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)

$-0.6\Delta_0$

Answer 2

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 Crystal field stabilisation energy
$= (- 3 \times 0.4 + 1 \times 0.6) \Delta_0$
$= (-1.2+ 0.6) \Delta_0 = - 0.6 \Delta_0$