A quick google search does reveal the molecular orbitals of this cyclic H3.

However, one source mentions there are are two nodes; while two of my professors said there's one node. Which is correct?

This seems to be a related question - I am not sure. There also is a question on linear H3+ - not cyclic.

  • 1
    $\begingroup$ There are three orbitals in the MO scheme of $\ce{H3}$. Which one are you interested in? $\endgroup$ – Jan Nov 14 '17 at 7:35
  • $\begingroup$ Particularly the two degenerate ones - the one with the lowest energy has zero nodes. $\endgroup$ – digikar Nov 14 '17 at 7:44

In linear $\ce{H_3}$ you have one MO with zero nodes, one with one node and one with two nodes. When you bend the molecule around to make a cycle, the two nodes in that last orbital merge to make a second orbital with one node. The two orbitals of cyclic $\ce{H_3}$ with one node are, of course, degenerate when the ring has threefold rotational symmetry.

  • $\begingroup$ And to see that we would need a detailed treatment; or is there any other way to "see" it? What I can see is: for the existence of two nodes, we would have required the two nodes to cross the opposite atoms, right? $\endgroup$ – digikar Nov 14 '17 at 11:56
  • $\begingroup$ A simple "rough sketch": start with two parallel planes between adjacent pairs of atoms in the linear molecule (these being the nodes of the highest orbital in the linear molecule). Bend the two bonds to make the outer two atoms come together, making the nodes follow along so they continue to pass through the molecule, and see what happens to them. You can't avoid the merger! $\endgroup$ – Oscar Lanzi Nov 14 '17 at 21:14
  • $\begingroup$ Are the two orbitals with one node really degenerate in H3 or only in H3+? $\endgroup$ – pH13 - Yet another Philipp Nov 16 '17 at 12:46
  • $\begingroup$ @pH13 the answer is edited to "neutralize" your q. $\endgroup$ – Oscar Lanzi Nov 16 '17 at 13:44

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