MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Use known oxidation states of oxygen and fluorine to find the unknown oxidation state of xenon.Explanation:Let the oxidation state of xenon be $x$.In $\mathrm{XeOF}_{4}$, oxygen has an oxidation state of $-2$.Fluorine always has an oxidation state of $-1$, and there are four fluorine atoms.The compound is neutral, so the total sum of oxidation states is $0$.Write the equation: $x + (-2) + 4(-1) = 0$.Simplify: $x - 2 - 4 = 0$.So, $x - 6 = 0$.Therefore, $x = +6$.Thus, the oxidation state of xenon in $\mathrm{XeOF}_{4}$ is $+6$.Answer:The oxidation state of Xe in $\mathrm{XeOF}_{4}$ is $+6$.Correct option: D.