Consider the following unbalanced redox reaction:
`H_(2)O+AX+BYrarrHA+OY+X_(2)B`
The oxidation number of `X` is `-2` and niether `X` nor water is involved in the redox process.
The elements(s) undergoing oxidation is`//`are

Options

A. `A`
B. `B`
C. `Y`
D. `B` or `Y` or both

Explanation

To solve the given unbalanced redox reaction:

**Reaction:**
\[ H_2O + AX + BY \rightarrow HA + OY + X_2B \]

**Step 1: Identify the oxidation states of the elements in the reaction.**
- The oxidation number of \( X \) is given as \(-2\).
- For \( A \) in \( AX \), we can assume its oxidation state is \( +2 \) since \( X \) is \(-2\) (to balance the overall charge to zero).
- For \( Y \) in \( BY \), let’s denote its oxidation state as \( y \). The overall charge must be zero, so we can set up the equation:
\[ -2 \times 2 + y = 0 \]
\[ -4 + y = 0 \]
\[ y = +4 \]
Thus, the oxidation state of \( B \) is \( +4 \).

**Step 2: Determine the oxidation states in the products.**
- In \( HA \), since \( H \) is \( +1 \) and \( A \) is \( -1 \) (to balance the charge), the oxidation state of \( A \) in the product is \( -1 \).
- In \( OY \), since \( O \) is typically \(-2\) and \( Y \) is \( +2 \) (to balance the charge), the oxidation state of \( Y \) is \( +2 \).
- In \( X_2B \), since \( X \) is \(-2\), the oxidation state of \( B \) remains \( +4 \).

**Step 3: Compare the oxidation states to identify oxidation and reduction.**
- For \( A \):
- Reactant side: \( +2 \)
- Product side: \( -1 \)
This indicates a reduction (decrease in oxidation state).

- For \( B \):
- Reactant side: \( 0 \)
- Product side: \( +4 \)
This indicates oxidation (increase in oxidation state).

- For \( Y \):
- Reactant side: \( 0 \)
- Product side: \( +2 \)
This also indicates oxidation (increase in oxidation state).

**Conclusion:**
The elements undergoing oxidation in this reaction are \( B \) and \( Y \).

**Final Answer:** The elements undergoing oxidation are \( B \) and \( Y \).

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