Q4.58P - Chemistry: Molecular Nature Of Matter And Change

Question

Describe how to determine the oxidation number of sulfur in

(a)  ${\mathrm{H}}_2\mathrm{S}$ and (b)${\mathbf{S}\mathbf{O}}_{\mathbf{3}}^{\mathbf{-}\mathbf{2}}$

Step-by-Step Solution

Answer

The oxidation number of sulfur in 

(a)  $H_{2}S is \left(-2\right)$

(b)  $S{O}_3^{-2} is \left(+4\right)$

Step 1: Defineoxidation number

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.

Step 2: Determine oxidation number of S in H 2 S

a. In  the sulfur atom is bonded with two hydrogen atoms with a single bond, since these hydrogensare less electronegative than sulfur which means the dipole moment is shifting towards the sulfur atom and we add a -1 charge tothe sulfur atomfrom each from two hydrogen atom. 

Let the oxidation number of sulphur be x.

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

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

Step 3: Determine oxidation number of S in S O 3 - 2

(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

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

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