Question

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

• A. $-1.8{\Delta }_0$
• B. $-1.6{\Delta }_0+P$
• C. $-1.2{\Delta }_0$
• D. $-0.6{\Delta }_0$

Answer 1

Answer: (D)

In case of high spin complex, ${\Delta }_0$ is small. Thus, the energy required to pair up the fourth electron with the electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital. Thus, pairing does not occur.
For high spin $d^4$ octahedral complex,

$=(-3\times 0.4+1\times 0.6){\Delta }_0$
$=(-1.2+0.6){\Delta }_0=-0.6{\Delta }_0$