Chemistry - Complete Pblock- Question with Solution | TestHub
ChemistryComplete Pblock-p-Block 17,18Hard2 minQB
ChemistryHardinteger
For the given reactions :
(i) XeF₄ + O₂F₂ →
(ii) XeF₂ + PF₅ →
(iii) XeF₆ + BF₃ →
(iv) XeF₆ + SiO₂ →
(v) XeF₄ + H₂O (excess) →
(vi) XeF₆ + 3H₂O →
Find the sum of x and y, where
x = number of reactions in which at least one gaseous product is formed
y = number of reactions in which the xenon fluoride acts as a fluoride DONOR
Answer:
5
Solution:
(i) XeF₄ + O₂F₂ → XeF₆ + O₂↑ (gas ✓) ; (ii) XeF₂ + PF₅ → [XeF]⁺[PF₆]⁻ (F⁻ donor ✓) ; (iii) XeF₆ + BF₃ → [XeF₅]⁺[BF₄]⁻ (F⁻ donor ✓) ; (iv) XeF₆ + SiO₂ → XeOF₄... + SiF₄↑ (gas ✓) ; (v) disproportionation giving Xe↑ and O₂↑ (gas ✓) ; (vi) → XeO₃ + 6HF (no gas, no transfer). x = 3 (i, iv, v), y = 2 (ii, iii) → x + y = 5.
Stream:JEESubject:ChemistryTopic:Complete Pblock-Subtopic:p-Block 17,18
⏱ 2mℹ️ Source: QB
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