Can the formation of the covalent bond in HCl be explained through the hybridization model?

We know for sure that it can be explained by the theory of molecular orbitals. In this way, it would be normal for the atomic orbital $s$ of the hydrogen to mix with one of the chlorine's atomic orbitals which has the same form or energy. Neither the symmetry of the form nor that of the energy can be preserved (totally), because the energy levels of the atomic orbitals of chlorine are much lower than that of the $s$ orbital of the hydrogen. Thus, the $s$ orbital of the hydrogen mixes with the $pz$ orbital of the chlorine and forms $2$ molecular orbitals (a bonding one and an antibonding one). The atomic orbitals $s$, $px$ and $py$ of the chlorine are nonbonding and remain at the initial energy.

According to the Lewis structure, between H and Cl we have a simple sigma bond and around the chlorine atom we are left with $3$ pairs of nonbonding electrons. The charges on H and Cl are null. With this in mind, we notice that $4$ hybrid orbitals would be needed to host the sigma bond and the $3$ pairs of nonbonding electrons of the chlorine, thus getting an $sp^3$ hybridization (the real geometry of the molecule is linear and the one in which are also involved the orbitals with pairs of nonbonding electrons is tetrahedral.

Considering the energy diffrence between the $s$ orbital of $H$ and the $s$ and $p$ orbitals of Cl (as we saw in the theory of molecular orbitals), is it possible to create hybrid orbitals in the HCl molecule?

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    $\begingroup$ Yes, you need 4 hybrid orbitals, but why should you assume that they must all be the same (hence sp3)? You could have sp2-hybridised chlorine (sp2 + sp2 + sp2 + p), which is still four orbitals, or sp-hybridised (sp + sp + p + p), or indeed completely unhybridised (s + p + p + p). Indeed, any combination of four atomic orbitals must give four hybrid orbitals, so knowing that we need four hybrid orbitals doesn't actually tell us anything about the resulting hybridisation. $\endgroup$ – orthocresol Jan 22 at 21:10
  • $\begingroup$ @orthocresol Oh, I understand. I didn't know that we can split them i.e. that sp+sp+p+p is the same as sp3, but it totally makes sense. Thank you! $\endgroup$ – ChemistryGeek Jan 22 at 21:28

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