Q30 / 150 · MHT CET 2026 (20 April Shift 2) · Chemistry
What is the oxidation state of Xe in $\text{XeOF}_4$?
- A$+4$
- B$-4$
- C$-6$
- Correct answerD$+6$
Solution
Let the oxidation state of Xe in $\text{XeOF}_4$ be $x$.
The oxidation state of oxygen (O) is $-2$ and that of fluorine (F) is $-1$.
Since the molecule is neutral, the sum of the oxidation states of all atoms must be zero.
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$
The oxidation state of Xe in $\text{XeOF}_4$ is $+6$.
Answer: $+6$
The oxidation state of oxygen (O) is $-2$ and that of fluorine (F) is $-1$.
Since the molecule is neutral, the sum of the oxidation states of all atoms must be zero.
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$
The oxidation state of Xe in $\text{XeOF}_4$ is $+6$.
Answer: $+6$
Hence, the correct answer is option: 'D'.