Question

Which of the following metal halides have more covalent character?

• A. \(\text{SnCl}_2\)
• B. \(\text{PbCl}_4\)
• C. \(\text{SbCl}_3\)
• D. \(\text{PbCl}_2\)

Hint:

Apply Fajans rules: a smaller, more highly charged cation polarises the anion more.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Fajans rules say covalent character grows when the cation is small and highly charged, and when the anion is large and easily polarised. All four compounds here have chloride as the anion, so the comparison depends on the cation.

Step 2: Key Formula or Approach:
Compare the charge and size of \(\text{Sn}^{2+}\), \(\text{Pb}^{4+}\), \(\text{Sb}^{3+}\) and \(\text{Pb}^{2+}\).

Step 3: Detailed Explanation:
\(\text{Pb}^{4+}\) has the highest charge in the list and is smaller than \(\text{Pb}^{2+}\).
A high charge on a small ion gives a very high polarising power, so it distorts the chloride electron cloud strongly.
\(\text{PbCl}_4\) is therefore a covalent liquid. The +2 chlorides \(\text{SnCl}_2\) and \(\text{PbCl}_2\) have a lower charge on the cation, so they are more ionic. \(\text{SbCl}_3\) has a +3 cation, which still polarises less than a +4 cation.

Final Answer:
\(\text{PbCl}_4\) has the most covalent character, option (B). \[ \boxed{\text{PbCl}_4} \]