In an exam paper it says that the lattice energy of NaCl is more exothermic than MgCl [sic] as the Na+ cation has a smaller ionic radius than Mg+.
Here are the resources given:
What is the reasoning for for Na+ having a smaller ionic radius even though Mg+ has a higher proton number?
We actually do mean $\ce{Mg^+}$ here. The idea is to show what factors are involved when an alkaline earth metal forms stable compounds with $\ce{M^{2+}}$ ions whereas an alkali metal favors $\ce{M^+}$. In this case, as pointed out in some of the comments, $\ce{Mg^+}$ is bulked up by the electron that remains in the relatively diffuse, loosely held $3s$ subshell. You need to remove that electron, making $\ce{Mg^{2+}}$, to get a compact ion that gives good lattice energies in ionic crystals.
Scandium monosulfide
Although $\ce{MgCl}$ and monatomic $\ce{Mg^+}$ are not real materials, the case of scandium monosulfude, $\ce{ScS}$ is instructive. A simple ionic model proposing $\ce{Sc^{2+}}$ and $\ce{S^{2-}}$ would require the scandium to retain a valence electron in an outer-shell orbital, apparently $3d$. In an early-group metal such orbitals are diffuse and, if occupied, would inhibit tight binding of the oppositely charged ions. Instead, according to Ref. [1] which identifies $\ce{ScS}$ as a metallic conductor, the extra electron per scandium atom is delocalized into a conduction band. This allows the sulfide ions to bind tightly with compact $\ce{Sc^{3+}}$. Thus scandium monosulfide is stable not as a conventional ionic salt, but as a compound that combines ionic and metallic characteristics.
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