Noble gases Questions in English

(B) The structure of $XeF_4$ is square planar due to the presence of two lone pairs on the central $Xe$ atom,which occupy the axial positions to minimize repulsion.
$XeF_4$ acts as a fluoride donor when it reacts with strong Lewis acids like $SbF_5$,$PF_5$,or $AsF_5$ to form $[XeF_3]^+$ and $[MF_6]^-$ ions.
Specifically,$XeF_4 + SbF_5 \rightarrow [XeF_3]^+[SbF_6]^-$.
Among the given options,$B$ correctly identifies the geometry as square planar and the property of acting as a fluoride donor with Lewis acids like $PF_5$.

(B) When $XeF_6$ dissolves in anhydrous $HF$,it acts as a fluoride ion donor.
The reaction is: $XeF_6 + HF \to XeF_5^+ + HF_2^-$.
This results in the formation of $XeF_5^+$ and $HF_2^-$ ions,which makes the solution conducting.

(D) Helium $(He)$ has a very small atomic size and high ionization energy,which makes it chemically inert. It does not form clathrate compounds because its size is too small to be trapped in the cavities of the host lattice (like quinol). Other noble gases like $Ar$,$Kr$,and $Xe$ can form clathrates.

(B) $XeF_4 + SbF_5 \to [XeF_3]^+ [SbF_6]^-$
The cation $[XeF_3]^+$ has $sp^3d$ hybridization with a $T$-shaped geometry.
The anion $[SbF_6]^-$ has $sp^3d^2$ hybridization with an octahedral shape.

(A) All three transformations are chemically correct and balanced:
$I.$ $XeF_6$ acts as a fluoride ion acceptor,reacting with $NaF$ to form the complex salt $Na^+[XeF_7]^-$.
$II.$ In the solid state,$PCl_5$ exists as an ionic lattice consisting of $[PCl_4]^+$ and $[PCl_6]^-$ ions.
$III.$ The hydrated aluminum ion $[Al(H_2O)_6]^{3+}$ acts as a weak acid in water,donating a proton to form $[Al(H_2O)_5OH]^{2+}$ and $H_3O^+$.
Therefore,all three transformations are correct.

(B) The hydrolysis of $XeF_{2}$ is given by: $XeF_{2} + H_{2}O \longrightarrow Xe + \frac{1}{2} O_{2} + 2 HF$.
The hydrolysis of $XeF_{4}$ is given by: $3 XeF_{4} + 6 H_{2}O \longrightarrow XeO_{3} + 2 Xe + \frac{3}{2} O_{2} + 12 HF$.
Comparing the products,$Xe$,$O_{2}$,and $HF$ are common products of the hydrolysis of both $XeF_{2}$ and $XeF_{4}$.
However,$XeO_{3}$ is formed only during the hydrolysis of $XeF_{4}$ and not $XeF_{2}$,making it an uncommon product in the context of both reactions.

(B) The reactions are:
$XeF_6 + 3 H_2O \rightarrow XeO_3 + 6 HF$ (where $X = XeO_3$)
$XeF_6 + 2 H_2O \rightarrow XeO_2F_2 + 4 HF$ (where $Y = XeO_2F_2$)
$1$. $XeO_3$ is a highly explosive solid.
$2$. $XeO_2F_2$ is an oxyfluoride of xenon,not an oxyacid.
$3$. Both reactions involve hydrolysis,which is a non-redox process.
$4$. $XeF_6$ undergoes partial hydrolysis to form $XeOF_4$ and $XeO_2F_2$.
Therefore,the statement that $Y$ is an oxyacid of xenon is incorrect.

(B) Noble gases with very small atomic sizes,specifically $He$ and $Ne$,are too small to be trapped in the cavities of the crystal lattices of host molecules. Therefore,they do not form clathrate compounds.

(B) Noble gases are monoatomic gases because they have a stable electronic configuration and do not need to form bonds with other atoms to achieve stability. Among the given options,$Neon$ $(Ne)$ is a noble gas.

(D) Due to their extremely stable electronic configuration (completely filled $ns^2 np^6$ shells),noble gases have very high ionization enthalpy and low electron gain enthalpy,making them chemically inert.

(B) Noble gases form fluorides with even oxidation states for Xenon,such as $+2, +4, +6,$ and $+8$.
$XeF_3$ is impossible because it would imply an oxidation state of $+3$ for Xenon,which is not stable for noble gases.

(A) $XeF_6$ has $sp^3d^3$ hybridization with one lone pair of electrons on the $Xe$ atom. Due to the presence of this lone pair,the geometry is distorted octahedral.

(C) Let the oxidation state of $Xe$ be $x$.
In $XeOF_2$,the oxidation state of $O$ is $-2$ and $F$ is $-1$.
The sum of oxidation states in a neutral molecule is $0$.
$x + (-2) + 2(-1) = 0$
$x - 2 - 2 = 0$
$x - 4 = 0$
$x = +4$.
Therefore,the oxidation state of $Xe$ is $+4$.

(C) The most abundant noble gas in the atmosphere is Argon $(Ar)$,which constitutes approximately $0.93\%$ by volume of dry air.

(B) The noble gas group (Group $18$) consists of elements $He, Ne, Ar, Kr, Xe, Rn,$ and $Og$. The last naturally occurring element in this group is Radon $(Rn)$,with atomic number $86$.

(D) The noble gases belong to Group $18$ of the periodic table.
The correct sequence of noble gases is $He$ (Helium),$Ne$ (Neon),$Ar$ (Argon),$Kr$ (Krypton),$Xe$ (Xenon),and $Rn$ (Radon).
Therefore,the correct option is $D$.

(B) The molar mass of air is approximately $29 \ g/mol$. The molar mass of Helium $(He)$ is $4 \ g/mol$. Since the molar mass of $He$ is significantly lower than that of air,it is lighter than air.

(A) The geometries of the given xenon compounds are as follows:
$1. \ XeF_4$: It has $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a $\text{Square planar}$ geometry $(A-3)$.
$2. \ XeF_6$: It has $sp^3d^3$ hybridization with $1$ lone pair,resulting in a $\text{Distorted octahedral}$ geometry $(B-1)$.
$3. \ XeO_3$: It has $sp^3$ hybridization with $1$ lone pair,resulting in a $\text{Pyramidal}$ geometry $(C-4)$.
$4. \ XeO_4$: It has $sp^3$ hybridization with $0$ lone pairs,resulting in a $\text{Tetrahedral}$ geometry $(D-2)$.
Therefore,the correct matching is $A-3, B-1, C-4, D-2$.

(B) Noble gases have completely filled valence shells $(ns^2 np^6)$,which makes them highly stable and results in very low chemical reactivity.

(C) In $XeF_2$,the central xenon atom has $2$ bond pairs and $3$ lone pairs of electrons.
Total electron pairs = $2 + 3 = 5$.
Therefore,the hybridization is $sp^3d$.

(A) $Argon$ was discovered by Lord Rayleigh.

(A) $XeO_3$ has three $\sigma$ bonds and three $\pi$ bonds. The structure of $XeO_3$ is pyramidal with one lone pair on $Xe$. In $XeO_4$,$Xe$ is $sp^3$ hybridized and has a tetrahedral geometry. Therefore,both statements $A$ and $B$ are incorrect,but $A$ is the most direct factual error regarding the bond count.

(A) The percentage of $Argon$ in the air is approximately $1\%$ by volume.

(C) Noble gases are generally inert due to their stable electronic configuration. However,it is incorrect to say they do not form any chemical compounds,as elements like $Kr$ and $Xe$ are known to form various chemical compounds (e.g.,$XeF_2$,$XeF_4$,$XeOF_4$). Therefore,statement $C$ is false.

(D) The boiling point of noble gases depends on the magnitude of interatomic forces of attraction. $He$ has the smallest atomic size and the lowest polarizability among all noble gases. Consequently,the van der Waals forces of attraction between $He$ atoms are extremely weak,resulting in the lowest boiling point of $4.2 \ K$.

(A) The reaction of $XeF_6$ with silica $(SiO_2)$ is a partial hydrolysis reaction.
The balanced chemical equation is:
$2XeF_6 + SiO_2 \to 2XeOF_4 + SiF_4$

(C) The solubility of noble gases in water increases down the group as the atomic size increases,which leads to stronger van der Waals forces of attraction between the gas molecules and water molecules.
Therefore,the correct order of solubility is $Xe > Kr > Ar > Ne > He$.

(D) To determine the number of lone pairs on the central atom $Xe$,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - M - C + A)$,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 $Xe$ (valence electrons $V = 8$):

Compound	Calculation	Lone Pairs
$XeO_3$	$\frac{1}{2}(8 - 0) = 4$ electron pairs; $3$ bonding pairs ($O$ is divalent) $\rightarrow 4 - 3 = 1$	$1$
$XeOF_4$	$\frac{1}{2}(8 - 4) = 2$ electron pairs; $5$ bonding pairs ($O$ is divalent,$F$ is monovalent) $\rightarrow 6 - 5 = 1$	$1$
$XeF_6$	$\frac{1}{2}(8 - 6) = 1$ electron pair	$1$

All three molecules $(i), (ii),$ and $(iii)$ have $1$ lone pair on the $Xe$ atom.

(A) Nitrogen obtained from air contains a small amount of $Argon$ $(Ar)$ as an impurity. Since the atomic mass of $Argon$ $(39.95 \ u)$ is higher than that of $Nitrogen$ $(28.01 \ u)$,the presence of $Argon$ makes the density of atmospheric nitrogen slightly higher than that of chemically pure nitrogen.

(C) Noble gases exist as monoatomic gases because they have a stable electronic configuration and do not need to form bonds to complete their octet. Therefore,they do not form diatomic molecules.

(C) $Neon$ is used in discharge tubes for advertising purposes because it emits a characteristic reddish-orange glow when an electric current is passed through it.

(A) Xenon $(Xe)$ is the most reactive noble gas among the given options,with an atomic number of $54$. It reacts with fluorine to form various xenon fluorides such as $XeF_2$,$XeF_4$,and $XeF_6$.

(A) Among the noble gases,reactivity increases down the group due to the decrease in ionization enthalpy. $Xe$ (Xenon) has the lowest ionization enthalpy among the given options,making it the most reactive noble gas.

(D) The solubility of noble gases in water increases down the group as the atomic size increases,which leads to stronger induced dipole-induced dipole interactions (London dispersion forces) with water molecules. Therefore,among the given options,$Kr$ is the most soluble.

(D) The polarizability of noble gases increases down the group as the atomic size and the number of electrons increase. Among the given options,$Xe$ has the largest atomic size and the highest number of electrons,leading to the highest polarizability.

(D) In clathrates of $Xe$ with water,the $Xe$ atoms are trapped in the cavities of the water lattice. The bonding between $Xe$ and water molecules is primarily due to $Dipole-induced \text{ } dipole \text{ } interaction$.

(A) Argon $(Ar)$ does not show any reactivity with fluorine $(F_2)$ under normal conditions.

(A) The reactivity of noble gases increases as we move down the group. $He$ is the most inert element among the given options,while $Xe$ is the most reactive and forms the maximum number of compounds. Therefore,$He$ forms compounds in the least amount (essentially none under normal conditions).

(C) The hydrolysis of $XeF_4$ is a violent reaction. The balanced chemical equation is:
$6XeF_4 + 12H_2O \to 2Xe + 4XeO_3 + 24HF + 3O_2$.
However,in many simplified contexts,the reaction is represented as:
$2XeF_4 + 3H_2O \to Xe + XeO_3 + 6HF + 0.5O_2$.
Given the options provided,the reaction products include $Xe$,$XeO_3$,$HF$,and $O_2$.

(C) $XeF_4$ has a square planar geometry. Due to its highly symmetrical structure,the bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(D) The polarizability of noble gases increases as we move down the group due to an increase in atomic size and the number of electrons.
Since $He$ has the smallest atomic size and the fewest electrons,it has the lowest polarizability.

(A) $XeF_2$ is a strong oxidizing agent,not a reducing agent. Therefore,the statement '$XeF_2$ is a strong reducing agent' is incorrect.

(B) Solid $XeF_6$ is an ionic compound that exists as $[XeF_5]^+[F]^-$. This structure is formed due to the transfer of a fluoride ion.

(A) Due to its very small atomic size,$He$ can occupy the interstitial sites in the crystal lattice of certain metals,thereby forming interstitial compounds.

(A) Liquid helium (boiling point $4.2 \ K$) is used in cryogenic studies because of its extremely low boiling point.

(C) $Rn$ (Radon) is the only radioactive noble gas among the given options. It is a decay product of radium.

(D) Krypton is a noble gas with a very high ionization energy. Among the noble gases,only Xenon forms stable fluorides like $XeF_2$,$XeF_4$,and $XeF_6$. Krypton does not form a stable hexafluoride $(KrF_6)$ under normal conditions.

(C) Noble gases are separated by the method of selective adsorption and desorption on coconut charcoal at different temperatures.

(B) Xenon reacts best with the most electronegative element,which is fluorine $(F_2)$. It forms fluorides directly through reaction.

(C) The reaction of Xenon with Fluorine in a $1:5$ ratio at $873 \ K$ and $7 \ bar$ gives $XeF_6$.
However,the reaction of Xenon with Fluorine in a $1:2$ ratio at $673 \ K$ and $5-6 \ bar$ gives $XeF_4$.
Thus,the reaction $Xe + 2F_2 \xrightarrow[673 \ K, 5-6 \ bar]{Ni} XeF_4$ produces $XeF_4$.