Xenon forms different types of compounds. One | vedclass

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(B) Given the ratio $n_1 : n_2 = 1$,the compound is $XeO_2F_2$.In $XeO_2F_2$,the central atom $Xe$ has $8$ valence electrons.It forms $2$ single bonds

Vedclass · Accepted Answer

It is a polar as well as planar compound

Vedclass · Answer

(B) Given the ratio $n_1 : n_2 = 1$,the compound is $XeO_2F_2$.In $XeO_2F_2$,the central atom $Xe$ has $8$ valence electrons.It forms $2$ single bonds with $F$ and $2$ double bonds with $O$,utilizing $6$ electrons for bonding.This leaves $2$ electrons,forming $1$ lone pair.The total number of electron domains is $5$ ($4$ bonding pairs + $1$ lone pair),resulting in a see-saw geometry.$XeO_2F_2$ is a polar and non-planar molecule.Therefore,the statement that it is a planar compound is incorrect.

Xenon forms different types of compounds. One of the compounds is represented as $XeO_{n_1}F_{n_2}$. If the ratio of $n_1$ to $n_2$ is $1$ for this compound,then which of the following options is incorrect?

Similar Questions

$XeF_2$ is hydrolyzed in the presence of a small amount of water. What are the gaseous products formed?

Which of the following statements is not correct for a noble gas?

Which factor is most responsible for the increase in boiling points of noble gases from $He$ to $Xe$?

Assertion $(A)$: The boiling points of noble gases increase from $He$ to $Xe$.
Reason $(R)$: The interatomic van der Waals' attractive forces increase from $He$ to $Xe$.
The correct answer is:

The inert gases are

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