# Can we figure out how many sigma and pi bonds are in a molecule without drawing it out?

Mar 25, 2016

It's extremely difficult without knowing the bonding preferences of each atom or making the structure. It can be hard to grasp at first, but it's something that you will find useful to know in future chemistry classes, so it's best to learn it as soon as possible.

For simpler molecules, you might be able to get by without drawing the structure, but sooner or later you will be considering the connection of atoms in space and eventually the overall structure.

EXAMPLE: DIATOMIC NITROGEN

Let's say that for some reason we don't want to draw out the structure until we figure out how many $\sigma$ and $\pi$ bonds there are first.

For example, ${\text{N}}_{2}$ is fairly simple; since nitrogen is atomic number $7$, it has an electron configuration of $1 {s}^{2} 2 {s}^{2} 2 {p}^{3}$, so it favorably shares its three $2 p$ electrons with three other $2 p$ electrons from another nitrogen to form a noble gas configuration ($1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$).

Since all three $2 p$ orbitals participate, we form one $2 {p}_{z} - 2 {p}_{z}$ head-on ($\sigma$) interaction, one $2 {p}_{x} - 2 {p}_{x}$ sidelong ($\pi$) interaction, and one $2 {p}_{y} - 2 {p}_{y}$ sidelong ($\pi$) interaction.

Thus, we form a triple bond, which consists of the one $\setminus m a t h b f \left(\sigma\right)$ and two $\setminus m a t h b f \left(\pi\right)$ bonds we just mentioned. The remaining two lone pairs of electrons are stored in each nitrogen's $2 s$ orbital. So, we get the following structure that illustrates this discussion:

EXAMPLE: ETHANE

To present another perspective, if we start from the structure, it is generally easier. I used to find that if I thought about $\sigma$ and $\pi$ bonds while drawing the structure, it slowed me down.

Ethane is a two-carbon alkane, and has the molecular formula ${\text{C"_2"H}}_{6}$, or the more useful structural formula ${\text{H"_3"C""CH}}_{3}$.

We can draw each ${\text{CH}}_{3}$ as one carbon surrounded by three hydrogens to begin with.

Since carbon is tetravalent (it has four valence electrons), it requires one more bond to form an octet. Conveniently, both carbons in both ${\text{CH}}_{3}$ need that, so they bond with each other.

Now that we have the structure, we simply say that:

• Each single bond contains one $\sigma$ bond
• Each double bond contains one $\sigma$ bond and one $\pi$ bond
• Each triple bond contains one $\sigma$ bond and two $\pi$ bonds

So, you can point to each single bond and say that this molecule has $3 + 1 + 3 = \setminus m a t h b f \left(7\right)$ $\setminus m a t h b f \left(\sigma\right)$ bonds.

EXAMPLE: SOMETHING CRAZY

And now, an example of something where there is no way anyone can ever figure out the number of $\sigma$ and $\pi$ bonds without knowing the structure first, or knowing how to visualize the structure.

This molecule has the structural formula "Fe"_2("CO")_6(mu_2-"CO")_3, which means that there are two irons in the middle that each are connected to three carbon monoxide molecules, and these irons are also bridged by three carbon monoxides (${\mu}_{2}$ means that the bridging $\text{CO}$ is connected to two irons).

That's insane! Now, if we expanded this structure to show all the bonds, we would get:

And it would be preposterous to try to figure out the number of $\sigma$ and $\pi$ bonds on this without having drawn it first!