The magnetic moment (spin only) of $$[\text{NiCl}_4]^{2-}$$ is:

Oxidation state of Ni in $$[\text{NiCl}_4]^{2-}$$:

$$\text{Ni} + 4(-1) = -2 \;\;\Rightarrow\;\; \text{Ni}^{2+}$$

Electronic configuration of the free ion $$\text{Ni}^{2+}$$ is $$[Ar]\,3d^{8}$$.

Chloride is a weak-field ligand and the complex is tetrahedral (hybridisation $$sp^{3}$$).
For a tetrahedral field the crystal-field splitting $$\Delta_t$$ is small, so electrons do not pair up beyond Hund’s rule.

Splitting pattern in a tetrahedral field: the lower set is $$e$$ (2 orbitals) and the upper set is $$t_2$$ (3 orbitals).

Filling $$3d^{8}$$ electrons:

  • $$e$$ set (2 orbitals): accommodates 4 electrons ⇒ both orbitals doubly occupied (no unpaired e⁻).
  • $$t_2$$ set (3 orbitals): accommodates the remaining 4 electrons.
- First three electrons occupy the three orbitals singly (↑ ↑ ↑).
    - The fourth electron pairs up in one of them (↑↓).
Thus, electrons in $$t_2$$ are distributed as ↑↓ ↑ ↑ giving 2 unpaired electrons overall.

Number of unpaired electrons, $$n = 2$$.

Spin-only magnetic moment formula: $$\mu = \sqrt{n(n+2)} \,\text{BM}$$.

$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \,\text{BM}$$

Therefore, the magnetic moment of $$[\text{NiCl}_4]^{2-}$$ is $$2.83 \,\text{BM}$$.

Option B which is: $$2.83 \, \text{BM}$$