Question

CFSE of high spin $d^5$ complex of $F{e}^{3+}$ in KJ is :

• A. 0
• B. 5
• C. 10
• D. None of these

Answer 1

The correct option is A 0

$F{e}^{3+}$ configuration- $\left[Ar\right]3d^5$

Since the compound is high spin $t_{2g}$ level will have 3 electrons and

$e_g$ level will have 2 electrons.

$CFSE$ = $\left[\right(-0.4)3+(0.6\left)2\right]$ ${\Delta }_o$

= $-1.2+1.2$

= $0$