This answer is not correct. You probably thought that the fluorine atom would be larger since there are fewer protons in the nucleus attracting the outer shell electrons. Your logic is good but remember Coulomb's law which describes the force of attraction between two charges. This attractive force is equal to k x Q₁Q₂/r². The distance "r" is squared and this makes an exponential reduction in the force as the distance increases. The magnitude of the charges, "Q", will show a linear effect at increasing the attraction.