In general when we talk about chemical bonding, we say that it is "nature's way to stabilize the systems" and the energy of the molecule formed by atoms will be lower than that of the individual atoms; hence leading to stabilization.

However, while reading about the Molecular Orbital Theory in my NCERT Chemistry Textbook, I came across a statement -

The total energy of the molecular orbitals formed remains same as that of the original atomic orbitals.

How can we then justify through the Molecular Orbital Theory that the process of chemical bonding leads to stabilization, if no energy is released?

  • 1
    $\begingroup$ Is true just for the orbitals They probably don't have made clear that the statement has no physical content, or when it happens it does not lead to a stable bound species. Indeed is valid, and under assumptions/approximation, only for molecules with bond order equal to zero. Take it as "the bonding and antibonding orbitals are roughly equally spaced from the energy levels of the atomic orbitals". It is all about the energy of empty orbitals. You can order them. Is the same as to say that eight 100 m has more potential energy than sea level, tough there is no stone at that altitude yet. $\endgroup$
    – Alchimista
    Dec 14 '19 at 8:40
  • $\begingroup$ @Alchemista thanks for the comment. However could you please elaborate your points further in more detail ?? I could not properly understand your point, especially about the stone. $\endgroup$ Dec 14 '19 at 8:46
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    $\begingroup$ They are speaking about energy of empty orbitals. A stone lift up would have more potential energy. But unless you actually bring it there, no physical energy is involved. The same for mo. The statement is roughly valid for the orbitals formed. If you equally fill both bonding and antibonding then the molecule does not occur. $\endgroup$
    – Alchimista
    Dec 14 '19 at 8:53
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    $\begingroup$ In other words molecular orbitals exists independently of the real existence of a molecule, they can be used to predict that the molecule does not exist /, is not stable. This should remove your doubt. No energy release no stable bonding. This is the use to be made, not the contrary. $\endgroup$
    – Alchimista
    Dec 14 '19 at 8:56
  • $\begingroup$ Thanks for the clarification @Alchemista. I do get your point now. $\endgroup$ Dec 14 '19 at 8:59

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