AP EAPCET 21 May 2024 Shift 1 Solved Paper

According to Fajans' rules, covalent character increases with smaller cation size, higher cation charge, larger and more polarizable anion, and pseudo-noble gas configuration of the cation.(i) $\mathrm{KF} < \mathrm{KI}$: Both have $\mathrm{K^+}$, but $\mathrm{I^-}$ is larger and more polarizable than $\mathrm{F^-}$, so $\mathrm{KI}$ is more covalent than $\mathrm{KF}$.Thus, (i) is correct.(ii) $\mathrm{LiF} < \mathrm{KF}$: $\mathrm{Li^+}$ is smaller than $\mathrm{K^+}$, so $\mathrm{Li^+}$ has greater polarizing power and $\mathrm{LiF}$ is more covalent than $\mathrm{KF}$.Thus, $\mathrm{LiF} < \mathrm{KF}$ is incorrect; the actual order is $\mathrm{LiF} > \mathrm{KF}$.(iii) $\mathrm{SnCl}_2 < \mathrm{SnCl}_4$: Sn is in $+2$ state in $\mathrm{SnCl}_2$ and $+4$ state in $\mathrm{SnCl}_4$.The higher charge on $\mathrm{Sn^{4+}}$ gives greater polarizing power, so $\mathrm{SnCl}_4$ is more covalent than $\mathrm{SnCl}_2$.Thus, (iii) is correct.(iv) $\mathrm{NaCl} < \mathrm{CuCl}$: $\mathrm{Cu^+}$ is smaller and has a pseudo-noble gas $3d^{10}$ configuration, so it polarizes $\mathrm{Cl^-}$ more than $\mathrm{Na^+}$ does.Hence, $\mathrm{CuCl}$ is more covalent than $\mathrm{NaCl}$.Thus, (iv) is correct.The correct orders are (i), (iii), and (iv) only, which is option i, iii, iv only.