The oxoacid of sulphur which has a lone pair of electron on sulphur atom is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(H_2SO_3\)
Sulphur has six valence electrons. In an oxoacid, the number of these electrons used in bonding equals the oxidation state of sulphur. At +6 all six are used and no lone pair is left; at +4 two electrons stay on sulphur as one lone pair.
In \( \ce{H2SO3} \): 2(+1) + x + 3(−2) = 0, so x = +4. Sulphur is bonded to one O by a double bond and to two OH groups and carries one lone pair (pyramidal around S). So sulphurous acid is the answer.
Why the other options are incorrect
Option (1) \(H_2SO_4\) — Incorrect. 2(+1) + x + 4(−2) = 0, so x = +6; all six valence electrons are used in two S=O and two S–OH bonds.
Option (3) \(H_2S_2O_8\) — Incorrect. Peroxodisulphuric acid has one O–O link, whose two O atoms are −1: 2(+1) + 2x + 6(−2) + 2(−1) = 0, so x = +6. Each S is bonded to four O atoms and has no lone pair.
Option (4) \(H_2S_2O_7\) — Incorrect. Pyrosulphuric acid (oleum): 2(+1) + 2x + 7(−2) = 0, so x = +6. Each S is tetrahedral, joined through an S–O–S bridge, with no lone pair.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Oxoacids of Sulphur
