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  • Crystal field stabilization energy for high spin d4 octahedral complex is :Option: 1 <img alt="-0.6\Delta _o" src="https://entrancecorner.

Crystal field stabilization energy for high spin d4 octahedral complex is :

Option: 1

-0.6\Delta _o


Option: 2

-1.8\Delta _o


Option: 3

-1.6\Delta _o\ +\ P


Option: 4

-1.2\Delta _o


Answers (1)

best_answer

CFSE =\left ( -\frac{2}{5} \times nt_2g+\frac{3}{5}n \: eg\right )\Delta_0
                 =\left ( -\frac{2}{5} \times3+\frac{3}{5}\times 1 \:\right )\Delta_0= -0.6\Delta _0

 

Crystal field stabilization energy (CFSE)=[-0.4(no.\ of\ electrons\ in\ t_2_g)+0.6(no.\ of\ electrons\ in\ e_g)]\Delta _o

for d^4 high spin complex, the electronic configuration is t_2_g^3e_g^1. Hence, the CFSE is 

[-0.4\times(3)+0.6\times(1)]\Delta _o = -0.6\Delta _o

The correct answer is option 1.

Posted by

sudhir.kumar

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