Question

What is the oxidation number of carbon monoxide?

Answer 1

Hint: We make use of Lewis Structure is a simplified model of a molecule's outermost shell electrons. It's used to demonstrate how electrons in a compound are organized around each of the atoms. Then we can utilize the electronic configuration to get the number of outer electrons. By the Lewis structure we will try to find the bonding mechanism and finally understand the overall charge that is the resultant oxidation number.

Complete step by step answer:
The compound under consideration here is carbon monoxide whose chemical formula looks like this: $CO$
Take the electronic configuration of carbon and oxygen to find out how many valence or outer electrons are there for each, only with that we can draw the Lewis structure.
Configurations:
$C \Rightarrow \;1{s^2}2{s^2}2{p^2}$, so clearly the outermost shell has $4$ valence electrons.
$C \Rightarrow \;1{s^2}2{s^2}2{p^4}$, so clearly the outermost shell has $6$ valence electrons.
Now let us form a Lewis structure for this compound: $CO$. Remember that by Lewis structure, the electrons are depicted as "dots" or as a line between the two atoms when they are bonded. The aim is to find the "right" electron configuration, which requires that the octet rule and formal charges be met.
Keep in mind that when a single pair of electrons get shared then it is calculated as one bond;
So here we see that there are three pairs of electrons getting shared here to form this bond between carbon and oxygen atoms, so it can be concluded to be a triple bond. This is because there is one pair of electrons given by carbon and two pairs given by oxygen.
So there are two electrons within the lone pair of carbon and two for oxygen. And the total count of bonding electrons is six.
Now to find the formal charge we require the formula:
$ \Rightarrow $Formal charge$ = no.\;of\;valence\;{e^ - } - [{e^ - }\;in\;lone\;pair + 0.5 \times bonding\;{e^ - }]$
So separately finding formal charge for carbon and oxygen will look like;
$
   \Rightarrow C = 4 - [2 + 0.5 \times 6] \\
   \Rightarrow C = - 1 \\
 $
So formal charge on carbon is $ - 1$
Then;
$
   \Rightarrow O = 6 - [2 + 0.5 \times 6] \\
   \Rightarrow O = 1 \\
 $
So formal charge on oxygen is $ + 1$
So total charge on this compound will be the oxidation number of it also;
$ \Rightarrow CO = 0$
$\therefore $We can conclude the oxidation number of carbon monoxide is $ \Rightarrow CO = 0$.

Note:
Take care that always when finding the oxidation state for carbon atom;
- If it bonds with any electronegative atom then the oxidation number goes up by one.
- If it bonds with another carbon then there will not be any change is its oxidation number.
- If it bonds with a hydrogen atom then the oxidation number reduces by one.