Is the class NP closed under complement?
First of all, the question you are asking is open, since an affirmative answer shows that $\sf NP = coNP$. In fact it is one of the most prominent open problems in computer science. 

If $\sf P= NP$, then the class $\sf NP$ is closed under complement since $\sf P$ is. If on the other hand $\sf P \not = NP$ then we cannot say whether $\sf NP = coNP$ or not. Notice that $\sf NP = coNP$ implies that the polynomial hierarchy collapses to the first level. However this would not imply that $\sf P= NP$.