Showing a problem on a specific graph is NP-hard
There's no such thing as a hard instance. You can always figure out the answer to that one instance and then hard-code that into an algorithm. For example, you can always have an algorithm for SAT that says

if input="(XvY)^(!Xv!YvZ)"
    return SATISFIABLE
else
    [loop through all possible variable assignments]


Incidentally, this is why nobody looks at "best-case complexity", since any algorithm can be adapted to run in best-case constant time in this way.  (Constant because you only have to read a fixed amount of the input to see if it's one of the ones that you have a hard-coded answer for.)

There's a category error in asking if an instance can be NP-complete. Recall the definition of what NP-completeness is:


  A language $L$ is NP-complete if is in NP and every other language in NP is polynomial-time many&ndash;one reducible to&nbsp;$L$.


First, that statement doesn't "type-check" if you try to make $L$ be a string rather than a language.  Second, the definition of many&ndash;one reductions requires that you have at least one "yes" instance and at least one "no" instance to use as targets for the reduction. One instance isn't enough to make these definitions work.