`CI_(2)` reacts with hot aqueous NaOH to give

Options

A. `NaCI`
B. `NaCIO_(3)`
C. `NaCIO_(2)`
D. `NaCIO_(4)`

Explanation

To solve the question regarding the reaction of \( Cl_2 \) with hot aqueous \( NaOH \), we can follow these steps:

### Step 1: Identify the Reaction Type
When \( Cl_2 \) reacts with hot aqueous \( NaOH \), it undergoes a disproportionation reaction. In this type of reaction, the same element is both oxidized and reduced.

**Hint:** Look for changes in oxidation states to identify disproportionation reactions.

### Step 2: Determine the Oxidation States
In \( Cl_2 \), the oxidation state of chlorine is 0. During the reaction:
- One chlorine atom is reduced to \( Cl^- \) (oxidation state -1).
- The other chlorine atom is oxidized to \( ClO_3^- \) (oxidation state +5).

**Hint:** Assign oxidation states to each species involved to track the changes.

### Step 3: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ 6 NaOH + Cl_2 \rightarrow 5 NaCl + NaClO_3 + 3 H_2O \]

**Hint:** Ensure that the number of atoms of each element is the same on both sides of the equation.

### Step 4: Identify the Products
From the balanced equation, the products formed are:
- Sodium chloride (\( NaCl \))
- Sodium chlorate (\( NaClO_3 \))
- Water (\( H_2O \))

**Hint:** List down all the products formed in the reaction.

### Step 5: Analyze the Options
Now, we can analyze the options given in the question:
1. \( NaCl \) - Correct
2. \( NaClO_3 \) - Correct
3. \( NaClO_2 \) - Incorrect
4. \( NaClO_4 \) - Incorrect

**Hint:** Compare the products obtained from the reaction with the provided options to determine which are correct.

### Final Answer
The correct products formed when \( Cl_2 \) reacts with hot aqueous \( NaOH \) are \( NaCl \) and \( NaClO_3 \).