XeF 4 reacts violently with water to giveA- Xe and O 2B- XeOF 4C- XeO 3 and HFD- XeO 3- O 2 and HF

The correct option is D XeO3, O2 and HF6XeF_4 + 12 H_2O → 4Xe + 2XeO_3 + 24HF + 3O_2 ...

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Standard XII

Chemistry

Fluorides of Xenon

XeF 4 reacts ...

Question

$X e F_{4}$ reacts violently with water to give

Xe and O₂

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XeOF₄

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XeO₃ and HF

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XeO₃, O₂ and HF

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Solution

The correct option is D XeO₃, O₂ and HF
$6 X e F_{4} + 12 H_{2} O \to 4 X e + 2 X e O_{3} + 24 H F + 3 O_{2}$

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Similar questions

Q. How many of the following is/are not obtained as one of the hydrolysis products of

X e F_{2}

with water?

X e O_{3}, X e, H F, O_{2}

X e F_{4}

reacts violently with water to give:

The noble gases have closed-shell electronic configuration and are monoatomic gases used normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation number +2, +4 and +6. $X e F_{4}$ reacts violently with water to give $X e O_{3}$ . The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

The structure of $X e O_{3}$ is

Q. Out of the following how many have Xe in

s p^{3} d

hybridization?

X e F_{2}, X e F_{4}, X e F_{6}, X e O F_{2}, X e O_{3}

Q. The oxidation state of

X e

X e O_{3}

and bond angle in it are:

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XeF4 reacts violently with water to give

The correct option is D XeO3, O2 and HF6XeF4 + 12H2O→4Xe + 2XeO3 + 24HF + 3O2

$X e F_{4}$ reacts violently with water to give

The correct option is D XeO₃, O₂ and HF
$6 X e F_{4} + 12 H_{2} O \to 4 X e + 2 X e O_{3} + 24 H F + 3 O_{2}$