Question

Which of the following reactions are correct with respect to the formation of products? I. \[ \mathrm{XeF_6+NaF\rightarrow Na[XeF_7]} \] II. \[ \mathrm{XeF_2+PF_5\rightarrow [XeF]^+[PF_6]^-} \] III. \[ \mathrm{XeF_4+SbF_5\rightarrow [XeF_3]^+[SbF_6]^-} \] The correct answer is

• A. II, III only
• B. I, III only
• C. I, II only
• D. I, II, III

Hint:

Strong Lewis acids such as \[ \boxed{\mathrm{PF_5}\ \text{and}\ \mathrm{SbF_5}} \] abstract fluoride ions from xenon fluorides, producing ionic xenon fluorides.

Answer 1

The Correct Option is D

Step 1: Check Statement-I. \[ \mathrm{XeF_6} \] acts as a Lewis acid and accepts \[ \mathrm{F^-} \] from NaF to form \[ \boxed{\mathrm{Na[XeF_7]}.} \] Hence, \[ \boxed{\text{Statement-I is correct}.} \]

Step 2: Check Statement-II. \[ \mathrm{PF_5} \] acts as a fluoride ion acceptor. Therefore, \[ \mathrm{XeF_2+PF_5\rightarrow[XeF]^+[PF_6]^-}. \] Hence, \[ \boxed{\text{Statement-II is correct}.} \]

Step 3: Check Statement-III. Similarly, \[ \mathrm{SbF_5} \] removes \[ \mathrm{F^-} \] from \[ \mathrm{XeF_4}, \] forming \[ \boxed{\mathrm{[XeF_3]^+[SbF_6]^-}.} \] Hence, \[ \boxed{\text{Statement-III is correct}.} \] Therefore, \[ \boxed{(D)} \] is the correct answer.