Let's take $\ce{NO}$ for example. Why is $\ce{O}$ lower in energy? I asked my teacher and he said that because it is more electronegative and this lowers the bonding energy of the MO. But why?

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    $\begingroup$ More strictly speaking it should be effective nuclear charge and not electronegativity, although there is a strong correlation between these two quantities. Very simply put, a larger effective nuclear charge means that the nucleus "pulls" on the electrons harder and therefore their energies are lower. $\endgroup$ – orthocresol Nov 5 '16 at 0:29
  • $\begingroup$ Nucleus pulls on the electrons harder so why the energy is lower? $\endgroup$ – Physics3067 Nov 5 '16 at 0:36
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    $\begingroup$ Do you study physics? It's the same as electric potential energy: $V = kq_1q_2/r$. In this case we could say $q_1$ is simply proportional to the effective nuclear charge and $q_2$ is the charge of an electron. They're opposite charges so $V$ is negative. Obviously for an atom classical electrostatics won't quite cut it but the physical idea is the same: the energy of an orbital is lower if $Z_\mathrm{eff}$ is larger. $\endgroup$ – orthocresol Nov 5 '16 at 0:37
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    $\begingroup$ Most strictly speaking it is the ionisation enthalpy that one can use to determine the energy of the outermost orbital (CC @ortho). $\endgroup$ – Jan Nov 5 '16 at 19:33
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    $\begingroup$ @orthocresol Maybe sum everything up and make it an answer (and hopefully prevent other non-answers). $\endgroup$ – Martin - マーチン Dec 13 '18 at 13:50

I am summarizing from the excellent comments while adding some glue:

The convention is that lower electron energy means electrons that are more tightly bound. An energy of zero would mean a free electron (not bound in the electrostatic field of a nucleus).

As you go across from left to right in a period of the periodic table, the number of protons in the nucleus increases and the number of electrons bound (in an atom) increases likewise. In the simplest explanation of electron energies, you conceptually separate the electrons into inner electrons and valence (or outer shell) electrons, and treat the inner electrons and the nucleus together as an entity with some charge (the effective nuclear charge) that the outer electrons "see".

Comparing oxygen to nitrogen, the effective nuclear charge increases by one (same number of inner electrons, one more proton in the nucleus), so valence electrons of oxygen are more tightly bound than valence electrons of nitrogen. Thus, they appear lower in an MO diagram.

Not the OP's question, but for completeness: If you go from one period to the next (compare F with Na, say), the effective nuclear charge decreases (because we now also count the n = 2 electrons as inner electrons), and valence electrons of Na are less tightly bound than those of F. Most atomic properties change in a periodic pattern when you plot them against atomic number for the same reason (e.g. ionization energies, atomic radii)

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