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

Explanation

To solve the problem of determining how the oxidation number of chlorine changes when Cl₂ gas reacts with hot and concentrated sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The reaction of chlorine gas with hot and concentrated sodium hydroxide can be represented as follows:

\[ 3 \text{Cl}_2 + 6 \text{NaOH} \rightarrow 5 \text{NaCl} + \text{NaClO}_3 + 3 \text{H}_2\text{O} \]

### Step 2: Identify the oxidation states of chlorine in the reactants and products.
- In Cl₂ (elemental chlorine), the oxidation state of chlorine is 0.
- In NaCl (sodium chloride), the oxidation state of chlorine is -1.
- In NaClO₃ (sodium chlorate), we need to calculate the oxidation state of chlorine.

### Step 3: Calculate the oxidation state of chlorine in NaClO₃.
To find the oxidation state of chlorine (let's denote it as \( x \)) in NaClO₃:

- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2. There are three oxygen atoms, contributing a total of -6.
- The overall charge of NaClO₃ is neutral (0).

Setting up the equation:

\[ x + 1 + (-6) = 0 \]

Solving for \( x \):

\[ x - 5 = 0 \implies x = +5 \]

### Step 4: Summarize the changes in oxidation states.
- Chlorine changes from an oxidation state of 0 in Cl₂ to -1 in NaCl.
- Chlorine also changes from an oxidation state of 0 in Cl₂ to +5 in NaClO₃.

### Conclusion:
The oxidation number of chlorine changes from 0 to -1 and from 0 to +5 during the reaction.

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

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