Electron affinity Questions in English

(B) The addition of the first electron to a neutral atom is generally exothermic (energy is released),as seen in options $A$,$C$,and $D$.
However,when an electron is added to a negatively charged ion,such as $O^{-}_{(g)}$,there is a strong electrostatic repulsion between the incoming electron and the existing negative charge.
To overcome this repulsion,energy must be supplied to the system.
Therefore,the process $O^{-}_{(g)} + e^{-} \rightarrow O^{2-}_{(g)}$ is endothermic and involves the absorption of energy.

(B) The electron gain enthalpy becomes more negative as we move across a period. However,in the case of halogens,Fluorine $(F)$ has a smaller size than Chlorine $(Cl)$.
Due to the small size of the $F$ atom,the incoming electron experiences significant inter-electronic repulsion from the electrons already present in the $2p$ subshell.
In contrast,the incoming electron in $Cl$ enters the $3p$ subshell,which is larger,resulting in less inter-electronic repulsion.
Therefore,$Cl$ has the highest value of negative electron gain enthalpy among the halogens.

(A) The electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom.
Noble gases like $Ne$ have a stable $ns^2 np^6$ electronic configuration.
Adding an electron to $Ne$ requires energy to place it in the next higher energy shell,making the process endothermic.
Therefore,$Ne$ has a positive electron gain enthalpy.

(B) The electron gain enthalpy becomes more negative as we move from left to right across a period and becomes less negative as we move down a group.
For the given elements:
$Cl$ (Group $17$,Period $3$) has the highest negative electron gain enthalpy.
$S$ (Group $16$,Period $3$) follows $Cl$.
$Li$ (Group $1$,Period $2$) and $Na$ (Group $1$,Period $3$) have low negative values,with $Li$ being more negative than $Na$ due to its smaller size.
Thus,the correct order of negative electron gain enthalpy is $Cl > S > Li > Na$.

(A) The electron gain enthalpy values for the given elements are:
$F$: $-328 \ kJ \ mol^{-1}$ (Matches $II$)
$Cl$: $-349 \ kJ \ mol^{-1}$ (Matches $IV$)
$O$: $-141 \ kJ \ mol^{-1}$ (Matches $I$)
$S$: $-200 \ kJ \ mol^{-1}$ (Matches $III$)
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(B) The electron gain enthalpy is the energy released when an electron is added to a neutral gaseous atom.
For the given elements,the values are:
$N$: $+7 \ kJ/mol$ (positive due to stable half-filled $2p^3$ configuration)
$O$: $-141 \ kJ/mol$
$Cl$: $-349 \ kJ/mol$
$Al$: $-43 \ kJ/mol$
Comparing the magnitudes,the order of electron gain enthalpy (most negative to least negative) is $Cl < O < Al < N$.
However,considering the standard convention where more negative values are considered 'higher' in terms of energy release,the order of increasing electron gain enthalpy is $Al < N < O < Cl$.

(A) The electron gain enthalpy generally becomes more negative (increases in magnitude) from left to right across a period and decreases down a group.
However,elements of the $2^{nd}$ period have very small atomic sizes,which leads to significant inter-electronic repulsion when an electron is added.
Due to this,the electron gain enthalpy of $O$ and $N$ is less negative than that of the corresponding elements in the $3^{rd}$ period ($S$ and $P$).
Fluorine $(F)$ has the highest electron gain enthalpy among these.
Comparing $S$,$O$,and $N$,the order is $S > O > N$.
Therefore,the correct order is $F > S > O > N$.

(C) is correct but $R$ is incorrect.
Due to the small size of the fluorine atom,the electron-electron repulsions in the $2p$ subshell are significantly higher compared to the larger chlorine atom.
Consequently,the incoming electron experiences less attraction in fluorine,making its electron gain enthalpy less negative than that of chlorine.

(B) The electron gain enthalpy (with negative sign) represents the energy released when an electron is added to a neutral gaseous atom.
For the given elements,the general trend is influenced by atomic size and electron-electron repulsion.
$1.$ Among halogens,$Cl$ has a higher (more negative) electron gain enthalpy than $F$ because the small size of $F$ leads to significant inter-electronic repulsion.
$2.$ Among group $15$ elements,$P$ has a higher electron gain enthalpy than $N$ due to the larger size and lower electron density of $P$ compared to $N$.
$3.$ Comparing the groups,$N$ has the lowest value due to its very small size and high electron density,while $Cl$ has the highest value.
Thus,the correct order is $N < P < F < Cl$.

(A) Electron gain enthalpy generally becomes more negative (increases in magnitude) as we move from left to right across a period due to decreasing atomic size.
However,there are exceptions due to electron-electron repulsions in smaller atoms.
For $F$ (Fluorine),it has the highest electron gain enthalpy among these.
For $O$ (Oxygen),the incoming electron experiences significant repulsion due to its small size,making its electron gain enthalpy less negative than that of $S$ (Sulphur).
$N$ (Nitrogen) has a half-filled $2p$ subshell $(2p^3)$,making it very stable and resulting in a very low (near zero) electron gain enthalpy.
Thus,the correct order of decreasing electron gain enthalpy is $F > S > O > N$.

(D) The process of formation of $O^{2-}$ in the gas phase is unfavorable even though $O^{2-}$ is isoelectronic with neon.
This is because the electronic repulsion between the incoming electron and the existing negative charge on $O^{-}$ outweighs the stability gained by achieving the noble gas configuration.
The formation of oxide ion,$O^{2-}_{(g)}$ from oxygen atom requires first an exothermic and then an endothermic step.
$O_{(g)} + e^{-} \longrightarrow O^{-}_{(g)} ; \Delta H^{\ominus} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^{-} \longrightarrow O^{2-}_{(g)} ; \Delta H^{\ominus} = +760 \ kJ \ mol^{-1}$
When an electron is added to the $O^{-}$ anion,there is strong electrostatic repulsion between the two negative charges; due to this,the second electron gain enthalpy of oxygen is positive.

(A) On moving down the group,the electron gain enthalpy generally decreases.
Oxygen,due to its small size,experiences significant electron-electron repulsions in its relatively small $2p$-subshell.
Consequently,an incoming electron is not accepted as easily as in the case of other elements in the same group,which is why oxygen has a lower electron gain enthalpy than sulfur.
Therefore,the correct order of electron gain enthalpy for the group $16$ elements is $S > Se > Te > O$.

(C) The process of electron gain is represented as: $Cl_{(g)} + e^{-} \xrightarrow{\Delta_{eg}H} Cl^{-}_{(g)}$
Energy of the product $Cl^{-}_{(g)}$ is equal to the sum of the energy of the reactant $Cl_{(g)}$ and the enthalpy change $\Delta_{eg}H$ of the reaction.
$\text{Energy of } Cl^{-}_{(g)} = \text{Energy of } Cl_{(g)} + \Delta_{eg}H$
Given that the energy of $Cl_{(g)}$ is $x \ kJ \ mol^{-1}$ and $\Delta_{eg}H = -349 \ kJ \ mol^{-1}$.
$\text{Energy of } Cl^{-}_{(g)} = x + (-349) = x - 349 \ kJ \ mol^{-1}$.

(A) The electron gain enthalpy becomes more negative as we move from left to right across a period and becomes less negative as we move down a group.
However,due to small size,the electron-electron repulsion in $F$ and $O$ makes their electron gain enthalpies less negative than those of $Cl$ and $S$ respectively.
The order of increasing electron gain enthalpy (more negative value means lower enthalpy) is:
$O < S < F < Cl$.
Thus,the correct order is $O < S < F < Cl$.

(B) The electron gain enthalpy $(\Delta_{eg} H)$ values for the given elements are:
$Cl = -349 \ kJ \ mol^{-1}$
$S = -200 \ kJ \ mol^{-1}$
$I = -295 \ kJ \ mol^{-1}$
Comparing these with the given values:
$X = -349 \ kJ \ mol^{-1} = Cl$
$Y = -200 \ kJ \ mol^{-1} = S$
$Z = -295 \ kJ \ mol^{-1} = I$
Therefore,$X, Y$ and $Z$ are $Cl, S$ and $I$ respectively.

(C) The electron gain enthalpy $(\Delta_{eg}H)$ values for the given elements are as follows:
$O$: $-141 \ kJ \ mol^{-1}$ $(A-III)$
$F$: $-328 \ kJ \ mol^{-1}$ $(B-IV)$
$Cl$: $-349 \ kJ \ mol^{-1}$ $(C-II)$
$S$: $-200 \ kJ \ mol^{-1}$ $(D-I)$
Therefore,the correct matching is $A-III, B-IV, C-II, D-I$.

(B) Electron gain enthalpy becomes less negative (or smaller in magnitude) as we move down a group due to an increase in atomic size.
Among the given elements,$Cl$ and $S$ are in the $3^{rd}$ period,while $O$ and $I$ are in the $2^{nd}$ and $5^{th}$ periods respectively.
$Cl$ has the most negative electron gain enthalpy.
$Iodine$ $(I)$ has the least negative electron gain enthalpy among the halogens listed due to its large atomic size,which reduces the attraction for an incoming electron.
Comparing $O$ and $S$,$O$ has a smaller electron gain enthalpy than $S$ due to inter-electronic repulsion in its small $2p$ orbital.
However,comparing all four,$Iodine$ $(I)$ has the least electron gain enthalpy.

(B) Noble gases have large positive electron gain enthalpies because the added electron has to enter the next higher principal quantum level,which leads to a very unstable electronic configuration.
Therefore,the set containing noble gases,$Kr, Xe, Rn$,will have positive electron gain enthalpies.
Thus,option $(B)$ is correct.

(C) The electron gain enthalpy becomes more negative as we move across a period,but there are exceptions due to electronic configuration and atomic size.
$Cl$ has the highest negative electron gain enthalpy among these elements because of its relatively larger size compared to $F$,which reduces inter-electronic repulsion.
$F$ has a lower negative electron gain enthalpy than $Cl$ due to its very small size,which causes significant inter-electronic repulsion.
Between $S$ and $P$,$S$ has a higher negative electron gain enthalpy than $P$ because $P$ has a stable half-filled $3p^3$ electronic configuration.
Therefore,the correct order of negative electron gain enthalpy is: $Cl > F > S > P$.

(D) The electron gain enthalpy values (in $kJ/mol$) for the given elements are: $O = -141$,$S = -200$,and $Se = -195$.
Due to the small size of the $2p$ orbital in oxygen,there is significant inter-electronic repulsion,which makes the addition of an electron less exothermic compared to sulfur and selenium.
Therefore,the correct order of electron gain enthalpy (magnitude) is $S > Se > O$.

(B) Electron affinity is the energy released when an electron is added to a neutral gaseous atom to form a negative ion.
$Cl$ has the highest electron affinity among the given elements.
Although $F$ is more electronegative than $Cl$,the small size of the $F$ atom leads to significant inter-electronic repulsion between the incoming electron and the existing electrons in the $2p$ subshell.
Consequently,the energy released upon adding an electron to $F$ $(-333 \ kJ \ mol^{-1})$ is less than that released for $Cl$ $(-349 \ kJ \ mol^{-1})$.
Therefore,$Cl$ has the greatest electron affinity.

(A) Electron gain enthalpy $\left(\Delta_{eg} H\right)$ is defined as the enthalpy change when an electron is added to a neutral gaseous atom in its ground state to form a negative ion (anion).
The process is represented by the equation: $X_{(g)} + e^{-} \longrightarrow X^{-}_{(g)}$.
Option $A$ correctly represents this process.

(C) Elements with half-filled or completely-filled orbitals,$i.e.$,having stable electronic configurations,have very low negative values of electron affinity.
Since the electron affinity of $B$ $(-60 \ kJ \ mol^{-1})$ is the lowest among the given values,it corresponds to the most stable electronic configuration,which is $3s^2 3p^3$ (half-filled $p$-orbital).

(B) The electron gain enthalpy of noble gases is positive due to their stable electronic configuration $(ns^2 np^6)$.
Among the given noble gases,$Ne$ has the highest electron gain enthalpy (most positive value).
As we move down the group,the atomic size increases,and the effective nuclear charge experienced by the incoming electron decreases,making it progressively easier to add an electron compared to the smaller atoms,although the values remain positive.
Therefore,$Ne$ has the highest electron gain enthalpy among the options.

(B) The electron affinity of elements in the second period is lower than that of the corresponding elements in the third period due to their small atomic size and high inter-electronic repulsion.
Thus,$S > O$ and $Cl > F$.

(B) Key Point:
$(i)$ Half-filled or fully-filled configurations have low electron affinity (i.e.,positive or less negative electron gain enthalpy).
$(ii)$ Smaller the size,generally the more negative the electron gain enthalpy.
On moving across a period,the size of atoms decreases due to an increase in nuclear charge.
$(i)$ Electronic configurations of the elements are:
$C (Z=6) = 1s^{2} 2s^{2} 2p^{2}$
$N (Z=7) = 1s^{2} 2s^{2} 2p^{3}$
$O (Z=8) = 1s^{2} 2s^{2} 2p^{4}$
$(ii)$ Due to the stable half-filled electronic configuration of nitrogen $(2p^{3})$,it has a very low electron affinity (positive electron gain enthalpy value of $30.9 \ kJ/mol$).
$(iii)$ As the size of $O$ is smaller than that of $C$,$O$ has a more negative electron gain enthalpy $(-141.1 \ kJ/mol)$ compared to $C$ $(-122.3 \ kJ/mol)$.
Thus,the correct order of electron affinity is $N < C < O$.

(A) The 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.
Among the given elements ($C$,$N$,$O$,$F$),$F$ has the highest effective nuclear charge and the smallest atomic size.
Although $Cl$ has the highest electron affinity in the entire periodic table,among the provided options,$F$ has the highest $1^{st}$ electron affinity.