So, one thing has been annoying me ever since I learned about orbital hybridization: you explain the shape of molecules by postulating that the orbitals of multi-electron atoms are linear combinations of the orbitals of the hydrogen atom. Fine.
Here's what gets me, and I've asked several chemistry professors about this, and I haven't gotten an answer. If I measure the orbital angular momentum of an electron in a hydrogen atom, I get the value $L = \hbar \sqrt{\ell(\ell +1)}$, so an s orbital has angular momentum $0$ and a p orbital has angular momentum $\hbar$.
So, what happens when I measure the orbital angular momentum of a sp${}^{3}$ electron? Do I get $\frac{3\hbar}{4}$ with probability 1? Or do I get $\hbar$ with probability 0.75, and $0$ with probability 0.25? Are the original orbitals really the fundamental thing, and the hybridized orbitals fundamentally just quantum superpositions of these things? Or does the multi-electron potential create a state that just looks like a linear combination of one s and 3 p orbitals?
Or is the actual answer more complicated and this a heuristic picture that I'm taking too seriously?
Best Answer
More like the latter than the former, although you've got one detail wrong. An $s$ orbital is $\ell=0$, so it corresponds to $L=0$, not $L=\hbar$. A $p$ orbital is $\ell=1$, corresponding to $L=\hbar\sqrt{2}$. If the electron is in a state that is a superposition of $s$ and $p$, then a measurement of angular momentum will yield either 0 or $\hbar\sqrt 2$, with probabilities given by the squares of the amplitudes of the different parts of the superposition. If it's an equally-weighted superposition (and I confess I don't know enough chemistry to know whether that's correct), then the probabilities are 25% and 75% as you say.