Why is the more electronegative atom lower in energy in an MO diagram?

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?

• 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. – orthocresol Nov 5 '16 at 0:29
• Nucleus pulls on the electrons harder so why the energy is lower? – Physics3067 Nov 5 '16 at 0:36
• 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. – orthocresol Nov 5 '16 at 0:37
• Most strictly speaking it is the ionisation enthalpy that one can use to determine the energy of the outermost orbital (CC @ortho). – Jan Nov 5 '16 at 19:33
• @orthocresol Maybe sum everything up and make it an answer (and hopefully prevent other non-answers). – Martin - マーチン Dec 13 '18 at 13:50