Question

Which of the following orders are correct regarding their covalent character?
(i) \( \text{KF} < \text{KI} \)
(ii) \( \text{LiF} < \text{KF} \)
(iii) \( \text{SnCl}_2 < \text{SnCl}_4 \)
(iv) \( \text{NaCl} < \text{CuCl} \)

• A. i,ii,iii Only
• B. ii,iii,iv Only
• C. i,iii,iv Only
• D. i,ii,iv Only

Hint:

Covalent character increases with the smaller size of the cation and higher charge on the central atom. The greater the polarization of the anion, the greater the covalent character.

Answer 1

The Correct Option is C

To determine the correct order regarding their covalent character for the given compounds, we need to understand the concept of covalent character in ionic compounds. The covalent character of an ionic bond is influenced by factors like the polarizing power of the cation and the polarizability of the anion.
1. Fajans’ Rules: According to Fajans' rules, increased covalent character in an ionic bond is observed with: a) Higher charge of the cation or anion. b) Smaller size of the cation. c) Larger size of the anion.
Now, let's analyze each statement:
(i) \( \text{KF} < \text{KI} \)
- The size of the iodide ion (\( \text{I}^- \)) is larger than the fluoride ion (\( \text{F}^- \)), leading to higher polarizability. Thus, \( \text{KI} \) has more covalent character than \( \text{KF} \).
(ii) \( \text{LiF} < \text{KF} \)
- Lithium ion (\( \text{Li}^+ \)) is smaller than potassium ion (\( \text{K}^+ \)), leading to higher polarization in \( \text{LiF} \) than \( \text{KF} \). Thus, \( \text{LiF} \) should actually have more covalent character than \( \text{KF} \). Hence this order is incorrect.
(iii) \( \text{SnCl}_2 < \text{SnCl}_4 \)
- \( \text{SnCl}_4 \) has tin in a higher oxidation state compared to \( \text{SnCl}_2 \), increasing the polarizing power of the tin ion and enhancing covalent character. Thus, \( \text{SnCl}_4 \) is more covalent than \( \text{SnCl}_2 \).
(iv) \( \text{NaCl} < \text{CuCl} \)
- The \( \text{Cu}^+ \) ion has a higher charge density and polarizing power than \( \text{Na}^+ \), so \( \text{CuCl} \) is more covalent than \( \text{NaCl} \).
Considering these analyses, the correct order for covalent character is in statements i, iii, and iv.

Answer 2

- KF<KI: Potassium iodide (KI) has a lower covalent character than potassium fluoride (KF) due to the larger size of iodide ion (\(I^-\)) compared to fluoride ion (\(F^-\)). Larger ions result in less polarizing ability, which leads to less covalent character. Hence, this order is correct.
- LiF<KF: Lithium fluoride (LiF) has a higher covalent character than potassium fluoride (KF) because lithium ion (\(Li^+\)) is smaller than potassium ion (\(K^+\)), and smaller cations have higher polarizing power. Hence, this order is incorrect.
- SnCl2<SnCl4: SnCl4 has a higher covalent character than SnCl2 because in SnCl4, the tin ion (\(Sn^{4+}\)) is more polarizing than in SnCl2, where tin is in the \(Sn^{2+}\) state. Higher charge on the central metal ion increases the covalent character. Hence, this order is correct.
- NaCl<CuCl: Copper(I) chloride (CuCl) exhibits higher covalent character than sodium chloride (NaCl) because copper(I) ion (\(Cu^+\)) has a higher polarizing power than sodium ion (\(Na^+\)). Therefore, this order is correct.
Thus, the correct option is \( {i, iii, iv only} \).