The oxidation state of xenon is $+ 6$ in:

X e O F_{2}

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X e F_{6}

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X e F_{4}

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X e O_{4}

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Solution

The correct option is C $X e F_{6}$
$X e F_{6}$
Let oxidation state be $x$
$x - 6 \times 1 = 0$
$∴ x = 6$
$X e O F_{2}$
$x - 2 - 1 \times 2 = 0$
$x = 4$
$X e F_{4}$
$x - 4 = 0$
$x = 4$
$X e O_{4}$
$x - 8 = 0$
$x = 8$

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