Electron affinity Questions in English

(C) Electron affinity is defined as the amount of energy released when an electron is added to a neutral gaseous atom to form a negative ion (anion).
It represents the attraction of the nucleus for the incoming electron in the outermost shell.
Therefore,the correct definition is the energy released when an electron is added to the outermost shell.

(D) The first electron affinity is the energy released when an electron is added to a neutral gaseous atom,which is often exothermic.
However,the second electron affinity involves adding an electron to a negatively charged ion (anion).
Due to the strong electrostatic repulsion between the incoming electron and the existing negative charge on the anion,energy must be supplied to overcome this repulsion.
Therefore,the second electron affinity is always endothermic.

(C) The electron affinity of halogens generally increases as we move up the group from $I$ to $F$ due to the decrease in atomic size.
However,$F$ has an exceptionally low electron affinity compared to $Cl$ due to its small size,which leads to strong inter-electronic repulsions between the incoming electron and the existing electrons in the $2p$ subshell.
Therefore,the correct order of electron affinity for halogens is $Cl > F > Br > I$.
Comparing the given options,$Br < Cl$ is correct.

(A) The energy released when an electron is added to a neutral gaseous atom is called electron gain enthalpy.
$Cl$ has a high electron affinity because it achieves a stable noble gas configuration $([Ne] 3s^2 3p^6)$ upon gaining one electron.
$P$ has a half-filled $p$-orbital,making it relatively stable,so adding an electron is less favorable.
$N$ has a half-filled $2p^3$ configuration,which is very stable,making electron addition highly unfavorable (energy is absorbed).
$C$ has a $2p^2$ configuration and is less electronegative than $Cl$.
Therefore,$Cl$ releases the maximum amount of energy upon electron gain.

(C) The electron gain enthalpy (electron affinity) depends on the electronic configuration of the elements.
$1$. $Be$ $(1s^2 2s^2)$: It has a fully filled $2s$ orbital,making it very stable. Therefore,it has a positive electron gain enthalpy (very low affinity).
$2$. $N$ $(1s^2 2s^2 2p^3)$: It has a half-filled $2p$ subshell,which is extra stable. Thus,it also has a very low electron affinity.
$3$. Comparing $B$ $(1s^2 2s^2 2p^1)$ and $C$ $(1s^2 2s^2 2p^2)$: $C$ has a higher effective nuclear charge than $B$,so it attracts electrons more strongly.
$4$. The order of electron affinity is $N < Be < B < C$. However,considering the stability of $Be$ and $N$,the experimental order is $N < Be < B < C$. Given the options provided,$N < Be < B < C$ is the most accurate representation of the trend.

(A) The electron gain enthalpy (or electron affinity) generally becomes less negative as we move down the group from $Cl$ to $I$. However,$F$ has an exceptionally low electron gain enthalpy compared to $Cl$ due to its small size and high inter-electronic repulsion between the incoming electron and the existing electrons in the $2p$ subshell. Therefore,the correct order is $F < Cl > Br > I$.

(B) The reaction $Cl_{(g)} + e^- \to Cl^-_{(g)}$ represents the electron gain enthalpy of chlorine.
Since energy is released when an electron is added to a neutral chlorine atom to form a stable chloride ion,the process is exothermic.
For an exothermic process,the change in enthalpy,$\Delta H$,is negative.

(C) Energy is released when an electron is added to a neutral atom to form an anion,a process known as electron gain enthalpy.
For the process $Cl + e^{-} \to Cl^{-}$,the electron gain enthalpy is negative,meaning energy is released.
In the case of $O^{-} + e^{-} \to O^{2-}$,energy is absorbed because of the strong electrostatic repulsion between the incoming electron and the negatively charged $O^{-}$ ion.
Processes $Cl \to Cl^{+} + e^{-}$ and $Ne + e^{-} \to Ne^{-}$ are endothermic processes requiring energy input.

(C) Electron affinity is the energy released when an electron is added to a neutral gaseous atom.
Elements with stable electronic configurations,such as half-filled or fully-filled orbitals,have very low or negative electron affinity because they are already stable and do not readily accept an additional electron.
Comparing the given configurations:
$ns^2 np^5$ (Halogen) has a high electron affinity as it needs only one electron to complete its octet.
$ns^2 np^2$ and $ns^2 np^4$ have intermediate electron affinities.
$ns^2 np^3$ represents a half-filled $p$-orbital $(p^3)$,which is exceptionally stable due to exchange energy and symmetry.
Therefore,the configuration $ns^2 np^3$ has the lowest electron affinity.

(B) Electron gain enthalpy is the energy released when an electron is added to a neutral gaseous atom.
Chlorine $(Cl)$ has the highest negative electron gain enthalpy among all elements due to its small size and high effective nuclear charge,which allows for strong attraction of the incoming electron.
Fluorine $(F)$ has a lower electron gain enthalpy than chlorine because of its very small size,which leads to significant inter-electronic repulsions between the incoming electron and the existing electrons in the $2p$ subshell.
Therefore,$Cl$ has the highest and $F$ has the lowest electron gain enthalpy among the given options.

(D) The process of adding an electron to an isolated gaseous atom to form a gaseous anion is defined as electron gain enthalpy.
$Cl(g) + e^- \rightarrow Cl^-(g) + 3.8 \ eV$.
Since energy is released,the electron gain enthalpy is $-3.8 \ eV$.
The electron affinity is defined as the negative of the electron gain enthalpy,which is $+3.8 \ eV$.
Conversely,the ionization potential of an ion is the energy required to remove an electron from it.
For $Cl^-(g) \rightarrow Cl(g) + e^-$,the energy required is equal to the electron affinity of the neutral atom,which is the ionization potential of the $Cl^-$ ion.
Therefore,the energy released during the formation of $Cl^-$ from $Cl$ is equal to the ionization potential of $Cl^-$.

(B) The energy required to remove an electron from a gaseous anion is known as the electron detachment enthalpy or the negative of the electron affinity.
For the given processes,we are looking at the reverse of electron gain enthalpy.
Among the given elements,$P$ has a half-filled $p$-orbital $(3s^2 3p^3)$,making it very stable.
Adding an electron to $P$ is unfavorable,and removing an electron from $P^-$ is relatively easy because the electron is held less tightly compared to the others.
Specifically,the electron affinity of $P$ is nearly zero or slightly negative,meaning the energy required to remove an electron from $P^-$ is the lowest among the options provided.

(B) The energy liberated during an electron affinity process is related to the stability of the resulting anion.
Option $B$ involves adding an electron to a $2p^2$ configuration to form a $2p^3$ configuration.
The $2p^3$ configuration is a half-filled $p$-orbital,which is exceptionally stable due to exchange energy and symmetry.
Therefore,the process $[He] \ 2s^2 \ 2p^2 + e^- \to [He] \ 2s^2 \ 2p^3$ (forming the $N^-$ ion) is highly exothermic compared to the others,as it achieves a stable half-filled subshell.

(D) Electron gain enthalpy $(\Delta H_{eg})$ is the energy released when an electron is added to a neutral gaseous atom or ion.
For species with a positive charge,the attraction for an incoming electron is significantly higher due to electrostatic forces.
Comparing the given species: $F^{-}$,$O$,$O^{-}$,and $Na^{+}$.
$Na^{+}$ is a cation with a high positive charge,which exerts a strong electrostatic attraction on the incoming electron,making the process highly exothermic.
Therefore,$Na^{+}$ has the most negative (maximum) electron gain enthalpy.

(B) The given process $X_{(g)} + e^- \to X^-_{(g)}$ represents the electron gain enthalpy of an atom $X$.
When an electron is added to a neutral gaseous atom to form a gaseous anion,energy is generally released,making the process exothermic.
For an exothermic process,the change in enthalpy $\Delta H$ is negative,i.e.,$\Delta H < 0$.

(B) The electron affinity of halogens generally decreases down the group due to an increase in atomic size and electron-electron repulsion.
However,$Cl$ has a higher electron affinity than $F$ due to its larger size reducing inter-electronic repulsion.
The order of electron affinity for halogens is $I < Br < F < Cl$.
Therefore,for the given elements $Cl, Br$ and $I$,the increasing order is $I < Br < Cl$.

(C) $(i)$ The electron gain enthalpy (or electron affinity) of the second period $p$-block elements is less negative than that of the corresponding third period $p$-block elements due to the small size and high inter-electronic repulsion in the $2p$ subshell. Thus,for the pair $F$ and $Cl$,$Cl$ releases more energy.
$(ii)$ Down a group,the electron gain enthalpy generally becomes less negative as the atomic size increases. Thus,for the pair $S$ and $Se$,$S$ releases more energy,and for the pair $Li$ and $Na$,$Li$ releases more energy.
Therefore,the correct elements are $Cl, S$,and $Li$.

(A) The electron gain enthalpy becomes less negative as we move down the group from $Cl$ to $I$.
However,$F$ has a less negative electron gain enthalpy than $Cl$ due to its small size and inter-electronic repulsions.
The correct values for $F, Cl, Br,$ and $I$ are $-333, -349, -325,$ and $-296 \ kJ/mol$ respectively.

(A) Electron gain enthalpy generally becomes more negative across a period as we move from left to right.
Within a group,electron gain enthalpy becomes less negative down a group.
However,adding an electron to the $2p$-orbital (as in $F$) leads to greater inter-electronic repulsion than adding an electron to the larger $3p$-orbital (as in $Cl$).
Therefore,$Cl$ has the most negative electron gain enthalpy.
Among the given elements,$P$ has the least negative electron gain enthalpy due to its stable half-filled $3p^3$ electronic configuration.

(A) $(i)$ $O$ and $F$ are in the same period. $F$ has a smaller atomic size and higher effective nuclear charge than $O$,allowing it to attract an incoming electron more strongly. Thus,$F$ has a more negative electron gain enthalpy than $O$.
$(ii)$ Although electron gain enthalpy generally becomes less negative down a group,$Cl$ has a more negative electron gain enthalpy than $F$. This is because the $2p$ subshell of $F$ is very compact,leading to significant inter-electronic repulsions for the incoming electron. In $Cl$,the incoming electron enters the larger $3p$ subshell,where repulsions are lower,making the process more exothermic.

(N/A) Halogens have the smallest size in their respective periods and therefore high effective nuclear charge. As a consequence,they readily accept one electron to acquire noble gas electronic configuration.

(N/A) The stability of an ionic compound is primarily determined by its lattice energy. $A$ higher lattice energy leads to greater stability of the compound.
Lattice energy is directly proportional to the product of the charges on the ions. When a metal reacts with oxygen,the lattice energy of an oxide containing $O^{2-}$ ions is significantly higher than that of an oxide containing $O^{-}$ ions.
Although the second electron gain enthalpy for oxygen is endothermic $(702 \ kJ \ mol^{-1})$,this energy is more than compensated for by the large amount of lattice energy released during the formation of the crystal lattice with $O^{2-}$ ions. Therefore,oxides containing $O^{2-}$ are more stable than those containing $O^{-}$.

(N/A) When an electron is added to a neutral gaseous atom $(X)$ to convert it into a negative ion,the enthalpy change accompanying the process is defined as the 'Electron Gain Enthalpy $\left( \Delta_{eg} H \right)$.'
Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form an anion,as represented by the equation:
$X_{(g)} + e^{-} \rightarrow X_{(g)}^{-}; \Delta_{eg} H$
Depending on the element,the process of adding an electron to the atom can be either endothermic or exothermic.
For many elements,energy is released when an electron is added to the atom,and the electron gain enthalpy is negative.
For example,group $17$ elements (the halogens) have very high negative electron gain enthalpies because they can attain stable noble gas electronic configurations by picking up an electron.
On the other hand,noble gases have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level,leading to a very unstable electronic configuration.
It may be noted that electron gain enthalpies have large negative values toward the upper right of the periodic table,preceding the noble gases.

(N/A) Electron gain enthalpy $(\Delta_{eg}H)$ shows less regularity compared to ionization enthalpy.
Across a period (left to right),electron gain enthalpy generally becomes more negative (increases in magnitude) with an increase in atomic number. This is because the effective nuclear charge increases,making it easier for an incoming electron to be added to the smaller atom,as the added electron is closer to the positively charged nucleus.

Group-$1$ Element $(\Delta_{eg} H)$	Group-$16$ Element $(\Delta_{eg} H)$	Group-$17$ Element $(\Delta_{eg} H)$	Group-$18$ Element $(\Delta_{eg} H)$
$H(-73)$	$-$	$-$	$He(+48)$
$Li(-60)$	$O(-141)$	$F(-328)$	$Ne(+116)$
$Na(-53)$	$S(-200)$	$Cl(-349)$	$Ar(+96)$
$K(-48)$	$Se(-195)$	$Br(-325)$	$Kr(+96)$
$Rb(-47)$	$Te(-190)$	$I(-295)$	$Xe(+77)$
$Cs(-46)$	$Po(-174)$	$At(-270)$	$Rn(+68)$

Down a group,the value of electron gain enthalpy generally becomes less negative (e.g.,$[S(-200), Se(-195), Te(-190), Po(-174)]$ and $[Cl(-349), Br(-325), I(-295), At(-270)]$). This occurs because the atomic volume increases,and the added electron is further from the nucleus.
Note: The electron gain enthalpy of oxygen is less negative than that of sulfur,and that of fluorine is less negative than that of chlorine. This is because when an electron is added to oxygen or fluorine ($n=2$ shell),it experiences significant inter-electronic repulsion due to the small size of the shell. In the $n=3$ shell (sulfur or chlorine),the added electron occupies a larger region of space,resulting in significantly less electron-electron repulsion.

(N/A) Electron gain enthalpy shows less irregularity compared to ionization enthalpy.
Across a period (left to right),electron gain enthalpy generally becomes more negative (increases in magnitude) as atomic number increases.
This is because the effective nuclear charge increases,causing the atomic radius to decrease. Consequently,the incoming electron is closer to the positively charged nucleus,resulting in a stronger attraction.

Group-$1$ Element $(\Delta_{eg} H)$	Group-$16$ Element $(\Delta_{eg} H)$	Group-$17$ Element $(\Delta_{eg} H)$	Group-$18$ Element $(\Delta_{eg} H)$
$H(-73)$	$-$	$-$	$He(+48)$
$Li(-60)$	$O(-141)$	$F(-328)$	$Ne(+116)$
$Na(-53)$	$S(-200)$	$Cl(-349)$	$Ar(+96)$
$K(-48)$	$Se(-195)$	$Br(-325)$	$Kr(+96)$
$Rb(-47)$	$Te(-190)$	$I(-295)$	$Xe(+77)$
$Cs(-46)$	$Po(-174)$	$At(-270)$	$Rn(+68)$

Down a group (top to bottom),the electron gain enthalpy becomes less negative. For example,in Group-$16$ [$S(-200), Se(-195), Te(-190), Po(-174)$] and Group-$17$ [$Cl(-349), Br(-325), I(-295), At(-270)$]. This occurs because the atomic size increases,and the added electron is further from the nucleus.
Note: The electron gain enthalpy of oxygen is less negative than that of sulfur,and that of fluorine is less negative than that of chlorine. This is because the incoming electron in $n=2$ (oxygen/fluorine) experiences significant inter-electronic repulsion due to the small size of the shell. In $n=3$ (sulfur/chlorine),the electron occupies a larger region,reducing repulsion.

(A) The first electron gain enthalpy of oxygen is negative because energy is released when an electron is added to a neutral oxygen atom to form an $O^-$ ion.
The second electron gain enthalpy of oxygen is positive. This is because after the addition of the first electron,the oxygen atom becomes a negatively charged $O^-$ ion.
The second electron is added to this negatively charged ion,and the incoming electron experiences strong electrostatic repulsion from the existing negative charge.
Therefore,energy must be supplied to overcome these coulombic repulsions to force the second electron into the anion,making the process endothermic (positive electron gain enthalpy).

(N/A) The electron gain enthalpy of fluorine is less negative than that of chlorine due to its small atomic size.
In fluorine,the incoming electron experiences significant inter-electronic repulsion because the $2p$ orbitals are compact.
In contrast,chlorine has a larger atomic size and the incoming electron enters the $3p$ orbital,which has more space available.
This results in less inter-electronic repulsion in chlorine,making the process more exothermic and leading to a more negative electron gain enthalpy compared to fluorine.

(N/A) When an electron is added to the valence shell of an isolated gaseous atom,the electron gain enthalpy of an element is equal to the energy released. $A_{(g)} + e^{-} \longrightarrow A^{-}_{(g)}$; $\Delta_{eg}H = \text{negative}$.
Factors affecting electron gain enthalpy:
$(i)$ Effective nuclear charge: As the attraction of the nucleus towards the incoming electron increases,the electron gain enthalpy becomes more negative (increases in magnitude) with an increase in effective nuclear charge.
$(ii)$ Size of an atom: As the atomic size increases,the electron gain enthalpy becomes less negative because the incoming electron is farther from the nucleus.
$(iii)$ Type of subshell: The ease of adding an electron depends on the subshell proximity. The general order of stability for adding an electron is $s > p > d > f$.
$(iv)$ Electronic configuration: Half-filled and fully-filled subshells have stable configurations,making the addition of an electron energetically unfavorable.
Variation in the periodic table:
Across a period: Generally,electron gain enthalpy becomes more negative from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
Down a group: Electron gain enthalpy becomes less negative from top to bottom because the atomic size increases,placing the added electron farther from the nucleus.
Note: Electron gain enthalpy of $O$ or $F$ is less negative than that of the successive elements ($S$ or $Cl$) because the added electron in $n=2$ experiences significant inter-electronic repulsion compared to the larger $n=3$ shell.

(N/A) $Pb^{4+}$ by gaining $2$ electrons changes into $Pb^{2+}$,which is more stable due to the inert pair effect. Thus,it acts as an oxidizing agent.
$Sn^{2+}$ is less stable than $Sn^{4+}$ and tends to lose $2$ electrons to form $Sn^{4+}$,thus acting as a reducing agent.
$(1)$ $Pb^{4+} + 2e^{-} \rightarrow Pb^{2+}$ (Reduction,hence oxidizing agent)
$Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Oxidation,hence reducing agent)
$(2)$ The electron gain enthalpy of fluorine is less negative than that of chlorine because of the small size of the fluorine atom. The inter-electronic repulsions in the compact $2p$ subshell of fluorine make the addition of an electron less favorable compared to the $3p$ subshell of chlorine.

(N/A) Elements of group $17$ accept $1$ electron and elements of group $16$ accept $2$ electrons to form anions and achieve the electronic configuration of the next noble gas. This is because they possess high negative electron gain enthalpy.

(N/A) Electron gain enthalpy $(\Delta_{eg}H)$ is defined as the change in enthalpy when an electron is added to a neutral gaseous atom $(X)_{(g)}$ to convert it into a negative ion $(X)_{(g)}^{-}$.
The process is represented as: $X_{(g)} + e^{-} \rightarrow X_{(g)}^{-}; \Delta_{eg}H$.
The value of $\Delta_{eg}H$ can be either negative or positive depending on the element.

(A) The noble gases have the most positive electron gain enthalpy because they have stable electronic configurations,making the addition of an electron energetically unfavorable.
Group-$17$ elements (halogens) have the most negative electron gain enthalpy because they are one electron short of a stable noble gas configuration,making the addition of an electron highly exothermic.

(A) $(i)$ $F$ $(ii)$ $Cl$
$O$ has $-141 \ kJ \ mol^{-1}$,$F$ has $-328 \ kJ \ mol^{-1}$,and $Cl$ has $-349 \ kJ \ mol^{-1}$ as their electron gain enthalpy $(\Delta_{eg}H)$.

(B) The electron gain enthalpy becomes more negative as we move from left to right across a period. However,due to the small size of the $F$ atom,the electron-electron repulsion is high,making the addition of an electron less favorable compared to $Cl$. Thus,$Cl$ has the highest (most negative) electron gain enthalpy value of $-349 \ kJ \ mol^{-1}$.

(A) In group $16$,the electron gain enthalpy of $S$ is more negative than that of $O$ due to the small size of the $O$ atom,which leads to inter-electronic repulsion.
In group $17$,the electron gain enthalpy of $Cl$ is more negative than that of $F$ because the small size of $F$ causes significant inter-electronic repulsion,making the addition of an electron less favorable compared to $Cl$.

(B) The $n = 3$ shell experiences less electron-electron repulsion compared to the $n = 2$ shell.
This is because,in the $n = 3$ shell,the incoming electron occupies a larger volume of space,which reduces the effective electron-electron repulsion.

(A-4, B-1, C-2, D-3) The electron gain enthalpy values are determined by the stability of the electronic configuration:
$(A) \ 1s^2 2s^2 2p^6$ represents Neon (a noble gas),which has a stable octet configuration,so it has a positive electron gain enthalpy: $(+53 \ kJ \ mol^{-1})$. Thus,$(A-4)$.
$(B) \ 1s^2 2s^2 2p^6 3s^1$ represents Sodium,which has a low electron gain enthalpy: $(-53 \ kJ \ mol^{-1})$. Thus,$(B-1)$.
$(C) \ 1s^2 2s^2 2p^5$ represents Fluorine,which has a very high negative electron gain enthalpy: $(-328 \ kJ \ mol^{-1})$. Thus,$(C-2)$.
$(D) \ 1s^2 2s^2 2p^4$ represents Oxygen,which has an electron gain enthalpy of: $(-141 \ kJ \ mol^{-1})$. Thus,$(D-3)$.
Therefore,the correct match is $(A-4, B-1, C-2, D-3)$.

(B) The process $H_{(g)} + e^{-} \rightarrow H_{(g)}^{-}$ represents the electron gain enthalpy of hydrogen,which is exothermic.
All other processes listed involve either the addition of an electron to an anion (which requires energy to overcome inter-electronic repulsion) or the removal of an electron (ionization energy),both of which are endothermic processes.

(C) The electron gain enthalpy is the energy released when an electron is added to a neutral gaseous atom.
For halogens,the expected trend based on size is $F > Cl > Br > I$.
However,due to the very small size of the fluorine atom,the inter-electronic repulsions are high,which makes the incoming electron experience less attraction compared to chlorine.
Therefore,the absolute value of electron gain enthalpy for chlorine is higher than that of fluorine.
The correct order of the absolute value of electron gain enthalpy is $Cl > F > Br > I$.

(A) The electron gain enthalpy becomes less negative as we move down the group from $S$ to $Te$ due to an increase in atomic size.
However,$O$ has an exceptionally low electron gain enthalpy because of its small size,which leads to strong inter-electronic repulsions when an electron is added.
Therefore,the correct order is $S > Se > Te > O$.

(D) The electron gain enthalpy $(\Delta_{eg}H)$ becomes less negative as we move down a group.
For Group $16$ ($Te$ and $Po$): The order is $Po < Te$ (in terms of magnitude of negative value,$Te$ is more negative than $Po$).
For Group $17$ ($F$ and $Cl$): Due to the small size of $F$,the electron-electron repulsion is high,making $Cl$ have a more negative electron gain enthalpy than $F$.
The values are:
$Cl: -349 \ kJ/mol$
$F: -328 \ kJ/mol$
$Te: -190 \ kJ/mol$
$Po: -174 \ kJ/mol$
Comparing the magnitude of negative values (more negative means more energy released,i.e.,lower enthalpy value):
$Po (-174) < Te (-190) < F (-328) < Cl (-349)$.
Thus,the correct order of electron gain enthalpies is $Po < Te < F < Cl$.

(C) The electron gain enthalpy $(\Delta_{eg}H)$ values for elements in the same group generally become less negative as we move down the group.
$1$. For $Rb$ and $Cs$ (Group $1$): The values are approximately $-47 \ kJ/mol$ and $-46 \ kJ/mol$ respectively,which are nearly identical.
$2$. For $Ar$ and $Kr$ (Group $18$): Noble gases have positive electron gain enthalpy values due to stable electronic configurations. Both $Ar$ and $Kr$ have values around $+96 \ kJ/mol$ to $+99 \ kJ/mol$,which are considered nearly identical.
Therefore,the pairs $A$ $(Rb, Cs)$ and $C$ $(Ar, Kr)$ have nearly identical electron gain enthalpies.

(B) The electron gain enthalpy $(\Delta_{eg} H)$ becomes less negative as we move down a group due to an increase in atomic size.
$1$. For $Cl$ and $F$: $\Delta_{eg} H (Cl) = -349 \ kJ/mol$ and $\Delta_{eg} H (F) = -328 \ kJ/mol$. Since $-349 < -328$,the statement $\Delta_{eg} H (Cl) < \Delta_{eg} H (F)$ is correct.
$2$. For $Se$ and $S$: $\Delta_{eg} H (Se) = -195 \ kJ/mol$ and $\Delta_{eg} H (S) = -200 \ kJ/mol$. Since $-195 > -200$,the statement $\Delta_{eg} H (Se) < \Delta_{eg} H (S)$ is incorrect.
$3$. For $I$ and $At$: $\Delta_{eg} H (I) = -295 \ kJ/mol$ and $\Delta_{eg} H (At) = -270 \ kJ/mol$. Since $-295 < -270$,the statement $\Delta_{eg} H (I) < \Delta_{eg} H (At)$ is correct.
$4$. For $Te$ and $Po$: $\Delta_{eg} H (Te) = -190 \ kJ/mol$ and $\Delta_{eg} H (Po) = -183 \ kJ/mol$. Since $-190 < -183$,the statement $\Delta_{eg} H (Te) < \Delta_{eg} H (Po)$ is correct.

(B) The electron gain enthalpy $(\Delta H_{eg})$ is the energy change when an electron is added to a neutral gaseous atom.
$Cl$ has the most negative (exothermic) electron gain enthalpy among all elements due to its high effective nuclear charge and small size.
$Ne$ is a noble gas with a stable octet configuration,making the addition of an electron highly unfavorable,resulting in a large positive (endothermic) electron gain enthalpy.
Therefore,the difference between the most positive value $(Ne)$ and the most negative value $(Cl)$ results in the maximum difference.

(D) Statement-$I$ is false because chlorine $(Cl)$ has the most negative electron gain enthalpy in its group due to its larger size compared to fluorine $(F)$,which minimizes inter-electronic repulsions.
Statement-$II$ is true because oxygen $(O)$ has the least negative electron gain enthalpy in its group (Group $16$) due to its small size,which leads to strong inter-electronic repulsions when an electron is added.
Therefore,Statement-$I$ is false and Statement-$II$ is true.

(A) The electron gain enthalpy $(\Delta_{eg}H)$ values for the given elements are:
$Ar$ (Noble gas): $ 96 \ kJ/mol$ (Positive due to stable configuration)
$S$ (Sulfur): $-200 \ kJ/mol$
$Br$ (Bromine): $-325 \ kJ/mol$
$F$ (Fluorine): $-333 \ kJ/mol$
Comparing the values: $ 96 > -200 > -325 > -333$.
Therefore,the correct sequence is $Ar > S > Br > F$,which corresponds to $A > D > B > C$.

(D) Electron affinity is the energy released when an electron is added to a neutral gaseous atom or ion.
$A$. $Be (2s^2)$ has a stable fully-filled configuration. Adding an electron is unfavorable,making the electron affinity negative (energy must be supplied).
$B$. $N (2s^2 2p^3)$ has a stable half-filled configuration. Adding an electron is unfavorable,making the electron affinity negative.
$C$. $O^{-} + e^{-} \rightarrow O^{2-}$. Adding an electron to a negatively charged ion $(O^{-})$ experiences strong inter-electronic repulsion,making the process endothermic (negative electron affinity).
$D$. $Na (3s^1)$ has a low energy orbital available. Adding an electron releases energy (positive electron affinity).
$E$. $Al (3s^2 3p^1)$ has an available $3p$ orbital. Adding an electron releases energy (positive electron affinity).
Therefore,the processes $A, B,$ and $C$ have negative electron affinity values.

(C) The electron affinity of $N$ is very low (nearly zero or negative) due to its stable half-filled $2p^3$ electronic configuration.
Comparing the given options:
$A$: $N < O < S$ is correct.
$B$: $S < F < Cl$ is correct ($Cl > F$ due to inter-electronic repulsion in $F$).
$C$: $O < N < S$ is incorrect because $N$ has a lower electron affinity than $O$.
$D$: $N < P < S$ is correct.

(A) The tendency to gain an electron is related to electron gain enthalpy and electronegativity.
Element $A$ $(1s^2 2s^2 2p^6)$ is Neon,a noble gas with a stable octet,so it has the least tendency to gain an electron.
Element $C$ $(1s^2 2s^2 2p^6 3s^1)$ is Sodium,an alkali metal with a strong tendency to lose an electron,so it has very low electron gain tendency.
Element $B$ $(1s^2 2s^2 2p^4)$ is Oxygen,which needs two electrons to complete its octet.
Element $D$ $(1s^2 2s^2 2p^5)$ is Fluorine,which needs only one electron to complete its octet and is the most electronegative element,thus having the highest tendency to gain an electron.
Comparing these,the order of increasing tendency to gain an electron is $A < C < B < D$.