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I know that the short answer is that "from bond formation" but as I think electron promotion occurs before formation of bonds between atom. So it seems like a paradox.

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    $\begingroup$ Hybridisation occurs in our heads. $\endgroup$ – Ivan Neretin Feb 10 '20 at 20:47
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Hybridation has an advantage : it is visual and it explains perfectly the geometry of the molecules. But hybridation is not necessary to explain these structures. You can explain the geometry of the molecules without hybridation, although it is more difficult to calculate.

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    $\begingroup$ This is exactly the other way around. You can determine the structure, then you can propose hybridisation of orbitals. $\endgroup$ – Martin - マーチン Feb 11 '20 at 1:33
  • $\begingroup$ You have to use VSEPR to determine the geometry and only then get the hybridisation; so hybridisation is really just an a posteriori justification. $\endgroup$ – ANZGC FlyingFalcon Feb 12 '20 at 3:43
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I think you can find plenty of discussion of the pros and cons of hyrbidization elsewhere on this site, so I'm going keep it fairly short.

First, the premise of your question shows confusion between the paper exercise of first creating hybrid orbitals and then filling them with bonding electrons and the physical event of molecule formation, which doesn't happen in those steps. Molecules can form in all sorts of ways, and the exercise of determining electronic structure by assembling the molecule from its constituent atoms is not related to its actual formation. So the only relevant comparison is between the two representations of the final molecule - one with each bond and lone pair represented by a localized hybrid orbital and one with delocalized molecular orbitals comprised of specific $s$ and $p$ (and $d$ and $f$ as necessary) atomic orbitals. These are often called "canonical" molecular orbitals.

A key point is that electrons do not actually reside in specific orbitals. Since electrons are indisinguishable and don't have specific locations, the wavefunction of a molecule involves all of the electrons collectively occupying all of the occupied orbitals. For practical purposes, we like to think of each electron in a specific orbital because it's simpler.

With that in mind, it may be easier to understand that as long as the bonds involved in the hybridization are all identical (eg the four C-H bonds in methane or the three C-O bonds in $\ce{CO3^2-}$), the electron density described by canonical MOs is exactly identical to that described by the hybrid $sp^3$ or $sp^2$ orbital bonds. On paper, there's a difference in the space covered by each electron, but since we said before that the electrons are indistinguishable from one another, that difference isn't physically real.

Where the simple hybridization model is inaccurate is when it is applied to different bonds/lone pairs. For example, describing the O of $\ce{H2O}$ as $sp^3$ hybridized is not mathematically rigorous. The degree of $s$ and $p$ contribution (ie hybridization) to the lone pairs is different than in the O-H bonds. This creates confusion when people try to make quantitative conclusions about things like bond angles based on this qualitative representation. But working out the actual non-integer hybridization values detracts from the powers of the hybridization approach, which is its intuitive qualitative simplicity.

So the take home message is that if you have a hybridization involving identical bonds/lone pairs, it is quantitatively equivalent to canonical MOs for the purposes of calculating electron densities. If the bonds/lone pairs are not identical, integer hybridization is not quantitatively equivalent, but with more work one can calculate the appropriate non-integer hybridizations.

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The important thing to note, really, is that hybridisation doesn’t really happen - what you’re learning as “Valence bond theory” is really just a dumbed-down and almost completely wrong explanation to chemical bonding that is outdated by at least 60 years.

In "proper" valence-bond theory, the many-electron wavefunction is described as a linear combination of multiple "resonance structures", with the energy variationally optimised. It can be shown that a full consideration of all possible such "structures" would be equivalent to a full configuration-interaction treatment of molecular orbital theory; however, if the Heitler-London wavefunction is the only one considered, VBT is "too covalent" while MOT is "too ionic".

In Pauling's day, it was thought that the lowest excited state of Carbon lay relatively close to the ground state, such that the promotion of one electron from the 2s to 2p orbital would be readily achievable. As that has been shown to be untrue, his model of hybridisation is hence invalid. Just keep in mind that what you're learning is both outdated and mostly wrong.

I’d suggest you look up molecular orbital theory if you’re interested in knowing more about Chemical bonding - otherwise, the book "A Chemist's Introduction to Valence Bond Theory" was pretty helpful for me to understand what modern VBT looks like.

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    $\begingroup$ It is important to note that valence bond theory is equal to molecular orbital theory when it is not dumbed down. $\endgroup$ – Martin - マーチン Feb 11 '20 at 1:35
  • $\begingroup$ I can't just leave it because of the syllabus but can you please help me to get out of this paradoxical thing. $\endgroup$ – Hemant Feb 11 '20 at 7:44
  • $\begingroup$ @Martin-マーチンthat's definitely true - I'll probably edit my answer to reflect that as soon as I have time. $\endgroup$ – ANZGC FlyingFalcon Feb 12 '20 at 3:30

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