Question

Calculate CFSE of the following complexes :

$\left[Fe\right(CN{)}_6{]}^{4-}$

• A. $-0.4{\Delta }_t$
• B. $-2.4{\Delta }_0$
• C. $0.4{\Delta }_0$
• D. $0.6{\Delta }_0$

Answer 1

Answer: (B)

$-2.4{\Delta }_0$

Option (B) is correct.

In $\left[Fe\right(CN{)}_6{]}^{4-}$, iron has $3d^6,4s^2$ system in ground state but in excited state it loses two electrons in the formation of ions and two electrons from $4s$, so thus Cobalt gets $3d^6$ configuration. Now it is of low spin complex due to $CN$ ligands so all 6 electrons will go to $t_{2g}$ orbitals. and 0 electrons will be in $e_g$ orbital. By applying formula,

$\Delta$ = no. of electrons in $t_{2g}$⋅(−0.4)+ no. of electrons in $e_g$(0.6)

= $6(-0.4)+0\left(0.6\right)$

= $-2.4{\Delta }_0$