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Why is B2 paramagnetic? It's just 2 borons with a triple bond between them. Aren't all the electrons paired?

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Look at the MO diagram for homonuclear diatomics. It shows that there are two unpaired electrons. –  brinnb Jul 27 at 21:24

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up vote 4 down vote accepted

For the same reason $\ce{O2}$ is: degeneracy. The filled orbitals are $\sigma$(1s)$^2$, $\sigma^*$(1s)$^2$, $\sigma$(2s)$^2$, $\sigma^*$(2s)$^2$. The half-filled orbitals are $\pi$(2p$_x$) and $\pi$(2p$_y$).

But you're right to think that this is unexpected: why are the $\pi$(2p) orbitals lower than the $\sigma$(2p)? That's the reverse of what we are told to expect! I've never encountered a rationalization that satisfied me.

See here for the expected and observed ordering:

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your link doesn't work. –  Jakob Weisblat Jul 27 at 21:25
    
fixed - thanks. –  Silvio Levy Jul 27 at 21:27

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