The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range.
Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour.
The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`.
Q. Which of the following compounds is (are) coloured due to charge transfer spectra and not due to d-d transition?

Options

A. `KMnO_4`
B. `K_2CrO_4`
C. `CrO_3`
D. all of these

Answer

To determine which compounds are colored due to charge transfer spectra and not due to d-d transitions, we need to analyze the oxidation states and electron configurations of the transition metal ions in each compound mentioned.

### Step-by-Step Solution:

1. **Understand the Concept of d-d Transition and Charge Transfer Spectra**:
- d-d transitions occur when electrons in d-orbitals absorb energy and move to higher energy d-orbitals. This is possible only when there are unpaired electrons in the d-orbitals.
- Charge transfer spectra occur when an electron is transferred between the ligand (like oxygen) and the metal ion, leading to color even in the absence of unpaired d-electrons.

2. **Analyze KMnO4**:
- In KMnO4, manganese (Mn) is in the +7 oxidation state (Mn^7+).
- The electron configuration of Mn^7+ is 3d^0 4s^0, meaning there are no d-electrons available for d-d transitions.
- However, the color arises from charge transfer when oxygen donates an electron to manganese, resulting in Mn^6+ for a brief moment. Thus, KMnO4 is colored due to charge transfer.

3. **Analyze K2Cr2O7 (Potassium Dichromate)**:
- In K2Cr2O7, chromium (Cr) is in the +6 oxidation state (Cr^6+).
- The electron configuration of Cr^6+ is also 3d^0 4s^0, indicating no d-electrons are present.
- Similar to KMnO4, the color comes from charge transfer when oxygen donates an electron to chromium, leading to Cr^5+. Therefore, K2Cr2O7 is also colored due to charge transfer.

4. **Analyze CrO3 (Chromium Trioxide)**:
- In CrO3, chromium is again in the +6 oxidation state (Cr^6+).
- The electron configuration is the same as in K2Cr2O7, with no d-electrons.
- The color in CrO3 is also due to charge transfer from oxygen to chromium, resulting in Cr^5+. Hence, CrO3 is colored due to charge transfer.

5. **Conclusion**:
- All the compounds analyzed (KMnO4, K2Cr2O7, and CrO3) are colored due to charge transfer spectra and not due to d-d transitions.

### Final Answer:
- The correct option is "All of these" as all the mentioned compounds are colored due to charge transfer spectra.