Question

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

• A. 2
• B. 1
• C. -2
• D. -1

Answer 1

The correct option is A 2

Here,
O. S. for $H_{2}O$ = 2(1) + (-2) = 0
O. S. for $N{H}_4^{+}$ = -3 + 4(1) = + 1
O. S. for $S{O}_4^{2-}$ = -2
Sum of O. S. for $(N{H}_4{)}_{2}S{O}_4$ = (+1)2 + (-2) = 0
Therefore,
$FeS{O}_4(N{H}_4{)}_2.6H_{2}0$ $\Rightarrow$ n+(-2) + 0 + 0 = 0
$\Rightarrow$ Fe = +2 Ans