- A$XeF_2$
- B$XeF_4$
- ✓$XeO_3$
- D$XeOF_2$
$Xe{O_3} = s{p^3}\frac{{\% \,{\text{Character = 25}}\,\% }}{{{\text{More EN}}}}$
The main reason for adding helium with oxygen to the breathing mix is to reduce the proportions of nitrogen and oxygen to allow the gas mix to be breathed safely on deep dives.
What is $A$
$2$. $XeF_2 \xrightarrow{H_2O} $
$3$. $XeF_4 \xrightarrow{H_2O} $
$4$. $PbO_2 \xrightarrow{\Delta } $
$5$. $XeF_6 \xrightarrow{H_2O} $
In how many of the above processes, $O_2$ will be one of the products
$\mathrm{XeF}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Xe}+2 \mathrm{HF}+\frac{1}{2} \mathrm{O}_{2}$
$\mathrm{XeF}_{4}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Xe}+\mathrm{XeO}_{3}+\mathrm{HF}+\mathrm{O}_{2}$
$\mathrm{PbO}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{Pb}+\mathrm{O}_{2}$
| Column $-I$ | Column $-II$ |
| $(a) \,Xe{F_6}$ | $(i)$ Distorted Octahedral |
| $(b)\, XeO_3$ | $(ii)$ Square planar |
| $(c)\, XeOF_4$ | $(iii)$ pyramidal |
| $(d)\, XeF_4$ | $(iv)$ Square pyramidal |

What is $A$
Hence option $\mathrm{C}$ is correct.
$Xe{F_6}\,\xrightarrow{{ + \,\,Excess\,\,{H_2}O}}\,'X'\, + \,HF$
$Xe{F_6}\,\xrightarrow{{ + 2\,\,\,{H_2}O}}\,'Y'\, + \,HF$
$Xe{F_6}\,\xrightarrow{{2\,\,{H_2}O}}\,4HF\, + \,Xe{O_2}{F_2}$

| $He$ | $Ne$ | $Ar$ | $Kr$ | $Xe$ | $Rn$ | |
| Boiling point of | $-269$ | $-246$ | $-186$ | $-153.6$ | $-108.1$ | $-62$ |
$x + ( - 2) + 2( - 1) = 0$ $⇒$ $x - 2 - 2 = 0$ $⇒$ $x = 4$
${T_C} \propto {\rm{B}}{\rm{.P}}{\rm{.}}$ and $B.P.$ $ \propto $ Molecular weight
So $Kr$ liquifies first.
$(1)$ Small atomic size
$(2)$ High Ionization energy
$(3)$ Absence of $d$ - orbitals
$Xe+{{F}_{2}}\,\underset{673\,K}{\mathop{\xrightarrow{Ni\text{ }\,\text{tube}}}}\,\,Xe{{F}_{2}}$ ;
$Xe+2{{F}_{2}}\,\underset{6\,\text{ atm}}{\mathop{\xrightarrow{673\,K}}}\,\,Xe{{F}_{4}}$
$Xe+3{{F}_{2}}\,\underset{50-60\,\text{atm}}{\mathop{\xrightarrow{523-573\,K}}}\,\,Xe{{F}_{6}}$
$Xe{{O}_{3}}$ is obtained by the hydrolysis of $Xe{{F}_{6}}$
$Xe{{F}_{6}}+3{{H}_{2}}O\,\to \,Xe{{O}_{3}}+6HF$
$_{88}R{a^{206}}\, \to {\,_{86}}R{n^{202}}{ + _2}H{e^4}$
Complete hydrolysis;
$2Xe{F_4} + 3{H_2}O\, \to $$Xe + Xe{O_3} + {F_2} + 6HF$
| Gas | (Abundance in air by volume ($ppm$)) |
| Helium | $5.2$ |
| Neon | $18.2$ |
| Argon | $93.4$ |
| Krypton | $1.1$ |
| Xenon | $ 0.09$ |