A xenon compound 'A' upon partial hydrolysis | vedclass

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(C) The partial hydrolysis of $XeF_{6}$ is given by the reaction: $XeF_{6} + 2H_{2}O \longrightarrow XeO_{2}F_{2} + 4HF$ Thus,compound $A$ is $XeF_{6}

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(C) The partial hydrolysis of $XeF_{6}$ is given by the reaction: $XeF_{6} + 2H_{2}O \longrightarrow XeO_{2}F_{2} + 4HF$ Thus,compound $A$ is $XeF_{6}$. The structure of $XeF_{6}$ is distorted octahedral with $1$ lone pair on the $Xe$ atom. Each fluorine $(F)$ atom has $3$ lone pairs of electrons. Total number of lone pairs in $XeF_{6} = 1 	ext{ (on } Xe) + (6 	imes 3 	ext{ on } F) = 1 + 18 = 19$.

$A$ xenon compound '$A$' upon partial hydrolysis gives $XeO_{2}F_{2}$. The number of lone pairs of electrons present in compound $A$ is ....... . (Round off to the nearest integer)

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