The pair in which both the species have the same magnetic moment (spi…

The correct answer is Option (2) → \([Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}\)

The spin-only magnetic moment depends only on the number of unpaired electrons n: μ = √n(n+2) BM. \( \ce{H2O} \) and Cl^– are weak field ligands, so all four complexes are high spin.

• [Cr(H₂O)₆]²⁺: Cr (Z = 24) is 3d⁵4s¹; \( \ce{Cr^{2+}} \) is 3d⁴. Octahedral, weak field: t_2g³e_g¹, n = 4, μ = √24 = 4.90 BM.

• [Fe(H₂O)₆]²⁺: Fe (Z = 26) is 3d⁶4s²; \( \ce{Fe^{2+}} \) is 3d⁶. Octahedral, weak field: t_2g⁴e_g², n = 4, μ = 4.90 BM.

• [Mn(H₂O)₆]²⁺: Mn (Z = 25) is 3d⁵4s²; \( \ce{Mn^{2+}} \) is 3d⁵. t_2g³e_g², n = 5, μ = √35 = 5.92 BM.

• [CoCl₄]^2–: Co (Z = 27) is 3d⁷4s²; \( \ce{Co^{2+}} \) is 3d⁷. Tetrahedral: e⁴t₂³, n = 3, μ = √15 = 3.87 BM.

NCERT (2026-27 edition):
Magnetic Properties – “… For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., μ = √n(n+2), where n is the number of unpaired electrons and μ is the magnetic moment in the units of Bohr magneton (BM).”

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