# How would you make a Bohr diagram for NaCl?

Feb 19, 2016

A Bohr diagram depicts an atom with a small, central nucleus and the electrons in their valence shells.

The first valence shell contains $2$ electrons, and the second and third shell have $8$ electrons each, and the number keeps growing.

To draw the Bohr diagram for $\text{NaCl}$, we should first draw the individual diagrams for both $\text{Na}$ and $\text{Cl}$.

The atomic number of $\text{Na}$ is $11$, so it has $11$ electrons.

The first and second valence shells are completely full, since their $2$ and $8$ electrons only take up the first $10$ of sodium's $11$ electrons. Thus there will be $1$ leftover electron in the third valence shell, so the Bohr diagram of $\text{Na}$ can be drawn as follows: We can follow a similar process for chlorine, which has $17$ electrons. The first two shells are full, taking up $10$ of $\text{Cl}$'s $17$ electrons, leaving $7$ electrons in the third valence shell (remember that it has a capacity of $8$). Now, we must draw the Bohr diagram for the $\text{NaCl}$ model. $\text{NaCl}$ is an ionic compound, meaning that electrons from each molecule that comprise the compound are ripped away from molecules and added to other molecules.

Notice that in $\text{Na}$, there is only $1$ electron in the third valence shell. Because of this, $\text{Na}$ wants to "shed" its single electron and form a ${\text{Na}}^{+}$ ion. Similarly, $\text{Cl}$ has $7$ electrons of $8$ possible in its third shell, so it will want to take an electron and form a ${\text{Cl}}^{-}$ ion.

Hence the chlorine atom will take the electron in sodium's third valence shell and add it to its own, so the ionic compound would be drawn as: Note:

${\text{Na}}^{+}$ has $10$ electrons: its shells have $2$ and $8$ electrons, respectively.

${\text{Cl}}^{-}$ has $18$ electrons: its shells have $2 , 8 ,$ and $8$ electrons, respectively.