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

To determine the increasing order of ionic radii for the ions \(O^{2-}\), \(S^{2-}\), \(N^{3-}\), and \(P^{3-}\), we need to understand how ionic radii are affected by the atomic structure and charge: 

First, consider the charge of the ions. When an atom gains electrons to form a negative ion, the ionic radius typically increases due to increased electron-electron repulsion in the valence shell.

Compare ions within the same group and neighboring groups in the periodic table:

• \(N^{3-}\) and \(P^{3-}\) are in Group 15.
• \(O^{2-}\) and \(S^{2-}\) are in Group 16.

Ionic radii increase down a group because of the addition of electron shells. Therefore, \(S^{2-}\) is larger than \(O^{2-}\) and \(P^{3-}\) is larger than \(N^{3-}\).

Between different groups, compare \(O^{2-}\) and \(N^{3-}\), and \(S^{2-}\) and \(P^{3-}\). Generally, the addition of more electrons leads to a larger ionic radius and group 15 ions like \(N^{3-}\) and \(P^{3-}\) have extra electrons compared to group 16 ions like \(O^{2-}\) and \(S^{2-}\).

Thus, comparing across groups, the order from smallest to largest ionic radii is:

• \(O^{2-}\) is smaller than \(N^{3-}\).
• \(N^{3-}\) is smaller than \(S^{2-}\).
• \(S^{2-}\) is smaller than \(P^{3-}\).

Therefore, the correct arrangement of ionic radii in increasing order is: \(O^{2-} < N^{3-} < S^{2-} < P^{3-}\).