Boron's outer electron structure is 2s²2p¹, giving it one less valence electron than valence orbitals. What oxidation state does boron typically exhibit in its stable compounds?

1. Boron has three valence electrons (2s²2p¹), and it typically forms three bonds by losing or sharing all three. 2. The text states: "boron exhibits an oxidation state of 3+ in most of its stable compounds." 3. This electron deficiency (three electrons but four valence orbitals) is also what allows boron compounds to act as electron-pair acceptors (Lewis acids). 4. So boron's typical oxidation state in stable compounds is 3+, not 1+, 2+, or 4+ (which is silicon's typical state, not boron's). _Source: OpenStax Chemistry Ch 18 "Representative Metals, Metalloids, and Nonmetals", p.978 §18.3 Structure and General Properties of the Metalloids_