Tin and lead form divalent Sn (II) and lead (II) and tetravalent i.e., Tin (IV) and lead. (IV) compounds. Tetravalent and dioxides of lead and Tin are amphoteric in nature. Lead tetra fluorite is ionic solid where as `PbCl_(4)` is covalent. `Pbl_(4)` does not exist. Lead (II) halides are white whereas `Pbl_(2)` is yellow in colour.
Q. Reaction of SnO with NaOH gives

Options

A. `Na_(2)SnO_(2)`
B. `H_(2)SnO_(2)`
C. `Na_(3)SnO_(2)`
D. `SnO_(2)`

Answer

To solve the question regarding the reaction of SnO (tin(II) oxide) with NaOH (sodium hydroxide), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- SnO (tin(II) oxide)
- NaOH (sodium hydroxide)

### Step 2: Understand the Nature of the Reaction
Tin(II) oxide is amphoteric, meaning it can react with both acids and bases. In this case, it reacts with a base (NaOH) to form a stannate.

### Step 3: Write the Reaction
The reaction of SnO with NaOH can be represented as follows:
\[ \text{SnO} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SnO}_2 + \text{H}_2\text{O} \]

### Step 4: Identify the Products
From the reaction, we can see that the products formed are:
- Sodium stannate (Na₂SnO₂)
- Water (H₂O)

### Step 5: Conclusion
Thus, the reaction of SnO with NaOH produces sodium stannate and water. The correct answer to the question is:
\[ \text{Na}_2\text{SnO}_2 \]

### Final Answer
The reaction of SnO with NaOH gives Na₂SnO₂ (sodium stannate).

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