Question

Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species $O^{2-},S^{2-},N^{3-},P^{3-}$?

• A. $O^{2-}<N^{3-}<S^{2-}<P^{3-}$
• B. $O^{2-}<P^{3-}<N^{3-}<S^{2-}$
• C. $N^{3-}<O^{2-}<P^{3-}<S^{2-}$
• D. $N^{3-}<S^{2-}<O^{2-}<P^{3-}$

Answer 1

The correct option is A $O^{2-}<N^{3-}<S^{2-}<P^{3-}$

The correct order representing the increasing trend (smallest to largest) of ionic radii of the given species is $O^{2-}$. Since, Sulphur and Phosphorus belongs to 3rd period, their size will be larger than those who belongs to 2nd period like Nitrogen and Oxygen. While comparing the size of $P^{3-}$ and $S^{2-}$, first we will compare the size of Sulphur and Phosphorus, the size of Phosphorus is greater than Sulphur as the inter nuclear charge increases as we go from left to right in periodic table. On increasing charge, the repulsion increases and hence, the size. The more charge on the atom more will be the atomic radii.