The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2, is:

🧠 Decode the μ Value

$\mu = 3.9$ BM ⟹ $n(n+2) = 15.2$ ⟹ $n \approx 3$ unpaired electrons.

In $[\mathrm{M(H_2O)_6}]\mathrm{Cl_2}$: $\mathrm{H_2O}$ is weak-field, M is M²⁺ ⟹ all complexes are high-spin octahedral. We need M²⁺ that gives 3 unpaired in HS oct.

🗺️ Test Each Option

| Ion | d-count | HS config (oct) | Unpaired | |---|---|---|---| | V²⁺ | d³ | $\mathrm{t_{2g}^3}$ | 3 ✓ | | Cr²⁺ | d⁴ | $\mathrm{t_{2g}^3 e_g^1}$ | 4 | | Mn²⁺ | d⁵ | $\mathrm{t_{2g}^3 e_g^2}$ | 5 | | Fe²⁺ | d⁶ | $\mathrm{t_{2g}^4 e_g^2}$ | 4 | | Co²⁺ | d⁷ | $\mathrm{t_{2g}^5 e_g^2}$ | 3 ✓ |

V²⁺ and Co²⁺ both have 3 unpaired → μ ≈ 3.87 BM ≈ 3.9 BM.

⚡ The d³ ↔ d⁷ HS μ-Match

Among first-row M²⁺ HS octahedral, d³ and d⁷ both give 3 unpaired — they share μ. This is the second classic μ-match (after d⁴ ↔ d⁶ HS, which both give 4).

⚠️ Symmetry of HS μ Curve

Across d¹ to d⁹ HS oct, the unpaired-count curve is: 1, 2, 3, 4, 5, 4, 3, 2, 1 — symmetric around d⁵. So d^k and d^(10−k) always match in unpaired count.

$\boxed{\text{Answer: (4) } \mathrm{V^{2+}}\text{ and } \mathrm{Co^{2+}}}$