JEE Mains Chemistry · The p-Block Elements

Group 16: Oxygen and Sulphur

Oxygen, sulphur, selenium, tellurium and polonium (ns² np⁴): oxygen is the small, anomalous member, the hydrides grow more acidic and more reducing down the group, and sulphur forms a family of oxoacids and a set of tests every salt analysis uses.

All The p-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Thirty PYQs, twenty-six of them multiple choice, and two from 2026. Thirteen test the group trends: oxygen's anomalies, oxidation states, the hydrides, the oxides, ozone and sulphur's allotropes; eight ask about the structures of sulphur's oxoacids; nine are sulphur redox reactions and the tests for sulphide and sulphite.

Concept 1 of 3: Group 16 trends: oxygen's anomalies, hydrides and oxides

Oxygen is small, highly electronegative and has no d orbitals, so it forms strong pπ–pπ double bonds and exists as O₂. Sulphur cannot do this well and bonds to itself with single bonds instead, giving S₈ rings. That difference in atomicity, not a small change in size, is why oxygen is a gas and sulphur a solid. Down the group the H–E bond grows longer and weaker, so the hydrides H₂E lose hydrogen more easily: they become stronger acids and stronger reducing agents from H₂O to H₂Te.

Definition

Oxygen is anomalous because of its small size and high electronegativity. Its covalency is usually 2 and never more than 4.
Oxygen's oxidation states: mostly −2, but −1 in $\mathrm{H_2O_2}$ , +1 in $\mathrm{O_2F_2}$ and +2 in $\mathrm{OF_2}$ .
S, Se, Te show −2, +2, +4 and +6. Down the group +4 becomes more stable and +6 less stable (inert pair).
Chalcogens: O, S, Se, Te, Po.
Sulphur allotropes: rhombic (α) is stable at room temperature; above 369 K it changes to monoclinic (β), and the change reverses on slow cooling. Both are $\mathrm{S_8}$ rings.
Oxides: $\mathrm{EO_2}$ and $\mathrm{EO_3}$ are formed by S, Se, Te and Po and are acidic. $\mathrm{SO_2}$ is mainly reducing (it can also oxidise, for example $\mathrm{H_2S}$ ); $\mathrm{TeO_2}$ is oxidising.
Ozone has 6 lone pairs in all: 1 on the central oxygen and 5 on the two end oxygens.

Hydride	Melting point (K)	H–E bond enthalpy (kJ/mol)	H–E–H angle (°)	Acid strength (Ka)
$\mathrm{H_2O}$	273	463	104	$1.8 \times 10^{-16}$ Hydrogen bonding makes water melt highest, though it is the lightest.
$\mathrm{H_2S}$	188	347	92	$1.3 \times 10^{-7}$
$\mathrm{H_2Se}$	208	276	91	$1.3 \times 10^{-4}$
$\mathrm{H_2Te}$	222	238	90	$2.3 \times 10^{-3}$

The bond enthalpy falls down the group, so acid strength and reducing power rise: H₂Te is the strongest acid and strongest reducing agent of the four.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q115Moderate

Example 1 · The p-Block Elements · Group 16: Oxygen and Sulphur

The nature of oxide

\left( TeO_{2} \right)

and hydride

\left( TeH_{2} \right)

formed by Te , respectively are :

Oxygen does not show only −2

Oxygen is −1 in peroxides, +1 in

\mathrm{O_2F_2}

and +2 in

\mathrm{OF_2}

. A statement that it shows only −2 is false.

Down group 16, +4 becomes MORE stable than +6

The inert pair effect makes the lower state more stable for the heavier elements. So the stability of +6 falls and that of +4 rises from S to Po.

Ozone has six lone pairs, not five

\mathrm{O_3}

the central oxygen has one lone pair and the two end oxygens have five between them (two on one, three on the other), six in all.

Rhombic sulphur is the room-temperature form

Monoclinic sulphur is stable only above 369 K. At room temperature the stable crystalline form is rhombic sulphur.

Drill 12 more on group 16 trends: oxygen's anomalies, hydrides and oxides

Concept 2 of 3: Oxoacids of sulphur: structures, S=O bonds and oxidation states

Every sulphur oxoacid is built from tetrahedral sulphur carrying S=O and S–OH groups. What changes is the link between units: an oxygen bridge S–O–S in pyrosulphuric acid, a peroxo O–O bridge in peroxodisulphuric acid, a direct S–S bond in dithionic acid, and a chain of sulphur atoms in the polythionic acids. Count π bonds by counting S=O: each double bond has one.

Definition

Oleum (fuming sulphuric acid) is pyrosulphuric acid, $\mathrm{H_2S_2O_7}$ : $\mathrm{SO_3 + H_2SO_4 \rightarrow H_2S_2O_7}$ . It has 7 oxygen atoms.
Marshall's acid, $\mathrm{H_2S_2O_8}$ , is made by electrolysing concentrated $\mathrm{H_2SO_4}$ (or a concentrated hydrogensulphate) at high current density: $\mathrm{2HSO_4^{-} \rightarrow HO_3SOOSO_3H + 2e^{-}}$ .
Peroxo link (O–O): only in $\mathrm{H_2S_2O_8}$ among the common oxoacids.
S in two different states: thiosulphuric acid, $\mathrm{H_2S_2O_3}$ .
Polythionic acids $\mathrm{H_2S_xO_6}$ : the two end S are +5, the chain S are 0.

Acid	Formula	Oxidation state of S	S=O bonds	Link between units
Sulphurous	$\mathrm{H_2SO_3}$	+4	1	One unit; a lone pair on S
Sulphuric	$\mathrm{H_2SO_4}$	+6	2	One unit, two S–OH
Thiosulphuric	$\mathrm{H_2S_2O_3}$	Average +2; the two S differ	1	A terminal S doubly bonded to the central S, in place of one O
Dithionic	$\mathrm{H_2S_2O_6}$	+5, both S alike	4	A direct S–S bond
Pyrosulphuric (oleum)	$\mathrm{H_2S_2O_7}$	+6	4	One S–O–S bridge
Peroxodisulphuric (Marshall's)	$\mathrm{H_2S_2O_8}$	+6	4	One O–O peroxo bridge Still +6: the two peroxo oxygens are −1 each.
Polythionic	$\mathrm{H_2S_xO_6}$	Ends +5, chain 0	4	A chain of S atoms between two $\mathrm{SO_3H}$ groups

Each S=O bond carries one π bond, so counting S=O counts the π bonds of every acid here except thiosulphuric, which also has an S=S.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q144Moderate

Example 2 · The p-Block Elements · Group 16: Oxygen and Sulphur

Sum of

\pi

- bonds present in peroxodisulphuric acid and pyrosulphuric acid is:

Pyrosulphuric acid has an S–O–S bridge, not a peroxo bond

\mathrm{H_2S_2O_7}

joins its two sulphurs through one oxygen. The O–O peroxo bond belongs to

\mathrm{H_2S_2O_8}

Marshall's acid needs concentrated sulphuric acid

Electrolysing dilute sulphuric acid or dilute sodium sulphate just splits water. Peroxodisulphuric acid forms at the anode only from a concentrated solution at high current density.

Drill 7 more on oxoacids of sulphur: structures, s=o bonds and oxidation states

Concept 3 of 3: Redox reactions of sulphur compounds and the tests for sulphide and sulphite

Sulphur spans −2 to +6, so its compounds are good at redox. Sulphur dioxide (+4) is usually a reducing agent: it turns orange dichromate green by reducing it to Cr³⁺. Elemental sulphur (0) sits in the middle, so in hot alkali it disproportionates to sulphide and thiosulphate. Thiosulphate is oxidised further by a strong oxidant than by a weak one: iodine stops it at tetrathionate, bromine takes it all the way to sulphate. These reactions are also the salt-analysis tests: hydrogen sulphide blackens lead acetate paper, and sulphur dioxide turns acidified dichromate paper green.

Definition

$\mathrm{SO_2}$ with acidified dichromate: $\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O}$ ; the green product is chromium(III) sulphate.
Sulphur in alkali: $\mathrm{S_8 + 12OH^{-} \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O}$ .
Thiosulphate: $\mathrm{2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^{-}}$ , but $\mathrm{S_2O_3^{2-} + 4Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 8Br^{-} + 10H^{+}}$ , because bromine is the stronger oxidant.
Ozone and lead sulphide: $\mathrm{PbS + 4O_3 \rightarrow PbSO_4 + 4O_2}$ .
Sulphide test: dilute $\mathrm{H_2SO_4}$ releases $\mathrm{H_2S}$ , which turns lead acetate paper black (PbS).
Sulphite test: dilute $\mathrm{H_2SO_4}$ releases $\mathrm{SO_2}$ , which turns acidified dichromate green.
Sulphide colours: $\mathrm{As_2S_3}$ and $\mathrm{As_2S_5}$ yellow, ammonium sulphide solution yellow, PbS and CuS black.

Sulphur redox reactions

\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O} \qquad \mathrm{S_8 + 12OH^{-} \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O} \qquad \mathrm{2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^{-}}

Worked example

Solid sulphur is boiled with sodium hydroxide solution. Identify the two sulphur-containing products, give the oxidation state of sulphur in each, and name the type of reaction.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q45Moderate

Example 3 · The p-Block Elements · Group 16: Oxygen and Sulphur

A paper dipped in a dil.

H_{2}SO_{4}

solution of ' X ' upon treatment with

SO_{2}

gas turns into green. The compound ' X ' is :

Lead acetate paper turns black from lead sulphide

Hydrogen sulphide gives black PbS. Lead sulphite is white, so a statement that the black colour is lead sulphite is false.

The green colour is chromium(III) sulphate, not Cr₂O₃

In acidified solution, dichromate is reduced to

\mathrm{Cr^{3+}}

, which stays in solution as

\mathrm{Cr_2(SO_4)_3}

\mathrm{Cr_2O_3}

is the green solid from heating ammonium dichromate.

Bromine takes thiosulphate further than iodine

Both oxidise thiosulphate. Iodine, the weaker oxidant, stops at

\mathrm{S_4O_6^{2-}}

; bromine, the stronger one, goes on to

\mathrm{SO_4^{2-}}

. Thiosulphate is oxidised in both cases, never reduced.

Drill 8 more on redox reactions of sulphur compounds and the tests for sulphide and sulphite

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.

Formulas (1)

Redox reactions of sulphur compounds and the tests for sulphide and sulphite
Sulphur redox reactions
$\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O} \qquad \mathrm{S_8 + 12OH^{-} \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O} \qquad \mathrm{2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^{-}}$

Reference tables (2)

Group 16 trends: oxygen's anomalies, hydrides and oxides4 rows

Hydride	Melting point (K)	H–E bond enthalpy (kJ/mol)	H–E–H angle (°)	Acid strength (Ka)
$\mathrm{H_2O}$	273	463	104	$1.8 \times 10^{-16}$ Hydrogen bonding makes water melt highest, though it is the lightest.
$\mathrm{H_2S}$	188	347	92	$1.3 \times 10^{-7}$
$\mathrm{H_2Se}$	208	276	91	$1.3 \times 10^{-4}$
$\mathrm{H_2Te}$	222	238	90	$2.3 \times 10^{-3}$

The bond enthalpy falls down the group, so acid strength and reducing power rise: H₂Te is the strongest acid and strongest reducing agent of the four.

Oxoacids of sulphur: structures, S=O bonds and oxidation states7 rows

Acid	Formula	Oxidation state of S	S=O bonds	Link between units
Sulphurous	$\mathrm{H_2SO_3}$	+4	1	One unit; a lone pair on S
Sulphuric	$\mathrm{H_2SO_4}$	+6	2	One unit, two S–OH
Thiosulphuric	$\mathrm{H_2S_2O_3}$	Average +2; the two S differ	1	A terminal S doubly bonded to the central S, in place of one O
Dithionic	$\mathrm{H_2S_2O_6}$	+5, both S alike	4	A direct S–S bond
Pyrosulphuric (oleum)	$\mathrm{H_2S_2O_7}$	+6	4	One S–O–S bridge
Peroxodisulphuric (Marshall's)	$\mathrm{H_2S_2O_8}$	+6	4	One O–O peroxo bridge Still +6: the two peroxo oxygens are −1 each.
Polythionic	$\mathrm{H_2S_xO_6}$	Ends +5, chain 0	4	A chain of S atoms between two $\mathrm{SO_3H}$ groups

Each S=O bond carries one π bond, so counting S=O counts the π bonds of every acid here except thiosulphuric, which also has an S=S.

Watch out for (9)

Oxygen does not show only −2
→ Group 16 trends: oxygen's anomalies, hydrides and oxides
Down group 16, +4 becomes MORE stable than +6
→ Group 16 trends: oxygen's anomalies, hydrides and oxides
Ozone has six lone pairs, not five
→ Group 16 trends: oxygen's anomalies, hydrides and oxides
Rhombic sulphur is the room-temperature form
→ Group 16 trends: oxygen's anomalies, hydrides and oxides
Pyrosulphuric acid has an S–O–S bridge, not a peroxo bond
→ Oxoacids of sulphur: structures, S=O bonds and oxidation states
Marshall's acid needs concentrated sulphuric acid
→ Oxoacids of sulphur: structures, S=O bonds and oxidation states
Lead acetate paper turns black from lead sulphide
→ Redox reactions of sulphur compounds and the tests for sulphide and sulphite
The green colour is chromium(III) sulphate, not Cr₂O₃
→ Redox reactions of sulphur compounds and the tests for sulphide and sulphite
Bromine takes thiosulphate further than iodine
→ Redox reactions of sulphur compounds and the tests for sulphide and sulphite

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