Question:

What is oxidation state of xenon in xenonmonooxytetrafluoride?

Show Hint

Xenon usually exhibits even oxidation states: +2, +4, +6, +8.
Updated On: Jun 19, 2026
  • +2
  • +4
  • +6
  • +8
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Step 1: Formula
Xenonmonooxytetrafluoride is $XeOF_{4}$.

Step 2: Analysis

Let oxidation state of $Xe$ be $x$. - $O = -2$ - $F = -1$ - Sum = 0

Step 3: Calculation

- $x + (-2) + 4(-1) = 0$ - $x - 2 - 4 = 0$ - $x = +6$

Step 4: Conclusion

Hence, the oxidation state is +6. Final Answer: (C)
Was this answer helpful?
0
0