Question

Which one of the following has the largest ionic radius?
A. $L{{i}^{+}}$
B. ${O}^{2-}$
C. ${{B}^{3+}}$
D. ${{F}^{-}}$

Answer 1

Hint : The ionic radius of the atoms or ions can be determined using $\dfrac{Z}{e}$ratio.
Here Z = Atomic number of the particular element or ion.
e = number electrons present in the element or atom.

Complete step by step answer:
Now we have to calculate the $\dfrac{Z}{e}$ ratio of all the given ions in the question.
For Option A, $L{{i}^{+}}$.
We know that the Atomic number of lithium = 3
We know that the Number of electrons in helium = 2.

$\dfrac{Z}{e}$ ratio of $L{{i}^{+}}$= $\dfrac{3}{2}$= 1.5

Coming to option B, $O_{2}^{2-}$.
We know that the Atomic number of oxygen = 8
We know that the Number electrons in ${O}^{2-}$= 10.

$\dfrac{Z}{e}$ ratio of $O_{2}^{2-}$= $\dfrac{8}{10}$= 0.8

Now for option C, ${{B}^{3+}}$.
We know that the Atomic number of boron = 5
We know that the Number of electrons in ${{B}^{3+}}$= 2

$\dfrac{Z}{e}$ ratio of ${{B}^{3+}}$= $\dfrac{5}{2}$= 2.5

Finally, looking at option D, ${{F}^{-}}$.
We know that the Atomic number of fluorine = 9
We know that the Number of electrons in ${{F}^{-}}$= 10.

$\dfrac{Z}{e}$ ratio of ${{F}^{-}}$= $\dfrac{9}{10}$= 0.9.

As the $\dfrac{Z}{e}$ ratio increases the ionic radius of the ion decreases and vice versa.
Therefore $\dfrac{Z}{e}$ ratio of $O^{2-}$is 0.8. Means ${O}^{2-}$has low $\dfrac{Z}{e}$ratio with nothing but high ionic radius.

So, the correct option is B.

Note: As the number of electrons are going to add to an atom in their valence atomic orbitals, the ionic radius of the atom increases. In the same way, as the number of electrons are removed from the atom, the atomic or ionic radius of the atom decreases since the same number of protons are pulling lesser electrons. In the periodic table the atomic radius increases as we move from top to bottom and decreases as we are moving from left to right.