Question

What is oxidation state of xenon in xenonmonooxytetrafluoride?

• A. +2
• B. +4
• C. +6
• D. +8

Hint:

Xenon usually exhibits even oxidation states: +2, +4, +6, +8.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to calculate the oxidation state of Xenon in the compound \( \text{XeOF}_4 \).

Step 2: Key Formula or Approach:
The sum of oxidation states in a neutral molecule is zero.
Common states: \( \text{O} = -2, \text{ F} = -1 \).

Step 3: Detailed Explanation:
Let the oxidation state of Xenon be \( \text{x} \).
The formula is \( \text{XeOF}_4 \).
\[ \text{x} + [1 \times (-2)] + [4 \times (-1)] = 0 \]
\[ \text{x} - 2 - 4 = 0 \]
\[ \text{x} - 6 = 0 \]
\[ \text{x} = +6 \]

Step 4: Final Answer:
The oxidation state of xenon in \( \text{XeOF}_4 \) is +6.