JEE Main 2020 — Coordination Compounds Question with Solution

Since spin only magnetic moment is 4.90 BM so number of unpaired electrons must be 4. There are two possibilities:

(1) If the complex is octahedral, then it has to be high spin complex with configuration ${{\mathrm{t}}_{2\mathrm{g}}}^{2,1,1}{\mathrm{e}}_{\mathrm{g}}^{1,1},$ in that case $\mathrm{CFSE}=4\times \left(-0.4{\Delta }_0\right)+2\times 0.6{\Delta }_0=-0.4{\Delta }_0$
(2) If the complex is tetrahedral then its electronic configuration will be $=e_g{{}^{2,1}\mathrm{t}_{2g}}^{1,1,1}$ and $\mathrm{CFSE}$ will $\mathrm{be}=3\mathrm{X}$ $\left(-0.6{\Delta }_t\right)+3X\left(0.4{\Delta }_t\right)=-0.6\Delta t$