Ncert Exercises & Solutions

The CFSE for octahedral ${[C o C l_{6}]}^{4 -}$ is 18,000 cm $^{- 1}$ . The CFSE for tetrahedral will be $\mathrm{CoCl}_4{ }^{2-}$: $[]$ $\mathrm{CoCl}_4{ }^{2-}$

1. 6000 cm $^{- 1}$

2. 16,000 cm $^{- 1}$

3. 18,000 cm $^{- 1}$

4. 8,000 cm $^{- 1}$

Hint: $\Delta_{t} =\frac{4}{9}\Delta_{o} $

The relation between $ \Delta_{t}$ and $\Delta_{o} $ is $\Delta_{t} =\frac{4}{9}\Delta_{o} $

Where,

∆_{0}

CFSE for octahedral complex,

∆_{t}

= CFSE for tetrahedral complex

According to question, $∆_{0} = 18, 000 c m^{- 1}$
$∆_{t} = \frac{4}{9} ∆_{0} = \frac{4}{9} \times 18, 000 c m^{- 1}$
$= 4 \times 2, 000 c m^{- 1} = 8, 000 c m^{- 1}$

Hence , correct choice is option fourth

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