Chemistry - Complete Pblock- Question with Solution | TestHub
ChemistryComplete Pblock-p-Block 17,18Hard2 minQB
ChemistryHardsingle choice
Xenon fluorides show a variety of reactions with water and hydrogen. Match List-I with List-II and select the correct answer using the code given below the lists :
List-I | List-II |
|---|---|
(P) XeF₂ + H₂O → | (1) Disproportionation |
(Q) XeF₄ + H₂O → | (2) No change in oxidation state of Xe |
(R) XeF₆ + excess H₂O → | (3) At least one gaseous product is formed |
(S) XeF₂ + H₂ → | (4) HF is one of the products |
| (5) Redox reaction |
The correct option is :
Options:
Answer:
A
Solution:
(P) 2XeF₂ + 2H₂O → 2Xe↑ + 4HF + O₂↑ : redox (Xe +2 → 0, O −2 → 0), gases, HF → 3,4,5. (Q) 3XeF₄ + 6H₂O → 2Xe↑ + XeO₃ + 12HF + 3/2O₂↑ : disproportionation (+4 → 0 and +6), redox, gases, HF → 1,3,4,5. (R) XeF₆ + 3H₂O → XeO₃ + 6HF : Xe stays +6, no gas → 2,4. (S) XeF₂ + H₂ → Xe↑ + 2HF : redox, gas, HF → 3,4,5.
Stream:JEESubject:ChemistryTopic:Complete Pblock-Subtopic:p-Block 17,18
⏱ 2mℹ️ Source: QB
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