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I understand (for example in the case of H2) that two s orbitals produce through constructive interference a bonding orbital and that the out of phase combination of two s orbitals leads to destructive interference, followed by the apparition of an antibonding orbital, but, as far as I know, these two orbitals form at the same time (the number of orbitals is conserved, although only the lower energy one is occupied in the first instance). How can this be possible? Are specific regions of the two s orbitals in-phase and others out of phase so as to produce two different outcomes at the same time? Do same-phase regions interact selectively with other same-phase regions or does the s orbital coexist in two phases at the same time, thereby simultaneously producing two different bonding orbitals (having to do with superpositions). Please help me, I am really confused and similar questions contain answers that rely on mathematical proofs - I would like to grasp it conceptually: are the two phases of the s orbital simultaneously occurring?

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    $\begingroup$ Hello. I'm afraid that deleting and reposting your question doesn't make it any less of a duplicate of this. You want something "conceptual", but in fact if you read it, the previous OP already asked for something similar. If the answers there don't satisfy you, then you should seek a new answer to that question, not a new question. $\endgroup$
    – orthocresol
    Jul 23 at 9:11
  • $\begingroup$ I (quite substantially) expanded on my answer there, with an analogy which you might hopefully find more helpful. You might want to check it out again. That's as far as I can help, though; this doesn't affect the fact that this is still a duplicate, so I'm not inclined to reopen, unless there is a convincing argument (and I don't see one now). $\endgroup$
    – orthocresol
    Jul 23 at 12:10
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I will try to answer without using any mathematical formula, as your question is just made of words, without any mathematical symbols.

You should know that the wave function of an electron in the $\ce{H}$ atom is normalized. It means that the integral of the wave function over the whole space is equal to $1$. And it must be equal to $1$.
If you add two of these wave functions from two $\ce{H}$ atoms, the integral of this sum will be $2$. This is forbidden. It must be $1$.

So you cannot simply add two normalized wave functions, like the $\mathrm{1s}$ in the $\ce{H}$ atom. You have to take one-half of the wave function of the first atom and to add it to one-half of the wave function of the second atom. This would produce a new molecular wave function whose integral is equal to $1$. It is nice. This new molecular wave function describes a sigma bond, designed as $\ce{H-H}$ in the Lewis model.

Now you have to use and combine the other half of these wave functions, without adding them. The only thing to do is to subtract them. But such a subtraction produces a point between the two nuclei where the obtained wave function is equal to zero. As a consequence, this molecular orbital is a antibonding orbital. And the integral of this antibonding is also equal to $1$.

Hopefully the pure theorists will pardon me for having simplified the description of the problem. I should have spoken of the factor $√2/2$ instead of $1/2$.

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    $\begingroup$ You basically answered exactly the same question as this. You might wish to consider answering there, instead of, or in addition to, this question. $\endgroup$
    – orthocresol
    Jul 23 at 9:17