Question

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

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

Answer 1

The Correct Option is C

The question tests understanding of ionic radii among isoelectronic species. Here, we are comparing the ionic radii of the following ions: N^{3-}, O^{2-}, and F^-\.

Concept: Isoelectronic species have the same number of electrons but different nuclear charges. As the nuclear charge increases, the attraction between the nucleus and the electrons increases, pulling the electrons closer to the nucleus and thereby decreasing the ionic radius.

For the given species:

• N^{3-} has 7 protons and 10 electrons.
• O^{2-} has 8 protons and 10 electrons.
• F^-\ has 9 protons and 10 electrons.

Since all three have the same number of electrons (10), they are isoelectronic. The nuclear charge is smallest in N^{3-} and largest in F^-\. Hence, the radii will be:

• N^{3-} with the least nuclear charge and largest radius.
• O^{2-} with a moderate nuclear charge.
• F^-\ with the highest nuclear charge and smallest radius.

Thus, the correct order of ionic radii from largest to smallest is:

• N^{3-} = 1.71 \, \text{\AA}
• O^{2-} = 1.40 \, \text{\AA}
• F^- = 1.36 \, \text{\AA}

Therefore, the correct answer is: 1.71, 1.40, and 1.36.