Ionisation energy Questions in English

(D) Ionization potential is the energy required to remove an electron from the outermost shell of an isolated gaseous atom.
As the atomic radius increases,the distance between the nucleus and the outermost electron increases.
This leads to a decrease in the electrostatic force of attraction between the nucleus and the valence electron.
Consequently,less energy is required to remove the electron,meaning the ionization potential decreases with an increase in atomic radii.

(B) The correct order for ionization energy is $B < Be < C < O < N$.
On moving from left to right in a period,with an increase in the atomic number,the ionization enthalpy generally increases.
However,there are two main exceptions due to electronic configuration:
$(1)$ Ionization energy of $B < $ ionization energy of $Be$. This is because $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ subshell,which is more stable than the $2p^1$ electron in $B$ $(1s^2 2s^2 2p^1)$.
$(2)$ Ionization energy of $O < $ ionization energy of $N$. Nitrogen $(1s^2 2s^2 2p^3)$ has a half-filled $2p$ subshell,which is extra stable. Therefore,more energy is required to remove an electron from $N$ compared to $O$ $(1s^2 2s^2 2p^4)$.

(C) In a period,the first ionization enthalpy generally increases from left to right. However,$N$ $(2s^2 2p^3)$ has a stable half-filled $p$-orbital configuration compared to $O$ $(2s^2 2p^4)$.
Due to this stability,the ionization enthalpy of $N$ is higher than that of $O$.
Therefore,the correct order for the first ionization enthalpy is $B < C < O < N$.
Thus,the arrangement $B < C < N < O$ is incorrect.

(C) The general trend for first ionization enthalpy across a period is that it increases from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
However,there are exceptions due to stable electronic configurations.
The electronic configurations of the second-period elements are:
$Li (1s^2 2s^1)$,$Be (1s^2 2s^2)$,$B (1s^2 2s^2 2p^1)$,$C (1s^2 2s^2 2p^2)$,$N (1s^2 2s^2 2p^3)$,$O (1s^2 2s^2 2p^4)$,$F (1s^2 2s^2 2p^5)$,$Ne (1s^2 2s^2 2p^6)$.
$Be$ has a fully filled $2s$ orbital,which is more stable than the $2p^1$ configuration of $B$.
$N$ has a half-filled $2p$ orbital,which is more stable than the $2p^4$ configuration of $O$.
Thus,the correct decreasing order for the elements $Be, B, C, N, F$ is $F > N > C > Be > B$.

(D) The first ionization energy of $N$ $(1s^2, 2s^2, 2p^3)$ is higher than that of $O$ $(1s^2, 2s^2, 2p^4)$ because $N$ has a stable,exactly half-filled $2p$ subshell.
Across a period,the effective nuclear charge increases,not decreases,due to the addition of protons in the same shell while shielding remains relatively constant.
Therefore,the Assertion is incorrect and the Reason is also incorrect.

(C) The Assertion is correct because elements tend to lose electrons to achieve a stable noble gas configuration.
The Reason is incorrect because ionization enthalpy is defined as the energy $required$ (absorbed) to remove an electron from an isolated gaseous atom,not the energy $released$.

(B) The first ionisation enthalpy generally increases across a period due to an increase in effective nuclear charge.
However,there are deviations due to stable electronic configurations.
The electronic configurations for second period elements are:
$Li (2s^1)$,$Be (2s^2)$,$B (2s^2 2p^1)$,$C (2s^2 2p^2)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$,$F (2s^2 2p^5)$,$Ne (2s^2 2p^6)$.
$Be$ has a fully filled $2s$ orbital,making its ionisation energy higher than $B$.
$N$ has a half-filled $2p$ subshell,making its ionisation energy higher than $O$.
The correct order is: $Li < B < Be < C < O < N < F < Ne$.

(A) The electronic configurations are:
$Na = [Ne] 3s^{1}$
$Mg = [Ne] 3s^{2}$
$Al = [Ne] 3s^{2} 3p^{1}$
$Si = [Ne] 3s^{2} 3p^{2}$
Ionization energy generally increases across a period,but $Mg$ has a higher ionization energy than $Al$ due to the stable,fully-filled $3s$ orbital in $Mg$ compared to the $3p$ electron in $Al$.
The values are:
$Na = 496 \ kJ/mol$
$Mg = 737 \ kJ/mol$
$Al = 577 \ kJ/mol$
$Si = 786 \ kJ/mol$
Thus,the correct order is $496, 737, 577, 786$.

(A) The $IE$ values indicate that the metal belongs to the $I^{st}$ group because the second $IE$ is very high,suggesting the presence of only one valence electron.
Therefore,the metal hydroxide will be of the type $MOH$.
The reaction with $HCl$ is:
$MOH + HCl \rightarrow MCl + H_{2}O$
From the stoichiometry,$1 \; mol$ of $MOH$ reacts with $1 \; mol$ of $HCl$.
The reaction with $H_{2}SO_{4}$ is:
$2MOH + H_{2}SO_{4} \rightarrow M_{2}SO_{4} + 2H_{2}O$
Dividing by $2$ to match $1 \; mol$ of $MOH$:
$MOH + \frac{1}{2} H_{2}SO_{4} \rightarrow \frac{1}{2} M_{2}SO_{4} + H_{2}O$
From the stoichiometry,$1 \; mol$ of $MOH$ reacts with $0.5 \; mol$ of $H_{2}SO_{4}$.
Thus,$1$ mole of $HCl$ and $0.5$ mole of $H_{2}SO_{4}$ are required.

(A) $Be$ has the electronic configuration $1s^{2} 2s^{2}$.
$B$ has the electronic configuration $1s^{2} 2s^{2} 2p^{1}$.
Statement $(I)$ is correct: It is easier to remove a $2p$ electron than a $2s$ electron because the $2p$ orbital is higher in energy and less stable.
Statement $(II)$ is correct: The $2p$ electron in $B$ experiences greater shielding from the $1s^{2} 2s^{2}$ core compared to the $2s$ electrons in $Be$,which are less shielded.
Statement $(III)$ is correct: The $2s$ orbital is closer to the nucleus and has higher penetration power than the $2p$ orbital.
Statement $(IV)$ is incorrect: $B$ has a smaller atomic radius than $Be$ due to the increase in effective nuclear charge across the period.
Therefore,statements $(I), (II),$ and $(III)$ are correct.

(A) The first ionization enthalpy for $Al$ will be closer to $575 \, kJ \, mol^{-1}$.
In the third period,the ionization enthalpy generally increases from left to right due to an increase in effective nuclear charge.
However,$Mg$ $(3s^2)$ has a fully filled orbital,making it more stable than $Al$ $(3s^2 3p^1)$.
Therefore,the first ionization enthalpy of $Al$ is lower than that of $Mg$ $(737 \, kJ \, mol^{-1})$.
Since $575 \, kJ \, mol^{-1}$ is less than $737 \, kJ \, mol^{-1}$,it is the correct prediction.

(N/A) The first ionization enthalpy of $Na$ is lower than that of $Mg$ due to two main reasons:
$1.$ The atomic size of $Na$ $([Ne] 3s^1)$ is larger than that of $Mg$ $([Ne] 3s^2)$.
$2.$ The effective nuclear charge of $Mg$ is higher than that of $Na$.
Consequently,the energy required to remove the first electron from $Mg$ is higher than that for $Na$.
However,the second ionization enthalpy of $Na$ is higher than that of $Mg$. After the loss of the first electron,$Na^+$ attains the stable noble gas configuration $(1s^2 2s^2 2p^6)$. Removing a second electron from this stable configuration requires a very high amount of energy.
In contrast,$Mg^+$ $([Ne] 3s^1)$ still has one electron in the $3s$-orbital. Removing this electron is relatively easier compared to removing an electron from the stable noble gas core of $Na^+$. Therefore,the second ionization enthalpy of $Na$ is significantly higher than that of $Mg$.

(N/A) The factors responsible for the decrease in ionization enthalpy of the main group elements down a group are as follows:
$(i)$ Increase in atomic size: As we move down a group,the number of electron shells increases,leading to a gradual increase in atomic size. As the distance between the nucleus and the valence electrons increases,the electrostatic attraction decreases,making it easier to remove the valence electrons.
$(ii)$ Increase in shielding effect: As we move down a group,the number of inner electron shells increases. This results in a greater shielding (or screening) effect of the valence electrons from the nuclear charge by the inner core electrons. Consequently,the effective nuclear charge experienced by the valence electrons decreases,requiring less energy to remove them.

(N/A) On moving down a group,ionization enthalpy generally decreases due to an increase in the atomic size and shielding effect.
Thus,on moving down group $13$,ionization enthalpy decreases from $B$ to $Al$.
However,$Ga$ has a higher ionization enthalpy than $Al$ because $Ga$ follows the $d$-block elements.
The shielding provided by $d$-electrons is poor,which leads to a higher effective nuclear charge experienced by the valence electrons in $Ga$.
Moving from $Ga$ to $In$,the ionization enthalpy decreases due to the increase in atomic size.
Finally,from $In$ to $Tl$,the ionization enthalpy increases because $Tl$ follows the $4f$ and $5d$ orbitals.
The poor shielding effect of $f$ and $d$ electrons results in a higher effective nuclear charge,holding the valence electron more strongly.

(N/A) The ionization enthalpy of an atom depends on the number of electrons and the nuclear charge (number of protons).
Since isotopes of the same element have the same atomic number,they possess the same number of protons and electrons.
Therefore,the electronic configuration and the effective nuclear charge experienced by the valence electrons remain identical for both isotopes.
Consequently,the first ionization enthalpy for two isotopes of the same element is the same.

(D) Electrons in orbitals with a lower principal quantum number $(n)$ are closer to the nucleus and experience a stronger electrostatic force of attraction compared to electrons in orbitals with a higher $n$ value.
Therefore,it requires more energy to remove an electron from an orbital with a lower $n$ value.
Conversely,the removal of an electron from an orbital with a higher $n$ value is easier.
Thus,statement $(d)$ is incorrect.

(N/A) The ionisation enthalpy of carbon (the first element of group $14$) is very high $(1086 \, kJ/mol)$.
This is expected due to its small atomic size.
However,on moving down the group to silicon,there is a sharp decrease in the ionisation enthalpy $(786 \, kJ/mol)$.
This is because of an appreciable increase in the atomic size of elements when moving down the group,which reduces the effective nuclear attraction on the valence electrons.

(N/A) Generally,when we move from left to right in the periodic table,the value of ionization enthalpy increases.
$(i)$ $\Delta_{i}H(Be) > \Delta_{i}H(B)$: $Be$ has a higher $\Delta_{i}H$ than $B$ because $Be$ has the electronic configuration $1s^2 2s^2$,while $B$ has the electronic configuration $1s^2 2s^2 2p^1$. The higher $\Delta_{i}H$ of $Be$ compared to $B$ is due to:
$\rightarrow$ The electronic configuration of $Be$ is more stable (completely filled $2s$ orbitals) than that of $B$.
$\rightarrow$ In $Be$,the electron to be removed is an $s$-electron,while in $B$,it is a $p$-electron.
$\rightarrow$ The penetration of a $2s$-electron to the nucleus is greater than that of a $2p$-electron. Consequently,the $2p$-electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$-electron of $Be$.
As a result,the $2s$-electron is more strongly attracted to the nucleus. Therefore,it is more difficult to remove a $2s$-electron from $Be$ than a $2p$-electron from $B$.
$(ii)$ $\Delta_{i}H(O) < \Delta_{i}H(N)$ and $\Delta_{i}H(O) < \Delta_{i}H(F)$: Oxygen has four electrons in $2p$-orbitals. Two of these electrons must occupy the same $2p$-orbital,resulting in increased electron-electron repulsion.
$\rightarrow$ Nitrogen $(N)$ has a stable half-filled configuration $(2p^3)$,which is highly stable.
$\rightarrow$ Fluorine $(F)$ has a higher nuclear charge than Oxygen,which holds its electrons more tightly.
Therefore,$O$ has a lower ionization enthalpy than both $N$ and $F$.

(N/A) In a period,the first ionisation enthalpy generally increases from left to right. This is because the atomic radius decreases and the effective nuclear charge increases as we move across a period,making it harder to remove an electron.
However,there are exceptions due to electronic configurations:
$1$. The ionisation enthalpy of $Be$ $(1s^2 2s^2)$ is higher than that of $B$ $(1s^2 2s^2 2p^1)$ because $Be$ has a fully filled $2s$ orbital,which is more stable.
$2$. The ionisation enthalpy of $N$ $(1s^2 2s^2 2p^3)$ is higher than that of $O$ $(1s^2 2s^2 2p^4)$ because $N$ has a half-filled $2p$ subshell,which provides extra stability.

(N/A) Ionisation enthalpy and atomic radius are closely related.
When moving from top to bottom in a group,the atomic radius increases as new electron shells are added.
As we descend a group,the outermost electron is placed at a greater distance from the nucleus,and the number of inner shells increases,which leads to a more pronounced shielding effect (or screening effect).
Although the nuclear charge increases,the effect of the increased distance and the shielding effect outweighs the increase in nuclear charge.
Consequently,the force of attraction between the nucleus and the valence electron decreases,making it easier to remove the electron.
Therefore,ionisation enthalpy decreases as we move from top to bottom in a group.

Within the main group elements,the ionization enthalpy decreases regularly as we move down the group. This is due to the following factors:
$(i)$ Atomic size: On moving down the group,there is a gradual increase in atomic size due to an additional main energy shell $(n)$.
$(ii)$ Shielding effect: There is an increase in shielding effect on the outermost electron due to an increase in the number of inner electrons.
$(iii)$ Nuclear charge: In going from top to bottom in a group,the nuclear charge increases.
The effect of the increase in atomic size and the shielding effect is much more than the effect of the increase in nuclear charge.
As a result,the electron becomes less tightly held to the nucleus as we move down the group.
Hence,there is a gradual decrease in the ionization enthalpies in a group.

(N/A) The order of ionisation enthalpy is $3^{rd} > 2^{nd} > 1^{st}$.
This is because,as an electron is removed,the remaining electrons experience a greater effective nuclear charge due to the decrease in electron-electron repulsion and the constant nuclear charge.
Consequently,the attraction between the nucleus and the remaining electrons increases,requiring more energy to remove subsequent electrons.

(A) The electronic configuration of $Na$ is $[Ne] 3s^{1}$ and that of $Mg$ is $[Ne] 3s^{2}$.
The configuration of $Mg$ is more stable (being completely filled) than that of $Na$.
Therefore,the first ionization enthalpy of $Mg$ is higher than that of $Na$.
After the loss of one electron from $Na$,it acquires the stable noble gas configuration of $Ne$ $(1s^{2} 2s^{2} 2p^{6})$.
On the other hand,for $Mg$,the configuration after the first ionization becomes $[Ne] 3s^{1}$.
Thus,the electronic configuration of $Na^{+}$ is more stable than that of $Mg^{+}$,making the removal of the second electron from $Na^{+}$ much more difficult than from $Mg^{+}$.
$Na ([Ne] 3s^{1}) \longrightarrow Na^{+} ([Ne]) + e^{-}$
$Mg ([Ne] 3s^{2}) \longrightarrow Mg^{+} ([Ne] 3s^{1}) + e^{-}$
$IE_{1}(Na) < IE_{1}(Mg)$
$IE_{2}(Na) > IE_{2}(Mg)$

(N/A) The term 'isolated gaseous atom' implies that the atom must be free from the influence of other atoms,existing in the gaseous phase. This ensures that the energy measured is solely due to the removal or addition of an electron,without interference from interatomic forces or bond dissociation energy.
'Ground state' refers to the lowest energy state of an atom. Defining these properties at the ground state ensures consistency,as measurements from excited states would yield different,non-standard values.

(N/A) Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
Since they have the same atomic number,they possess the same number of electrons and the same nuclear charge (number of protons).
Ionization enthalpy depends on the electronic configuration and the effective nuclear charge experienced by the valence electrons.
Since both isotopes have identical electronic configurations and effective nuclear charges,their first ionization enthalpies are expected to be the same.

(N/A) The electronic configuration of $Na$ is $[Ne] 3s^1$ and $Mg$ is $[Ne] 3s^2$.
Since $Na$ has only one electron in its outermost shell,it is easier to remove the first electron from $Na$ than from $Mg$,which has a stable fully-filled $3s^2$ configuration. Thus,the first ionisation enthalpy of $Na$ is lower than that of $Mg$.
After the removal of the first electron,$Na^+$ attains the stable noble gas configuration of $Ne$ $(1s^2 2s^2 2p^6)$. Removing the second electron from $Na^+$ requires breaking this stable octet,which is very difficult.
In contrast,$Mg^+$ has the configuration $[Ne] 3s^1$. Removing the second electron from $Mg^+$ is easier because it leads to the stable noble gas configuration of $Ne$. Therefore,the second ionisation enthalpy of $Na$ is significantly higher than that of $Mg$.

(N/A) The general trend in a period is that the first ionization enthalpy increases from left to right due to an increase in effective nuclear charge and a decrease in atomic radius.
However,there are deviations in this trend:
$1$. $Be$ $(1s^2 2s^2)$ has a higher ionization enthalpy than $B$ $(1s^2 2s^2 2p^1)$. This is because $Be$ has a fully filled $2s$ orbital,which is more stable and requires more energy to remove an electron compared to the $2p$ electron in $B$.
$2$. $N$ $(1s^2 2s^2 2p^3)$ has a higher ionization enthalpy than $O$ $(1s^2 2s^2 2p^4)$. This is because $N$ has a half-filled $2p$ subshell,which provides extra stability. In $O$,the removal of an electron leads to a more stable half-filled $p$-orbital configuration,making it easier to remove an electron compared to $N$.

(N/A) Ionisation enthalpy: The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom to convert it into a gaseous cation is called its ionisation enthalpy. It is represented by $\Delta_{i}H$.
Factors affecting ionisation enthalpy of the elements: It depends upon the following factors:
$(i)$ Nuclear charge: As nuclear charge increases,ionisation enthalpy increases. This is because,with an increase in nuclear charge,the electrons of the outer shell are more tightly bonded by the nucleus,and thus a greater amount of energy is required to remove that electron from the atom.
For example,the ionisation enthalpy increases as we move from left to right in a period due to an increase in nuclear charge.

Element of $2^{nd}$ period	$Li$	$Be$	$B$	$C$	$N$	$O$	$F$	$Ne$
Nuclear charge	$+3$	$+4$	$+5$	$+6$	$+7$	$+8$	$+9$	$+10$
$1^{st}$ ionisation enthalpy $(kJ \ mol^{-1})$	$520$	$899$	$801$	$1086$	$1402$	$1314$	$1681$	$2080$

$(ii)$ Atomic size or radius: As the size of the atom increases,ionisation enthalpy decreases. The distance between the outer electron and the nucleus increases with an increase in atomic radius,and the attractive force on the outer electron decreases. As a result,outer electrons are held less tightly,and hence less energy is required to remove that electron. Thus,ionisation enthalpy decreases on moving from top to bottom in a group.

Element (alkali metals)	$Li$	$Na$	$K$	$Rb$	$Cs$
First ionisation enthalpies $(kJ \ mol^{-1})$	$520$	$496$	$419$	$403$	$374$

$(iii)$ Penetration effect of the electrons: As the penetration effect of the electrons increases,the ionisation enthalpy increases. In multi-electron atoms,the $s$-orbital electrons' probability is maximum near the nucleus,and this probability decreases for $p$,$d$,and $f$ orbitals.

(N/A) The ionisation enthalpy of carbon (the first element of group $14$) is very high $(1086 \ kJ/mol)$. This is expected due to its small atomic size.
However,on moving down the group to silicon,there is a sharp decrease in the ionisation enthalpy $(786 \ kJ/mol)$.
This significant decrease is primarily due to the appreciable increase in the atomic size of elements as we move down the group,which reduces the effective nuclear attraction on the valence electrons.

(A) Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.
Mathematically,it is represented as: $X(g) + \text{Energy} \rightarrow X^+(g) + e^-$.

(N/A) The ionisation enthalpy decreases regularly as we move from top to bottom in any group.
This trend is explained by the following factors:
$(i)$ Atomic size: On moving from top to bottom,the atomic size increases due to the addition of a new principal energy shell at each succeeding element. This increases the distance between the valence electrons and the nucleus,reducing the force of attraction,thus decreasing the ionisation enthalpy.
$(ii)$ Shielding effect: With the addition of new shells,the shielding effect increases,which decreases the effective nuclear charge felt by the valence electrons,leading to a decrease in ionisation enthalpy.
$(iii)$ Nuclear charge: Although the nuclear charge increases with atomic number,the combined effect of the increase in atomic size and the screening effect more than compensates for it. Consequently,valence electrons are less firmly held by the nucleus,and ionisation enthalpy decreases down the group.
Comparison: Group-$1$ elements (alkali metals) have the lowest ionisation enthalpies in their respective periods because they have the largest atomic size. Group-$17$ elements (halogens) have the highest ionisation enthalpies in their respective periods due to their small atomic size and high effective nuclear charge.

Group-$1$ elements and $1^{st}$ ionisation enthalpies $(kJ \ mol^{-1})$	Group-$17$ elements and $1^{st}$ ionisation enthalpies $(kJ \ mol^{-1})$
$H$: $1312$	$F$: $1681$
$Li$: $520$	$Cl$: $1255$
$Na$: $496$	$Br$: $1142$
$K$: $419$	$I$: $1009$
$Rb$: $403$	$At$: $917$
$Cs$: $374$	-

(N/A) The quantitative measure of the tendency of an element to lose an electron is given by its ionization enthalpy.
"The energy required to remove an electron from an isolated gaseous neutral atom $(X)$ in its ground state is called ionization enthalpy."
$X_{(g)} + \Delta_{i}H = X_{(g)}^{+} + e^{-}$
Where $\Delta_{i}H$ is the first ionization enthalpy.
The $SI$ unit of ionization enthalpy is $kJ \ mol^{-1}$ or $J \ mol^{-1}$.

(B) The second ionization enthalpy is defined as the energy required to remove the second most loosely bound electron from an isolated gaseous atom or ion.
It is represented by the process: $M^+(g) \rightarrow M^{2+}(g) + e^-$.

(A) For any element,the energy required to remove electrons increases successively because the effective nuclear charge increases as electrons are removed.
Therefore,the order is $IE_1 < IE_2 < IE_3 < \dots$
This is because after the removal of the first electron,the remaining electrons are more strongly attracted by the nucleus,making it harder to remove subsequent electrons.

(C) The first ionization enthalpy is the energy required to remove the first electron from a neutral atom.
After the removal of the first electron,the remaining electrons are more strongly attracted by the nucleus because the effective nuclear charge per electron increases.
Consequently,removing the second and third electrons requires significantly more energy,leading to the order $1^{st} < 2^{nd} < 3^{rd}$.

(A) In any given period,the ionization enthalpy increases from left to right due to the increase in effective nuclear charge and decrease in atomic size.
Noble gases have the maximum ionization enthalpy due to their stable $ns^2 np^6$ electronic configuration.
Alkali metals have the minimum ionization enthalpy because they have the largest atomic size and the lowest effective nuclear charge in their respective period.

(N/A) The factors to be considered for understanding the trends in ionization enthalpy are:
$(i)$ The attraction of electrons towards the nucleus (effective nuclear charge).
$(ii)$ The repulsion between electrons (shielding effect or screening effect).

(D) The ionization enthalpy depends on the following factors:
$1$. Atomic size: As the atomic size increases,the distance between the nucleus and the valence electrons increases,leading to a decrease in ionization enthalpy.
$2$. Nuclear charge: As the nuclear charge increases,the attraction between the nucleus and the valence electrons increases,leading to an increase in ionization enthalpy.
$3$. Screening effect: Inner shell electrons shield the valence electrons from the nuclear charge,decreasing the effective nuclear charge and thus decreasing the ionization enthalpy.
$4$. Electronic configuration: Atoms with stable configurations (like fully filled or half-filled subshells) have higher ionization enthalpies.

(A) Ionization enthalpy generally increases across a period and decreases down a group.
$(i)$ $Ne > Ar$ (Down the group,size increases,$IE$ decreases).
$(ii)$ $F > Cl$ (Down the group,size increases,$IE$ decreases).
$(iii)$ $F > O$ (Across the period,effective nuclear charge increases,$IE$ increases).
$(iv)$ $N > O$ ($N$ has a stable half-filled $2p^3$ configuration,making its $IE$ higher than $O$).
$(v)$ $Na > K$ (Down the group,size increases,$IE$ decreases).
$(vi)$ $Cl > S$ (Across the period,effective nuclear charge increases,$IE$ increases).
$(vii)$ $Kr > Xe$ (Down the group,size increases,$IE$ decreases).
$(viii)$ $P > S$ ($P$ has a stable half-filled $3p^3$ configuration,making its $IE$ higher than $S$).

(A) The correct value is $800 \ kJ \ mol^{-1}$.
In the second period,the ionization enthalpy increases from $Li$ to $Ne$ with some exceptions.
Specifically,$Be$ $(1s^2 2s^2)$ has a higher ionization enthalpy than $B$ $(1s^2 2s^2 2p^1)$ because $Be$ has a stable fully-filled $2s$ orbital.
Therefore,the order of first ionization enthalpy for these elements is $Li < B < Be < C$.
Since $B$ must be lower than $Be$ $(899 \ kJ \ mol^{-1})$,$800 \ kJ \ mol^{-1}$ is the appropriate value.

(A) The correct value is $1310 \, kJ \, mol^{-1}$.
In the periodic table,the first ionization enthalpy generally increases across a period from left to right.
However,due to the stable half-filled $2p^3$ electronic configuration of Nitrogen $(1s^2 2s^2 2p^3)$,it has a higher ionization enthalpy than Oxygen $(1s^2 2s^2 2p^4)$.
Therefore,the ionization enthalpy of Oxygen must be lower than that of Nitrogen $(1402 \, kJ \, mol^{-1})$.
Thus,$1310 \, kJ \, mol^{-1}$ is the correct value.

(B) The process $(i) X_{(g)} \to X^+_{(g)} + e^-$ represents the first ionisation enthalpy,which is the energy required to remove the first electron from a neutral gaseous atom.
The process $(ii) X^+_{(g)} \to X^{2+}_{(g)} + e^-$ represents the second ionisation enthalpy,which is the energy required to remove the second electron from a unipositive gaseous ion.

(A) The electronic configuration of Beryllium ($Be$,$Z=4$) is $1s^{2} 2s^{2}$. The first ionization enthalpy involves the removal of an electron from the $2s$ orbital.
The electronic configuration of Boron ($B$,$Z=5$) is $1s^{2} 2s^{2} 2p^{1}$. The first ionization enthalpy involves the removal of an electron from the $2p$ orbital.
Therefore,the electrons are removed from $2s$ and $2p$ respectively.

(B) In oxygen $(Z=8)$,the electronic configuration is $1s^2 2s^2 2p^4$.
In nitrogen $(Z=7)$,the electronic configuration is $1s^2 2s^2 2p^3$.
In oxygen,the $2p^4$ configuration results in greater electron-electron repulsion due to the pairing of electrons in one of the $2p$ orbitals.
This increased repulsion makes it easier to remove an electron from oxygen compared to nitrogen,resulting in a lower first ionization enthalpy for oxygen compared to nitrogen.

(N/A) Both hydrogen and sodium have one electron in the valence shell. However,the atomic size of hydrogen is much smaller than that of sodium. Consequently,the valence electron in hydrogen is much closer to the nucleus and experiences a stronger electrostatic force of attraction. Therefore,the ionisation enthalpy of hydrogen $(1312 \ kJ \ mol^{-1})$ is significantly higher than that of sodium $(496 \ kJ \ mol^{-1})$.

(B) The successive ionization enthalpies are given as $IE_1 = 800 \ kJ \ mol^{-1}$,$IE_2 = 2427 \ kJ \ mol^{-1}$,$IE_3 = 3658 \ kJ \ mol^{-1}$,$IE_4 = 25024 \ kJ \ mol^{-1}$,and $IE_5 = 32824 \ kJ \ mol^{-1}$.
$A$ large jump in ionization enthalpy indicates the removal of an electron from a stable,noble gas-like inner shell configuration.
Comparing the values,we observe a significant increase (jump) between $IE_3$ and $IE_4$ $(25024 - 3658 = 21366 \ kJ \ mol^{-1})$.
This indicates that the first $3$ electrons are valence electrons,and the $4^{th}$ electron is removed from a stable inner shell.
Therefore,the number of valence electrons in the element is $3$.

(D) The first ionization energy $(IE_1)$ of an element is the energy required to remove the first electron,and the second ionization energy $(IE_2)$ is the energy required to remove the second electron.
For alkali metals like potassium $(K)$,the electronic configuration is $[Ar] 4s^1$.
After the removal of the first electron,it achieves the stable noble gas configuration of argon $([Ar])$.
Removing the second electron requires breaking this stable octet,which demands a very high amount of energy.
Therefore,the difference between $IE_2$ and $IE_1$ is exceptionally large for alkali metals compared to alkaline earth metals or transition metals.

(A) For element $X$,the jump from $IE_1$ $(495 \ kJ/mol)$ to $IE_2$ $(4563 \ kJ/mol)$ is very large,indicating that the second electron is removed from a stable noble gas core. This corresponds to $Na$ $([Ne] 3s^1)$.
For element $Y$,the values $IE_1$ $(731 \ kJ/mol)$ and $IE_2$ $(1450 \ kJ/mol)$ are relatively close,which is characteristic of an alkaline earth metal like $Mg$ $([Ne] 3s^2)$,where both valence electrons are removed easily compared to the third electron.
Therefore,$X=Na$ and $Y=Mg$.

(C) The first ionization energy $(IE_1)$ of $3^{rd}$ period elements follows the order: $Na < Al < Mg < Si < S < P < Cl < Ar$.
Magnesium $(Mg)$ has a higher $IE_1$ than $Na$ $(Z)$ but lower than $Cl$ and $Ar$ ($X$ and $Y$).
Therefore,$X$ and $Y$ are $Ar$ and $Cl$,and $Z$ is $Na$.