A compound of vanadium has a magnetic moment of 1.73 BM. The electronic configuration of vanadium io — Atomic Structure Chemistry Question

💡 Solution & Explanation

Magnetic moment $\mu = 1.73$ BM corresponds to $n=1$ unpaired electron ($\sqrt{1(1+2)} = \sqrt{3} \approx 1.73$). Vanadium ($Z=23$) is $[Ar] 3d^3 4s^2$. To have 1 unpaired electron in the d-subshell, it must be the $3d^1$ configuration, which corresponds to V$^{4+}$. Therefore, correct answer is B.