Match List-I with List-II: (A) [Cr(CN)6]3- (B) [Fe(H2O)6]2+ (C) [Co(NH3)6]3+ (D) [Ni(NH3)6]2+ Number of unpaired…

🧠 Walk Each Complex, Count Unpaired

🗺️ One at a Time

(A) $[\mathrm{Cr(CN)_6}]^{3-}$: Cr(III), $\mathrm{d^3}$. With strong CN⁻ — but $\mathrm{d^3}$ has the same configuration ($\mathrm{t_{2g}^3}$) high-spin or low-spin → 3 unpaired → (II).

(B) $[\mathrm{Fe(H_2O)_6}]^{2+}$: Fe(II), $\mathrm{d^6}$. H₂O is weak → high-spin: $\mathrm{t_{2g}^4 e_g^2}$ → 4 unpaired → (IV).

(C) $[\mathrm{Co(NH_3)_6}]^{3+}$: Co(III), $\mathrm{d^6}$ + N-donor → low-spin: $\mathrm{t_{2g}^6}$ → 0 unpaired → (I).

(D) $[\mathrm{Ni(NH_3)_6}]^{2+}$: Ni(II), $\mathrm{d^8}$ → $\mathrm{t_{2g}^6 e_g^2}$ → 2 unpaired → (III).

A→II, B→IV, C→I, D→III → option (1).

⚡ The "Co(III) + N-Donor = 0 Unpaired" Anchor

Repeated theme in JEE: $[\mathrm{Co(NH_3)_6}]^{3+}, [\mathrm{Co(en)_3}]^{3+}, [\mathrm{Co(CN)_6}]^{3-}$ — all low-spin Co(III), all 0 unpaired.

⚠️ d³ Doesn't Care About Field Strength

For Cr(III) $\mathrm{d^3}$, the configuration $\mathrm{t_{2g}^3 e_g^0}$ is the same in low-spin and high-spin scenarios. So the unpaired count (3) is field-independent. The CN⁻ here doesn't change anything.

$\boxed{\text{Answer: (1) A-II, B-IV, C-I, D-III}}$