Video: Features of Molecular Orbital Theory

Which of the following statements is false? [A] When atomic orbitals overlap to form molecular orbitals, the total number of orbitals does not change. [B] Molecules containing an even number of electrons can be paramagnetic. [C] When two p orbitals overlap to form a 𝜎 bond, most of the electron density is concentrated between the nuclei. [D] The greatest overlap between an atomic orbital and a πœ‹ antibonding orbital is produced when the atomic orbital approaches at 90Β° to the πœ‹ bond axis. [E] The greatest overlap between an atomic orbital and a 𝜎 antibonding orbital is produced when the atomic orbital approaches parallel to the 𝜎 bond axis.

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Video Transcript

Which of the following statements is false? A) When atomic orbitals overlap to form molecular orbitals, the total number of orbitals does not change. B) Molecules containing an even number of electrons can be paramagnetic. C) When two p orbitals overlap to form a 𝜎 bond, most of the electron density is concentrated between the nuclei. D) The greatest overlap between an atomic orbital and a πœ‹ antibonding orbital is produced when the atomic orbital approaches at 90 degrees to the πœ‹ bond axis. Or E) the greatest overlap between an atomic orbital and a 𝜎 antibonding orbital is produced when the atomic orbital approaches parallel to the 𝜎 bond axis.

Let’s have a look at each of these potential answers in turn. First, we’ll look at, when atomic orbitals overlap to form molecular orbitals the total number of orbitals does not change. This statement is true and is really important to remember when drawing molecular orbital diagrams. In this diagram for example, we have two atomic orbitals, one on the left and one on the right. Here, they’re both two s orbitals. In the centre, we can see that these two have combined to form two molecular orbitals. The molecular orbital that’s lower in energy than the starting atomic orbitals is a bonding orbital.

The molecular orbital, which is higher in energy than the original atomic orbitals, is antibonding in nature. So we can see that the number of orbitals does not change between atomic orbitals and molecular orbitals. Since the question is asking us which of these statements is false, we can rule out A since it’s true. Statement B says that molecules containing an even number of electrons can be paramagnetic. Remember that paramagnetic means something which contains unpaired electrons. The opposite of this is diamagnetic where all electrons are paired.

At first glance, you might think that a molecule with an even number of electrons must be all paired up and therefore diamagnetic. But let’s look at an example. The example we’ll look at is Oβ‚‚, oxygen. Each oxygen atom has six electrons in its valence shell. Two reside in the two s orbital and the rest in two p orbitals. Remember from statement one that the number of molecular orbitals formed has to be the same as the number of atomic orbitals used to form them.

Here, we have four atomic orbitals from each oxygen, giving us a total of eight atomic orbitals. So these combine to form eight molecular orbitals. The molecular orbitals formed from single lines are 𝜎 orbitals. And those formed of two lines are πœ‹ orbitals. We can label them with numbers, letters, and either 𝑒 or 𝑔 from gerade and ungerade. This denotes the symmetry of the orbital. Now, let’s start filling our molecular orbitals with electrons. Both of the one 𝜎 orbitals are completely filled by the two two s shells. We then have eight electrons to put into the upper molecular orbitals. Two go in the two πœŽπ‘” orbital. Four fit in the one πœ‹π‘’ orbital, remembering that each orbital on its own can hold two electrons. So this πœ‹ orbital can hold four.

Now, we have two electrons left to place in our one πœ‹π‘” orbital. Remember, however, that each line is a separate orbital. So we could draw it a bit like this, where one electron is in one of the orbitals and the other is in a different orbital. What this means is that these two electrons are not paired. Each electron resides in a different molecular orbital. And they both have parallel spins. This gives Oβ‚‚ a net spin angular momentum of one, meaning it’s in a triplet state. This therefore means it is paramagnetic because we do have unpaired electrons. So statement B is true. Therefore, we can rule it out as a correct answer.

The last three potential answers all discuss orbital overlap. So let’s look more closely at this. We can think of bonds forming in areas where electrons accumulate and where atomic orbitals overlap constructively. Remember that orbitals can either interact constructively or destructively. To illustrate this point, we can look at the case of a hydrogen molecule where two hydrogen one s orbitals overlap. Here, we can see the wave functions for each of the atomic orbitals. In the top case where we form a bonding molecular orbital, we can see that the two atomic orbitals interfere constructively. This means that there’s a region in the centre of the two nuclei where electrons can accumulate and where the atomic orbitals overlap.

In the antibonding case, when the two atomic orbitals combine, they interfere destructively. This means that the resultant molecular orbital doesn’t have a convenient place in between the nuclei where electrons can accumulate. This is why we describe this type of molecular orbital as antibonding in nature. Now, let’s specifically consider p orbitals. p orbitals can be aligned along each of the perpendicular axes. When the p orbital is aligned along the π‘₯-axis, it is referred to as a p π‘₯ orbital. If it is aligned along the 𝑦-axis, it is a p 𝑦 orbital. Or along the 𝑧-axis, we have a p 𝑧 orbital.

Each p orbital has two lobes. And they have opposing phases denoted by being shaded in or not. In order for two p orbitals to overlap in a constructive manner, they need to have the same phase, for example, like this. Statement C specifically refers to p orbitals overlapping to form a 𝜎 bond. 𝜎 bonds are always built from orbitals which have an appropriate symmetry. This symmetry means that the atomic orbitals need to be cylindrically symmetrical about the internuclear axis.

So if this is our molecule, the atomic p orbitals that we’re going to combine need to be symmetrical along the axis of the bond. The example drawn here is indeed symmetrical about this bond axis. Both s orbitals and p 𝑧 orbitals conform to this criterion. So when considering where the electron density is concentrated in such a bond, we can see from the diagram that this is in between the two nuclei. So statement C is true and therefore not a correct answer. The last two statements consider something similar. Statement D says that the greatest overlap between an atomic orbital and a πœ‹ antibonding orbital is produced when the atomic orbital approaches at 90 degrees to the bond axis.

Let’s draw a diagram to investigate this. We’ve already established that 𝜎 orbitals are built along the bond axis. This means that πœ‹ orbitals are built perpendicular to the bond axis. Here, we’re looking at the combination of p π‘₯ and p 𝑦 atomic orbitals. Statement D asks about an atomic orbital and a πœ‹ antibonding orbital. Remember that antibonding means that the two atomic orbitals which built that molecular orbital interfere destructively. So let’s build our πœ‹ antibonding orbital to start.

We have our molecule in the centre and we have two p orbitals perpendicular to the bond axis. Note that they have opposing phases which means that they’re going to combine and interfere destructively to form an antibonding orbital. This will form a πœ‹ 𝑔 orbital, 𝑔 standing for gerade. We can tell this is gerade because if we carry out an inversion through the centre of symmetry, there is no change to the lobe sign. If there was a change, this would be an ungerade orbital. The nodal plane for this orbital will be down the centre.

Now that we have our πœ‹ antibonding orbital, we can look at what happens when an atomic orbital approaches at 90 degrees. So here, we have another p orbital approaching at a 90-degree angle to our πœ‹ bond. Statement D says that this arrangement produces the greatest overlap. You can see that in approaching at 90 degrees to our πœ‹ bond axis, we’re actually going to have very poor overlap. As it approaches, it’s going to be perpendicular to our πœ‹ bond which means that there’s really no where for the overlap to have occurred.

The greatest overlap would occur if the p orbital actually approached at the same orientation. So statement D is false. This makes it a potentially correct answer to our question. But let’s check the last statement just to be safe. Statement D talks about the overlap between an atomic orbital and a 𝜎 antibonding orbital. Remember that 𝜎 orbitals lie along the bond axis. 𝜎 antibonding orbitals can be formed from s or p 𝑧 atomic orbitals. Note, the phase mismatch in the centre means that this is antibonding.

Statement E says that the greatest overlap between an approaching atomic orbital with this 𝜎 antibonding orbital occurs when the atomic orbital approaches parallel to the 𝜎 bond axis. Approaching parallel to the bond axis means that it will be approaching along the orbital. This continues with the criterion that the atomic orbitals need to be cylindrically symmetrical about the internuclear axis.

This will result in good overlap between the approaching atomic orbital and the 𝜎 antibonding orbital. So this statement is true. So the only one of these statements which is false is D that the greatest overlap between an atomic orbital and a πœ‹ antibonding orbital is produced when the atomic orbital approaches at 90 degrees to the πœ‹ bond axis. So this is our correct answer.

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