Which of the following inert gases liquefies most easily?

What are the elements at the end of periods $1$ to $7$ called? Which are these elements?

Why does the following reaction occur?
$XeO_{6(aq)}^{4-} + 2F_{(aq)}^{-} + 6H_{(aq)}^{+} \to XeO_{3(g)} + F_{2(g)} + 3H_2O_{(l)}$
What conclusion about the compound $Na_4XeO_6$ (of which $XeO_6^{4-}$ is a part) can be drawn from the reaction?

The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers $+2, +4$ and $+6$. $XeF_4$ reacts violently with water to give $XeO_3$. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.
$1.$ Argon is used in arc welding because of its
$A.$ low reactivity with metal
$B.$ ability to lower the melting point of metal
$C.$ flammability
$D.$ high calorific value
$2.$ The structure of $XeO_3$ is
$A.$ linear
$B.$ planar
$C.$ pyramidal
$D.$ $T$-shaped
$3.$ $XeF_4$ and $XeF_6$ are expected to be
$A.$ oxidizing
$B.$ reducing
$C.$ unreactive
$D.$ strongly basic
Give the answer for questions $1, 2$ and $3$.

$XeF_{6}$ on partial hydrolysis gives a compound $X$,which has square pyramidal geometry. '$X$' is

The type of hybridisation and number of lone pair$(s)$ of electrons of $Xe$ in $XeOF_4$ respectively are