In the reaction: `3Br_(2)+6OH^(Theta)rarr 5Br^(Theta)+BrO_(3)^(Theta)+3H_(2)O,Br_(2)` is

Options

A. Oxidised
B. Reduced
C. Disintegrated
D. Disproportionted

Explanation

To solve the question regarding the reaction:

\[ 3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O \]

we need to analyze the changes in the oxidation states of bromine (Br) in the reaction.

### Step 1: Identify the oxidation states of bromine in the reactants and products.
- In \( Br_2 \), the oxidation state of bromine is 0 (since it is in its elemental form).
- In \( Br^- \), the oxidation state of bromine is -1.
- In \( BrO_3^- \), we can calculate the oxidation state of bromine. The oxidation state of oxygen is -2. Therefore, for \( BrO_3^- \):
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
So, the oxidation state of bromine in \( BrO_3^- \) is +5.

### Step 2: Determine the changes in oxidation states.
- Bromine in \( Br_2 \) (0) is reduced to \( Br^- \) (-1).
- Bromine in \( Br_2 \) (0) is oxidized to \( BrO_3^- \) (+5).

### Step 3: Classify the reaction.
Since bromine is undergoing both reduction (gaining electrons) and oxidation (losing electrons), this type of reaction is known as a **disproportionation reaction**.

### Conclusion:
In the given reaction, \( Br_2 \) is undergoing disproportionation, where it is both oxidized and reduced simultaneously.

### Final Answer:
Br in the reaction is undergoing disproportionation.

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