Which compound is coloured due to charge transfer spectra and not due to d–d transitions?

All of these

The correct answer is option 4. all of these.

Charge Transfer vs. d-d Transitions

To explain why the compounds \( \text{K}_2\text{Cr}_2\text{O}_7 \), \( \text{KMnO}_4 \), and \( \text{CrO}_3 \) are colored due to charge transfer rather than d–d transitions, we need to understand the difference between these two types of electronic transitions.

d–d Transitions:

Occur within the d orbitals of transition metal ions. Require unpaired electrons in the d orbitals. Typical for complexes where the metal ion has a partially filled d subshell. Common in transition metal complexes with lower oxidation states.

Charge Transfer Transitions:

Involve the transfer of an electron from a ligand to a metal (ligand-to-metal charge transfer, LMCT) or from a metal to a ligand (metal-to-ligand charge transfer, MLCT). Typically occur in compounds where the metal is in a high oxidation state. Result in intense colors compared to d–d transitions because they have higher molar absorptivities.

Analysis of the Given Compounds

1. \( \text{K}_2\text{Cr}_2\text{O}_7 \) (Potassium dichromate):

Chromium is in the +6 oxidation state. The \( \text{Cr}_2\text{O}_7^{2-} \) ion exhibits color primarily due to LMCT transitions, where electrons are transferred from the oxygen atoms (ligands) to the empty d orbitals of the chromium atoms. There are no unpaired d electrons in \( \text{Cr}^{6+} \), so d–d transitions are not possible.

2. \( \text{KMnO}_4 \) (Potassium permanganate):

Manganese is in the +7 oxidation state. The \( \text{MnO}_4^- \) ion shows color due to LMCT transitions, where electrons are transferred from the oxygen atoms to the empty d orbitals of manganese. Like chromium in \( \text{K}_2\text{Cr}_2\text{O}_7 \), manganese in \( \text{KMnO}_4 \) has no unpaired d electrons, so d–d transitions are not possible.

3. \( \text{CrO}_3 \) (Chromium trioxide): Chromium is in the +6 oxidation state. The color is due to charge transfer transitions, as there are no unpaired d electrons in \( \text{Cr}^{6+} \).

Conclusion

All three compounds are colored due to charge transfer transitions because:

They all contain transition metals in high oxidation states (Cr in +6, Mn in +7). They have no unpaired d electrons, which rules out d–d transitions. The intense colors are a result of charge transfer transitions, which involve significant electron movement between the ligands and the metal ions, creating strong absorption bands in the visible region.

Therefore, the correct answer is: 4. All of these.