The reaction of $X e F_{6}$ with silica $(S i O_{2})$ gives:

X e O_{3}

and

S i F_{4}

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X e F_{4}

and

S i F_{4}

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X e O F_{2}

and

S i F_{4}

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X e O F_{4}

and

S i F_{4}

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Solution

The correct option is C $X e O F_{4}$ and $S i F_{4}$
The product of the given reaction is as follows:

$2 X e F_{6} + S i O_{2} \to 2 X e O F_{4} + S i F_{4}$

Hence, option D is correct.

Suggest Corrections

Similar questions

S i O_{2} + 4 H F ⟶ S i F_{4} + 2 H_{2} O; Δ H = - 10.17 k c a l

S i O_{2} + 4 H C l ⟶ S i {C l}_{4} + 2 H_{2} O; Δ H = + 36.7 k c a l H C l

2 X e O F_{4} + S i O_{2} \to 2 X e O_{2} F_{2} + S i F_{4}

S i F_{4} + H_{2} O \to A 1000^{o} C - --- \to B N a_{2} C O_{3} - ---- \to C

S i O_{2} + 4 H F \to S i F_{4} + 2 H_{2} O, Δ H = - 10.17 K c a l

S i O_{2} + 4 H C l \to S i C l_{4} + 2 H_{2} O, Δ H = 36.7 K c a l

S i O_{2} + 4 H F \to S i F_{4} + 2 H_{2} O, Δ H = - 10.17 k c a l

S i O_{2} + 4 H C l \to S i C l_{4} + 2 H_{2} O, Δ H = 36.7 k c a l

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