Question

The correct order of the ionic radii of ${\mathrm{O}}^{2-}$, ${\mathrm{N}}^{3-}$,${\mathrm{F}}^{-}$, ${\mathrm{Mg}}^{2+}$, ${\mathrm{Na}}^{+}$ and${\mathrm{Al}}^{3+}$ is:

• A. ${\mathrm{N}}^{3-}<{\mathrm{O}}^{2-}<{\mathrm{F}}^{-}<{\mathrm{Na}}^{+}<{\mathrm{Mg}}^{2+}<{\mathrm{Al}}^{3+}$
• B. ${\mathrm{N}}^{3-}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}<{\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{Al}}^{3+}$
• C. ${\mathrm{Al}}^{3+}<{\mathrm{Na}}^{+}<{\mathrm{Mg}}^{2+}<{\mathrm{O}}^{2-}<{\mathrm{F}}^{-}<{\mathrm{N}}^{3-}$
• D. ${\mathrm{Al}}^{3+}<{\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}<{\mathrm{N}}^{3-}$

Answer 1

The correct option is D

${\mathrm{Al}}^{3+}<{\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}<{\mathrm{N}}^{3-}$

The correct option is (D):

Explanation for the correct answer:

Ionic radii:-

1. Basically ionic radii is the distance that is from the nucleus of an ion and up to which it will have an influence on the electron cloud.
2. It is actually explained as the effective distance between the nucleus as well as the outer shell of the ion that will have the electron.
3. As the cation is formed by the loss of one or more than one electron, so because of this loss the radii would be smaller as compared with the original atom. So, $\mathrm{ionic}\mathrm{radii}\propto \frac{1}{\mathrm{positive}\mathrm{charge}\mathrm{on}\mathrm{the}\mathrm{cation}}$.
4. As such the anion is formed by the gain of one or more than one electron, so because of this gain, the radii here would be greater or larger than that of the original atom. So, $\mathrm{ionic}\mathrm{radii}\propto \mathrm{negative}\mathrm{charge}\mathrm{on}\mathrm{the}\mathrm{anion}$.
5. And in case if all the species are isoelectronic which means that the group of ions, atoms or molecules will have the same arrangement of the electrons then,

Therefore, the correct option is (D):- ${\mathrm{Al}}^{3+}<{\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}<{\mathrm{N}}^{3-}$.