ionic radius decreases across a period. which ion will have the smaller radius? fluoride f⁻¹ oxide o⁻²

Explanation:

Step1: Recall ionic - radius trend

Ionic radius decreases across a period for isoelectronic species. Fluoride ($F^{-}$) and oxide ($O^{2 -}$) are isoelectronic.

Step2: Analyze nuclear charge

$F^{-}$ has a higher nuclear charge (9 protons) compared to $O^{2 -}$ (8 protons). Higher nuclear charge pulls the electrons closer.

Answer:

Fluoride $F^{-1}$