# Volume of orbitals

As I was learning about atomic structure, the lecturer made a seemingly dubious claim:

The volume of a p orbital is one-third that of the s orbital. Thus, inter-electronic repulsions are particularly significant in p orbitals.

She was explaining the lower first ionisation energy of oxygen compared to that of nitrogen and of course, she invoked the idea of inter-electronic, intra-orbital repulsion in her explanation. However, she then went on to ask why is it that we do not observe this trend when looking at $\ce {ns^{1}}$ and $\ce {ns^{2}}$ configurations and answered with the above claim.

I was rather intrigued by her claim as I've never heard it ever before and intuitively, I feel that it is wrong. Thus, I would like to clarify if this claim is indeed wrong?

I would also like to ask how can the volume of orbitals be determined?

## 1 Answer

It would most likely be impossible to come up with a reliable measure of the volume of an orbital.

Orbitals are defined over all space, so an integral like this $$\iiint\phi(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$$ is not guaranteed to be finite, and also doesn't really mean anything. Clearly, the orbitals are decaying, but are they decaying quickly enough to have this integral be finite? I would suspect they probably are because the hydrogen atom orbitals decay like $$\exp(-r)$$, and this integral from $$0\rightarrow\infty$$ is finite. Regardless, $$\phi$$, is an atomic orbital here, but the integral over the space of an atomic orbital, molecular orbital, or wavefunction doesn't mean anything physically that I am aware of. Only the integral over the magnitude squared.

One could think about picking a fixed distance from the nucleus, and then integrating over all space inside this distance from the nucleus, but this would always be arbitrary. I doubt that a $$p$$ orbital would ever be one third the volume of an $$s$$ orbital for a reasonable choice of distance.

Third, any measure of the volume of an orbital which gives a finite answer is going to be spoiled when you consider that the shape or orbitals is not unique in the sense that a unitary transformation of the molecular orbitals gives another set of valid molecular orbitals. These transformations will almost certainly not maintain the same volume within some distance of the nucleus or anything like this.

It is perhaps possible to quantify what this lecturer is getting at in a more concrete way, however. For instance, there are different schemes of creating localized molecular orbitals in the context of Hartree-Fock. The Edmiston-Reudenberg method of orbital localization maximizes the electron-electron repulsion. So, one could carry out this procedure to generate orbitals which look like the typical $$s$$ and $$p$$ orbitals we think about, then look at the result of this maximization procedure to verify that the electron-electron repulsion is larger in the $$p$$-like orbitals than the $$s$$-like orbitals. This might work, but because the shape of the orbitals can easily be changed, I doubt this measure has any chemical significance.

So to summarize, the lecturer's reasoning may not be strictly wrong, but she is being very loose with objects that are poorly defined, such as the volume of an orbital. I also explained why the volume of an orbital is poorly defined, and there is really no way to fix this completely.

As to the question she was actually trying to answer, inter-electronic repulsion in $$p$$ orbitals versus $$s$$ orbitals is completely unnecessary. Here is a graph of ionization energy with atomic number It is quite clear that the dips in the graph correspond to time when we go from a half-filled shell to one extra electron (nitrogen to oxygen) or from a full shell to the first electron in an empty shell (beryllium to boron). The beryllium to boron case says this cannot possibly be due to inter-electronic repulsion being stronger in $$p$$ orbitals because there is only one electron in the $$p$$ orbital of boron. I would say most of the explanation of this phenomenon is answered in this question. Basically, it boils down to the fact that these electrons which are added to a new shell or half-full shell must be the opposite spin of the other electrons, so they are not stabilized by exchange, and are higher in energy despite the greater attraction to a more positively charged nucleus. Notice this effect decreases as you go down the periodic table. This is understood by noting that exchange is a somewhat distance-dependent phenomenon, so as the orbitals become larger, this loss of stabilization from losing exchange is not as competitive with the increased nuclear charge.

• My take is, you could say that the average distance $r = |\vec{r}_1 - \vec{r}_2|$ between two electrons in the same p orbital is smaller than that between two electrons in the same s orbital. Hence the repulsion - which scales as $1/r$ - is greater for p electrons. Volume is indeed a poor choice of word. – orthocresol Mar 14 '18 at 18:06
• @orthocresol I agree, but even this remains to be shown quantitatively. It may very well change as you go down the periodic table and so on. Also, the measure you suggest $|\vec{r}_1-\vec{r}_2|$ is the very thing which is minimized in the Edmiston-Reudenberg localization scheme as this does indeed maximize electron-electron repulsion. I guess the bigger point is that these orbitals aren't unique, so reasoning based off of any particular set of orbitals is always somewhat questionable. – jheindel Mar 14 '18 at 18:23
• Why do you claim that the integral above is not guaranteed to be finite? I think it is – user1420303 Mar 14 '18 at 22:57
• @user1420303 Ya, it probably is, but the only thing that really must be finite is the integral over the square of that orbital, and there are certainly functions whose squares are convergent when integrated over all space, even though that is not true of the function. The most obvious example is $1/x$, which famously diverges and $1/x^2$ which converges. There is probably a physical argument that all atomic orbitals decay faster than $1/x$, but I don't know this is actually true since atomic orbitals are not strictly real things (in the physics sense). – jheindel Mar 14 '18 at 23:03
• @Tan Yong Boon I would say that is because the ns1 electron does not benefit from exchange, so there is no stabilization lost when that next electron is added, so the ionization energy increases because nuclear attraction just be larger than the electronic repulsion or else the electron wouldn't bind at all. – jheindel Mar 15 '18 at 1:35