The correct order of decreasing oxidising power is:

Options

A. `F_(2) gt Cl_(2) gt Br_(2) gt l_(2)`
B. `F_(2) lt Cl_(2) lt Br_(2) lt l_(2) `
C. `F_(2) lt Cl_(2) gt Br_(2) gt l_(2) `
D. `l_(2) lt F_(2) lt Br_(2) lt Cl_(2)`

Explanation

To determine the correct order of decreasing oxidizing power among the halogens (F2, Cl2, Br2, I2), we can follow these steps:

### Step 1: Understand Oxidizing Power
Oxidizing power refers to the ability of a substance to gain electrons and thereby oxidize another substance. The higher the oxidizing power, the more readily the substance can accept electrons.

### Step 2: Know the Standard Reduction Potentials
The standard reduction potential (E°) values indicate the tendency of a species to be reduced. A higher E° value means a stronger oxidizing agent. For halogens, the standard reduction potentials are:
- F2 + 2e⁻ → 2F⁻; E° = +2.87 V
- Cl2 + 2e⁻ → 2Cl⁻; E° = +1.36 V
- Br2 + 2e⁻ → 2Br⁻; E° = +1.08 V
- I2 + 2e⁻ → 2I⁻; E° = +0.54 V

### Step 3: Compare the E° Values
From the E° values, we can see that:
- F2 has the highest E° value (+2.87 V), indicating it is the strongest oxidizing agent.
- Cl2 follows with +1.36 V.
- Br2 is next with +1.08 V.
- I2 has the lowest E° value (+0.54 V), making it the weakest oxidizing agent.

### Step 4: Arrange in Decreasing Order of Oxidizing Power
Based on the E° values, we can arrange the halogens in order of decreasing oxidizing power:
1. F2
2. Cl2
3. Br2
4. I2

### Final Answer
The correct order of decreasing oxidizing power is: **F2 > Cl2 > Br2 > I2**

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