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 $E$.

(D) $1$. Calculate the steric number for $XeOF_4$: $Steric \ number = \frac{1}{2} (V + M - C + A) = \frac{1}{2} (8 + 4 + 0 - 0) = 6$.
$2$. $A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization.
$3$. In $XeOF_4$, the central atom $Xe$ is bonded to $4$ fluorine atoms and $1$ oxygen atom (double bond), leaving $1$ lone pair.
$4$. Due to the presence of $5$ bond pairs and $1$ lone pair, the geometry is octahedral, but the shape is square pyramidal.

(D) Let the oxidation state of $Xe$ be $x$.
The oxidation state of $O$ is $-2$ and $F$ is $-1$.
Sum of oxidation states in a neutral molecule is $0$.
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$
Therefore, the oxidation state of $Xe$ is $+6$.

(D) $1$. $XeF_4$: The central atom $Xe$ has $2$ lone pairs and $4$ bond pairs, resulting in a square planar geometry (ii).
$2$. $XeF_6$: The central atom $Xe$ has $1$ lone pair and $6$ bond pairs, resulting in a distorted octahedral geometry (iii).
$3$. $XeOF_4$: The central atom $Xe$ has $1$ lone pair and $5$ bond pairs, resulting in a square pyramidal geometry (iv).
$4$. $XeO_3$: The central atom $Xe$ has $1$ lone pair and $3$ bond pairs, resulting in a pyramidal geometry $(i)$.
Therefore, the correct matching is $(a)$-(ii), $(b)$-(iii), $(c)$-(iv), $(d)$-$(i)$.

(D) $1$. Noble gases are generally inert due to their stable electronic configuration.
$2$. Among the noble gases, $Xe$ (Xenon) has the lowest ionization enthalpy due to its large atomic size.
$3$. Because of this, $Xe$ can easily form compounds with highly electronegative elements like $F$ and $O$, exhibiting higher oxidation states such as $+2, +4, +6,$ and $+8$ (e.g., in $XeF_6$ and $XeO_4$).