Question

The complex $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$ is formed in the brown ring test for nitrates when freshly prepared $FeS{O}_4$ solution is added to the aqueous solution of $N{O}_3^{-}$ followed by the addition of concentrated $H_{2}S{O}_4$.

The correct statement above this complex is :

• A. The colour change is due to charge transfer.
• B. It has iron in $+1$ oxidation state and nitrosyl as $N{O}^{+}$.
• C. It has magnetic moment of 3.87 BM, confirming three unpaired electrons in $Fe$.
• D. All of the above

Answer 1

The correct option is D All of the above

In this complex $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$, the oxidation state of Fe is +1 and that of NO is +1.

Here, $F{e}^{+}$ has $d^7$ configuration, which is a higher spin complex, so it will have 3 unpaired electrons.

So, the magnetic moment of the complex is $\mu =\sqrt{n(n+2)}=\sqrt{3\times 5}=3.87$ BM. It shows colour change due to charge transfer.

Hence, the option $D$ is correct.