The nitrogen atoms in ${\mathrm{NH}}_3,{\mathrm{NH}}_2^{-}$ and ${\mathrm{NH}}_4^{+}$ are all surrounded by eight electrons. When these three species are arranged in order of increasing $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle, what is the correct order?

More so than lone pair-bond pair or bond pair-bond pair repulsion is lone pair-lone pair repulsion.
${\mathrm{NH}}_2^{-}$

 has two lone pairs and two bond pairs, however the angle $\mathrm{H}-\mathrm{N}-\mathrm{H}$ is less than $190°$ because of the lone pairs.
Three bond pairs and one single pair make up

${\mathrm{NH}}_3$. 

The angle is greater than in the situation above because there are fewer lone pairs.
Four bond pairs made up

 ${\mathrm{NH}}_4^{+}$. 

Angle is $190°$. ${\mathrm{NH}}_2^{-},{\mathrm{NH}}_3,{\mathrm{NH}}_4^{+}$ is the proper increasing order of the $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle.