JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Coordination Compounds

A high-spin $d^6$ ion in an octahedral field takes the configuration $t_{2g}^4{e}_g^2$, so $CFSE=4(-0.4{\Delta }_o)+2(+0.6{\Delta }_o)=-0.4{\Delta }_o$. In a tetrahedral field the high-spin $d^6$ arrangement is $e^3{t}_2^3$, so $CFSE=3(-0.6{\Delta }_t)+3(+0.4{\Delta }_t)=-0.6{\Delta }_t$. Quoting the tetrahedral value as $-0.27$ mixes the two scales, because $-0.6{\Delta }_t$ equals $-0.27{\Delta }_o$ only after converting with ${\Delta }_t=\frac{4}{9}{\Delta }_o$.