Question

Find the magnetic moment (in B.M.) of ${}_{23}{V}^{4+}$ and ${}_{30}Z{n}^{2+}$ respectively

• A. $\sqrt{3}$ B.M., $\sqrt{0}$ B.M.
• B. $\sqrt{4}2$ B.M., $\sqrt{3}$ B.M.
• C. $\sqrt{12}$ B.M., $\sqrt{72}$ B.M.
• D. $\sqrt{72}$ B.M., $\sqrt{12}$ B.M.

Answer 1

The correct option is A $\sqrt{3}$ B.M., $\sqrt{0}$ B.M.

Electronic configuration of V = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^3$
Electronic configuration of $V^{4+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$
So, number of unpaired electrons 'n' = 1
Magnetic moment = $\sqrt{n(n+2)}$ B.M = $\sqrt{3}$ B.M.

Electronic configuration of Zn = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$
Electronic configuration for $Z{n}^{2+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^{10}$
So. number of unpaired electrons 'n' = 0
Magnetic moment = $\sqrt{n(n+2)}$ B.M = $\sqrt{0}$ B.M.