XeF₄ reacts with SbF₅ to form
[XeFₘ]ⁿ⁺[SbF_y]ᶻ⁻.
m+n+y+z= __________

Solution

Chemistry ma’am · 3 steps

SbF₅ is a strong fluoride-ion acceptor, so it pulls one F⁻ off XeF₄.

1. Fluoride transfer. SbF₅ is a strong Lewis acid and grabs a fluoride ion. XeF₄ gives up one F⁻, so it is left as a cation with one fluorine fewer.

   XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]⁻

2. Read off values. The cation is [XeF₃]⁺, so m = 3 and n = 1. The anion is [SbF₆]⁻, so y = 6 and z = 1. The charges balance, +1 and -1, as they must.
3. So. m + n + y + z = 3 + 1 + 6 + 1 = 11.

The common slip: 13. Writing the XeF₆ product [XeF₅]⁺ as the cation takes m = 5 and gives 13, not 11. XeF₄ loses one fluorine, so m = 3.

The idea to keep. Xenon fluorides react with SbF₅, PF₅ or AsF₅ by losing one F⁻: XeF₂ → XeF⁺, XeF₄ → XeF₃⁺, XeF₆ → XeF₅⁺. The acceptor always becomes MF₆⁻.

Answer from the NTA final answer key. Solution by PrepWiser.