To understand which ion boron is most likely to form based on the octet rule, we need to look at its position in the periodic table and its electron configuration.
1. Element and Electron Configuration:
- Boron (B) is the 5th element in the periodic table.
- Its electron configuration is \( 1s^2 \, 2s^2 \, 2p^1 \).
2. Valence Electrons:
- The valence electrons are those in the outermost shell. For boron, the valence electrons are in the \( 2s \) and \( 2p \) orbitals.
- Therefore, boron has 3 valence electrons.
3. Octet Rule:
- The octet rule states that atoms tend to gain, lose, or share electrons to achieve a full set of 8 valence electrons, similar to the electron configuration of a noble gas.
4. Achieving a Stable Configuration:
- Boron, having 3 valence electrons, will need to either gain 5 more electrons to reach 8 valence electrons (which would form \(B^{5-}\) but that's quite unstable and unlikely due to electrostatic repulsion) or lose 3 electrons to empty its valence shell, achieving the stable configuration of the noble gas helium (with 2 electrons).
- Hence, the most favorable and simplest way for boron to achieve a stable electron configuration is by losing its 3 valence electrons.
5. Formation of Ions:
- By losing 3 valence electrons, boron will form a \(3+\) charge, becoming \(B^{3+}\).
Thus, based on the octet rule, boron most likely forms a [tex]\(B^{3+}\)[/tex] ion. The correct choice is [tex]\(B^{3+}\)[/tex].
