A) \[XeO{{F}_{4}}(+6)\]and \[Xe{{O}_{2}}{{F}_{2}}(+6)\]
B) \[XeO{{F}_{4}}(+6)\] and \[Xe{{O}_{3}}(+6)\]
C) \[Xe{{O}_{2}}{{F}_{2}}(+6)\] and \[Xe{{O}_{2}}(+4)\]
D) \[Xe{{O}_{2}}(+4)\] and \[Xe{{O}_{3}}(+6)\]
Correct Answer: A
Solution :
Xenon hexafluoride on partial hydrolysis produces compounds \[XeO{{F}_{4}}\] and \[Xe{{O}_{2}}{{F}_{2}}\]. In the compounds \[XeO{{F}_{4}}\] and \[Xe{{O}_{2}}{{F}_{2}}\], the oxidation states of Xe are +6 and +6 respectively. \[Xe{{F}_{6}}+{{H}_{2}}O\to XeO{{F}_{4}}+2HF\] \[XeO{{F}_{4}}+{{H}_{2}}O\to Xe{{O}_{2}}{{F}_{2}}+2HF\]
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