Predict the formula of the ionic compound formed from the reaction of rubidium and oxygen atoms.
The correct answer and explanation is:
When rubidium (Rb) reacts with oxygen (O), an ionic compound is formed. Rubidium is an alkali metal found in Group 1 of the periodic table. Alkali metals like rubidium have one electron in their outermost shell, which they tend to lose easily to achieve a stable electron configuration. In contrast, oxygen is a nonmetal located in Group 16 of the periodic table and has six electrons in its outer shell. To achieve a full octet (eight electrons), oxygen atoms need two additional electrons.
During the reaction, rubidium atoms lose one electron each to form Rb+\text{Rb}^+ ions. Oxygen, needing two electrons, gains two electrons to form an O2−\text{O}^{2-} ion. The resulting ionic compound is formed by the electrostatic attraction between the positively charged rubidium ions and the negatively charged oxide ions.
To balance the charges, two rubidium ions are required for each oxide ion. This is because the Rb+\text{Rb}^+ ion has a charge of +1, while the O2−\text{O}^{2-} ion has a charge of -2. Therefore, two Rb+\text{Rb}^+ ions will balance out the -2 charge of one O2−\text{O}^{2-} ion, leading to the empirical formula Rb2O\text{Rb}_2\text{O}.
In summary, the formula of the ionic compound formed from the reaction of rubidium and oxygen is Rb2O\text{Rb}_2\text{O}. This compound is known as rubidium oxide. The reaction involves the transfer of electrons from rubidium to oxygen, resulting in the formation of Rb+\text{Rb}^+ and O2−\text{O}^{2-} ions that combine in a 2:1 ratio to form the ionic compound.