# If a neutral atom has 12 electrons, how many can be used for creating bonds?

Jul 13, 2016

Since a neutral atom was requested, its charges must cancel, and therefore, its atomic number must have been $12$.

The neutral atom $\text{Mg}$ has atomic number $Z = 12$, and it indeed has $12$ electrons:

• two $1 s$ electrons
• two $2 s$ electrons
• six $2 p$ electrons
• two $3 s$ electrons

Since $\text{Mg}$ is on group 2, it has two valence electrons that it can donate to form the ${\text{Mg}}^{2 +}$ ion at most (before it becomes energetically unfavorable to donate more, as it is now noble-gas-like).

These donated valence electrons go into making bonds.

A suitable atom, such as---but not limited to---$\text{Cl}$ (from group 17), can accept $\text{Mg}$'s electrons.

Since each $\text{Cl}$ needs one valence electron, two $\text{Cl}$ atoms can each accept one valence electron to account for $\text{Mg}$'s two $3 s$ valence electrons.

Hence, a compound containing $\text{Mg}$ and $\text{Cl}$ would be $\setminus m a t h b f \left({\text{MgCl}}_{2}\right)$, magnesium chloride, where $\text{Mg}$ makes two single bonds by donating its two total valence electrons in conjunction with one valence electron from each $\text{Cl}$:

$\stackrel{. .}{\text{Cl"-"Mg"-stackrel(..)"Cl}} :$
${\text{^(..) color(white)(iiiiiiiiiiiii)}}^{. .}$