On addition of excess of sodium hydroxide solution to stannous chloride solution, we obtain

Options

A. `Sn(OH)_(2)`
B. `SnO_(2).H_(2)O`
C. `Na_(2)SnO_(2)`
D. None of these

Explanation

To solve the question regarding the reaction of stannous chloride (SnCl2) with excess sodium hydroxide (NaOH), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have stannous chloride (SnCl2) and sodium hydroxide (NaOH).

2. **Write the Initial Reaction**:
- When stannous chloride reacts with sodium hydroxide, the reaction can be written as:
\[ \text{SnCl}_2 + 2 \text{NaOH} \rightarrow \text{Sn(OH)}_2 + 2 \text{NaCl} \]
- Here, stannous chloride reacts with sodium hydroxide to form tin hydroxide (Sn(OH)2) and sodium chloride (NaCl).

3. **Consider the Excess Sodium Hydroxide**:
- Since we have excess NaOH, the tin hydroxide (Sn(OH)2) will further react with the excess sodium hydroxide.
- The reaction can be represented as:
\[ \text{Sn(OH)}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SnO}_2 + 2 \text{H}_2\text{O} \]
- This shows that tin hydroxide reacts with sodium hydroxide to form sodium stannite (Na2SnO2) and water.

4. **Final Product**:
- The final product of the reaction when excess sodium hydroxide is added to stannous chloride is sodium stannite (Na2SnO2).

5. **Conclusion**:
- Therefore, the correct answer to the question is **Na2SnO2**.

### Final Answer:
**Na2SnO2 (Sodium Stannite)**

---