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Le Chatelier's Principle for Del N Equals to 0

[CoCl4]2- blu...

Question

$[C o C l_{4}]^{2 -}$ (blue) and $[C o (H_{2} O)_{6}]^{2 +}$ (pink) have different colours because of $Δ_{t e t} < Δ_{o c t}$ .
If true enter 1, else enter 0.

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Solution

The correct answer is 1.
$[C o C l_{4}]^{2 -}$ (blue) and $[C o (H_{2} O)_{6}]^{2 +}$ (pink) have different colours because of $Δ_{t e t} < Δ_{o c t}$ .
The $[C o (H_{2} O)_{6}]^{2 +}$ complex is pink due to $d - d$ transition, $Δ_{o c t}$ absorbs blue colour and the complimentary color red/pink is seen( weak absorbance).
The $[C o C l_{4}]^{2 -}$ complex is a deeper blue as $d - d$ transition requires smaller energy. $Δ_{t e t}$ absorbs red (more strongly) and it appears blue.

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