The pair in which both the species have the same magnetic moment (spin only) is:

🧠 Same μ ⟺ Same Unpaired-Electron Count

Spin-only μ depends only on the number of unpaired electrons. So scan each pair for matching unpaired counts.

🗺️ Walk the Options

| Pair | Ion 1 (config, n) | Ion 2 (config, n) | |---|---|---| | (1) $[\mathrm{Cr(H_2O)_6}]^{2+}$, $[\mathrm{Fe(H_2O)_6}]^{2+}$ | Cr²⁺ d⁴ HS, n=4 | Fe²⁺ d⁶ HS, n=4 ✓ | | (2) $[\mathrm{Co(OH)_4}]^{2-}$, $[\mathrm{Fe(NH_3)_6}]^{2+}$ | Co²⁺ d⁷ Td, n=3 | Fe²⁺ d⁶ (NH₃ borderline), n=4 | | (3) $[\mathrm{Mn(H_2O)_6}]^{2+}$, $[\mathrm{Cr(H_2O)_6}]^{2+}$ | Mn²⁺ d⁵, n=5 | Cr²⁺ d⁴, n=4 | | (4) $[\mathrm{Cr(H_2O)_6}]^{2+}$, $[\mathrm{CoCl_4}]^{2-}$ | n=4 | Co²⁺ d⁷ Td, n=3 |

Only pair (1) matches: 4 = 4.

⚡ The HS d⁴ ↔ d⁶ Coincidence

$\mathrm{H_2O}$ is a weak-field ligand, so both Cr²⁺ ($\mathrm{t_{2g}^3 e_g^1}$) and Fe²⁺ ($\mathrm{t_{2g}^4 e_g^2}$) are high-spin and both have 4 unpaired electrons. This is the canonical HS d⁴ = HS d⁶ μ-match.

$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM for both.

⚠️ Don't Forget Geometry Affects d-count Splitting

In option (4), $[\mathrm{CoCl_4}]^{2-}$ is tetrahedral (Cl⁻ is weak, Co²⁺ d⁷ Td → 3 unpaired). The geometry change matters — never assume octahedral by default.

$\boxed{\text{Answer: (1) } [\mathrm{Cr(H_2O)_6}]^{2+}\text{ and } [\mathrm{Fe(H_2O)_6}]^{2+}}$