Noble gases Questions in English

(C) $XeF_6$ reacts with $SiO_2$ present in Pyrex glass.
$2XeF_6 + SiO_2 \rightarrow 2XeOF_4 + SiF_4$
$2XeOF_4 + SiO_2 \rightarrow 2XeO_2F_2 + SiF_4$

(D) Noble gases are not obtained from sea water.
This is due to the extremely low solubility of noble gases in water.

(B) The ease of liquefaction of noble gases increases down the group because the magnitude of van der Waals forces of attraction increases with an increase in the size of the atoms.
Therefore,the correct decreasing order of the ease of liquefaction is $Xe > Kr > Ar > Ne > He$.

(D) Noble gases have very low melting and boiling points due to the presence of weak van der Waals forces of attraction between their atoms.

(B) Although noble gases are chemically inert,some compounds of noble gas elements,particularly of $Xe$ and $Kr$,are known.

(B) The reaction of Xenon with Fluorine in a $1:2$ ratio at $673 \ K$ and $1 \ bar$ pressure in the presence of a $Ni$ catalyst yields Xenon difluoride:
$Xe(g) + F_2(g) \xrightarrow[673 \ K, 1 \ bar]{Ni} XeF_2(s)$
Thus,the correct option is $B$.

(B) Among the noble gases,$Xe$ shows the highest reactivity.
This is due to its low ionization enthalpy,which allows it to lose electrons more easily to form bonds with highly electronegative elements like fluorine.

(B) Sir William Ramsay is credited with the discovery of the first noble gas,$Argon$ $(Ar)$,in $1894$.

(D) Radon $(Rn)$ is not present in the atmosphere. It is a radioactive element formed by the radioactive decay of radium $(Ra)$.

(B) $Ne$ is present in the atmosphere in free state. It is not obtained from radioactive disintegration. $Rn$ is a radioactive element formed by the decay of $Ra$ and has a half-life of $3.8$ days. Thus,the statement that $Ne$ is obtained during radioactive disintegration is incorrect.

(D) $He$ has only two electrons in its outermost shell,which corresponds to a doublet rather than an octet.

(A) In noble gases,as we move down the group,the addition of successive electronic shells increases the atomic size,which results in a decrease in the ionization enthalpy.

(D) Helium was first discovered in the sun's atmosphere and subsequently found on the earth.

(A) The statement that $Helium$ is used in balloons because it is lighter than $Hydrogen$ is incorrect. $Helium$ is actually heavier than $Hydrogen$ ($Molar \ mass \ of \ He = 4 \ g/mol$ while $Molar \ mass \ of \ H_2 = 2 \ g/mol$). However,it is preferred over $Hydrogen$ because it is non-inflammable and safe,whereas $Hydrogen$ is highly inflammable.

(B) The number of lone pairs on the central $Xe$ atom can be calculated using the formula: $\text{Lone pairs} = \frac{1}{2} (V - M)$,where $V$ is the number of valence electrons of the central atom and $M$ is the number of monovalent atoms attached.
For $XeF_2$: $V = 8$,$M = 2$. $\text{Lone pairs} = \frac{1}{2} (8 - 2) = 3$.
For $XeF_4$: $V = 8$,$M = 4$. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.
For $XeF_6$: $V = 8$,$M = 6$. $\text{Lone pairs} = \frac{1}{2} (8 - 6) = 1$.
Thus,the number of lone pairs is $3, 2, 1$.

(A) Let us analyze the given reactions:
$1$. $XeO_3 + 6HF \to XeF_6 + 3H_2O$ is an incorrect reaction because $XeO_3$ is a stable oxide and does not react with $HF$ to form $XeF_6$.
$2$. $3XeF_4 + 6H_2O \to 2Xe + XeO_3 + 12HF + 1.5O_2$ is a correct disproportionation reaction of $XeF_4$ with water.
$3$. $2XeF_2 + 2H_2O \to 2Xe + 4HF + O_2$ is a correct reaction of $XeF_2$ with water.
$4$. $XeF_6 + RbF \to Rb[XeF_7]$ is a correct reaction showing the fluoride ion donor-acceptor behavior of $XeF_6$.

(C) Electronegativity of an atom increases with an increase in the $s$-character of its hybrid orbitals.
In $XeF_2$,$Xe$ is $sp^3d$ hybridized ($20\% \ s$-character).
In $XeF_4$,$Xe$ is $sp^3d^2$ hybridized ($16.6\% \ s$-character).
In $XeO_3$,$Xe$ is $sp^3$ hybridized ($25\% \ s$-character).
In $XeOF_2$,$Xe$ is $sp^3d$ hybridized ($20\% \ s$-character).
Since $sp^3$ hybridization has the highest $s$-character $(25\%)$,the $Xe$ atom in $XeO_3$ exhibits the highest electronegativity.

(C) $(a) P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$ ($3$ products)
$(b) P_4 + 10SO_2Cl_2 \rightarrow 4PCl_5 + 10SO_2$ ($2$ products)
$(c) 6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$ ($4$ products)
$(d) NH_4NO_3 + 4Zn + 7NaOH \rightarrow NH_3 + 4Na_2ZnO_2 + 2H_2O$ ($3$ products)
Comparing the number of products,option $(c)$ yields the maximum number of products.

(B) The hydrolysis reactions of Xenon fluorides are as follows:
$1. XeF_2 + H_2O \longrightarrow Xe + \frac{1}{2}O_2 + 2HF$ (Does not produce $XeO_3$)
$2. 6XeF_4 + 12H_2O \longrightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$ (Produces $XeO_3$)
$3. XeF_6 + 3H_2O \longrightarrow XeO_3 + 6HF$ (Produces $XeO_3$)
$4. XeO_2F_2 + H_2O \longrightarrow XeO_3 + 2HF$ (Produces $XeO_3$)
Thus,$XeF_2$ is the compound that does not produce explosive $XeO_3$ upon complete hydrolysis.

(B) The partial hydrolysis of $XeF_6$ proceeds as follows:
$1$. $XeF_6 + H_2O \rightarrow XeOF_4 (A) + 2HF$
$2$. $XeOF_4 + H_2O \rightarrow XeO_2F_2 (B) + 2HF$
$3$. $XeO_2F_2 + H_2O \rightarrow XeO_3 (C) + 2HF$
In $XeOF_4 (A)$,the hybridization is $sp^3d^2$ (square pyramidal).
In $XeO_2F_2 (B)$,the hybridization is $sp^3d$ (see-saw).
In $XeO_3 (C)$,the hybridization is $sp^3$ (pyramidal).
Option $B$ states $B$ is $XeOF_2$,which is incorrect as $B$ is $XeO_2F_2$.

(A) The reaction of Xenon with Fluorine under specific conditions of temperature and pressure yields different Xenon fluorides.
When $Xe$ is reacted with $F_2$ in a $1:5$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure,the product formed is $XeF_6$.
The balanced chemical equation is:
$Xe(g) + 3F_2(g) \xrightarrow{573 \ K, 60-70 \ bar} XeF_6(s)$.

(B) The ease of liquefaction of gases is directly proportional to their critical temperature $(T_c)$.
As we move down the group from $He$ to $Rn$,the atomic size increases,leading to stronger van der Waals forces of attraction,which results in a higher critical temperature.
The critical temperatures of noble gases are: $He$ $(5.1 \ K)$,$Ne$ $(44.4 \ K)$,$Ar$ $(150.7 \ K)$,$Kr$ $(209.4 \ K)$,and $Xe$ $(289.7 \ K)$.
Therefore,the order of ease of liquefaction is $Xe > Kr > Ar > Ne > He$.

(C) For $XeF_5^-$,the number of valence electrons on $Xe$ is $8$. The charge is $-1$,so total valence electrons $= 8 + 5 + 1 = 14$.
Number of electron pairs $= 14 / 2 = 7$.
This corresponds to $sp^3d^3$ hybridisation.
The geometry is pentagonal planar with two non-bonding electron pairs,one above the plane and the other below the plane.

(D) Noble gases (Group $18$ elements) have a general valence shell electronic configuration of $ns^2 np^6$ (except Helium which is $1s^2$).
This configuration represents a stable octet (or duplet for Helium),which is highly stable.
Due to this stable electronic configuration,they have very high ionization enthalpy and low electron gain enthalpy,making them chemically inert under normal conditions.

(B) Xenon is a noble gas that forms fluorides with fluorine. The known stable fluorides of xenon are $XeF_2$,$XeF_4$,and $XeF_6$.
$XeF_3$ is not a stable compound because xenon typically forms fluorides with an even number of fluorine atoms due to the nature of its bonding and the promotion of electrons into $d$-orbitals.
Therefore,$XeF_3$ is impossible.

(A) Clathrates are a type of inclusion compound in which molecules of one substance are trapped within the crystal lattice of another substance.
These are often referred to as $Cage \ compounds$ because the host molecules form a cage-like structure that encloses the guest molecules.
Noble gases like $Ar$,$Kr$,and $Xe$ form such compounds with substances like quinol or water.

(C) The atmosphere contains several noble gases,but $Argon$ $(Ar)$ is the most abundant among them.
It constitutes approximately $0.93\%$ by volume of the Earth's atmosphere.

(C) The noble gases are present in the atmosphere in very small amounts. Among them,Argon $(Ar)$ is the most abundant noble gas in the Earth's atmosphere,accounting for approximately $0.93\%$ by volume.

(B) Noble gases (Group $18$ elements) have completely filled valence shells ($ns^2 np^6$ configuration,except $He$ which is $1s^2$).
Due to their stable electronic configuration,they have very high ionization enthalpy and zero electron gain enthalpy.
Consequently,they show very low chemical reactivity under normal conditions.

(A) $1$. In $XeO_3$,the structure is pyramidal with $Xe$ at the center. It has three $Xe=O$ double bonds and one lone pair on $Xe$. Each double bond consists of one $\sigma$ and one $\pi$ bond. Thus,$XeO_3$ has three $\sigma$ and three $\pi$ bonds. The statement in option $A$ is incorrect.
$2$. In $XeF_4$,$Xe$ has $8$ valence electrons,$4$ are used for bonding with $F$ atoms,and $2$ lone pairs remain. Steric number = $4$ (bond pairs) + $2$ (lone pairs) = $6$,which corresponds to $sp^3d^2$ hybridization. This is correct.
$3$. Argon is the most abundant noble gas in the atmosphere (approx $0.93\%$ by volume). This is correct.
$4$. Liquid $He$ has a very low boiling point $(4.2 \ K)$ and is widely used as a cryogenic agent. This is correct.

(A) Dry air consists of various gases. The composition of noble gases in dry air is dominated by Argon $(Ar)$.
The percentage of Argon $(Ar)$ in dry air by volume is approximately $0.93 \%$.

(C) Noble gases have stable electronic configurations $(ns^2 np^6)$,which results in very high ionization energy and almost zero electron affinity. Due to weak van der Waals forces,they have low boiling points and do not liquefy easily. However,the statement that they do not form any chemical compounds is incorrect,as noble gases like $Xe$ and $Kr$ are known to form compounds with highly electronegative elements like $F$ and $O$.

(D) Helium $(He)$ has the lowest boiling point $(4.2 \ K)$ among all known substances. This is because helium atoms are very small and have a very stable electronic configuration $(1s^2)$. Due to this,the interatomic forces of attraction,known as van der Waals forces,are extremely weak. These weak forces are easily overcome by thermal energy,resulting in a very low boiling point.

(C) The reaction of xenon hexafluoride $(XeF_6)$ with silica $(SiO_2)$ is a partial hydrolysis reaction.
The balanced chemical equation is:
$2XeF_6 + SiO_2 \rightarrow 2XeOF_4 + SiF_4$
Thus,the products formed are xenon oxytetrafluoride $(XeOF_4)$ and silicon tetrafluoride $(SiF_4)$.

(D) To determine the number of lone pairs on $Xe$,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$(i) XeO_3$: $Xe$ has $8$ valence electrons. Oxygen is divalent,so $M=0$. $\text{Lone pairs} = \frac{1}{2} (8 - 0) = 4$ electrons,which corresponds to $1$ lone pair.
$(ii) XeOF_4$: $Xe$ has $8$ valence electrons. $F$ is monovalent $(M=4)$,$O$ is divalent. $\text{Lone pairs} = \frac{1}{2} (8 - 4) = 2$ electrons,which corresponds to $1$ lone pair.
$(iii) XeF_6$: $Xe$ has $8$ valence electrons. $F$ is monovalent $(M=6)$. $\text{Lone pairs} = \frac{1}{2} (8 - 6) = 2$ electrons,which corresponds to $1$ lone pair.
All three molecules $(i), (ii),$ and $(iii)$ have $1$ lone pair on the $Xe$ atom. Therefore,the correct option is $(D)$.

(A) Lord Rayleigh observed that nitrogen obtained from the atmosphere is slightly denser than nitrogen obtained from chemical sources like ammonia or urea.
This difference in density is due to the presence of noble gases,primarily $Argon$ $(Ar)$,in atmospheric nitrogen.
Since $Argon$ has a higher atomic mass $(39.95 \ u)$ compared to $Nitrogen$ $(28.01 \ u)$,its presence increases the overall density of the atmospheric nitrogen sample.

(B) Among the noble gases,$Xe$ (Xenon) has the lowest ionization enthalpy due to its large atomic size.
This allows it to react with highly electronegative elements like fluorine and oxygen to form a wide variety of stable compounds such as $XeF_2$,$XeF_4$,$XeF_6$,$XeO_3$,and $XeOF_4$.
Therefore,$Xe$ forms the largest number of compounds among noble gases.

(C) Neon gas is widely used in discharge tubes for advertising purposes because it emits a characteristic bright red-orange glow when an electric current is passed through it at low pressure.

(B) To determine the geometry of the given $Xe$ compounds,we look at their hybridization and lone pairs:
$1$. $XeO_4$: $sp^3$ hybridization,tetrahedral geometry (non-planar).
$2$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,square planar geometry (planar).
$3$. $XeOF_4$: $sp^3d^2$ hybridization with $1$ lone pair,square pyramidal geometry (non-planar).
$4$. $XeO_3$: $sp^3$ hybridization with $1$ lone pair,pyramidal geometry (non-planar).
Therefore,$XeF_4$ is the only planar molecule.

(C) The hydrolysis of $XeF_4$ with water is a disproportionation reaction.
The chemical equation is: $6XeF_4 + 12H_2O \rightarrow 2Xe + 4XeO_3 + 24HF + 3O_2$.
Thus,the products formed are $Xe$,$XeO_3$,$HF$,and $O_2$.

(D) $1$. Calculate the moles of each element in $100 \ g$ of the compound:
Moles of $Xe = \frac{63.8}{133} \approx 0.48 \ mol$.
Moles of $F = \frac{100 - 63.8}{19} = \frac{36.2}{19} \approx 1.90 \ mol$.
$2$. Determine the molar ratio:
Ratio of $Xe : F = 0.48 : 1.90 \approx 1 : 4$.
$3$. The empirical formula is $XeF_4$.
$4$. In $XeF_4$,the oxidation state of $Xe$ is $x + 4(-1) = 0$,so $x = +4$.

(A) In the perxenate ion $[XeO_6]^{4-}$,the oxidation state of Xenon $(Xe)$ is calculated as follows: $x + 6(-2) = -4$,which gives $x = +8$.
Since the maximum oxidation state of Xenon is $+8$,all oxygen atoms are bonded to Xenon via single bonds.
The structure of $[XeO_6]^{4-}$ is an octahedral geometry where Xenon is at the center and six oxygen atoms are attached to it.
There are no oxygen-oxygen $(O-O)$ bonds present in this structure.
Therefore,the number of peroxide bonds is $0$.

(B) In $[XeO_6]^{4-}$,the oxidation state of $Xe$ is $x + 6(-2) = -4$,so $x = +8$.
In $XeF_8$,the oxidation state of $Xe$ would be $+8$,but this compound is sterically hindered and does not exist.
In $OsO_4$,the oxidation state of $Os$ is $x + 4(-2) = 0$,so $x = +8$.
In $RuO_4$,the oxidation state of $Ru$ is $x + 4(-2) = 0$,so $x = +8$.
Since $XeF_8$ is not a stable compound,it does not exhibit the $+8$ oxidation state.

(C) In $XeO_4$,the central atom $Xe$ is in its $8^{th}$ oxidation state.
It undergoes $sp^{3}$ hybridization involving one $5s$ and three $5p$ orbitals.
These four $sp^{3}$ hybrid orbitals form four $\sigma$ bonds with four oxygen atoms.
Additionally,the four unpaired electrons in the $5d$ orbitals of $Xe$ form four $p\pi-d\pi$ bonds with the $2p$ orbitals of the oxygen atoms.
Therefore,the molecule contains four $p\pi-d\pi$ bonds.

(C) Complete hydrolysis of $XeF_6$ yields $XeO_3$.
$XeF_6 + 3H_2O \to XeO_3 + 6HF$

(D) Activated charcoal is a very good adsorbent of gases due to its high surface area and porous structure.
In contrast,anhydrous $CaCl_2$,$Fe(OH)_3$,and conc. $H_2SO_4$ are primarily used as dehydrating agents to remove moisture.

(B) The central atom in $XeOF_4$ is Xenon $(Xe)$.
$Xe$ has $8$ valence electrons.
In $XeOF_4$,$Xe$ forms $4$ single bonds with $4$ Fluorine atoms and $1$ double bond with $1$ Oxygen atom.
Total electrons used in bonding = $4 + 2 = 6$ electrons.
Remaining electrons on $Xe$ = $8 - 6 = 2$ electrons.
These $2$ electrons form $1$ lone pair.
Therefore,the total number of lone pair of electrons on the central atom $Xe$ is $1$.