When `Cl_(2)` is passed through hot NaOH , oxidation number of Cl changes from

Options

A. `-1 " to " 0`
B. `0 " to " -1`
C. `0 " to " +5`
D. `0 " to " +7`

Answer

To determine the change in oxidation number of chlorine when \( Cl_2 \) is passed through hot sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Write the Reaction
When chlorine gas (\( Cl_2 \)) is passed through hot NaOH, it reacts to form sodium chloride (\( NaCl \)), sodium chlorate (\( NaClO_3 \)), and water (\( H_2O \)). The balanced chemical equation for this reaction is:

\[ Cl_2 + 6 NaOH \rightarrow 5 NaCl + NaClO_3 + 3 H_2O \]

### Step 2: Identify the Oxidation States
Next, we need to identify the oxidation states of chlorine in the reactants and products.

- In \( Cl_2 \), the oxidation state of chlorine is 0 (elemental form).
- In \( NaCl \), the oxidation state of chlorine is -1 (since sodium is +1).
- In \( NaClO_3 \), we can calculate the oxidation state of chlorine. The overall charge of the compound is neutral (0). Sodium is +1, and each oxygen is -2. Therefore, we have:

\[ +1 + X + 3(-2) = 0 \implies X - 6 + 1 = 0 \implies X = +5 \]

So, the oxidation state of chlorine in \( NaClO_3 \) is +5.

### Step 3: Determine the Change in Oxidation State
Now we can summarize the changes in oxidation states:
- Chlorine changes from 0 in \( Cl_2 \) to -1 in \( NaCl \) (reduction).
- Chlorine changes from 0 in \( Cl_2 \) to +5 in \( NaClO_3 \) (oxidation).

### Step 4: Conclusion
Thus, the oxidation number of chlorine changes from 0 to -1 and from 0 to +5. The correct answer is that the oxidation number of chlorine changes from 0 to -1 and 0 to +5.

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

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