Particle in a box with different barriers between two molecules I'm trying to understand different ways to solve a PIB model.

I'm given a molecule with a tetrahedral geometry and there are two related ground states - electron pair pointing up or down. I'm being asked to draw the n=1 and n=2 states in three separate cases of a finite potential barrier in the middle of the box -> VBarrier = 0, VBarrier = V0, and VBarrier = ∞. So that left of the barrier are the wavefunctions of the up orientation and right is the down orientation.

I know that for the case of VBarrier = ∞, it'll just be the same sin(n pi x/L) wavefunctions on both sides of the barrier. I don't understand how they would change for V=0 and V=V0, and why these barriers would have different heights? I'd be grateful for any help.
for the infinite barrier I'd put an electron on just one side of the box with length L/2, meaning that electron is trapped on one side/face of the ammonia.

For the no barrier part treat the electron as a box with length L meaning it can easily go from one side to the other without barrier (for example maybe the ammonia is at the planar transition state). 

For the finite barrier, I'd show tunneling from one side to the second, however, normally for PIB the tunneling region has finite length. 

Assuming you have a tetrahedral geometry, according to the Born-Oppenheimer separation of electron and nuclear motion the finite barrier would extend from L/2 to L (or equivalently from 0 to L/2). Therefore the tunneling would be from the middle to the edge. 

Image source