Explanation:
The ionization enthalpy is the energy required to remove an electron from an atom or ion in the gaseous state. It is a measure of the tendency of an atom or ion to lose an electron.

Isoelectronic ions:
Isoelectronic ions are ions that have the same number of electrons. In this question, we are comparing the ionization enthalpies of four isoelectronic ions: K, Ca2+, Cl-, and S2-.

Ionization enthalpy trend:
The ionization enthalpy generally increases across a period and decreases down a group in the periodic table.

Analysis:
Let's analyze the given ions and their electronic configurations:

a) K: [Ar] 4s1 (1 valence electron)
b) Ca2+: [Ar] (no valence electrons)
c) Cl-: [Ne] (18 valence electrons)
d) S2-: [Ne] 3s23p6 (18 valence electrons)

Among the given options, K has the lowest ionization enthalpy. This can be explained by the following factors:

Effective nuclear charge:
The effective nuclear charge is the net positive charge experienced by an electron in an atom or ion. It increases from left to right across a period. As K is on the leftmost side of the period, it experiences the lowest effective nuclear charge among the given ions.

Shielding effect:
The shielding effect is the reduction in the effective nuclear charge experienced by an electron due to the presence of other electrons between it and the nucleus. K has only one valence electron, which is shielded by the filled inner electron shells. This reduces the attraction between the nucleus and the valence electron, making it easier to remove.

Distance from the nucleus:
K is the largest atom/ion among the given options. The valence electron in K is farther from the nucleus compared to the other ions, which decreases the attraction between the nucleus and the valence electron.

These factors contribute to K having the lowest ionization enthalpy among the given options.