Final answer:
The force on the outermost electrons in Argon would be expected to be weaker than that in Chlorine, due to Argon's complete valence electron shell and Chlorine's tendency to gain an additional electron, which leads to stronger effective nuclear charge in Chlorine.
Step-by-step explanation:
The force on the outermost electrons in an atom is largely dependent on the effective nuclear charge and the distance of the electrons from the nucleus. For Argon (Ar), the outermost electrons experience a force that is weaker than that of Chlorine (Cl) as we consider the effective nuclear charge and the shielding effect. Chlorine has a total of 17 protons compared to Argon's 18, but because Chlorine has one less electron in its valence shell than Argon, it is more inclined to attract an additional electron and thus holds its remaining electrons more tightly. Furthermore, since Chlorine will gain an electron to form the anion Cl−, its radii may be larger than the neutral atom, which slightly reduces the force on its outer electrons compared to its neutral state. However, with Argon being a noble gas with a filled valence shell, its outer electrons are shielded by a full set of inner electrons, hence reducing the effective nuclear charge experienced by the outermost electrons, making Argon's outermost electrons less tightly held compared to Chlorine. The implications of these atomic structures explain why the London Dispersion Forces (LDF) are generally stronger in larger atoms like Argon, as larger atoms tend to have greater polarizability due to more electrons and further distance from the nucleus, leading to stronger intermolecular forces in compounds they form.