Question: For a $\mathrm{d}^{4}$ metal ion in an octahedral field, the correct electronic configuration is :
$\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{0}$ when $\Delta_{\mathrm{O}}<\mathrm{P}$
$\mathrm{e}_{\mathrm{g}}^{2} \mathrm{t}_{2 \mathrm{~g}}^{2}$ when $\Delta_{\mathrm{O}}<\mathrm{P}$
$\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}^{1}$ when $\Delta_{\mathrm{O}}<\mathrm{P}$
$\mathrm{t}_{2 \mathrm{~g}}^{3} \mathrm{e}_{\mathrm{g}}^{1}$ when $\Delta_{\mathrm{O}}>\mathrm{P}$
Correct Option: , 3
Solution:
back pairing is not possible because pairing energy
$>\Delta_{\mathrm{O}}$
