
2. Multiple Bond Formation: Carbon readily forms double and triple bonds (C=C, C≡C, C=O, C≡N) due to its small size and effective p-orbital overlap. Other group 14 elements rarely form stable multiple bonds due to poor p-orbital overlap.
3. Allotropy: Carbon has two important allotropic forms — diamond (sp³, 3D network, hardest substance) and graphite (sp², layered, conductor) — plus fullerene (C₆₀). Other elements show limited allotropy.
4. Tetravalency with Small Size: Carbon's small size allows four bond pairs to be accommodated around it without repulsion, giving stable tetrahedral geometry.
5. Hybridisation: Carbon shows sp³, sp², and sp hybridisation, enabling formation of single, double, and triple bonds with directional properties.
6. Formation of Extremely Stable Compounds: CO₂ is a gas (molecular), while SiO₂ is a solid (network). This is because CO₂ forms discrete molecules with C=O double bonds, whereas SiO₂ forms a 3D network of single bonds.
Common Mistakes to Avoid
- ❌ Saying boron is a metal. ✅ Boron is a non-metal (metalloid), unlike other group 13 elements which are metals.
- ❌ Confusing orthoboric acid as a Bronsted acid. ✅ H₃BO₃ is a Lewis acid; it accepts OH⁻ from water, not donates H⁺.
- ❌ Writing that group 14 catenation increases down the group. ✅ Catenation decreases down the group: C >> Si > Ge > Sn > Pb.
- ❌ Forgetting to mention the inert pair effect for thallium and lead. ✅ The inert pair effect explains why +1 is more stable for Tl and +2 for Pb.
- ❌ Confusing silicates with silicones. ✅ Silicates are minerals with SiO₄ units; silicones are synthetic polymers with Si-O backbones and organic side groups.
- ❌ Writing unbalanced equations for borax bead test or boric acid decomposition. ✅ Always balance equations and show intermediate products.
- ❌ Confusing coordination number of boron with aluminium. ✅ Boron has a maximum coordination number of 4; aluminium can have 6.
- ❌ Forgetting that diborane has 3-centre 2-electron bonds. ✅ The bridging B-H-B bonds in diborane are unique 3-centre 2-electron (banana) bonds.
Exam Checklist
- [ ] Can you list group 13 and group 14 trends in atomic size, ionisation energy, and metallic character?
- [ ] Can you explain the inert pair effect with examples (Tl, Pb)?
- [ ] Can you write equations for borax bead test with at least two metal cations?
- [ ] Can you show H₃BO₃ acting as a Lewis acid with the equation?
- [ ] Can you draw/explain the structure of diborane (terminal and bridging bonds)?
- [ ] Can you differentiate between diamond and graphite structure and properties?
- [ ] Can you write the Bayer's process equations for bauxite concentration?
- [ ] Can you explain the Hall-Heroult process with electrode reactions?
- [ ] Can you classify silicate structures (ortho, pyro, chain, sheet, framework)?
- [ ] Can you define zeolites and their uses?
- [ ] Can you explain why carbon shows anomalous behaviour (catenation, multiple bonds)?
- [ ] Can you write balanced equations for alum formation and uses?