# Hybridization and excitation of electrons, how is it possible?

When studying the structures of molecules by hybridization why we take into consideration of excitation of electron if the excited electron only stays in the upper shell by absorbing for only 10^-8 seconds{approximately }?

• We don't.$\mathstrut$ – Ivan Neretin Aug 28 '18 at 9:53
• Hybridization is a rather simple concept used for educational purposes mostly. The description of electronic excitations goes way beyond the scope of the model. Nobody (that I'm aware of) really quantitatively studies the structure of molecules using hybridization. – Raditz_35 Aug 28 '18 at 14:32

We don't really consider the excitation of electrons. If it appears like this, then this is probably due to an incorrect understanding of what an orbital actually is: The attempt to describe each single electron by its own wave function which is independent of all the other electrons (which it is not).

An orbital is defined as single electron wave function $\varphi(\vec r_i)$ where $\vec r_i$ are the coordinates of electron $i$. We then try to approximate the many electron wave function as an (anti-symmetrized) product of orbitals

\begin{equation} \Psi(\vec r_1, \vec r_2, \dots \vec r_N) \approx \prod_{i=1}^N \varphi(\vec r_i) \end{equation}

But this is strictly speaking not true, because the electrons are correlated with each other, due to the Coulomb repulsion $V\propto \frac{1}{|r_i-r_j|}$. The product ansatz does not work for the kind of differential equation we need to solve. The Hartree-Fock method tries to do so anyway, and gives therefore wrong (or at least only approximate) result.

This means the whole concept of orbitals is an approximation, which may give a qualitative understanding, but are failing when trying to get quantitative results.

One approach to correct this approximation is expanding the many electron wave function in some basis and solve the resulting eigenvalue problem. This basis is not unique, but one possible choice is to use the Hartree-Fock determinant and the excited configurations you can generate from it. But these excited configurations are more of an abstract mathematical thing and have limited physical meaning, because they are based on the above approximation. The real physics/chemistry is in the solutions of such methods. And those solution cannot be describe by the above ansatz for $\Psi$.

• Indeed, one of the surprising things about the universe is how good the single electron solutions are even for atoms with many electrons. One might argue on 'good', but the fact we can identify noble gases all the way down the periodic table means they are fantastically good at some level. – Jon Custer Aug 28 '18 at 14:36
• @JonCuster Good point. This is nothing one would expect from a purely mathematical perspective. I think it is consequence of nature preferring locality. – Feodoran Aug 28 '18 at 15:11