Among the following molecules, in which does bromine show the maximum oxidation state?

Options

A. `Hg_(2)(BrO_(3))_(2)`
B. `Br-Cl`
C. `KBrO_(4)`
D. `Br_(2)`

Answer

To determine the maximum oxidation state of bromine in the given molecules, we will analyze each option step by step.

### Step 1: Analyze the first molecule, Hg2BrO3 (whole twice)
1. **Identify oxidation states:**
- Mercury (Hg) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of bromine (Br) be \( x \).

2. **Set up the equation for the overall charge:**
- The molecule is neutral, so the sum of the oxidation states must equal 0:
\[ 2(+1) + x + 3(-2) = 0 \]
\[ 2 + x - 6 = 0 \]
\[ x - 4 = 0 \implies x = +4 \]

3. **Conclusion for Hg2BrO3:**
- The oxidation state of bromine is +4.

### Step 2: Analyze the second molecule, BrCl
1. **Identify oxidation states:**
- Chlorine (Cl) has an oxidation state of -1.
- Let the oxidation state of bromine (Br) be \( y \).

2. **Set up the equation for the overall charge:**
- The molecule is neutral:
\[ y + (-1) = 0 \]
\[ y = +1 \]

3. **Conclusion for BrCl:**
- The oxidation state of bromine is +1.

### Step 3: Analyze the third molecule, KBrO4
1. **Identify oxidation states:**
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of bromine (Br) be \( z \).

2. **Set up the equation for the overall charge:**
- The molecule is neutral:
\[ +1 + z + 4(-2) = 0 \]
\[ 1 + z - 8 = 0 \]
\[ z - 7 = 0 \implies z = +7 \]

3. **Conclusion for KBrO4:**
- The oxidation state of bromine is +7.

### Step 4: Analyze the fourth molecule, Br2
1. **Identify oxidation states:**
- In Br2, the oxidation state of bromine is 0.

2. **Conclusion for Br2:**
- The oxidation state of bromine is 0.

### Final Conclusion:
- The oxidation states of bromine in the given options are:
- Hg2BrO3: +4
- BrCl: +1
- KBrO4: +7
- Br2: 0

Among these, the maximum oxidation state of bromine is **+7 in KBrO4**.

### Answer:
The molecule in which bromine shows the maximum oxidation state is **KBrO4**.