During the hybridization of certain elements such as phosphorus during the formation of $\rm PCl_5$, the atom first gets changed to the excited state. In the case of $\rm PCl_5$, the paired electron in the 3s orbital gets promoted to 3d orbital. Similarly in $\rm SF_6$, the paired electrons in 3s and 3p orbitals get promoted to the 3d orbital. Why does this happen? Shouldn't the electrons get promoted to the 4s orbital as it is lower in energy than the 3d orbital, and also nearer to the ground state orbitals?
Electron Promotion – Why Electron in 3p Orbital Promotes to 3d Orbital Before Hybridisation
atomic-physicselectronsorbitalsphysical-chemistryquantum mechanics
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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.
Hartree-Fock is a groundstate method. The equations are constructed to optimize the groundstate energy and you will not obtain reasonable excited state energies by replacing an occupied orbital with an unoccupied one in your Slater determinant. This yields a very poor approximation for an excited state. Virtual orbitals do not correspond to excited states in such a simple manner.
But the results from a Hartree-Fock calculation can be used in Post-Hartree-Fock Methods to obtain excited states.
The virtual orbital eigenvalues can be interpreted as electron attachment energy/electron affinity. By looking at the energy of $E(N) - E(N+1)=-\epsilon_v$ where N stands for the number of electrons. $E(N)$ stands for the energy of your groundstate calculated within HF with $N$ occupied orbitals. $E(N+1)$ stands for the energy of a determinant that you obtain by occupying one additional virtual orbital with index $v$ while keeping the former $N$ occupied orbitals as they were. The difference of the energies of these two determinants corresponds to the electron affinity of the unoccupied orbital and is given by its negative orbital energy eigenvalue $\epsilon_v$. This is basically the same as Koopmans theorem but now for electron affinity instead of ionization potential.
But the energies calculated using Koopmans theorem are usually not good approximations. The determinant energies with $N\pm 1$ are bad approximations, when we keep using orbitals that were optimized for a $N$ electron system. If you did a HF calculation starting witn $N+1$ electrons, you would get different orbitals since your effective potential that is part of the Fock Operator changes with the number of electrons. The orbitals of a $N$ and a $N\pm1$ calculation are often similar but not the same. This difference is also often called orbital relaxation when we take the $N$ to $N-1$ case. Since the electrons feel the repulsion of one electron less and thus can "relax" compared to the system with one electron more. And to get these "relaxed" orbitals you would need to do new HF calculation with just $N-1$ electrons, instead of using the orbitals from the $N$ calculation.
Best Answer
The ground state of phosphorus is $\mathrm{[Ne] 3s^2 3p^3}$.
In order for $\mathrm{PCl_5}$ to exist with the molecular shape we know from experiment, a hybrisation occurs in $\mathrm{P}$'s valence electrons, that is:
$$\mathrm{sp^4}$$
These are five atomic orbitals, identical in all respects.
This explains the molecule's trigonal bipyridal shape, shown at the bottom of that (an answer of mine) link.
The $\mathrm{d}$ orbital doesn't enter the equation, if we assume a set of five $\mathrm{sp^4}$ hybridised orbitals. The $\mathrm{sp^4}$ have energies in between the $\mathrm{s}$ and $\mathrm{p}$ energy levels.