It is not possible to measure the atomic radius prectsely since the electron cloub surrouding the atom does not have a sharp boundary. One practial approach to estimate the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound togther by a signle bond in a covalent molecule and then dividing by two. For metals we define the term "matellic radius" which is taken as half the internuclear distance separating the metal cores in the metallic crystal. Then van der waal's radius represents the over all size of the atoms which includes its valence shell in a non bonded situation. It is the half of the distance between two similar atoms in separate molecules in a solid. The atomis radius decreases across a period and increases down the group. Same trends are observed in case of ionic radius. Ionic radius of the species having same number of electrons depends on the number of protons in their nuclei. Sometimes, atomic and ionic radii give unexpected trends due to poor shielding of nuclear charge by `d`-and `f-`orbital electrons.
Now answer the following three questions:
`C^(+), CI^(-), Ca^(2+), S^(2-)` ions are isoelectronic. The decreasing order of their size is:

Options

A. `Ca^(2+) gt K^(+) gt CI^(-) gt S^(2-)`
B. `S^(2-) gt CI^(-) gt K^(+) gt Ca^(2+)`
C. `K^(+) gt CI^(-) gt Ca^(2+) gt S^(2-)`
D. `S^(2-) gt CI^(-) gt Ca^(2+) gt K^(+)`

Answer

B

Explanation

Ionic size `prop (1)/("Nuclear charge")` for isoelectronic species.