When `Cl_(2)` gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

Options

A. zero to +1 and zero to -5
B. zero to -1 and zero to +5
C. zero to -1 and zero to +3
D. zero to +1 and zero to -3

Explanation

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

### Step 1: Identify the Reactants and Products
When chlorine gas (Cl₂) reacts with hot and concentrated sodium hydroxide, the products formed are sodium chloride (NaCl) and sodium chlorate (NaClO₃), along with water (H₂O).

### Step 2: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ 3 \text{Cl}_2 + 6 \text{NaOH} \rightarrow 5 \text{NaCl} + \text{NaClO}_3 + 3 \text{H}_2\text{O} \]

### Step 3: Determine the Oxidation States
- In Cl₂, the oxidation state of chlorine is 0 (since it is in its elemental form).
- In NaCl, the oxidation state of chlorine is -1 (as it is in the chloride ion, Cl⁻).
- In NaClO₃, the oxidation state of chlorine is +5.

### Step 4: Identify the Changes in Oxidation States
- The oxidation state of chlorine changes from 0 in Cl₂ to -1 in NaCl (reduction).
- The oxidation state of chlorine changes from 0 in Cl₂ to +5 in NaClO₃ (oxidation).

### Step 5: Conclusion
Thus, the oxidation number of chlorine changes from 0 to -1 and from 0 to +5 during the reaction. This indicates that chlorine undergoes a disproportionation reaction, where it is both oxidized and reduced.

### Final Answer
The oxidation number of chlorine changes from 0 to -1 and from 0 to +5.

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