$\mathrm{XeF}_{4}$ reacts with $\mathrm{SbF}_{5}$ to form

$[\mathrm{XeF}_{m}]^{\mathrm{n}+}\left[\mathrm{SbF}_{y}\right]^{z-}$.

$\mathrm{m}+\mathrm{n}+\mathrm{y}+\mathrm{z}=$ __________

<p>The reaction of $XeF_4$ with $SbF_5$ is a known reaction that forms a complex compound $XeF_3^+$ and $SbF_6^−$. Here, the $XeF_4$ is providing one fluoride ion to $SbF_5$, resulting in $XeF_3^+$ and $SbF_6^−$.</p> <p>In the form $[XeF_m]^{n+}[SbF_y]^{z-}$, the parameters m, n, y, and z correspond to the following:</p> <ul> <li>m is the number of fluoride ions attached to Xe.</li><br/> <li>n is the charge of the $XeF_m$ cation.</li><br/> <li>y is the number of fluoride ions attached to Sb.</li><br/> <li>z is the charge of the $SbF_y$ anion.</li> </ul> <p>Given the formation of $XeF_3^+$ and $SbF_6^−$, the values would be as follows:</p> <ul> <li>m = 3 (from $XeF_3$)</li><br/> <li>n = 1 (from $XeF_3^+$)</li><br/> <li>y = 6 (from $SbF_6$)</li><br/> <li>z = 1 (from $SbF_6^−$)</li> </ul> <p>Therefore, $m + n + y + z = 3 + 1 + 6 + 1 = 11$.</p>