# Question 23ae1

Jun 9, 2014

${\left[2 , 8\right]}^{2 +}$

A magnesium atom has atomic number 12, so 12 protons in the nucleus and therefore 12 electrons. These are arranged 2 in the innermost (n=1) shell, then 8 in the next (n=2) shell, and the last two in the n=3 shell. Therefore a magnesium atom is [2,8,2]

The magnesium ion $M {g}^{2 +}$ is formed when the magnesium atom loses the two electrons from its outer shell! to form a stable ion with a noble gas configuration. With the two electrons lost, the electron configuration becomes ${\left[2 , 8\right]}^{2 +}$, the charge on the brackets reminding us that this is an ion not an atom, and that the number of electrons now is NOT the same as the number if protons in the nucleus. Nov 1, 2015

${\text{Mg}}^{2 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$

#### Explanation:

Your starting point for finding the electron configuration of the magnesiu mcation, ${\text{Mg}}^{2 +}$, will be the electron configuration of the neutral magnesium atom, $\text{Mg}$.

You know that magnesium is located in period 3, group 2 of the periodic table, and that it has an atomic number equal to $12$.

This means that a neutral magnesium atom will have a total of $12$ protons in its nucleus and $12$ electrons surrounding its nucleus. Its electron configuration would look like this

$\text{Mg: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2}$

Now, when the magnesium cation is formed, two electrons, hence the $2 +$ charge of the cation, are lost by the magnesium atom. These electrons, which are called valence electrons, come from magnesium's outermost energy level. As you can see, magnesium has its outermost electrons located on the third energy level, $n = 3$.

This means that when the cation is formed, its electron configuration will look like this

${\text{Mg: " 1s^2 2s^2 2p^6 color(red)(cancel(color(black)(3s^2))) implies "Mg}}^{2 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$

Using noble gas shorthand notation, you will get

"Mg: " ["Ne"]color(red)(cancel(color(black)(3s^2))) implies "Mg"^(2+): ["Ne"]# 