Question

The ionic radii of $O^{2-},F^{-},N{a}^{+}$ and $M{g}^{2+}$ are 1.35, 1.34, 0.95 and 0.66 $\mathring{A}$ respectively. The radius of the $Ne$ atom is:

• A. 1.54 $\mathring{A}$
• B. 1.12 $\mathring{A}$
• C. 0.85 $\mathring{A}$
• D. 0.50 $\mathring{A}$

Answer 1

The correct option is B 1.12 $\mathring{A}$

Here all these species are isoelectronic species.

The species which has maximum negative charge will have maximum ionic radii and the species which have maximum positive charge will have minimum ionic radii.

Thus, the order of ionic radii will be:

$O^{2-}>F^{-}>Ne>N{a}^{+}>M{g}^{2+}$

Here Ne has greater ionic radii than sodium and magnesium ion as Ne has a van der Waals radius which is larger than the ionic radii of these cations.

Thus radii of $Ne$ should be between $F^{-}$ and $N{a}^{+}$.

i.e. radii of $Ne$ should be greater than 0.95 $\mathring{A}$ and less than 1.34 $\mathring{A}$.

Thus ionic radii of $Ne$ should be 1.12$A^0$.

Hence, the correct option is $\text{B}$