What is oxidation state of xenon in xenonmonooxytetrafluoride?

The oxidation state of xenon in xenonmonooxytetrafluoride ($XeOF_4$) is determined by considering the known oxidation states of oxygen and fluorine. Oxygen typically has an oxidation state of $-2$, and fluorine is always $-1$ in compounds. Since $XeOF_4$ is neutral, the sum of the oxidation states is zero.

Let $x$ be the oxidation state of xenon. The equation becomes:

$x + (-2) + 4(-1) = 0$

Simplifying: $x - 2 - 4 = 0$

Thus, $x - 6 = 0$ and $x = +6$.

The oxidation state of xenon is +6, corresponding to answer choice C.