Which compound is prepared by the following reaction?
Xe + $F_{2}$ (2:1 volume ratio) $N i \to$ at 673 K

X e F_{4}

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X e F_{2}

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X e F_{6}

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None of these

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Solution

The correct option is B $X e F_{2}$
$X e + F_{2} 673 K ⟶ N i X e F_{2} (1 : 3$ $R a t i o)$
$X e + 2 F_{2} 873 K ⟶ N i X e F_{4} (1 : 5$ $R a t i o)$
$X e + 3 F_{2} 673 K ⟶ N i X e F_{6} (1 : 20$ $R a t i o)$

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