# Question #bc265

Apr 20, 2017

By getting rid of this one electron, the electrons left form a "noble gas" configuration, that is more stable. The alkali will become a positive ion by doing this, e.g. $K \to {K}^{+} + {e}^{-}$
The electron will be gladly accepted by e.g. a halogen, because these lack one electron as compared to the "noble gas" configuration: $B r + {e}^{-} \to B {r}^{-}$