The maximum change in the oxidation number of `Cl_(2)` in the reaction of hot and conc. NaOH with `Cl_(2)` is `"___"`

Answer

To determine the maximum change in the oxidation number of chlorine (Cl) in the reaction of chlorine gas (Cl₂) with hot and concentrated sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction of chlorine gas with hot and concentrated NaOH can be represented as:
\[ 6 \text{NaOH} + 3 \text{Cl}_2 \rightarrow 5 \text{NaCl} + \text{NaClO}_3 + 3 \text{H}_2\text{O} \]

### Step 2: Identify the oxidation states of chlorine in the reactants
In the reactant \( \text{Cl}_2 \), the oxidation state of chlorine is 0 (since it is in its elemental form).

### Step 3: Identify the oxidation states of chlorine in the products
In the products:
- In \( \text{NaCl} \), the oxidation state of chlorine is -1.
- In \( \text{NaClO}_3 \), we need to calculate the oxidation state of chlorine. The oxidation state of sodium (Na) is +1 and that of oxygen (O) is -2. The overall charge of the compound is neutral (0).

Using the equation:
\[ \text{Na} (+1) + \text{Cl} (x) + 3 \times \text{O} (-2) = 0 \]
This simplifies to:
\[ 1 + x - 6 = 0 \]
\[ x - 5 = 0 \]
Thus, \( x = +5 \). Therefore, the oxidation state of chlorine in \( \text{NaClO}_3 \) is +5.

### Step 4: Calculate the change in oxidation states
Now we can calculate the changes:
1. From 0 (in \( \text{Cl}_2 \)) to -1 (in \( \text{NaCl} \)):
- Change = \( 0 - (-1) = 1 \)

2. From 0 (in \( \text{Cl}_2 \)) to +5 (in \( \text{NaClO}_3 \)):
- Change = \( 5 - 0 = 5 \)

### Step 5: Determine the maximum change
The maximum change in oxidation number of chlorine in this reaction is the larger of the two changes calculated:
- Change to -1: 1
- Change to +5: 5

Thus, the maximum change in the oxidation number of \( \text{Cl}_2 \) is **5**.

### Final Answer
The maximum change in the oxidation number of \( \text{Cl}_2 \) in the reaction of hot and concentrated NaOH with \( \text{Cl}_2 \) is **5**.

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