P orbital, π bond, and lots of questions

[Dr. Nick]

p orbital, π bond, and lots of questions :)

I’m a little bit confused, and I APOLOGIZE if I’m asking too simple and stupid questions.

OK, we have two atoms, each with one p orbital (8), occupied with one electron and they form π bond.

You all know how π bond looks 8=8, and I wondered why it isn’t cylindrical. Then I remembered that solution of wave function for p orbital has +- ψ and only ++ and -- combination will give bonding orbital. So if we now have nodal plane in π bond, as we had nodal plane in p orbital, can I conclude that semi orbitals of π bond have inherited sign of function (+ψ or –ψ) from p orbitals ?
If it is so, I can’t see it from the explanation I’ve been given, which is LCAO: (aψ + bψ)^2 = aψ ^2 +aψ bψ + bψ^2. Could it be that this explanation is for s orbitals and sigma bond only?

And this bothers me too, is there correlation between electron spin and sign of wave function (I think not)? And is it easy and frequent for electron to change spin on room temperature in semi populated orbital? Because, if have two orbitals ready for bonding there won't bond until electrons obtain opposite spins, even dough orbitals have same wave function sing ?
And this too, ψ^2 (probability density) gave us an idea how does orbital looks like, but at the same time in my books p orbital lobes are assigned with signs + and -, how can they do this, when I know that probability density is non-negative everywhere. And later when we combine orbitals into bond we deal with ψ but not with ψ^2 ? thank you, thank you, for reading this, and eventually helping me !

 

[Gokul43201]

I’m a little bit confused, and I APOLOGIZE if I’m asking too simple and stupid questions.

OK, we have two atoms, each with one p orbital (8), occupied with one electron and they form π bond.

You all know how π bond looks 8=8, and I wondered why it isn’t cylindrical. Then I remembered that solution of wave function for p orbital has +- ψ and only ++ and -- combination will give bonding orbital. So if we now have nodal plane in π bond, as we had nodal plane in p orbital, can I conclude that semi orbitals of π bond have inherited sign of function (+ψ or –ψ) from p orbitals ?
If it is so, I can’t see it from the explanation I’ve been given, which is LCAO: (aψ + bψ)^2 = aψ ^2 +aψ bψ + bψ^2. Could it be that this explanation is for s orbitals and sigma bond only?

No, LCAO is valid for pi bonding too.

You are making an algebraic error in the above expansion. For simplicity, let's assume the two atoms are identical, and look at :

$$|\psi_{\pi}(\mathbf{r})|^2 = |\psi_{\2p_y,A}(\mathbf{r})+ \psi_{\2p_y,B}(\mathbf{r})|^2 = |\psi_{\2p_y,A}(\mathbf{r})|^2 + |\psi_{\2p_y,B}(\mathbf{r})|^2 + 2|\psi_{\2p_y,A}(\mathbf{r}) ||\psi_{\2p_y,B}(\mathbf{r})|cos[\phi(\mathbf{r})]$$

where $\phi(\mathbf{r})$ is the phase difference at the point $\mathbf{r}$, between $\psi_{\2p_y,A}$ and $\psi_{\2p_y,B}$. Now you know that the two "lobes" of each p-orbital have opposite phase, so we choose +1 and -1 as the phase factors in the wavefunction for the two lobes. Next, if you'll allow me to state (without proving) that the phases in the lobes of the $p_y$ orbital of the other atom can also be chosen as +1 or -1, you can only have phase differences or 0 (bonding) or $\pi$ (anti-bonding). Now with this picture, the above expansion becomes :

$$|\psi_{\pi}(\mathbf{r})|^2 = |\psi_{\2p_y,A}(\mathbf{r})|^2 + |\psi_{\2p_y,B}(\mathbf{r})|^2 \pm 2|\psi_{\2p_y,A}(\mathbf{r}) ||\psi_{\2p_y,B}(\mathbf{r})|$$

This above expression correctly defines the density of the pi-cloud.

For instance, in the anti-bonding orbital, at any point $\mathbf{r}$ that is equidistant from A and B, you would have $|\psi_{\2p_y,A}(\mathbf{r}_{mid})|=|\psi_{\2p_y,B}(\mathbf{r}_{mid})| = |\psi_{mid}|$, say. Hence :

$$|\psi_{\pi}(\mathbf{r}_{mid-plane})|^2 = |\psi_{mid}|^2 + |\psi_{mid}|^2 - 2|\psi_{mid}|^2 = 0$$

This predicts correctly, that the density vanishes at all points on x=0 (setting the origin at the mid-point between the atoms).

PS : 1. I've been a little sloppy with notation

2. The phase of the wavefunction in the lobes is not related to the spin of the electron.

 

[denian]

Hello there,

First of all, there is no such thing as a stupid question in science. It is important to ask questions and seek clarification in order to fully understand a concept.

To answer your first question, the shape of the π bond is not cylindrical because it is formed by the overlap of two p orbitals, which have a dumbbell shape. The nodal plane in the π bond comes from the overlap of the two p orbitals, not from the individual p orbitals themselves. Therefore, the semi-orbitals of the π bond do not inherit the sign of the function from the p orbitals.

The LCAO (Linear Combination of Atomic Orbitals) explanation is used for both s and p orbitals in forming σ and π bonds. In this explanation, the wave functions (ψ) are combined to form the molecular orbital. The square of the wave function (ψ^2) is used to determine the probability density, which gives us an idea of the shape and orientation of the orbital.

There is no correlation between electron spin and the sign of the wave function. The spin of an electron is a quantum property and is not related to the orbital shape or orientation.

It is possible for electrons to change spin at room temperature, but it is not easy or frequent. This process is known as spin-flipping and requires a significant amount of energy. In most cases, electrons in semi-populated orbitals will have opposite spins, allowing for bonding to occur.

In terms of assigning signs (+/-) to p orbital lobes, this is done for convenience and to help visualize the orbital. The signs do not represent the actual charge or probability density in those regions.

I hope this helps clarify some of your questions. Don't hesitate to ask for further clarification if needed. Keep exploring and asking questions – that's what makes a great scientist!