# Question #28bb4

So, ammonia ($N {H}_{3}$) has a total of $8$ valence electrons, $5$ from $N$ and $1$ from each $H$ atom.
Each $H$ atom bonds with the $N$ in a single bond - 2 electrons shared. The 3 single bonds between $N$ and the $H$ atoms make up for $6$ of the valence electrons, the remaining $2$ being set as a lone pair around the $N$ atom.
A quick way to determine the type of bonds formed is to take into account the fact that $H$ can only form single bonds, since it only has one orbital filled with one electron; therefore, it only requires one electron more to form a stable doublet configuration.