When `Al` is added to `NaOH` solution

Options

A. No action takes place
B. `NaAlO_(2)` is formed and `H_(2)` is evolved.
C. `Al(OH)_(3)` is formed and `H_(2)` is evolved
D. `Na_(2)AlO_(2)` is formed and `H_(2)` is evolved.

Explanation

To solve the question "When Al is added to NaOH solution," we can follow these steps:

### Step 1: Identify the Reactants
We start with aluminum (Al) and sodium hydroxide (NaOH) in an aqueous solution.

### Step 2: Write the Reaction
When aluminum reacts with sodium hydroxide, the reaction can be represented as:
\[ \text{Al} + \text{NaOH} + \text{H}_2\text{O} \rightarrow \text{NaAlO}_2 + \text{H}_2 \uparrow \]

### Step 3: Identify the Products
From the reaction, we can see that two products are formed:
1. Sodium metaluminate (\( \text{NaAlO}_2 \))
2. Hydrogen gas (\( \text{H}_2 \))

### Step 4: Analyze the Options
Now, let's analyze the options provided in the question:
1. No action takes place - This is incorrect since a reaction occurs.
2. \( \text{NaAlO}_2 \) is formed - This is correct as we derived it from the reaction.
3. \( \text{Al(OH)}_3 \) is formed - This is incorrect; aluminum hydroxide is not formed in this reaction.
4. \( \text{Na}_2\text{AlO}_2 \) is formed - This is also incorrect; the correct product is \( \text{NaAlO}_2 \).

### Conclusion
The correct answer is that when aluminum is added to sodium hydroxide solution, sodium metaluminate (\( \text{NaAlO}_2 \)) is formed along with the evolution of hydrogen gas.

### Final Answer
The correct option is: **Sodium metaluminate (\( \text{NaAlO}_2 \)) is formed, and hydrogen gas is evolved.**

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