The reaction, `3Br_(2)+6CO_(3)^(2-)+3H_(2)Oto5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-)`

Options

A. decolourization of reddish brown colour of `Br_(2)`
B. addition of `H_(2)SO_(4)` will increase intensity of reddish brown colour of `Br_(2)`
C. bromine is reduced
D. it is disproportionation reaction or autoredox change

Explanation

To solve the given reaction and analyze its characteristics, 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: Determine the oxidation states of each element
1. **Bromine in \(Br_2\)**: The oxidation state is 0 (elemental form).
2. **Carbon in \(CO_3^{2-}\)**: Let the oxidation state of carbon be \(x\).
\[ x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4 \]
3. **Bromine in \(Br^{-}\)**: The oxidation state is -1 (due to the negative charge).
4. **Bromine in \(BrO_3^{-}\)**: Let the oxidation state of bromine be \(y\).
\[ y + 3(-2) = -1 \implies y - 6 = -1 \implies y = +5 \]
5. **Hydrogen in \(H_2O\)**: The oxidation state is +1.
6. **Carbon in \(HCO_3^{-}\)**: Let the oxidation state of carbon be \(z\).
\[ z + 3(-2) + 1 = -1 \implies z - 6 + 1 = -1 \implies z = +4 \]

### Step 3: Summarize the oxidation states
- \(Br_2\): 0
- \(CO_3^{2-}\): C = +4
- \(H_2O\): H = +1, O = -2
- \(Br^{-}\): -1
- \(BrO_3^{-}\): Br = +5
- \(HCO_3^{-}\): C = +4, H = +1, O = -2

### Step 4: Analyze the changes in oxidation states
- **Bromine**:
- From \(Br_2\) (0) to \(Br^{-}\) (-1): Reduction (gain of electrons)
- From \(Br_2\) (0) to \(BrO_3^{-}\) (+5): Oxidation (loss of electrons)

- **Carbon**: The oxidation state of carbon remains +4 in both reactants and products.

### Step 5: Conclusion about the reaction
Since bromine is undergoing both oxidation and reduction, this reaction is classified as a **disproportionation reaction**.

### Step 6: Identify the correct statement
From the analysis, the correct statement regarding the reaction is that it is a disproportionation reaction.

### Final Answer
The correct option is that this reaction is a disproportionation reaction.

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