"cspringer" is exactly correct. However, I'm going to present an alternative way for you to get some easy MCAT marks.
You proposed H2-N-[triple bond]-N-H2 which means that each nitrogen has 5 bonds (the triple + the 2 H's).
Setting aside d-orbitals (knowledge), whenever you are presented with something new on the MCAT, consider what you know with great confidence. Nitrogen in NH4 is maximally bonded (4) and it has a positive charge (you never saw NH5 and I feel strange typing it!). Note that the lone pair on N is given to H to form that 4th bond (happens to be called a coordinate covalent bond). Nitrogen in NH3 is neutral (conclusion: N with 3 bonds is neutral). Nitrogen in NH2 is negatively charged.
Yes, you have seen N2 where N[triple bond]N which means each has a lone pair and each is neutral (notice each N is in the middle of an octet; notice it follows the same basic rules of ammonia, NH3).
So when you are presented with neutral N2H4, your first thought: like molecules I know, N2 or NH3, I'm expecting that the neutral form with have 3 bonds to each nitrogen and one bond to each hydrogen.
The other situation that comes up often? 2 bonds to oxygen: neutral (H2O), 3 bonds (H3O): positive, 1 bond: negative (OH).
Of course the exam will choose molecular arrangements that initially feel awkward, you can use VSEPR, or you can get easier marks by just going back to the fundamentals.
