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I read something recently that said "BN has been observed in the gas phase, is paramagnetic, and has a vibrational frequency lower than N2. These properties can be explained by the molecular orbital diagram of BN".

However, when I drew such a diagram, I failed to see how it described any of these properties, particularly paramagnetism, because there were no lone pairs of electrons. Any ideas?

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This is the general MO diagram you need to fill with the valence electrons of BN

enter image description here

Boron has 3 valence electrons, and nitrogen has 5 valence electrons, this makes 8 electrons.

You have to start filling the orbitals from those with lowest energy to those with higher energy.

So, 2 electrons on $\sigma_\mathrm{2s}$, two electrons on $\sigma_\mathrm{2s}^*$, two electrons on $\sigma_\mathrm{2p}$. You have now 2 electrons left, and two orbitals of the same energy.

In this case, you need to follow Hund's rule, which states that if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs. So you will have to put one electron on the $\pi_x$ orbital and another one on the $\pi_y$ orbital. So you end up with 2 umpaired electrons, and paramagnetism of the molecule is explained.

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    $\begingroup$ Does interference lead to sigma/pi 2p order switch when some atoms are n <= 7? $\endgroup$
    – Hanry Hu
    Commented Oct 13, 2017 at 1:24

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