The electronic configuration of Mg 2+ is

Analysis

To find the electronic configuration of an ion like $\text{Mg}^{2+}$, we must first determine the configuration of the neutral atom and then account for the loss of electrons.

Concept

1. Atomic Number: Magnesium ($\text{Mg}$) has an atomic number of 12. This means a neutral magnesium atom has 12 protons and 12 electrons.
2. Neutral Configuration: Following the Aufbau principle, we fill the orbitals in order of increasing energy: $1s \rightarrow 2s \rightarrow 2p \rightarrow 3s$. • For neutral $\text{Mg}$ (12 electrons): $1s^2 2s^2 2p^6 3s^2$.
3. Ion Formation: The $\text{Mg}^{2+}$ ion indicates that the neutral atom has lost 2 electrons. • We remove the electrons from the outermost shell (the highest energy level), which is the $3s$ orbital.

Step-by-Step Solution

1. Start with neutral Magnesium: $1s^2 2s^2 2p^6 3s^2$ (Total = 12 electrons)
2. Remove 2 electrons: Since the $3s$ orbital contains 2 electrons, removing them leaves us with: $1s^2 2s^2 2p^6$ (Total = 10 electrons)

Comparing this result to your options:

• Option A: $1s^2 2s^2 2p^6 3s^2 3p^2$ (Incorrect, this is Silicon)
• Option B: $1s^2 2s^2 2p^6$ (Correct)
• Option C: $1s^2 2s^2 2p^6 3s^2$ (Incorrect, this is neutral Magnesium)
• Option D: $1s^2 2s^2 2p^4$ (Incorrect)

Exam Tip

When dealing with cations (positive ions), always remove electrons from the shell with the highest principal quantum number ($n$) first. For transition metals, remember that the $4s$ electrons are removed before the $3d$ electrons!