Noble gases Questions in English

(C) The reaction between $Xe$ and $O_2F_2$ is: $Xe + 2 O_2F_2 \rightarrow XeF_4 + 2 O_2$.
Thus,the compound $P$ is $XeF_4$.
The complete hydrolysis of $XeF_4$ proceeds as follows: $XeF_4 + 2 H_2O \rightarrow XeO_2F_2 + 4 HF$.
Therefore,the complete hydrolysis of $1 \ mol$ of $XeF_4$ produces $4 \ moles$ of $HF$.

(A) $XeO_3$: Hybridization is $sp^3$ with $1$ lone pair,resulting in a pyramidal geometry.
$XeF_2$: Hybridization is $sp^3 d$ with $3$ lone pairs,resulting in a linear geometry.
$XeOF_4$: Hybridization is $sp^3 d^2$ with $1$ lone pair,resulting in a square pyramidal geometry.
$XeF_6$: Hybridization is $sp^3 d^3$ with $1$ lone pair,resulting in a distorted octahedral geometry.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(B) Given the ratio $n_1 : n_2 = 1$,the compound is $XeO_2F_2$.
In $XeO_2F_2$,the central atom $Xe$ has $8$ valence electrons.
It forms $2$ single bonds with $F$ and $2$ double bonds with $O$,utilizing $6$ electrons for bonding.
This leaves $2$ electrons,forming $1$ lone pair.
The total number of electron domains is $5$ ($4$ bonding pairs + $1$ lone pair),resulting in a see-saw geometry.
$XeO_2F_2$ is a polar and non-planar molecule.
Therefore,the statement that it is a planar compound is incorrect.

(B) Noble gases have stable electronic configurations ($ns^2 np^6$,except $He$ which is $1s^2$),which makes them chemically inert.
Due to their stable configuration,they have weak dispersion forces between their atoms,resulting in very low melting and boiling points.
They are sparingly soluble in water.
They have large positive values of electron gain enthalpy because they do not have any tendency to accept an electron.
Therefore,the statement that noble gases have very high melting and boiling points is incorrect.

(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.
As the atomic size increases down the group,the electron cloud becomes more diffuse and the polarisability of the atoms increases.
Greater polarisability leads to stronger instantaneous dipole-induced dipole interactions,which results in higher boiling points.

(A) $XeO_3$ has a trigonal pyramidal shape,not trigonal bipyramidal.
$ClF_3$ has a $T$-shape.
$XeOF_4$ has a square pyramidal shape.
$XeF_2$ has a linear shape.
Therefore,the incorrect match is $XeO_3$.

(D) $At$ (Astatine),$Po$ (Polonium),and $Rn$ (Radon) are radioactive elements belonging to the heavier groups of the periodic table.
$Ar$ (Argon) is a noble gas and is a stable,non-radioactive element.

(D) The valence shell of $Xe$ has $8$ electrons.
In $XeF_2$,$Xe$ forms $2$ bonds with $F$ atoms,leaving $6$ electrons ($3$ lone pairs).
In $XeF_4$,$Xe$ forms $4$ bonds with $F$ atoms,leaving $4$ electrons ($2$ lone pairs).
In $XeOF_2$,$Xe$ forms $2$ bonds with $F$ and $2$ bonds with $O$ (total $4$ bonds),leaving $4$ electrons ($2$ lone pairs).
In $XeOF_4$,$Xe$ forms $4$ bonds with $F$ and $2$ bonds with $O$ (total $6$ bonds),leaving $2$ electrons ($1$ lone pair).
Therefore,$XeOF_4$ is the compound where $Xe$ has one lone pair of electrons.

(A) Liquid $He$ (Helium) is used as a coolant in the superconducting magnets of $MRI$ (Magnetic Resonance Imaging) scanners to maintain the extremely low temperatures required for superconductivity.

(C) The chemical formula for xenon monooxytetrafluoride is $XeOF_4$.
Let the oxidation state of xenon be $x$.
The oxidation state of oxygen is $-2$ and the oxidation state of fluorine is $-1$.
Setting the sum of oxidation states equal to the charge of the molecule $(0)$:
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation state of xenon in $XeOF_4$ is $+6$.

(D) The structure of $ClF_5$ is square pyramidal,which is based on $sp^3d^2$ hybridization with one lone pair of electrons on the central chlorine atom.
$XeOF_4$ has the central xenon atom bonded to four fluorine atoms and one oxygen atom (double bond).
Using the $VSEPR$ theory,the steric number for $Xe$ in $XeOF_4$ is $5$ (four single bonds + one double bond) plus one lone pair,resulting in $sp^3d^2$ hybridization.
Thus,$XeOF_4$ also exhibits a square pyramidal geometry,making it isostructural with $ClF_5$.

(A) Xenon $(Xe)$ has a large atomic size and lower ionization enthalpy compared to $He$,$Ne$,$Ar$,and $Kr$.
Due to its lower ionization energy,it can easily form compounds with highly electronegative elements like fluorine and oxygen.
Consequently,xenon exhibits the highest number of different oxidation states (e.g.,$+2, +4, +6, +8$) among the noble gases.

(A) Electric bulbs are filled with chemically inert gases to prevent the oxidation of the tungsten filament at high temperatures. $Ar$ (Argon) and $N_2$ (Nitrogen) are commonly used for this purpose because they are non-reactive and help prolong the life of the filament.

(C) Ionization enthalpy decreases on moving down the group,and $Xe$ has the lowest ionization enthalpy among the given noble gases.
Due to this low ionization energy,$Xe$ can easily form stable crystalline fluorides such as $XeF_2$,$XeF_4$,and $XeF_6$ upon reaction with fluorine.

(C) Argon is used in the welding of specialty alloys as well as in the welding of automobile frames,mufflers,and other automotive parts. It is called a shield gas because it does not react with the gases and metals in the vicinity of the welding process. It merely takes up space and prevents unwanted reactions from occurring due to reactive gases such as $N_2$ and $O_2$.

(C) The filament in an electric bulb is made using tungsten,which would burn quickly when it comes in contact with oxygen in the air.
This is the reason why unreactive gases such as $Ar$ and $N_2$ are used for filling electric bulbs.
These gases are chemically inactive and prevent the oxidation of the tungsten filament,thereby increasing the lifespan of the bulb.

(D) Radon $(Rn)$ is a radioactive element.
Due to its radioactive nature,it is used in radiotherapy for the treatment of cancer.

(B) Krypton $(Kr)$ is the noble gas used in miner's cap lamps because it provides a bright,white light and has a long filament life.

(B) The inert gases $Xe$ (Xenon) and $Kr$ (Krypton) are commonly used in high-intensity flash bulbs for photography because they produce a bright white light when an electric discharge is passed through them.

(C) $Helium$ $(He)$ is the inert gas used for filling balloons because it is much lighter than air and is chemically inert,making it safe to use.

(D) Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater.
Increased pressure increases the solubility of atmospheric gases in the blood.
When divers ascend towards the surface,the pressure gradually decreases,which can cause nitrogen bubbles to form in the blood,a condition known as 'bends' or decompression sickness.
To avoid this,scuba divers use tanks filled with air diluted with helium (typically $11.7 \% \text{ helium}, 56.2 \% \text{ nitrogen, and } 32.1 \% \text{ oxygen}$).
Helium is used because it is less soluble in blood than nitrogen.
Thus,option $(d)$ is correct.

(A) Argon $(Ar)$ is a noble gas belonging to Group $18$.
Noble gases exist as monoatomic gases because they have a stable valence shell configuration $(ns^2 np^6)$.
Oxygen $(O_2)$,Nitrogen $(N_2)$,and Bromine $(Br_2)$ exist as diatomic molecules in their standard states.

(B) Argon $(Ar)$ is the most abundant noble gas present in the atmosphere,comprising approximately $0.93\%$ by volume of dry air.

(C) Xenon forms the following stable fluorides: $XeF_{2}$,$XeF_{4}$,and $XeF_{6}$.
$XeF_{5}$ is not a known stable compound of xenon.

(B) The incorrectly matched pair is $XeF_4$: Tetrahedral.
$XeF_4$ has a square planar structure due to $sp^3d^2$ hybridization with two lone pairs of electrons occupying the axial positions.

(C) The nuclear reaction between ${}_{98}^{249}Cf$ and ${}_{20}^{48}Ca$ is given by:
${}_{98}^{249}Cf + {}_{20}^{48}Ca \longrightarrow {}_{118}^{294}Og + 3{}_0^1n$
Here,${}_{118}^{294}Og$ represents Oganesson,which is a synthetic radioactive noble gas with atomic number $Z = 118$.

(C) The group $18$ elements (noble gases) have completely filled valence subshells ($ns^2 np^6$,except $He$ which is $1s^2$).
Due to the presence of paired electrons in all orbitals,these elements are diamagnetic.
The naturally occurring group $18$ elements are $He, Ne, Ar, Kr, Xe,$ and $Rn$.
Since $Rn$ is radioactive,the number of naturally occurring stable $p$-block elements that are diamagnetic is $5$ $(He, Ne, Ar, Kr, Xe)$.

(D) In Ramsay and Rayleigh's isolation of noble gases from air,the nitrogen of the air is finally converted into $NaNO_{2}$ and $NaNO_{3}$.
$N_{2} + O_{2} \xrightarrow{\text{Electric discharge}} 2 NO$
$2 NO + O_{2} \longrightarrow 2 NO_{2(g)}$
$2 NO_{2} + 2 NaOH \longrightarrow NaNO_{2} + NaNO_{3} + H_{2}O$

(D) Among the noble gases,$He$ (Helium) has the least tendency to form compounds.
This is because $He$ has the highest first ionization enthalpy among all elements in the periodic table,making it extremely stable and chemically inert.

(C) $XeF_{6}$ on partial hydrolysis with water gives $XeOF_{4}$.
The reaction involved is as follows:
$XeF_{6} + H_{2}O \rightarrow XeOF_{4} + 2HF$
The structure of $XeOF_{4}$ is square pyramidal.

(A) The molar mass of $H_2O$ is $18 \ g/mol$.
Given mass of $H_2O = 1.8 \ g$.
Number of moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$.
The balanced chemical equation for the reaction of $XeF_6$ with $H_2O$ in a $1:1$ molar ratio is:
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$.
Since $0.1 \ mol$ of $XeF_6$ reacts with $0.1 \ mol$ of $H_2O$,the product formed is $XeOF_4$.

(B) The molecule $XeF_6$ has $7$ electron pairs around the central $Xe$ atom ($6$ bond pairs and $1$ lone pair),which corresponds to $sp^3d^3$ hybridization.
According to $VSEPR$ theory,the presence of one lone pair on the $Xe$ atom causes distortion in the regular octahedral geometry.
Therefore,the shape of $XeF_6$ is described as distorted octahedral.

(C) The hybridization of $Xe$ in $XeO_3$ is $sp^3$,which involves three bond pairs and one lone pair.
Due to the presence of one lone pair,$XeO_3$ adopts a pyramidal geometry.
$XeOF_4$ has a square pyramidal geometry,$XeF_2$ is linear,and $XeF_4$ is square planar.

(D) Xenon $(Xe)$ is used in discharge tubes for producing a high-speed flash of bluish light,which is used in high-speed photography.

(A) Noble gases exist as monoatomic molecules.
Since a monoatomic molecule consists of only a single atom,it has only translational degrees of freedom and lacks rotational or vibrational degrees of freedom.
Therefore,noble gases do not possess vibrational energy.

(C) Argon is an inert gas. It is primarily used in high temperature welding and other metallurgical operations that require a non-oxidising atmosphere and the absence of nitrogen to prevent unwanted chemical reactions.

(A) Helium $(He)$ is a non-flammable (incombustible) gas.
Its lifting power is $93 \%$ as compared to flammable hydrogen gas.
Due to these reasons,it is safer and preferred for filling balloons and other lighter-than-air crafts.

(D) The separation of noble gases is based on the principle of selective adsorption on coconut charcoal at different temperatures.
At $173 \ K$,$Ar$,$Kr$,and $Xe$ are adsorbed on the surface of the coconut charcoal.
However,$He$ and $Ne$ have very low boiling points and weak van der Waals forces,so they are not adsorbed at this temperature.
Therefore,the gases that remain in the gaseous phase are $He$ and $Ne$.

(D) $He$ (helium) has an unusual property of diffusing through materials such as rubber,glass,or plastic.
This is due to its very small atomic size.
The increasing order of diffusibility of noble gases is $Xe < Kr < Ar < Ne < He$.

(A) $Xe$ is the most easily liquefiable rare gas.
This is because the magnitude of van der Waals forces (interatomic interactions) increases with an increase in atomic size and atomic number,which facilitates easier liquefaction.

(D) The $6$th period corresponds to the filling of the $6s, 4f, 5d,$ and $6p$ orbitals.
The last element of the $6$th period is the noble gas Radon $(Rn)$,which has an atomic number of $86$.
The electronic configuration of Radon is $[Xe] 4f^{14} 5d^{10} 6s^2 6p^6$.
Therefore,the outermost electronic configuration is $4f^{14}, 5d^{10}, 6s^2, 6p^6$.

(A) For $XeF_2$: The central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds with $F$ atoms and has $3$ lone pairs. The steric number is $2 + 3 = 5$,which corresponds to $sp^3d$ hybridization. Due to the presence of $3$ lone pairs in the equatorial positions,the shape is linear.
For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to the presence of $2$ lone pairs in the axial positions,the shape is square planar.
Thus,$XeF_2$ is linear and $XeF_4$ is square planar.

(A) The complete hydrolysis of $XeF_6$ is represented by the following chemical equation:
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
Here,the compound $X$ is $XeO_3$.
In $XeO_3$,the central atom $Xe$ has $8$ valence electrons.
It forms $3$ double bonds with $3$ oxygen atoms,utilizing $6$ electrons,and has $1$ lone pair of electrons.
Total electron pairs = $3$ (bond pairs) + $1$ (lone pair) = $4$.
Therefore,the hybridisation of $Xe$ in $XeO_3$ is $sp^3$.

(B) Xenon reacts with fluorine under different conditions to form $XeF_2$,$XeF_4$,and $XeF_6$.
The reaction for the formation of $XeF_4$ is given by:
$Xe_{(g)} + 2 F_{2(g)} \xrightarrow{873 \text{ K}, 7 \text{ bar }} XeF_{4(s)}$
In this reaction,$1 \text{ mole}$ of $Xe$ reacts with $2 \text{ moles}$ of $F_2$ to form $XeF_4$.
However,in the industrial preparation of $XeF_4$,a mixture of $Xe$ and $F_2$ in a $1 : 5$ ratio is used to ensure the complete conversion of $Xe$ to $XeF_4$ and to prevent the formation of $XeF_2$ or $XeF_6$ as impurities.
Therefore,the required ratio is $1 : 5$.

(A) The hydrolysis reaction of $XeF_4$ is given by:
$6XeF_4 + 12H_2O \longrightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$
Here,'$X$' is $XeO_3$.
In $XeO_3$,the central atom $Xe$ has $8$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ lone pair.
According to $VSEPR$ theory,the hybridization is $sp^3$ with one lone pair,resulting in a pyramidal geometry.

(D) In the reaction $XeF_6 + 3 H_2O \rightarrow XeO_3 + 6 HF$,the products are $XeO_3$ and $HF$. Oxygen is not released as a product in this reaction. In the other reactions,oxygen is produced as a byproduct.

(A) $Xe$ forms fluorides with an even number of fluorine atoms.
This is because the promotion of one,two,or three electrons from the $5p$-orbital to the $5d$-orbital results in two,four,or six half-filled orbitals,respectively.
These half-filled orbitals form bonds with fluorine atoms,leading to the formation of $XeF_2$,$XeF_4$,or $XeF_6$.
Therefore,$XeF_3$ does not exist.

(B) All noble gases have completely filled valence shells,which makes them stable and chemically inert. Therefore,they do not form bonds with other atoms and exist as monoatomic gases.
Noble gases have very weak van der Waals forces of attraction between their atoms,which leads to their very low melting and boiling points.
While both statements are scientifically correct,the monoatomic nature is due to the stable electronic configuration,not due to their melting or boiling points. Thus,$R$ is not the correct explanation of $A$.

(A) Liquid helium $(b.p. \ 4.2 \ K)$ is used as a cryogenic agent for carrying out various experiments at low temperatures.
It is used to produce and sustain powerful superconducting magnets,which are essential components of modern $NMR$ spectrometers and Magnetic Resonance Imaging $(MRI)$ systems for clinical diagnosis.
Helium is a light,inert,and non-flammable gas.
Therefore,statements $(i)$ and $(ii)$ are correct.

(B) $Xenon$ is a noble gas with a high ionization enthalpy.
It reacts primarily with highly electronegative elements like fluorine and oxygen to form compounds such as $XeF_2$,$XeF_4$,$XeF_6$,and $XeO_3$.
Neil Bartlett first demonstrated this by reacting $Xenon$ with $PtF_6$,a very strong oxidizing agent,to form $[Xe]^+[PtF_6]^-$.
Therefore,$Xenon$ reacts best with the most electronegative elements.