[Rosalind Franklin]

Hey guys!

When we derive radius ratio rules, we assume that atoms are hard spheres(same goes for ionic, covalent radius calculation as well) but we know that atoms aren't really spherical in shape. So, how does one modify these rules depending on the shape of the atom? How different are the results when we don't make that assumption?

[Arkcon]

Well, what shapes do the atoms actually have?  If we don't really know, then we can't really do anything.  Start with an example.

[Rosalind Franklin]

For example, if we consider a Sodium atom, which has an electronic configuration 1s² 2s² 2p⁶ 3s¹, we know the s orbitals are spherically symmetrical but the p orbital is dumbbell in shape. Clearly, the shape is much more complicated and not spherical. Also, these boundaries are not well defined and they are only probability distributions. How will our result change if we take this into consideration?

[P]

with that specific Na example the 3s orbital is outside the 2p ones anyway - so the outer most orbital is spherical.

If you had a atom with p orbitals as the outermost orbitals then the atom can still approximate to a sphere for looking at distances and shapes I think - as far as I am aware you can still define a radius for the size of the atom.

I think, in terms of 'looking' at an atom though AFM or some other microscopy method then, they appear spherical (from what I have seen).   I am sorry if you want more detailed/academic explanations then someone else can probably give you more than I can.

[sjb]

At some level, I think the orbital shapes fall out of the maths used to describe them. I seem to recall there are different ways to describe the f-orbitals, for instance.

[Enthalpy]

It's absolutely right that packing spheres to imagine a crystal is only a mental representation. In addition, distances between atoms vary depending on the neighbours, the crystal structure and so on. A consequence is that the atomic radius, which is anyway fuzzy, depends a lot on the convention used to define it.

If you take the example of diamond, silicon and germanium, they all have the diamond crystal structure, but their band diagram is quite different. Spherical atoms wouldn't explain this.

My suggestion is: don't try to stretch this simplistic model beyond its capability. Even if you take atomic radii for Ni and Ti deduced from the densities, you get a very wrong result for NiTi - same for Invar or for bell bronze. So refining it with non-spherical ions would be an exaggeration, to my opinion.

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One 2p orbital is not spherical. But 2p⁶, in your sodium example, fills all three 2p orbitals. I believe (I didn't check) that 2p⁶ is perfectly spherical - and if not, it's closely spherical. At least for independent electrons, cos² (angle to 0x) + cos² (angle to 0y) is independent of the direction in the 0xy plane. In transition elements you might find an incomplete shell, possibly non-spherical, below the valence shell.

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2p orbitals are peacock shaped... or not. As long as 2p_x, 2p_y and 2p_z have exactly the same energy, any linear combination of them is still a stationary solution of old Erwin's linear equation. That is, the linear combination is an orbital, and a 2p one.

Starting arbitrarily from the peacock set of 2p, you could define an other set like
(2p_x + j*2p_y)/sqrt(2)
(2p_y + j*2p_z)/sqrt(2)
(2p_z + j*2p_x)/sqrt(2)
which is equally good and is even orthogonal. With other coefficients, you can get ellipsoidal 2p orbitals, maybe less convenient, but valid too, and some sets are orthogonal.

Chemists often use the peacock-shaped 2p set because this one is natural for chemical bonds. But if you study the Zeeman effect, with an external magnetic field, then the doughnut-shaped set of 2p is more convenient.

All sets of 2p are equally valid, none is more fundamental than the others. And if you deduce one from an other by an adequate linear transformation, the shape of the full 2p⁶ remains the same.