Noble gases Questions in English

(B) When excess of xenon gas reacts with fluorine gas at a temperature of $673 \ K$ and a pressure of $1 \ bar$,xenon difluoride $(XeF_2)$ is formed.
$Xe \text{ (excess)} + F_{2(g)} \xrightarrow{673 \ K, 1 \ bar} XeF_2$
Hence,the correct option is $B$.

(D) Although $Xe$ and $Ne$ are both noble gases,$Xe$ can form compounds with highly electronegative elements like $O$ and $F$ due to its larger atomic size and lower ionization energy.
$Ne$ has a very small atomic size and a very high ionization energy,making it extremely inert.
Therefore,$Ne$ does not form stable chemical bonds with other atoms,and $NeF_2$ does not exist.

(D) Liquid helium is used as a cryogenic agent for carrying out various experiments at low temperatures.
Therefore,statement $(d)$ is incorrect.
Helium is lighter than air but heavier than hydrogen,and it is non-flammable,making it safer for balloons.
It is used in gas-cooled nuclear reactors and for cooling superconducting magnets.

(A) Helium $(He)$ has the lowest boiling point of $4.2 \ K$ among all known substances.
This is because the interatomic forces of attraction between $He$ atoms are extremely weak London dispersion forces due to the small size and stable electronic configuration of $He$ atoms.
Therefore,both assertion $(A)$ and reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(D) . $Xe + 3F_2 \xrightarrow{573K, 60-70 bar} XeF_6$. This statement is correct.
$B$. $Ar$ is used in electric bulbs to provide an inert atmosphere. This statement is correct.
$C$. In $XeF_2$,$Xe$ has $8$ valence electrons. Two are used in bonding with $F$,leaving $6$ electrons,which form $3$ lone pairs. This statement is correct.
$D$. $He$ has the lowest boiling point $(4.2 K)$ among all noble gases due to weak van der Waals forces. Thus,this statement is incorrect.

(A) The hydrolysis of $XeF_2$ occurs according to the following reaction:
$2XeF_2(s) + 2H_2O(l) \rightarrow 2Xe(g) + 4HF(aq) + O_2(g)$
In this reaction,$XeF_2$ reacts with water to produce xenon gas $(Xe)$,hydrogen fluoride $(HF)$,and oxygen gas $(O_2)$.
Therefore,the gaseous products formed are $Xe$ and $O_2$.

(A) The boiling point of noble gases increases down the group due to the increase in the magnitude of van der Waals forces as the atomic size increases.
Helium $(He)$ is the first element in Group $18$ and has the smallest atomic size.
Due to its extremely weak interatomic van der Waals forces,Helium has the lowest boiling point of all known substances,which is approximately $4.2 \ K$.

(B) The reaction of $Xe$ with $F_2$ in a $1:20$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure yields $XeF_6$ $(A)$.
$Xe_{(g)} + 3F_{2(g)} \xrightarrow{573 \ K, 60-70 \ bar} XeF_{6(s)}$
Complete hydrolysis of $XeF_6$ produces $XeO_3$ $(B)$ and $HF$.
$XeF_{6(s)} + 3H_2O_{(l)} \rightarrow XeO_{3(s)} + 6HF_{(aq)}$
Therefore,$A$ is $XeF_6$ and $B$ is $XeO_3$.

(D) The preparation of $XeF_6$ involves the reaction of xenon with excess fluorine.
The reaction is given by: $Xe(g) + 3F_2(g) \xrightarrow{573 \ K, 60-70 \ bar} XeF_6(s)$.
To ensure the formation of $XeF_6$,$F_2$ is taken in a large excess,typically in a molar ratio of $1:20$ $(Xe:F_2)$.

(D) The hydrolysis reactions of xenon fluorides are as follows:
$1$. $2XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2$ (Liberates $O_2$)
$2$. $6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$ (Liberates $O_2$)
$3$. $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$ (Does not liberate $O_2$)
$4$. $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$ (Liberates $O_2$)
Therefore,the reaction $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$ does not involve the liberation of oxygen.

(D) Assertion $(A)$ is true because noble gases are monoatomic and possess only weak van der Waals forces of attraction between their atoms,leading to very low boiling points.
Reason $(R)$ is also true because the general electronic configuration of noble gases (group $18$) is $ns^2 np^6$ (with $He$ being $1s^2$).
However,the electronic configuration is not the direct reason for the low boiling points; the weak intermolecular forces are. Therefore,$(R)$ is not the correct explanation of $(A)$.

(C) The complete hydrolysis of $XeF_6$ is given by the reaction:
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
Here,the oxide of xenon '$O$' is $XeO_3$.
In the structure of $XeO_3$,the xenon atom is bonded to three oxygen atoms by double bonds $(Xe=O)$.
Each $Xe=O$ bond consists of one $\sigma$ bond and one $\pi$ bond.
Therefore,there are $3$ $\sigma$ bonds and $3$ $\pi$ bonds.
Total number of $\sigma$ and $\pi$ bonds $= 3 + 3 = 6$.

(A) In $XeO_3$,the central atom $Xe$ is $sp^3$ hybridized with one lone pair. It forms $3$ double bonds with oxygen atoms. Each double bond consists of one $\sigma$ bond and one $p \pi-d \pi$ $\pi$ bond. Thus,there are $3$ $p \pi-d \pi$ $\pi$ bonds.
In $XeO_4$,the central atom $Xe$ is $sp^3$ hybridized with no lone pair. It forms $4$ double bonds with oxygen atoms. Each double bond consists of one $\sigma$ bond and one $p \pi-d \pi$ $\pi$ bond. Thus,there are $4$ $p \pi-d \pi$ $\pi$ bonds.
Therefore,the number of $p \pi-d \pi$ $\pi$ bonds in $XeO_3$ and $XeO_4$ are $3$ and $4$ respectively.

(A) The reactions are:
$Xe + 3F_2 \rightarrow XeF_6$ $(X)$
$XeF_6 + H_2O \rightarrow XeOF_4$ $(Y)$ $+ 2HF$
$XeF_6 + 3H_2O \rightarrow XeO_3$ $(Z)$ $+ 6HF$
The structure of $XeOF_4$ $(Y)$ is square pyramidal with $sp^3d^2$ hybridization and $1$ lone pair.
The structure of $XeO_3$ $(Z)$ is pyramidal with $sp^3$ hybridization,$3 \sigma$ bonds,$3 \pi$ bonds,and $1$ lone pair of electrons on the central atom.
Therefore,statements $I$ and $III$ are correct.

(A) The valence shell of Xenon $(Xe)$ contains $8$ electrons. The number of lone pairs can be calculated using the formula: $\text{Lone pairs} = \frac{1}{2} (V - B)$,where $V$ is the number of valence electrons and $B$ is the number of bonding electrons.
$1$. In $XeOF_4$: $Xe$ forms $4$ single bonds with $F$ and $1$ double bond with $O$,using $4 + 2 = 6$ electrons. Lone pairs = $\frac{1}{2} (8 - 6) = 1$.
$2$. In $XeF_4$: $Xe$ forms $4$ single bonds with $F$,using $4$ electrons. Lone pairs = $\frac{1}{2} (8 - 4) = 2$.
$3$. In $XeF_2$: $Xe$ forms $2$ single bonds with $F$,using $2$ electrons. Lone pairs = $\frac{1}{2} (8 - 2) = 3$.
$4$. In $XeF_6$: $Xe$ forms $6$ single bonds with $F$,using $6$ electrons. Lone pairs = $\frac{1}{2} (8 - 6) = 1$.
Thus,the number of lone pairs are $1, 2, 3, 1$ respectively.

(B) The reaction is: $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$.
Here,$X = XeF_6$.
The hydrolysis reaction is: $XeF_6 + H_2O \rightarrow XeOF_4 + 2 HF$.
Here,$Y = XeOF_4$.
The shape of $XeF_6$ is distorted octahedral due to the presence of a lone pair.
The shape of $XeOF_4$ is square pyramidal.

(B) In $XeO_3$,the $Xe$ atom is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
In $XeOF_4$,the $Xe$ atom is $sp^3d^2$ hybridized with one lone pair,resulting in a square pyramidal geometry.

(A) To determine the hybridisation $(H)$ of the central atom $Xe$ in $XeO_3$,we use the formula:
$H = \frac{1}{2}(V + M - C + A)$
Where:
$V$ = number of valence electrons of the central atom $(Xe)$ = $8$
$M$ = number of monovalent atoms attached = $0$ (Oxygen is divalent)
$C$ = number of cationic charge = $0$
$A$ = number of anionic charge = $0$
Substituting the values:
$H = \frac{1}{2}(8 + 0 - 0 + 0) = 4$
$A$ value of $4$ corresponds to $s p^3$ hybridisation.

(D) $XeF_2$ has $2$ bond pairs and $3$ lone pairs on the central $Xe$ atom,resulting in a steric number of $5$.
Therefore,the hybridisation of $Xe$ in $XeF_2$ is $s p^3 d$,not $d^2 s p^3$.
Thus,Assertion $(A)$ is false.
$XeF_2$ has $10$ electrons around the central $Xe$ atom,which is an expanded octet,meaning it does not follow the octet rule.
Thus,Reason $(R)$ is true.
Therefore,the correct option is $(D)$.

(C) In $XeOF_4$,the central atom $Xe$ has $8$ valence electrons.
It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom.
This accounts for $6$ electrons,leaving $2$ electrons as one lone pair.
The total number of electron pairs is $5 + 1 = 6$ (where $5$ are bonding pairs and $1$ is a lone pair),corresponding to $sp^3d^2$ hybridization.
Due to the presence of one lone pair,the geometry is square pyramidal.

(C) The hybridization of the central atom $Xe$ is determined by the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $XeF_2$: $H = \frac{1}{2}(8 + 2) = 5$,which corresponds to $sp^3d$ hybridization.
For $XeF_4$: $H = \frac{1}{2}(8 + 4) = 6$,which corresponds to $sp^3d^2$ hybridization.
Therefore,the correct pair is $XeF_2, sp^3d$.

(B) Among the given compounds,$NeF_2$ does not exist.
Neon $(Ne)$ belongs to the $2nd$ period and does not have $d$-orbitals available for bonding.
Furthermore,the valence electrons of Neon are held very strongly by the nucleus due to its small size and high effective nuclear charge,making it energetically unfavorable to excite electrons for bond formation.

(A) Noble gases are monoatomic and possess the largest atomic radius in their respective periods.
The atomic size of noble gases increases down the group from $He$ to $Rn$ due to the addition of new shells.
Among the noble gases,$He$ has the smallest atomic size.
Due to its extremely small size,$He$ atoms can easily diffuse through the pores of silica glass.
The atomic radius of noble gases is referred to as the van der Waals radius.

(C) The partial hydrolysis of $XeF_6$ is represented by the following chemical equation:
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$
Here,'$X$' is $XeOF_4$.
The central atom $Xe$ in $XeOF_4$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom.
Total electron pairs around $Xe = 5$ (bonding pairs) + $1$ (lone pair) = $6$ electron pairs.
This corresponds to $sp^3d^2$ hybridization with an octahedral electron geometry.
Due to the presence of one lone pair,the shape of $XeOF_4$ is square pyramidal.

(B) $1$. Helium $(He)$ is used as a diluent for oxygen in modern diving apparatus because of its low solubility in blood.
$2$. Argon $(Ar)$ is used to provide an inert atmosphere in high-temperature metallurgical processes.
$3$. The question asks for '$Y$' and '$X$' respectively.
$4$. Since '$Y$' is $Ar$ and '$X$' is $He$,the correct pair is $(Ar, He)$.

(C) Neon $(Ne)$ is a noble gas,which is colorless,odorless,and chemically inert.
Option $C$ is incorrect because $Ne$ is not a fluorescent green gas and does not have a rotten egg smell.
$PH_3$ has a rotten fish smell,$Cl_2$ is a greenish-yellow gas with a pungent smell,and $SO_2$ is a colorless gas with a pungent smell.

(B) The oxidizing and fluorinating power of xenon fluorides depends on the ability of the $Xe$ atom to get reduced,which is directly related to the oxidation state of $Xe$ in the compound.
In $XeF_6$,the oxidation state of $Xe$ is $+6$.
In $XeF_4$,the oxidation state of $Xe$ is $+4$.
In $XeF_2$,the oxidation state of $Xe$ is $+2$.
As the oxidation state increases,the tendency to get reduced increases,making the compound a stronger oxidizing and fluorinating agent.
Therefore,the correct decreasing order is $XeF_6 > XeF_4 > XeF_2$.

(C) The balanced chemical equation for the hydrolysis of $XeF_4$ is:
$6 XeF_4 + 12 H_2O \rightarrow 4 Xe + 2 XeO_3 + 24 HF + 3 O_2$
Comparing this with the given equation $a XeF_4 + b H_2O \rightarrow p Xe + q XeO_3 + r HF + s O_2$,we get:
$a = 6, b = 12, p = 4, q = 2, r = 24, s = 3$.

(A) The enthalpy of vaporisation of noble gases depends on the magnitude of van der Waals forces of attraction. \\ As the atomic size and atomic weight increase down the group,the magnitude of van der Waals forces increases. \\ Therefore,the enthalpy of vaporisation increases in the order: $He < Ne < Ar < Kr < Xe$.

(C) The complete hydrolysis of $XeF_4$ is given by the reaction: $6XeF_4 + 12H_2O \rightarrow 2XeO_3 + 4Xe + 24HF + 3O_2$. Thus,$P$ is $XeO_3$.
The complete hydrolysis of $XeF_6$ is given by the reaction: $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$. Thus,$Q$ is $XeO_3$.
Therefore,both $P$ and $Q$ are $XeO_3$.

(C) Noble gases consist of monoatomic molecules with stable electronic configurations.
Due to the lack of permanent dipoles or chemical bonding between atoms,the only forces of attraction between them are weak van der Waals' forces (also known as London dispersion forces).
These weak forces require very little energy to overcome,resulting in low melting and boiling points.

(A) The complete hydrolysis of $XeF_4$ is represented by the following chemical equation:
$6 XeF_4 + 12 H_2O \longrightarrow 2 XeO_3 + 4 Xe + 3 O_2 + 24 HF$
As shown in the equation,the products formed are $Xe$,$XeO_3$,$O_2$,and $HF$.
Therefore,option $(A)$ is the correct answer.

(A) Complete hydrolysis of $XeF_6$ is given by the reaction:
$XeF_6 + 3 H_2O \longrightarrow XeO_3 + 6 HF$
Thus,$X = XeO_3$.
Partial hydrolysis of $XeF_6$ occurs in two steps:
$XeF_6 + H_2O \longrightarrow XeOF_4 + 2 HF$
$XeF_6 + 2 H_2O \longrightarrow XeO_2 F_2 + 4 HF$
Thus,$Y$ and $Z$ are $XeOF_4$ and $XeO_2 F_2$ respectively.
Therefore,the products $X, Y, Z$ are $XeO_3, XeOF_4, XeO_2 F_2$.

(A) $XeF_4$ is obtained by heating a mixture of xenon and fluorine in the molar ratio of $1: 5$ at $873 \ K$ and $7 \ bar$ pressure in an enclosed nickel vessel for a few hours.
The reaction proceeds as:
$Xe_{(g)} + 2 \ F_{2(g)} \xrightarrow{873 \ K, 7 \ bar} XeF_4$
The extra fluorine is added to ensure the complete conversion of xenon to $XeF_4$.

(D) Liquid xenon $(Xe)$ is used in bubble chambers for the detection of $\gamma$-photons and neutral mesons due to its high density and high atomic number,which increases the probability of interaction with these particles.

(A) Noble gases are monoatomic and held together by weak van der Waals' forces.
As the atomic size increases from $He$ to $Xe$,the magnitude of van der Waals' forces increases due to an increase in the number of electrons and polarizability.
Consequently,more energy is required to overcome these forces,leading to an increase in the boiling point.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(D) Krypton is used in miner's cap lamps because it provides a high-intensity light and has a long filament life.

(B) The abundance of noble gases in the atmosphere by volume (which is proportional to weight for these gases) is as follows:
$Ar$ $(0.934\%)$ > $Ne$ $(0.0018\%)$ > $Kr$ $(0.00011\%)$.
Therefore,the correct order is $Ar > Ne > Kr$.

(A) In Fischer-Ringe's method,air that is free from moisture and $CO_2$ is passed over a heated mixture $(800^{\circ}C)$ of $90 \% CaC_2 + 10 \% CaCl_2$ in an iron tube.
This process removes nitrogen and oxygen from the air mixture.
The chemical reactions involved are:
$CaC_2 + N_2 \xrightarrow{800^{\circ}C} CaCN_2 + C$
$2C + O_2 \rightarrow 2CO$
$2CaC_2 + 3CO_2 \rightarrow 2CaCO_3 + 5C$
$CuO + CO \rightarrow Cu + CO_2$
The remaining $CO_2$ is then absorbed by a $KOH$ solution,leaving behind the mixture of noble gases.

(C) In the Dewar's method,coconut charcoal is used to adsorb noble gases at $173 \ K$ $(-100^{\circ}C)$.
At this temperature,gases with higher boiling points are adsorbed,while those with very low boiling points remain in the gaseous phase.
$He$ (boiling point $\approx 4 \ K$) and $Ne$ (boiling point $\approx 27 \ K$) have boiling points much lower than $173 \ K$,so they are not adsorbed on the coconut charcoal.
Therefore,the mixture of $He$ and $Ne$ remains unadsorbed.

(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 $103^{\circ}$.