Question:

The correct order of wavelength ($\lambda_{max}$) of the halide to metal charge-transfer band of [Co(NH$_3$)$_5$Cl]$^{2+}$ (I), [Co(NH$_3$)$_5$Br]$^{2+}$ (II) and [Co(NH$_3$)$_5$I]$^{2+}$ (III), is

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When dealing with metal-ligand charge-transfer bands, larger halide ions tend to result in absorption at longer wavelengths due to lower energy transitions.
Updated On: Dec 11, 2025
  • III < II < I
  • I < II < III
  • II < III < I
  • I < III < II
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the question.
This question concerns the charge-transfer bands in coordination complexes. The wavelength of a charge-transfer band in metal complexes can be influenced by the halide ligands (Cl$^{-}$, Br$^{-}$, and I$^{-}$).
Step 2: Analyzing the halide effect.
In general, the charge-transfer band tends to shift to longer wavelengths (redshift) as the halide ligand size increases. Iodine is the largest halide ion, followed by bromine and then chlorine. This is due to the larger atomic size of iodine, which lowers the energy required for the charge transfer and shifts the absorption to longer wavelengths.
Step 3: Conclusion.
Thus, the correct order of $\lambda_{max}$ is I (largest halide) < II < III. The correct answer is (B) I < II < III.
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