Chemistry - Complete d-block Question with Solution | TestHub

The spin-only magnetic moment is given by the formula $\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. The cation with the lowest number of unpaired electrons will have the lowest spin-only magnetic moment.

Let us determine the number of unpaired electrons for each given cation:

For $V^{3+}$ (Z = 23): The electronic configuration of V is $[\mathrm{Ar}]3d^{3}4s^2$. Removing three electrons gives $V^{3+}$ as $[\mathrm{Ar}]3d^2$. Number of unpaired electrons, $n=2$.

For ${\mathrm{Cr}}^{3+}$ (Z = 24): The electronic configuration of Cr is $[\mathrm{Ar}]3d^{5}4s^1$. Removing three electrons gives ${\mathrm{Cr}}^{3+}$ as $[\mathrm{Ar}]3d^3$. Number of unpaired electrons, $n=3$.

For ${\mathrm{Mn}}^{2+}$ (Z = 25): The electronic configuration of Mn is $[\mathrm{Ar}]3d^{5}4s^2$. Removing two electrons gives ${\mathrm{Mn}}^{2+}$ as $[\mathrm{Ar}]3d^5$. Number of unpaired electrons, $n=5$.

For ${\mathrm{Fe}}^{2+}$ (Z = 26): The electronic configuration of Fe is $[\mathrm{Ar}]3d^{6}4s^2$. Removing two electrons gives ${\mathrm{Fe}}^{2+}$ as $[\mathrm{Ar}]3d^6$. Number of unpaired electrons, $n=4$.

Since $V^{3+}$ has the lowest number of unpaired electrons ($n=2$), it develops the lowest value of spin-only magnetic moment ($\mu =\sqrt{2(2+2)}=\sqrt{8}\approx 2.83$ BM).

Answer: $V^{3+}$