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MOT is really confusing me.

I was taught that the no. of orbitals remains conserved.That means, the number of atomic orbitals used is equal in number to the molecular orbitals formed.

But how is it possible in the following case?

For example in Case 1, I take a positive lobe of a p orbital and combine it head on with the positive lobe of another p orbital I get a bonding MO.

In a different scenario i.e Case 2, I take a positive lobe of a p orbital and combine it head on with the negative lobe of another p orbital to get an antibonding MO.

So overall I have used 4 atomic orbitals (2 in each case) to get 2 MO( one bonding and one antibonding). So what is happening here? Am I missing something?

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    $\begingroup$ You don't take lobes separately. There is no such thing. There is an orbital as a whole: take it or leave it. $\endgroup$ – Ivan Neretin Nov 10 '15 at 11:05
  • $\begingroup$ I am not too sure of this, but isn't only one lobe actively involved in bonding.In pictures,the other one is usually behind. $\endgroup$ – Karan Singh Nov 10 '15 at 11:07
  • $\begingroup$ Possibly related: chemistry.stackexchange.com/q/39223/186. $\endgroup$ – Wildcat Nov 10 '15 at 11:07
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    $\begingroup$ @KaranSingh, no, you don't combine lobes, you combine the whole orbitals. But you combine them in two different ways: one way gives rives to bonding MO, another to antibonding MO. $\endgroup$ – Wildcat Nov 10 '15 at 11:11
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    $\begingroup$ @KaranSingh, that is because there is no any time out there. AOs are combined into MOs only on a sheet of paper, it is not a real physical process taking place in space-time. Rather, it is just a mathematical formalism. $\endgroup$ – Wildcat Nov 10 '15 at 12:03
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In the linear combination of atomic orbitals (LCAO) there would be a bonding orbital and an antibonding orbital when two atomic orbitals combine to form a bond. The bonding orbital would be lower in energy that either atom contributing an orbital, and the antibonding higher than either.

One twist comes from the number of electrons. If there are just two electrons, then they both go into a bonding orbital. If there are three electrons then two go into a bonding orbital and one into the antibonding orbital.

There are three p orbitals so two atoms could be bonding with 1-3 p orbitals. If two p orbital bonds were being used then both p molecule bonding orbitals should be lower in energy than either antibonding orbital.

The whole concept of having a bonding orbital and an antibonding orbital is necessary to explain for instance electronic transitions. Imagine an oxygen atom in free space. If an electron is added, the oxygen molecule will accept the electron to form an ion $\ce{O^{-}_2}$. The electron doesn't go to one oxygen or the other, it is shared between both in a molecular orbital. Look at the diagram below for $\ce{O2}$. We'd predict that the electron will fall into one of the antibonding pi orbitals ($\pi^*$). From this we'd predict that a free oxygen atom could actually take three extra electrons and be stable. However the fourth electron would fill all the orbitals and $\ce{O^{-4}_2}$ would have no more electron binding energy than two free $\ce{O^{-2}}$ atoms.

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