Noble gases Questions in English

(B) Helium was the first noble gas to be discovered,and it was identified by William Ramsay in $1895$ using the spectroscopic method.

(A) Helium is the only noble gas that does not crystallize in a $ccp$ structure because it has the smallest atomic size and very weak interatomic forces.

(A) In noble gases,as we move down the group,the number of electronic shells increases,which leads to an increase in the atomic size.
Due to the increase in atomic size and the shielding effect,the attraction between the nucleus and the valence electrons decreases.
Consequently,the energy required to remove an electron,known as the $IE$ (Ionization Energy),decreases.

(D) The geometry of $XeF_4$ is square planar.
In $XeF_4$,the central atom $Xe$ has $8$ valence electrons.
It forms $4$ bonds with $F$ atoms and has $2$ lone pairs of electrons.
According to $VSEPR$ theory,the steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridization with an octahedral electron geometry and square planar molecular geometry.

(B) In $XeOF_4$,the central atom is Xenon $(Xe)$.
$Xe$ has $8$ valence electrons.
It forms $4$ single bonds with $4$ Fluorine $(F)$ atoms and $1$ double bond with $1$ Oxygen $(O)$ atom.
Total electrons involved in bonding = $4 + 2 = 6$.
Number of electrons remaining = $8 - 6 = 2$.
These $2$ electrons constitute $1$ lone pair.
Thus,there is $1$ lone pair of electrons on the $Xe$ atom.

(A) The central atom $Xe$ in $XeO_4$ has $8$ valence electrons. It forms $4$ double bonds with $4$ oxygen atoms.
The steric number is $4 + 0 = 4$,which corresponds to $sp^3$ hybridization.
Thus,the geometry of $XeO_4$ is tetrahedral,not square planar.
Each $Xe=O$ bond consists of one $\sigma$ bond (formed by $sp^3-p$ overlap) and one $\pi$ bond (formed by $p\pi-d\pi$ overlap).
Therefore,there are four $\sigma$ bonds and four $p\pi-d\pi$ $\pi$ bonds.
Statement $A$ is incorrect.

(D) The reaction $XeO_3 + 6HF \rightarrow XeF_6 + 3H_2O$ is not feasible.
$XeF_6$ is highly reactive and undergoes hydrolysis in the presence of water to form $XeO_3$ and $HF$. Therefore,the reverse reaction is not spontaneous under standard conditions.
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$

(B) As we move down the group in the noble gases,the atomic size increases.
This leads to an increase in the magnitude of the van der Waals forces of attraction between the atoms.
Consequently,the boiling point increases as we move down the group.
Among the given options ($He$,$Ne$,$Kr$,$Xe$),$Xe$ is at the bottom of the group,thus it has the highest boiling point.

(A) In the reaction $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$,the oxidation state of $Xe$ changes from $+4$ to $+6$ (oxidation) and the oxidation state of $O$ changes from $+1$ to $0$ (reduction).
Since both oxidation and reduction occur,this is a redox reaction.
$\stackrel{+4}{Xe}F_4 + \stackrel{+1}{O_2}F_2 \rightarrow \stackrel{+6}{Xe}F_6 + \stackrel{0}{O_2}$

(C) The complete hydrolysis of $XeF_6$ is represented by the following chemical equation:
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
Thus,the product obtained is $XeO_3$.

(B) $Xe$ has $8$ valence electrons.
In $XeF_6$,$6$ electrons are involved in bonding with $6$ fluorine atoms,and $2$ electrons remain as a lone pair.
Steric number = (Number of bond pairs) + (Number of lone pairs) = $6 + 1 = 7$.
$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridisation.
Due to the presence of one lone pair,the geometry is distorted octahedral.

(B) The reactions of $XeF_6$ with water are as follows:
$1$. $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$. Here,$X$ is $XeO_3$.
$2$. $XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF$. Here,$Y$ is $XeO_2F_2$.
$XeO_3$ is a highly explosive solid. $XeO_2F_2$ is an oxyfluoride,not an oxyacid. Both reactions involve hydrolysis,which is a non-redox process. Therefore,the statement '$Y$ is an oxyacid' is incorrect.

(B) Noble gases like $Ne$ are chemically inert due to their stable electronic configuration and small atomic size.
$Xe$ has a larger atomic size and lower ionization energy,allowing it to form compounds with highly electronegative elements like $F$ and $O$.
$Ne$ does not form stable compounds like $NeF_2$ because its ionization energy is very high and it lacks vacant $d$-orbitals for bonding.
Therefore,$NeF_2$ does not exist.

(C) $1$. The reaction is $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ [SbF_6]^-$.
$2$. The cationic part is $[XeF_3]^+$. The central atom $Xe$ has $8$ valence electrons. It forms $3$ bonds with $F$ and has $2$ lone pairs. Total electron pairs = $5$,so hybridization is $sp^3d$.
$3$. The anionic part is $[SbF_6]^-$. The central atom $Sb$ has $5$ valence electrons. It forms $6$ bonds with $F$ and has $0$ lone pairs. Total electron pairs = $6$,so hybridization is $sp^3d^2$.
$4$. Option $A$: $XeOF_2$ has $sp^3d$ hybridization. $[SbF_6]^-$ has $sp^3d^2$. Incorrect.
$5$. Option $B$: $I_3^-$ has $3$ lone pairs on the central $I$ atom. $[XeF_3]^+$ has $2$ lone pairs on the central $Xe$ atom. Incorrect.
$6$. Option $C$: The anionic part of $[XeF]^+ [PF_6]^-$ is $[PF_6]^-$. $[PF_6]^-$ is octahedral $(sp^3d^2)$,and $[SbF_6]^-$ is also octahedral $(sp^3d^2)$. They are isostructural. Correct.

(D) $1$. $2XeF_{2} + 2H_{2}O \longrightarrow 2Xe + 4HF + O_{2}$ is a redox reaction because the oxidation states of $Xe$ and $O$ change.
$2$. $4H_{3}PO_{3} \stackrel{\Delta}{\longrightarrow} PH_{3} + 3H_{3}PO_{4}$ is a disproportionation reaction,which is a type of redox reaction.
$3$. $6XeF_{4} + 12H_{2}O \longrightarrow 2Xe + 4XeO_{3} + 24HF + 3O_{2}$ is a redox reaction.
$4$. $XeF_{6} + 3H_{2}O \longrightarrow XeO_{3} + 6HF$ is a non-redox reaction because the oxidation states of all elements ($Xe$,$F$,$H$,$O$) remain unchanged in the products compared to the reactants.

(B) $XeF_2$ has a linear geometry and $XeF_4$ has a square planar geometry. Both are planar molecules.
The hydrolysis reactions are:
$2 XeF_2 + 2 H_2O \rightarrow 2 Xe + 4 HF + O_2$
$6 XeF_4 + 12 H_2O \rightarrow 2 XeO_3 + 4 Xe + 3 O_2 + 24 HF$
In both reactions,the oxidation state of $Xe$ changes (from $+2$ to $0$ in $XeF_2$ and from $+4$ to $0$ and $+6$ in $XeF_4$),meaning they undergo redox hydrolysis. Therefore,option $C$ is incorrect.
Bond lengths in $XeF_2$ and $XeF_4$ are different due to different hybridization and electronic environments. Therefore,option $A$ is incorrect.
Since both $XeF_2$ and $XeF_4$ are planar,option $B$ is the correct statement.

(D) 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 $6$ bonding electron pairs.
Including $1$ lone pair,the total number of electron domains is $7$ (steric number $7$),which corresponds to $sp^3d^3$ hybridization.
However,due to the presence of one lone pair and the double bond,the geometry is distorted.
The molecule adopts a square pyramidal shape,where the $Xe$ atom is at the center of a square base formed by $4$ $F$ atoms,and the $O$ atom occupies the axial position.
$\therefore$ $(d)$ is the correct answer.

(A) Neil Bartlett in $1962$ prepared a compound $O_2PtF_6$ by reacting $O_2$ with a powerful oxidising agent,platinum hexafluoride $(PtF_6)$.
He observed through $X$-ray examination that this compound contains $O_2^+$ and $PtF_6^-$ ions.
Since the first ionisation enthalpy of xenon $(1170 \ kJ \ mol^{-1})$ is quite close to that of oxygen $(1175 \ kJ \ mol^{-1})$,he replaced oxygen with xenon.
This led to the synthesis of $XePtF_6$,which was the first real compound of any noble gas.

(A) The reaction between $XeF_2$ and $SbF_5$ is a Lewis acid-base reaction where $XeF_2$ acts as a fluoride donor and $SbF_5$ acts as a fluoride acceptor.
The chemical equation for the reaction is: $XeF_2 + SbF_5 \rightarrow [XeF]^+ [SbF_6]^-$.
Thus,the product formed is $[XeF]^+ [SbF_6]^-$,which is an ionic adduct.

(C) Bartlett observed that the first ionisation enthalpy of $Xe$ $(1170 \ kJ \ mol^{-1})$ is very close to that of the $O_2$ molecule $(1175 \ kJ \ mol^{-1})$.
He used this similarity to predict that $PtF_6$ would also oxidise $Xe$ to $Xe^+$,similar to how it oxidises $O_2$ to $O_2^+$.
Hence,option $C$ is correct.

(A) The reaction of Xenon with Fluorine in a $1:20$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure leads to the formation of Xenon hexafluoride $(XeF_6)$.
The chemical equation is: $Xe(g) + 3F_2(g) \xrightarrow{573 \ K, 60-70 \ bar} XeF_6(s)$.

(D) The reaction of xenon difluoride with water is a redox reaction where xenon is reduced and oxygen is oxidized.
The balanced chemical equation is:
$XeF_2 + H_2O \to Xe + \frac{1}{2}O_2 + 2HF$
Thus,the products formed are xenon $(Xe)$,oxygen $(O_2)$,and hydrogen fluoride $(HF)$.

(C) $He$ (Helium) is a noble gas and is chemically inert,meaning it is non-inflammable. Therefore,the statement that $He$ is an inflammable gas is incorrect.

(C) In the hydrolysis of $XeF_6$,the reaction is: $XeF_6 + 3H_2O \to XeO_3 + 6HF$.
In this reaction,the oxidation state of $Xe$ remains $+6$ in both $XeF_6$ and $XeO_3$.
Since there is no change in the oxidation state of any element,it is not a redox reaction.
In contrast,the hydrolysis of $Cl_2$,$XeF_4$,and $XeF_2$ involves changes in oxidation states,making them redox reactions.

(A) The structures of the given Xenon compounds are as follows:
$(a) XeF_6$: It has $7$ electron pairs ($6$ bond pairs and $1$ lone pair) around the central $Xe$ atom,resulting in a distorted octahedral geometry.
$(b) XeO_3$: It has $4$ electron pairs ($3$ bond pairs and $1$ lone pair) around the central $Xe$ atom,resulting in a pyramidal shape.
$(c) XeOF_4$: It has $6$ electron pairs ($5$ bond pairs and $1$ lone pair) around the central $Xe$ atom,resulting in a square pyramidal shape.
$(d) XeF_4$: It has $6$ electron pairs ($4$ bond pairs and $2$ lone pairs) around the central $Xe$ atom,resulting in a square planar shape.
Thus,the correct matching is: $a-i, b-iii, c-iv, d-ii$.

(D) Clathrates are cage-like structures where guest molecules are trapped within the crystal lattice of host molecules.
In the case of $Xe$ clathrates,$Xe$ atoms are trapped in the cavities formed by water molecules or other host molecules like quinol.
Since $Xe$ is a noble gas,it is non-polar.
When it approaches a polar molecule (like $H_2O$),it induces a dipole in the $Xe$ atom.
This results in a weak attractive force known as a dipole-induced dipole interaction.

(B) The partial hydrolysis of $XeF_6$ with water leads to the formation of xenon oxytetrafluoride $(XeOF_4)$ and hydrogen fluoride $(HF)$.
The chemical equation is: $XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$.

(D) The reaction of Xenon $(Xe)$ with Fluorine $(F_2)$ depends on the molar ratio and reaction conditions.
When $Xe$ reacts with $F_2$ in a $1:2$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure,it produces Xenon tetrafluoride $(XeF_4)$.
The balanced chemical equation is: $Xe(g) + 2F_2(g) \xrightarrow{573 \ K, \ 60-70 \ bar} XeF_4(s)$.
Therefore,$A$ is $XeF_4$.

(C) Xenon is a noble gas that forms compounds primarily with highly electronegative elements like fluorine and oxygen.
$XeBr_4$ does not exist because bromine is not sufficiently electronegative to stabilize the xenon atom in a tetra-bromide configuration.

(B) The solubility of noble gases in water is due to the interaction between the permanent dipole of water molecules and the induced dipole in the noble gas atoms. This type of interaction is known as $Debye$ forces (or dipole-induced dipole interaction).

(C) Neil Bartlett observed that the first ionization enthalpy of $Xe$ $(1170 \ kJ \ mol^{-1})$ is very close to that of the $O_2$ molecule $(1175 \ kJ \ mol^{-1})$.
Based on this observation,he successfully prepared the first noble gas compound by reacting $Xe$ with $PtF_6$,which was previously used to oxidize $O_2$ to $O_2^+PtF_6^-$.
Thus,the similarity in ionization enthalpy was the key factor.

(B) The hydrolysis of $XeF_6$ proceeds as follows:
$1$. $XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$ $(A = XeOF_4)$
$2$. $XeOF_4 + H_2O \rightarrow XeO_2F_2 + 2HF$ $(B = XeO_2F_2)$
$3$. $XeO_2F_2 + H_2O \rightarrow XeO_3 + 2HF$ $(C = XeO_3)$
Evaluating the options:
- $A$ is $XeOF_4$,which has $sp^3d^2$ hybridization (square pyramidal geometry). This is correct.
- $B$ is $XeO_2F_2$. Option $B$ states $B$ is $XeOF_2$,which is incorrect.
- $C$ is $XeO_3$,which is pyramidal. This is correct.
- In all these compounds $(XeOF_4, XeO_2F_2, XeO_3)$,the oxidation state of $Xe$ is $+6$. This is correct.
Therefore,the incorrect statement is $B$.

(C) The partial hydrolysis of $XeF_6$ occurs in steps:
$1. XeF_6 + H_2O \to XeOF_4 + 2HF$
Here,$X$ is $XeOF_4$. The oxidation state of $Xe$ is $x + (-2) + 4(-1) = 0$,so $x = +6$.
$2. XeOF_4 + H_2O \to XeO_2F_2 + 2HF$
Here,$Y$ is $XeO_2F_2$. The oxidation state of $Xe$ is $x + 2(-2) + 2(-1) = 0$,so $x = +6$.
Thus,the compounds are $XeOF_4 (+6)$ and $XeO_2F_2 (+6)$.

(C) The partial hydrolysis of $XeF_6$ is given by the reaction: $XeF_6 + H_2O \to XeOF_4 + 2HF$.
Similarly,$XeF_6$ reacts with silica $(SiO_2)$ to form $XeOF_4$ and $SiF_4$: $2XeF_6 + SiO_2 \to 2XeOF_4 + SiF_4$.
Thus,the compound '$X$' is $XeOF_4$ (Xenon oxytetrafluoride).

(D) The hydrolysis of xenon fluorides ($XeF_4$ and $XeF_6$) yields various xenon-oxo compounds.
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$
$XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF$
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
$XeO_4$ (Xenon tetroxide) is an unstable compound and cannot be prepared by the direct hydrolysis of xenon fluorides. It is typically prepared by the reaction of barium perxenate $(Ba_2XeO_6)$ with concentrated sulfuric acid $(H_2SO_4)$.

(A) $XeF_4$ has a square planar structure because it has $4$ bond pairs and $2$ lone pairs on the central $Xe$ atom,resulting in $sp^3d^2$ hybridization.
$NH_4^+$,$BF_4^-$,and $CCl_4$ all have $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridization and a tetrahedral structure.

(A) To determine the hybridisation and lone pairs of $Xe$ in $XeOF_4$:
$1$. The central atom is $Xe$,which has $8$ valence electrons.
$2$. $Xe$ forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom.
$3$. Total number of electron pairs around $Xe$ = (Number of sigma bonds) + (Number of lone pairs).
$4$. $Xe$ forms $5$ sigma bonds ($4$ with $F$ and $1$ with $O$) and $1$ pi bond with $O$.
$5$. The number of lone pairs on $Xe$ = (Total valence electrons - Electrons used in bonding) / $2$ = $(8 - (4 \times 1 + 1 \times 2)) / 2 = (8 - 6) / 2 = 1$ lone pair.
$6$. Steric number = (Number of sigma bonds) + (Number of lone pairs) = $5 + 1 = 6$.
$7$. $A$ steric number of $6$ corresponds to $sp^3d^2$ hybridisation.
$8$. Therefore,the hybridisation is $sp^3d^2$ and the number of lone pairs is $1$.

(B) The noble gases are $He$,$Ne$,$Ar$,$Kr$,$Xe$,and $Rn$.
All noble gases except $Rn$ (Radon) are present in the atmosphere.
$Rn$ is a radioactive element formed by the radioactive decay of radium $(^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^{4}_{2}He)$.
Therefore,$Rn$ does not occur in the atmosphere.

(A) The structures of the given xenon compounds are as follows:
$A$. $XeF_4$: It has $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar geometry.
$B$. $XeF_6$: It has $sp^3d^3$ hybridization with $1$ lone pair,resulting in a distorted octahedral geometry.
$C$. $XeO_3$: It has $sp^3$ hybridization with $1$ lone pair,resulting in a pyramidal geometry.
$D$. $XeOF_2$: It has $sp^3d$ hybridization with $2$ lone pairs,resulting in a $T$-shaped geometry.
Therefore,the correct matching is $A \to 4, B \to 3, C \to 1, D \to 2$.

(B) Neil Bartlett observed that the first ionization enthalpy of $Xe$ is quite close to that of $O_2$. He prepared a red-colored compound by reacting $Xe$ with $PtF_6$,which he formulated as $Xe^+[PtF_6]^-$. Thus,$XePtF_6$ was the first noble gas compound prepared by Bartlett.

(B) $PF_5$ acts as a fluoride ion acceptor.
It reacts with $XeF_2$ to form an ionic adduct.
The reaction is: $XeF_2 + PF_5 \rightarrow [XeF]^+ [PF_6]^-$.

(B) Solid $XeF_6$ exists as an ionic compound $[XeF_5]^+ [F]^-$.
In the cationic part $[XeF_5]^+$,the central $Xe$ atom is bonded to $5$ fluorine atoms and has $1$ lone pair.
The total number of electron pairs around $Xe$ is $5 + 1 = 6$.
Therefore,the hybridisation is $sp^3d^2$.

(C) The central atom in $I_3^-$ is $I$,which has $7$ valence electrons. It forms $2$ bonds with other $I$ atoms and carries a negative charge,resulting in $3$ lone pairs on the central $I$ atom.
$Xe$ has $8$ valence electrons.
In $XeF_2$,the $Xe$ atom forms $2$ covalent bonds with $F$ atoms,leaving $8 - 2 = 6$ electrons,which form $3$ lone pairs.
In $XeF_4$,$Xe$ forms $4$ bonds,leaving $2$ lone pairs.
In $XeO_3$,$Xe$ forms $6$ bonds (double bonds with $O$),leaving $1$ lone pair.
In $XeO_4$,$Xe$ forms $8$ bonds,leaving $0$ lone pairs.
Therefore,$XeF_2$ has the same number of lone pairs as $I_3^-$.

(A) $XeF_4$ has $sp^3d^2$ hybridisation and has $2$ lone pairs.
Due to the presence of $2$ lone pairs and $4$ bond pairs,the geometry is octahedral,but the shape is square planar.

(D) The boiling points of noble gases increase from $He$ to $Xe$ due to the increase in the magnitude of van der Waals forces of attraction.
These forces are directly proportional to the size of the atom.
As the atomic size increases from $He$ to $Xe$,the polarisability of the electron cloud increases.
Therefore,the increase in polarisability leads to stronger dispersion forces,resulting in higher boiling points.

(C) Xenon is a noble gas that forms compounds primarily with highly electronegative elements like fluorine and oxygen.
$XeF_4$,$XeO_3$,and $XeO_2F_2$ are well-known stable compounds of xenon.
However,$XeCl_4$ cannot be formed because chlorine is not sufficiently electronegative to stabilize the xenon-chlorine bond,and the large size of the chlorine atom leads to steric hindrance and instability in the coordination sphere of xenon.
Therefore,the correct option is $C$.

(C) The noble gases are separated from a mixture by using coconut charcoal at different temperatures.
This method relies on the principle of selective adsorption,where the extent of adsorption on coconut charcoal increases with the increase in the atomic weight of the noble gases.
Helium $(He)$,having the lowest atomic weight,is adsorbed the least,whereas Xenon $(Xe)$,having the highest atomic weight,is adsorbed the most strongly.
By varying the temperature,the gases can be desorbed sequentially,allowing for their separation.

(C) In the $I_3^-$ ion,the central $I$ atom has $3$ lone pairs of electrons.
$Xe$ has $8$ valence electrons.
In $XeF_2$,the central $Xe$ atom forms $2$ covalent bonds with $F$ atoms,leaving $6$ electrons as $3$ lone pairs.
In $XeF_4$,the central $Xe$ atom forms $4$ covalent bonds,leaving $4$ electrons as $2$ lone pairs.
In $XeO_3$,the central $Xe$ atom forms $6$ bonds with $3$ oxygen atoms,leaving $2$ electrons as $1$ lone pair.
In $XeO_4$,the central $Xe$ atom forms $8$ bonds with $4$ oxygen atoms,leaving $0$ lone pairs.
Therefore,$XeF_2$ has the same number of lone pairs as $I_3^-$.

(C) The hydrolysis reactions are as follows:
$XeF_2 + H_2O \rightarrow Xe + 2HF + \frac{1}{2} O_2$
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
From the reactions,it is clear that $XeF_2$ produces $O_2$ gas upon hydrolysis,whereas $XeF_6$ produces $XeO_3$ and $HF$ without the evolution of $O_2$.

(C) The reaction between an alkali metal fluoride $(MF)$ and xenon tetrafluoride $(XeF_4)$ results in the formation of the complex salt $M[XeF_5]$.
$MF + XeF_4 \to M^{+}[XeF_5]^{-}$.
In the $[XeF_5]^{-}$ anion,the central xenon atom has $5$ bond pairs and $2$ lone pairs of electrons,resulting in a total of $7$ electron pairs.
This corresponds to $sp^3d^3$ hybridisation.
The shape of the $[XeF_5]^{-}$ ion is pentagonal planar.