Concept: Atomic and Ionic Sizes

To determine which species has the largest size, we need to compare their atomic or ionic radii. All the given species are isoelectronic (they have the same number of electrons, which is 18). However, their sizes differ due to their different nuclear charges (number of protons).

Step 1: Identify the Number of Electrons

First, let's find the number of electrons in each species:

S^2-: Sulfur (atomic number 16) gains 2 electrons → 16 + 2 = 18 electrons
K⁺: Potassium (atomic number 19) loses 1 electron → 19 - 1 = 18 electrons
Ca²⁺: Calcium (atomic number 20) loses 2 electrons → 20 - 2 = 18 electrons
Ar: Argon (atomic number 18) has 18 electrons

All have 18 electrons, so they are isoelectronic with the electron configuration of argon: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$

Step 2: Compare Nuclear Charges

The size of an isoelectronic series decreases as the nuclear charge (number of protons) increases. This is because a higher nuclear charge pulls the electron cloud closer, making the ion smaller.

List the nuclear charges (atomic numbers):

S^2-: 16 protons
K⁺: 19 protons
Ca²⁺: 20 protons
Ar: 18 protons

Step 3: Order by Size

The species with the smallest nuclear charge will have the weakest attraction for the electron cloud and thus the largest size. Ordering by increasing nuclear charge (and thus decreasing size):

S^2- (16) < Ar (18) < K⁺ (19) < Ca²⁺ (20)

Therefore, S^2- has the largest size.

Final Answer

S^2- has the largest size.

Related Topics

Isoelectronic Series: Atoms or ions that have the same number of electrons.
Effective Nuclear Charge (Z_eff): The net positive charge experienced by an electron in an atom. It increases across a period, leading to a decrease in atomic size.
Periodic Trends: Atomic and ionic sizes generally decrease across a period and increase down a group.

Key Formulae and Theory

Size Trend in Isoelectronic Ions: For a series of isoelectronic ions, the size decreases as the nuclear charge (atomic number, Z) increases.

Example Generalization: For two isoelectronic ions, the one with the smaller Z will be larger.

Coulomb's Law Application: The force of attraction (F) between the nucleus and electrons is given by: $F \propto \frac{(Z_{nucleus} \times Z_{electron})}{r^{2}}$ A higher nuclear charge (Z_nucleus) increases the attractive force, pulling the electrons closer and decreasing the radius (r).