Xenon fluorides are colourless crystalline solids and are readily hydrolysed even by traces of water. From the following given products, select the one which is not common to the hydrolysis of $X e F_{2} and X e F_{4}$ .

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X e O_{3}

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O_{2}

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X e

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Solution

The correct option is B $X e O_{3}$
$2 X e F_{2} + 2 H_{2} O + 2 X e + 4 H F + O_{2}$

$6 X e F_{4} = 12 H_{2} O \to 4 X e + 2 X e O_{3} + 24 H F + 3 O_{2}$

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Similar questions

Xenon forms three binary fluorides $X e F_{2}, X e F_{4}$ and $X e F_{6}$ . All the three fluorides are colourless crystalline solids sublimating at 295K. They are powerful fluorinating agents and they are readily hydrolysed by water.
$X e O_{3}$ is a colorless explosive solid acid, prepared by

Q. Which of the following hydrolysis product(s) are common in hydrolysis of

X e F_{2} and X e F_{4}

respectively?

Q. Which of the following hydrolysis product is not common in hydrolysis of

X e F_{2} and X e F_{4}

respectively?

How are xenon fluorides $X e F_{2}, X e F_{4}$ and $X e F_{6}$ obtained?

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Xenon fluorides are colourless crystalline solids and are readily hydrolysed even by traces of water. From the following given products, select the one which is not common to the hydrolysis of XeF2 and XeF4.

The correct option is B XeO32XeF2+2H2O+2Xe+4HF+O2 6XeF4=12H2O→4Xe+2XeO3+24HF+3O2

Xenon fluorides are colourless crystalline solids and are readily hydrolysed even by traces of water. From the following given products, select the one which is not common to the hydrolysis of $X e F_{2} and X e F_{4}$ .

The correct option is B $X e O_{3}$
$2 X e F_{2} + 2 H_{2} O + 2 X e + 4 H F + O_{2}$

$6 X e F_{4} = 12 H_{2} O \to 4 X e + 2 X e O_{3} + 24 H F + 3 O_{2}$