Iodide of which of the following is most covalent ?

Question

Accepted Answer

The iodide of lithium (LiI) is the most covalent among the given options. This can be explained using Fajans' rules, which help predict whether a chemical bond will be more ionic or covalent. According to Fajans' rules, the covalent character of a compound increases with:

1. **Small Cation Size**: Smaller cations have a higher charge density, which allows them to polarize the electron cloud of the anion more effectively, leading to increased covalent character. Lithium (Li) is the smallest cation among the given options (Li, Na, K, Rb), which means it has the highest charge density and can polarize the iodide ion (I-) more effectively.

2. **Large Anion Size**: Larger anions are more easily polarized. Iodide (I-) is a large anion, and when paired with a small cation like lithium, the polarization is significant, enhancing the covalent character of the bond.

3. **High Charge on Cation**: While all the cations here have the same charge (+1), the size and polarizing power of lithium make it more effective in creating covalent character.

Now, let's explain why the other options are less covalent:

- **Option 2: NaI (Sodium Iodide)**: Sodium is larger than lithium, which means it has a lower charge density and less ability to polarize the iodide ion compared to lithium. This results in a more ionic character than LiI.

- **Option 3: KI (Potassium Iodide)**: Potassium is even larger than sodium, further decreasing the charge density and polarizing power. Consequently, KI is more ionic than both LiI and NaI.

- **Option 4: RbI (Rubidium Iodide)**: Rubidium is the largest cation among the options provided. Its large size results in the lowest charge density and polarizing power, making RbI the most ionic and least covalent among the options.

In summary, due to its small size and high polarizing power, lithium iodide (LiI) exhibits the most covalent character among the given iodides.