The reaction
`5H_(2)O_(2)+XClO_(2)+2OH^(-) rarr XCl^(-)+YO_(2)+6H_(2)O`
is balanced if

Options

A. `x=5`, `y=2`
B. `x=2`, `y=5`
C. `x=4`, `y=10`
D. `x=5`, `y=5`

Explanation

To solve the given redox reaction and find the values of X and Y, we will follow these steps:

### Step 1: Write the unbalanced reaction
The unbalanced reaction is:
\[ 5H_2O_2 + XClO_2 + 2OH^- \rightarrow XCl^- + YO_2 + 6H_2O \]

### Step 2: Identify oxidation and reduction half-reactions
1. **Oxidation Half-Reaction**:
- Hydrogen peroxide (\(H_2O_2\)) is oxidized to oxygen (\(O_2\)).
- The half-reaction can be written as:
\[ H_2O_2 \rightarrow O_2 \]

2. **Reduction Half-Reaction**:
- Chlorine in \(ClO_2\) is reduced to \(Cl^-\).
- The half-reaction can be written as:
\[ ClO_2 + 4H^+ + 5e^- \rightarrow Cl^- + 2H_2O \]

### Step 3: Balance the oxidation half-reaction
- The oxidation half-reaction is:
\[ H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^- \]

### Step 4: Balance the reduction half-reaction
- The reduction half-reaction is:
\[ ClO_2 + 4H_2O + 5e^- \rightarrow Cl^- + 2H_2O + 4OH^- \]

### Step 5: Equalize the number of electrons transferred
- To balance the electrons, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
1. **Oxidation Half-Reaction** (multiplied by 5):
\[ 5H_2O_2 + 10OH^- \rightarrow 5O_2 + 10H_2O + 10e^- \]

2. **Reduction Half-Reaction** (multiplied by 2):
\[ 2ClO_2 + 8H^+ + 10e^- \rightarrow 2Cl^- + 4H_2O \]

### Step 6: Combine both half-reactions
Now we can combine both half-reactions:
\[ 5H_2O_2 + 2ClO_2 + 2OH^- \rightarrow 2Cl^- + 5O_2 + 6H_2O \]

### Step 7: Identify X and Y
From the balanced equation:
- \(X\) (coefficient of \(ClO_2\)) = 2
- \(Y\) (coefficient of \(O_2\)) = 5

### Final Answer
Thus, the values are:
- \(X = 2\)
- \(Y = 5\)