JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
The ionic radii of K⁺, Na⁺, Al³⁺ and Mg²⁺ are in the order:
- (a) Na⁺ < K⁺ < Mg²⁺ < Al³⁺
- (b) Al³⁺ < Mg²⁺ < K⁺ < Na⁺
- (c) Al³⁺ < Mg²⁺ < Na⁺ < K⁺
- (d) K⁺ < Al³⁺ < Mg²⁺ < Na⁺
Answer and solution
Answer (c) Al³⁺ < Mg²⁺ < Na⁺ < K⁺
Solution
Chemistry ma’am · 4 steps
Spot the isoelectronic trio first, then remember that K⁺ has one extra shell over all of them.
- Isoelectronic set. Na⁺, Mg²⁺ and Al³⁺ each have 10 electrons in the same two shells. The only difference is the number of protons pulling on those electrons: 11, 12 and 13.
- More protons, smaller ion. With the same electron cloud, a larger nuclear charge pulls it in more tightly. So the size falls as the positive charge rises: Al³⁺ < Mg²⁺ < Na⁺.
- Where K+ fits. K⁺ has 18 electrons spread over three shells, one more shell than the other three ions. The extra shell outweighs its higher nuclear charge, so K⁺ is the largest.
Ion Protons Electrons Shells Radius (approx.) Al³⁺ 13 10 2 54 pm Mg²⁺ 12 10 2 72 pm Na⁺ 11 10 2 102 pm K⁺ 19 18 3 138 pm - So. The order is Al³⁺ < Mg²⁺ < Na⁺ < K⁺, which is option (c).
The idea to keep. For ions, compare shells first: an extra shell means a bigger ion. Within an isoelectronic series, the ion with more protons is smaller.
Answer from the NTA final answer key. Solution by PrepWiser.