Question

The ionic radii in (\({Å}\)) of \({N}^{3-}\), \({O}^{2-}\) and \({F}^-\) are respectively.

• A. 1.71, 1.40 and 1.36
• B. 1.71, 1.36 and 1.40
• C. 1.36, 1.40 and 1.71
• D. 1.36, 1.71 and 1.40

Hint:

In isoelectronic species, the ionic radius decreases with increasing nuclear charge because the greater positive charge attracts the electrons more strongly, causing the ion to contract.

Answer 1

The Correct Option is A

Step 1: {Isoelectronic Species}
The ions \( {N}^{3-} \), \( {O}^{2-} \), and \( {F}^- \) are isoelectronic, possessing an identical electron count. Each ion contains 10 electrons. Variations in nuclear charge, stemming from differing atomic numbers (N: Z=7, O: Z=8, F: Z=9), cause differences in their ionic radii. Nuclear charge increases from N to O to F across the periodic table.
Step 2: {Ionic Radii Order}
An elevated nuclear charge exerts a stronger attraction on electrons, leading to a diminished ionic radius. Consequently, ionic radii decrease sequentially from \( {N}^{3-} \) to \( {O}^{2-} \) to \( {F}^- \). The resulting order of ionic radii is: \[ {N}^{3-}>{O}^{2-}>{F}^- \] Thus, option (A) is correct.