Question

The oxidation number of underlined element in $\left[\underset{–}{Fe}\left(NO\right){\left(H_{2}O\right)}_5\right]S{O}_4$ compound is :

• A. $+1$
• B. $+3$
• C. $+5$
• D. None of these

Answer 1

The correct option is A $+1$

Let $x$ be the oxidation number of $Fe$ in $\left[\underset{–}{Fe}\left(NO\right){\left(H_{2}O\right)}_5\right]S{O}_4$.
Since, the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.

We know that the Oxidation no. of $NO=+1$, $H_{2}O=0$, and $S{O}_4=-2$
$\therefore x+1+5\left(0\right)+(-2)=0$
$\Rightarrow x-1=0$
$\Rightarrow x=+1$
Thus, the oxidation number of $Fe$ in $\left[\underset{–}{Fe}\left(NO\right){\left(H_{2}O\right)}_5\right]S{O}_4$ is $+1$.

Hence, the correct option is A.