Question

If ${△}_0<P,$ where $P$ is pairing energy and ${△}_0$ is the crystal field spliting energy, the correct electronic configuration for $d^4$ system will be:

• A. $t_{2g}^4{e}_g^0$
• B. $t_{2g}^3{e}_g^1$
• C. $t_{2g}^0{t}_g^4$
• D. $t_{2g}^2{e}_g^2$

Answer 1

The correct option is B $t_{2g}^3{e}_g^1$

If ${△}_0$ is lower than pairing energy therefore, pairing will not take place and system will have $t_{2g}^3{e}_g^1$ electronic configuration.