Question

How many unpaired electrons are present in the Brown Ring complex
$\left[Fe\right(H_{2}O{)}_5(NO\left)\right]{SO}_4$?

Answer 1

The spin of the compound is $3/2$ as $N{o}^{+}$ is a paramagnetic substance$.$

So that unpaired electron it can give up to $Fe$ to reduce it from $F{e}^{2+}$ to $+1$ and no becomes $N{O}^{+}$

Hence$,$ No$.$ of unpaired electron present in $\left[Fe{\left(H_{2}O\right)}_5\left(NO\right)\right]S{O}_4$ is $6.$