Question

Crystal field stabilization energy for low spin $d^4$ octahedral complex is:

• A. $-0.6{\Delta }_0$
• B. $-1.8{\Delta }_0$
• C. $-1.6{\Delta }_0+P$
• D. $-1.2{\Delta }_0$

Answer 1

The correct option is C $-1.6{\Delta }_0+P$

Low spin $d^4$ octahedral complex means electrons will get paired here.
Thus the configuration would be $t_{2g}^4{e}_g^0$.

$\therefore$ CFSE $=[-\frac{2}{5}\times 4\times {\Delta }_0+\frac{3}{5}\times 0\times {\Delta }_0]+P$
$=-1.6{\Delta }_0+P$
(Here P = pairing energy)