In the reacion ,
`3Br_(2)+6CO_(3)^(2-)+3H_(2)Orarr5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-)`

Options

A. Bromine is reduced and carbonate ion is oxidised.
B. Bromine undergoes disproportionation.
C. Bromine is reduced and water is oxidised.
D. Only water is oxidised to carbonic acid.

Explanation

To analyze the given redox reaction and determine the changes in oxidation states, we will follow these steps:

### Step 1: Write down the reaction
The reaction given is:
\[ 3Br_2 + 6CO_3^{2-} + 3H_2O \rightarrow 5Br^{-} + BrO_3^{-} + 6HCO_3^{-} \]

### Step 2: Assign oxidation states
Next, we assign oxidation states to each element in the reactants and products.

- For \( Br_2 \): Each bromine (Br) has an oxidation state of 0. Therefore, \( Br_2 \) is 0.
- For \( CO_3^{2-} \): The oxidation state of carbon (C) is +4, and each oxygen (O) is -2.
- For \( H_2O \): The oxidation state of hydrogen (H) is +1, and oxygen (O) is -2.

In the products:
- For \( Br^{-} \): The oxidation state of bromine (Br) is -1.
- For \( BrO_3^{-} \): The oxidation state of bromine (Br) is +5 (since O is -2 and there are three O atoms, contributing -6 total, making Br +5 to balance the -1 charge).
- For \( HCO_3^{-} \): The oxidation state of carbon (C) is +4, and each oxygen (O) is -2.

### Step 3: Determine changes in oxidation states
Now we will summarize the changes in oxidation states for bromine:

- In \( Br_2 \) (0) to \( Br^{-} \) (-1): This is a reduction (gain of electrons).
- In \( Br_2 \) (0) to \( BrO_3^{-} \) (+5): This is an oxidation (loss of electrons).

### Step 4: Identify the type of reaction
Since the same element (bromine) is undergoing both oxidation and reduction, this is a disproportionation reaction.

### Conclusion
Thus, in the given reaction, bromine is both reduced and oxidized, confirming that it undergoes disproportionation.

### Final Answer
Bromine undergoes disproportionation in the reaction.

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