question_answer

Crystal field stabilization energy for high spin \[{{d}^{4}}\] octahedral complex is

A) \[-1.8{{\Delta }_{o}}\]                   

B) \[-1.6{{\Delta }_{o}}+P\]

C) \[-1.2{{\Delta }_{o}}\]   

D)        \[-0.6{{\Delta }_{o}}\]

Correct Answer: D

Solution :

Key Idea In case of high spin complex, \[{{\Delta }_{o}}\] 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, \[\therefore \]Crystal field stabilization energy                 \[=(-3\times 0.4+1\times 0.6){{\Delta }_{o}}\]                 \[=(-1.2+0.6)\,{{\Delta }_{o}}\]                 \[=-0.6\,{{\Delta }_{o}}\]