Certainly! Let's determine the ion that boron will most likely form based on the octet rule.
The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a full outer shell of 8 electrons, similar to the electron configuration of the noble gases.
1. Understand Boron's Electron Configuration:
- Boron (B) has an atomic number of 5, which means a neutral boron atom has 5 protons and 5 electrons.
- The electron configuration for a neutral boron atom is \(1s^2 2s^2 2p^1\).
2. Determine the Electron Gain or Loss:
- Boron has 3 electrons in its outermost shell (2s^2 2p^1).
- To achieve a stable octet, boron can either gain 5 more electrons to fill the outer shell or lose 3 electrons to empty the outer shell.
- Gaining 5 electrons is highly unfavorable due to the high energy required to add such a large number of electrons, especially since it's easier for boron to lose electrons.
3. Forming a Stable Ion:
- By losing 3 outermost electrons (2s^2 2p^1), boron can achieve the electron configuration of helium \(1s^2\), which is a stable, noble gas configuration.
- Losing 3 electrons will result in a positively charged ion.
4. Charge of the Ion:
- When boron loses 3 electrons, it ends up with 3 more protons than electrons, leading to a \(\text{B}^{3+}\) charge.
Thus, based on the octet rule, boron will most likely form a [tex]\(\text{B}^{3+}\)[/tex] ion.
