# How do Lewis acids react?

Dec 31, 2014

According to the Lewis definition, Lewis acids are substances that accept electron pairs - electron acceptors, while Lewis bases are substance that donate electons - electron donors.

A classic example of a Lewis acid-base reaction takes place between ammonia ($N {H}_{3}$) and boron trifluoride ($B {F}_{3}$).

Since boron ($B$) is in group 13, it only has 3 valence electrons, which means it can only form three bonds. $B$ bonds with three $F$ atoms and adds 3 more electrons to its outermost shell, for a total of 6 electrons (3 it owns and 3 come from the covalent bonds it has with $F$).

Notice that it still needs 2 more electrons to complete its octet, i.e. to have 8 electrons in its outermost shell, therefore can accept a pair of electrons. This is where $N {H}_{3}$ comes into play.

$N$ is in group 15, which means it has 5 valence electrons and can form three bonds (this is because out of the five valence electrons, only 3 are unpaired) - it can therefore donate a pair of electrons.

Here's what happens during their reaction. Since it is bonded to three other atoms, $B$ is $s {p}^{2}$ hybridized. This means that one of its three 2p-orbitals ($2 {p}_{z}$) is empty. The lone pair of electrons on $N$ is picked up by $B$'s empty p-orbital and a covalent bond is formed between the two molecules.

$N {H}_{3}$ acted a Lewis base - it donated a pair of electrons, while $B {F}_{3}$ acted as a Lewis acid - it accepted a pair of electrons.