If group IV the ppt of `Zn(OH)_(2)` dissolve in excess of `NaOH` due to the formation of

Options

A. `Na_(2)ZnO_(2)`
B. `NaZnO_(2)`
C. `Zn`
D. `Na_(3)ZnO_(2)`

Answer

To solve the question regarding the dissolution of zinc hydroxide in excess sodium hydroxide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**: The compound in question is zinc hydroxide, which has the chemical formula `Zn(OH)₂`. It is known to be a white precipitate.

2. **Understand the Reaction with NaOH**: When excess sodium hydroxide (NaOH) is added to zinc hydroxide, a reaction occurs. The hydroxide ions from NaOH can react with the zinc ions from zinc hydroxide.

3. **Write the Reaction**: The reaction can be represented as follows:
\[ Zn(OH)_2 (s) + 2 NaOH (aq) \rightarrow Na_2ZnO_2 (aq) + 2 H_2O (l) \]
Here, zinc hydroxide reacts with sodium hydroxide to form sodium zincate and water.

4. **Identify the Product**: The product formed in this reaction is sodium zincate, which can be represented as `Na₂ZnO₂`. This compound is soluble in water.

5. **Conclusion**: Therefore, the answer to the question is that the precipitate of zinc hydroxide dissolves in excess sodium hydroxide due to the formation of sodium zincate.

### Final Answer:
The precipitate of `Zn(OH)₂` dissolves in excess `NaOH` due to the formation of sodium zincate, `Na₂ZnO₂`.

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