`2KMnO_(4) +3H_(2)SO_(4) rarr K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[O]`
`X+O rarr Y`
X can't be:

Options

A. CO
B. `SO_(4)^(2-)`
C. `SO_(3)^(2-)`
D. `C_(2)O_(4)^(2-)`

Explanation

To solve the given question, we need to identify which compound X cannot be when it is oxidized by the oxidizing agent KMnO4 in the presence of H2SO4.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction provided is:
\[ 2KMnO_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 3H_2O + 5[O] \]
Here, KMnO4 acts as an oxidizing agent.

2. **Identifying Oxidation**:
Oxidation involves an increase in the oxidation state of a substance. In this case, we need to determine the oxidation states of the elements in the potential compounds that could be X.

3. **Analyzing Possible Candidates for X**:
We need to check various candidates for X to see which one cannot be oxidized.

- **Candidate 1: \( SO_4^{2-} \)** (Sulfate Ion)
- The oxidation state of sulfur (S) in \( SO_4^{2-} \):
\[ \text{Let oxidation state of S be } x. \]
\[ x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6 \]
- Since sulfur is in its highest oxidation state (+6), it cannot be oxidized further. Therefore, \( SO_4^{2-} \) cannot be X.

- **Candidate 2: \( SO_3^{2-} \)** (Sulfite Ion)
- The oxidation state of sulfur (S) in \( SO_3^{2-} \):
\[ \text{Let oxidation state of S be } x. \]
\[ x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4 \]
- The oxidation state of sulfur is +4, which can be oxidized to +5 or +6. Therefore, \( SO_3^{2-} \) can be X.

4. **Conclusion**:
Based on the analysis, the only candidate that cannot be oxidized by KMnO4 is \( SO_4^{2-} \). Thus, the answer is:
\[ \text{X cannot be: } SO_4^{2-} \]