Step-by-Step Solution

1. (A) Ionization Energy: Generally, ionization energy decreases down a group. However, due to the smaller size and higher electron density of Fluorine, it experiences greater interelectronic repulsion, making it slightly easier to remove an electron from Fluorine than from Chlorine. Hence, the correct order is $F>Cl$. Therefore, option (A) is incorrect.

2. (B) Basic Nature of Oxides: Basic nature of oxides decreases from left to right across a period and increases down a group. $K_{2}O$ is more basic than $Na_{2}O$ because K is below Na in the group. $Na_{2}O$ is more basic than $Al_{2}O_{3}$ because Na is to the left of Al in the period. Therefore, the correct order is $K_{2}O>Na_{2}O>Al_{2}O_{3}$. Hence, option (B) is correct.

3. (C) Metallic Character: Metallic character increases down a group and decreases across a period from left to right. Therefore, the correct order is $K>Na>Mg>Al$. Hence, option (C) is incorrect.