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There is a rule for diatomic homo-nuclear species that says that if $Z<8$ then $$E(\sigma_{2p})>[E(\pi_{2p_y})=E(\pi_{2p_z})]$$ and the inverse if $Z\geq8$ and in both case we have energy of $$E(\sigma_{2s}^{*})>[E(\pi_{2p_y}^{*})=E(\pi_{2p_z}^{*})].$$

This is understandable but what if we have diatomic heteronuclear species like $\ce{NO}$ or $\ce{CO}$?

How can we determine which of the molecular orbitals $\sigma$ or $\pi$ is higher in energy? I.e. how should we know which to place below which in the molecular orbital diagram (sigma or pi orbitals)?

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Actually, the gist idea is the difference of energy between the orbitals $E_{2s(\mathrm{A})}-E_{2p(\mathrm{B})}$ and $E_{2s(\mathrm{B})}-E_{2p(\mathrm{A})}$. If the difference is less than 10 eV in absolute value, there are interactions between the orbitals $2s$ and $2p$ and the diagram for diatomic homo-nuclear species with $Z<8$ is valid. (with a slight difference is that the levels of energy of the atomic orbitals of the more electronegative atom are more negative than the those of the other one). This is the case for the molecules you propose, i.e. $\ce{CO}$ and $\ce{NO}$.

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  • $\begingroup$ I am taking a general chemistry course and in any given question i am not given the values of the energies you mentioned. Is there another method for solving this problem? $\endgroup$ – YAZO Jan 15 '16 at 23:37
  • $\begingroup$ To the best of my knowledge, it is the only criterion. But, in general for the molecules of type AB, the molecular diagram of $\ce{N2}$ is valid (Z<8). Otherwise, the teacher has to report an indication or a hint in his question to use the other diagram (Z>8). $\endgroup$ – Yomen Atassi Jan 16 '16 at 7:27

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