Do that. Then count the electrons around H, O, and N.
I count two for each of the N-H bonds, 2 for the O-H bond, then for N I see 12. And that's too many if you are trying to obey the octet rule. Even if you aren't trying to obey the octet rule, N doesn't have the holes to accommodate 12 electrons. It has a
1s2 2s2 2p3 so it has "holes" for three bonds only (to fill the 2p3 orbitals and make 6. The 2 electrons in the 1s2 orbital are the lone pair around N if you draw the structure correctly.
You must write a Lewis structure that obeys the Octet Rule. You start by counting the number of valence electrons on the two central atoms, N(5), O(6), and the three H's (1 each). Electrons must be shared so that N and O end up with 8 electrons around them (shared and unshared total). Each hydrogen shares 2 electrons with O or N. Hydrogen is unique in that it does not go beyond 2 lectrons through sharing (no octet). We can't draw Lewis structures here but all bonds are single in H2N-O-H. The nitrogen has one unshared pair and the O two unshared pairs.
What I just described is determined by trial and error and it is the only way to obey the Octet Rule. The basic skill you need is writing Lewis structures.
okay i understand so writing the Lewis structures is just the math not the drawing?
No, I don't think so. It's the math, true enough, but it's also the drawing. We want to know that there are single bonds in H2NOH and that there are no double bonds or triple bonds. In other words, You can't place the electrons helter skelter (just anywhere) that looks good.
" ..It's the math, true enough, but it's also the drawing..". I think that is equivalent to my terse Both. The procedure for writing Lewis structures can't be fully demonstrated on a bulletin board posting like this one. Using some trial and error within the proper constraints, and producing a "helter skelter" electron dot pattern are not logically equivalent. My students usually can tell the difference.