Question

The number of unpaired electrons present in the species $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$ which is formed during 'brown ring' test is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

The oxidation state of Fe in the complex is +1 and Coordination number of Fe is 6. The configuration of Fe is $3d^{6}4s^2$ and that of $F{e}^{+}$ is $3d^{6}4s^1$. Now, since the $N{O}^{+}$ is a strong field ligand so only one pairing of 3d and 4s electron occurs and hence the number of unpaired electron corresponds to n=3.Hence the correct option is B.