Question

48. Oxidation number of Fe in $FeS{O}_4.{\left(N{H}_4\right)}_{2}S{O}_4.6H_{2}O$

Answer 1

Step 1:

Oxidation number of $S{O}_4=-2$

Oxidation number of $N{H}_4=+1$

Water is electrically neutral. So, oxidation number will be taken as zero.

There is no charge on the formula of the compound. So, charge will be taken as zero.

Step 2:

$x-2+2\left(1\right)-2+6\left(0\right)=0$

$x=+2$

Therefore, the oxidation number of Fe is +2.