Question 6b789

Jan 4, 2017

Here's why that is the case.

Explanation:

For starters, a calcium has an atomic number equal to $20$, which means that it contains $20$ protons inside its nucleus. This means that a neutral calcium atom has $20$ electrons surrounding its nucleus.

The electron configuration of a neutral calcium atom looks like this

$\text{Ca: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{3} 3 {p}^{6} \textcolor{red}{4} {s}^{\textcolor{b l u e}{2}}$

As you can see, a neutral calcium atom has $\textcolor{b l u e}{2}$ valence electrons. You can tell by looking at the number of electrons present in the outermost energy shell of the atom, i.e. in the valence shell.

In this case, calcium's outermost energy level, which corresponds to $n = \textcolor{red}{4}$, contains $\textcolor{b l u e}{2}$ electrons.

Now, in order to complete its octet, i.e. get $8$ electrons in its outermost shell, calcium loses those $\textcolor{b l u e}{2}$ electrons. The resulting calcium ion will have a $2 +$ charge because it contains

• the same number of protons as the neutral atom
• two electrons less than the neutral atom The electron configuration of the ${\text{Ca}}^{2 +}$ ion looks like this

${\text{Ca}}^{2 +} : \textcolor{w h i t e}{.} 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} \textcolor{red}{3} {s}^{2} \textcolor{red}{3} {p}^{6}$

The outermost energy shell for a calcium ion has $n = \textcolor{red}{3}$ and contains

overbrace("2 e"^(-))^(color(blue)("in the 3s orbital")) + overbrace("6 e"^(-))^(color(blue)("in the 3p orbitals")) = "8 e"^(-) -># a complete octet