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Question

For the following two complexes
A:[NiCl6]4 and [NiCl4]2 the ratio of CFSE will be nearly:

A
32
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B
49
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C
94
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D
278
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Solution

The correct option is A 32
[NiCl6]4 octahedral complex (t2g=6,eg=4)

[NiCl4]2Tetrahedral complex (eg=4,t2g=4)

for octahedral CFSE =0.4×6+0.6×2=1.2

for tetradidral CFSE=+0.4×40.6×4=0.8

CFSE(1)CFSE(2)=1.20.8=32
Option A is correct.

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