Match List-I with List-II: (A) [Cr(NH3)6]3+ (B) [NiCl4]2- (C) [CoF6]3- (D) [Ni(CN)4]2- (I) 4.90 BM (II) 3.87 BM (III)…

🧠 Compute μ for Each, Then Match

Spin-only formula: $\mu = \sqrt{n(n+2)}$ BM, where $n$ = unpaired electrons.

🗺️ Walk Each Complex

(A) $[\mathrm{Cr(NH_3)_6}]^{3+}$: Cr(III), $\mathrm{d^3}$, octahedral. With $\mathrm{NH_3}$ (mid-field), $\mathrm{d^3}$ stays $\mathrm{t_{2g}^3}$ → 3 unpaired → $\mu = \sqrt{15} \approx 3.87$ BM → (II).

(B) $[\mathrm{NiCl_4}]^{2-}$: Ni²⁺, $\mathrm{d^8}$, tetrahedral (Cl⁻ is weak; Ni²⁺ tends tetrahedral with halides). Tetrahedral d⁸: $\mathrm{e^4 t_2^4}$ → 2 unpaired → $\mu = \sqrt{8} \approx 2.83$ BM → (IV).

(C) $[\mathrm{CoF_6}]^{3-}$: Co(III), $\mathrm{d^6}$. F⁻ is weak field → high-spin: $\mathrm{t_{2g}^4 e_g^2}$ → 4 unpaired → $\mu = \sqrt{24} \approx 4.90$ BM → (I).

(D) $[\mathrm{Ni(CN)_4}]^{2-}$: Ni²⁺, $\mathrm{d^8}$. CN⁻ is strong field → forces square-planar Ni(II): all 8 d-electrons paired in lower 4 orbitals → 0 unpaired → $\mu = 0$ BM → (III).

A→II, B→IV, C→I, D→III → option (2).

⚡ The "Halide vs CN⁻ on Ni²⁺" Showdown

| Ligand | Geometry | Magnetic | |---|---|---| | Cl⁻, Br⁻, I⁻ | tetrahedral | paramagnetic (μ = 2.83) | | CN⁻, CO | square-planar | diamagnetic (μ = 0) |

Same Ni²⁺, opposite spin states — driven entirely by ligand field strength.

⚠️ F⁻ on 3d: Always High-Spin

F⁻ is weak. With $\mathrm{Co^{3+}}\,(\mathrm{d^6})$, you might be tempted by the "Co³⁺ + N-donor = low-spin" shortcut — but F⁻ is not an N-donor; it's at the bottom of the spectrochemical series. High-spin, 4 unpaired.

$\boxed{\text{Answer: (2) A-II, B-IV, C-I, D-III}}$