NTA Abhyas JEE Main2020Chemistryp Block Elements (Group 15, 16, 17 & 18)Practice
The reaction of solid X e F 2 with A s F 5 in 1:1 ratio affords
Options
- AX e F 4 and A s F 3
- BX e F 6 and A s F 3
- CX e F + A s F 6 -
- D[ X e 2 F 3 ] + [ A s F 6 ] -
Correct answer
C. X e F + A s F 6 -
Step-by-step solution
X e F 2 (Xenon difluoride) acts as a fluoride donor and thus, forms complex when mixed with covalent pentafluorides like A s F 5 . X e F 2 1 + A s F 5 1 → X e F + A s F 6 -