Question

What is the oxidation number for sulphur? Give an example.

Answer 1

Oxidation number: Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element.

Example: Sulphuric acid$\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$

Calculating the oxidation number of Sulphur(S) in Sulphuric acid$\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$:

The oxidation state of Oxygen(O) = -2

The oxidation state of Hydrogen(H) = +1

Consider the oxidation state of Sulphur(S) = x

Now,

$$\begin{gathered} 2(+1)+\mathrm{x}+\left(-2\times 4\right)=0 \\ +2+\mathrm{x}-8=0 \\ \mathrm{x}=+6 \end{gathered}$$

Hence, the oxidation number for Sulphur in Sulphuric acid is $+6$.