JEE Mains Chemistry · The p-Block Elements

Group 14: Carbon, Silicon, Tin and Lead

Carbon, silicon, germanium, tin and lead (ns² np²): carbon alone forms strong multiple bonds and long chains, silicon and the heavier elements use d orbitals, and the inert pair makes Sn²⁺ a reducing agent and Pb⁴⁺ an oxidising agent.

All The p-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Sixteen PYQs, all multiple choice, and two from 2026. Six test carbon, silicon and the group trends: the structure of C₆₀, silicones, which hexahalide ions exist and the nature of the oxides; four ask which tin and lead ions oxidise and which reduce; six are the lead tests of salt analysis.

Concept 1 of 3: Group 14 trends, carbon's allotropes and silicones

Carbon is small, has no d orbitals and forms strong pπ–pπ multiple bonds with itself. Those two gifts, catenation and multiple bonding, give it allotropes such as graphite and the fullerenes. From silicon down, the atoms are too big for pπ–pπ bonds but have d orbitals, so they can take more than four groups: silicon forms

\mathrm{[SiF_6]^{2-}}

. Silicon's chemistry also gives silicones, chains and sheets of alternating Si and O, and the number of OH groups on the starting silicon decides how far the chain can grow.

Definition

Covalent radius rises from C to Pb; electronegativity falls from C (2.5) to Si, then stays near 1.8–1.9.
First ionisation enthalpy of each group 14 element is higher than that of the group 13 element in the same period.
Maximum covalency of carbon is 4; Si, Ge, Sn and Pb can exceed 4 using d orbitals. Heavier elements do not form $p\pi$ – $p\pi$ bonds. Carbon also shows negative oxidation states.
Hexahalide ions: $\mathrm{[SiF_6]^{2-}}$ , $\mathrm{[GeCl_6]^{2-}}$ and $\mathrm{[Sn(OH)_6]^{2-}}$ exist; $\mathrm{[SiCl_6]^{2-}}$ does not, because six large chlorides cannot fit round small silicon.
Oxides: $\mathrm{CO_2}$ , $\mathrm{SiO_2}$ and $\mathrm{GeO_2}$ acidic; $\mathrm{SnO}$ , $\mathrm{SnO_2}$ , $\mathrm{PbO}$ and $\mathrm{PbO_2}$ amphoteric.
$\mathrm{C_{60}}$ : 20 six-membered and 12 five-membered rings; every carbon $sp^2$ with three σ bonds; a five-membered ring is fused only to six-membered rings.
Silicones by number of OH on Si: $\mathrm{R_3SiOH}$ gives a dimer, $\mathrm{R_2Si(OH)_2}$ a linear chain, $\mathrm{RSi(OH)_3}$ a cross-linked (2D) silicone; $\mathrm{R_4Si}$ has no OH and stays a silane.

Element	Covalent radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity	What sets it apart
C	77	1086	2.5	Catenation and $p\pi$ – $p\pi$ bonds; maximum covalency 4; allotropes
Si	118	786	1.8	Uses d orbitals: $\mathrm{[SiF_6]^{2-}}$ exists; $\mathrm{SiO_2}$ is acidic; forms silicones
Ge	122	761	1.8	$\mathrm{GeO_2}$ acidic; $\mathrm{[GeCl_6]^{2-}}$ exists
Sn	140	708	1.8	+4 more stable than +2; oxides amphoteric
Pb	146	715	1.9	+2 more stable than +4; oxides amphoteric Lead's ionisation enthalpy is a little HIGHER than tin's: poor shielding by 4f and 5d electrons.

The ionisation enthalpy falls from C to Sn, then rises slightly at Pb.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 June 2022 · Q39Moderate

Example 1 · The p-Block Elements · Group 14: Carbon, Silicon, Tin and Lead

Match List-I with List-II List-I List-II (Si-Compounds) (Si-Polymeric/other products) (a)

\left( CH_{3} \right)_{4}Si

(I) Chain silicone (b)

\left( CH_{3} \right)Si(OH)_{3}

(II) Dimeric silicone (c)

\left( CH_{3} \right)_{2}Si(OH)_{2}

(III) Silane (d)

\left( CH_{3} \right)_{3}Si(OH)

(IV)

2D

- Silicone Choose the correct answer from the options given below:

Carbon's allotropy comes from pπ–pπ bonds, not pπ–dπ

Carbon has no d orbitals. Its allotropes arise from catenation and its ability to form

p\pi

–

p\pi

multiple bonds with itself.

C₆₀ has 20 six-membered and 12 five-membered rings

It is easy to swap the numbers. Buckminsterfullerene has twelve pentagons, each surrounded only by hexagons, and twenty hexagons.

Covalent radius increases down group 14

From C to Pb each element adds a shell, so the covalent radius increases. A statement that it decreases in a regular manner is false.

Drill 5 more on group 14 trends, carbon's allotropes and silicones

Concept 2 of 3: Inert pair effect in tin and lead: which ions oxidise and which reduce

Tin prefers +4, so

\mathrm{Sn^{2+}}

readily gives up two electrons and is a reducing agent. Lead prefers +2 because its 6s pair is held tightly (the inert pair effect), so

\mathrm{Pb^{4+}}

readily takes two electrons back and is a strong oxidising agent. The rule works across groups: the ion that is NOT in the element's preferred state is the reactive one.

Definition

Tin: +4 more stable, so $\mathrm{Sn^{2+}}$ is reducing. $\mathrm{SnCl_2}$ reduces $\mathrm{HgCl_2}$ .
Lead: +2 more stable, so $\mathrm{Pb^{4+}}$ is strongly oxidising; $\mathrm{E^\circ(Pb^{4+}/Pb^{2+}) = +1.67\ V}$ .
$\mathrm{PbO_2}$ is a strong oxidising agent, is amphoteric and is the cathode material of the lead storage battery; it oxidises HCl to chlorine: $\mathrm{PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O}$ .
Across groups 13 and 14, the oxidising ions are the ones above the preferred state: $\mathrm{Tl^{3+}}$ and $\mathrm{Pb^{4+}}$ .

Ion	Preferred state of the element	Behaves as	Evidence
$\mathrm{Sn^{2+}}$	+4	Reducing agent	$\mathrm{E^\circ(Sn^{4+}/Sn^{2+}) = +0.15\ V}$ : $\mathrm{Sn^{2+}}$ is easily oxidised
$\mathrm{Sn^{4+}}$	+4	Stable; a very weak oxidant at most	Same small potential, $+0.15\ V$
$\mathrm{Pb^{2+}}$	+2	Stable	The 6s pair stays out of bonding
$\mathrm{Pb^{4+}}$	+2	Strong oxidising agent	$\mathrm{E^\circ(Pb^{4+}/Pb^{2+}) = +1.67\ V}$ , the most positive here The strongest oxidant among these p-block ions.
$\mathrm{Tl^{3+}}$	+1	Strong oxidising agent	$\mathrm{Tl^{3+}}$ reduced to $\mathrm{Tl^{+}}$ : $+1.26\ V$
$\mathrm{Tl^{+}}$	+1	Stable	The 6s pair stays out of bonding

The more positive the reduction potential, the stronger the oxidising agent.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 7 Apr 2025 · Q38Moderate

Example 2 · The p-Block Elements · Group 14: Carbon, Silicon, Tin and Lead

The group 14 elements A and B have the first ionisation enthalpy values of 708 and

715\text{ }kJ{\text{ }mol}^{- 1}

respectively. The above values are lowest among their group members. The nature of their ions

A^{2 +}

B^{4 +}

respectively is.

The lower state is not always the reducing one

\mathrm{Sn^{2+}}

is a reducing agent but

\mathrm{Pb^{2+}}

is not. What matters is each element's preferred state: +4 for tin, +2 for lead.

Pb⁴⁺ is not stable like Sn⁴⁺

Both are +4, but lead's 6s pair resists bonding, so

\mathrm{Pb^{4+}}

is a strong oxidising agent while

\mathrm{Sn^{4+}}

is stable.

Drill 3 more on inert pair effect in tin and lead: which ions oxidise and which reduce

Concept 3 of 3: Tests for the lead ion in salt analysis

Lead salts are identified by a chain of precipitates. Lead chloride is only slightly soluble in cold water but dissolves in hot water, which separates it from silver chloride. The coloured tests confirm it: yellow lead chromate, yellow lead iodide and white lead sulphate. Two of these dissolve again in a second reagent, and the product is asked often: lead chromate in sodium hydroxide, because lead is amphoteric, and lead sulphate in ammonium acetate.

Definition

Lead chloride is white, sparingly soluble in cold water, soluble in hot water.
Lead sulphide is black; it dissolves in hot dilute nitric acid: $\mathrm{3PbS + 8HNO_3 \rightarrow 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O}$ .
Lead chromate is yellow and dissolves in NaOH: $\mathrm{PbCrO_4 + 4NaOH \rightarrow Na_2[Pb(OH)_4] + Na_2CrO_4}$ . The product is a dianionic complex with coordination number 4.
Lead sulphate is white and dissolves in ammonium acetate as soluble lead acetate; JEE writes it as the complex $\mathrm{(NH_4)_2[Pb(CH_3COO)_4]}$ .
Lead nitrate is soluble, so it is never a confirmatory precipitate.

Reagent added to Pb²⁺	Product	Colour	What happens next
Dilute HCl	$\mathrm{PbCl_2}$	White	Dissolves on heating the water
$\mathrm{H_2S}$	$\mathrm{PbS}$	Black	Dissolves in hot dilute $\mathrm{HNO_3}$ to give $\mathrm{Pb(NO_3)_2}$
$\mathrm{K_2CrO_4}$	$\mathrm{PbCrO_4}$	Yellow	Dissolves in NaOH as $\mathrm{Na_2[Pb(OH)_4]}$ Charge 2−, four OH groups: coordination number 4.
KI	$\mathrm{PbI_2}$	Yellow	Dissolves in hot water and returns as golden spangles on cooling
Dilute $\mathrm{H_2SO_4}$	$\mathrm{PbSO_4}$	White	Dissolves in ammonium acetate solution

Chloride, sulphate and nitrate of lead are the white or colourless ones; chromate and iodide are yellow; sulphide is black.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q26Moderate

Example 3 · The p-Block Elements · Group 14: Carbon, Silicon, Tin and Lead

Consider the following reactions.

PbCl_{2}+K_{2}CrO_{4}\rightarrow\text{ }A + 2KCl

(Hot solution)

{A + NaOH \rightleftharpoons B +Na_{2}CrO_{4} }{PbSO_{4}+ 4CH_{3}COONH_{4}\rightarrow\left( NH_{4} \right)_{2}SO_{4}+ X }

In the above reactions,

A,B

and X are respectively.

Lead chromate in NaOH gives a 2− complex with four OH

The product is

\mathrm{Na_2[Pb(OH)_4]}

: lead(II) with four hydroxide groups, coordination number 4, overall charge 2−. It is not neutral and not six-coordinate.

Lead nitrate is not a confirmatory test

Every confirmatory test for

\mathrm{Pb^{2+}}

makes an insoluble salt: chromate, iodide or sulphate. Lead nitrate is soluble, so its formation confirms nothing.

Drill 5 more on tests for the lead ion in salt analysis

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (3)

Group 14 trends, carbon's allotropes and silicones5 rows

Element	Covalent radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity	What sets it apart
C	77	1086	2.5	Catenation and $p\pi$ – $p\pi$ bonds; maximum covalency 4; allotropes
Si	118	786	1.8	Uses d orbitals: $\mathrm{[SiF_6]^{2-}}$ exists; $\mathrm{SiO_2}$ is acidic; forms silicones
Ge	122	761	1.8	$\mathrm{GeO_2}$ acidic; $\mathrm{[GeCl_6]^{2-}}$ exists
Sn	140	708	1.8	+4 more stable than +2; oxides amphoteric
Pb	146	715	1.9	+2 more stable than +4; oxides amphoteric Lead's ionisation enthalpy is a little HIGHER than tin's: poor shielding by 4f and 5d electrons.

The ionisation enthalpy falls from C to Sn, then rises slightly at Pb.

Inert pair effect in tin and lead: which ions oxidise and which reduce6 rows

Ion	Preferred state of the element	Behaves as	Evidence
$\mathrm{Sn^{2+}}$	+4	Reducing agent	$\mathrm{E^\circ(Sn^{4+}/Sn^{2+}) = +0.15\ V}$ : $\mathrm{Sn^{2+}}$ is easily oxidised
$\mathrm{Sn^{4+}}$	+4	Stable; a very weak oxidant at most	Same small potential, $+0.15\ V$
$\mathrm{Pb^{2+}}$	+2	Stable	The 6s pair stays out of bonding
$\mathrm{Pb^{4+}}$	+2	Strong oxidising agent	$\mathrm{E^\circ(Pb^{4+}/Pb^{2+}) = +1.67\ V}$ , the most positive here The strongest oxidant among these p-block ions.
$\mathrm{Tl^{3+}}$	+1	Strong oxidising agent	$\mathrm{Tl^{3+}}$ reduced to $\mathrm{Tl^{+}}$ : $+1.26\ V$
$\mathrm{Tl^{+}}$	+1	Stable	The 6s pair stays out of bonding

The more positive the reduction potential, the stronger the oxidising agent.

Tests for the lead ion in salt analysis5 rows

Reagent added to Pb²⁺	Product	Colour	What happens next
Dilute HCl	$\mathrm{PbCl_2}$	White	Dissolves on heating the water
$\mathrm{H_2S}$	$\mathrm{PbS}$	Black	Dissolves in hot dilute $\mathrm{HNO_3}$ to give $\mathrm{Pb(NO_3)_2}$
$\mathrm{K_2CrO_4}$	$\mathrm{PbCrO_4}$	Yellow	Dissolves in NaOH as $\mathrm{Na_2[Pb(OH)_4]}$ Charge 2−, four OH groups: coordination number 4.
KI	$\mathrm{PbI_2}$	Yellow	Dissolves in hot water and returns as golden spangles on cooling
Dilute $\mathrm{H_2SO_4}$	$\mathrm{PbSO_4}$	White	Dissolves in ammonium acetate solution

Chloride, sulphate and nitrate of lead are the white or colourless ones; chromate and iodide are yellow; sulphide is black.

Watch out for (7)

Carbon's allotropy comes from pπ–pπ bonds, not pπ–dπ
→ Group 14 trends, carbon's allotropes and silicones
C₆₀ has 20 six-membered and 12 five-membered rings
→ Group 14 trends, carbon's allotropes and silicones
Covalent radius increases down group 14
→ Group 14 trends, carbon's allotropes and silicones
The lower state is not always the reducing one
→ Inert pair effect in tin and lead: which ions oxidise and which reduce
Pb⁴⁺ is not stable like Sn⁴⁺
→ Inert pair effect in tin and lead: which ions oxidise and which reduce
Lead chromate in NaOH gives a 2− complex with four OH
→ Tests for the lead ion in salt analysis
Lead nitrate is not a confirmatory test
→ Tests for the lead ion in salt analysis

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