Noble gases Questions in English

(A) Inert gases (noble gases) have a stable electronic configuration with completely filled valence shells (octet configuration,$ns^2 np^6$,except for $He$ which is $1s^2$).
Since all orbitals are completely filled,there are no unpaired electrons present.
Therefore,the number of unpaired electrons is $0$.

(D) $Krypton$ $(Kr)$ has zero valency because it is a noble gas with a stable octet configuration,meaning it contains $8$ electrons in its outermost shell.

(A) Inert gases (noble gases) belong to group $18$ of the periodic table.
Their outermost shell contains a stable octet configuration,which is represented by the general electronic configuration $ns^2 np^6$ (except for Helium,which is $1s^2$).

(B) $XeF_6$ has $sp^3d^3$ hybridization.
Due to the presence of one lone pair of electrons on the $Xe$ atom,the geometry is not perfectly octahedral.
Therefore,the shape of $XeF_6$ is distorted octahedral.

(B) The oxidation state of $Xe$ in $XeO_3$ is calculated as: $x + 3(-2) = 0 \implies x = +6$.
The oxidation state of $Xe$ in $XeF_6$ is calculated as: $x + 6(-1) = 0 \implies x = +6$.
Thus,in both compounds,the oxidation state of $Xe$ is $+6$.

(B) The elements with atomic numbers $10$ $(Ne)$,$18$ $(Ar)$,$36$ $(Kr)$,$54$ $(Xe)$,and $86$ $(Rn)$ belong to Group $18$ of the periodic table.
These elements have a stable valence shell electronic configuration of $ns^2 np^6$,which makes them chemically inert.
Therefore,they are known as inert gases or noble gases.

(D) Inert elements,also known as noble gases,are chemically stable elements with a fully filled outer electronic configuration.
Because their valence shells are complete,they do not have the affinity to participate in chemical reactions.
Among the given options,$He$ (Helium) has the electronic configuration $1s^2$,which is a stable duplet,making it an inert element.

(D) The electronic configuration $ns^2np^6$ represents a completely filled valence shell,which corresponds to the stable octet configuration.
Elements with this configuration are chemically inert and are known as noble gases (Group $18$ elements).

(C) The correct answer is $(C)$.
Helium was first detected in the solar spectrum by Pierre Janssen and Norman Lockyer in $1868$.
Edward Frankland and Norman Lockyer named the element 'Helium' (derived from the Greek word 'Helios' meaning Sun).

(D) All the noble gases are monoatomic,colourless and odourless gases.
Their monoatomic nature is due to the stable outer configuration $ns^2 np^6$ of their atoms.
As a result,they do not enter into chemical combination even amongst themselves.

(A) Every noble gas atom has a saturated outermost shell. The valence shell configuration is $ns^2 np^6$ (except for $He$,which is $1s^2$),making them chemically inert due to their stable electronic configuration.

(D) The noble gas argon was discovered by Sir William Ramsay in $1894$.

(C) Among noble gases,$Xe$ (Xenon) has the lowest ionization enthalpy,which allows it to form a large number of compounds with highly electronegative elements like fluorine and oxygen.
Examples of such compounds include $XeF_2, XeF_4, XeF_6, XeOF_2, XeOF_4, XeO_3,$ and $XeO_4$.

(C) The abundance of noble gases in the atmosphere by volume is as follows:

Gas	Abundance in air by volume $(ppm)$
$Helium$	$5.2$
$Neon$	$18.2$
$Argon$	$9340$
$Krypton$	$1.1$
$Xenon$	$0.09$

Among the given options,$Argon$ has the highest abundance in the atmosphere ($0.934\%$ or $9340 \ ppm$). Therefore,the correct option is $(C)$.

(B) Noble gases are chemically inert due to their stable electronic configuration $(ns^2 np^6)$.
$He$ (Helium) has a very small size and high ionization energy,and it lacks $d$-orbitals,making it impossible to form stable compounds like $HeF_4$.
Therefore,$HeF_4$ does not exist.

(B) Coloured discharge tubes used for advertisement purposes are primarily filled with $Neon$ gas.
When an electric discharge is passed through $Neon$ gas,it emits a characteristic reddish-orange glow.
Different colours can be produced by mixing $Neon$ with other gases or by using coloured glass tubes.

(B) Among the given options,$Argon$ is a noble gas.
Noble gases are the least reactive because they are the most stable.
Their outer orbitals are fully filled,and so they have a stable electronic configuration.

(A) Noble gases have a stable electronic configuration with completely filled valence shells ($ns^2 np^6$ for all except Helium which is $1s^2$).
This stable configuration leads to high ionization enthalpy and low electron gain enthalpy,making them chemically inert under normal conditions.

(A) Monazite is a phosphate mineral containing rare earth elements and thorium. It is a significant natural source of helium $(He)$ gas,which is trapped within the mineral lattice due to the radioactive decay of thorium and uranium present in it.

(B) The partial hydrolysis of $XeF_4$ is represented by the following chemical equation:
$XeF_4 + H_2O \to XeOF_2 + 2HF$
In contrast,complete hydrolysis of $XeF_4$ is represented by:
$6XeF_4 + 12H_2O \to 2Xe + 4XeO_3 + 3O_2 + 24HF$
Therefore,the correct product of partial hydrolysis is $XeOF_2$.

(D) Polarizability depends on the size of the electron cloud. Larger atoms have more loosely held electrons,making them more polarizable.
Since $He$ has the smallest atomic size among the given noble gases,its electrons are most tightly held by the nucleus.
Therefore,$He$ is the least polarizable.

(A) The correct option is $(A)$.
$Rn$ (Radon) is not found in the atmosphere because it is a radioactive element.
It is obtained by the radioactive disintegration of radium $(_{88}Ra^{226})$.
The nuclear reaction is: $_{88}Ra^{226} \to _{86}Rn^{222} + _{2}He^{4}$.

(B) Noble gases are generally inert.
$Ar$ is used in electric bulbs to prevent the oxidation of the filament.
$He$ is used in cryogenic applications to produce very low temperatures.
$Rn$ is a radioactive element with a half-life of $3.8 \ days$.
$Kr$ is primarily obtained from the atmosphere via fractional distillation of liquid air,not from radioactive disintegration.
$Rn$ is the noble gas that is formed as a byproduct of the radioactive decay of radium $(^{226}Ra)$.
Therefore,the statement that $Kr$ is obtained during radioactive disintegration is incorrect.

(D) Noble gases have a stable electronic configuration with a complete octet,resulting in zero valency.
Among the given options,$Sodium$ $(Na)$,$Beryllium$ $(Be)$,and $Aluminium$ $(Al)$ are reactive metals.
$Krypton$ $(Kr)$ is a noble gas belonging to group $18$,which possesses zero valency.

(C) The noble gases belong to Group $18$ of the periodic table.
The correct sequence in increasing order of their atomic numbers is:
$He (Z=2), Ne (Z=10), Ar (Z=18), Kr (Z=36), Xe (Z=54), Rn (Z=86)$.
Thus,the correct sequence is $He, Ne, Ar, Kr, Xe, Rn$.

(A) Noble gases have completely filled valence shells and belong to group $18$ of the periodic table.
Their general outermost electronic configuration is $ns^2 np^6$,where $n$ represents the highest principal quantum number.
Option $A$ represents the configuration of Xenon $(Xe)$,which is a noble gas,as its outermost shell $(n=5)$ is completely filled with $5s^2 5p^6$ electrons.

(A) The complete hydrolysis of $XeF_6$ results in the formation of xenon trioxide $(XeO_3)$ and hydrogen fluoride $(HF)$.
The balanced chemical equation is:
$XeF_6 + 3H_2O \to XeO_3 + 6HF$

(C) The solubility of noble gases in water increases with an increase in atomic size and polarizability.
As we move down the group from $He$ to $Xe$,the atomic size increases,leading 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) The central atom $Xe$ has $8$ valence electrons.
For $XeF_2$: $Xe$ forms $2$ bonds with $F$ atoms,using $2$ electrons. Remaining electrons = $8 - 2 = 6$. Number of lone pairs = $6 / 2 = 3$.
For $XeF_4$: $Xe$ forms $4$ bonds with $F$ atoms,using $4$ electrons. Remaining electrons = $8 - 4 = 4$. Number of lone pairs = $4 / 2 = 2$.
For $XeF_6$: $Xe$ forms $6$ bonds with $F$ atoms,using $6$ electrons. Remaining electrons = $8 - 6 = 2$. Number of lone pairs = $2 / 2 = 1$.
Thus,the number of lone pairs on $Xe$ in $XeF_2, XeF_4, XeF_6$ are $3, 2, 1$ respectively.

(B) Noble gases belong to Group $18$ of the periodic table. They possess a stable electronic configuration with a complete octet $(ns^2 np^6)$,except for Helium which has a duplet $(1s^2)$. Due to this stable configuration,they have very little tendency to gain,lose,or share electrons,resulting in very low chemical activity.

(D) The solubility of noble gases in water increases with an increase in atomic size due to the increase in polarizability and the magnitude of van der Waals forces.
Among the given noble gases,$Xe$ has the largest atomic size and the highest polarizability.
Therefore,$Xe$ is the most soluble in water.

(A) As the number of electronic shells increases down the group in noble gases,the atomic size increases.
Due to the increase in atomic size,the distance between the nucleus and the outermost valence electrons increases,which reduces the effective nuclear charge experienced by the valence electrons.
Consequently,the energy required to remove an electron,known as the $Ionization energy$ $(I.E.)$,decreases.

(B) The low chemical reactivity of the noble gases is primarily due to their stable electronic configuration,which results in high ionization energies and low electron affinity. This makes it difficult to remove electrons or add electrons to their valence shells.

(A) The percentage of $Ar$ (Argon) in the atmosphere is approximately $0.93\%$.
Rounding this value to the nearest whole number,it is about $1\%$.

(C) $XeF_2$,$XeF_4$,and $XeF_6$ can be prepared by the direct reaction of $Xe$ and $F_2$.
$Xe + F_2 \xrightarrow{673 \ K} XeF_2$
$Xe + 2F_2 \xrightarrow{673 \ K, 6 \ atm} XeF_4$
$Xe + 3F_2 \xrightarrow{573 \ K, 50-60 \ atm} XeF_6$
$XeO_3$ is obtained by the hydrolysis of $XeF_6$:
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$

(D) Noble gases are generally inert due to their stable electronic configuration. However,$Xe$ (Xenon) has a lower ionization enthalpy compared to other noble gases like $He$,$Ne$,and $Ar$. Due to this,$Xe$ can react with highly electronegative elements like fluorine to form compounds such as $XeF_2$,$XeF_4$,and $XeF_6$.

(A) The correct answer is $A$.
$He$ (Helium) forms the least number of compounds among the given noble gases.
This is due to the following reasons:
$1$. It has a very small atomic size.
$2$. It possesses a very high ionization enthalpy.
$3$. It lacks $d$-orbitals in its valence shell,making it chemically inert under normal conditions.

(B) Xenon forms several fluorides such as $XeF_2$,$XeF_4$,and $XeF_6$. Among the given options,$XeF_4$ is a stable compound formed by Xenon.

(D) To determine the number of lone pairs on the central $Xe$ atom in each molecule,we use the formula: $\text{Lone pairs} = \frac{1}{2} \times (V - M - C + A)$,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.
$1.$ For $XeO_3$:
$Xe$ has $8$ valence electrons. Oxygen is divalent,so $M = 0$.
$\text{Lone pairs} = \frac{1}{2} \times (8 - 0) = 4$ electrons,which corresponds to $1$ lone pair.
$2.$ For $XeOF_4$:
$Xe$ has $8$ valence electrons. $F$ is monovalent $(M = 4)$,$O$ is divalent $(M = 0)$.
$\text{Lone pairs} = \frac{1}{2} \times (8 - 4) = 2$ electrons,which corresponds to $1$ lone pair.
$3.$ For $XeF_6$:
$Xe$ has $8$ valence electrons. $F$ is monovalent $(M = 6)$.
$\text{Lone pairs} = \frac{1}{2} \times (8 - 6) = 2$ electrons,which corresponds to $1$ lone pair.
Since all three molecules ($XeO_3$,$XeOF_4$,and $XeF_6$) have $1$ lone pair on the $Xe$ atom,the correct option is $(D)$.

(D) Neil Bartlett prepared the first noble gas compound,which is Xenon hexafluoroplatinate $(IV)$,represented as $XePtF_6$.

(D) The noble gases (Group $18$ elements) are Helium $(He)$,Neon $(Ne)$,Argon $(Ar)$,Krypton $(Kr)$,Xenon $(Xe)$,and Radon $(Rn)$.
Radon $(Rn)$ is the last naturally occurring member of the inert gas group in the periodic table.

(D) $He$,$Ne$,and $Kr$ are all found in very small amounts in the atmosphere,therefore they are all classified as rare gases.

(A) $(A)$. The ease of liquefaction of a gas is directly related to its critical temperature $(T_C)$.
$1$. Gases with higher boiling points have higher critical temperatures and thus liquefy more easily.
$2$. The boiling point of noble gases increases down the group due to an increase in the magnitude of van der Waals forces as the atomic size and molecular weight increase.
$3$. The order of boiling points is $He < Ne < Ar < Kr < Xe$.
$4$. Among the given options, $Kr$ has the highest boiling point and therefore liquefies most easily.

(C) Let the oxidation state of $Xe$ be $x$ in $XeOF_2$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$x + (-2) + 2(-1) = 0$
$x - 2 - 2 = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation number of xenon is $4$.

(A) The boiling point of noble gases increases down the group due to an increase in the magnitude of van der Waals forces as the atomic size increases.
Among the given options,$Xe$ (Xenon) has the largest atomic size and thus the strongest van der Waals forces,resulting in the highest boiling point.

$Gas$	$He$	$Ne$	$Ar$	$Kr$	$Xe$
Boiling point $(^{\circ}C)$	$-269$	$-246$	$-186$	$-153.6$	$-108.1$

(D) Inert gases,also known as noble gases,belong to Group $18$ of the periodic table.
They have a stable electronic configuration with a complete octet (except Helium,which has a complete duplet).
Among the given options,Argon $(Ar)$ is a noble gas,while $H_2$,$O_2$,and $N_2$ are diatomic molecules that are reactive under certain conditions.

(D) The polarizability of an atom depends on its size and the strength of the hold the nucleus has on its valence electrons.
As we move down the group in the periodic table,the atomic size increases and the valence electrons are further away from the nucleus.
Therefore,the valence electrons are more loosely held,making the atom more polarizable.
Among the given options,$Xe$ has the largest atomic size and the lowest ionization energy,making it the most polarizable.

(B) The central atom $Xe$ has $8$ valence electrons.
In $XeOF_4$,$Xe$ forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom.
Total electrons involved in bonding $= 4 + 2 = 6$.
Number of lone pair electrons on $Xe = (8 - 6) / 2 = 1$ lone pair.
Thus,the total number of lone pairs on the central atom $Xe$ is $1$.

(A) The correct answer is $(A)$.
Helium is used as a diluent for oxygen in deep-sea diving and for medical purposes because it is less soluble in blood than nitrogen.
$A$ mixture of helium and oxygen is used in artificial respiration for asthma patients to facilitate easier breathing.

(C) The correct answer is $(C)$.
Iridium ($Ir$,$Z = 77$) is a transition metal belonging to group $9$ and period $6$,not a noble gas.
Noble gases are elements of group $18$ such as Helium $(He)$,Neon $(Ne)$,Argon $(Ar)$,Krypton $(Kr)$,Xenon $(Xe)$,and Radon $(Rn)$.