Do hybrid orbitals exist in unbonded molecules? What would they look like?

Question

For example, the ground state of a neutral carbon atom could be notated as:

$$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$

I know $\ce{s}$ orbitals are shaped like spheres so the two electrons in the $\ce{s}$ orbital should probably form a spherical shape.

But how should the electrons in the $\ce{p}$ orbitals be distributed?

I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.

So do hybrid orbitals exist in unbonded molecules? What would they look like?

I think this would have interesting results for collisions between atoms and how probable certain reactions are.

Answer 1

Hybrid orbitals do not exist. Individual atoms have electronic configurations which can be explained by considering atomic orbitals. Molecules have electronic configurations which can be explained by considering molecular orbitals. Hybrid orbitals are just one mathematical way of arriving at molecular orbitals from combining atomic orbitals. They are a pure mathematical fiction and have no physical meaning (this does not mean they aren't useful).

Some schools or textbooks teach that in order for a reaction to occur energy is required to 'promote' electrons into hybrid orbitals to create some sort of excited atom prior to bonding. This does not happen in real life and is not a useful way of looking at bonding. What does happen is that as atoms approach each other, their atomic orbitals start to overlap significantly, giving rise to a new set of molecular orbitals (in quantum mechanics you are changing the potential in which the electrons reside and so the allowed energy states change). If the electrons in the new molecular orbitals are lower in energy than in the atomic orbitals then the reaction will proceed.

As for the electron distribution of atoms - it is spherically symmetric. The two $\ce{2p}$ electrons will reside in different orbitals because this does not require pairing of their spins (which is a higher energy state). However, it is meaningless to ask which $\ce{p}$ orbital they are in because all three orbitals are degenerate and so are indistinguishable.

Answer 2

... the ground state of a neutral carbon atom could be notated as: $$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$ I know $\ce{s}$ orbitals ... form a spherical shape. But how should the electrons in the $\ce{p}$ orbitals be distributed?

The answer from a physically point of view is unambiguous. With the help of the mathematics for spherical harmonics and in the limitation of the Cartesian coordinate system the shapes of the probability to find the electron in some volume around the nucleus (with some percentage of say 90%] are computated for the p-subshells of the Neon- and respectively Argon-elements as follows:

Despite their name, spherical harmonics take their simplest form in Cartesian coordinates, where they can be defined as homogeneous polynomials of degree l in (x,y,z) that obey Laplace's equation. (Wikipedia)

So according the quote from Wikipedia some other solutions in noncarthesian coordinates should exist. In the German version of WP some forms are outlined, of which the lower left might be of interest:

There are 8 volumes computed, all of the same symmetricity to the center of the sphere, 4 with a sign “+” and 4 with a sign “-“. Isn’t that a closer solution to the $sp^{x}$ hybridizations? The sign shows the direction of both the electrons magnetic dipole moments and the electrons spin. The electrons of the noble gases Ne and Ar are in a perfect equilibration of their 8 electrons of the “$sp^{3}$” shells. It is obvious that such a solution doesn’t correspond with the founded solutions in the scientific literature and we have to reject this solution.

Your question, how the electrons of single atoms are distributed around the nucleus is only of theoretical interest. Has anyone ever measured the shape of single atoms respectively the indisturbed outer electrons from these atoms? A the end you are interested in the electron distribution of molecules. From the form of the lattice one could conclude about the electron distribution for molecules and from this knowledge one can predict the existence and compound structure of new molecules.

With this in mind the above working assumption (Arbeitshypothese) about noble gas configuration with 4 (spin up) + 4 (spin down) perfectly balanced electrons on the edges of a cube has some advantages. A good example is the crystal structures of cubic lattices. Diamond as a form of carbon has this lattice too:

Simply take the $sp^{x}$- hybridization as the standard AND imagine the molecules you are interested in as compounds between their atoms with electron configurations of filled cubes for each atom. Take for example the Noble gas compounds. For some the imagination about cubic - or octahedral (which is the same, see Platonic solides) - distribution works perfect and for others such model doesn’t fit the founded geometries:

Images from Wikipedia

I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.

From the point of physics the the answer for a carbon atom is clear: two electrons are in the s-subshell and two electrons are in atoms he p-subshell.

So do hybrid orbitals exist in unbonded molecules? What would they look like?

I think this would have interesting results for collisions between atoms and how probable certain reactions are.

If you want to go in the direction of a scientist this could be your area of work. The question is a astonishing good.