To determine which pairs of elements are most likely to form a stable ionic bond, we should consider their positions in the periodic table and their tendency to lose or gain electrons to achieve a stable electron configuration.
An ionic bond typically forms between a metal and a nonmetal, where the metal donates electrons and becomes a cation, while the nonmetal accepts electrons and becomes an anion.
Now, let's evaluate each option:
a. Carbon (Period IVA) and Oxygen (Period VIA): Both elements are nonmetals. Carbon typically shares electrons rather than forms an ionic bond, and oxygen also tends to gain electrons to form covalent bonds. Therefore, this pair is unlikely to form an ionic bond.
b. Lithium (Period IA) and Fluorine (Period VIIA): Lithium is a metal that readily loses one electron to form Li⁺, while fluorine is a nonmetal that readily gains one electron to form F⁻. This pair is very likely to form a stable ionic bond (LiF).
c. Magnesium (Period IIA) and Iron (Period VIIIB): Magnesium is a metal that can form Mg²⁺ by losing two electrons, but iron can exist in multiple oxidation states (often +2 or +3) and is less likely to form a stable ionic bond with magnesium. They tend to form metallic bonds or complex structures, so this pair is less likely to form a simple ionic bond.
d. Potassium (Period IA) and Tin (Period IVA): Potassium is a metal that loses one electron to form K⁺, while tin, typically a metalloid, does not readily form anions and can share electrons. It's not likely for potassium and tin to form a stable ionic bond.
Based on this analysis, the pair that is most likely to form a stable ionic bond is b. Lithium (Period IA) and Fluorine (Period VIIA).