Zinc(II) ion on reaction with NaOH first give a white precipitate which dissolves in excess of NaOH due to the formation of :

Options

A. `ZnO`
B. `Zn(OH)_(2)`
C. `[Zn(OH)_(4)]^(2-)`
D. `[Zn(H_(2)O)_(4)]^(2-)`

Answer

To solve the question regarding the reaction of Zinc(II) ion with NaOH, we will follow these steps:

### Step 1: Understand the Reaction
When Zinc(II) ions (Zn²⁺) react with sodium hydroxide (NaOH), they initially form a precipitate. This is a common reaction in inorganic chemistry involving metal ions and hydroxides.

### Step 2: Write the Initial Reaction
The reaction of Zinc(II) ions with NaOH can be represented as follows:

\[ \text{Zn}^{2+} + 2 \text{NaOH} \rightarrow \text{Zn(OH)}_2 \downarrow + 2 \text{Na}^+ \]

Here, Zinc(II) hydroxide (Zn(OH)₂) is formed, which is a white precipitate.

### Step 3: Identify the Behavior of the Precipitate
The white precipitate of Zinc(II) hydroxide (Zn(OH)₂) can dissolve in excess NaOH. This is a characteristic behavior of amphoteric hydroxides, which can react with both acids and bases.

### Step 4: Write the Reaction in Excess NaOH
When excess NaOH is added, Zinc(II) hydroxide reacts further to form a soluble complex ion:

\[ \text{Zn(OH)}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2[\text{Zn(OH)}_4] \]

This results in the formation of the tetrahydroxozincate(II) ion, \([\text{Zn(OH)}_4]^{2-}\).

### Step 5: Conclusion
Thus, the white precipitate formed initially dissolves in excess NaOH due to the formation of the complex ion \([\text{Zn(OH)}_4]^{2-}\).

### Final Answer
The white precipitate formed is Zinc(II) hydroxide (Zn(OH)₂), which dissolves in excess NaOH to form the complex ion \([\text{Zn(OH)}_4]^{2-}\).

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