Shoot. Thanks Len. Fixed!

Well, you’d have 7 p orbitals. The lowest energy MO (pi-1) would have all of the lobes facing the same way (zero nodes). The highest energy MO (pi 7) would have all p-orbital lobes alternating (6 nodes).
It will resemble allyl and pentadienyl in that the second-lowest orbital (pi 2) will have a node right on the central carbon. In the pi 4 and pi 6 orbitals there will also be a node on the central carbon, in addition to additional nodes.

The rest of the pi orbitals you can build up based on the principle that the number of nodes increases by 1 for each energy level, and also that the nodes should be placed symmetrically about the pi system.

Once you build up the seven pi orbitals of the heptatrienyl system, then you just need to fill it with orbitals. The cation will have 6 pi electrons, the radical 7 pi electrons, and the anion 8 pi electrons.

No it will not be the same for 1,2-butadiene. If you look at the orbitals in 1,2-butadiene they are at right angles to each other, and are not conjugated. They therefore behave much like two separate pi bonds, not a conjugated system. The analysis in this post does not apply to 1,2-butadiene.
See: https://www.masterorganicchemistry.com/2015/01/13/chiral-allenes-and-chiral-axes/

Great question. MO calculations, which I’m skipping over here, tell us that any odd-numbered MO in a system of odd-numbered orbitals (e.g. pi3 in the pentadienyl system) will have nodes coinciding with the positions of at least one of the carbons.

Hi Fayeshun – it does tend to get treated rather quickly, I thought it might be helpful to develop this concept very slowly. If you found it helpful, great!

Thanks Shuaib – glad you found this article helpful.

Hi Ravi, I’m glad you found this post helpful. Thanks for using the site.