# Why does Boron only need 6 valence electrons unlike the standard 8?

So I was a bit confused. Hydrogen and Boron seem to be the only outliers to the octet rule. Hydrogen makes sense because it has only one shell and 2 electrons complete its shell. Boron's configuration, however, is $1s^2 2s^2 2p^1$. Having 6 valence electrons would mean having a configuration of $1s^2 2s^2 2p^4$ which doesn't seem as stable as $1s^2 2s^2 2p^6$ or even as stable as $1s^2 2s^2 2p^3$ since one would think that there is repulsion going on between electrons in the first orbital of p in $1s^2 2s^2 2p^4$. Thanks in advance!

• Boron wants the standard octet as badly as anyone else, and indeed reaches it, albeit sometimes by unusual means. – Ivan Neretin Apr 29 '16 at 20:06
• Also, lithium dimerizes in the gas phase. Just like hydrogen, it fills its s shell, and adds no more. Another outlier. – Lighthart Apr 29 '16 at 20:16
• By this, I would assume that helium is also an outlier. – SendersReagent Apr 29 '16 at 20:33
• – Mithoron Apr 29 '16 at 21:20
• – Mithoron Apr 29 '16 at 21:21