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 option does not change the color of acidified \( K_2Cr_2O_7 \), we need to analyze the oxidation states of the compounds provided in the options and their ability to act as reducing agents.

### Step-by-Step Solution:

1. **Understanding the Color of \( K_2Cr_2O_7 \)**:
- \( K_2Cr_2O_7 \) (potassium dichromate) is orange in color when in acidic solution due to the presence of \( Cr^{6+} \) ions.

2. **Identifying the Reaction**:
- \( K_2Cr_2O_7 \) acts as an oxidizing agent. When it is reduced, it changes color (typically to green due to \( Cr^{3+} \)). Therefore, we need to find a substance that does not change the oxidation state of chromium in \( K_2Cr_2O_7 \).

3. **Analyzing Each Option**:
- **Option 1: \( H_2O_2 \)** (Hydrogen peroxide)
- \( H_2O_2 \) can act as a reducing agent, reducing \( Cr^{6+} \) to \( Cr^{3+} \) and thus changing the color of \( K_2Cr_2O_7 \).
- **Conclusion**: This option changes the color.

- **Option 2: \( SnO_2 \)** (Tin(IV) oxide)
- \( SnO_2 \) can also act as a reducing agent, reducing \( Cr^{6+} \) to \( Cr^{3+} \).
- **Conclusion**: This option changes the color.

- **Option 3: \( HF \)** (Hydrofluoric acid)
- Fluorine has a high electronegativity and does not change its oxidation state when interacting with \( K_2Cr_2O_7 \). Therefore, it does not reduce \( Cr^{6+} \).
- **Conclusion**: This option does not change the color.

- **Option 4: \( HBr \)** (Hydrobromic acid)
- \( HBr \) can act as a reducing agent as bromine can be oxidized from -1 to higher oxidation states. Thus, it can reduce \( Cr^{6+} \) to \( Cr^{3+} \).
- **Conclusion**: This option changes the color.

4. **Final Answer**:
- The color of acidified \( K_2Cr_2O_7 \) is not changed by **Option 3: \( HF \)**.