XeF$$_4$$ reacts with SbF$$_5$$ to form [XeF$$_m$$]$$^{n+}$$[SbF$$_y$$]$$^{2-}$$. $$m + n + y + z$$ = ?

XeF₄ reacts with SbF₅ to form [XeF$$_m$$]$$^{n+}$$[SbF$$_y$$]$$^{z-}$$, and we seek the value of $$m + n + y + z$$.

SbF₅ acts as a strong Lewis acid (electron pair acceptor) while XeF₄ donates a fluoride ion (Lewis base), transferring one F⁻ ion to SbF₅ as shown by the following equilibria:

$$XeF_4 \rightarrow XeF_3^+ + F^-$$

$$SbF_5 + F^- \rightarrow SbF_6^-$$

Combining these steps gives the overall reaction:

$$XeF_4 + SbF_5 \rightarrow [XeF_3]^+[SbF_6]^-$$

In the cation [XeF₃]⁺, xenon retains three fluorine atoms after donating one F⁻, resulting in a +1 charge. In the anion [SbF₆]⁻, antimony accepts the fluoride ion to form six fluorine bonds, carrying a −1 charge.

From this we identify that $$m = 3$$, $$n = 1$$, $$y = 6$$, and $$z = 1$$.

Therefore, adding these values yields

$$m + n + y + z = 3 + 1 + 6 + 1 = 11$$

The correct answer is 11.