Use Lewis theory to determine the chemical formula for the compound formed between Rb and O.

Question

Rb₃O₂

RbO₂

RbO

Rb₂O

Rb₂O₃

_{I have narrowed it down to} Rb₂O_{3, it has an even number of Valence Electrons at 20, but drawing the Lewis structure is where I am slightly uncertain and struggling.} Rb₂O has 8 valence electrions as well, but that structure doesn't seem to work. I am not sure what I am missing.

Answer 1

Hello Mike

Rb (rubedium) is a Group 1 metal, and O (oxygen) is a Group 16 nonmetal. Metals and nonmetals combine to form ionic compounds where the metal will lose its valence electron(s) to form a metal cation while the nonmetal will gain electrons to form a nonmetal anion.

Since Rb is in Group 1, it has 1 valence electron that it loses to form a rubedium cation and achieve a noble gas configuration. The formation of the cation is denoted below:

Rb --> Rb⁺ + e^-

Since Oxygen is in Group 16, it needs to gain 2 electrons to complete its octet and achieve a noble gas configuration. The formation of the anion is denoted below:

O + 2e^- --> O^2-

When a metal cation and nonmetal anion combine to form an ionic compound, the sum of their charges must equal zero since the compound is neutral. For the formation of rubedium oxide, you need two (2) Rb cations and one (1) oxygen anion for the sum of their charges to equal zero. The resultant chemical formula is therefore:

Rb₂O (answer choice D)