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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $m + n + y + z = .......$.

Similar Questions

Ionisation potential values of noble gases decrease down the group with increase in atomic size. Xenon forms binary fluorides by the direct reaction of elements. Identify the correct statement$(s)$ from below.

Which of the following Xenon compounds does not produce explosive $XeO_3$ on its complete hydrolysis?

Which of the following compounds is not known?

Helium was discovered by

Which group of elements are all gases under normal conditions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module