p-Block Elements I — Boron & Carbon Families (Groups 13–14)

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Groups 13 and 14 are where the p-block begins, and both start with an element that refuses to behave like the rest. Boron is a non-metal in a group of metals; carbon chains endlessly while lead barely bonds to itself. Learn why the first member is different and why the heaviest members prefer a lower oxidation state, and the rest of the chapter is description.

What you will be able to do

• Explain trends in size, ionisation enthalpy and oxidation states in groups 13 and 14, including the inert pair effect
• Explain the Lewis acidity of boron halides and the structure of diborane
• Write the preparation and key reactions of borax, boric acid and diborane
• Compare the allotropes of carbon and the chemistry of CO, CO₂ and SiO₂
• Describe silicones, silicates and zeolites

The big picture

First member odd, last member lazy.

In one line: In groups 13 and 14 the first element (B, C) is small, non-metallic and limited to four bonds; going down, metallic character rises and the ns² pair becomes reluctant to bond, so the lower oxidation state (+1 for Tl, +2 for Pb) becomes the stable one.

Property	Group 13 (B → Tl)	Group 14 (C → Pb)
Valence configuration	ns² np¹	ns² np²
Common oxidation states	+3; +1 grows down (Tl⁺ most stable)	+4; +2 grows down (Pb²⁺ most stable)
Character	B non-metal; Al to Tl metals	C, Si non-metals; Ge metalloid; Sn, Pb metals
Oxides	B₂O₃ acidic → Al₂O₃, Ga₂O₃ amphoteric → In₂O₃, Tl₂O₃ basic	CO₂, SiO₂, GeO₂ acidic; SnO₂, PbO₂ amphoteric

Think of it like this: Picture the ns² pair as two old residents of a building who get more comfortable — and less willing to move out — as the building (atom) grows taller. In boron and carbon they join every bond. By thallium and lead they mostly stay home, and only the p electrons go out to bond.

Group 13: trends

Data first, then the exceptions.

Property	B	Al	Ga	In	Tl
Atomic radius / pm	85	143	135	167	170
ΔiH₁ / kJ mol⁻¹	801	577	579	558	589
Electronegativity (Pauling)	2.0	1.5	1.6	1.7	1.8
m.p. / K	2453	933	303	430	576
b.p. / K	3923	2740	2676	2353	1730

• Ga is smaller than Al — the ten poorly shielding 3d electrons raise the effective nuclear charge (d-block contraction).
• Ionisation enthalpies are irregular for the same reason: Ga’s ΔiH₁ is slightly above Al’s, and Tl’s is raised by poor 4f/5d shielding.
• Electronegativity falls from B to Al, then rises slightly — again because of d and f contraction.
• Ga melts at 303 K (it melts in the hand) but boils at 2676 K — the widest liquid range of any element, so it is used in high-temperature thermometers.
• Boron is very hard and high-melting because it forms a giant covalent network (B₁₂ icosahedra).

Oxidation states and the inert pair effect

• Down the group the +1 state becomes more stable: Tl⁺ is more stable than Tl³⁺, so Tl³⁺ is a strong oxidising agent.
• B and Al are almost always +3; Ga, In show +1 and +3; Tl mainly +1.
• Cause: the ns² electrons in the heavier elements are held more tightly (poor shielding by d and f electrons), and the bonds formed by the heavier atoms are weaker, so the energy gained from two extra bonds no longer pays for unpairing the ns² pair.

Group 13: compounds and Lewis acidity

Electron-deficient halides and amphoteric aluminium.

• Trihalides MX₃ have only six electrons around M → Lewis acids (electron-pair acceptors): BF₃ + NH₃ → F₃B←NH₃.
• Lewis acid strength of boron halides: BF₃ < BCl₃ < BBr₃ < BI₃ — the opposite of what electronegativity suggests. In BF₃ the small F atom back-donates a lone pair into boron’s empty 2p orbital (pπ–pπ back bonding), partly filling it. The overlap is poorer for the larger halogens.
• AlCl₃ exists as the dimer Al₂Cl₆ in the vapour and in non-polar solvents (each Al completes its octet using a lone pair from a bridging Cl). It is the Lewis-acid catalyst of Friedel–Crafts reactions.
• Boron cannot exceed four bonds ([BF₄]⁻, no d orbitals); aluminium can form [AlF₆]³⁻.
• B₂O₃ is acidic, Al₂O₃ amphoteric, Tl₂O₃ basic — metallic character rises down the group.

Aluminium

2Al + 6HCl → 2AlCl₃ + 3H₂
2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂

Amphoteric metal: dissolves in both acid and alkali

• Aluminium is rendered passive by conc. HNO₃ (a protective oxide film).
• Al₂O₃ is amphoteric; Al(OH)₃ dissolves in excess NaOH to give [Al(OH)₄]⁻.
• Uses: light alloys (duralumin), wires, foil, packaging; thermite welding (Al reduces Fe₂O₃ with great heat).

Solved example 01 — Ranking Lewis acidity

Arrange BF₃, BCl₃ and BBr₃ in increasing Lewis acid strength and explain.

1. All three have an empty 2p orbital on boron.
2. F’s 2p orbital matches B’s 2p closely, so back-donation from F is strongest and partly fills B’s empty orbital.
3. Cl (3p) and Br (4p) overlap less well, so boron stays more electron-deficient.

Answer: BF₃ < BCl₃ < BBr₃ (back bonding is strongest in BF₃).

Check yourself

• C1. Why is AlCl₃ a dimer in the vapour phase?
• C2. Why is Tl³⁺ a good oxidising agent?

Answers at the end.

Borax and boric acid

Two boron compounds every exam asks about.

Borax

• Formula Na₂B₄O₇·10H₂O, more correctly Na₂[B₄O₅(OH)₄]·8H₂O — the anion contains two tetrahedral and two trigonal boron atoms.
• Aqueous borax is alkaline: Na₂B₄O₇ + 7H₂O → 2NaOH + 4H₃BO₃ (a salt of a strong base and weak acid). It is used as a primary standard for titrating acids.
• On heating it loses water and swells, then melts to a clear glass: Na₂B₄O₇·10H₂O → Na₂B₄O₇ → 2NaBO₂ + B₂O₃ (transparent bead).
• Borax bead test: the glassy NaBO₂ + B₂O₃ bead fused with a transition-metal salt gives a coloured metaborate. Co gives a blue bead (Co(BO₂)₂); Cu blue-green; Ni brown; Cr green.

Boric acid, H₃BO₃

• Made by acidifying borax: Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4H₃BO₃.
• A weak monobasic Lewis acid — it does not give H⁺ itself; it accepts OH⁻ from water: B(OH)₃ + 2H₂O → [B(OH)₄]⁻ + H₃O⁺.
• Layered crystal: planar B(OH)₃ units joined by hydrogen bonds, so it is soft and slippery.
• On heating: H₃BO₃ → HBO₂ (metaboric acid, about 370 K) → B₂O₃ (red heat).
• Its acid strength increases in the presence of polyhydroxy compounds (glycerol, mannitol), which bind [B(OH)₄]⁻ — this allows it to be titrated with NaOH.

Solved example 02 — Why boric acid is monobasic

H₃BO₃ has three O–H groups, yet it is a monobasic acid. Explain.

1. Boron has only six electrons around it in B(OH)₃ and an empty p orbital.
2. It accepts an OH⁻ ion from water: B(OH)₃ + H₂O → [B(OH)₄]⁻ + H⁺.
3. Only one H⁺ is released per molecule, and not from its own O–H groups.

Answer: It behaves as a Lewis acid that accepts one OH⁻; it is monobasic.

Diborane

The molecule that broke the two-electron-bond rule.

Preparation

• Lab: 2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂.
• From BF₃: 4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiF + 3AlF₃ (in ether).
• Industrial: 2BF₃ + 6NaH → B₂H₆ + 6NaF (450 K).

Structure

• Two BH₂ units are joined by two bridging H atoms above and below the plane of the four terminal H atoms.
• Terminal B–H bonds (4): ordinary two-centre two-electron bonds, 119 pm.
• Bridging B–H–B bonds (2): three-centre two-electron (3c–2e) “banana” bonds, 134 pm — one electron pair holds three atoms together.
• Each boron is roughly sp³ hybridised. B₂H₆ has 12 valence electrons — too few for 8 ordinary bonds, which is why the 3c–2e bonds exist (it is electron-deficient).

Reactions

Reagent	Product	Note
O₂	B₂O₃ + 3H₂O	ΔH = −1976 kJ mol⁻¹; burns spontaneously
H₂O	2B(OH)₃ + 6H₂	Hydrolysis
Lewis bases (CO, NH₃, N(CH₃)₃)	Adducts: BH₃·CO, BH₃·N(CH₃)₃	Symmetric cleavage of the bridges
NH₃ (heated)	B₃N₃H₆ (borazine) via B₂H₆·2NH₃	Borazine is “inorganic benzene”
NaH (in ether)	2NaBH₄	Sodium borohydride, a reducing agent
LiH	2LiBH₄	Lithium borohydride

Solved example 03 — Counting bonds in diborane

How many electrons does B₂H₆ have for bonding, and how are they used?

1. Valence electrons: 2 × 3 (B) + 6 × 1 (H) = 12.
2. Four terminal B–H bonds use 8 electrons.
3. The remaining 4 electrons form two 3c–2e B–H–B bridges.

Answer: 12 electrons: 4 normal bonds (8 e⁻) + 2 banana bonds (4 e⁻).

Check yourself

• C3. What is the colour of the cobalt borax bead?
• C4. How many 3-centre-2-electron bonds are in diborane?

Answers at the end.

Group 14: trends

From non-metal carbon to metallic lead.

Property	C	Si	Ge	Sn	Pb
Covalent radius / pm	77	118	122	140	146
ΔiH₁ / kJ mol⁻¹	1086	786	761	708	715
Electronegativity	2.5	1.8	1.8	1.8	1.9
m.p. / K	4373 (sublimes)	1693	1218	505	600
Element–element bond enthalpy / kJ mol⁻¹	C–C 348	Si–Si 297	Ge–Ge 260	Sn–Sn 240	—

• Catenation (self-linking into chains and rings) falls sharply: C ≫ Si > Ge ≈ Sn; Pb does not catenate. The C–C bond is strong; Si–Si and below are much weaker.
• pπ–pπ multiple bonds: carbon forms C=C, C≡C, C=O easily; Si and heavier elements hardly do (poor sideways overlap of large orbitals). That is why CO₂ is a molecule but SiO₂ is a giant network.
• +2 state stabilises down the group: Sn²⁺ is a reducing agent (it goes to Sn⁴⁺); Pb⁴⁺ is a strong oxidising agent (PbO₂). PbI₄ does not exist — Pb⁴⁺ would oxidise I⁻ to I₂.
• Halides: all form MX₄ (except PbBr₄ and PbI₄). CCl₄ does not hydrolyse, but SiCl₄ hydrolyses readily (SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl): silicon has vacant 3d orbitals to accept a lone pair from water, while carbon has none.
• Oxides: CO neutral; CO₂, SiO₂, GeO₂ acidic; SnO, PbO, SnO₂, PbO₂ amphoteric; GeO is acidic.

Solved example 04 — Hydrolysis of tetrachlorides

Why does SiCl₄ fume in moist air while CCl₄ does not react with water?

1. Hydrolysis begins with water donating a lone pair to the central atom.
2. Si has empty 3d orbitals and is larger, so it can accept the lone pair and expand its coordination.
3. Carbon has no d orbitals in its valence shell and is crowded by four Cl atoms.

Answer: SiCl₄ hydrolyses to Si(OH)₄ + HCl (the fumes are HCl); CCl₄ cannot.

Solved example 05 — PbO₂ as an oxidant, SnCl₂ as a reductant

Write one reaction showing PbO₂ as an oxidising agent and one showing SnCl₂ as a reducing agent, and explain both with the inert pair effect.

1. PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O — Pb(IV) is reduced to the more stable Pb(II).
2. SnCl₂ + 2HgCl₂ → SnCl₄ + Hg₂Cl₂ — Sn(II) is oxidised to the more stable Sn(IV).

Answer: Pb prefers +2, Sn prefers +4: the inert pair effect is strong for Pb but weak for Sn.

Allotropes of carbon

Same element, very different solids.

Allotrope	Hybridisation	Structure	Key properties
Diamond	sp³	3D network of tetrahedra; C–C 154 pm	Hardest natural substance; insulator; high refractive index
Graphite	sp²	Hexagonal layers; C–C 141.5 pm in a layer; layers 340 pm apart, held by weak forces	Soft, slippery (lubricant); conducts along layers (delocalised π electrons)
Fullerene C₆₀	sp²	Cage of 20 hexagons and 12 pentagons (“buckyball”)	Molecular; soluble in organic solvents; pure form of carbon
Graphene	sp²	A single graphite layer	Very strong, excellent conductor
Carbon nanotubes	sp²	Graphene rolled into tubes	Very strong, light; electronics

• Graphite is the thermodynamically most stable allotrope (ΔfH° = 0); diamond has ΔfH° = +1.9 kJ mol⁻¹.
• Impure forms (amorphous): coke, charcoal, carbon black — made by heating carbon-containing materials with little air. Activated charcoal adsorbs gases and poisons.

Carbon monoxide and carbon dioxide

Two oxides, very different behaviour.

	CO	CO₂
Laboratory preparation	HCOOH → H₂O + CO (conc. H₂SO₄, 373 K)	CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Industrial	Water gas: C + H₂O → CO + H₂ (473–1273 K); producer gas: 2C + O₂ + 4N₂ → 2CO + 4N₂ (1273 K)	Burning carbon fuels; heating limestone
Nature	Neutral oxide; colourless, odourless, very poisonous	Acidic oxide; CO₂ + H₂O ⇌ H₂CO₃
Key use or role	Reducing agent in metallurgy (Fe₂O₃ + 3CO → 2Fe + 3CO₂); ligand in Ni(CO)₄	Photosynthesis; solid CO₂ (dry ice) as refrigerant; fire extinguishers; greenhouse gas

Key idea: Why CO is so toxic: it binds to the iron of haemoglobin about 300 times more strongly than O₂, forming carboxyhaemoglobin, which cannot carry oxygen. Why CO is a good ligand: it donates a lone pair from carbon (σ) and accepts metal d electrons back into its empty π∗ orbitals (π back-bonding).

• CO has a triple bond (one σ, two π) with a lone pair on each atom; CO₂ is linear (O=C=O, sp carbon) with zero dipole moment.
• The carbonic acid–bicarbonate buffer keeps blood pH between 7.26 and 7.42.

Silica, silicones, silicates and zeolites

Silicon’s chemistry is built on Si–O bonds.

Silicon dioxide, SiO₂

• A 3D network: each Si is bonded tetrahedrally to four O atoms, each O shared by two Si. That is why SiO₂ is a hard, high-melting solid while CO₂ is a gas.
• Crystalline forms: quartz, cristobalite, tridymite. Quartz is piezoelectric (used in clocks, pressure gauges). Amorphous forms: silica gel (drying agent, chromatography), kieselguhr.
• Very unreactive, but attacked by HF and by hot NaOH: SiO₂ + 4HF → SiF₄ + 2H₂O; SiO₂ + 2NaOH → Na₂SiO₃ + H₂O.

Silicones

• Polymers with a –Si–O–Si– backbone and organic groups on Si: general formula (R₂SiO)ₙ.
• Made from alkylchlorosilanes: 2CH₃Cl + Si → (CH₃)₂SiCl₂ (Cu catalyst, 573 K); hydrolysis gives (CH₃)₂Si(OH)₂, which condenses into chains.
• (CH₃)₃SiCl caps the chain (chain terminator); CH₃SiCl₃ gives cross-linked (3D) polymers.
• Water-repellent (outer alkyl groups), heat-resistant, electrical insulators, chemically inert, non-toxic: used in sealants, greases, electrical insulation, water-proofing fabrics and surgical implants.

Silicates and zeolites

Silicate type	Unit	Example
Orthosilicate	Discrete SiO₄⁴⁻	Zircon ZrSiO₄, olivine
Pyrosilicate	Si₂O₇⁶⁻ (two tetrahedra share one O)	Thortveitite
Cyclic / chain	(SiO₃)ₙ²ⁿ⁻ — two O shared per tetrahedron	Beryl (cyclic); pyroxenes (chains)
Sheet	(Si₂O₅)ₙ²ⁿ⁻ — three O shared	Mica, talc, clays
3D	All four O shared (SiO₂); some Si replaced by Al in aluminosilicates	Quartz; feldspar, zeolites

• Zeolites are aluminosilicates with a honeycomb of pores; they act as shape-selective catalysts and ion exchangers. ZSM-5 converts alcohols directly into gasoline; hydrated zeolites soften hard water.

Solved example 06 — Why SiO₂ is a solid but CO₂ is a gas

Explain the difference in physical state between CO₂ and SiO₂.

1. Carbon forms strong pπ–pπ double bonds with oxygen, so CO₂ is a small discrete molecule held to others only by weak van der Waals forces.
2. Silicon is too large for effective pπ–pπ overlap, so it forms four single Si–O bonds instead.
3. Each O bridges two Si atoms, building a giant covalent network.

Answer: CO₂ is molecular (gas); SiO₂ is a covalent network solid (m.p. about 1983 K).

Solved example 07 — Role of chain terminators

What is the effect of adding (CH₃)₃SiCl to the hydrolysis mixture of (CH₃)₂SiCl₂ in silicone manufacture?

1. (CH₃)₂Si(OH)₂ has two –OH groups and links at both ends, growing a chain.
2. (CH₃)₃SiOH has only one –OH, so when it joins a chain end, no further growth is possible at that end.

Answer: It caps the chains and controls the chain length (and hence viscosity) of the silicone.

Check yourself

• C5. Which allotrope of carbon conducts electricity, and why?
• C6. Why does SiCl₄ hydrolyse but CCl₄ does not?

Answers at the end.

Advanced problems

Each needs two ideas from the chapter.

Solved example 08 — Hydrogen from aluminium

What volume of H₂ at STP is released when 5.4 g of Al dissolves completely in excess NaOH solution?

1. 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂: 2 mol Al → 3 mol H₂.
2. 5.4 g Al = 0.20 mol → 0.30 mol H₂.
3. 0.30 × 22.4 L.

Answer: 6.72 L of H₂ at STP.

Solved example 09 — Borax as a primary standard

Write the reaction of borax with HCl and explain why borax is used to standardise acids.

1. Na₂B₄O₇·10H₂O + 2HCl → 2NaCl + 4H₃BO₃ + 5H₂O.
2. Boric acid is too weak to affect methyl orange, so the end point reflects only the HCl used.
3. Borax is a pure, stable, non-hygroscopic crystalline solid of high molar mass (381 g mol⁻¹).

Answer: One mole of borax neutralises exactly two moles of HCl; its purity and stability make it a primary standard.

Solved example 10 — Why BCl₃ is a monomer but AlCl₃ dimerises

BCl₃ exists as a monomer, while AlCl₃ forms Al₂Cl₆. Explain.

1. Boron is small; four large Cl atoms cannot fit comfortably around it, and B–Cl back bonding partly relieves its electron deficiency.
2. Al is larger: it can accommodate four Cl atoms, and it completes its octet by accepting a lone pair from a bridging Cl.

Answer: Size: small B cannot dimerise through bulky Cl bridges; larger Al can.

Solved example 11 — Graphite and diamond: conductivity

Explain why graphite conducts electricity while diamond does not, though both are pure carbon.

1. Diamond: every carbon is sp³ and uses all four electrons in localised σ bonds — no mobile electrons.
2. Graphite: each carbon is sp², bonded to three others; the fourth electron is in a p orbital delocalised over the whole layer.

Answer: Delocalised π electrons in graphite carry current along the layers; diamond has none.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	BF₃ is the strongest Lewis acid among the boron halides (F is most electronegative).	BF₃ is the weakest: BF₃ < BCl₃ < BBr₃ < BI₃.	pπ–pπ back bonding partly fills boron’s empty p orbital in BF₃.
2	Boric acid is a tribasic protonic acid.	It is a monobasic Lewis acid.	It accepts OH⁻ from water to give [B(OH)₄]⁻ + H⁺.
3	Diborane has 8 two-centre two-electron bonds.	4 two-centre bonds and 2 three-centre two-electron bonds.	Only 12 valence electrons are available.
4	Al is smaller than Ga.	Ga (135 pm) is smaller than Al (143 pm).	Poor shielding by 3d electrons in Ga.
5	Tl³⁺ is more stable than Tl⁺.	Tl⁺ is more stable (inert pair effect).	Tl³⁺ is therefore an oxidising agent.
6	Pb⁴⁺ compounds are more stable than Pb²⁺.	Pb²⁺ is more stable; PbO₂ is a strong oxidant; PbI₄ does not exist.	Inert pair effect.
7	CCl₄ hydrolyses like SiCl₄.	CCl₄ does not hydrolyse.	Carbon has no vacant d orbitals.
8	Diamond is the most stable allotrope of carbon.	Graphite is the most stable (ΔfH° = 0).	Diamond is metastable at room conditions.
9	Diamond conducts because it is carbon like graphite.	Diamond is an insulator.	All four electrons are in localised σ bonds (sp³).
10	CO is an acidic oxide.	CO is neutral.	It does not react with water or bases under ordinary conditions.
11	(CH₃)₃SiCl gives cross-linked silicones.	It is a chain terminator; CH₃SiCl₃ gives cross-links.	One –OH caps; three –OH groups branch.
12	Aqueous borax is acidic because it contains boric acid units.	Aqueous borax is alkaline.	It hydrolyses to NaOH (strong base) + H₃BO₃ (weak acid).

Memory hook: “F back-feeds boron” — BF₃ is the weakest boron-halide Lewis acid. · “Twelve electrons, two bananas” — diborane. · “Lazy pair at the bottom” — Tl⁺, Pb²⁺. · “Si says yes to water, C says no” — tetrachloride hydrolysis.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The most stable oxidation state of thallium is: (Easy)

• (a) +1
• (b) +2
• (c) +3
• (d) +4

Q02. Boric acid is: (Easy)

• (a) a tribasic acid
• (b) a monobasic Lewis acid
• (c) a strong acid
• (d) a base

Q03. Colour of the cobalt borax bead: (Easy)

• (a) green
• (b) blue
• (c) brown
• (d) red

Q04. The hardest natural substance is: (Easy)

• (a) graphite
• (b) diamond
• (c) fullerene
• (d) quartz

Q05. CO is: (Easy)

• (a) acidic
• (b) basic
• (c) neutral
• (d) amphoteric

Q06. Which element does not show catenation? (Easy)

• (a) C
• (b) Si
• (c) Ge
• (d) Pb

Q07. Formula of borax: (Easy)

• (a) Na₂B₄O₇·10H₂O
• (b) Na₂B₄O₇·5H₂O
• (c) NaBO₂
• (d) H₃BO₃

Q08. Correct order of Lewis acid strength: (Moderate)

• (a) BF₃ > BCl₃ > BBr₃
• (b) BBr₃ > BCl₃ > BF₃
• (c) BCl₃ > BF₃ > BBr₃
• (d) all equal

Q09. Number of 3c–2e bonds in diborane: (Moderate)

• (a) 1
• (b) 2
• (c) 4
• (d) 6

Q10. AlCl₃ exists as a dimer because: (Moderate)

• (a) Al has d orbitals
• (b) Al completes its octet using a lone pair of bridging Cl
• (c) it is ionic
• (d) of hydrogen bonding

Q11. Which is amphoteric? (Moderate)

• (a) B₂O₃
• (b) Al₂O₃
• (c) Tl₂O₃
• (d) CO₂

Q12. Ga has a smaller atomic radius than Al because of: (Moderate)

• (a) inert pair effect
• (b) poor shielding by 3d electrons
• (c) lanthanoid contraction
• (d) higher electronegativity only

Q13. Diborane reacts with NH₃ on heating to give: (Moderate)

• (a) B₂O₃
• (b) borazine
• (c) NaBH₄
• (d) boric acid

Q14. Hybridisation of carbon in graphite: (Moderate)

• (a) sp
• (b) sp²
• (c) sp³
• (d) sp³d

Q15. Fullerene C₆₀ contains: (Moderate)

• (a) 20 pentagons and 12 hexagons
• (b) 12 pentagons and 20 hexagons
• (c) 30 hexagons
• (d) 60 pentagons

Q16. Which hydrolyses readily? (Moderate)

• (a) CCl₄
• (b) SiCl₄
• (c) both
• (d) neither

Q17. Producer gas is a mixture of: (Moderate)

• (a) CO + H₂
• (b) CO + N₂
• (c) CO₂ + H₂
• (d) CH₄ + H₂

Q18. Which does not exist? (Moderate)

• (a) PbCl₄
• (b) PbF₄
• (c) PbI₄
• (d) SnCl₄

Q19. ZSM-5 is used to: (Moderate)

• (a) soften water only
• (b) convert alcohols to gasoline
• (c) make silicones
• (d) purify CO

Q20. Terminal and bridging B–H bond lengths in B₂H₆ are: (Tough)

• (a) 134 and 119 pm
• (b) 119 and 134 pm
• (c) both 119 pm
• (d) both 134 pm

Q21. Aqueous solution of borax is: (Tough)

• (a) acidic
• (b) neutral
• (c) alkaline
• (d) amphoteric

Q22. Cross-linked silicones are obtained by hydrolysis of: (Tough)

• (a) (CH₃)₃SiCl
• (b) (CH₃)₂SiCl₂
• (c) CH₃SiCl₃
• (d) SiCl₄ only

Q23. In the silicate Si₂O₇⁶⁻, the number of oxygen atoms shared is: (Tough)

• (a) 0
• (b) 1
• (c) 2
• (d) 3

Q24. CO is a good ligand because it is: (Tough)

• (a) a σ-donor only
• (b) a π-acceptor only
• (c) both a σ-donor and π-acceptor
• (d) ionic

Q25. Which is the strongest reducing agent? (Tough)

• (a) SnCl₂
• (b) PbCl₂
• (c) CCl₄
• (d) SiCl₄

Answer key with explanations

1. (a) +1 — Inert pair effect.
2. (b) a monobasic Lewis acid — Accepts OH⁻.
3. (b) blue — Co(BO₂)₂.
4. (b) diamond — sp³ network.
5. (c) neutral — Neutral oxide.
6. (d) Pb — Weak Pb–Pb bond.
7. (a) Na₂B₄O₇·10H₂O — Decahydrate.
8. (b) BBr₃ > BCl₃ > BF₃ — Back bonding strongest in BF₃.
9. (b) 2 — Two B–H–B bridges.
10. (b) Al completes its octet using a lone pair of bridging Cl — Electron-deficient Al.
11. (b) Al₂O₃ — Reacts with acid and base.
12. (b) poor shielding by 3d electrons — d-block contraction.
13. (b) borazine — “Inorganic benzene”.
14. (b) sp² — Planar layers.
15. (b) 12 pentagons and 20 hexagons — Soccer-ball cage.
16. (b) SiCl₄ — Vacant 3d orbitals on Si.
17. (b) CO + N₂ — Air used instead of steam.
18. (c) PbI₄ — Pb⁴⁺ oxidises I⁻.
19. (b) convert alcohols to gasoline — Shape-selective zeolite catalyst.
20. (b) 119 and 134 pm — Bridging bonds are longer.
21. (c) alkaline — NaOH + weak H₃BO₃.
22. (c) CH₃SiCl₃ — Three –OH groups.
23. (b) 1 — Pyrosilicate.
24. (c) both a σ-donor and π-acceptor — Synergic bonding.
25. (a) SnCl₂ — Sn²⁺ → Sn⁴⁺ readily.

Check-yourself answers

• C1. Each Al has only six electrons in AlCl₃; a lone pair from a Cl of the other molecule completes its octet, so two units bridge through Cl.
• C2. Tl⁺ is the more stable state (inert pair effect), so Tl³⁺ readily gains two electrons.
• C3. Blue.
• C4. Two.
• C5. Graphite: each carbon uses three electrons in σ bonds; the fourth is delocalised over the layer and carries current.
• C6. Silicon has vacant 3d orbitals that accept a lone pair from water; carbon has none.

One-page revision

Read this page the night before the exam.

Group 13 trends

• ns²np¹; +3, and +1 down (Tl⁺ stable)
• Ga (135) < Al (143) — d contraction
• Ga m.p. 303 K, b.p. 2676 K
• B₂O₃ acidic, Al₂O₃ amphoteric, Tl₂O₃ basic

Boron compounds

• BF₃ < BCl₃ < BBr₃ (back bonding)
• Borax Na₂[B₄O₅(OH)₄]·8H₂O; alkaline
• Co bead blue; H₃BO₃ monobasic Lewis acid
• B₂H₆: 4 terminal + 2 banana bonds

Diborane and Al

• 2NaBH₄ + I₂ → B₂H₆
• B₂H₆ + NH₃ → borazine; + NaH → NaBH₄
• Al dissolves in HCl and NaOH; passive to HNO₃
• AlCl₃ dimer Al₂Cl₆

Group 14 trends

• ns²np²; +4 and +2 (Pb²⁺ stable)
• Catenation C ≫ Si > Ge ≈ Sn; Pb none
• SiCl₄ hydrolyses, CCl₄ does not
• PbI₄ does not exist

Carbon

• Diamond sp³, graphite sp² (most stable)
• C₆₀: 12 pentagons, 20 hexagons
• CO neutral, toxic; σ-donor π-acceptor
• CO₂ acidic, linear; dry ice

Silicon

• SiO₂ 3D network; attacked by HF, NaOH
• Silicones (R₂SiO)ₙ; R₃SiCl caps; RSiCl₃ cross-links
• Silicates: ortho, pyro, chain, sheet, 3D
• Zeolites; ZSM-5 alcohol → gasoline