Question Video: Orbital Configurations in Butadiene

Video Transcript

Butadiene contains two occupied 𝜋 bonding orbitals and two unoccupied 𝜋* antibonding orbitals. The 𝜋 electrons are delocalized between the four carbon atoms. Which of the following diagrams illustrates the 𝜋 bonding orbital with the highest energy?

To answer this question, let’s begin by thinking about all of the molecular orbitals in butadiene. The question is asking specifically about 𝜋 bonding and 𝜋* antibonding. So we can ignore all sigma bonding. Each of the carbons in butadiene is sp2 hybridized. sp2 hybridization is the mixing of 1s orbital and 2p atomic orbitals. This, of course, leaves each carbon with a p atomic orbital which can take part in 𝜋 bonding. If we begin with four p atomic orbitals, we’re going to end up with four molecular orbitals.

Remember that the number of orbitals doesn’t change when they are combined. So now that we know we have for molecular orbitals, we can begin to draw our energy diagram. When we linearly combine our four 2p orbitals on the carbon atoms, we create four 𝜋 orbitals. These are molecular orbitals. Each carbon contributes one electron to these orbitals. So we can fill in four electrons.

Electrons will fill the orbital at the lowest energy first. So those at the bottom of the diagram are the lowest energy and those at the top, the highest energy. 𝜋 orbitals which sit at an energy lower than the C2p atomic orbitals are considered bonding orbitals. Those at a higher energy than the constituent atomic orbitals are mostly antibonding in nature. But how does this relate to the diagrams in the question?

Each of these diagrams shows the arrangement of the atomic p orbitals, which will be combined to form molecular orbitals. The key thing we’re going to be looking for are nodes or nodal planes. The more internuclear nodes there are, the higher the energy of the molecular orbital. So we can arrange these diagrams according to the number of nodes they contain.

So what does a node look like on one of these diagrams? Let’s take a closer look at diagram (A). The shading of the orbital lobes denotes their phase. Regardless of whether the lobes are shaded or not, we’re really looking at the pattern they produce. Orbitals of the same phase are considered to overlap in a constructive manner, whereas orbitals of opposing phases overlap in a destructive manner. So we could consider these atomic orbitals as combining to form something like this. On the left-hand side, the two orbitals in the same phase have combined constructively. The same is true of the combination of the two orbitals is on the right. However, the problem comes in the center.

These center two orbitals have opposite phases, which means that they combine destructively. This leaves us with a node between the two. A node is somewhere where there is no probability of finding an electron. So as we look at our diagrams, we can count the number of nodes by counting how many times the orbitals invert their phases from left to right. In diagram (A), we have one node. In diagram (B), we also have one node. In diagram (C), we have three nodes as each of the orbitals is the opposite phase to the ones either side of it. (D) has two nodes. And finally, (E) has no nodes at all since all of the orbitals are lined up in the same phase pattern.

We can use the numbers of nodes to now arrange these diagrams in energy order, from lowest to highest. Diagram (E) has no nodes. So it’s the lowest energy diagram. This forms our first 𝜋 bonding orbital. Both (A) and (B) have only one node. It turns out that the correct arrangement for the second 𝜋 orbital is that of (A) with a node in the center. This is for symmetry reasons. Next, we have diagram (D), which has two nodes. And this forms the three 𝜋 orbital, which is our first antibonding orbital. This leaves us with diagram (C), the one which has the most nodes, three nodes. And this forms our highest energy antibonding 𝜋 orbital.

Now, let’s go back to the original question. The question is, which of the following diagrams illustrates the 𝜋 bonding orbital with the highest energy? We know that the bottom two 𝜋 orbitals are our bonding orbitals. We also know that the energy increases as the number of nodes increases. So of these two diagrams, (A) and (E), both of which are bonding, (A) has one node. So this has the highest energy of the bonding orbitals. So our answer for this question is (A).

An electron is added to a butadiene molecule. Which of the following diagrams illustrates the 𝜋* antibonding orbital occupied by the additional electron?

We’ve already done a lot of the working out for this question in our previous answer. So we can reuse the same energy level orbital diagram. We’re told to add an additional electron to our butadiene molecule. We fill up our orbitals from bottom to top. So our additional electron is going to go into the three 𝜋 orbital. This is actually our first antibonding orbital, a 𝜋* orbital. And this matches the information we’re given in the question. This antibonding 𝜋* orbital is represented by diagram (D). So the answer to this question is diagram (D).