Arrange the following ions in order of increasing ionic radius: K+, P3-, S2-, Cl-.

K+, Cl-, S2-, P3-

K+, P3-, S2-, Cl-

P3-, S2-, Cl-, K+

Cl-, S2-, P3-, K+

Cl-, S2-, K+, P3-

K + < Cl - < S2 - < P3-.

Explanation:

The electronic configuration of K is 2,8,8,1. So, K has 19 electrons. Hence, K + will have 18 electrons.

Similarly, electronic configuration of P is 2,8,5. So, P has 15 electrons. Hence, P^3 - will have 18 electrons.

Similarly, electronic configuration of S is 2,8,6. So, S has 16 electrons. Hence, S^2 - will have 18 electrons.

Similarly, electronic configuration of Cl is 2,8,7. So, Cl has 17 electrons. Hence, Cl^ 1 - will have 18 electrons.

As, all of these species are isoelectronic, so the smallest radius among these will be of that which possesses more amount of + ve charge on it. The reason is that a species having + ve charge on it will be effectively able to attract the outermost valence electrons to the nucleus.

Hence, the order of increasing order of radii is as follows:

K + < Cl - < S2 - < P3-.