In order for sodium to have 8 valence electrons, it would need to either gain _ electrons or lose _ electron.
The Correct Answer and Explanation is:
In order for sodium (Na) to have 8 valence electrons, it would need to either gain 7 electrons or lose 1 electron.
Explanation:
Sodium is an alkali metal located in group 1 of the periodic table. Its atomic number is 11, which means it has 11 electrons when neutral. The electron configuration of sodium is (1s^2 2s^2 2p^6 3s^1). This shows that sodium has a total of 11 electrons: 2 in the first shell, 8 in the second shell, and 1 in the third shell. The first two shells can hold a maximum of 2 and 8 electrons, respectively, while the third shell can hold more than 8 electrons.
The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a full outer shell of 8 electrons, which is energetically favorable and leads to increased stability. For sodium, the outermost shell (the third shell) has only 1 electron. To reach an octet, sodium has two main options:
- Gain 7 Electrons: If sodium were to gain 7 electrons, it would have a total of 8 electrons in its outer shell. However, this is energetically unfavorable for sodium due to its position on the periodic table and the high energy required to add such a large number of electrons. Gaining multiple electrons would also create a very negatively charged ion, which is less likely for an alkali metal.
- Lose 1 Electron: The more favorable option for sodium is to lose its single valence electron. By losing this 1 electron, sodium achieves a stable electron configuration similar to that of neon (the nearest noble gas), which has a full outer shell of 8 electrons. This loss results in the formation of a positively charged ion, Na⁺. This process is energetically favorable because it allows sodium to achieve a more stable electron configuration with less energy required than gaining multiple electrons.
In summary, while sodium could theoretically gain 7 electrons to achieve a full octet, the more practical and favorable path is for it to lose 1 electron, resulting in a stable Na⁺ ion with a complete octet in the previous electron shell.