Which of the following can be oxidised by acidified `K_2Cr_2O_7` ?

Options

A. Nitrate ions
B. Sulphite ions
C. Ferric ions
D. Borate ions

Answer

To determine which of the given ions can be oxidized by acidified \( K_2Cr_2O_7 \), we need to analyze each option provided: nitrate ion, sulfite ion, ferric ions, and borate ions.

### Step-by-Step Solution:

1. **Understanding the Oxidizing Agent**:
- Acidified \( K_2Cr_2O_7 \) (potassium dichromate) is a strong oxidizing agent. In acidic medium, it can oxidize various substances.

2. **Analyzing Each Ion**:
- **Nitrate Ion (\( NO_3^- \))**:
- Nitrate is already in a high oxidation state (+5). It is generally stable and does not get oxidized further.
- **Sulfite Ion (\( SO_3^{2-} \))**:
- The oxidation state of sulfur in sulfite is +4. When oxidized, it can convert to sulfate (\( SO_4^{2-} \)), where sulfur has an oxidation state of +6. This indicates that sulfite can be oxidized.
- **Ferric Ion (\( Fe^{3+} \))**:
- Ferric ion is already in a +3 oxidation state. It does not get oxidized further as it is already in a high oxidation state.
- **Borate Ion (\( BO_3^{3-} \))**:
- Borate does not participate in redox reactions in the same way as the other ions listed. It is stable and does not get oxidized by \( K_2Cr_2O_7 \).

3. **Conclusion**:
- Among the given options, only the sulfite ion can be oxidized by acidified \( K_2Cr_2O_7 \) to sulfate ion.

### Final Answer:
- The correct answer is **sulfite ion**.