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Question: The radii of \(N,{N^{3 - }},O{\text{ and }}{O^{2 - }}\) are in the order: A) \({O^{2 - }} > {N^{3 ...

The radii of N,N3−,O and O2−N,{N^{3 - }},O{\text{ and }}{O^{2 - }} are in the order:
A) O2−>N3−>O>N{O^{2 - }} > {N^{3 - }} > O > N
B) O2−>N3−>N>O{O^{2 - }} > {N^{3 - }} > N > O
C) N3−>O2−>N>O{N^{3 - }} > {O^{2 - }} > N > O
D) O2−>N3−>N>O{O^{2 - }} > {N^{3 - }} > N > O

Explanation

Solution

As atomic number increases, atomic radii decreases. Also, recall that among isoelectronic species, more is the negative charge, more will be the radii of the anion and also, size of anion is always greater than its parent or neutral atom. Here, N3−{N^{3 - }} and O2−{O^{2 - }} are anions of their neutral atom N and O respectively and are also isoelectronic species because they both have 10 electrons in total.

Complete step by step answer:
Nitrogen has atomic number 7 and atomic mass 14 while oxygen has atomic number 8 and atomic mass 16. Nitrogen (N) and oxygen (O) both belong to the same period number 2. The atomic size or atomic radii generally decreases across a period. It is because as we move from left to right within the period, the outer electrons enter in the same valence shell and the effective nuclear charge increases as the atomic number increases. Thus, atomic radii of NN would be larger than that of OO .
In general, size of anion is always greater than its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge. Thus, radii of N3−{N^{3 - }} will be greater than N and radii of O2−{O^{2 - }} will be greater than OO .
Atoms or ions which have the same number of electrons are called isoelectronic species. Here, N3−{N^{3 - }}and O2−{O^{2 - }} are isoelectronic species because both have 10 electrons. Among the isoelectronic species, anion with the greater negative charge will have the larger radii. N3−{N^{3 - }} has larger negative charge than O2−{O^{2 - }} ,therefore, radii of N3−{N^{3 - }} is larger than O2−{O^{2 - }}.
On summarising all the above reasons, we conclude that the order of radii of N,N3−,O and O2−N,{N^{3 - }},O{\text{ and }}{O^{2 - }}will be: N3−>O2−>N>O{N^{3 - }} > {O^{2 - }} > N > O.
So, the correct answer is “Option C”.

Note: Size of cation is always smaller than its parent atom while size of anion is greater than its parent atom. Among the isoelectronic species, the cation with the greater positive charge will have a smaller radius and an anion with the greater negative charge will have the larger radius.