PSPACE-complete problems can't be in NL using the space hierarchy theorem?
This question is currently open, since a positive answer (i.e. no complete problems for PSPACE can lie in NL) would imply $\mathsf{P}\neq\mathsf{PSPACE}$. This statement is actually equivalent to $\mathsf{P}\neq\mathsf{PSPACE}$ (as a PSPACE complete problem in NL obviously implies P=PSPACE).

Examine the equivalent statement, that $\mathsf{P}=\mathsf{PSPACE}$ implies there exists a PSPACE complete problem in NL. To see why this holds, note that if $\mathsf{P}=\mathsf{PSPACE}$ then every non trivial language $L\in\mathsf{PSPACE}$ (i.e. $L\neq \emptyset,\Sigma^*$) is complete for PSPACE. Now let $L$ be some non trivial language in NL (which is a subset of PSPACE), then $L$ is a PSPACE complete problem which lies in NL.