The correct sequence of reducing power of halide ions are

Options

A. `Cl^(-) gt Br^(-) gt I^-`
B. `Br^(-) gt I^(-) gt Cl^(-)`
C. `I^(-) gt Br^(-) gt Cl^(-)`
D. `Cl^(-) gt I^(-) gt Br^(-)`

Answer

To determine the correct sequence of reducing power of halide ions, we need to analyze the properties of the halide ions based on their size and tendency to lose electrons. Here’s a step-by-step solution:

### Step 1: Understand the Halide Ions
The halide ions we are considering are:
- Fluoride (F⁻)
- Chloride (Cl⁻)
- Bromide (Br⁻)
- Iodide (I⁻)

### Step 2: Analyze the Size of the Ions
As we move down the group in the periodic table from fluorine to iodine, the size of the halide ions increases:
- F⁻ (smallest)
- Cl⁻
- Br⁻
- I⁻ (largest)

### Step 3: Relate Size to Reducing Power
The reducing power of an ion is related to its ability to lose electrons. Larger ions have a greater tendency to lose electrons because their outer electrons are farther from the nucleus and are less tightly held. Thus, the reducing power increases with the size of the ion.

### Step 4: Establish the Sequence
Based on the size and tendency to lose electrons:
- I⁻ (largest size, highest reducing power)
- Br⁻ (next largest size)
- Cl⁻ (smaller than Br⁻)
- F⁻ (smallest size, lowest reducing power)

### Step 5: Write the Correct Sequence
The correct sequence of reducing power of halide ions from highest to lowest is:
I⁻ > Br⁻ > Cl⁻ > F⁻

### Final Answer
Thus, the correct sequence of reducing power of halide ions is:
I⁻ > Br⁻ > Cl⁻ > F⁻