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 species, meaning they all have the same number of electrons. Specifically, each of these ions has 10 electrons. Despite having the same number of electrons, their ionic radii will differ because of the varying nuclear charges. The nuclear charge increases from nitrogen to oxygen to fluorine as we move across the periodic table: - \( {N} \) has atomic number \( Z = 7 \), - \( {O} \) has atomic number \( Z = 8 \), - \( {F} \) has atomic number \( Z = 9 \).
Step 2: {Ionic Radii Order} 
As the nuclear charge increases, the electrons are attracted more strongly by the nucleus, which results in a decrease in ionic radius. Therefore, the ionic radii will decrease as we move from \( {N}^{3-} \) to \( {O}^{2-} \) to \( {F}^- \). This gives the following order of ionic radii: \[ {N}^{3-}>{O}^{2-}>{F}^- \] Hence, the correct answer is (A).