Calculate the spin only magnetic moment of a $Ni^{2+}$ in $[Ni(CN)_4]^{2-}$ (atomic number of Ni is 28).

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The correct answer is Option (3) → 0

Step 1: Determine the oxidation state and electron configuration

• Ni has atomic number 28 → electron configuration:

$[Ar] 3d^8 4s^2$

• Ni in $[Ni(CN)_4]^{2-}$ is $Ni^{2+}$:

$Ni^{2+} : [Ar] 3d^8$

Step 2: Determine the geometry and crystal field

• $[Ni(CN)_4]^{2-}$ is a strong field ligand complex (because CN⁻ is strong field).
• Strong field ligands cause low-spin square planar geometry for d⁸ metal ions.
• Low-spin d⁸ in square planar:

$t_{2g}^6 e_g^2$ (all electrons paired in the lower orbitals)

So, there are no unpaired electrons.

Step 3: Spin-only magnetic moment formula

The spin-only magnetic moment is:

$\mu_\text{so} = \sqrt{n(n+2)} \ \text{BM}$

Where $n$ is the number of unpaired electrons.

• Here $n=0$ (all electrons are paired).

$\mu_\text{so} = \sqrt{0(0+2)} = 0 \ \text{BM}$