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Question:

Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species: O²⁻, S²⁻, N³⁻, P³⁻?

O²⁻<P³⁻<N³⁻<S²⁻

N³⁻<O²⁻<P³⁻<S²⁻

N³⁻<S²⁻<O²⁻<P³⁻

O²⁻<N³⁻<S²⁻<P³⁻

Solution:

The correct order representing the increasing trend (smallest to largest) of ionic radii of the given species is O²⁻ < N³⁻ < S²⁻ < P³⁻. Since Sulphur and Phosphorus belong to the 3rd period, their size will be larger than those that belong to the 2nd period like Nitrogen and Oxygen. While comparing the size of P³⁻ and S²⁻, first we will compare the size of Sulphur and Phosphorus; the size of Phosphorus is greater than Sulphur as the internuclear charge increases as we go from left to right in the periodic table. On increasing charge, the repulsion increases and hence, the size. The more the charge on the atom, the more will be the atomic radii.