Arrange decreasing order of K^(+) , Cl^(-) , S^(2-) , Ca^(2+)with explanation.

Arrange decreasing order of K^(+) , Cl^(-) , S^(2-) , Ca^(2+)with expl

1.	Arrange decreasing order of K^(+) , Cl^(-) , S^(2-) , Ca^(2+)with explanation.
Answer» Solution :`K^(+) , Cl^(-) , S^(2-) , Ca^(2+)`ions having same e- but its radius are different because they have different radius. Positive ion having positive charge so effective nuclear charge increases, so radius decreases Negative ion having negative charge so effective nuclear charge decreases, so radius increases. S contain `16e^(-)` and 16 p but `S^(2-)` contain 16 p and `18e^(-)` . Cl contain 17P and `19e^(-) " but " K^(+)` contain17p and `18e^(-)`. K contain 19p and `19e^(-)" but " K^(+)` contain 19p and `18e^(-)` . Ca having 20p and `20e^(-), " but " Ca^(+2)` having 20p and `18e^(-)` . So, `S^(2-) , Cl^(-) , K^(+) , Ca^(2+)` contain `18e^(-)` but nuclear charges are different. In `Ca^(2+)` ion ATTRACTION between proton and ELECTRON is MAXIMUM and in `S^(-2)` ion attraction between proton and electron is minimum so, ORDER of ionic radius is as follows : ` S^(2-) gt Cl^(-) gt K^(+) gt Ca^(2+)`