# Question #4e4a3

Dec 14, 2014

The total number of valence electrons, since we are now dealing with 3 O atoms, is up to $3 \cdot 6 = 18$, all of which are accounted for by the lone pairs - 1 for the top atom, and 2 and 3, respectively, for the other two - and the atoms shared in the bonds.

#### Explanation:

Lewis dot structures diagram the number of valence electrons outside of an atom. Since oxygen is in group VIA, it has six valence electrons that need to be accounted for when drawing its Lewis dot structure.

The reaction describes part of the ozone-oxygfen cycle (more here:http://www.theozonehole.com/ozonecreation.htm)

${O}_{\left(g\right)} + {O}_{2 \left(g\right)} \to {O}_{3 \left(g\right)}$

Here's a diagram of the Lewis structures for the reactans - top row - and for the product - bottom row - Upper left side is where the LEwsi structure for $O$ is. Since there is only one atom, the total number of valence electrons will be equal to 6.

The upper right structure represents ${O}_{2}$; since we are dealing with 2 $O$ atoms, the total number of valence electrons will be $2 \cdot 6 = 12$; in order for both $O$ atoms to complete their octet, 4 electrons will be shared in a double-bond.

The last two Lewis structures represent equivalent representations of the ozone (${O}_{3}$) molecule, and they are called resonance structures. The total number of valence electrons, since we are now dealing with 3 $O$ atoms, is up to $3 \cdot 6 = 18$, all of which are accounted for by the lone pairs - 1 for the top atom, and 2 and 3, respectively, for the other two - and the atoms shared in the bonds.