JEE Advanced 2022 Paper 2 Q04 Chemistry Inorganic Chemistry p-Block Elements (Groups 15-18) Medium

JEE Advanced 2022 Paper 2 · Q04 · p-Block Elements (Groups 15-18)

The reaction of Xe and O$_2$F$_2$ gives a Xe compound P. The number of moles of HF produced by the complete hydrolysis of 1 mol of P is _____.

Reveal answer + step-by-step solution

Correct answer:2

Solution

Step 1 — identify P. Dioxygen difluoride (O$_2$F$_2$) is a powerful fluorinating-oxidising agent; with Xe it transfers an O and two F to give xenon oxydifluoride: Xe + O$_2$F$_2$ $\rightarrow$ XeOF$_2$ + (1/2) O$_2$ (after balancing $\rightarrow$ Xe + O$_2$F$_2$ $\rightarrow$ XeOF$_2$ + (1/2)O$_2$). So P = XeOF$_2$.

Step 2 — hydrolysis of XeOF$_2$. The two Xe-F bonds are hydrolysed liberating HF; the Xe=O stays as XeO$_2$ (or XeO$_3$ depending on conditions, but the F atoms are the only source of HF): XeOF$_2$ + H$_2$O $\rightarrow$ XeO$_2$ + 2 HF.

1 mol P releases 2 mol HF.

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