Sn is dissolved in excess of NaOH solution, the compound obtained is

Options

A. `Sn(OH)_(2)`
B. `Na_(2)SnO_(3)`
C. `Na_(2)SnO_(2)`
D. `SnO_(2)`

Explanation

To solve the question, "Sn is dissolved in excess of NaOH solution, the compound obtained is," we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have tin (Sn) and sodium hydroxide (NaOH).

2. **Understand the Reaction**:
- When tin reacts with sodium hydroxide, especially in excess, it forms a compound known as sodium stannate.

3. **Write the Chemical Equation**:
- The reaction can be represented as:
\[ \text{Sn} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SnO}_2 + \text{H}_2 \]
- In this case, sodium stannate (Na2SnO3) is formed along with hydrogen gas.

4. **Identify the Compound**:
- The compound formed from the reaction of tin with excess sodium hydroxide is sodium stannate, which can be represented as Na2SnO3.

5. **Conclusion**:
- Therefore, the compound obtained when Sn is dissolved in excess NaOH solution is sodium stannate (Na2SnO3).

### Final Answer:
The compound obtained is **Na2SnO3** (sodium stannate).

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