Question

The brown ring complex is formulated as$\left[{Fe\left(H_{2}O\right)}_{5}NO\right]{SO}_4$. The oxidation state of iron is:
(A) +1 (B) +2 (C) +3 (D) 0

Answer 1

The correct option is A

The oxidation number of iron in the above compound can be calculated as follow

$\left[fe\right(No\left)\right(H_{2}0{)}_5]S{o}_4$ dissociates in water to liberate $\left[fe\right(No\left)\right(H_{2}o{)}_5{]}^{2+}$ ions

The charge on No is +1 that is $N{o}^{+}$ whereas $H_{2}o$ ligand is neutral

Hence the oxidation state of iron in above complex is

$x+(+1)+5\left(0\right)=+2$

$x=+1$

Therefore iron is present in $+1$ oxidation state in the above complex