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 ##”NaCl”##, we should first draw the individual diagrams for both ##”Na”## and ##”Cl”##.

The of ##”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 ##”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 ##”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 ##”NaCl”## model. ##”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 ##”Na”##, there is only ##1## electron in the third valence shell. Because of this, ##”Na”## wants to “shed” its single electron and form a ##”Na”^+## ion. Similarly, ##”Cl”## has ##7## electrons of ##8## possible in its third shell, so it will want to take an electron and form a ##”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:

##”Na”^+## has ##10## electrons: its shells have ##2## and ##8## electrons, respectively.

##”Cl”^-## has ##18## electrons: its shells have ##2,8,## and ##8## electrons, respectively.

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