The crystal field stabilization energy for a | vedclass

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Question

(B) In a high spin octahedral complex,the crystal field splitting energy $\Delta_o$ is small. Therefore,the energy required to pair the fourth electro

Vedclass · Accepted Answer

$- 0.6 \Delta_o$

Vedclass · Answer

(B) In a high spin octahedral complex,the crystal field splitting energy $\Delta_o$ is small. Therefore,the energy required to pair the fourth electron in the lower energy $t_{2g}$ orbitals is greater than the energy required to place it in the higher energy $e_g$ orbital. Thus,the electronic configuration for a high spin $d^4$ complex is $t_{2g}^3 e_g^1$. The crystal field stabilization energy $(CFSE)$ is calculated as: $	ext{CFSE} = (3 	imes -0.4 \Delta_o) + (1 	imes 0.6 \Delta_o)$ $= (-1.2 + 0.6) \Delta_o$ $= -0.6 \Delta_o$

The crystal field stabilization energy for a high spin $d^4$ octahedral complex is:

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