JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Chemical Bonding and Molecular Structure

Covalent character follows Fajans' rules: a smaller or more highly charged cation and a larger, more polarisable anion raise covalency, and a cation with an eighteen-electron pseudo-noble-gas shell polarises an anion more than one with a noble-gas shell. For the potassium halides the cation is common, so the larger iodide is polarised more strongly, giving $\mathrm{KF}<\mathrm{KI}$; for the fluorides the anion is common, so the smaller lithium ion polarises more, giving $\mathrm{LiF}>\mathrm{KF}$. Raising the charge on tin makes $\mathrm{SnC}{\mathrm{l}}_4>\mathrm{SnC}{\mathrm{l}}_2$, while the pseudo-noble-gas $\mathrm{C}{\mathrm{u}}^{+}$ polarises chloride more than $\mathrm{N}{\mathrm{a}}^{+}$, giving $\mathrm{CuCl}>\mathrm{NaCl}$. The relations that hold are therefore $\mathrm{KF}<\mathrm{KI}$ with $\mathrm{LiF}>\mathrm{KF}$, $\mathrm{SnC}{\mathrm{l}}_4>\mathrm{SnC}{\mathrm{l}}_2$ with $\mathrm{CuCl}>\mathrm{NaCl}$, and $\mathrm{KF}<\mathrm{KI}$ with $\mathrm{CuCl}>\mathrm{NaCl}$, so the correct statements are B, C and E only.