Question

Which of the following order is not correct for the 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. \( {N}^{3-} < {S}^{2-} < {O}^{2-} < {P}^{3-} \)

Hint:

Ionic radii increase as we move down a group in the periodic table. The greater the negative charge on an ion, the larger its radius.

Answer 1

The Correct Option is A

Step 1: Ionic radii increase as we move down a group in the periodic table because the number of electron shells increases, making the ion larger. 
Step 2: Among the given ions, the ionic radii should follow the order: \[ {N}^{3-}>{O}^{2-}>{P}^{3-}>{S}^{2-}, \] because: - \( {N}^{3-} \) has the smallest ionic radius due to the high effective nuclear charge acting on the electrons. 
- \( {S}^{2-} \) has the largest ionic radius because sulfur is larger in size and has fewer protons to hold the electrons tightly. 
Step 3: The order in option (A) is incorrect, as \( {O}^{2-} \) should have a larger ionic radius than \( {N}^{3-} \).