Final answer:
To draw the Lewis structure for BH3, show B in the center with three single bonds to hydrogen atoms. When B reacts with Rb, they likely form an ionic compound with the ions Rb+ and B-. BH3 is an exception to the octet rule, as B only has six valence electrons in this molecule.
Step-by-step explanation:
To draw the Lewis structure for BH3, and to predict the compound B + Rb might form, we need to apply the general strategy for drawing Lewis structures.
Boron (B) has three valence electrons, and in its compound BH3, it bonds with three hydrogen atoms. The Lewis structure for BH3 shows boron in the center with three single bonds, each to a hydrogen atom (H). Each H atom has two electrons (one from B and one from itself) completing its valence shell. Boron achieves an incomplete octet with only six valence electrons in this case, which is an exception to the octet rule.
When B reacts with rubidium (Rb), a metal, it's likely to form an ionic compound due to the transfer of electrons from Rb to B. Rubidium, with one valence electron, is likely to lose that electron to boron, forming Rb+ and B- ions. However, a precise prediction requires more information on the reaction conditions and stoichiometry.