Ionisation energy Questions in English

(D) Ionization energy increases with the removal of successive electrons from the same atom because the effective nuclear charge increases.
Comparing the options:
$A$ and $B$ represent the first ionization energy $(IE_1)$ of $Ba$ and $Ca$ respectively. Since $IE_1$ decreases down the group,$IE_1(Ca) > IE_1(Ba)$.
$C$ and $D$ represent the sum of the first and second ionization energies $(IE_1 + IE_2)$ for $Ca$ and $Mg$ respectively.
Since $Mg$ is above $Ca$ in the periodic table,$IE_1(Mg) > IE_1(Ca)$ and $IE_2(Mg) > IE_2(Ca)$.
Therefore,the total energy required for $Mg \to Mg^{2+} + 2e^-$ is the highest among the given options.

(A) The first ionization enthalpy decreases as we move down the group in the periodic table due to an increase in atomic size and shielding effect.
Since $Mg, Ca, Sr,$ and $Ba$ belong to Group $2$ (alkaline earth metals) and are arranged in the order $Mg < Ca < Sr < Ba$ down the group,the decreasing order of their first ionization enthalpy is $Mg > Ca > Sr > Ba$.

(A) The first ionization energy of alkaline earth metals is higher than that of alkali metals because of the increased nuclear charge and smaller atomic size compared to alkali metals in the same period.

(B) The electronic configurations of the given elements are:
$V (Z=23): [Ar] 3d^3 4s^2$
$Cr (Z=24): [Ar] 3d^5 4s^1$
$Mn (Z=25): [Ar] 3d^5 4s^2$
$Fe (Z=26): [Ar] 3d^6 4s^2$
For the second ionization enthalpy,we remove one electron from the $M^+$ ion:
$V^+: [Ar] 3d^3 4s^1$
$Cr^+: [Ar] 3d^5$
$Mn^+: [Ar] 3d^5 4s^1$
$Fe^+: [Ar] 3d^6 4s^1$
Removing the second electron from $Cr^+$ is difficult because it has a stable half-filled $3d^5$ configuration. Therefore,$Cr$ has the highest second ionization enthalpy.

(C) The electronic configurations of the elements are: $C (2s^2 2p^2)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$,$F (2s^2 2p^5)$.
After the removal of the first electron,the configurations become: $C^+ (2s^2 2p^1)$,$N^+ (2s^2 2p^2)$,$O^+ (2s^2 2p^3)$,$F^+ (2s^2 2p^4)$.
The second ionization potential involves removing an electron from these cations.
$O^+$ has a stable half-filled $2p^3$ configuration,making it the most difficult to remove an electron from.
Following the periodic trend and stability,the order of second ionization potential is $O > F > N > C$.

(A) The ionisation potential refers to the energy required to remove an electron from a gaseous atom or ion.
Comparing the species:
$Li^{+}$ $(1s^2)$ has a stable noble gas configuration (isoelectronic with $He$).
$Mg^{+}$ $([Ne] 3s^1)$ has one electron in the $3s$ orbital.
$Al^{+}$ $([Ne] 3s^2)$ has two electrons in the $3s$ orbital.
$Ne$ $(1s^2 2s^2 2p^6)$ is a neutral noble gas.
Among these,$Li^{+}$ has the highest effective nuclear charge relative to its size and a very stable $1s^2$ configuration,making it the most difficult to remove an electron from.
Therefore,$Li^{+}$ has the highest ionisation potential.

(B) The general trend for ionisation energy $(IE)$ increases across a period from left to right due to an increase in effective nuclear charge.
However,there are exceptions due to stable electronic configurations.
$Be$ $(1s^2 2s^2)$ has a fully filled $2s$ orbital,making its $IE$ higher than $B$ $(1s^2 2s^2 2p^1)$.
$N$ $(1s^2 2s^2 2p^3)$ has a stable half-filled $p$-orbital,making its $IE$ higher than $O$ $(1s^2 2s^2 2p^4)$.
Therefore,the correct order is $B < Be < C < O < N$.

(D) Ionisation energy generally increases across a period due to a decrease in atomic size and effective nuclear charge.
Comparing the configurations:
$A: [Ne] 3s^2 3p^2$ (Period $3$,Group $14$)
$B: [Ar] 3d^{10} 4s^2 4p^3$ (Period $4$,Group $15$)
$C: [Ne] 3s^2 3p^1$ (Period $3$,Group $13$)
$D: [Ne] 3s^2 3p^3$ (Period $3$,Group $15$)
Elements in Period $3$ have smaller atomic radii than those in Period $4$,so elements in Period $3$ generally have higher ionisation energies than those in Period $4$.
Among the Period $3$ elements $(A, C, D)$,the configuration $[Ne] 3s^2 3p^3$ corresponds to a half-filled $p$-orbital,which is exceptionally stable.
Therefore,$[Ne] 3s^2 3p^3$ has the highest ionisation energy.

(B) Ionisation enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
As the principal quantum number $(n)$ increases,the distance of the valence electron from the nucleus increases,leading to a decrease in the effective nuclear charge experienced by the electron.
Option $B$ $(1s^2 \ 2s^2 \ 2p^5 \ 3s^1)$ represents an atom with its valence electron in the $n=3$ shell,which is further from the nucleus compared to the electrons in the $n=2$ shell in the other options.
Therefore,the electron in the $3s$ orbital is held least tightly,resulting in the lowest ionisation enthalpy.

(C) The first ionization enthalpy $(IE_1)$ generally increases across a period due to an increase in effective nuclear charge and decreases down a group due to an increase in atomic size and shielding effect.
$1$. Comparing elements in the same period: $S$ (Group $16$,Period $3$) and $Ar$ (Group $18$,Period $3$). $Ar$ has a stable noble gas configuration,so it has the highest $IE_1$ among the given elements.
$2$. Comparing elements in the same group: $Ca$ (Group $2$,Period $4$) and $Ba$ (Group $2$,Period $6$). Since $Ba$ is below $Ca$,$IE_1$ of $Ba < Ca$.
$3$. Comparing $S$ and $Se$: $S$ (Period $3$) and $Se$ (Period $4$). Since $Se$ is below $S$,$IE_1$ of $Se < S$.
$4$. Combining the trends: $Ba < Ca < Se < S < Ar$.
Therefore,the correct order is $Ba < Ca < Se < S < Ar$.

(B) The first ionization process is represented as: $Na \rightarrow Na^{+} + e^{-}$,where $IE = 5.1 \, eV$.
The electron gain enthalpy of $Na^{+}$ corresponds to the reverse process: $Na^{+} + e^{-} \rightarrow Na$.
Since the second reaction is the exact inverse of the first reaction,the energy change is equal in magnitude but opposite in sign.
Therefore,the electron gain enthalpy of $Na^{+}$ is $\Delta H_{eg} = -IE = -5.1 \, eV$.

(B) The first ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
$K$ $(Z=19)$ and $Na$ $(Z=11)$ are alkali metals (Group $1$),which have the lowest ionization energies in their respective periods.
$Rb$ $(Z=37)$ is also an alkali metal with an even lower ionization energy than $K$ due to its larger atomic size.
$Sc$ $(Z=21)$ is a $d$-block transition metal.
Transition metals generally have higher first ionization energies compared to alkali metals in the same or adjacent periods because of their smaller atomic radii and higher effective nuclear charge,which holds the valence electrons more tightly.

(A) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.
$B$ (Boron,group $13$) has the lowest ionization enthalpy among the given elements.
Comparing $P$ (Phosphorus,group $15$) and $S$ (Sulfur,group $16$),$P$ has a higher ionization enthalpy than $S$ because $P$ has a stable half-filled $p$-orbital configuration $(3s^2 3p^3)$.
$F$ (Fluorine,group $17$) has the highest ionization enthalpy due to its small size and high effective nuclear charge.
Thus,the order is $B < S < P < F$.

(C) The $IE$ increases along a period and decreases down the group.
Group $15$ elements have a half-filled $p$-subshell $(ns^2 np^3)$,which provides extra stability compared to group $16$ elements $(ns^2 np^4)$.
Comparing the given electronic configurations:
Option $A$ is $Al$ $(3s^2 3p^1)$,Option $B$ is $S$ $(3s^2 3p^4)$,Option $C$ is $P$ $(3s^2 3p^3)$,and Option $D$ is $Ge$ $(4s^2 4p^2)$.
Since $P$ has a stable half-filled configuration,it possesses the highest $IE$ among the given options.

(D) According to the Bohr model,the energy of an electron in an orbit is given by $E_n = -13.6 \times \frac{Z_{eff}^2}{n^2} \, eV$.
For $Na$ $(Z=11)$,the valence electron is in the $3s$ orbital,so $n=3$.
The ionisation energy $(I.E.)$ is the energy required to remove the electron from $n=3$ to $n=\infty$,which is equal to the magnitude of the energy of the $n=3$ state.
$I.E. = 13.6 \times \frac{Z_{eff}^2}{n^2} \, eV$.
Given $Z_{eff} = 1.84$ and $n = 3$:
$I.E. = 13.6 \times \frac{(1.84)^2}{3^2} = 13.6 \times \frac{3.3856}{9} \approx 5.11 \, eV$.

(C) Electronic configurations: $Mg (Z=12): [Ne] 3s^2$,$Al (Z=13): [Ne] 3s^2 3p^1$.
$1$. First ionization potential $(IE_1)$: $Mg$ has a stable fully-filled $3s$ orbital,so $IE_1(Mg) > IE_1(Al)$. Thus,statement $A$ is correct.
$2$. Second ionization potential $(IE_2)$: $Mg^+$ is $[Ne] 3s^1$,$Al^+$ is $[Ne] 3s^2$. Removing an electron from $Al^+$ is harder due to the stable $3s^2$ configuration,so $IE_2(Al) > IE_2(Mg)$. Thus,statement $B$ is correct.
$3$. Third ionization potential $(IE_3)$: $Mg^{2+}$ is $[Ne]$,$Al^{2+}$ is $[Ne] 3s^1$. $Mg^{2+}$ has a noble gas configuration,making it extremely difficult to remove the third electron. Thus,$IE_3(Mg) > IE_3(Al)$. Thus,statement $D$ is correct.
$4$. Comparing $IE_1(Mg)$ and $IE_3(Al)$: $IE_1(Mg)$ is relatively low (removing from $3s^2$),while $IE_3(Al)$ is very high (removing from $3s^2$ after $3p^1$ and $3s^1$ are gone). Therefore,$IE_3(Al) > IE_1(Mg)$. Statement $C$ is incorrect.

(C) The second ionization potential $(IP_2)$ corresponds to the energy required to remove an electron from a unipositive ion $(M^+ \rightarrow M^{2+} + e^-)$.
$1$. For $Mn$ $(3d^5 4s^2)$,$Mn^+$ is $3d^5 4s^1$. For $Cr$ $(3d^5 4s^1)$,$Cr^+$ is $3d^5$. Since $Cr^+$ has a stable half-filled $d$-subshell,removing the second electron from $Cr^+$ is harder than from $Mn^+$. Thus,$Cr > Mn$.
$2$. For $Zn$ $(3d^{10} 4s^2)$,$Zn^+$ is $3d^{10} 4s^1$. For $Cu$ $(3d^{10} 4s^1)$,$Cu^+$ is $3d^{10}$. $Cu^+$ has a stable fully-filled $d$-subshell,making $IP_2$ of $Cu$ higher than $Zn$. Thus,$Cu > Zn$.
$3$. For $Na$ $([Ne] 3s^1)$,$Na^+$ is $[Ne]$ (stable noble gas configuration). For $Cu$ $([Ar] 3d^{10} 4s^1)$,$Cu^+$ is $[Ar] 3d^{10}$. The removal of an electron from the stable noble gas core of $Na^+$ requires significantly more energy than from $Cu^+$. Thus,$Na > Cu$.
$4$. For $Cr$ $([Ar] 3d^5 4s^1)$,$Cr^+$ is $3d^5$. For $K$ $([Ar] 4s^1)$,$K^+$ is $[Ar]$. $K^+$ has a stable noble gas configuration,so $IP_2$ of $K$ is very high. Thus,$K > Cr$.
Comparing the options,$Na > Cu$ is the correct order.

(B) Ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
$A$. $[Ne] 3s^2 3p^1$ (Aluminum): Has one electron in the $p$-orbital,relatively easy to remove.
$B$. $[Ne] 3s^2 3p^3$ (Phosphorus): Has a half-filled $p$-orbital $(p^3)$,which is highly stable,making it difficult to remove an electron.
$C$. $[Ne] 3s^2 3p^2$ (Silicon): Has a partially filled $p$-orbital,easier to remove than $p^3$.
$D$. $[Ar] 3d^{10} 4s^2 4p^3$ (Arsenic): Also has a half-filled $p$-orbital $(4p^3)$,but since it is in the $4^{th}$ shell,the electron is further from the nucleus compared to the $3p$ electron in Phosphorus.
Comparing $B$ and $D$,the $3p$ electrons are more strongly attracted to the nucleus than $4p$ electrons due to smaller atomic size. Therefore,the ionization energy of $[Ne] 3s^2 3p^3$ is higher than that of $[Ar] 3d^{10} 4s^2 4p^3$.

(A) The ionization energy $(I.E.)$ generally increases across a period from left to right due to an increase in effective nuclear charge.
For the $2^{nd}$ period elements $(Li, Be, B, C, N, O, F, Ne)$,the $1^{st}$ $I.E.$ is highest for $Ne$ (noble gas).
The $2^{nd}$ $I.E.$ corresponds to the energy required to remove an electron from a unipositive ion $(M^+)$.
For $Ne$,the $1^{st}$ $I.E.$ is very high,and the $2^{nd}$ $I.E.$ is even higher because the electron is being removed from a stable octet configuration.
Comparing the sum of $1^{st}$ and $2^{nd}$ $I.E.$,$Ne$ has the highest value because both its $1^{st}$ and $2^{nd}$ ionization energies are significantly higher than those of other elements in the same period.

(D) The energy required to remove a particle from an atom depends on the binding energy of that particle within the atom.
$1$. An $\alpha$ particle,a proton,and a neutron are located within the nucleus and are held together by strong nuclear forces,which require a very high amount of energy to overcome.
$2$. An electron is located in the extranuclear region and is held by electrostatic forces of attraction with the nucleus.
$3$. The energy required to remove an electron (ionization energy) is significantly lower than the energy required to remove nucleons from the nucleus.
Therefore,an electron can be removed with the least energy change.

(C) The ionization potential $(I.P.)$ decreases as we move down a group in the periodic table.
This is due to the increase in atomic size,which reduces the electrostatic force of attraction between the nucleus and the valence electrons.
Additionally,the shielding or screening effect increases,which further decreases the ionization potential.
Since potassium $(K)$ is below sodium $(Na)$ in Group $1$,its $I.P.$ must be lower than $5.48 \ eV$.
The experimental value for the $I.P.$ of potassium is approximately $4.34 \ eV$.

(A) The electronic configurations are: $K (Z=19): [Ar] 4s^1$,$Ca (Z=20): [Ar] 4s^2$,$Ba (Z=56): [Xe] 6s^2$.
After the first ionization,the configurations become: $K^+: [Ar]$,$Ca^+: [Ar] 4s^1$,$Ba^+: [Xe] 6s^1$.
The second ionization potential involves removing an electron from these ions.
For $K^+$,the electron is removed from the stable noble gas configuration $([Ar])$,which requires a very high amount of energy.
For $Ca^+$ and $Ba^+$,the second electron is removed from the $ns^1$ orbital.
Comparing $Ca^+$ and $Ba^+$,$Ca$ is above $Ba$ in the group,so $Ca^+$ has a smaller size and higher effective nuclear charge,making its second $IP$ higher than that of $Ba^+$.
Therefore,the order is $K > Ca > Ba$.

(A) The general electronic configuration of an alkali metal is $ns^1$. After the removal of the first electron (first ionization energy),it attains a stable noble gas configuration (octet). Therefore,the removal of the second electron requires a very high amount of energy (second ionization energy).
In contrast,the general electronic configuration of an alkaline earth metal is $ns^2$. After the removal of the first electron,it attains an $ns^1$ configuration. The removal of the second electron is relatively easier compared to the alkali metal because it does not involve breaking a stable noble gas core.
Thus,the second ionization energy of an alkali metal $(X)$ is significantly higher than that of an alkaline earth metal $(Y)$ of the same period.
Therefore,the correct relation is $X > Y$.

(D) reaction proceeds spontaneously if the ionization energy of the product species is lower than that of the reactant species,or if the electron affinity of the product is higher than that of the reactant.
In option $(D)$,$Kr + He^{+} \to Kr^{+} + He$,the ionization energy of $Kr$ $(13.99 \ eV)$ is significantly lower than the ionization energy of $He$ $(24.58 \ eV)$.
Since $He^{+}$ has a higher electron affinity (or ionization potential) than $Kr$,the electron will transfer from $Kr$ to $He^{+}$ to form the more stable $Kr^{+}$ and $He$ atoms.
Thus,the reaction $Kr + He^{+} \to Kr^{+} + He$ is spontaneous.

(B) To determine the highest sum,we evaluate the ionization energies $(IE)$:
$1$. $IE_1$ of $He$ is very high $(2372 \ kJ/mol)$.
$2$. $IE_2$ of $Li$ is very high $(7298 \ kJ/mol)$ because it involves removing an electron from a stable $1s^2$ configuration.
$3$. $IE_3$ of $Mg$ is also very high $(7733 \ kJ/mol)$ as it involves removing an electron from a stable $2p^6$ configuration.
Comparing the sets:
Set $A$: $IE_1(He) + IE_2(Li) + IE_2(Ca) = 2372 + 7298 + 1145 = 10815 \ kJ/mol$.
Set $B$: $IE_1(He) + IE_2(Li) + IE_3(Mg) = 2372 + 7298 + 7733 = 17403 \ kJ/mol$.
Set $C$: $IE_1(Li) + IE_1(He) + IE_1(K) = 520 + 2372 + 419 = 3311 \ kJ/mol$.
Set $D$: $IE_2(Na) + IE_2(Ba) + IE_2(Sr) = 4562 + 965 + 1064 = 6591 \ kJ/mol$.
Thus,Set $B$ has the highest sum.

(A) The electronic configurations are as follows:
$Pd (46): [Kr] \, 4d^{10} 5s^{0}$
$Ag (47): [Kr] \, 4d^{10} 5s^{1}$
$Rb (37): [Kr] \, 5s^{1}$
$Y (39): [Kr] \, 4d^{1} 5s^{2}$
Ionisation energy depends on the stability of the electronic configuration and the effective nuclear charge.
$Pd$ has a completely filled $4d^{10}$ subshell and no electrons in the $5s$ orbital,making it highly stable.
Removing an electron from the stable $4d^{10}$ configuration of $Pd$ requires the highest amount of energy compared to the others,where electrons are removed from $5s$ or $4d$ orbitals with less stability or lower effective nuclear charge.
Therefore,$Pd$ has the maximum first ionisation energy.

(B) The second ionisation energy $(IE_2)$ refers to the energy required to remove an electron from a unipositive ion.
$1$. For $Te$ $(5s^2 5p^4)$ and $Sb$ $(5s^2 5p^3)$: The $IE_2$ of $Te$ is higher than $Sb$ because $Sb^+$ has a stable half-filled $p^2$ configuration,but $Te^+$ has a $p^3$ configuration. Actually,$Sb$ has a stable half-filled $p^3$ configuration,so removing the second electron from $Sb^+$ is harder than from $Te^+$. Thus,$Te > Sb$ is correct.
$2$. For $In$ $(5s^2 5p^1)$ and $Sr$ $(5s^2)$: $Sr^+$ has a stable $s^1$ configuration,while $In^+$ has a $s^2$ configuration. Removing an electron from $Sr^+$ is harder,so $Sr > In$. The comparison $In > Sr$ is incorrect.
$3$. $He$ has a very high $IE_2$ due to its noble gas configuration.
$4$. $Fe < Fe^+$ is a comparison of ionisation energy,which is conceptually incorrect as $IE$ refers to neutral atoms,but $Fe^+$ is already an ion. The question asks for the incorrect comparison among the options provided regarding $IE_2$ trends.

(D) The first ionization energy $(IE_1)$ of a molecule is generally higher than that of its constituent atom because the electrons in a molecule are held more strongly due to the formation of stable covalent bonds and the increased effective nuclear charge experienced by the bonding electrons.
For $F_2$,$O_2$,and $N_2$,the bond dissociation energy and the stability of the molecular orbitals result in a higher energy requirement to remove an electron compared to the isolated atoms ($F$,$O$,and $N$ respectively).
Therefore,for all the given pairs,the molecule has a higher $IE_1$ value than its corresponding atom.

(A) For Triad-$I$ $[N^{3-}, O^{2-}, Na^{+}]$: These are isoelectronic species with $10$ electrons. The $IP$ increases as the nuclear charge $(Z)$ increases. The $Z$ values are $N=7, O=8, Na=11$. Thus,the $IP$ order is $N^{3-} < O^{2-} < Na^{+}$. The species with the lowest $IP$ is $N^{3-}$.
For Triad-$II$ $[N^{+}, C^{+}, O^{+}]$: These are ions of elements in the same period. $IP$ generally increases from left to right across a period. The order of $Z$ is $C(6) < N(7) < O(8)$. Thus,the $IP$ order is $C^{+} < N^{+} < O^{+}$. The species with the highest $IP$ is $O^{+}$.

(C) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which is a stable half-filled $p$-orbital configuration.
Oxygen $(O)$ has the configuration $1s^2 2s^2 2p^4$.
Although oxygen has a higher effective nuclear charge than nitrogen,the $2p^4$ configuration of oxygen experiences greater inter-electron repulsion between two electrons in the same $p$-orbital.
This repulsion makes it easier to remove an electron from oxygen compared to the stable half-filled configuration of nitrogen,thus resulting in a lower first ionisation energy for oxygen.

(A) The electronic configuration of Nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which contains a half-filled $p$-orbital.
This half-filled $p$-orbital configuration is exceptionally stable.
Therefore,more energy is required to remove an electron from the stable half-filled $p$-orbital of Nitrogen compared to the $2p^4$ configuration of Oxygen.

(D) The first ionization potentials $(eV)$ of $Be$ and $B$ are $9.32 \ eV$ and $8.29 \ eV$ respectively.
Generally,as we move from left to right in a period,the ionization potential increases due to an increase in effective nuclear charge.
However,in the case of $Be$ $(Z=4)$ and $B$ $(Z=5)$,$Be$ has a higher ionization potential than $B$ because $Be$ has a stable,completely filled $2s$ orbital configuration $(1s^2 2s^2)$,which requires more energy to remove an electron compared to the $2p^1$ electron in Boron.
Therefore,option $D$ is correct.

(C) The successive ionization energies are $IE_1 = 191 \ kcal$,$IE_2 = 578 \ kcal$,$IE_3 = 872 \ kcal$,and $IE_4 = 5962 \ kcal$.
$A$ large jump in ionization energy occurs between the third and fourth ionization energies $(IE_4 - IE_3 = 5962 - 872 = 5090 \ kcal)$.
This indicates that the first three electrons are removed from the valence shell,while the fourth electron is removed from a stable inner shell.
Therefore,the element has $3$ valence electrons.

(A) As we move across the period,nuclear charge increases and atomic size decreases,hence ionization enthalpy generally increases.
For $Al$ $(3s^{2} 3p^{1})$,the electron is removed from a partially filled $3p$ orbital.
For $Mg$ $(3s^{2})$,the electron is removed from a stable,fully filled $3s$ orbital.
Removal of an electron from a stable,fully filled orbital requires more energy than removal from a partially filled orbital.
Thus,the ionization enthalpy of $Mg$ is greater than that of $Al$.
The correct order of first ionization enthalpies is: $Na < Mg > Al < Si$.

(C) The electronic configurations of the elements are:
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
$F (Z=9): 1s^2 2s^2 2p^5$
For the second ionization energy,we remove one electron to form the $M^+$ ion:
$C^+: 1s^2 2s^2 2p^1$
$N^+: 1s^2 2s^2 2p^2$
$O^+: 1s^2 2s^2 2p^3$ (Half-filled $p$-orbital,highly stable)
$F^+: 1s^2 2s^2 2p^4$
Due to the stable half-filled $2p^3$ configuration of $O^+$,it has the highest second ionization energy.
Comparing the remaining,$F^+$ has a higher effective nuclear charge than $N^+$,and $N^+$ is higher than $C^+$.
Therefore,the correct order is $O > F > N > C$.

(B) The general trend is that ionization energy increases from left to right across a period due to an increase in effective nuclear charge $(Z_{eff})$.
However,there are exceptions due to stable electronic configurations.
For the pair $Mg$ $([Ne] 3s^2)$ and $Al$ $([Ne] 3s^2 3p^1)$,the ionization energy of $Mg$ is higher than that of $Al$ because $Mg$ has a fully filled $s$-orbital,which is more stable.
Therefore,the statement is not correct for the pair $Mg, Al$.

(C) The $2^{nd}$ ionisation potential corresponds to the energy required to remove an electron from a unipositive ion $(M^+)$.
The electronic configurations of the unipositive ions are:
$C^+ (1s^2 2s^2 2p^1)$
$N^+ (1s^2 2s^2 2p^2)$
$O^+ (1s^2 2s^2 2p^3)$
$F^+ (1s^2 2s^2 2p^4)$
$O^+$ has a stable half-filled $2p^3$ configuration,which makes it very difficult to remove the second electron,resulting in the highest $2^{nd}$ ionisation potential.
Comparing the others,$F^+$ has a higher effective nuclear charge than $N^+$,and $N^+$ has a higher effective nuclear charge than $C^+$.
Thus,the correct order is $O > F > N > C$.

(B) The electronic configurations of the elements in their valence shells are:
$B: 2s^2 2p^1$
$C: 2s^2 2p^2$
$N: 2s^2 2p^3$
$O: 2s^2 2p^4$
Ionization potential generally increases across a period from left to right.
However,$N$ has a stable half-filled $p$-orbital configuration $(2p^3)$,which requires more energy to remove an electron compared to $O$ $(2p^4)$.
Thus,the order of first ionization potential is $B < C < O < N$.
Therefore,$N$ has the maximum first ionization potential.

(B) The electronic configurations for the species formed after the first ionisation are:
$Li^+: 1s^2$ (Stable noble gas configuration)
$Be^+: 1s^2 2s^1$
$B^+: 1s^2 2s^2$ (Stable fully filled $2s$ subshell)
To remove the second electron:
For $Li^+$,we must remove an electron from the stable $1s^2$ shell,which requires very high energy.
For $B^+$,we must remove an electron from the stable $2s^2$ subshell.
For $Be^+$,we remove an electron from the $2s^1$ orbital,which is easier than removing from $B^+$.
Therefore,the order of second ionisation potential is $Li > B > Be$.

(D) The electronic configuration of $Be$ is $1s^2 2s^2$. Due to the fully filled $2s$ orbital,$Be$ has a very high ionization potential.
$1$. $Be$ vs $B$: $Be$ $(1s^2 2s^2)$ has a higher $IP$ than $B$ $(1s^2 2s^2 2p^1)$ because removing an electron from a fully filled $s$-orbital is more difficult than removing it from a $p$-orbital.
$2$. $Be$ vs $Mg$: $Be$ is in the $2^{nd}$ period and $Mg$ is in the $3^{rd}$ period. $IP$ decreases down the group,so $IP$ of $Be > Mg$.
$3$. $Be$ vs $Li$: $Be$ $(1s^2 2s^2)$ has a higher $IP$ than $Li$ $(1s^2 2s^1)$ due to higher effective nuclear charge.
$4$. $Be$ vs $Al$: $Be$ is in the $2^{nd}$ period and $Al$ is in the $3^{rd}$ period. $Be$ has a higher $IP$ than $Al$.
$5$. $Be$ vs $K$: $K$ is an alkali metal in the $4^{th}$ period with a very low $IP$. Thus,$IP$ of $Be > K$.
Therefore,the $IP$ of $Be$ is higher than all the given elements $(a, b, c, d, e)$.

(B) In the arrangement of option $B$,the order is not correct according to the property indicated against it.
The correct order for increasing first ionisation energy $(IE_1)$ is $B < C < O < N$.
The ionisation energy of $N$ is higher than that of $O$. This is because $N$ $(2s^2 2p^3)$ has a stable half-filled $p$-orbital configuration,which requires more energy to remove an electron compared to $O$ $(2s^2 2p^4)$.

(B) The electronic configuration of $Mg$ is $[Ne] 3s^2$ and $Na$ is $[Ne] 3s^1$.
After the removal of one electron,$Mg^+$ becomes $[Ne] 3s^1$ and $Na^+$ becomes $[Ne] 2s^2 2p^6$ (stable noble gas configuration).
Since $Na^+$ has a stable noble gas configuration,it is extremely difficult to remove a second electron from it.
Therefore,the second ionization potential of $Na$ is significantly higher than that of $Mg$.
Thus,the statement that the second ionization potential of $Mg$ is greater than that of $Na$ is incorrect.

(A) The $1^{st}$ Ionization Potential $(IP)$ trends are as follows:
$1$. For $Al$ $(13)$ and $Ga$ $(31)$,$Ga$ has a higher $IP$ than $Al$ due to the poor shielding effect of $d$-electrons,so $Al < Ga$ is correct.
$2$. For $P$ $(15)$ and $S$ $(16)$,$P$ has a stable half-filled $p$-orbital configuration $(3s^2 3p^3)$,making its $IP$ higher than $S$ $(3s^2 3p^4)$. Thus,$P > S$ is correct.
$3$. For $Cu$ and $Zn$,$Zn$ has a fully filled $d^{10}s^2$ configuration,making its $IP$ higher than $Cu$. Thus,$Cu < Zn$ is correct.
$4$. For $Zr$ and $Hf$,due to the lanthanoid contraction,the $IP$ of $Hf$ is slightly higher than $Zr$. Thus,$Zr < Hf$ is correct.
Wait,re-evaluating the options: $Al$ $(577 \ kJ/mol)$ vs $Ga$ $(579 \ kJ/mol)$,so $Al < Ga$. $P$ $(1012 \ kJ/mol)$ vs $S$ $(1000 \ kJ/mol)$,so $P > S$. $Cu$ $(745 \ kJ/mol)$ vs $Zn$ $(906 \ kJ/mol)$,so $Cu < Zn$. $Zr$ $(640 \ kJ/mol)$ vs $Hf$ $(658 \ kJ/mol)$,so $Zr < Hf$.
Actually,in many textbooks,$Al > Ga$ is considered the incorrect order because $Ga$ is expected to be lower but is actually higher. However,looking at the options provided,$P > S$ is a standard trend. Let's re-check $Al > Ga$. $Al$ is $577$ and $Ga$ is $579$. So $Al < Ga$. The statement $Al > Ga$ is indeed incorrect.

(C) The graph shows the variation of the $1^{st}$ Ionisation Energy $(I.E.)$ with respect to atomic number for elements in the $3^{rd}$ period.
Alkali metals have the lowest $1^{st}$ $I.E.$ in their respective periods because they have a stable noble gas configuration after losing one electron.
In the $3^{rd}$ period,the alkali metal is Sodium ($Na$,atomic number $11$).
Looking at the graph,the point $S$ represents the element with the lowest $1^{st}$ $I.E.$ in the sequence,which corresponds to the alkali metal of the $3^{rd}$ period.

(B) The elements in the $2$nd period generally show an increase in ionization enthalpy from left to right.
However,nitrogen $(N)$ has a higher first ionization enthalpy than oxygen $(O)$ due to its stable half-filled $p$-orbital configuration $(2s^{2} 2p^{3})$.
Among the given values $(8.3, 11.3, 14.5, 13.6)$,the value $14.5 \ eV$ corresponds to nitrogen because it is the peak in the trend before the drop at oxygen.
Therefore,the first ionization enthalpy of nitrogen is $14.5 \ eV$.

(B) The difference $(IP_2 - IP_1)$ is related to the energy required to remove the second electron compared to the first.
For $Mg$ $(Z=12)$,the electronic configuration is $[Ne] 3s^2$.
$IP_1$ involves removing an electron from the stable $3s^2$ orbital,while $IP_2$ involves removing an electron from the $3s^1$ orbital.
For $Al$ $(Z=13)$,the configuration is $[Ne] 3s^2 3p^1$.
$IP_1$ involves removing the $3p^1$ electron,and $IP_2$ involves removing a $3s^2$ electron.
Generally,the jump $(IP_2 - IP_1)$ is significantly higher when the second electron is removed from a more stable or inner shell.
For $Mg$,$(IP_2 - IP_1)$ is approximately $15.03 \ eV - 7.64 \ eV = 7.39 \ eV$,which is $< 11 \ eV$.
For $Al$,$(IP_2 - IP_1)$ is approximately $18.82 \ eV - 5.98 \ eV = 12.84 \ eV$,which is $> 11 \ eV$.
Thus,the condition holds for $Mg$.

(A) The $3^{rd}$ ionisation energy involves the removal of an electron from the $M^{2+}$ ion. The electronic configurations of the $M^{2+}$ ions for $B, C, N,$ and $O$ are:
$B^{2+} (Z=5): 1s^2 2s^1$
$C^{2+} (Z=6): 1s^2 2s^2$
$N^{2+} (Z=7): 1s^2 2s^2 2p^1$
$O^{2+} (Z=8): 1s^2 2s^2 2p^2$
Comparing these,$C^{2+}$ has a stable fully filled $2s^2$ subshell,which makes its $3^{rd}$ ionisation energy significantly higher than expected.
$N^{2+}$ and $O^{2+}$ have higher effective nuclear charge than $B^{2+}$ and $C^{2+}$.
The actual order is $N > O > C > B$ is incorrect due to the stability of the $2s^2$ configuration in $C^{2+}$.
Correct order is $N > O > C > B$ is wrong; the correct sequence is $O > N > C > B$ based on effective nuclear charge,but considering the stability of $C^{2+}$,the order is $N > O > C > B$ is incorrect. The correct order is $O > N > C > B$.

(D) The ionization enthalpy depends on the distance of the electron from the nucleus.
Orbitals with a lower principal quantum number $(n)$ are closer to the nucleus,resulting in a stronger electrostatic attraction between the nucleus and the electron.
Therefore,removing an electron from an orbital with a lower $n$ value requires more energy and is more difficult,not easier.
Thus,the statement in option $D$ is incorrect.