The ratio of magnetic moments of $\mathrm{Fe}(\mathrm{III})$ and $\mathrm{Co}(\mathrm{II}) \mathrm{s}$
The ratio of magnetic moments of $\mathrm{Fe}(\mathrm{III})$ and $\mathrm{Co}(\mathrm{II}) \mathrm{s}$
$7: 3$
$3: 7$
$\sqrt{7}: \sqrt{3}$
$\sqrt{3}: \sqrt{7}$
Solution
Electronic configuration of \(\mathrm{Fe}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^{\circ}\)
So number of unpaired elctrons is \(\mathrm{n}=5\)
Hence, \(\mu=\sqrt{n(n+2)}=\sqrt{35}\)
Now,
Electronic configuration of \(\mathrm{Co}^{2+}:\left[\mathrm{Ar}^3 \mathrm{~d}^7 4 \mathrm{~s}^{\circ}ight.\)
So number of unpaired electros is \(\mathrm{n}=3\)
Hence, \(\mu=\sqrt{n(n+2)}=\sqrt{15}\)
So, Ratio of Magnetic moments of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Co}^{2+}\) is: \(\frac{\sqrt{7}}{\sqrt{3}}\)
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