Electron affinity Questions in English

(C) The electron affinity $(EA)$ generally increases across a period from left to right and decreases down a group.
Comparing the given elements: $N$ (Group $15$,Period $2$),$O$ (Group $16$,Period $2$),$Cl$ (Group $17$,Period $3$),and $Al$ (Group $13$,Period $3$).
$1$. $Al$ is a metal and has the lowest tendency to gain electrons,so it has the lowest electron affinity.
$2$. Among $N$,$O$,and $Cl$: $N$ has a stable half-filled $p$-orbital $(2p^3)$,making its electron affinity very low (even lower than $O$).
$3$. $Cl$ has the highest electron affinity among these due to its smaller size and effective nuclear charge compared to $Al$,and it is higher than $O$ because $O$ has high inter-electronic repulsion in its small $2p$ orbital.
The correct order is $N < Al < O < Cl$.

(A) The electron affinity of $Cl$ $(349 \ kJ \ mol^{-1})$ is higher than that of $F$ $(322 \ kJ \ mol^{-1})$.
This is because the $2p$ subshell of $F$ is very small,leading to significant inter-electronic repulsion when an incoming electron is added.
In contrast,the $3p$ subshell of $Cl$ is larger,which results in weaker electron-electron repulsion,making it easier for $Cl$ to accept an electron.

(B) Electron affinity is the energy released when an electron is added to a gaseous species. It is directly related to the stability of the species formed after electron gain.
$N^{+} (1s^{2} 2s^{2} 2p^{2})$: Adding an electron gives $N$ $(1s^{2} 2s^{2} 2p^{3})$,which has a stable half-filled $p$-orbital configuration.
$O^{+} (1s^{2} 2s^{2} 2p^{3})$: Adding an electron gives $O$ $(1s^{2} 2s^{2} 2p^{4})$,which is less stable than the half-filled state.
$NO^{+}$: This is an isoelectronic species with $N_{2}$ ($14$ electrons). Adding an electron to $NO^{+}$ leads to $NO$ ($15$ electrons),where the electron enters the antibonding $\pi^{*}$ orbital,which is less favorable.
Comparing the effective nuclear charge $(Z_{eff})$,$O^{+}$ has a higher $Z_{eff}$ than $N^{+}$,but the stability of the resulting neutral atom is the dominant factor here. The correct order of electron affinity is $O^{+} > N^{+} > NO^{+}$.

(B) The $CORRECT$ order of electron gain enthalpy $(\Delta H_{eg})$ is $O < S < F < Cl$.
Generally,electron gain enthalpy becomes more negative (more exothermic) as we move from left to right across a period due to a decrease in atomic size and an increase in effective nuclear charge.
However,for elements in the second period $(O, F)$,the incoming electron experiences significant inter-electronic repulsion due to the small size of the $2p$ orbitals. As a result,elements of the third period $(S, Cl)$ have more negative electron gain enthalpy values than their corresponding second-period counterparts.
Therefore,the order is $O < S < F < Cl$ (where the values are increasingly negative).

(C) The process $O_{(g)}^{-} + e^{-} \to O_{(g)}^{2-}$ involves the addition of an electron to a negatively charged ion.
Due to the strong interelectronic repulsion between the incoming electron and the existing electrons in the $O^{-}$ ion,energy must be supplied to overcome this repulsion.
Therefore,this process is endothermic (energy is absorbed).
In contrast,the other processes involve the removal of an electron (ionization) or the addition of an electron to a positive ion,which are generally exothermic or require different energy considerations.
Hence,option $C$ is the correct answer.

(A) The electronic configurations correspond to the following elements:
$I$: $1s^2 2s^2 2p^6 3s^2 3p^5$ is Chlorine $(Cl)$.
$II$: $1s^2 2s^2 2p^3$ is Nitrogen $(N)$.
$III$: $1s^2 2s^2 2p^5$ is Fluorine $(F)$.
$IV$: $1s^2 2s^2 2p^6 3s^2 3p^1$ is Aluminium $(Al)$.
Electron affinity generally increases across a period and decreases down a group.
However,Nitrogen $(N)$ has a half-filled $p$-orbital $(2p^3)$,making it very stable,resulting in a very low (near zero or negative) electron affinity.
Fluorine $(F)$ has a smaller size than Chlorine $(Cl)$,leading to inter-electronic repulsions,which makes the electron affinity of $Cl$ higher than $F$.
Aluminium $(Al)$ is a metal with a low electron affinity.
Thus,the order is $N < Al < F < Cl$,which corresponds to $II < IV < III < I$.

(D) The energy required to remove an electron from a uninegative ion is known as the second electron affinity or the ionization energy of the anion.
Comparing the given processes,we are looking at the removal of an electron from $F^-$,$P^-$,$S^-$,and $Cl^-$.
$F^-$ has a very small size and high electron density,making it difficult to remove an electron.
$P^-$ has a stable half-filled $p$-orbital configuration ($3p^4$ after losing one electron,but the parent $P^-$ is $3p^4$),which is relatively stable.
$S^-$ has a configuration of $3p^5$,and removing an electron leads to $3p^4$.
$Cl^-$ has a stable noble gas configuration $([Ne] 3s^2 3p^6)$. Removing an electron from $Cl^-$ requires the least energy because the resulting $Cl$ atom is highly stable due to its half-filled $p$-subshell and the large size of the $Cl^-$ ion compared to $F^-$.
Therefore,the process $Cl_{(g)}^{-} \to Cl_{(g)} + e^{\Theta}$ requires the least energy.

(C) The process $Na \rightarrow Na^{+} + e^{-}$ represents the Ionization Potential $(IP)$ of $Na$,which is given as $5 \ eV$.
The process $Na + e^{-} \rightarrow Na^{-}$ represents the Electron Affinity $(EA)$ of $Na$.
The electron affinity of $Na$ is approximately $0.55 \ eV$.
Therefore,the value of $X$ is $0.55 \ eV$.

(D) The addition of an electron to an atom is generally exothermic,but it becomes endothermic if the atom has a stable electronic configuration.
$Ne$ $(1s^2 2s^2 2p^6)$ has a stable noble gas configuration.
$N$ $(1s^2 2s^2 2p^3)$ has a stable half-filled $p$-orbital configuration.
$Be$ $(1s^2 2s^2)$ has a stable fully-filled $s$-orbital configuration.
Since all three elements have stable electronic configurations,adding an electron to them requires energy,making all these processes endothermic.

(C) The formation of the oxide ion,$O^{2-}_{(g)}$,from an oxygen atom involves two steps.
The first step is exothermic because energy is released when an electron is added to a neutral oxygen atom.
The second step,adding an electron to the $O^{-}$ anion,is highly endothermic because of the strong electrostatic repulsion between the incoming electron and the negatively charged $O^{-}$ ion.
Even though the $O^{2-}$ ion achieves a stable noble gas configuration (isoelectronic with neon),the energy required to overcome the inter-electronic repulsion is much greater than the stability gained.
Therefore,the overall process is unfavorable in the gas phase.

(B) The electronic configuration of $Li$ $(Z=3)$ is $1s^2 \, 2s^1$. Adding one electron forms $Li^-$ $(1s^2 \, 2s^2)$,which is stable due to a fully filled $2s$ subshell.
The electronic configuration of $Be$ $(Z=4)$ is $1s^2 \, 2s^2$. Adding one electron forms $Be^-$ $(1s^2 \, 2s^2 \, 2p^1)$. This process disrupts the stable,fully filled $2s^2$ configuration of $Be$,making $Be^-$ the least stable ion.
The electronic configuration of $B$ $(Z=5)$ is $1s^2 \, 2s^2 \, 2p^1$. Adding one electron forms $B^-$ $(1s^2 \, 2s^2 \, 2p^2)$,which is relatively stable.
The electronic configuration of $C$ $(Z=6)$ is $1s^2 \, 2s^2 \, 2p^2$. Adding one electron forms $C^-$ $(1s^2 \, 2s^2 \, 2p^3)$,which is stable due to the half-filled $p$-subshell.

(A) The electron affinity $(EA)$ is defined as the energy released when an electron is added to a neutral gaseous atom or ion.
Neutral oxygen $(O)$ has the highest tendency to accept an electron because it is not negatively charged.
Anions like $O^-$ and $O^{2-}$ experience strong electrostatic repulsion from the incoming electron due to their negative charge,making the process endothermic or less favorable.
Therefore,the neutral atom $O$ $(x=0)$ has the maximum electron affinity.

(C) The molar mass of $Cl$ is $35.5 \ g/mol$.
Given that the electron affinity $(EA)$ for $1 \ mole$ of $Cl$ is $348 \ kJ/mol$,this means $348 \ kJ$ of energy is released when $35.5 \ g$ of $Cl$ is converted into $Cl^-$ gas.
To find the energy released for $1.0 \ g$ of $Cl$:
$\text{Energy} = \frac{348 \ kJ}{35.5 \ g} \times 1.0 \ g$
$\text{Energy} \approx 9.802 \ kJ$
Therefore,the energy released is approximately $9.8 \ kJ$.

(D) The processes $Cu \to Cu^{+}$ and $Li \to Li^{+}$ represent ionization,which involves the removal of an electron. These elements are metals and have relatively low ionization energies.
In contrast,$Br \to Br^{-}$ and $I \to I^{-}$ represent electron gain processes. The energy change associated with electron gain is known as electron gain enthalpy.
For halogens,as we move down the group from $Br$ to $I$,the atomic size increases,which decreases the effective nuclear attraction for the incoming electron.
However,the question asks for the process where the highest energy is absorbed. In the case of $I \to I^{-}$,the electron gain enthalpy is less negative (or more positive/endothermic in some contexts compared to others) due to inter-electronic repulsion in the larger $I$ atom compared to $Br$.
Comparing all options,the formation of $I^{-}$ is the least favorable process among the electron gain options,requiring the most energy input to overcome repulsion,making it the process where the highest energy is absorbed.

(B) The electron affinity of elements in a group generally decreases as we move down the group due to an increase in atomic size and shielding effect. However,for the carbon family (Group $14$),the electron affinity of $Si$ is higher than that of $C$. This is because the $2p$ orbitals in $C$ are very small,leading to significant inter-electronic repulsion,which makes it harder for an incoming electron to be added compared to the larger $3p$ orbitals of $Si$. The general order for Group $14$ is $Si > C > Ge > Sn > Pb$. Therefore,the correct order for $C$,$Si$,and $Ge$ is $Si > C > Ge$,which corresponds to $C < Si > Ge$.

(D) The electron affinity (or electron gain enthalpy) generally increases (becomes more negative) across a period and decreases down a group.
However,due to the small size of the $F$ atom,the incoming electron experiences significant inter-electronic repulsion in the $2p$ subshell.
In contrast,the $Cl$ atom has a larger $3p$ subshell,which reduces this repulsion,making the addition of an electron more favorable.
Therefore,the magnitude of electron affinity follows the order $Cl > F > O$.

(C) The electron gain enthalpy order for halogens is $Cl > F > Br > I$.
Due to the small size of the fluorine atom,the incoming electron experiences significant inter-electronic repulsion from the existing electrons in the compact $2p$ subshell.
Therefore,the energy released upon adding an electron to fluorine is less than that for chlorine,resulting in a lower electron gain enthalpy value for fluorine.

(D) The molar mass of chlorine $(Cl)$ is $35.5 \ g \ mol^{-1}$.
Number of moles of $Cl = \frac{1 \ g}{35.5 \ g \ mol^{-1}} = \frac{1}{35.5} \ mol$.
The energy released for $1 \ mol$ of $Cl$ is $3.7 \ eV \times 23.06 \ kcal \ mol^{-1} \ eV^{-1} = 85.322 \ kcal \ mol^{-1}$.
Energy released for $\frac{1}{35.5} \ mol = \frac{1}{35.5} \times 85.322 \ kcal \approx 2.4 \ kcal$.

(D) 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 $O^-$.
However,the second electron gain enthalpy is positive because energy must be supplied to overcome the strong electrostatic repulsion between the incoming electron and the negatively charged $O^-$ ion.

(A) The electronic configurations are: $A = 1s^2\, 2s^2\, 2p^4$ (Oxygen atom,$O$),$B = 1s^2\, 2s^2\, 2p^5$ (Oxygen anion,$O^-$),$C = 1s^2\, 2s^2\, 2p^6$ (Oxygen dianion,$O^{2-}$).
$1$. The process $A + e^- \to B$ $(O + e^- \to O^-)$ is exothermic because energy is released when an electron is added to a neutral atom.
$2$. The process $B + e^- \to C$ $(O^- + e^- \to O^{2-})$ is endothermic because energy is required to overcome the electrostatic repulsion between the incoming electron and the negatively charged $O^-$ ion.
$3$. Therefore,the statement '$B_{(g)} + e^- \to C_{(g)}$ is exothermic' is incorrect.
$4$. Ionization energy $(A_{(g)} \to A_{(g)}^{+} + e^-)$ is always endothermic,so option $B$ is incorrect.
$5$. The process $C_{(g)} \to A_{(g)} + 2e^-$ is the reverse of the two electron gain processes. Since the first step is exothermic and the second is endothermic,the overall process $C \to A + 2e^-$ is endothermic.
$6$. The process $B_{(g)} \to A_{(g)} + e^-$ is the reverse of the first electron gain process $(A + e^- \to B)$. Since $A + e^- \to B$ is exothermic,the reverse process $B \to A + e^-$ must be endothermic.
$7$. Re-evaluating the options based on standard chemical principles,the provided image suggests $A + e^- \to B$ is exothermic and $B + e^- \to C$ is endothermic. None of the options provided in the prompt are technically correct based on standard definitions. However,if we assume the question implies identifying the nature of electron gain,$A + e^- \to B$ is exothermic.

(C) The second electron gain enthalpy is the energy change when an electron is added to a uninegative ion $(X^-)$.
For oxygen $(O)$ and sulphur $(S)$,the addition of the second electron is an endothermic process due to strong inter-electronic repulsion between the incoming electron and the negative ion.
The values for the $2^{nd}$ electron gain enthalpy of $O$ and $S$ are $+ 780 \ kJ/mol$ and $+ 590 \ kJ/mol$ respectively.
Therefore,the correct option is $C$.

(A) Electron affinity generally increases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
For elements in the same period,the order is:
$Pnicogen (Group \ 15) < Chalcogen (Group \ 16) < Halogen (Group \ 17)$.
Halogens have the highest electron affinity in their respective periods because they only need one electron to achieve a stable noble gas configuration.
Therefore,the correct order is $Pnicogen < Chalcogen < Halogen$.

(B) Moving down the group $16$ elements $(O, S, Se, Te)$,the electron gain enthalpy generally becomes less negative (decreases) due to the increase in atomic size.
However,oxygen $(O)$ belongs to period $2$ and has a very small,compact atomic size.
This small size leads to significant inter-electronic repulsion between the incoming electron and the existing electrons in the $2p$ subshell,making the electron gain enthalpy of oxygen less negative than that of sulfur $(S)$.
Therefore,the correct order of electron gain enthalpy $(\Delta H_{eg})$ is $O < Se < S$.

(C) $1$. Carbon $(C)$ has lower $EA$ than Fluorine $(F)$. Statement $A$ is False $(F)$.
$2$. Sulphur $(S)$ has lower $EA$ than Fluorine $(F)$. Statement $B$ is True $(T)$.
$3$. Iodine $(I)$ has lower $EA$ than Bromine $(Br)$. Statement $C$ is False $(F)$.
$4$. Chlorine $(Cl)$ has higher $EA$ than Sulphur $(S)$. Statement $D$ is True $(T)$.
Thus,the sequence is $F, T, F, T$.

(C) An exothermic reaction is one where energy is released,meaning the electron gain enthalpy is negative.
$1$. Option $A$ represents the ionization energy of $F^-$,which is endothermic.
$2$. Option $B$ involves adding an electron to $Be$. Since $Be$ has a stable $2s^2$ configuration,adding an electron is endothermic.
$3$. Option $C$ represents the first electron gain enthalpy of oxygen,which is exothermic ($O_{(g)} + e^- \to O_{(g)}^{-}$,$\Delta H = -141 \ kJ/mol$).
$4$. Option $D$ represents the formation of $O^{2-}$ from $O$. The addition of the second electron to $O^-$ is endothermic due to inter-electronic repulsion.

(C) Electron affinity $(EA)$ is defined as the energy released when an electron is added to a neutral gaseous atom to form a negative ion.
However,for ions,we consider the electron gain enthalpy of the species.
Among the given options,$F^{-}$,$Cl^{-}$,$Li^{+}$,and $Na^{+}$ are all ions.
Anions like $F^{-}$ and $Cl^{-}$ already have a stable octet configuration and adding another electron would require energy (endothermic process),meaning they have negative electron affinity.
Cations like $Li^{+}$ and $Na^{+}$ have a strong electrostatic attraction for electrons due to their positive charge.
$Li^{+}$ has a smaller ionic radius compared to $Na^{+}$,resulting in a higher effective nuclear charge density,which makes it easier for $Li^{+}$ to attract an electron.
Therefore,$Li^{+}$ has the highest electron affinity among the given options.

(C) $(A) \, S_{(g)}^{-} \rightarrow S_{(g)}^{2-}$: The addition of an electron to an anion is endothermic due to inter-electronic repulsion.
$(B) \, Na_{(g)}^{+} + Cl_{(g)}^{-} \rightarrow NaCl_{(s)}$: This represents the formation of an ionic lattice,which is an exothermic process.
$(C) \, N_{(g)} \rightarrow N_{(g)}^{-}$: Nitrogen has a stable half-filled $p$-subshell $(2s^2 2p^3)$. Adding an electron to it is endothermic.
$(D) \, Al_{(g)}^{2+} \rightarrow Al_{(g)}^{3+}$: The removal of an electron (ionization) is always an endothermic process.

(B) The energy required for the process $X^{-}_{(g)} \to X_{(g)} + e^-$ is equal to the magnitude of the electron affinity (or electron gain enthalpy) of the atom $X$.
Let us consider the electron gain enthalpy process for each:
$(a) \, F_{(g)} + e^- \to F^{-}_{(g)} \, (\Delta H_1)$
$(b) \, P_{(g)} + e^- \to P^{-}_{(g)} \, (\Delta H_2)$
$(c) \, S_{(g)} + e^- \to S^{-}_{(g)} \, (\Delta H_3)$
$(d) \, Cl_{(g)} + e^- \to Cl^{-}_{(g)} \, (\Delta H_4)$
The energy released during electron gain follows the order: $|\Delta H_4| (Cl) > |\Delta H_1| (F) > |\Delta H_3| (S) > |\Delta H_2| (P)$.
Since the energy required for the reverse process is equal to the energy released in the forward process,the transformation requiring the least energy is the one with the lowest electron gain enthalpy magnitude,which is $P^{-}_{(g)} \to P_{(g)} + e^-$.

(D) The energy change that occurs when an electron is added to a neutral gaseous atom to form a negative ion is known as electron gain enthalpy.
When energy is released during this process,the electron gain enthalpy is negative.
Ionization enthalpy is the energy required to remove an electron.
Hydration enthalpy is the energy released when ions are hydrated.
Electronegativity is the tendency of an atom to attract a shared pair of electrons.

(A) The difficulty of electron attachment is related to the electron gain enthalpy. $A$ positive or very low negative electron gain enthalpy indicates that the process is difficult.
$A)$ Radon $(Rn)$ is a noble gas with a stable,fully filled valence shell configuration: $[Xe] 4f^{14} 5d^{10} 6s^2 6p^6$. It has no tendency to accept an electron,making the process highly endothermic and most difficult.
$B)$ Nitrogen $(N)$ has a half-filled $2p$ subshell: $[He] 2s^2 2p^3$. This provides extra stability,making electron attachment difficult,but less so than for a noble gas.
$C)$ Oxygen $(O)$ has a $2p^4$ configuration and readily accepts an electron to achieve a more stable state.
$D)$ Radium $(Ra)$ is an alkaline earth metal with a $7s^2$ configuration. While it has a low electron affinity,it is still more likely to accept an electron than a noble gas.
Therefore,the attachment of an electron is most difficult for Radon.

(B) The process of adding an electron to a neutral atom (first electron gain enthalpy) is generally exothermic,meaning energy is released.
However,adding an electron to an already negatively charged ion (like $S^-$) requires overcoming the electrostatic repulsion between the incoming electron and the negative ion.
Therefore,energy must be supplied (absorbed) to force the second electron onto the anion.
In the reaction $S^{-}_{(g)} + e^- \to S^{2-}_{(g)}$,energy is absorbed.

(A) The electron affinity of elements generally increases across a period and decreases down a group.
For the given elements,$N$ has a stable half-filled $2p^3$ configuration,making its electron affinity very low (near zero or negative).
Oxygen $(O)$ has a smaller size,which leads to significant inter-electronic repulsion when an electron is added to its $2p$ orbital,resulting in a lower electron affinity compared to Sulfur $(S)$.
Sulfur $(S)$ has a larger $3p$ orbital,which reduces inter-electronic repulsion,giving it the highest electron affinity among the three.
Therefore,the correct order of decreasing electron affinity is $S > O > N$.

(B) The electron affinity generally increases across a period from left to right as the effective nuclear charge increases and the size of the atom decreases.
Option $A$ represents $Al$ $([Ne] \, 3s^2, \, 3p^1)$.
Option $B$ represents $Cl$ $([Ne] \, 3s^2, \, 3p^5)$.
Option $C$ represents $S$ $([Ne] \, 3s^2, \, 3p^4)$.
Option $D$ represents $Cr$ $([Ne] \, 3s^2, \, 3p^6, \, 3d^5, \, 4s^1)$.
Among these,$Cl$ is a halogen with a high tendency to gain an electron to achieve a stable noble gas configuration. Therefore,it has the highest electron affinity.

(A) First,identify the elements based on their electronic configurations:
$I. 1s^2 2s^2 2p^6 3s^2 3p^5$ is Chlorine $(Cl)$,which has a high electron affinity.
$II. 1s^2 2s^2 2p^3$ is Nitrogen $(N)$,which has a half-filled $p$-orbital,making it very stable and resulting in a very low (near zero) electron affinity.
$III. 1s^2 2s^2 2p^5$ is Fluorine $(F)$. Although $F$ is more electronegative than $Cl$,its electron affinity is slightly lower than $Cl$ due to high inter-electronic repulsion in the small $2p$ subshell.
$IV. 1s^2 2s^2 2p^6 3s^1$ is Sodium $(Na)$,which is an alkali metal with a low electron affinity,but higher than Nitrogen $(N)$ because it can form a stable $3s^2$ configuration.
Comparing these: Nitrogen $(II)$ has the lowest electron affinity due to its stable half-filled configuration. Sodium $(IV)$ is next. Fluorine $(III)$ is higher than Sodium,and Chlorine $(I)$ has the highest electron affinity among these.
Therefore,the increasing order is $II < IV < III < I$.

(B) When an electron is added to a neutral atom,energy is released,making the first electron gain enthalpy negative.
However,when a second electron is added to a uninegative ion $(X^-)$,there is a strong electrostatic repulsion between the incoming electron and the negatively charged ion.
To overcome this repulsion,energy must be supplied to the system.
Therefore,the second electron gain enthalpy is always positive.

(A) The correct order of electron affinity is $Cl > F > S > O$.
Due to the small size of $2^{nd}$ period elements ($F$ and $O$),the incoming electron experiences greater inter-electronic repulsion compared to the $3^{rd}$ period elements ($Cl$ and $S$).
Consequently,the electron affinity of $2^{nd}$ period non-metals is lower than that of their respective $3^{rd}$ period counterparts.

(A) The process of adding an electron to a neutral gaseous atom is called electron gain enthalpy.
For most non-metals,this process is exothermic (releases energy).
However,for Nitrogen $(N)$,the process $N + e^- \to N^-$ is endothermic (requires absorption of energy).
This is because Nitrogen has a stable half-filled $2p^3$ electronic configuration,and the incoming electron experiences significant inter-electronic repulsion due to the small size of the atom.
Therefore,the correct option is $A$.

(B) The $Be^{-}$ ion is the most unlikely to exist because $Be$ has a stable fully filled $2s^2$ electronic configuration.
Adding an electron to $Be$ would require placing it into the $2p$ orbital,which is higher in energy and destabilizes the atom.
Furthermore,$Be$ has a positive electron gain enthalpy,meaning it has a strong tendency to lose electrons rather than gain them.

(A) Electron affinity generally increases across a period and decreases down a group. However,due to the small size of $F$ and $O$ atoms,the incoming electron experiences significant inter-electronic repulsion,making their electron affinity values lower than those of the elements below them in the same group ($Cl$ and $S$ respectively). The correct order of increasing electron affinity is $O < S < F < Cl$.

(D) The first electron gain enthalpy is exothermic,but the second electron gain enthalpy is always endothermic because of the strong electrostatic repulsion between the incoming electron and the negatively charged ion.
For oxygen $(O^-)$,the second electron gain enthalpy is approximately $+780 \ kJ \ mol^{-1}$.
For sulphur $(S^-)$,the second electron gain enthalpy is approximately $+590 \ kJ \ mol^{-1}$.
Thus,the values are $+780 \ kJ \ mol^{-1}$ and $+590 \ kJ \ mol^{-1}$ respectively.

(B) The second electron affinity is always endothermic (positive) because an electron is added to a negatively charged ion,leading to inter-electronic repulsion.
For oxygen $(O^-)$,the electron is added to a smaller $2p$ orbital,resulting in higher repulsion compared to sulphur $(S^-)$ where the electron is added to a larger $3p$ orbital.
Therefore,the magnitude of the second electron affinity of sulphur is less than that of oxygen.
Additionally,the first electron affinity of chlorine is greater than that of fluorine due to the small size and high inter-electronic repulsion in the $2p$ subshell of fluorine.
The correct order is $Cl > F > Br > I$.

(C) Electron gain enthalpy is the energy released when an electron is added to an isolated gaseous atom. Its value is negative,and the more easily an atom accepts an electron,the more negative its electron gain enthalpy becomes (i.e.,its magnitude increases).
With an increase in nuclear charge,the electron gain enthalpy generally increases.
Due to the extra stability of nitrogen (it has a half-filled valence shell $[He] \, 2s^2 \, 2p^3$),it is very difficult to add an electron to it,resulting in an almost zero electron gain enthalpy.
In general,on moving down a group,electron gain enthalpy decreases due to an increase in atomic size. However,from $F$ to $Cl$,the electron gain enthalpy increases.
Due to the very small size of the $F$ atom and the absence of a $3^{rd}$ shell,the incoming electron experiences inter-electronic repulsion,making the electron gain enthalpy of $F$ less negative than that of $Cl$. Thus,$Cl$ has the highest electron gain enthalpy in the periodic table. Therefore,the statement that electron gain enthalpy decreases from fluorine to iodine is incorrect,as it increases from $F$ to $Cl$ and then decreases.

(C) The formation of the oxide ion $O^{2-}_{(g)}$ occurs in two steps.
In the first step,an electron is added to a neutral oxygen atom,which is an exothermic process $(\Delta H = -142 \, kJ \, mol^{-1})$ due to the attraction between the nucleus and the incoming electron.
In the second step,an electron is added to the negatively charged $O^{-}$ ion.
Since both the $O^{-}$ ion and the incoming electron are negatively charged,there is a strong electrostatic repulsion between them.
To overcome this repulsion and force the electron into the ion,energy must be supplied,making the process endothermic $(\Delta H = 844 \, kJ \, mol^{-1})$.
Therefore,the $O^{-}$ ion resists the addition of another electron.

(B) The process $O^- + e^- \to O^{2-}$ involves the addition of an electron to a negatively charged ion.
Due to strong interelectronic repulsion between the incoming electron and the existing electrons in the $O^-$ ion,energy must be supplied to overcome this repulsion.
Therefore,this process is endothermic,meaning energy is absorbed.

(B) The electron affinity $(EA)$ generally increases across a period and decreases down a group.
However,there are exceptions due to electronic configuration.
For the given elements $(C, N, O, Cl)$:
$1$. $Cl$ (Chlorine) has the highest electron affinity among these because it is a halogen with a high effective nuclear charge and a small size.
$2$. $O$ (Oxygen) has a higher electron affinity than $C$ (Carbon) and $N$ (Nitrogen).
$3$. $N$ (Nitrogen) has a very low (near zero or positive) electron affinity because it has a stable half-filled $2p^3$ subshell.
$4$. Comparing $C$ and $N$,$C$ has a higher electron affinity than $N$.
Thus,the correct order is $Cl > O > C > N$.

(D) The electron gain enthalpy values for the given processes are as follows:
$1$. For $(i)$,$(ii)$,and $(iii)$,the addition of an electron to a neutral atom is an exothermic process,so $\Delta H_1, \Delta H_2, \text{ and } \Delta H_3$ are negative.
$2$. The order of electron gain enthalpy (magnitude) is $Cl > F > O$. Thus,$\Delta H_3$ (for $Cl$) is more negative than $\Delta H_2$ (for $F$) and $\Delta H_1$ (for $O$).
$3$. For $(iv)$,the addition of an electron to a negatively charged ion $(O^-)$ experiences strong inter-electronic repulsion,making the process endothermic. Thus,$\Delta H_4$ is positive.
$4$. Comparing the options,statement $(D)$ is incorrect because $\Delta H_2$ is negative,not positive.

(A) In a group,the electron affinity generally decreases down the group. However,due to the small size of the oxygen atom,the inter-electronic repulsion is high,which makes the electron affinity of oxygen lower than that of sulfur. Therefore,the order is $S > Se > O$.

(D) The electron affinity of halogens follows the order: $Cl > F > Br > I$.
Due to the small size of the $F$ atom,there is strong inter-electronic repulsion among the lone pairs,which makes the incoming electron experience less attraction compared to $Cl$.
Therefore,$F > I$ is a correct statement among the given options.

(B) The electron affinity of halogens decreases as we move down the group in the periodic table.
Therefore,the correct increasing order of electron affinity for these halogens is $I < Br < Cl$.

(B) Electron affinity is the energy released when an electron is added to a neutral gaseous atom.
Generally,electron affinity increases across a period and decreases down a group.
However,due to small size and inter-electronic repulsion,the electron affinity of $O$ is less than $S$,and $N$ has a very low (near zero) electron affinity due to its stable half-filled $2p^3$ configuration.
Thus,the correct order is $N < O < S < Cl$.