Chemistry - Complete Pblock- Question with Solution | TestHub

Finding $x$ (Gaseous products): At least one gaseous product is formed in 4 reactions:

$Xe{F}_4+O_2{F}_2\to Xe{F}_6+{\mathbf{O}}_2(\mathbf{g})$

$Xe{F}_6+Si{O}_2\to XeO{F}_4+\mathbf{Si}{\mathbf{F}}_4(\mathbf{g})$

$Xe{F}_4+H_{2}O(excess)\to \mathbf{Xe}(\mathbf{g})+Xe{O}_3+HF+{\mathbf{O}}_2(\mathbf{g})$

$Xe{F}_6+O{H}^{-}(excess)\to Xe{O}_6^{4-}+\mathbf{Xe}(\mathbf{g})+{\mathbf{O}}_2(\mathbf{g})+H_{2}O$

Thus, $x=4$.

Finding $y$ (Fluoride acceptor): In reactions with $Sb{F}_5$ and $B{F}_3$, the xenon halides ($Xe{F}_2$ and $Xe{F}_6$) act as fluoride donors (forming $[XeF{]}^{+}$ and $[Xe{F}_5{]}^{+}$). None of the xenon halides act as a Lewis acid fluoride acceptor here, so $y=0$.

Final Sum: Therefore, the sum of $x+y=4+0=4$.