What is the oxidation number of sulphur in $\mathrm{H}_2\mathrm{SO}_4$? Apply the standard rules: H is $+1$, O is $-2$ (in non-peroxides), and the sum of oxidation numbers equals the charge of the molecule.

1. $\mathrm{H}_2\mathrm{SO}_4$ is a neutral molecule, so the sum of oxidation numbers of all atoms must equal $0$. 2. Let the oxidation number of S be $x$. 3. Contributions: H atoms give $2(+1) = +2$. O atoms give $4(-2) = -8$. S gives $x$. 4. Sum to zero: $(+2) + x + (-8) = 0$. 5. Solve: $x = +6$. 6. Option B is the oxidation number of S in $\mathrm{H}_2\mathrm{SO}_3$ (sulphurous acid). Option A is sulphide S in compounds like FeS. Option D is impossible since S has only 6 valence electrons. _Source: NCERT Class 11 Chemistry Part 2, Ch 7, §7.3 (Oxidation Number rules), p. 4 ¶1._