Q35 / 180 · KCET 2022 · Chemistry
Crystal field splitting energy (CFSE) for $\left[\mathrm{CoCl}_6\right]^{4-}$ is $18000 \mathrm{~cm}^{-1}$. The crystal field splitting energy (CFSE) for $\left[\mathrm{CoCl}_4\right]^{2-}$ will be
- A$16000 \mathrm{~cm}^{-1}$
- Correct answerB$8000 \mathrm{~cm}^{-1}$
- C$10,000 \mathrm{~cm}^{-1}$
- D$18000 \mathrm{~cm}^{-1}$
Solution
$\left[\mathrm{CoCl}_6\right]^{4-}$ is an octahedral complex, its CFSE value, $\Delta_0=18000 \mathrm{~cm}^{-1}$
$\left[\mathrm{CoCl}_4\right]^{2-}$ is a tetrahedral complex, $\Delta_t=$ ?
We know that,
$$
\Delta_t=\frac{4}{9} \Delta_0=\frac{4}{9} \times 18000=8000 \mathrm{~cm}^{-1}
$$
Hence, the correct answer is option: 'B'.