Question

The oxidation state of $Fe$ in $FeS{O}_4$.$(N{H}_4{)}_{2}S{O}_4$.$6H_{2}O$ is .

• A. 2
• B. 3
• C. 1

Answer 1

The correct option is A 2

Here, Sum of O. N. for $H_{2}O$ = 2(1) + (-2) = 0

Sum of O.N. for $N{H}_4$ = -3 + 4(1) = +1

O.N. for $S{O}_4$ = -2

Sum of O.N. for $(N{H}_4{)}_{2}S{O}_4$ = (+1)2 + (-2) = 0

Therefore, $FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O$ $\Rightarrow$ n + (-2) + 0 + 0 = 0

$\Rightarrow$ n = +2