question_answer

Crystal field stabilization 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)        \[-\,0.6\,{{\Delta }_{0}}\]

Correct Answer: D

Solution :

\[{{\operatorname{d}}^{4}}\] in high spin octahedral complex \[\operatorname{CFSE}\,\,\left( -\,0.4\,x\,\,+\,\,0.6\,y \right){{\Delta }_{0}}\] Where, \[\operatorname{x}\to \,\,electrons\,\,in\,\,{{t}_{2}}_{g}\,orbital\] \[\operatorname{y}\to  electrons in {{e}_{g}} orbital\,\] \[\operatorname{CFSE} = \left[ 0.6 \times  1 \right] + \left[ -\,0.4 \times  3 \right] = - 0.6aq\]