p-Block Elements Class 11 Notes

• Groups 13 B, 14 C, 15 N, 16 O, 17 F (halogens), 18 noble gases.
• The group shows the full spread: metals (Al), metalloids (Si, Ge), and non-metals (C, N, O).
• Oxidation states run from +3 to −3; the highest oxidation state equals the group number (sum of s + p electrons).

• Atomic and ionic radii increase down the group, but increase less than in s-block because more electrons share the same shell.
• Ionisation enthalpy is high for the lightest member, then tends to fall; Group 15 has an anomalous upward jump due to half-filled orbitals.
• Metallic character increases down each group — B is a metalloid, Al and Tl are metals.
• Electronegativity falls down the group but not smoothly; the heavy elements show the inert pair effect.
• Noble gases have maximum ionisation enthalpy, making them the least reactive.

• The ns² pair is held tight by the poor shielding of the intervening d and f electrons, so it is not easily lost.
• PbCl₂ and TlCl are stable because the lower oxidation state is preferred.
• The +1 oxidation state of Tl, +2 of Pb and Sn is more stable than the group number oxidation state.

• B, Al, Ga, In, Tl; general configuration ns²np¹ and oxidation states +3 and +1.
• Boron is a metalloid with a very high melting point; Al onwards are soft metals with lower melting points.
• Boron has a unique ability to form electron-deficient compounds and stable covalent bonds (3-centre-2-electron bonding in diborane).
• Boric acid H₃BO₃ is a weak monobasic acid that acts as a Lewis acid by accepting OH⁻ from water; it gives a green flame test with methanol.
• Borax Na₂B₄O₇·10H₂O is the common sodium borate — the classic bead test gives coloured beads with transition metals.

• In the gas phase diborane burns with a green flame to give boric oxide and water.
• It adds across alkenes — hydroboration — which is a cornerstone reaction in organic chemistry.
• Boron's diagonal relationship with silicon shows in the acid character of B(OH)₃ and Si(OH)₄ and their reaction with NaOH.

• C, Si, Ge, Sn, Pb; configuration ns²np² with oxidation states +4 and +2.
• Carbon is non-metallic, silicon and germanium are metalloids, tin and lead are metals.
• Carbon bonds catenate — forms long chains — giving the vast chemistry of organic compounds.
• Allotropic forms: diamond (network covalent, hardest), graphite (layered, conducts electricity) and fullerenes (molecular cages); graphite is the thermodynamically stable form.
• Inert pair effect again: the stability of +2 outweighs +4 for Sn and Pb.

Example: Why is PbCl₂ stable but PbCl₄ unstable?

Solution: The inert pair effect makes the ns² electrons of lead reluctant to bond, so the +2 state is preferred. Pb⁴⁺ is strongly oxidising and PbCl₄ tends to lose chlorine, whereas PbCl₂ is covalent and stable.

Example: Which factor decides metallic character within the p-block?

Solution: Metallic character increases down a group because the ionisation enthalpy falls and the atomic size increases, so electrons are lost more easily. Metals give way to metalloids and then non-metals as you move right across the p-block.

Exercises (NCERT Q11.1–Q11.38)

• 11.1) Discuss the pattern of variation in the oxidation states of (i) B to Tl and (ii) C to Pb.
• 11.2) How can you explain higher stability of BCl₃ as compared to TlCl₃?
• 11.3) Why does boron triflouride behave as a Lewis acid?
• 11.4) Consider the compounds, BCl₃ and CCl₄. How will they behave with water? Justify.
• 11.5) Is boric acid a protic acid? Explain.
• 11.6) Explain what happens when boric acid is heated.
• 11.7) Describe the shapes of BF₃ and BH₄⁻. Assign the hybridisation of boron in these species.
• 11.8) Write reactions to justify amphoteric nature of aluminium.
• 11.9) What are electron deficient compounds? Are BCl₃ and SiCl₄ electron deficient species? Explain.
• 11.10) Write the resonance structures of CO₃²⁻ and HCO₃⁻.
• 11.11) What is the state of hybridisation of carbon in (a) CO₃²⁻ (b) diamond (c) graphite?
• 11.12) Explain the difference in properties of diamond and graphite on the basis of their structures.
• 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.
• 11.14) Suggest reasons why the B–F bond lengths in BF₃ (130 pm) and BF₄⁻ (143 pm) differ.
• 11.15) If B–Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.
• 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.
• 11.17) Suggest a reason as to why CO is poisonous.
• 11.18) How is excessive content of CO₂ responsible for global warming?
• 11.19) Explain structures of diborane and boric acid.
• 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?
• 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.
• 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.
• 11.23) Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
• 11.24) How would you explain the lower atomic radius of Ga as compared to Al?
• 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?
• 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.
• 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.
• 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.
• 11.29) What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?
• 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.
• 11.31) Write balanced equations for: (i) BF₃ + LiH → (ii) B₂H₆ + H₂O → (iii) NaH + B₂H₆ → (iv) H₃BO₃ → (Δ) (v) Al + NaOH → (vi) B₂H₆ + NH₃ →
• 11.32) Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.
• 11.33) An aqueous solution of borax is (a) neutral (b) amphoteric (c) basic (d) acidic
• 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
• 11.35) The type of hybridisation of boron in diborane is (a) sp (b) sp² (c) sp³ (d) dsp²
• 11.36) Thermodynamically the most stable form of carbon is (a) diamond (b) graphite (c) fullerenes (d) coal
• 11.37) 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
• 11.38) If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.

Answers at a glance

1. 1
   Q11.1 → (i) Group 13: +3 is the common state, but its stability falls from B to Tl while +1 (inert pair effect) becomes more stable — Tl is mainly +1. (ii) Group 14: +4 dominates for C, Si, Ge but grows less stable to Pb, while +2 becomes increasingly stable; Pb is chiefly +2.
2. 2
   Q11.2 → The inert pair effect makes Tl⁺ the stable state, so TlCl₃ is a strong oxidising agent and tends to lose a chlorine to give TlCl; boron has no inert pair effect and is stable in the +3 state as BCl₃.
3. 3
   Q11.3 → Boron in BF₃ has only six electrons around it (electron deficient — one vacant p orbital), so it readily accepts a lone pair from a Lewis base — exactly the definition of a Lewis acid.
4. 4
   Q11.4 → BCl₃ hydrolyses readily (boron is electron deficient and water attacks it): BCl₃ + 3H₂O → B(OH)₃ + 3HCl. CCl₄ does not hydrolyse — carbon is saturated (octet complete) with no vacant orbital to accept water's electrons.
5. 5
   Q11.5 → No. Boric acid does not donate a proton directly; it behaves as a Lewis acid, accepting OH⁻ from water: B(OH)₃ + 2HOH ⇌ [B(OH)₄]⁻ + H₃O⁺.
6. 6
   Q11.6 → On heating it loses water stepwise: B(OH)₃ → HBO₂ (metaboric acid) → H₂B₄O₇ (tetraboric acid) → B₂O₃ (boric oxide, obtained on strong/red heating).
7. 7
   Q11.7 → BF₃ is trigonal planar (B is sp² hybridised); BH₄⁻ is tetrahedral (B is sp³ hybridised).
8. 8
   Q11.8 → With acid: 2Al + 6HCl → 2AlCl₃ + 3H₂. With base: 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂ (or Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O).
9. 9
   Q11.9 → Compounds in which the central atom has fewer than eight valence electrons. BCl₃ is electron deficient (B has only 6 electrons, a vacant p orbital); SiCl₄ is not — silicon's octet is complete.
10. 10
    Q11.10 → CO₃²⁻ has three equivalent resonance forms in which the C=O double bond is delocalised over each of the three oxygens (each O carries −2/3 charge); HCO₃⁻ similarly has three resonance structures, the double bond moving among the three oxygens (with one −OH).
11. 11
    Q11.11 → (a) CO₃²⁻ — sp² (planar). (b) diamond — sp³ (tetrahedral 3-D network). (c) graphite — sp² (planar hexagonal sheets).
12. 12
    Q11.12 → Diamond is a rigid 3-D network of sp³ carbon atoms — the hardest natural substance, a poor conductor, transparent, high density (3.53 g/cm³). Graphite has flat sp² hexagonal layers held weakly by van der Waals forces, so it is soft (layers slide, hence a lubricant), a good conductor, opaque and lighter (2.25 g/cm³).
13. 13
    Q11.13 → PbCl₂ + Cl₂ → PbCl₄ (Pb²⁺ is oxidised to Pb⁴⁺ by Cl₂). PbCl₄ →Δ PbCl₂ + Cl₂ (Pb⁴⁺ is strongly oxidising, so it reverts easily). PbI₄ is not formed because I⁻ is a reducing agent that would immediately reduce Pb⁴⁺ back — PbI₄ → PbI₂ + I₂.
14. 14
    Q11.14 → In planar BF₃, boron's vacant p orbital participates in pπ–pπ back bonding, giving the B–F bonds partial double-bond character (shorter, 130 pm); in tetrahedral BF₄⁻, boron is saturated — no vacant p orbital, no π bonding — so all bonds are normal single bonds (longer, 143 pm).
15. 15
    Q11.15 → BCl₃ is trigonal planar with the three equal B–Cl dipoles set at 120°; they are symmetrically arranged, so their vectors cancel exactly, giving the molecule a zero net dipole moment.
16. 16
    Q11.16 → Anhydrous HF is a weak acid with no free F⁻ ions, so AlF₃ cannot dissolve. NaF supplies F⁻, which complexes Al³⁺: AlF₃ + 3NaF → Na₃AlF₆ (soluble cryolite). Bubbling BF₃ (a stronger Lewis acid than Al³⁺) pulls the fluoride back: Na₃AlF₆ + BF₃ → 3NaBF₄ + AlF₃↓, precipitating AlF₃.
17. 17
    Q11.17 → CO binds to haemoglobin about 300 times more strongly than O₂, blocking the transport of oxygen in the blood and causing oxygen starvation (anoxia) — even small amounts prove fatal.
18. 18
    Q11.18 → CO₂ is a greenhouse gas — it lets sunlight in but traps outgoing infrared (heat) radiation from the Earth, raising the average temperature (global warming) as its atmospheric level climbs.
19. 19
    Q11.19 → Diborane B₂H₆: each boron is sp³, held by four terminal 2-centre–2-electron B–H bonds and two bridging hydrogen atoms in 3-centre–2-electron B–H–B (banana) bonds, so the molecule is like two BH₂ units joined through two H bridges. Boric acid: planar B(OH)₃ molecules linked by strong O–H···O hydrogen bonds into a layered polymeric (sheet) structure.
20. 20
    Q11.20 → (a) Na₂B₄O₇·10H₂O →Δ Na₂B₄O₇ → 2NaBO₂ + B₂O₃ (first swells, then gives a clear glassy bead of metaborate + boric oxide). (b) Boric acid ionises by accepting OH⁻: B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺ (mildly acidic). (c) 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂. (d) BF₃ + NH₃ → BF₃·NH₃ (a donor–acceptor adduct).
21. 21
    Q11.21 → (a) Si + CH₃Cl →(Cu, Δ) organosilicon chlorides (e.g. CH₃SiCl₃ etc.), the monomers of silicones. (b) SiO₂ + 4HF → SiF₄ + 2H₂O; the SiF₄ dissolves in more HF: SiF₄ + 2HF → H₂SiF₆. (c) ZnO + CO → Zn + CO₂ (CO reduces the oxide). (d) Al₂O₃·xH₂O + 2NaOH → 2NaAlO₂ + (x + 1)H₂O (dissolves as sodium meta-aluminate).
22. 22
    Q11.22 → (i) Conc. HNO₃ passivates aluminium — it forms a protective oxide film that stops further attack. (ii) NaOH + Al gives hydrogen and heat (and aluminate ions) that help break up clogging matter in drains. (iii) Graphite's hexagonal layers slide over one another easily. (iv) Diamond is the hardest known natural substance. (v) Al alloys are light yet strong — a high strength-to-weight ratio. (vi) Aluminium reacts with (impure) water and corrodes if left immersed overnight. (vii) Aluminium is light and a good conductor of electricity, so it is used for overhead transmission cables.
23. 23
    Q11.23 → Silicon's outer electron lies in a fresh principal shell (3p), much farther from the nucleus and well shielded by inner electrons, so it is far easier to remove than carbon's 2p electron — a large drop in ionisation enthalpy.
24. 24
    Q11.24 → The intervening 3d¹⁰ electrons of gallium shield the valence electrons very poorly (d-block contraction), so the higher effective nuclear charge pulls the outer electrons closer — making Ga smaller than Al.
25. 25
    Q11.25 → Allotropes are different structural forms of the same element with different physical properties. Diamond: 3-D network of sp³ carbons → hardest, insulating. Graphite: planar sp² hexagonal layers → soft (slippery), conducting. Structure dictates that diamond is an abrasive while graphite is a lubricant/conductor.
26. 26
    Q11.26 → (a) CO — neutral; B₂O₃, SiO₂, CO₂ — acidic; Al₂O₃, PbO₂ — amphoteric; Tl₂O₃ — basic. (b) B₂O₃ + 3H₂O → 2B(OH)₃; SiO₂ + 2NaOH → Na₂SiO₃ + H₂O; CO₂ + H₂O → H₂CO₃; Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O and Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O; PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O; Tl₂O₃ + 6HCl → 2TlCl₃ + 3H₂O.
27. 27
    Q11.27 → In the +3 state Tl resembles Al: Tl(OH)₃, like Al(OH)₃, is amphoteric and dissolves in both acid and alkali. In the +1 state it resembles group 1: monovalent Tl₂O and TlOH are strong bases (like KOH), and Tl⁺ salts behave like alkali metal salts, e.g. TlCl/Tl₂CO₃ are water soluble and Tl forms an alum.
28. 28
    Q11.28 → X = Al; A = Al(OH)₃ (white ppt); B = Na[Al(OH)₄]; C = AlCl₃; D = Al₂O₃. 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂; Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O; 2Al(OH)₃ →Δ Al₂O₃ + 3H₂O (Al₂O₃ is electrolysed to extract aluminium).
29. 29
    Q11.29 → (a) The tendency of the ns² pair not to take part in bonding, so the lower oxidation state becomes more stable down a group (e.g. Tl⁺, Pb²⁺). (b) Existence of the same element in more than one structural form (e.g. carbon as diamond/graphite). (c) Self-linking of atoms of an element into chains and rings, most pronounced in carbon (silicon shows it weakly).
30. 30
    Q11.30 → X = borax Na₂B₄O₇·10H₂O; Y = a glassy bead (NaBO₂ + B₂O₃); Z = boric acid H₃BO₃. (i) Borate ion hydrolyses to give alkaline solution. (ii) Na₂B₄O₇·10H₂O →Δ Na₂B₄O₇ → 2NaBO₂ + B₂O₃. (iii) Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃.
31. 31
    Q11.31 → (i) 2BF₃ + 6LiH → B₂H₆ + 6LiF (ii) B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂ (iii) B₂H₆ + 2NaH → 2NaBH₄ (iv) H₃BO₃ →Δ HBO₂ →Δ B₂O₃ (via H₂B₄O₇) (v) Al + NaOH + 2H₂O → Na[Al(OH)₄] + (3/2)H₂ (vi) B₂H₆ + 2NH₃ → 2BH₃·NH₃ (on heating → borazine B₃N₃H₆).
32. 32
    Q11.32 → CO — industrial: incomplete combustion/water gas (C + H₂O → CO + H₂); laboratory: dehydration of formic acid, HCOOH →(conc. H₂SO₄, Δ) CO + H₂O. CO₂ — industrial: lime kiln, CaCO₃ →Δ CaO + CO₂; laboratory: action of dilute HCl on marble, CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.
33. 33
    Q11.33 → (c) basic — the tetraborate ion hydrolyses to give OH⁻.
34. 34
    Q11.34 → (b) the presence of hydrogen bonds — O–H···O bridges link planar B(OH)₃ units into sheets.
35. 35
    Q11.35 → (c) sp³ — each boron in B₂H₆ is sp³ hybridised, using two orbitals for terminal bonds and two for the bridge bonds.
36. 36
    Q11.36 → (b) graphite — it is the standard state of carbon (ΔH_f° = 0).
37. 37
    Q11.37 → (b) they exhibit oxidation states of +2 and +4 — the +2 state becoming more stable down the group by the inert pair effect.
38. 38
    Q11.38 → Hydrolysis of RSiCl₃ and condensation gives a cross-linked three-dimensional silicone polymer of the type (RSiO₃/₂)ₙ, in which each silicon is connected to three oxygen atoms (an R–SiO₃/₂ network).

MCQ Questions

1 Mark Questions

1. 1
   What is the inert pair effect? (Ans: the tendency of the ns² pair to remain uninvolved in bonding, making the lower oxidation state more stable down a group, e.g. Tl⁺, Pb²⁺)
2. 2
   Name the allotropes of carbon. (Ans: diamond, graphite, fullerenes)
3. 3
   Why is BF₃ a Lewis acid? (Ans: boron is electron-deficient with a vacant p orbital, so it accepts a lone pair of electrons)
4. 4
   What is catenation? (Ans: self-linking of atoms of an element into chains or rings, most shown by carbon)

2 Mark Questions

1. 1
   How do BCl₃ and CCl₄ behave towards water? (Ans: BCl₃ hydrolyses to B(OH)₃ and HCl; CCl₄ does not hydrolyse as carbon has no vacant orbital)
2. 2
   Explain why the B–Cl bond dipoles in BCl₃ cancel out. (Ans: BCl₃ is trigonal planar with the three equal bond dipoles at 120°, so the vector sum is zero)
3. 3
   Why is CO poisonous? (Ans: it binds to haemoglobin ~300 times more strongly than O₂ and blocks oxygen transport)
4. 4
   Compare the structures of diamond and graphite. (Ans: diamond is a 3-D sp³ network (hard, insulator); graphite is planar sp² sheets (soft, conductor))

3 Mark Questions

1. 1
   Explain the structure of diborane, including the 3-centre–2-electron B–H–B bridge bonds. (Ans: two sp³ B atoms, four terminal B–H bonds, two bridging hydrogens in banana bonds)
2. 2
   Classify the following oxides as neutral, acidic, basic or amphoteric, with equations: CO, B₂O₃, SiO₂, CO₂, Al₂O₃, PbO₂, Tl₂O₃. (Ans: CO neutral; B₂O₃/SiO₂/CO₂ acidic; Al₂O₃/PbO₂ amphoteric; Tl₂O₃ basic)
3. 3
   Explain the inert pair effect with the stability order of +3 oxidation state in group 13 and +4/+2 in group 14. (Ans: lower states become more stable; Tl mainly +1 and Pb mainly +2)
4. 4
   Describe what happens when boric acid is heated, and give the equations. (Ans: B(OH)₃ → HBO₂ → H₂B₄O₇ → B₂O₃ on successive heating)

5 Mark Questions

1. 1
   Rationalise the higher stability of BCl₃ over TlCl₃, the instability of PbCl₄ on heating, and why PbI₄ is not formed, with balanced equations. (Ans: inert pair effect; PbCl₄ → PbCl₂ + Cl₂; I⁻ reduces Pb⁴⁺)
2. 2
   A salt X gives an alkaline aqueous solution, swells to a glassy material Y on strong heating, and with hot conc. H₂SO₄ gives crystals of an acid Z. Identify X, Y and Z with equations. (Ans: X = borax Na₂B₄O₇·10H₂O; Y = NaBO₂ + B₂O₃ glass; Z = H₃BO₃ borax + H₂SO₄)
3. 3
   With the help of structures, explain why diamond is the hardest and graphite the softest form of carbon, and how the structure accounts for their electrical conductivity, density and use as an abrasive/lubricant.
4. 4
   Describe the chemistry of borax and boric acid — the action of heat, their monomeric/polymeric structures, and the reactions of diborane with H₂O, NaH and NH₃ — with balanced equations.

Assignments walk the chapter's topics in order; the tests follow the current CBSE blueprint (MCQ/VSA, 2M, 3M and 5M questions) with answer keys. They will appear here as direct download links.