Colour of acidified `K_(2)Cr_(2)O_(7)` is not changed by:

Options

A. `H_(2)O_(2)`
B. `Sn^(2+)(aq.)`
C. `HF`
D. `HB r`

Answer

To determine which substance does not change the color of acidified potassium dichromate (K₂Cr₂O₇), we need to analyze the properties of the given options in relation to their ability to act as reducing agents.

### Step-by-Step Solution:

1. **Understanding Potassium Dichromate**:
- Potassium dichromate is a strong oxidizing agent. Its color is orange due to the presence of Cr in the +6 oxidation state.

2. **Effect of Reducing Agents**:
- When a reducing agent is added to K₂Cr₂O₇, it reduces Cr from the +6 state to a lower oxidation state (usually +3), which changes the color from orange to green.

3. **Analyzing the Options**:
- We need to identify which of the given options can act as a reducing agent. A reducing agent is a substance that can be oxidized (increase its oxidation state) while reducing another substance.

4. **Evaluating Each Option**:
- **H₂O₂ (Hydrogen Peroxide)**:
- It can be oxidized (the oxidation state of oxygen can increase), thus it can act as a reducing agent and change the color of K₂Cr₂O₇.
- **Sn²⁺ (Tin(II) Ion)**:
- It can be oxidized to Sn⁴⁺, thus it can also act as a reducing agent and change the color of K₂Cr₂O₇.
- **HF (Hydrofluoric Acid)**:
- Fluorine in HF is in the -1 oxidation state and does not undergo oxidation. Therefore, HF cannot act as a reducing agent and will not change the color of K₂Cr₂O₇.
- **HBr (Hydrobromic Acid)**:
- Bromide ions (Br⁻) can be oxidized to Br₂ (0 oxidation state), thus HBr can act as a reducing agent and change the color of K₂Cr₂O₇.

5. **Conclusion**:
- The only substance that does not change the color of acidified K₂Cr₂O₇ is **HF**.

### Final Answer:
The color of acidified K₂Cr₂O₇ is not changed by **HF**.

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