Question

The ionic radii (in ${{\text{A}}^{\text{o}}}$ ) of ${{\text{N}}^{{\text{3 - }}}}{\text{, }}{{\text{O}}^{{\text{2 - }}}}$ and ${{\text{F}}^{\text{ - }}}$ are respectively:
(A) ${\text{1}}{\text{.36, 1}}{\text{.40, and 1}}{\text{.71}}$
(B) ${\text{1}}{\text{.36, 1}}{\text{.71, and 1}}{\text{.40}}$
(C) ${\text{1}}{\text{.71, 1}}{\text{.40, and 1}}{\text{.36}}$
(D) ${\text{1}}{\text{.71, 1}}{\text{.36, and 1}}{\text{.40}}$

Answer 1

In an ion, the distance from the nucleus of the ion to the electron from that last shell is known as ionic radii of that ion. When an electron is added or removed from the shell it forms an anion or a cation respectively. More is the negative charge on an anion, more will be the electron electron repulsion and thus, less will be the effective nuclear charge.

Complete step-by-step solution
Every atom or ion has a positively charged nucleus and negatively charged electrons in the shells. The number of electrons in an atom is equal to the number of neutrons present in the nucleus. But, when an atom loses an electron, it becomes a cation with an extra positive charge. Similarly, when an atom gains an electron it becomes an anion with an extra negative charge.
Ionic radius is the distance between the nucleus and electron of the last shell in an ion. In the question, three species ${{\text{N}}^{{\text{3 - }}}}{\text{, }}{{\text{O}}^{{\text{2 - }}}}$ and ${{\text{F}}^{\text{ - }}}$ given and we have to find their ionic radii. Let us first find out the number of electrons present in these ions. Nitrogen (N) has an atomic number of 7, hence the number of electrons in ${{\text{N}}^{{\text{3 - }}}}$ will be 7 + 3 that is 10 electrons.
Similarly, for ${{\text{O}}^{{\text{2 - }}}}$ it will be $8 + 2 = 10$ . For ${{\text{F}}^{\text{ - }}}$ , $9 + 1 = 10$ . All three have the same number of electrons, that is they are isoelectronic.
In isoelectronic ions, the ionic radii are said to be larger as much as a large number of negative charges on it. That is species with a more negative charge will have large ionic radii. Therefore, as ${{\text{N}}^{{\text{3 - }}}}$ has more number of negative charges it will have a larger ionic radius, followed by ${{\text{O}}^{{\text{2 - }}}}$ and then ${{\text{F}}^{\text{ - }}}$ .
Thus, ionic radii of ${{\text{N}}^{{\text{3 - }}}}$ is ${\text{1}}{\text{.71}}$ ${{\text{A}}^{\text{o}}}$ , ${{\text{O}}^{{\text{2 - }}}}$ is ${\text{1}}{\text{.40}}$ ${{\text{A}}^{\text{o}}}$ , and ${{\text{F}}^{\text{ - }}}$ is ${\text{1}}{\text{.36}}$ ${{\text{A}}^{\text{o}}}$ .
Hence, the correct option will be option C - ${\text{1}}{\text{.71, 1}}{\text{.40, and 1}}{\text{.36}}$

Note
When an electron is added to the last shell, it starts repelling from electrons present on other shells increasing the size of the ion. However, in cation when an electron is lost, the electrons are attracted more by the nucleus reducing the size of the ion.