For a d^8 ion such as Ni(II) with the same ligands, use Delta_t=tfrac49Delta_o and ignore pairing energy. By how much (in units of Delta_o) is the octahedral CFSE larger in magnitude than the tetrahedral CFSE?

JEE Advanced levelSingle correct optionDifficulty 3 of 3cc049

For a $d^{8}$ ion such as Ni(II) with the same ligands, use $Δ_{t} = \frac{4}{9} Δ_{o}$ and ignore pairing energy. By how much (in units of $Δ_{o}$ ) is the octahedral CFSE larger in magnitude than the tetrahedral CFSE?

Octahedral

d^{8}

t_{2 g}^{6} e_{g}^{2}

: CFSE

= - 2.4 + 1.2 = - 1.2 Δ_{o}

. Tetrahedral

d^{8}

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

: CFSE

= - 0.8 Δ_{t} = - 0.8 \times \frac{4}{9} Δ_{o} = - 0.356 Δ_{o}

. Difference

= 1.2 - 0.356 = 0.844 \approx 0.84 Δ_{o}

, which is why Ni(II) strongly prefers octahedral coordination.