Question

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

• A. $-1.2{\Delta }_o$
• B. $-0.6{\Delta }_o$
• C. $-0.8{\Delta }_o$
• D. $-1.6{\Delta }_o$

Answer 1

Answer: (B)

Step 1:

We know that for high spin, $d^4=t_{2g}^3{e}_g^1$
$CFSE=(–0.4x+0.6y){\Delta }_o$
where, $x=$ Number of electrons occupying $t_{2g}$ orbitals
$y=$ Number of electrons occupying $e_g$ orbitals

Step 2:
$\therefore CFSE=(–0.4\times 3+0.6\times 1){\Delta }_o$
$=(–1.2+0.6){\Delta }_o=–0.6{\Delta }_o$