The CFSE for octahedral $[C o C l_{6}]^{4 -}$ is $18, 000 c m^{- 1}$ .
The CFSE for tetrahedral $[C o C l_{4}]^{2 -}$ will be

18, 000 c m^{- 1}

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16, 000 c m^{- 1}

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8, 000 c m^{- 1}

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20, 000 c m^{- 1}

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Solution

The correct option is C $8, 000 c m^{- 1}$
The crystal field splitting in the tetrahedral field is intrinsically smaller than in the octahedral field. For most purposes the relationship may be represented as $Δ_{t} = \frac{4}{9} Δ_{o}$
$Δ_{t} = \frac{4}{9} (18000 c m^{- 1}) = 8000 c m^{- 1}$

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Similar questions

The CFSE for octahedral $[C o C l_{6}]^{4 -}$ is 18000 $c m^{- 1}$ . The CFSE for tetrahedral $[C o C l_{4}]^{2 -}$ will be

[C o C l_{6}]^{4 -}

18, 000 c m^{- 1}

[C o C l_{4}]^{2 -}

The CFSE for octahedral $[{CoCI}_{6}]^{4 -} is 18, 000 {cm}^{- 1}$ The CFSE for tetrahedral $[{CoCI}_{4}]^{2 -}$ will be

C F S E

] (C o C l)_{6}]^{4 -}

18000 c m^{- 1}

△

[C o C l_{4}]^{2 -}

[C o C l_{6}]^{4 -}

c m^{- 1}

Δ

[C o C l_{4}]^{2 -}

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