Using CFSE_tet=(-0.6 n_e+0.4 n_t_2)Delta_t and ignoring pairing energy, the CFSE of the tetrahedral complex [NiCl4]^2- is

JEE Main levelSingle correct optionDifficulty 2 of 3cc028

Using CFSE $_{t e t} = (- 0.6 n_{e} + 0.4 n_{t_{2}}) Δ_{t}$ and ignoring pairing energy, the CFSE of the tetrahedral complex $[NiCl X_{4}] X^{2 -}$ is

Ni(II) is

d^{8}

, and tetrahedral

d^{8}

e^{4} t_{2}^{4}

. CFSE

= - 0.6 \times 4 + 0.4 \times 4 = - 2.4 + 1.6 = - 0.8 Δ_{t}