Q20 / 150 · MHT CET 2025 (22 Apr Shift 2) · Chemistry

A compound of Xe and F is found to have atomic ratio $\mathrm{Xe}: \mathrm{F}$ as $0 \cdot 4: 2 \cdot 4$, Find the oxidation number of Xe ?

  • A
    $-4$
  • B
    Zero
  • C
    $+4$
  • Correct answer
    D
    $+6$

Solution

The atomic ratio $\mathrm{Xe:F} = 0.4:2.4$ simplifies to $1:6$ by dividing both values by $0.4$, giving the formula $\mathrm{XeF}_6$.

Since fluorine has oxidation state $-1$ and the compound is neutral, the oxidation number $x$ of xenon satisfies:

$x + 6 \times (-1) = 0$

$x = +6$

The oxidation number of Xe is +6, corresponding to option D.

Hence, the correct answer is option: 'D'.

Relevant Questions