The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (A) Ni(CO)₄ (B)…

🧠 Identify Each Complex's Spin State

| Complex | Metal/OS | d-count | Geometry | Unpaired | |---|---|---|---|---| | (A) $\mathrm{Ni(CO)_4}$ | Ni⁰ | d¹⁰ | Td | 0 | | (B) $[\mathrm{Ni(H_2O)_6}]\mathrm{Cl_2}$ | Ni²⁺ | d⁸ | Octahedral | 2 | | (C) $\mathrm{Na_2[Ni(CN)_4]}$ | Ni²⁺ | d⁸ | sq. planar | 0 | | (D) $\mathrm{PdCl_2(PPh_3)_2}$ | Pd²⁺ | d⁸ | sq. planar | 0 |

Only (B) is paramagnetic; A, C, D are all diamagnetic.

🗺️ Order

A ≈ C ≈ D (all 0) < B (μ ≈ 2.83 BM).

⚡ The d⁸ Geometry Decision

Ni²⁺ d⁸ has two stable geometries with very different magnetism:

• Octahedral with weak ligand (H₂O, NH₃ etc.): $\mathrm{t_{2g}^6 e_g^2}$ → 2 unpaired, paramagnetic.
• Square planar with strong ligand (CN⁻, dmg, etc.): all 8 paired in dxy, dxz, dyz, dz² → diamagnetic.

Pd²⁺ and Pt²⁺ d⁸ default to square planar regardless of ligand (large Δ for 4d/5d).

⚠️ Ni(CO)₄ ≠ Ni²⁺

In $\mathrm{Ni(CO)_4}$, Ni is zerovalent (CO is neutral). Ni⁰ is d¹⁰ — automatically diamagnetic regardless of geometry.

$\boxed{\text{Answer: (4) (A) ≈ (C) ≈ (D) < (B)}}$