XeF_4 reacts with SbF_5 to form [XeF_m]^{n+}[ | vedclass

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Question

(B) The reaction between $XeF_4$ and $SbF_5$ is a fluoride ion transfer reaction where $XeF_4$ acts as a fluoride donor and $SbF_5$ acts as a fluoride

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) The reaction between $XeF_4$ and $SbF_5$ is a fluoride ion transfer reaction where $XeF_4$ acts as a fluoride donor and $SbF_5$ acts as a fluoride acceptor.The balanced chemical equation is: $XeF_4 + SbF_5 \rightarrow [XeF_3]^+[SbF_6]^-$.Comparing this with the general form $[XeF_m]^{n+}[SbF_y]^{z-}$,we get:$m = 3$$n = 1$$y = 6$$z = 1$Therefore,$m + n + y + z = 3 + 1 + 6 + 1 = 11$.

$XeF_4$ reacts with $SbF_5$ to form $[XeF_m]^{n+}[SbF_y]^{z-}$. Calculate $m + n + y + z = .......$.

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