What is the criterion for filling up of electrons in molecular orbitals of increasing energy in simple diatomic species?

So far I have found different ways in different books. For e.g. One says

"In MO electronic configuration of diatomic species up to 14 electrons, $\sigma2p_z$ orbital (considering $z$ as the internuclear axis) is greater in energy than $\pi2p_y$ and $\pi2p_x$ orbitals. For species with more than 14 electrons, $\sigma2p_z$ orbital is lower in energy than $\pi2p_y$ and $\pi2p_x$ orbitals."

Whereas the other book used 16 as the no. of electrons instead of 14. This leads to different ways of filling electrons in molecules with 15 electrons such as $O_2^{+}, NO$ etc. I want to know which one is correct.

What is known for sure (and is mentioned in all the books I've seen so far) is that for homonuclear diatomics the "$\pi \rightarrow \sigma$ switch" happens when going from $\ce{N2}$ to $\ce{O2}$: for 14-electron case $\pi_{2p_{x,y}}$ orbitals are lower in energy then $\sigma_{2p_{z}}$, for 16-electon one it is other way around. The usual "explanation" is that moving from $\ce{Li2}$ to $\ce{F2}$ the successive decrease of the $s-p$ interaction between the $2s$ orbital of one atom and the $2p_z$ orbital of another atom decreases the energy of the $\sigma_{2p_{z}}$. But you have to understand that this is merely a manifestation of a known fact: we know how orbitals of these molecules are arranged from the molecular orbital theory calculations and use the notion of $s-p$ interaction to rationalize this knowledge.
The problem is that no information about 15-electron case can be inferred from the above mentioned knowledge about molecular orbitals of homonuclear diatomics from $\ce{Li2}$ to $\ce{F2}$. A priori we could not tell will $\sigma_{2p_{z}}$ orbital be lower in energy than $\pi_{2p_{x,y}}$ orbitals or higher then them for a 15-electron case. Besides, things coul be different for different 15-electron cases, such as $\ce{N2-}$, $\ce{O2+}$, $\ce{NO}$, etc.: for some $\sigma_{2p_{z}}$ orbital might be lower than $\pi_{2p_{x,y}}$ orbitals, for others - higher. There is no simple rule to use, one have to resort to molecular orbital theory calculations to tell the truth.
• See also the related question on MO diagram for $\ce{N2-}$. The answer here is basically the same. – Wildcat Mar 14 '16 at 19:15
• @Phill2, it depends on what the nuclear axis is. If it assumed to be $z$, then surely $\mathrm{p}_{z}$ and $\mathrm{s}$ orbitals can overlap to form a $\sigma$ bond. – Wildcat Mar 16 '16 at 18:19