Question

The ratio of magnetic moments of Fe(III) and Co(II) is
(a) 7 : 3
(b) 3 : 7
(c) $\sqrt{7} : \sqrt{3}$
(d) $\sqrt{3} : \sqrt{7}$

Answer 1

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

Explanation:

Formula used for magnetic moment :

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

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

Now put all the given values in the above formula, we get the unpaired electrons.

For $Fe$ with atomic no of 26, the configuration is:

$1s^22s^22p^63s^23p^63d^64s^2$

The electronic configuration of $Fe^{3+}$ is,  

$1s^22s^22p^63s^23p^63d^5$ containing 5 unpaired electrons.

$\mu=\sqrt{5(5+2)}=\sqrt{5\times 7}$

For $Co$ with atomic no of 27, the configuration is:

$1s^22s^22p^63s^23p^63d^74s^2$

The electronic configuration of $Co^{2+}$ is,  

$1s^22s^22p^63s^23p^63d^7$ containing 3 unpaired electrons.

$\mu=\sqrt{3(3+2)}=\sqrt{5\times 3}$

The ratio of magnetic moments of Fe(III) and Co(II) is $\sqrt{7}$ : $\sqrt{3}$.