Question

In the brown ring complex $\left[Fe\right(H_{2}O{)}_5\left(NO\right)]S{O}_4$ nitric oxide behaves as:

• A. $N{O}^{+}$
• B. neutral $NO$ molecule
• C. $N{O}^{-}$
• D. $N{O}_2^{-}$

Answer 1

Answer: (A)

$N{O}^{+}$

The O.S. of $Fe$ in Brown ring is +1. The complex formed is $\left[Fe{\left(H_{2}O\right)}_{5}NO\right]S{O}_4$. The charge on anion is -2. Hence charge on the cationic complex has to be +2 as the ring is electrically neutral. The charge on $Fe$ on calculation comes out to be +2. But electron transfer take place from $NO$ to $Fe$. Hence $NO$ and $Fe$ acquire a charge of +1 each. So the oxidation state of $NO$ will be in +1 state.