Delocalized bonding!

[parwana]

Consider the delocalized P orbitals of butadiene (above). Which of the following statements are true ?
C4H6, it has 2 double bonds



a) Butadiene has 2 P orbitals

b) Butadiene has 8 P electrons

c) The highest energy P orbital in butadiene has no nodes besides the plane of the molecule

d) Butadiene has two filled P orbitals

e) The highest energy filled P orbital in butadiene has one node, besides the plane of the molecule

f) The highest energy filled P orbital in butadiene is antibonding between carbons 2 and 3




which of those are right??

[parwana]

please help, i know it will have 4 electrons, and 4 pi orbitals(2 bonding and 2 antibonding), but what about the rest?? Also i know the bonding orbitals will be filled, while the antibonding will be empty.

[chem_tr]

First off, I would speculate about presence of only two P orbitals, as there are four carbons with 2P orbitals each, so a, b, and d don't seem to be correct. It is possible though, b may be correct nevertheless, since the CH carbons may contribute to delocalization, I am not sure about this.

Maybe molecular orbital diagrams may help you about deciding which orbitals are filled.

[parwana]

so is it c, e, f, cause iam sure a and b are not correct, but isnt d correct, since there are 2 pi bonds, and 4 electrons total, and two pi bonding orbitals will be filled with 2 electrons each and the antibonding ones will be left out.

[chem_tr]

According to your reasoning, d seems to be correct. You may also want to view "P orbitals" thread, in Atoms, Molecules and Solids division.

[parwana]

i dont think a, b, and c are correct, please help me about e and f as iam confused on them, please tell me what u think would be correct. Also look at my attachment of the lewis structure of butadeine

[movies]

Do you mean p orbitals or pi orbitals. P orbitals are atomic orbitals, pi orbitals are molecular orbitals. It makes a big difference!

[parwana]

so which of the answers are right then??

[chem_tr]

As Movies said, there are not P orbitals in butadiene anymore, let's treat them as if it reads "pi orbitals", since the other doesn't make sense.

About nodal planes, frankly, I have not much ideas. But I know that two P orbitals with the same symmetry combines with each other to produce a pi orbital. This molecular orbital has a geometry in which the bonds are inside and outside of the plane.

The final answer lies on finding the molecular orbital diagram or at least the nodal representation of the molecule. Sorry I haven't been of much help.

[parwana]

so u dont know which of those could be correct?

[movies]

If it did say pi orbitals then I think all of the last three would be correct, unless my recollection is wrong.

[parwana]

u are right it is d, e, and f, but why e???

[chem_tr]

There are pi molecular orbitals of butadiene, and they are parallel to the plane, by combination of two p atomic orbitals. This is probably what is intended by saying "besides the plane of the molecule"

[movies]

So there are 4 pi orbitals, right? They are made up of a p orbital (atomic orbital) on each carbon. Each p orbital has a node in the plane of the butadiene, right? Other than that the lowest energy molecular orbital will have no additional nodes, the second lowest will have 1 node, the next highest 2 nodes, and the highest will have 3 nodes. The nodes arise when the p orbitals are out of phase with the orbital next to it. So if you think of the phase of the orbitals as X or O then you would have these four orbitals:

XOXO (highest E)
XOOX
XXOO
XXXX (lowest E)

There are only 4 electrons to distribute and each orbital can "hold" two electrons. So the highest occupied molecular orbital (HOMO) is the one with one node (in addition to the one in the plane of the butadiene).

[chem_tr]

This is a great explanation, thank you Movies.