`I_(2)+NaOH to NaI +NaOI`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To determine the type of reaction for the given equation \( I_2 + NaOH \rightarrow NaI + NaOI \), we will analyze the oxidation states of the elements involved and identify the nature of the reaction.

### Step 1: Identify the oxidation states
1. **Iodine (I2)** is in its elemental form, so its oxidation state is 0.
2. **Sodium (Na)** in sodium hydroxide (NaOH) has an oxidation state of +1.
3. **Oxygen (O)** in NaOH has an oxidation state of -2.
4. **Hydrogen (H)** in NaOH has an oxidation state of +1.

### Step 2: Determine the oxidation states in the products
1. In **sodium iodide (NaI)**, sodium (Na) remains +1, and iodine (I) in NaI has an oxidation state of -1.
2. In **sodium hypoiodite (NaOI)**, sodium (Na) is +1, oxygen (O) is -2, and iodine (I) must balance the equation. Let’s calculate the oxidation state of iodine (I) in NaOI:
- The sum of oxidation states in NaOI must equal 0:
\[ +1 + (-2) + X = 0 \implies X = +1 \]
So, iodine in NaOI has an oxidation state of +1.

### Step 3: Analyze the changes in oxidation states
- Iodine (I2) goes from an oxidation state of 0 to -1 in NaI (reduction).
- Iodine (I2) also goes from an oxidation state of 0 to +1 in NaOI (oxidation).

### Step 4: Identify the type of reaction
Since one species (iodine) is both oxidized and reduced in the same reaction, this is classified as a **disproportionation reaction**.

### Conclusion
The reaction \( I_2 + NaOH \rightarrow NaI + NaOI \) is a **disproportionation reaction**.