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Question: Arrange the following species in order of increasing size. \(Ar\), \({K^ + }\), \(C{l^ - }\), \({S...

Arrange the following species in order of increasing size.
ArAr, K+{K^ + }, Cl−C{l^ - }, S2−{S^{2 - }}, Ca2+C{a^{2 + }}.
(A) Ca2+<K+<Ar<Cl−<S2−C{a^{2 + }} < {K^ + } < Ar < C{l^ - } < {S^{2 - }}
(B) Ca2+<Ar<K+<Cl−<S2−C{a^{2 + }} < Ar < {K^ + } < C{l^ - } < {S^{2 - }}
(C) Ca2+<K+<Ar<S2−<Cl−C{a^{2 + }} < {K^ + } < Ar < {S^{2 - }} < C{l^ - }
(D) None of these

Explanation

Solution

The atomic radius and ionic radius are different for different elements because of their different nuclear charge so we can arrange them according to their size. Cation is smaller than the parent atom due to greater positive charge and effective nuclear charge and opposite is the case of anion.

Complete step by step solution:
As we know that the ionic radius is the distance between the center of nucleus of an atom and the electron in outermost orbit of an ion. It has different values for different elements because of their different nuclear charges so we can arrange them according to their size. Cation is smaller than the parent atom due to greater positive charge and effective nuclear charge and anion is larger than the parent atom due to the low values of effective nuclear charge.
We should know that as the positive charge on the cation increases, the ionic radius decreases and the greater the positive charge, smaller will be the size of the atom and as the value of negative charge increases on an anion, the ionic radius increases and the greater the negative charge value, larger will be the atom size.
In the given species all have the same number of electrons but their radius is different, the net repulsion in this case will outweigh the nuclear charge and ion will expand in size, so according to their positive and negative charge we can see that sulphur will have the largest size followed by chlorine, then comes argon with no charge followed by potassium and calcium.
So the increasing order will be: Ca2+<K+<Ar<Cl−<S2−C{a^{2 + }} < {K^ + } < Ar < C{l^ - } < {S^{2 - }}.

Hence, the correct answer is (A).

Note: when we move from top to bottom in periodic table, the ionic radius of elements increases as new shells are added and while moving from left to right the ionic radius of elements decreases as the number of shells remains the same.