Molecular Orbital Theory: Explaining a "Dot" Representing No Interaction

[1bird9stones]

Question on Molecular Orbital Theory (and the "dot" that represents no interaction)

I have a question about the "dot" that represents no interaction between orbitals. For example, in the molecular orbitals of H3, there is the lowest energy molecular orbital that has two bonding interactions, the highest energy molecular orbital with two antibonding interactions, and a middle energy molecular orbital with no interaction (and a "dot" representing where the middle H orbital is).

Why then, in say BeH2, does the molecular orbitals that have the "dot" not show up as one of the accepted molecular orbitals? The MO's of BeH2 I'm thinking of are the ones where you combine the 2s orbital of Be with the antibonding \phi of H2. (and I guess the same question applies to when you combine the 2px of Be with the bonding \phi of H2).

Thanks for reading!

 

[JohnRC]

I have a question about the "dot" that represents no interaction between orbitals. For example, in the molecular orbitals of H3, there is the lowest energy molecular orbital that has two bonding interactions, the highest energy molecular orbital with two antibonding interactions, and a middle energy molecular orbital with no interaction (and a "dot" representing where the middle H orbital is).

Why then, in say BeH2, does the molecular orbitals that have the "dot" not show up as one of the accepted molecular orbitals? The MO's of BeH2 I'm thinking of are the ones where you combine the 2s orbital of Be with the antibonding \phi of H2. (and I guess the same question applies to when you combine the 2px of Be with the bonding \phi of H2).

Thanks for reading!

I am not familiar with any notation that uses a "dot" in this way. Could you perhaps link a document that provides an example?

 

[1bird9stones]

Hmm I guess the dot isn't a crucial part to my question so let me try to rephrase:In the molecular orbitals of H3, the middle energy orbital is, let's say, one positive orbital on the left, nothing in the middle (this is where the "dot" would be in the other convention), and a negative orbital on the right.

How come then, in BeH2, does the orbitals made up from the 2s orbital of Be with the antibonding ϕ of H2 and the orbitals made up from the 2px of Be with the bonding ϕ of H2 not appear as a molecular orbital? For example the 2s orbital of Be with the antibonding ϕ of H2 would be say, a positive orbital on the left, a negative orbital on the right, and a positive or negative 2s orbital in between. Why then, does this non interacting orbital exist in H3, but not in BeH2?

My initial guess is that it is solely because at this level of theory we can assume only orbitals closest to energies can interact, but I was just wondering if there was actually a different reason.

Thanks again!

 

[JohnRC]

No, that is not quite right.

In linear H₃ there are only 3 valence orbitals on the atoms involved in the bonding. That means 3 molecular orbitals. All are cylindrically symmetric. Two of them are symmetric end-for-end -- the bonding orbital made up of
0.5 ψ(H1 1s) +√0.5 ψ(H2 1s) + 0.5 ψ(H3 1s)
and the antibonding orbital made up of
0.5 ψ(H1 1s) – √0.5 ψ(H2 1s) + 0.5 ψ(H3 1s)

The other is antisymmetric end-for-end:
√0.5 ψ(H1 1s) – √0.5 ψ(H3 1s)

BeH₂ is quite different.

In this case we have to account for 6 valence orbitals on the atoms. But ψ(Be 2px) and ψ(Be 2py) are not cylindrically symmetric, so they cannot take part in bonding with the cylindrically symmetric hydrogen atomic orbitals; they remain as unoccupied, non-bonding atomic orbitals.

That leaves 4 orbitals. The end-to-end symmetric ones, one bonding and one antibonding, involve mixing of the two hydrogen atomic orbitals with ψ(Be 2s), and the end-to-end antisymmetric orbitals, again one bonding and one antibonding mix ψ(Be 2pz) with the hydrogen atomic orbitals.

In the molecule BeH₂, there are only 4 valence electrons, and so only the two bonding orbitals are filled.

 

[1bird9stones]

Thanks for the reply!

I understand how to get the 6 molecular orbitals of BeH2, but what I am still confused about is something else:

I guess another way to show my misunderstanding is this:

How come the orbital that is theoretically made up of the 2s of Be and the anti-bonding molecular orbital of H2 not exist as one of the orbitals? In theory, this molecular orbital would have a net 0 of bonding or antibonding interactions - similar to the net 0 interactions of the "antisymmetric end for end" H3 orbital you described above. Yet in H3 this molecular orbital is present while in BeH2 it is not.

That is my main misunderstanding. I understand where the 6 correct orbitals of BeH2 come from, but I am not sure why some others do no exist while a similar orbital in H3 can.

Another way to phrase my question:

How come (when we combine atomic orbitals for BeH2) we disregard

ψ(H1 1s)-ψ(Be 2s)-ψ(H2 1s)

because it violates symmetry of the molecule and has partial bonding and antibonding characteristics, when a similar MO in H3 also has partial bonding and antibonding characteristics?

 

[DrDu]

As Be has also 2p orbitals, the wavefunction antisymmetric in H1 and H2 can be written as
ψ(H1 1s)+aψ(Be 2s)}+bψ(Be 2p_x)-ψ(H2 1s)
Like in H_3, the coefficient a of the 2s orbital of Be will be zero. So there is no difference between the two molecules. In fact, also in H_3 the orbital will have some p2_x admixture on the central hydrogen atom. However, as 2p_x is energetically much higher than 1s, this admixture will be small.

 

[JohnRC]

As Be has also 2p orbitals, the wavefunction antisymmetric in H1 and H2 can be written as
ψ(H1 1s)+aψ(Be 2s)}+bψ(Be 2p_x)-ψ(H2 1s)
Like in H_3, the coefficient a of the 2s orbital of Be will be zero. So there is no difference between the two molecules. In fact, also in H_3 the orbital will have some p2_x admixture on the central hydrogen atom. However, as 2p_x is energetically much higher than 1s, this admixture will be small.

A possible point of confusion here -- DrDu has used a different axis system where he has taken the axis of cylindrical symmetry -- the axis along the line of the three atoms in the molecule as the x-axis. I have taken that direction as the z-axis. His 2px is my 2pz (the usual convention is to take the axis of maximum symmetry as the z-axis).

from 1bird9stones:

Another way to phrase my question:

How come (when we combine atomic orbitals for BeH2) we disregard

ψ(H1 1s)-ψ(Be 2s)-ψ(H2 1s)

because it violates symmetry of the molecule and has partial bonding and antibonding characteristics, when a similar MO in H3 also has partial bonding and antibonding characteristics?

Another way to phrase my answer:

Unlike the case for BeH₂, there is no valence orbital on H2 in H₃ that is antisymmetric end-to-end, so there is no other orbital to interact with ψ(H1 1s)-ψ(H3 1s)

The antisymmetric orbital in H₃ involves zero contribution from ψ(H2 1s), just as the antisymmetric orbitals in BeH₂ involve zero contribution from ψ(Be 2s).

 

[DrDu]

Yes you are right, I have assumed the axis that passes through the three nuclei to be the x-axis.