Which of the following statements are true?

Only type of interactions between particles of noble gases are due to weak dispersion forces.

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Ionization enthalpy of molecular oxygen is very close to that of xenon.

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Xenon fluorides are not reactive.

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Hydrolysis of

X e F_{6}

is a redox reaction.

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Solution

The correct option is B Ionization enthalpy of molecular oxygen is very close to that of xenon.
Analyzing the Options

Option (A):

All noble gases have weak intermolecular forces i.e., dispersion or Vander Waals forces.

As noble gases exist as monoatomic gases, they can have only weak dispersion forces between them.

Option (B):

The Ionization energy of $O_{2} i s 1170 k J / m o l$ and ionization energy of $X e i s 1170 k J / m o l$ as observed by Neil Bartlett and they are nearly the same.

This particular property was actually used to form the first compound of Xe which was $X e^{+} P t F_{6}^{-}$

Option (C):

Hydrolsis reaction of $X e F_{6}$ is $X e F_{6} + 3 H_{2} O \to X e O_{3} + 6 H F$
The Oxidation state of $X e i n X e F_{6} and X e O_{3}$ is same i.e., +6.

Since the oxidation number of Xe remains same, the given reaction is not a redox reaction.

Option (D):

Xe fluorides are $X e F_{2}, X e F_{4} a n d X e F_{6} .$
All are powerful fluorinating agents.
They can also easily get hydrolysed even by traces of water.

$2 X e F_{2} + 2 H_{2} O \to 2 X e + 4 H F + O_{2}$
Therefore, Xenon fluorides are highly reactive in nature.

Hence, options (A) & (B) are the correct options.

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