Question

The spin only magnetic moment of $[NiCl_4]^{2-}$ is_ _ _ _ (Nearest integer).

Hint:

Tetrahedral complexes are always high spin due to small crystal field splitting. For \(d^8\) tetrahedral, unpaired electrons = 2.

Answer 1

Correct Answer: 3

Concept: Spin-only magnetic moment: \[ \mu = \sqrt{n(n+2)} \text{ BM} \] where \(n\) = number of unpaired electrons.

Step 1: Oxidation state of Ni. Let oxidation state of Ni be \(x\): \[ x + 4(-1) = -2 \Rightarrow x - 4 = -2 \Rightarrow x = +2 \]

Step 2: Electronic configuration of Ni\(^{2+}\). Ni (Z = 28): \([Ar]\,3d^8 4s^2\) Ni\(^{2+}\): \([Ar]\,3d^8\)

Step 3: Geometry and ligand field. Cl\(^-\) is a weak field ligand. \([NiCl_4]^{2-}\) is tetrahedral (Ni\(^{2+}\) with weak field ligands forms tetrahedral complexes). Tetrahedral complexes are always high spin.

Step 4: Crystal field splitting for tetrahedral. For tetrahedral \(d^8\): \(e^4\, t_2^4\) - \(e\) orbital (lower energy): 4 electrons → completely paired (2 pairs) - \(t_2\) orbital (higher energy): 4 electrons → 2 pairs Number of unpaired electrons: \[ n = 2 \]

Step 5: Magnetic moment. \[ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.828 \text{ BM} \] Nearest integer = 3.