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Question: The CFSE for \({\left[ {{\text{CoC}}{{\text{l}}_6}} \right]^{4 - }}\) complex is \(18000\,{\text{c}}...

The CFSE for [CoCl6]4{\left[ {{\text{CoC}}{{\text{l}}_6}} \right]^{4 - }} complex is 18000cm118000\,{\text{c}}{{\text{m}}^{ - 1}}. The for [CoCl4]2{\left[ {{\text{CoC}}{{\text{l}}_4}} \right]^{2 -}} will be

A. 18000cm118000\,{\text{c}}{{\text{m}}^{ - 1}}
B. 16000cm116000\,{\text{c}}{{\text{m}}^{ - 1}}
C. 6000cm16000\,{\text{c}}{{\text{m}}^{ - 1}}
D. 2000cm12000\,{\text{c}}{{\text{m}}^{ - 1}}

Explanation

Solution

The octahedral crystal field splitting is equal to the 4/94/9 of octahedral crystal field splitting. The octahedral crystal field splitting is larger than the tetrahedral crystal field splitting.
The following formula can be used-
 td = 49 oh {{\text{ }}_{{\text{td}}}}\,{\text{ = }}\,\dfrac{{\text{4}}}{{\,{\text{9}}}}{{\text{ }}_{{\text{oh}}}}

Step by step answer: he octahedral field splitting is represented as follows:
Valence electronic configuration of Co2+{\text{C}}{{\text{o}}^{2 + }} = 3d7{\text{3}}{{\text{d}}^7}

The tetrahedral field splitting is represented as follows:

The relation between the energy difference of tetrahedral and octahedral field is as follows:
 td = 49 oh{{\text{ }}_{{\text{td}}}}\,{\text{ = }}\,\dfrac{{\text{4}}}{{\,{\text{9}}}}{{\text{ }}_{{\text{oh}}}}
Where,
 td{{\text{ }}_{{\text{td}}}}is the tetrahedral field splitting energy.
oh{ _{{\text{oh}}}} is the octahedral field splitting energy.

In the octahedral complex, six ligands split the energy level of the metal and the tetrahedral complex four ligands split the energy level of the metal, so the value of octahedral crystal field splitting is larger than the tetrahedral crystal field splitting.

Substitute 18000cm118000\,{\text{c}}{{\text{m}}^{ - 1}} for oh{ _{{\text{oh}}}}.
 td = 49×18000cm1{{\text{ }}_{{\text{td}}}}\,{\text{ = }}\,\dfrac{{\text{4}}}{{\,{\text{9}}}} \times 18000\,{\text{c}}{{\text{m}}^{ - 1}}
 td = 8000cm1{{\text{ }}_{{\text{td}}}}\,{\text{ = }}\,8000\,{\text{c}}{{\text{m}}^{ - 1}}
So, the tetrahedral field splitting energy for [CoCl4]2{\left[ {{\text{CoC}}{{\text{l}}_4}} \right]^{2 - }}complex is 8000cm18000\,{\text{c}}{{\text{m}}^{ - 1}}.
So, option (A), (B) and (D) are incorrect.
Therefore, option (C) 8000cm18000\,{\text{c}}{{\text{m}}^{ - 1}} is correct.

Additional information: The oh{ _{{\text{oh}}}} per ligand is determined by dividing the oh{ _{{\text{oh}}}} by six and  td{{\text{ }}_{{\text{td}}}} per ligand is determined by dividing the  td{{\text{ }}_{{\text{td}}}} by four.

Note: Both the values for octahedral as well as tetrahedral should be in the same unit. As the octahedral crystal field splitting is always larger than the tetrahedral crystal field splitting so, here, the answer of  td{{\text{ }}_{{\text{td}}}} will be less than18000cm118000\,{\text{c}}{{\text{m}}^{ - 1}}. For the calculation of oh{ _{{\text{oh}}}} from the given  td{{\text{ }}_{{\text{td}}}}, the formula used will be  oh = 94 td\,{{\text{ }}_{{\text{oh}}}}{\text{ = }}\,\dfrac{9}{{\,4}}{{\text{ }}_{{\text{td}}}}.