91Ó°ÊÓ

Each atom in a molecule (or ionic compound) is assigned an oxidation number (O.N.), or oxidation state, the charge the atom would have if electrons were not shared but were transferred completely.

The oxidation number for each element in a binary ionic compound equals the ionic charge.

Let the oxidation number of sulphur be x.

$$\begin{gathered} \left(2×1\right) + x = 0 \\ x=-2 \end{gathered}$$

Hence in $H_{2}S$sulfur has a -2 oxidation number.

(b) The charge on one oxygen atom is -2. The oxidation number of sulphur in $S{O}_3^{-2}$is calculated below.

Let the oxidation number of sulphur be x.

Since, the molecule has a charge of -2

$\begin{array}{rcl}x + \left(-2×3\right) & =&  -2\\ x + \left(-6\right) & =& -2\\ x & =&  6-2\\ x & =& 4\\ & & \end{array}$

Thus, according to its structure oxidation number of sulfur indata-custom-editor="chemistry" ${\mathrm{SO}}_3^{-2}$is +4.