Arrange the following for reducing power :
(a) `I^(-)`
(b) `Cl^(-)`
(c ) `Br^(-)`

Options

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

Answer

To arrange the given halide ions (I^(-), Cl^(-), and Br^(-)) in order of reducing power, we need to consider their ability to lose electrons. The reducing power of an element is determined by its tendency to be oxidized, which is influenced by its atomic size and electronegativity.

### Step-by-Step Solution:

1. **Understanding Reducing Power**:
- Reducing power refers to the ability of an ion to donate electrons. The greater the tendency to lose electrons, the stronger the reducing agent.

2. **Group Trend in Halogens**:
- The halogens are located in Group 17 of the periodic table. As we move down the group from fluorine (F) to iodine (I), the atomic size increases due to the addition of electron shells.

3. **Atomic Size and Electronegativity**:
- As the atomic size increases (from Cl to Br to I), the distance between the nucleus and the outermost electrons also increases. This results in a decrease in electronegativity, which is the ability of an atom to attract shared electrons.

4. **Comparison of Ions**:
- **I^(-)**: Iodide ion has the largest atomic size among the three, making it easier for it to lose an electron.
- **Br^(-)**: Bromide ion is smaller than iodide but larger than chloride, thus it has moderate reducing power.
- **Cl^(-)**: Chloride ion has the smallest atomic size, making it the least likely to lose an electron compared to the others.

5. **Arranging the Ions**:
- Based on the above reasoning, the order of reducing power from highest to lowest is:
- I^(-) > Br^(-) > Cl^(-)

### Final Arrangement:
- **I^(-) > Br^(-) > Cl^(-)**