HS Chemistry - Chemical Bonding

Sigma & Pi Bonds

Overview of The Page

This page will cover:

• What is electron subshell hybridization?
• What are sigma bonds?
• What are pi bonds?

To adequately explain sigma and pi bonds, we'll start off by briefly reviewing how covalent bonds work.

Covalent Bonds

A covalent bond forms between two non-metal atoms. Hydrogen, despite formally being a metal, also bonds covalently. We'll use Hydrogen to demonstrate how it works, for simplicity's sake.

A hydrogen molecule $H~2~$ consists of two Hydrogen atoms in a covalent bond, with a single bond between them $H\-H$. When separate, the two Hydrogen atoms look something like this $not drawn to scale$:

But that doesn't show everything, so let's draw the atoms again

It's not a very good drawing $it shouldn’t be used as a diagram$, but it serves our purpose. The area that's covered in blue lines represents the area where the proton's electrical force will act on an electron - that is, any electron in the shaded area will be pulled by the proton in the Hydrogen atom's nucleus. Therefore, the electron isn't stuck in an orbital; it moves all around in its subshell $this is where picking Hydrogen makes things easy; the s subshell is just one spherical orbital surrounding the nucleus, so it's easy to draw$.

This gives way to a thought experiment. What if we move the atoms close enough together so that the shaded blue areas overlap?

What if we move them even closer, so that the electrons are right on top of each other? And let's color the borders of the nuclei's electrical attraction area:

As you can see, they overlap, and the electrons are in the middle of that overlap. Each nucleus is pulling on both of the electrons, trying to complete its own outer shell. Yet while the nuclei are bound together by the shared electrons, they also repel each other, as they are both positively charged, resulting in a fixed molecule shape, as they don't move any closer or any farther from each other than they currently are.

With the fixed molecule shape, we get the molecular orbital.

The boundaries of the electron shells have disappeared. They no longer exist - both electrons in the bond are free to move in the combined area, called the molecular orbital, outlined in red here.

The final shape of the hydrogen molecule ends up being

with the electrons being somewhere in the molecular orbital. To make the notation $note: this isn't a formal notation$ easier, we can just draw:

The red oval represents the molecular orbital. The atoms have bonded together in a covalent bond. Note that the nuclei are not free-floating - they remain in fixed positions relative to each other, repelling each other.

Hybridization

Making a hydrogen molecule was easy, because Hydrogen atoms only have one s subshell. But how do you make other molecules?

Let's say we want methane $CH~4~$. Each hydrogen atom will share its electron with one of carbon's outer electrons, and thus both atoms will complete their valence shell.

Except it can't do that. Carbon has both s subshell electrons and p subshell electrons in its outer shell, and therefore its valence electrons have different energies. It can't just form bonds like this, because otherwise the bonds won't be in balance.

So what happens?

The outer subshells, the ones that contain the electrons that will bond, are going to hybridize to an intermediate energy level in order to gain equal energies. The p orbitals and the s orbitals of the outermost shell are going to hybridize into spⁿ orbitals, with each spⁿ orbital having the same energy.

If one s orbital and one p orbital hybridize, we get two $one \+ one$ s¹p¹ orbitals $we don't write the ^1^ in notation so we'll just call it an sp orbital$. If one s orbital and two p orbitals hybridize, we get three $one + two$ s¹p² orbitals $sp^2^ orbitals$. If one s orbital and three p orbitals hybridize, we get four sp³ orbitals. And so on. Later on, the d subshell orbitals also hybridize sometimes, leading to sp³d² orbitals and the like. In general, when hybridization occurs, all the outermost s and p orbitals will hybridize unless forced not to. The only time some p orbitals won't hybridize while others will is when those p orbitals are needed to make pi bonds; in that case, only the p orbitals needed to make pi bonds will remain unhybridized. This is covered later in this page. Additionally, none of the outermost d orbitals will hybridize unless forced to. The reason for this has to do with the arrangement of the orbitals; something that won't be covered here.

Once hybridized, the s and p orbitals $and d orbitals if they are used$ that are used in the hybridization no longer exist, as they have changed into hybridized orbitals. If, say, Sulfur formed three sp² orbitals, its 3s orbital $the outermost s subshell$, as well as two of its 3p orbitals $the outermost p subshell$, would be gone, producing three sp² orbitals. However, it would still have one 3p orbital left, as that wasn't used in the hybridization.

When four sp³ orbitals are formed, all of them have the same energy. They thus all repel each other with equal force, forming the tetrahedral Electron Geometry shape. Even when there are three sp² orbitals or two sp orbitals, the orbitals that are hybridized repel each other equally, as they have equal energies. This is how the electron geometry shapes are formed; each hybridized orbital repels all the other hybridized orbitals equally, causing the orbitals to assume specific positions.

This has an important implication, which we'll get to in a bit.

One last thing: here's how a hybridized orbital looks. The black circle represents the nucleus, while the red shape is the shape of the orbital.

Bonding

Now that the orbitals are hybridized, they can finally bond. $We assume that both atoms in the bonding have hybridized their orbitals by this point, except hydrogen, which can't hybridize because it only has one orbital. If they haven't both hybridized, they must both do so before they can bond$.

In a sigma $σ$ bond, a hybridized orbital from one atom overlaps with a hybridized orbital from the other atom to form a bond. However, there are some rules as to how this happens.

If there are two electrons in a hybridized orbital, then that orbital doesn’t have space to take any more electrons. It therefore can't bond. This pair of electrons is called a lone pair. Only a hybridized orbital with one electron can bond and include a second electron. This goes for both atoms involved in bonding.

In the atom shown above, the bottom-right hybridized orbital is the only one with space for an extra electron. Therefore, this is the only orbital that will form a sigma bond.

After bonding, the electrons can travel between both overlapping hybridized orbitals. Since there are only two electrons in that bond, each sigma bond is a single bond. And since each hybridized orbital in an atom points in a fixed, different direction as per its electron geometry, and the nuclei of the bonded atoms will repel each other $they also have to stay far enough away from each other$, each pair of atoms can only have one sigma bond.

The drawing above shows this. None of the other orbitals are in the correct positions to overlap. As only one pair of orbitals can overlap, only one sigma bond can be formed.

One thing to note is saying that each pair of atoms can only have one sigma bond assumes we're still talking about the same two atoms. An atom that has a sigma bond with one atom can still have a sigma bond with another atom, it just can't have a second sigma bond with the same atom that it has already formed a sigma bond with. Thus, for methane $CH~4~$, there are 4 sigma bonds. $Remember, hydrogen doesn't have hybridized orbitals$

But then how do some pairs of atoms have double or triple bonds between them?

Pi Bond

Let's look at an example with ethene $C~2~H~4~$.

Each of the hydrogen atoms is connected to a carbon atom with a sigma bond, as there is only one bond between each hydrogen and carbon atom. Between the two carbon atoms, however, the first bond is a sigma bond, but the second is a pi (π) bond. If we had, say, an N₂ molecule $N≡N$, the first bond would be a sigma bond, and the second and third bond between the nitrogen atoms would both be pi bonds.

Pi bonds are created by unhybridized p orbitals. Remember how earlier it was said that the only time a p orbital will remain unhybridized while the others undergo hybridization is when that p orbital is needed for a pi bond? That's what happens here. Since they're unhybridized, they don't have the same energy as the hybridized orbitals, and therefore they are not repelled by them.

When a p orbital is unhybridized, it exists in a plane perpendicular to the hybridized orbitals. Since there are only three or two hybridized orbitals left now, the hybridized orbitals will form either the trigonal planar shape $if there are three$ or linear shape $if there are two$, repelling each other as much as possible. This causes all the hybridized orbitals to be on the same plane, allowing the p orbital to be perpendicular to them.

When three sp² orbitals are formed, and one p orbital remains unhybridized, the three sp² orbitals assume the trigonal planar shape, and the p orbital goes through their center, perpendicular to them. When two sp orbitals are formed, and two p orbitals remain unhybridized, the sp orbitals assume the linear shape, and the two p orbitals go through their center, perpendicular to them, but the two p orbitals are also perpendicular to each other. Graphically, we can assume it looks something like this:

In the above drawing, you'll notice that the key references the p orbital lobes, rather than just the p orbital. Unlike the s orbitals, each p orbital consists of two lobes, one above and one below the nucleus $or in front and behind, or to the left and right$. The two lobes combined make one orbital, and it's this shape that explains the shape of the pi bond.

Pi bonds are formed by unhybridized p orbitals that are parallel to each other. This is probably most easily explained through a couple drawings/diagrams. Here's how the pi bond forms in an ethene $C~2~H~4~$ molecule.

Each CH₂ $one half of a C~2~H~4~ molecule$ has three sp² orbitals, and one p orbital. Right now, we can't see the p orbital, since in this view it goes through the screen. Let's rotate the CH₂ so we can easily see the p orbitals.

There are actually 2 hydrogen atoms and 3 Carbon sp² orbitals in each CH₂ molecule, but we don't see that because they're hidden behind each other from this view. Each of the CH₂ has 6 valence electrons, 2 from the 2 Hydrogen atoms and 4 from the Carbon atom. We therefore completely fill the orbitals connected to the Hydrogen, and put the remaining 2 electrons in the remaining 2 orbitals.

Since we're not focusing on the Hydrogen atoms or the sp² orbitals they're connected to, let's remove them from the picture. They're still there, we just won't show them:

And finally, let's bring them close together, overlap the spare sp² orbitals, and create a sigma bond between them.

Now all that remains is for the second bond between the carbon atoms, the pi bond, to form.

Remember how in a covalent bond, the nuclei of both atoms will attract the shared electrons? Here, the electrons are in the p orbital, and they're not touching.

But the atoms have already created a sigma bond. They're close enough to create a sigma bond, which means they're already pulling on the shared electrons in the sigma bond.

That means they're also close enough to pull on the electrons in the p orbitals.

The nucleus of the atom on the left is going to pull on both p orbital lobes of the atom on the right.

And both p orbital lobes of the atom on the right are going to be attracted towards the nucleus of the atom on the left.

Notice that rather than get pulled along, the p orbital lobes expanded towards the other nucleus. This is because they are still attracted towards the nucleus of their own atom, and therefore they won't leave it. Instead, they just expand towards the other nucleus.

How can they expand? The p orbital isn't something made of matter, or something that takes up a definite amount of space. It's an area for electrons that's determined by electrostatic forces. Any change to those forces is going to cause the p orbital's shape to change. When the nucleus of the atom on the left pulled on the p orbital of the atom on the right, it applied a force of electrostatic attraction on it. That caused the p orbital's shape to change.

The nucleus of the atom on the right is also going to pull on both p orbital lobes of the atom on the left.

And both p orbital lobes of the atom on the left are going to be attracted towards the nucleus of the atom on the right.

And now both p orbitals overlap. They have bonded, producing the following shape of the bond:

The p orbitals of both Carbon atoms have combined to produce the shape shown above, creating the second bond between the Carbon atoms. This bond is called a pi bond. A pi bond, as shown in the structure, can only be made by two parallel unhybridized p orbitals.

Now let's draw how the whole molecule looks :

There's two things to note here: 1. The areas where the bonding orbitals overlap has been erased. The orbitals have combined, and there is no longer a distinction between them. 2. Due to our view, we can't see the third and fourth Hydrogen atoms $they are behind the two shown here$. However, they are still there.

Triple Bonds

There's just one last thing to note: remember how some molecules, such as N₂, have triple bonds? In those cases, there's only one sigma bond, and two pi bonds. The pi bonds for N₂ form just like the ones for C₂H₄, except since the p orbitals are perpendicular to each other, the pi bonds are also perpendicular to each other.

Practice

1. Try explaining how the atoms in a CO₂ go through each step to create CO₂ (O=C=O).

   There's one sigma bond between each Oxygen atom and the central Carbon atom, and then there's also a pi bond between each Oxygen atom and the central Carbon atom.

   If it helps, first determine the electron and molecular geometry for the CO₂ molecule.

   Try drawing a diagram $like the one of C~2~H~4~ above$ for CO₂, showing the sigma and pi bonds and their structure, and any lone pairs in hybridized orbitals, as well as where the lone pairs would be.