1. Find the oxidation state of Ni in $[NiCl_4]^{2-}$

   Let the oxidation state of Ni be $x$. Since each $Cl^-$ has charge $-1$, $x + 4(-1) = -2$ $x - 4 = -2$ $x = +2$

   So, the metal ion is $Ni^{2+}$.

2. Write the electronic configuration of $Ni^{2+}$

   Atomic number of Ni = 28.

   Neutral Ni: $[Ar]3d^8 4s^2$

   For $Ni^{2+}$, remove two electrons from $4s$ first: $Ni^{2+} = [Ar]3d^8$

3. Determine the geometry and pairing

   In $[NiCl_4]^{2-}$, $Cl^-$ is a weak field ligand. For $Ni^{2+}(3d^8)$ with weak field ligands, the complex is typically tetrahedral.

   In a tetrahedral field, the crystal field splitting is small, so electrons remain unpaired as much as possible.

   For tetrahedral $d^8$, the number of unpaired electrons is: $n = 2$

4. Use the spin-only magnetic moment formula

   $\mu = \sqrt{n(n+2)}\, \text{BM}$

   Substituting $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{8}$ $\mu = 2.828 \approx 2.82\, \text{BM}$

5. Check options

   • A: $5.46$ BM ❌
   • B: $2.82$ BM ✅
   • C: $1.41$ BM ❌
   • D: $1.82$ BM ❌

Therefore, the correct answer is B.