Using the following Latimer diagram for bromine,
`pH=0, BrO_(4)^(-) overset(1.82 V)rarr BrO_(3)^(-)overset(1.50 V)rarr HBrO overset(1.595 V) rarr Br_(2) overset(1.06552 V) rarr Br^(-)`
the species undergoing disproportionation is

Options

A. `BrO_(4)^(-)`
B. `BrO_(3)^(-)`
C. HBrO
D. `Br_(2)`

Explanation

To determine which species undergoes disproportionation using the provided Latimer diagram for bromine, we will follow these steps:

### Step 1: Understand the Latimer Diagram
The Latimer diagram shows the reduction potentials for different oxidation states of bromine. The species listed are:
- \( \text{BrO}_4^- \) (oxidation state +7)
- \( \text{BrO}_3^- \) (oxidation state +5)
- \( \text{HBrO} \) (oxidation state +1)
- \( \text{Br}_2 \) (oxidation state 0)
- \( \text{Br}^- \) (oxidation state -1)

### Step 2: Identify the Reduction Potentials
The reduction potentials given in the diagram are:
- \( \text{BrO}_4^- \) to \( \text{BrO}_3^- \): 1.82 V
- \( \text{BrO}_3^- \) to \( \text{HBrO} \): 1.50 V
- \( \text{HBrO} \) to \( \text{Br}_2 \): 1.595 V
- \( \text{Br}_2 \) to \( \text{Br}^- \): 1.06552 V

### Step 3: Determine Oxidation States
Calculate the oxidation states for each species:
- For \( \text{BrO}_4^- \): Let \( x \) be the oxidation state of Br.
\[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \]
- For \( \text{BrO}_3^- \):
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
- For \( \text{HBrO} \):
\[ x + 1(-2) + 1 = 0 \implies x - 2 + 1 = 0 \implies x = +1 \]
- For \( \text{Br}_2 \):
\[ \text{Oxidation state} = 0 \]
- For \( \text{Br}^- \):
\[ \text{Oxidation state} = -1 \]

### Step 4: Analyze Disproportionation
Disproportionation occurs when a species can simultaneously be oxidized and reduced. We will check each species:
- **\( \text{BrO}_4^- \)**: Can be reduced to \( \text{BrO}_3^- \) but does not undergo disproportionation.
- **\( \text{BrO}_3^- \)**: Can be reduced to \( \text{HBrO} \) but does not undergo disproportionation.
- **\( \text{HBrO} \)**: The potential to the left (1.50 V) is less than the potential to the right (1.595 V). This indicates that \( \text{HBrO} \) can be both reduced to \( \text{Br}_2 \) and oxidized to \( \text{BrO}_3^- \), thus it undergoes disproportionation.
- **\( \text{Br}_2 \)**: Can be reduced to \( \text{Br}^- \) but does not undergo disproportionation.

### Conclusion
The species that undergoes disproportionation is \( \text{HBrO} \).

### Final Answer
The species undergoing disproportionation is \( \text{HBrO} \).

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