Consider the following perhalate ions in acidic medium `ClO_(4)^(-)(I),BrO_(4)^(-)(II),IO_(4)^(-)(III)`
Arrange these in the decreasing order of oxidizing power

Options

A. I > II > III
B. I > III > II
C. II > I > III
D. II > III > I

Answer

To determine the decreasing order of oxidizing power for the perhalate ions \( ClO_4^{-} \), \( BrO_4^{-} \), and \( IO_4^{-} \) in acidic medium, we will analyze their standard reduction potentials. The higher the reduction potential, the stronger the oxidizing agent.

### Step-by-Step Solution:

1. **Identify the Reduction Reactions**:
We first write the reduction reactions for each of the perhalate ions:
- For \( ClO_4^{-} \):
\[ ClO_4^{-} + 2H^+ + 2e^{-} \rightarrow ClO_3^{-} + H_2O \]
- For \( BrO_4^{-} \):
\[ BrO_4^{-} + 2H^+ + 2e^{-} \rightarrow BrO_3^{-} + H_2O \]
- For \( IO_4^{-} \):
\[ IO_4^{-} + 2H^+ + 2e^{-} \rightarrow IO_3^{-} + H_2O \]

2. **Find the Standard Reduction Potentials**:
Next, we look up the standard reduction potentials for these reactions:
- \( E^{\circ} \) for \( ClO_4^{-} \) is +1.19 V
- \( E^{\circ} \) for \( BrO_4^{-} \) is +1.74 V
- \( E^{\circ} \) for \( IO_4^{-} \) is +1.65 V

3. **Compare the Reduction Potentials**:
Now we compare the reduction potentials:
- \( BrO_4^{-} \) has the highest reduction potential (+1.74 V).
- \( IO_4^{-} \) is next (+1.65 V).
- \( ClO_4^{-} \) has the lowest reduction potential (+1.19 V).

4. **Determine the Order of Oxidizing Power**:
Since a higher reduction potential indicates a stronger oxidizing agent, we can arrange the ions in decreasing order of oxidizing power:
\[ BrO_4^{-} > IO_4^{-} > ClO_4^{-} \]

5. **Final Arrangement**:
Thus, the final arrangement of the perhalate ions in decreasing order of oxidizing power is:
\[ BrO_4^{-} (II) > IO_4^{-} (III) > ClO_4^{-} (I) \]
This corresponds to the order: **2 > 3 > 1**.

### Conclusion:
The correct answer is \( BrO_4^{-} > IO_4^{-} > ClO_4^{-} \) or in terms of their labels, **II > III > I**.