Which of the following has the highest percentage of ionic character in its bonding?

Options

A. LiI
B. `MgCl_(2)`
C. `CsF`
D. `CsI`

Explanation

To determine which of the given compounds has the highest percentage of ionic character in its bonding, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compounds**:
We need to analyze the given options, which include pairs of cations and anions. For example, we may have options like LiI, MgCl2, CsI, and CsF.

2. **Understand Ionic Character**:
Ionic character in a bond is determined by the difference in electronegativity between the two atoms involved in the bond. The greater the difference, the more ionic the bond is.

3. **List the Electronegativity Values**:
- Fluorine (F) has the highest electronegativity (approximately 4.0).
- Chlorine (Cl) has a lower electronegativity (approximately 3.0).
- Iodine (I) has an even lower electronegativity (approximately 2.5).
- Lithium (Li) has an electronegativity of about 1.0.
- Magnesium (Mg) has an electronegativity of about 1.2.
- Cesium (Cs) has the lowest electronegativity (approximately 0.7).

4. **Calculate Electronegativity Differences**:
- For LiI: |1.0 - 2.5| = 1.5
- For MgCl2: |1.2 - 3.0| = 1.8
- For CsI: |0.7 - 2.5| = 1.8
- For CsF: |0.7 - 4.0| = 3.3

5. **Compare the Differences**:
Among the calculated differences, CsF has the highest difference of 3.3, indicating it has the highest ionic character.

6. **Conclusion**:
Therefore, the compound with the highest percentage of ionic character is **CsF (Cesium Fluoride)**.

### Final Answer:
The compound with the highest percentage of ionic character in its bonding is **CsF (Cesium Fluoride)**.