Question

The increasing order of the ionic radii of the given isoelectronic species is:
(A) $C{l^ - },C{a^{2 + }},{K^ + },{S^{2 - }}$
(B) ${S^{2 - }},C{l^ - },C{a^{2 + }},{K^ + }$
(C) $C{a^{2 + }},{K^ + },C{l^ - },{S^{2 - }}$
(D) ${K^ + },{S^{2 - }},C{a^{2 + }},C{l^ - }$

Answer 1

In the case of isoelectronic species, we can say that as the atomic number of the species increases the atomic radii of the ions decreases gradually and as the atomic number decreases the atomic radii of the ions increases.

Complete step by step solution:
We know that in isoelectronic species, the number of electrons is the same. However, their nuclear configurations are not same as they are not the same elements.
- Here, we need to compare the ionic radii of the isoelectronic species.
- So, the only difference will be the positive charge in the nucleus that keeps the electrons closer. We can say that as the atomic number increases, the number of protons present in the nucleus also increases. This increases the positive charge on the nucleus.
- As the positive charge increases, it keeps the electrons more closer and it results in a small ionic radius.
- Out of the given ions, $C{a^{2 + }}$ has the highest atomic number which is 20. So, the attraction with the electrons will be highest here and so it will have the lowest ionic radii.
- Atomic number of potassium in ${K^ + }$ is 19. So, the positive charge will be less in comparison with calcium ions. So, its ionic radius will be larger than calcium ions.
- Atomic number of Cl is 17. So, $C{l^ - }$ ions will have a larger ionic radius than potassium ions.
- Atomic number of S is 16. So, ${S^{2 - }}$ has the lowest atomic number of the given ions. So, it will have the largest atomic radius.
So, if we arrange the atomic radius of the given ions, the increasing order will be $C{a^{2 + }},{K^ + },C{l^ - },{S^{2 - }}$.

So, the correct answer is (C).

Note: Note that here in the case of isoelectronic species, the charge on the ions does not have any effect on the ionic radii of the ions. The deciding factor is the positive charge present in the nucleus in the form of protons.