In noble-gas notation, what is the electron configuration of the magnesium ion $M^2+$ (formed by losing two electrons)?

• A. $\left[Ar\right]$
• B. $\left[H\right]s^2$
• C. $\left[Ne\right]$
• D. $\left[Ne\right]s^2$

Answer: (C)

Verified step by step guidance

1. Identify the atomic number of magnesium (Mg), which is 12, meaning a neutral Mg atom has 12 electrons.
2. Write the electron configuration of neutral magnesium using noble-gas notation: start with the noble gas preceding Mg, which is neon (Ne), representing 10 electrons, then add the remaining electrons in the 3s orbital, so the configuration is \([\mathrm{Ne}]\,3s^{2}\).
3. Understand that the magnesium ion \(\mathrm{Mg}^{2+}\) is formed by losing two electrons from the neutral atom, so remove two electrons from the outermost shell, which is the 3s orbital.
4. After removing two electrons from the 3s orbital, the electron configuration of \(\mathrm{Mg}^{2+}\) becomes the same as the noble gas neon, which is \([\mathrm{Ne}]\).
5. Therefore, the electron configuration of \(\mathrm{Mg}^{2+}\) in noble-gas notation is simply \([\mathrm{Ne}]\), indicating it has 10 electrons like neon.