In the reaction
`HAsO_(2)+Sn^(2+) rarr As+Sn^(4+)+H_(2)O` oxidising agent is

Options

A. `Sn^(2+)`
B. `Sn^(4+)`
C. `As`
D. `HAsO_(2)`

Explanation

To determine the oxidizing agent in the reaction

\[ \text{HAsO}_2 + \text{Sn}^{2+} \rightarrow \text{As} + \text{Sn}^{4+} + \text{H}_2\text{O} \]

we need to analyze the oxidation states of the elements involved in the reaction.

### Step-by-Step Solution:

1. **Identify the oxidation states of the reactants and products**:
- In HAsO2, arsenic (As) has an oxidation state of +3.
- In Sn²⁺, tin (Sn) has an oxidation state of +2.
- In the product, elemental arsenic (As) has an oxidation state of 0.
- In Sn⁴⁺, tin (Sn) has an oxidation state of +4.

2. **Determine the changes in oxidation states**:
- Arsenic (As) changes from +3 in HAsO2 to 0 in the product As. This indicates a reduction (gain of electrons).
- Tin (Sn) changes from +2 in Sn²⁺ to +4 in Sn⁴⁺. This indicates an oxidation (loss of electrons).

3. **Identify the roles of the substances**:
- The substance that is reduced (gains electrons) is the oxidizing agent. In this case, HAsO2 is reduced as arsenic goes from +3 to 0.
- The substance that is oxidized (loses electrons) is the reducing agent. Here, Sn²⁺ is oxidized as it goes from +2 to +4.

4. **Conclusion**:
- Since HAsO2 is reduced, it acts as the oxidizing agent in this reaction.

Thus, the oxidizing agent in the reaction is **HAsO2**.