Question

A complex of a certain metal has the magnetic moment of 4.9 BM whereas another complex of the same metal with the same oxidation state has zero magnetic moment. The metal ion would be:

• A. $C{o}^{2+}$
• B. $M{n}^{2+}$
• C. $F{e}^{2+}$
• D. $F{e}^{3+}$

Answer 1

The correct option is C $F{e}^{2+}$

Given, magnetic moment = 4.9 BM
$\Rightarrow \sqrt{n(n+2)}=\sqrt{24}$
$\Rightarrow n=4$
$C{o}^{2+}$ has 8 valence electrons, $M{n}^{2+}$ and $F{e}^{3+}$ has 5 valence electrons and $F{e}^{2+}$ has 6 valence electrons. So, for 4 unpaired electrons the only possible option is that, the metal must have 6 valence electrons. Hence, the required metal ion is $F{e}^{2+}$.
In the presence of a strong field ligand, all the valence electrons will get paired up and the electronic configuration would be $t_{2g}^6{e}_g^0$. Thus, there is no unpaired electron. Hence, zero magnetic moment.