Q.Select the smaller ion from each of the following pairs:
a. K⁺, Li⁺
b. N³⁻, F⁻

Step 1a. Compare K⁺ and Li⁺. Both are group-1 cations, but K is period 4 (K⁺ retains 3 shells) while Li is period 2 (Li⁺ retains only 1 shell) — the down-group radius trend applies (section 7.5.2.1), so Li⁺ is far smaller.

Step 2a. Answer. Li⁺ is the smaller ion.

Step 1b. Compare N³⁻ and F⁻. Electron counts: N³⁻ has $7+3=10$ electrons; F⁻ has $9+1=10$ electrons — the two are isoelectronic species. Among isoelectronic species, the one with the LARGER actual nuclear charge is smaller (section 7.5.2.2); F's nuclear charge (+9) is larger than N's (+7).

Step 2b. Answer. F⁻ is the smaller ion.

Applying the down-group radius trend (K⁺/Li⁺) and the isoelectronic-species nuclear-charge rule (N³⁻/F⁻)

• Treating N³⁻ vs F⁻ as an ordinary same-group or same-period comparison instead of recognising them as an isoelectronic pair requiring the nuclear-charge rule.
• Assuming a triply-charged anion (N³⁻) must automatically be larger than a singly-charged one (F⁻) without first checking they are isoelectronic and then comparing nuclear charge.