 Solution

The halting problem is an unconditional example:Reduce SAT to it as follows. Given a Boolean formula $\phi$, construct a Turing machine $M_{\phi }$ that ignores its input, enumerates all Boolean valuations of $\phi$, and halts if it finds a satisfying one. If none satisfies $\phi$, it enters an infinite loop after the enumeration. ThenThe construction of the machine description takes polynomial time; the potentially exponential running time of the constructed machine is irrelevant to the time needed to compute the polynomial-time reduction. Since SAT is NP-complete, this proves NP-hardness of the halting problem.

Every problem in NP is decidable, by enumerating the finitely many polynomial-length certificates and running the verifier. The halting problem is undecidable, so it is outside NP and therefore cannot be NP-complete.