Among the isoelectronic ions
`(O^(2-), N^(3-), Mg^(2+),Na^+)`, the ions with least and highest ionic radius are respectively

Options

A. `Mg^(2+), N^(3-)`
B. `Mg^(2+),O^(2-)`
C. `Na^(+),N^(3-)`
D. `Na^(+),O^(2-)`

Answer

A

Explanation

If the ions derived from different atoms are isoelectronic, that means they all have same number of electrons in their electronic shells. Also, they will have get same electronic configuration but their nuclear charge will differ because of their difference in number of protons in the nucleus. With increase in number of protons in the nucleus the electrons are more attracted towards nucleus thereby causing the decrease in ionic radius.
The given ions are :
`Mg^(2+)`-number of protons = 12 and number of electrons = 10
`Na^+`-number of protons = 11 and number of electrons = 10
`O^(2-)`-number of protons = 8 and number of electrons = 10
`N^(3-)`-number of protons = 7 and number of electrons = 10
Hence, the correct order of ionic radius of given ions will be
`{:(Mg^(2+)),("Least ionic"):}" "{:(ltNa^+),("radius"):}" "{:(ltO^(2-)),():}" "{:(ltN^(3-)),("Highest ionic radius"):}`