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Need explanation for: Xenon hexafluoride on partial hydrolysis produces compounds ‘X’ and ‘Y’. Compounds ‘X’ and ‘Y’ and the oxidation state of Xe are respectively :

Xenon hexafluoride on partial hydrolysis produces compounds ‘X’ and ‘Y’. Compounds ‘X’ and ‘Y’ and the oxidation state of Xe are respectively :

  • Option 1)

    XeO_{2}\left ( +4 \right )\: and\: XeO_{3}\left ( +6 \right )

     

     

     

  • Option 2)

    XeOF_{4}\left ( +6 \right )\: and\: XeO_{3}\left ( +6 \right )

  • Option 3)

    XeO_{2}F_{2}\left ( +6 \right )\: and\: XeO_{2}\left ( +4 \right )

  • Option 4)

    XeOF_{4}\left ( +6 \right )\: and\: XeO_{2}F_{2}\left ( +4 \right )

 
Answers (1)
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As we learned 

 

Oxy fluoride of Xenon -

XeF4+H2O\rightarrow X_{e}OF_{2}+2HF

XeF6+H2O\rightarrow X_{e}OF_{4}+2HF

XeF6+2H2O\rightarrow X_{e}O_{2}F_{2+4HF

- wherein

Fluoride of Xe partially hydrolyse by water to give oxy  fluoride of Xe

 

 

 


Option 1)

XeO_{2}\left ( +4 \right )\: and\: XeO_{3}\left ( +6 \right )

 

 

 

Option 2)

XeOF_{4}\left ( +6 \right )\: and\: XeO_{3}\left ( +6 \right )

Option 3)

XeO_{2}F_{2}\left ( +6 \right )\: and\: XeO_{2}\left ( +4 \right )

Option 4)

XeOF_{4}\left ( +6 \right )\: and\: XeO_{2}F_{2}\left ( +4 \right )

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