Question

In the formula of brown ring complex $\left[Fe{\left(H_{2}O\right)}_5\left(NO\right)\right]S{O}_4$, the magnetic moment is 3.87 B.M. The oxidation state of Fe and number of unpaired electrons present respectively are :

• A. $1+,3$
• B. $2+,3$
• C. $3+,3$
• D. $3+,5$

Answer 1

The correct option is A $1+,3$

Magnetic moment $\mu =\sqrt{n(n+2)}$ , where n is no. of unpaired electron.
Here $\mu =3.87$ so n$=$3.
So electronic configuration of Fe $d^7,F{e}^{+}$.