Order of Covalent bond;
A.$$KF > KI; LiF > KF$$
B. $$KF < KI; LiF > KF$$
C. $$SnCl_4 > SnCl_2$$; CuCl > NaCl
D. $$LiF > KF; CuCl > NaCl$$
E. $$KF < KI; CuCl > NaCl$$

Covalent character in an ionic compound is assessed using Fajans' rules. Covalent character increases when (i) the cation is small and highly charged, (ii) the anion is large and easily polarisable, and (iii) the cation has a pseudo-noble-gas configuration (e.g., $$Cu^{+}$$ with $$18$$ electrons in the outermost shell) rather than a noble-gas configuration.

Statement A: $$KF > KI;\; LiF > KF$$. Since $$I^{-}$$ is larger and more polarisable than $$F^{-}$$, we have $$KI > KF$$, not $$KF > KI$$. The first part is wrong, so A is incorrect.

Statement B: $$KF < KI;\; LiF > KF$$. As argued above, $$KI > KF$$ because $$I^{-}$$ is more polarisable. Also, $$Li^{+}$$ is much smaller than $$K^{+}$$ and therefore more polarising, so $$LiF > KF$$. Both parts are correct, so B is correct.

Statement C: $$SnCl_{4} > SnCl_{2};\; CuCl > NaCl$$. In $$SnCl_{4}$$, tin is in the $$+4$$ oxidation state, which is smaller and more highly charged than $$Sn^{2+}$$, so $$SnCl_{4}$$ has greater covalent character than $$SnCl_{2}$$. For $$CuCl$$ vs $$NaCl$$, $$Cu^{+}$$ has a pseudo-noble-gas configuration $$(3d^{10})$$ and is smaller than $$Na^{+}$$, giving $$CuCl$$ much greater covalent character. Both parts are correct, so C is correct.

Statement D: $$LiF > KF;\; CuCl < NaCl$$. While $$LiF > KF$$ is correct, $$CuCl < NaCl$$ is wrong (it should be $$CuCl > NaCl$$). So D is incorrect.

Statement E: $$KF < KI;\; CuCl > NaCl$$. Both parts follow directly from the reasoning above and are correct. So E is correct.

The correct statements are B, C, and E.

Hence, the correct answer is Option 3.