XeF4 reacts with SbF5 to form: [XeFm]^n+[SbFy]^z- What is m+n+y+z… — Solution | JEE Chemistry

MCQEasyJEE 2023Group 18 Elements

XeF $_4$ reacts with SbF $_5$ to form: $[XeF_m]^{n+}[SbF_y]^{z-}$ What is $m+n+y+z$ ?

Answer

Correct answer:B

Standard Method

Given: XeF $_4$ reacts with SbF $_5$ to form $[XeF_m]^{n+}[SbF_y]^{z-}$ .

Find: The value of $m+n+y+z$ .

From the reaction shown in the solution:

\text{XeF}_4 + \text{SbF}_5 \to [\text{XeF}_3]^+ + (\text{SbF}_6)^-

So, comparing with $[XeF_m]^{n+}[SbF_y]^{z-}$ :

Therefore,

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

The correct option is B.

Taking the cation as $[\text{XeF}_4]^+$ is incorrect because the extracted reaction in the solution shows formation of $[\text{XeF}_3]^+$ . Use the species actually given in the reaction before assigning $m$ and $n$ .
Ignoring the magnitude of charge on the ions is wrong because $n$ and $z$ come from the superscripts. Here the ions are $+1$ and $-1$ , so both $n$ and $z$ are counted as $1$ in the sum.
Using $\text{SbF}_5^-$ for the anion is incorrect. Antimony pentafluoride accepts an extra fluoride to form $[\text{SbF}_6]^-$ , so $y=6$ , not $5$ .

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