Question

Crystal field stabilization energy for $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ octahedral complex excluding pairing energy is $x{\Delta }_o$. Then $\mid x\mid$ is:

Answer 1

In $\left[Co\right(N{H}_3{)}_6{]}^{3+}$, $Co$ is in +3 oxidation state. So due to higher oxidation state of $Co$, though $N{H}_3$ is not a strong field ligand but still electron will get paired up. So here the electronic configuration of $C{o}^{3+}$ would be $t_{2g}^6{e}_g^0$.
So the CFSE of the complex excluding pairing energy = $6\times -0.4{\Delta }_o=-2.4{\Delta }_o$
$\therefore \mid x\mid =2.4{\Delta }_o$