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

- 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 \( {i, iii, iv only} \).