Question

The ratio of magnetic moments of $Fe\left(III\right)$ and $Co\left(II\right)$ is:

• A. $7:3$
• B. $3:7$
• C. $\sqrt{7}:\sqrt{3}$
• D. $\sqrt{3}:\sqrt{7}$

Answer 1

Answer: (C)

$\sqrt{7}:\sqrt{3}$

Electronic configuration of $F{e}^{3+}:\left[Ar\right]3d^{5}4s^0$

So number of unpaired elctrons is $n=5$

Hence, $\mu =\sqrt{n(n+2)}=\sqrt{35}$

Now,

Electronic configuration of $C{o}^{2+}:\left[Ar\right]3d^{7}4s^0$

So number of unpaired electros is $n=3$

Hence, $\mu =\sqrt{n(n+2)}=\sqrt{15}$

So, Ratio of Magnetic moments of $F{e}^{3+}$ and $C{o}^{2+}$ is$:\frac{\sqrt{7}}{\sqrt{3}}$