Noble gases Questions in English

(C) The oxidation state of $Xe$ in $XeO_3$ is calculated as follows:
$x + 3(-2) = 0$
$x - 6 = 0$
$x = +6$
In $XeO_3$,$Xe$ undergoes $sp^3$ hybridization with one lone pair,resulting in a pyramidal geometry.
The bond angle in $XeO_3$ is approximately $103^{\circ}$.

(A) The ionisation potential of noble gases decreases down the group as the atomic size increases.
Because of the lower ionisation energy of heavier noble gases like $Xe$,they can be oxidised by highly electronegative elements like fluorine.
Therefore,only the heavier noble gases form such compounds,and this occurs because they form compounds with electronegative ligands like $F$.
While the octet rule provides stability,it is not the reason why $Xe$ forms fluorides; rather,it is the ease of ionisation of heavier noble gases.

(A) The structure of $XeO_{2}F_{2}$ is based on a trigonal bipyramidal geometry with one lone pair in the equatorial position. This results in a see-saw shape.
Analysis of statements:
$A$. $TRUE$: The molecule has a see-saw shape.
$B$. $FALSE$: $Xe$ has $7$ electron pairs in its valence shell ($5$ bonding pairs and $1$ lone pair,where double bonds count as $1$ electron pair for geometry).
$C$. $FALSE$: The $O-Xe-O$ bond angle is close to $106^{\circ}$ (equatorial-equatorial).
$D$. $TRUE$: The $F-Xe-F$ bond angle is close to $170^{\circ}-180^{\circ}$ (axial-axial).
$E$. $FALSE$: $Xe$ has $14$ valence electrons in $XeO_{2}F_{2}$ ($8$ from $Xe$ + $2$ from $2F$ + $4$ from $2O = 14$).
Therefore,the statements that are $NOT$ $TRUE$ are $B, C$ and $E$.