(solved)Question 3.12 of NCERT Class XI Chemistry Chapter 3

$\displaystyle \underline{\underline{\text{Part (a)}}}$

N^3- has N(Z=7) + 3 = 10 electrons

O^2- has O(Z=8) + 2 = 10 electrons

F^- has F(Z=9) + 1 = 10 electrons

Na⁺ has Na(Z=11) - 1 = 10 electrons

Mg²⁺ has N(Z=12) - 2 = 10 electrons

Al³⁺ has N(Z=13) - 3 = 10 electrons

All are iso-electronic with 10 electrons $\displaystyle \underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (b)}}}$

Point 1. All have the same electrons but nuclear charge is increasing from 7 (N3-) to 13 (Al3+)

Point 2: So there is a greater attraction between the nucleus and its electrons that decreases the radius

Hence the cation Al3+ is the smallest and the anion N3- the largest

Al3+ < Mg2+ < Na+ < F- < O2- < N3- $\displaystyle \underline{Ans}$

Note: Noble gas Ne is also isolectronic but it doesn't form covalent bonds like others, but its vanderwal radius rather than ionic/atomic radius is more meaningful.