At 143 K,the reaction of XeF_4 with O_2F_2 p | vedclass

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(A) The chemical reaction is: $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$.The compound $Y$ is $XeF_6$.In $XeF_6$,there are $6$ fluorine atoms. Each fluor

Vedclass · Accepted Answer

$19$

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(A) The chemical reaction is: $XeF_4 + O_2F_2 ightarrow XeF_6 + O_2$.The compound $Y$ is $XeF_6$.In $XeF_6$,there are $6$ fluorine atoms. Each fluorine atom has $3$ lone pairs of electrons,contributing $6 	imes 3 = 18$ lone pairs.The xenon atom in $XeF_6$ has $1$ lone pair of electrons.Therefore,the total number of lone pairs of electrons in the $XeF_6$ molecule is $18 + 1 = 19$.

At $143 \ K$,the reaction of $XeF_4$ with $O_2F_2$ produces a xenon compound $Y$. The total number of lone pair$(s)$ of electrons present on the whole molecule of $Y$ is . . . . .

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