A compound of vanadium has a magnetic moment of 1.73 B.M. The electronic configuration of vanadium i — Atomic Structure Chemistry Question

💡 Solution & Explanation

According to the spin-only formula μ = √(n * (n + 2)) B.M., a magnetic moment of 1.73 B.M. corresponds to exactly n = 1 unpaired electron (since √(1 * 3) = √(3) = 1.732 B.M.). Neutral vanadium (Z = 23) has the electronic configuration [Ar] 3d^3 4s^2. To have exactly 1 unpaired electron in its d-subshell, the vanadium ion must have lost four electrons (V^4+), leaving 1 electron in the d-orbitals: [Ar] 3d^1. Therefore, the electronic configuration is [Ar] 3d^1.