The correct order of the calculated spin-only magnetic moments of complexs (A) to (D) is:
(A) Ni(CO)₄
(B) [Ni(H₂O)₆]Cl₂
(C) Na₂[Ni(CN)₄]
(D) PdCl₂(PPh₃)₂

Solution

Chemistry ma’am · 3 steps

Count the d electrons, then ask if the ligand forces pairing: only the aqua complex keeps unpaired electrons.

1. The formula. Spin-only moment depends only on the number of unpaired electrons n: μ = √n(n+2) BM. If n = 0, then μ = 0.
2. Find n. For each complex, find the metal's oxidation state and d-electron count. Then check whether a strong-field ligand or a square planar geometry pairs the electrons up.

   Complex	Metal, d count	What happens	Unpaired, μ
   (A) Ni(CO)₄	Ni(0), 3d⁸ 4s²	Strong-field CO pushes the 4s electrons into 3d, giving 3d¹⁰. sp³, tetrahedral	0, 0 BM
   (B) [Ni(H₂O)₆]²⁺	Ni²⁺, 3d⁸	Weak-field H₂O, octahedral sp³d². The two e_g electrons stay unpaired	2, 2.83 BM
   (C) [Ni(CN)₄]²⁻	Ni²⁺, 3d⁸	Strong-field CN⁻ pairs all eight. dsp², square planar	0, 0 BM
   (D) PdCl₂(PPh₃)₂	Pd²⁺, 4d⁸	A 4d metal has a large splitting, so d⁸ Pd(II) is square planar. dsp², all paired	0, 0 BM

3. So. (A), (C) and (D) all have μ = 0, and only (B) has 2.83 BM. The order is (A) ≈ (C) ≈ (D) < (B), option (c).

The idea to keep. For d⁸ ions, strong-field ligands or 4d/5d metals mean square planar with zero unpaired electrons. Weak-field octahedral Ni²⁺ keeps 2 unpaired.

Answer from the NTA final answer key. Solution by PrepWiser.