Question

Arrange the following in increasing order of covalent character: (A) LiF
(B) LiBr
(C) LiCl
(D) LiI

• A. (A), (B), (C), (D)
• B. (A), (C), (B), (D)
• C. (B), (A), (D), (C)
• D. (C), (B), (A), (D)

Hint:

The covalent character of the bond increases as the size of the halide anion increases.

Answer 1

The Correct Option is A

Step 1: Understanding Covalent Character.
Covalent character refers to the sharing of electrons between atoms in a molecule. The ionic character of a compound generally decreases as the size of the anion increases, leading to a greater covalent character in the bond. Similarly, the more the electronegativity difference between the two atoms, the higher the ionic character, and the lower the covalent character. Therefore, the smaller the anion, the more ionic the compound is, leading to a lower covalent character.

Step 2: Analyzing the Compounds.
- (A) LiF: Fluorine is the smallest halogen, and its high electronegativity creates a highly ionic bond with lithium, leading to the lowest covalent character. - (B) LiBr: Bromine is larger than fluorine, and hence the bond is less ionic, resulting in higher covalent character than LiF. - (C) LiCl: Chlorine is larger than bromine, so the ionic character is further reduced, and the covalent character increases. - (D) LiI: Iodine is the largest halogen, leading to the most covalent character.

Step 3: Conclusion.
Thus, the increasing order of covalent character is: \[ \text{LiF} < \text{LiBr} < \text{LiCl} < \text{LiI} \] So, the correct answer is option (1): (A), (B), (C), (D).