To rank Na-S, Cs-F, and Rb-Cl from least ionic to most ionic, consider the electronegativity differences. The correct order is Na-S < Rb-Cl < Cs-F, with Cs-F being the most ionic due to fluorine's high electronegativity.
The question asks to rank the bonds Na-S, Cs-F, and Rb-Cl from least ionic to most ionic based on the difference in electronegativity between the atoms forming each bond. Generally, as the difference in electronegativity between two atoms increases, the bond becomes more ionic. Electronegativity tends to increase across a period and decrease down a group in the periodic table.
Sodium (Na) is less electronegative than both rubidium (Rb) and cesium (Cs), but sulfur (S) is more electronegative than both chlorine (Cl) and fluorine (F). Since fluorine is the most electronegative element, Cs-F would be expected to be the most ionic bond. Between Na-S and Rb-Cl, Rb is more electropositive than Na, suggesting that Rb-Cl would be more ionic than Na-S.
Therefore, the bonds would be ranked from least ionic to most ionic as follows: Na-S < Rb-Cl < Cs-F.