Final answer:
For isoelectronic ions, the ionic radius is inversely proportional to the nuclear charge. The ions K+, S2-, and Se2-, which have the same [Ar] electron configuration, can be arranged in order of increasing ionic radius based on their nuclear charges, resulting in K+ having the smallest radius and S2- the largest.
Step-by-step explanation:
Atoms and ions that have the same electron configuration are known as isoelectronic. When comparing the sizes of isoelectronic ions, the number of protons in the nucleus, or the nuclear charge, plays a fundamental role. The more protons there are, the smaller the ionic radius becomes, because a greater nuclear charge will pull the electrons closer to the nucleus due to stronger electrostatic forces.
In the isoelectronic series K+, S2-, and Se2-, which share the same closed-shell electron configuration of [Ar], we can order them by increasing radius based on their nuclear charges. Potassium ion (K+) with 19 protons has the smallest radius, while sulfide ion (S2-) with 16 protons has a larger radius. Following this trend, selenide ion (Se2-), being directly below sulfur in the periodic table with the same number of electrons but one more proton, will have a slightly smaller radius than S2-, but still larger than K+.
Hence, the correctly arranged order of increasing radius for these isoelectronic ions is: K+ < Se2- < S2-.