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I am slightly confused about how non-bonding MOs form, for example in H3+ ion as shown below.

Molecular orbital diagram for triatomic hydrogen

Why does the central atom not contribute an orbital?

The source mentions symmetry, but I can only see that the central orbital has a positive or negative phase, so it will inevitably cancel somewhat on one side and add with the wavefunction on the other side. This orbital will have a distinct appearance.

Could someone explain why there is no contribution, despite us adding s orbitals whose symmetry always allows them to be added (and even if we had p orbitals, surely the p orbital which can contribute will contribute?)

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  • $\begingroup$ Could the down-voter please let me know what is wrong with my question and how I could edit it? $\endgroup$ – 21joanna12 May 19 '17 at 21:39
  • $\begingroup$ H3 cation is triangular not linear, also @Not. $\endgroup$ – Mithoron May 20 '17 at 11:35
  • $\begingroup$ I know Mithoron — its quite a common undergrad exercise to compare linear and triangular however (cf Flemings MO book ) $\endgroup$ – NotEvans. May 20 '17 at 11:52
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In order to draw the MO diagram for the H3+ system, it helps to first thing about the MO diagram for a simple diatomic case, and then mix in the additional s-orbital.

The diagram below hopefully illustrates this technique (and can be found in many introductory textbooks).


First consider two s-orbitals (the right hand side of the diagram). They can either mix in a symmetric (in phase) or anti-symmetric (out of phase) combination.

The two atomic orbitals we initially combine can give rise to two molecular orbitals, as shown on the right hand side.

We can give them symmetry labels to help us identify what kinds of things these molecular orbitals can then combine with to form the new molecular orbitals for the triatomic system.

The labels g and u are related to whether the molecular orbital is symmetric or anti-symmetric with respect to an inversion centre.

Once our two molecular orbitals for the diatomic system are formed, we can mix these orbitals with another hydrogen s-orbital in order to generate the final MO diagram as per your question.

Only the g molecular orbital from the diatomic system is of the correct symmetry to interact. This mixing creates two molecular orbitals (the top and bottom ones) for the triatomic system. The middle, non-bonding molecular orbital is simply the orbital of u symmetry from the initial diatomic system, which is unable to interact since there is nothing of the correct symmetry for it to do so.

Alternative MO diagram for H3

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