Question

Identify metal halide from following having highest ionic character? (M = metal atom)

• A. MF
• B. MBr
• C. MI
• D. MCl

Hint:

Fajan's Rules made simple: Small cation + Large anion = Highly Covalent. Large cation + Small anion = Highly Ionic. Fluorine is the smallest anion, so it almost always forms the most ionic compounds!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Given a generic metal M, we must determine which of its corresponding halides (fluoride, bromide, iodide, or chloride) exhibits the maximum ionic bonding character.

Step 2: Key Formula or Approach:
The ionic versus covalent character of a bond can be predicted using Fajan's Rules, which evaluate the polarizing power of the cation and the polarizability of the anion.
According to Fajan's Rules, covalent character increases as the size of the anion increases. Conversely, ionic character increases as the size of the anion decreases.

Step 3: Detailed Explanation:
In this set of options, the metal cation ($\text{M}^+$) remains constant, so we only need to compare the halide anions: $\text{F}^-$, $\text{Cl}^-$, $\text{Br}^-$, and $\text{I}^-$.
The ionic radii of the halogens increase down Group 17: $\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-$.
The iodide ion ($\text{I}^-$) is the largest and its electron cloud is the most easily distorted (highest polarizability) by the metal cation, making MI the most covalent.
The fluoride ion ($\text{F}^-$) is the smallest and holds its electrons very tightly (lowest polarizability). It strongly resists distortion, resulting in bonds that are highly ionic.

Step 4: Final Answer:
The metal halide with the highest ionic character is MF, which corresponds to option (A).