# Question #2e8c2

Jun 29, 2016

Here's what I got.

#### Explanation:

You assign oxidation numbers to two bonded atoms by assuming that all the bonding electrons are being taken by the most electronegative of the two atoms.

In your case, carbon dioxide, ${\text{CO}}_{2}$, has the following Lewis structure

Let's take the double bond between the carbon atom and the oxygen on the left first.

In a double bond, a total of $4$ bonding electrons are shared between two atoms, $2$ coming from one of the atoms and $2$ coming from the other.

Now, oxygen is more electronegative than carbon. This means that when you assign oxidation numbers, you assume that oxygen will take all $4$ bonding electrons.

This is equivalent to saying that oxygen will take the $2$ electrons that carbon shares. As a result, carbon will have an oxidation state of $+ 2$, since it's now "missing" $2$ electrons.

Oxygen, on the other hand, "gained" $2$ electrons, so its oxidation number is $- 2$.

The exact same thing happens for the second double bond between carbon and the oxygen on the right side.

Once again, oxygen will take all $4$ bonding electrons and will have a $- 2$ oxidation number. Carbon will "lose" $2$ electrons and have an oxidation number of $+ 2$.

Since carbon "loses" a total of $4$ electrons, $2$ to each oxygen atom, it will have a total oxidation number of $+ 4$.

Therefore, carbon has a $+ 4$ oxidation number and oxygen has a $- 2$ oxidation number in the ${\text{CO}}_{2}$ molecule.

${\stackrel{\textcolor{b l u e}{+ 4}}{\text{C") stackrel(color(blue)(-2))("O}}}_{2}$