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Suppose we have a hydrogen molecule $\ce{H2}.$ At room temperature the average distance of the nuclei is $r_0$ (without applying pressure).

If we put $n$ hydrogen molecules in a box with volume $V_\mathrm b < nr_0^3$ and the box isn't compressible, then the hydrogen atoms will be under pressure and will be forced to come closer than $r_0.$

Does that change the energy of the molecular orbitals, and if so, how?

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    $\begingroup$ Yes, by complicated way, due complicated potential field. $\endgroup$
    – Poutnik
    Jun 21 at 12:43
  • $\begingroup$ Think about the other extreme - if enough pressure is applied you will get solid H2, and so the individual orbital energies of the isolated molecules have morphed into the bands of the solid. In between you will get something in between, so as pressure is increased the orbital energies will start to spread under inter-molecular interactions into wider and wider bands as the pressure increases $\endgroup$
    – Ian Bush
    Jun 21 at 16:27
  • $\begingroup$ You would have a particle in a box with infinitely high walls inside which is the potential corresponding to the H2 molecule (H$_2^+$ would be simpler) the walls come into the calculation as boundary conditions on the integrals. There are many text-book examples of PIB's with potentials inside. $\endgroup$
    – porphyrin
    Jun 22 at 8:52

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