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XeF2 when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest H-bond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?
  • a)
    A is XeO3
  • b)
    B is F2
  • c)
    C is Xe
  • d)
    D is HF2-
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
XeF2when dissolved in water produces three compound A, B, C. A is iner...
Reaction mentioned is given as below
2XeF2 +2H2O →2Xe + 4HF + O2
HF forms H-bond with F- and exists as HF2-
Hence option (d) is correct.
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Most Upvoted Answer
XeF2when dissolved in water produces three compound A, B, C. A is iner...
Explanation:
When XeF2 is dissolved in water, it undergoes hydrolysis to produce three compounds, namely A, B, and C. Let's analyze each compound individually.

A:
The statement mentions that compound A is inert. In chemistry, the term "inert" is often used to describe substances that are unreactive or do not undergo chemical reactions. Based on this information, we can rule out compounds that are known to be reactive. One possibility for compound A is XeO3 (xenon trioxide), which is relatively inert and stable.

B:
The statement states that compound B forms the strongest hydrogen bond with its anion. In order for a compound to form strong hydrogen bonds, it typically needs to have a hydrogen atom bonded to a highly electronegative atom, such as oxygen or nitrogen. F2 (fluorine gas) does not fit this description, as it consists of two fluorine atoms bonded together. However, HF2 (hydrogen difluoride) does have a hydrogen atom bonded to a highly electronegative fluorine atom and can form strong hydrogen bonds. Therefore, B is likely HF2.

C:
The statement mentions that compound C is used in combustion. Combustion reactions typically involve the reaction of a substance with oxygen to produce heat and light. Xenon, being a noble gas, is generally unreactive and does not readily undergo combustion reactions. Therefore, it is unlikely that compound C is xenon-based. Instead, compound C could be a substance that readily undergoes combustion, such as a hydrocarbon.

D:
Compound D is described as the form of compound B that exists when it is bonded to its anion. Since compound B is likely HF2, compound D would be the form of HF2 when it is bonded to its anion. In this case, the anion would be F- (fluoride). Therefore, D can be identified as HF2-.

In summary, based on the given information, the correct answer is option D, where A is XeO3, B is HF2, C is a hydrocarbon used in combustion, and D is HF2-.
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XeF2when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest H-bond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?a)A is XeO3b)B is F2c)C is Xed)D is HF2-Correct answer is option 'D'. Can you explain this answer?
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XeF2when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest H-bond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?a)A is XeO3b)B is F2c)C is Xed)D is HF2-Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about XeF2when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest H-bond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?a)A is XeO3b)B is F2c)C is Xed)D is HF2-Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for XeF2when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest H-bond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?a)A is XeO3b)B is F2c)C is Xed)D is HF2-Correct answer is option 'D'. Can you explain this answer?.
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