Which of the following are most likely to form a stable ionic bondaCarbon (Period IVA) and Oxygen (Period VIA)bLithium (Period IA) and Fluorine (Period VIIA)cMagnesium (Period IIA) and Iron (Period VIIIB)dPotassium (Period IA) and Tin (Period IVA)

Answer 1

To determine which pairs of elements are most likely to form stable ionic bonds, we need to look for combinations of metals and nonmetals, as ionic bonds typically form between these types of elements.

1. Carbon (Period IVA) and Oxygen (Period VIA): Both carbon and oxygen are nonmetals, and they are more likely to form covalent bonds rather than ionic bonds.

2. Lithium (Period IA) and Fluorine (Period VIIA): Lithium is a metal, and fluorine is a nonmetal. This combination is very likely to form a stable ionic bond because lithium can lose one electron to form Li⁺, and fluorine can gain one electron to form F⁻.

3. Magnesium (Period IIA) and Iron (Period VIIIB): Both magnesium and iron are metals. They do not typically form ionic bonds with each other; instead, metallic bonding or alloying would occur.

4. Potassium (Period IA) and Tin (Period IVA): Potassium is a metal, but tin is also a metal (though it can exhibit some nonmetallic character). However, they are unlikely to form a stable ionic bond as both are metals.

Based on the above analysis, the pair most likely to form a stable ionic bond is b) Lithium (Period IA) and Fluorine (Period VIIA).