The magnetic moment of a transition metal compound has been calculated to be 3.87 BM. The metal ion is:

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ BM

Given $\mu = 3.87$ BM: $n(n+2) \approx 15 \Rightarrow n = 3 \quad (\sqrt{15} = 3.873 \text{ BM})$

Need ion with 3 unpaired electrons (in +2 state):

| Ion | M²⁺ Config | Unpaired e⁻ | $\mu$ (BM) | |-----|------------|-------------|------------| | Cr²⁺ | $3d^4$ | 4 | 4.90 | | Mn²⁺ | $3d^5$ | 5 | 5.92 | | V²⁺ | $3d^3$ | 3 | 3.87 ✓ | | Ti²⁺ | $3d^2$ | 2 | 2.83 |

$\text{V}^{2+}$: $[\text{Ar}]\, 3d^3$ → 3 unpaired electrons → $\mu = \sqrt{15} = 3.87$ BM ✓

Answer: Option (3) — $\text{V}^{2+}$