Oxoacids of sulphur — structures and properties of H₂SO₃, H₂SO₄, H₂S₂O₇

Question

Draw the structures of H $_2$ SO $_3$ , H $_2$ SO $_4$ , and H $_2$ S $_2$ O $_7$ (oleum). Discuss their key properties and the oxidation state of sulphur in each.

(JEE Main 2023, similar pattern)

Solution — Step by Step

Oxidation state of S: +4

Structure: Sulphur is bonded to two OH groups and one oxygen atom (double bond). The lone pair on S gives it a pyramidal geometry.

S has one $\text{S}=\text{O}$ double bond and two $\text{S}-\text{OH}$ bonds
It is a dibasic acid (two replaceable H atoms, both attached to OH)
It is a reducing agent (S is in +4 state, can be oxidised to +6)
Aqueous SO $_2$ acts as H $_2$ SO $_3$ (it exists mainly in solution, rarely isolated pure)

Oxidation state of S: +6

Structure: Sulphur is at the centre with tetrahedral geometry, bonded to two OH groups and two oxygen atoms (double bonds).

S has two $\text{S}=\text{O}$ bonds and two $\text{S}-\text{OH}$ bonds
It is a strong dibasic acid, a powerful dehydrating agent, and an oxidising agent (when hot and concentrated)
Manufactured by the Contact process: SO $_2$ + $\frac{1}{2}$ O $_2$ $\xrightarrow{\text{V}_2\text{O}_5}$ SO $_3$ , then SO $_3$ + H $_2$ SO $_4$ → H $_2$ S $_2$ O $_7$

Oxidation state of S: +6 (same as in H $_2$ SO $_4$ )

Structure: Two tetrahedral SO $_4$ units joined by a bridging oxygen atom. Each S has one $\text{S}=\text{O}$ , one $\text{S}-\text{OH}$ , and two S-O bonds (one bridging, one double bond).

Formed when SO $_3$ dissolves in concentrated H $_2$ SO $_4$
Dilution with water gives back H $_2$ SO $_4$ : H $_2$ S $_2$ O $_7$ + H $_2$ O → 2H $_2$ SO $_4$
It is a stronger acid and more reactive than H $_2$ SO $_4$ — used industrially for sulfonation reactions

Property	H $_2$ SO $_3$	H $_2$ SO $_4$	H $_2$ S $_2$ O $_7$
Oxidation state	+4	+6	+6
S-O bonds	1 double + 2 single	2 double + 2 single	Bridging + double bonds
Nature	Reducing agent	Oxidising + dehydrating	Stronger than H $_2$ SO $_4$
Basicity	Dibasic	Dibasic	Dibasic

Why This Works

Sulphur can form oxoacids with oxidation states +4 and +6 because of its ability to expand its octet using 3d orbitals. The number of S=O bonds increases with oxidation state: H $_2$ SO $_3$ has one, H $_2$ SO $_4$ has two, and the pyro- acid links two sulphate units through an oxygen bridge.

The key structural principle: every H that can ionise is attached to oxygen as an OH group. Direct S-H bonds do not ionise as acids. Count OH groups to determine basicity.

Alternative Method — Identifying Structures by Bond Counting

For any oxoacid of sulphur: Total bonds from S = 4 (tetrahedral) or 3 (pyramidal for +4 state). Double-bonded oxygens (S=O) don’t carry H. Only OH groups contribute to basicity.

JEE frequently asks to identify the number of S-O-S bridges in polythionic acids and other sulphur oxoacids. The rule: “pyro” means one bridging oxygen between two central atoms. H $_2$ S $_2$ O $_7$ has exactly one S-O-S bridge. Peroxo acids (like H $_2$ SO $_5$ , Caro’s acid) have an O-O peroxo linkage instead.

Common Mistake

Students draw H directly bonded to S in the structure of H $_2$ SO $_3$ or H $_2$ SO $_4$ . In oxoacids, hydrogen is always bonded to oxygen (as -OH), never directly to sulphur. Drawing S-H bonds gives the wrong structure and wrong basicity prediction. The acidic hydrogens come from O-H bond breaking, not S-H.

Question

Solution — Step by Step

Why This Works

Alternative Method — Identifying Structures by Bond Counting

Common Mistake

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