Question

Crystal field stabilisation energy for high spin $d^4$ octahedral complex is

• A. $-1.8{\Delta }_o+P$
• B. $-1.8{\Delta }_o$
• C. $-1.2{\Delta }_o$
• D. $-0.6{\Delta }_o$

Answer 1

The correct option is D $-0.6{\Delta }_o$

In case of high spin complex, ${\Delta }_o$ is small. Thus, the energy required to pair up the fourth electron with the electrons of lower energy $d-$orbitals would be higher than required to place the electrons in the higher $d-$orbital. Thus, pairing does not occur.
For high spin $d^4$ octahedral complex Crystal field stabilisation energy
$=(3\times -0.4+1\times +0.6){\Delta }_o$
$=(-1.2+0.6){\Delta }_o=-0.6{\Delta }_o$