At $143 \ K$,the reaction of $XeF_4$ with $O_2F_2$ produces a xenon compound $Y$. The total number of lone pair$(s)$ of electrons present on the whole molecule of $Y$ is . . . . .

Which combination gives the maximum number of products?

Which of the following is the correct sequence of the noble gases in their group in the periodic table?

$XeF_2$ is hydrolyzed in the presence of a small amount of water. What are the gaseous products formed?

In the preparation of compounds of $Xe$,Bartlett had taken $O^{+}_2 Pt F^{-}_6$ as a base compound. This is because

In reaction $(1)$, $XeF_6$ undergoes complete hydrolysis to form $HF$ and $X$. In reaction $(2)$, $XeF_6$ undergoes partial hydrolysis to form $HF$, $Y$, and $Z$. The products $X, Y, Z$ respectively, are