The formula for the resulting ion when a boron atom loses its valence electrons is B 3+. Boron achieves this by losing all three valence electrons, resulting in a trivalent cation with +3 charge.
When a boron atom loses its valence electrons, the formula for the resulting ion is B 3+. Boron is a metalloid with an electronic configuration of 2s²2p¹, which means it has three valence electrons. In order to achieve a full valence shell, boron tends to lose these three electrons, resulting in a trivalent cation with a charge of +3.
Boron commonly makes only three covalent bonds because it has three valence electrons available for bonding, leaving it with six valence electrons in compounds like BF3 (boron trifluoride). Therefore, when forming an ion, boron achieves stability by losing all three of its valence electrons, giving it a 3+ oxidation state in its stable ionic form.