# P orbital, π bond, and lots of questions

1. Feb 10, 2006

### 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 wont 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 ?

Sincerely, thank you, thank you, for reading this, and eventually helping me !

2. Feb 10, 2006

### Gokul43201

Staff Emeritus
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.

Last edited: Feb 10, 2006