KCET · Chemistry · Coordination Compounds
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 \(16000 \mathrm{~cm}^{-1}\)
- B \(8000 \mathrm{~cm}^{-1}\)
- C \(10,000 \mathrm{~cm}^{-1}\)
- D \(18000 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(B) \(8000 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{CoCl}_6\right]^{4-}\) 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,
\[
\Delta_t=\frac{4}{9} \Delta_0=\frac{4}{9} \times 18000=8000 \mathrm{~cm}^{-1}
\]
\(\left[\mathrm{CoCl}_4\right]^{2-}\) is a tetrahedral complex, \(\Delta_t=\) ?
We know that,
\[
\Delta_t=\frac{4}{9} \Delta_0=\frac{4}{9} \times 18000=8000 \mathrm{~cm}^{-1}
\]
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