Question

Calculate crystal field stabilization energy for the complex $\left[Cu\right(N{H}_3{)}_6{]}^{2+}$.

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

Answer 1

The correct option is B $-0.6{\Delta }_0$

The complex given here is $\left[Cu\right(N{H}_3{)}_6{]}^{2+}$.
So it is an octahedral complex.
Here, $Cu$ is $+2$ oxidation state.
So its electronic configuration is $\left[Ar\right]3d^9$
According to the crystal field spliting theory, the electronic configuration in $d$-orbital will be $t_{2g}^6{e}_g^3$
So, crystal field stabilization energy will be
$=(-0.4\times 6+0.6\times 3){\Delta }_o=(-2.4+1.8){\Delta }_o=-0.6{\Delta }_o$