-6
$\begingroup$

i am a high school student,,and i am confused in molecular orbital theory,,,my confusion is that how bonding and non bonding orbitals are formed at the same time my teacher taught me that the two wave functions interfere constructively forming bonding orbital and if they interfere destructively the form anti bonding orbitale.g when we fill the electrons in say O2 molecule then why do we fill electrons in bonding as well as anti bonding at the same time,,,please explain this in a very simple language without going into mathematics???does really bonding and anti bonding orbitals are formed from constructive and destructive interference???

$\endgroup$
1
  • 2
    $\begingroup$ Your question is really hard to comprehend. Please use proper spelling and punctuation. For what I took away from it: in general terms (between two atomic centres) a bonding type molecular orbital will form by constructive interference. An anti-bonding orbital will form by destructive interference. When talking about more bonding partners, the definitions are not that straight anymore... $\endgroup$ – Martin - マーチン Jun 23 '20 at 10:11
1
$\begingroup$

First of all, please keep in mind, that molecular orbital theory is just a model. Much like the Bohr model is valid yet incomplete, orbitals are not the whole truth. They are just closer to the truth than the Bohr model.

Orbitals are a way of representing how electrons are arranged in atoms and molecules. They give some idea of where the electrons are located in space and how much energy they have, meaning how easily they can be removed from the molecule.

As lower (more negative) energies mean the electron is harder to remove, those arrangements (orbitals) are preferred. As the name implies, bonding orbitals typically have lower energy than their corresponding anti-bonding orbitals and are therefore occupied first. If more electrons are available and need to be distributed in the molecule, energetically higher lying orbitals will be occupied.

The different available orbitals may be constructed as linear combinations of atomic orbitals (LCAO). In mathematical terms this is just a basis set expansion and nothing special in quantum mechanics. The choice of atomic orbitals as the basis set is arbitrary but valid, and physically motivated. The rule is: you can form as many molecular orbitals as you use atomic orbitals to form them. The simplest case is the one you described: two atomic orbitals will form one bonding molecular orbitals and one anti-bonding molecular orbital, with constructive and destructive interference respectively.

If enough electrons are available, both types of orbitals will be occupied at the same time. The anti-bonding orbitals will counter the effect of bonding orbitals, which is why $\ce{N2}$ is more stable than $\ce{O2}$ and $\ce{Ne2}$ will dissociate into $2\ \ce{Ne}$.

$\endgroup$
7
  • 1
    $\begingroup$ Related: How can antibonding orbitals be more antibonding than bonding orbitals are bonding? Please note that the terms you are using for bonding, non-bonding, and anti-bonding, are not congruent with how they are used within LCAO. $\endgroup$ – Martin - マーチン Jun 23 '20 at 10:22
  • $\begingroup$ @Martin-γƒžγƒΌγƒγƒ³ what exactly do you mean? $\endgroup$ – Feodoran Jun 23 '20 at 10:30
  • 1
    $\begingroup$ I've interpreted your statement bonding orbitals typically have lower energy than anti-bonding or non-bonding orbitals and are therefore occupied first in terms of orbital eigenvalues, I'm not sure you've meant it that way. Anti-bonding LCAO can be (for most diatomics are) occupied before bonding LCAO. The definition in terms of energy is not necessarily straightforward, and it depends on the knowledge of the AO eigenfunctions. You can also define bonding and anti-bonding from the mathematical description of the MO, i.e. (not) having a nodal surface perpendicular to the bonding interaction. $\endgroup$ – Martin - マーチン Jun 23 '20 at 10:45
  • $\begingroup$ Do you mean situations like $2\sigma^*$ vs $3\sigma$? Does formed from different sets of AOs, therefore I did not consider them here. $\endgroup$ – Feodoran Jun 23 '20 at 11:01
  • $\begingroup$ Yes, but I only picked up this nuance when reading it again, after your comment. I might be worthwhile clarifying this more, since the OP's question style asks for 'very simple language'. $\endgroup$ – Martin - マーチン Jun 23 '20 at 11:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.