Ionisation energy Questions in English

(B) $1$. The first ionization potential $(IE_1)$ of $Mg$ $(3s^2)$ is higher than $Al$ $(3s^2 3p^1)$ due to the stable fully-filled $s$-orbital. Statement $A$ is correct.
$2$. $Na^+$ has a stable noble gas configuration $(1s^2 2s^2 2p^6)$. Removing a second electron from $Na^+$ requires significantly more energy than removing a second electron from $Mg^+$ $(3s^1)$. Thus,the second ionization potential $(IE_2)$ of $Na$ is much greater than that of $Mg$. Statement $B$ is incorrect.
$3$. $IE_1$ of $Na$ $([Ne] 3s^1)$ is less than $Mg$ $([Ne] 3s^2)$. Statement $C$ is correct.
$4$. $Mg^{2+}$ has a stable noble gas configuration $(1s^2 2s^2 2p^6)$. Removing a third electron from $Mg^{2+}$ requires much higher energy than removing a third electron from $Al^{2+}$ $(3s^1)$. Statement $D$ is correct.

(D) The first ionization enthalpy values for the given elements are: $Si = 786 \ kJ \ mol^{-1}$,$P = 1012 \ kJ \ mol^{-1}$,$S = 999 \ kJ \ mol^{-1}$,and $Cl = 1256 \ kJ \ mol^{-1}$.
Across a period,ionization enthalpy generally increases due to increasing effective nuclear charge and decreasing atomic size.
However,$P$ $(3s^2 3p^3)$ has a stable half-filled $p$-orbital configuration,which makes its ionization enthalpy higher than that of $S$ $(3s^2 3p^4)$.
Therefore,the correct order is $Si < S < P < Cl$,and the corresponding values are $786, 999, 1012, 1256$.

(D) The electronic configuration of Boron $(Z=5)$ is $1s^2 2s^2 2p^1$.
After losing $3$ electrons,Boron attains the stable noble gas configuration of Helium $(1s^2)$.
Removing the third electron from Boron requires breaking the stable $2s^2$ shell,which involves a very high amount of energy compared to the other elements listed.
Therefore,Boron has the maximum third ionization energy.

(D) Given that $A^{+}$,$B^{2+}$,and $C^{3+}$ have zero electrons,these ions must be $H^{+}$,$He^{2+}$,and $Li^{3+}$.
Thus,the neutral atoms are $A = H$ $(1s^1)$,$B = He$ $(1s^2)$,and $C = Li$ $(1s^2 \ 2s^1)$.
Comparing the ionization energies:
$A_1 = IE_1(H) = 13.6 \ eV$
$B_1 = IE_1(He) = 24.6 \ eV$
$B_2 = IE_2(He) = 54.4 \ eV$
$C_1 = IE_1(Li) = 5.4 \ eV$
$C_2 = IE_2(Li) = 75.6 \ eV$
$C_3 = IE_3(Li) = 122.4 \ eV$
Comparing the values:
$C_3 (122.4) > C_2 (75.6) > B_2 (54.4) > B_1 (24.6) > A_1 (13.6) > C_1 (5.4)$.
Evaluating the options:
$A$. $C_3 > B_2 > A_1$ $(122.4 > 54.4 > 13.6)$ is correct.
$B$. $B_1 > A_1 > C_1$ $(24.6 > 13.6 > 5.4)$ is correct.
$C$. $C_3 > C_2 > B_2$ $(122.4 > 75.6 > 54.4)$ is correct.
$D$. $B_2 > C_3 > A_1$ $(54.4 > 122.4 > 13.6)$ is incorrect.
Therefore,the incorrect order is $D$.

(A) The electronic configuration of $Se$ $(Z=34)$ is $[Ar] 3d^{10} 4s^2 4p^4$. The second ionisation energy involves removing an electron from $Se^+$. Since $As$ $(Z=33)$ has a stable half-filled $4p^3$ configuration,removing the second electron from $As^+$ is harder than from $Se^+$. Thus,the statement is incorrect.
$(b)$ $C^{2+}$ has a stable $2s^2$ configuration,making its ionisation energy higher than $N^{2+}$ $(2p^1)$.
$(c)$ $F^{2+}$ has a stable $2p^3$ half-filled configuration,making its third ionisation energy higher than $O^{2+}$ $(2p^2)$.
$(d)$ Noble gases,not halogens,have the highest $I.E.$ in their respective periods due to their stable octet configuration.

(C) For element $P$,the large jump between $IE_1$ $(495.8)$ and $IE_2$ $(4562)$ indicates it has $1$ valence electron,identifying it as an alkali metal.
For element $Q$,the large jump between $IE_2$ $(1451)$ and $IE_3$ $(7733)$ indicates it has $2$ valence electrons,identifying it as an alkaline earth metal.
For element $R$,the jumps between $IE_1, IE_2,$ and $IE_3$ are relatively small,suggesting it has $3$ valence electrons (e.g.,Aluminum),making it a $p-block$ element.
Since $P, Q,$ and $R$ represent elements with $1, 2,$ and $3$ valence electrons respectively,they are likely in the same period (e.g.,$Na, Mg, Al$).
Thus,the statement '$R$: $s-block$ element' is incorrect.

(A) The ionization energy of mercury $(Hg)$ is very high,which prevents the sharing of electrons for metallic bonding.
Consequently,the metallic bonds between mercury atoms are very weak.
Due to these weak metallic bonds,mercury remains in a liquid state at $0 \, ^{\circ}C$.

(A) 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 $1s^2 2s^2 2p^4$.
Due to the extra stability of the half-filled $2p$ subshell in nitrogen,it requires more energy to remove an electron compared to oxygen.
Therefore,the first ionization potential of nitrogen $(14.6 \ eV)$ is higher than that of oxygen $(13.6 \ eV)$.

(D) The ionization energy increases across a period and decreases down a group.
Noble gases have the highest ionization energy in their respective periods due to their stable electronic configuration $(ns^2 np^6)$.
Therefore,elements like $He, Ne, Ar, Kr$ represent the peaks of ionization energy in the periodic table.

(C) Noble gases have stable electronic configurations $(ns^2 np^6)$,which makes it very difficult to remove an electron from their valence shell. Therefore,they possess very high ionization energy values.

(C) $1$. $Pb$ $(I.E.)$ > $Sn$ $(I.E.)$: Due to the inert pair effect and poor shielding of $4f$ and $5d$ electrons,$Pb$ has a higher effective nuclear charge than $Sn$,making its $I.E.$ higher. This is correct.
$2$. $Na^{+}$ $(I.E.)$ > $Mg^{+}$ $(I.E.)$: $Na^{+}$ has a stable noble gas configuration $(1s^2 2s^2 2p^6)$,while $Mg^{+}$ has a $3s^1$ configuration. Removing an electron from a stable octet requires significantly more energy. This is correct.
$3$. $Li^{+}$ $(I.E.)$ < $O^{+}$ $(I.E.)$: $Li^{+}$ $(1s^2)$ is a stable noble gas configuration,whereas $O^{+}$ $(1s^2 2s^2 2p^3)$ has a half-filled $p$-orbital. However,the $I.E.$ of $Li^{+}$ is extremely high due to the removal of an electron from the $1s$ shell. Thus,$Li^{+}$ $(I.E.)$ > $O^{+}$ $(I.E.)$. This is incorrect.
$4$. $Be^{+}$ $(I.E.)$ < $C^{+}$ $(I.E.)$: $Be^{+}$ $(1s^2 2s^1)$ has a lower $I.E.$ than $C^{+}$ $(1s^2 2s^2 2p^1)$ because $C^{+}$ has a higher effective nuclear charge. This is correct.

(C) The energy absorbed for the conversion of $M_{(g)}$ to $M_{(g)}^{+}$ is the first ionization energy $(IE_1)$.
For the elements $Li, Be, B, C, N$ (period $2$),the general trend of $IE_1$ increases across the period due to increasing effective nuclear charge.
However,there are exceptions due to stable electronic configurations:
$Li (1s^2 2s^1) = a$
$Be (1s^2 2s^2) = b$ (fully filled $s$-orbital,higher than $B$)
$B (1s^2 2s^2 2p^1) = c$
$C (1s^2 2s^2 2p^2) = d$
$N (1s^2 2s^2 2p^3) = e$ (half-filled $p$-orbital,higher than $O$)
The order is $Li < B < Be < C < N$,which corresponds to $a < c < b < d < e$.

(B) The ionization potential $(I.P.)$ increases across a period and decreases down a group.
Among the given elements,$He$ (Helium) is a noble gas located at the top of the periodic table.
It has the smallest atomic size and a stable electronic configuration $(1s^2)$,which makes it extremely difficult to remove an electron.
Therefore,$He$ has the highest $I.P.$ value among all elements in the periodic table.

(C) Let us analyze each option:
$A$. $N (2s^2 2p^3) > O (2s^2 2p^4)$: Nitrogen has a stable half-filled $p$-orbital,making its $I.P.$ higher than Oxygen. This is correct.
$B$. $Zr < Hf$: Due to Lanthanide contraction,the size of $Hf$ is similar to $Zr$,but $Hf$ has a higher effective nuclear charge,leading to a higher $I.P.$ for $Hf$. Thus,$Zr < Hf$ is correct.
$C$. $Na > K$: As we move down a group,the size increases and $I.P.$ decreases. $Na$ is above $K$,so $Na > K$ is correct. However,the reason given is $Z_{eff}$. While $Z_{eff}$ plays a role,the primary reason for the trend in a group is the increase in the number of shells and shielding effect. This is often considered a mismatch in context.
$D$. $Al < Ga$: $Ga$ has a higher $I.P.$ than $Al$ due to poor shielding by $d$-electrons (Transition contraction). This is correct.
Therefore,the statement $C$ is the incorrect match because the primary reason for $Na > K$ is the increase in atomic size/number of shells,not just $Z_{eff}$.

(A) The electronic configuration of Nitrogen ($N$,$Z=7$) is $1s^2 2s^2 2p^3$.
Because of the stable half-filled $2p$ subshell,it has a higher ionization enthalpy.
The electronic configuration of Oxygen ($O$,$Z=8$) is $1s^2 2s^2 2p^4$.
Due to the removal of an electron from a more stable $2p^3$ configuration in Nitrogen compared to the $2p^4$ configuration in Oxygen,Nitrogen has a higher first ionization enthalpy $(14.6 \ eV)$ than Oxygen $(13.6 \ eV)$.

(B) The first ionization enthalpy depends on the stability of the electronic configuration and the effective nuclear charge.
$ns^2 np^6$ represents a noble gas configuration,which is highly stable and has the highest ionization enthalpy.
$ns^2 np^3$ represents a half-filled p-orbital,which is also extra stable.
Comparing $ns^2 np^4$ and $ns^2 np^5$,the $ns^2 np^4$ configuration has a lower effective nuclear charge and less stability compared to the half-filled or fully-filled states,making it easier to remove an electron compared to the others listed in the context of periodic trends.
However,among the given options,$ns^2 np^4$ has the lowest ionization energy because it is neither half-filled nor fully-filled,and it is further to the left in the period compared to $ns^2 np^5$ and $ns^2 np^6$.

(A) The first ionization enthalpy $(IE_1)$ 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 increases while the number of electrons decreases.
Additionally,the cation formed is smaller than the neutral atom,which increases the electrostatic force of attraction between the nucleus and the remaining electrons.
Therefore,more energy is required to remove the second electron,making the second ionization enthalpy $(IE_2)$ always greater than the first $(IE_1)$.

(B) The first ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Noble gases,such as $He$,have a completely filled valence shell ($1s^2$ for $He$),which makes them exceptionally stable.
Due to this stable electronic configuration and high effective nuclear charge,removing an electron from $He$ requires the highest amount of energy compared to the other elements listed ($Be$,$Li$,$Fe$).

(D) The first ionization enthalpy $(IE_1)$ generally increases across a period,but there are exceptions due to stable electronic configurations (fully filled or half-filled orbitals).
$(i) \ Li < B < Be < C$: The correct order is $Li < B < Be < C$. $Be$ $(1s^2 2s^2)$ has a higher $IE_1$ than $B$ $(1s^2 2s^2 2p^1)$ because of the stable fully filled $2s$ orbital. Thus,the statement is correct.
$(ii) \ O < N < F$: The correct order is $O < N < F$. $N$ $(1s^2 2s^2 2p^3)$ has a higher $IE_1$ than $O$ $(1s^2 2s^2 2p^4)$ due to the stable half-filled $2p$ subshell. Thus,the statement is correct.
$(iii) \ Be < N < Ne$: $Be$ $(IE_1 \approx 899 \ kJ/mol)$,$N$ $(IE_1 \approx 1402 \ kJ/mol)$,and $Ne$ $(IE_1 \approx 2080 \ kJ/mol)$. The order $Be < N < Ne$ is correct.
Therefore,all three statements are correct.

(B) The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$.
It has a stable half-filled $p$-orbital configuration.
The electronic configuration of $O$ $(Z=8)$ is $1s^2 2s^2 2p^4$.
Removing an electron from the stable half-filled $2p^3$ orbital of $N$ requires more energy than removing an electron from the $2p^4$ orbital of $O$.
Therefore,the first ionization enthalpy of $O$ is less than that of $N$.

(C) Ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.
However,stability of electronic configurations plays a significant role.
Option $A$ $([Ne]\, 3s^2\, 3p^3)$ represents a half-filled $p$-orbital,which is exceptionally stable.
Option $D$ $([Ne]\, 3s^2)$ represents a fully-filled $s$-orbital,which is also stable,but it belongs to a lower group $(Group\, 2)$ compared to the $p$-block elements.
Comparing elements in the same period $(Period\, 3)$: $3s^2$ (Magnesium),$3s^2\, 3p^3$ (Phosphorus),$3s^2\, 3p^4$ (Sulfur),and $3s^2\, 3p^5$ (Chlorine).
Chlorine $(3s^2\, 3p^5)$ has the highest effective nuclear charge among these,making it the most difficult to remove an electron from.
Therefore,$[Ne]\, 3s^2\, 3p^5$ has the highest ionization enthalpy.

(B) The successive ionization energies are given as: $\Delta _iH_1 = 7.5 \ eV$,$\Delta _iH_2 = 25.6 \ eV$,$\Delta _iH_3 = 48.6 \ eV$,and $\Delta _iH_4 = 170.6 \ eV$.
$A$ large jump in ionization energy indicates the removal of an electron from a stable,fully-filled or half-filled shell (noble gas configuration).
Here,the jump from $\Delta _iH_3$ $(48.6 \ eV)$ to $\Delta _iH_4$ $(170.6 \ eV)$ is very large,which suggests that the atom has $3$ valence electrons.
Among the given options,the configuration $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$ (Aluminum,$Z=13$) has $3$ valence electrons in the $3s$ and $3p$ orbitals.
Therefore,the correct electronic configuration is $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$.

(B) The enthalpy change for the process $Mg \to Mg^{2+} + 2e^-$ is the sum of the first and second ionization enthalpies.
Given:
$\Delta_iH_1 = 178 \, kcal \, mol^{-1}$
$\Delta_iH_2 = 348 \, kcal \, mol^{-1}$
Total enthalpy change $\Delta H = \Delta_iH_1 + \Delta_iH_2$
$\Delta H = 178 + 348 = 526 \, kcal \, mol^{-1}$
Since ionization is an endothermic process,the value is positive.

(C) The ionization enthalpy $(IE)$ is the energy required to remove an electron from a gaseous atom or ion.
$(i)$ Removing an electron from a positively charged ion $(Be^{+})$ requires more energy than removing it from a neutral atom $(Be)$ because the effective nuclear charge is higher in the ion. Thus,$IE(Be^{+}) > IE(Be)$ is correct.
$(ii)$ $Be > Be^{+}$ is incorrect based on the logic above.
$(iii)$ Across a period from left to right,$IE$ generally increases. Since $C$ is to the right of $Be$,$IE(C) > IE(Be)$ is correct.
$(iv)$ $B$ has a lower $IE$ than $Be$ because $Be$ has a stable fully-filled $2s^{2}$ configuration. Thus,$IE(B) < IE(Be)$ is correct,making $B > Be$ incorrect.
Therefore,the correct statements are $(i)$ and $(iii)$.

(D) The ionization energy increases significantly when an electron is removed from a stable,noble gas-like configuration or a completely filled shell.
For option $D$,the configuration is $1s^2\, 2s^2\, 2p^6\, 3s^2$ (Magnesium,$Z=12$).
The first ionization energy removes one $3s$ electron.
The second ionization energy removes the second $3s$ electron.
After removing two electrons,the configuration becomes $1s^2\, 2s^2\, 2p^6$,which is the stable neon noble gas configuration.
Therefore,removing the third electron requires a very high amount of energy,leading to a large difference between the second and third ionization energies.

(B) The ionization energy values are given as: $IP_1 = 7.1 \ eV$,$IP_2 = 14.3 \ eV$,$IP_3 = 34.5 \ eV$,$IP_4 = 46.8 \ eV$,and $IP_5 = 162.2 \ eV$.
To identify the element,we look for the largest jump in successive ionization energy values.
The jump from $IP_4$ $(46.8 \ eV)$ to $IP_5$ $(162.2 \ eV)$ is very large,which indicates that the $5^{th}$ electron is being removed from a stable,noble gas-like inner shell configuration.
This implies the element has $4$ valence electrons.
Among the given options,$Si$ (Silicon) has an electronic configuration of $[Ne] 3s^2 3p^2$,which has $4$ valence electrons.
Therefore,the element is $Si$.

(B) The ionization potential $(IP)$ generally increases across a period from left to right and decreases down a group.
Comparing the elements $B$ (Boron,group $13$,period $2$),$Li$ (Lithium,group $1$,period $2$),and $K$ (Potassium,group $1$,period $4$):
$1$. $B$ is in period $2$ and group $13$,so it has a higher $IP$ than $Li$ (group $1$,period $2$).
$2$. $Li$ is in period $2$ and $K$ is in period $4$ (same group $1$),so $Li$ has a higher $IP$ than $K$.
Therefore,the correct order is $B > Li > K$.

(C) Alkali metals have a general electronic configuration of $ns^1$.
They have a low first ionization energy $(IE_1)$ because removing the single valence electron results in a stable noble gas configuration.
However,the second ionization energy $(IE_2)$ is significantly higher because it involves removing an electron from a stable,fully filled inner shell (noble gas core).
Comparing the options:
$X: IE_2/IE_1 = 1.1$
$Y: IE_2/IE_1 = 1.26$
$Z: IE_2/IE_1 = 2.56$
$M: IE_2/IE_1 = 1.25$
Option $Z$ shows a large jump between $IE_1$ and $IE_2$,which is characteristic of an alkali metal.

(C) The successive ionization enthalpy values are given as: $\Delta_i H_1 = 899 \ kJ/mol$,$\Delta_i H_2 = 1757 \ kJ/mol$,$\Delta_i H_3 = 14847 \ kJ/mol$,and $\Delta_i H_4 = 17948 \ kJ/mol$.
We observe a very large jump in energy between the second and third ionization enthalpies ($\Delta_i H_2$ to $\Delta_i H_3$).
This indicates that the element has $2$ valence electrons in its outermost shell,as removing the third electron requires breaking a stable noble gas configuration.
Elements with $2$ valence electrons belong to Group $2$,which is the alkaline earth metal group.

(A) As we move down a group in the periodic table,the atomic number increases.
$1$. The atomic radius increases due to the addition of new shells.
$2$. The ionization energy decreases because the valence electrons are further from the nucleus and more shielded.
$3$. Electronegativity decreases as the attraction between the nucleus and shared pair of electrons decreases.
$4$. Electron affinity generally decreases as the size of the atom increases,making it harder to add an electron.
Therefore,the statement 'Ionization energy increases' is incorrect.

(D) The electronic configuration of $Mg$ $(Z=12)$ is $[Ne] 3s^2$,which is a stable,completely filled orbital configuration.
The electronic configuration of $Al$ $(Z=13)$ is $[Ne] 3s^2 3p^1$.
Removing an electron from $Mg$ requires breaking the stable,fully filled $3s$ subshell,which requires more energy.
In $Al$,the electron is removed from the $3p$ orbital,which is easier as it is shielded by the $3s$ electrons and is less stable than the fully filled $3s$ subshell.
Therefore,the first ionization enthalpy of $Al$ is lower than that of $Mg$ due to the stable electronic configuration of $Mg$.

(D) The electronic configuration of $Na$ is $[Ne] 3s^1$ and $Mg$ is $[Ne] 3s^2$.
For the first ionization enthalpy $(I)$,$Mg$ has a stable fully-filled $s$-orbital,so $I(Mg) > I(Na)$.
For the second ionization enthalpy $(II)$,$Na^+$ has the stable noble gas configuration $[Ne]$,making it very difficult to remove the second electron.
Conversely,$Mg^+$ has the configuration $[Ne] 3s^1$,from which the second electron is relatively easier to remove.
Therefore,$II(Na) > II(Mg)$.

(A) To match the ionization energies $(IE)$,we look at the valence electrons and the jump in energy after removing all valence electrons:
$1$. $H$ $(1s^1)$: Only $1$ electron,so $IE_1 = 1312 \ kJ \ mol^{-1}$ (matches $A$ approx $1510$).
$2$. $Li$ $(1s^2 2s^1)$: $1$ valence electron,large jump after $IE_1$. $IE_1=520, IE_2=7300$ (matches $B$ $495, 6500, 10200$).
$3$. $Be$ $(1s^2 2s^2)$: $2$ valence electrons,large jump after $IE_2$. $IE_1=899, IE_2=1757, IE_3=14848$ (matches $C$ $840, 1630, 13100$).
$4$. $B$ $(1s^2 2s^2 2p^1)$: $3$ valence electrons,large jump after $IE_3$. $IE_1=801, IE_2=2427, IE_3=3660$ (matches $D$ $600, 2050, 3100$).
Thus,the correct match is $A-1, B-2, C-3, D-4$.

(D) The second ionisation potential $(IE_2)$ corresponds to the energy required to remove an electron from a unipositive ion $(M^+ \rightarrow M^{2+} + e^-)$.
Electronic configurations of the elements:
$C (Z=6): 1s^2 2s^2 2p^2 \rightarrow C^+: 1s^2 2s^2 2p^1$
$N (Z=7): 1s^2 2s^2 2p^3 \rightarrow N^+: 1s^2 2s^2 2p^2$
$O (Z=8): 1s^2 2s^2 2p^4 \rightarrow O^+: 1s^2 2s^2 2p^3$
$F (Z=9): 1s^2 2s^2 2p^5 \rightarrow F^+: 1s^2 2s^2 2p^4$
Among these,$N^+$ has a stable half-filled $2p^2$ configuration,but $O^+$ has a $2p^3$ configuration which is half-filled and exceptionally stable.
However,the second ionisation potential generally increases across a period due to increasing effective nuclear charge.
Comparing $N^+$ and $O^+$,the removal of an electron from $N^+$ $(2p^2)$ is easier than from $O^+$ $(2p^3)$.
Comparing $O^+$ and $F^+$,$F^+$ has a higher effective nuclear charge than $O^+$,making it harder to remove an electron from $F^+$.
Thus,$F$ has the highest second ionisation potential among the given options.

(A) In a group,the Ionization Potential $(I.P.)$ generally decreases from top to bottom due to an increase in atomic size and shielding effect.
However,the graph shows an irregular trend $(A > B < C > D < E)$.
This specific zig-zag pattern of $I.P.$ values is characteristic of elements in group $13$ (e.g.,$B, Al, Ga, In, Tl$).
In group $13$,the $I.P.$ decreases from $B$ to $Al$,then increases from $Al$ to $Ga$ due to poor shielding of $d$-electrons,decreases from $Ga$ to $In$,and increases from $In$ to $Tl$ due to poor shielding of $f$-electrons.
Therefore,the group is $13$.

(C) The Ionization Potential $(IP)$ decreases down a group as the atomic size increases and the valence electrons are further from the nucleus.
For noble gases,the $IP$ values are: $He (2372 \ kJ/mol)$,$Ne (2080 \ kJ/mol)$,$Ar (1520 \ kJ/mol)$,$Kr (1351 \ kJ/mol)$,$Xe (1170 \ kJ/mol)$.
Calculating the ratios:
$A) He/Ne = 2372/2080 \approx 1.14$
$B) Ne/Ar = 2080/1520 \approx 1.37$
$C) He/Xe = 2372/1170 \approx 2.03$
$D) Kr/Xe = 1351/1170 \approx 1.15$
The highest ratio is obtained for the pair $He : Xe$.

(B) The successive ionization energies are $IE_1 = 20 \ eV, IE_2 = 45 \ eV, IE_3 = 150 \ eV, IE_4 = 900 \ eV, IE_5 = 1800 \ eV$.
There is a large jump in ionization energy between $IE_3$ and $IE_4$ $(900 - 150 = 750 \ eV)$.
This indicates that the fourth electron is being removed from a stable noble gas core,meaning the element $(A)$ has $3$ valence electrons.
Therefore,the stable oxidation state of $(A)$ is $+3$.
Thus,the formula of the halide of $(A)$ is $AX_3$.

(A) The given species $S^{2-}$,$Cl^{-}$,$Ar$,and $K^{+}$ are isoelectronic,meaning they all have $18$ electrons.
For isoelectronic species,the ease of losing an electron (ionization energy) depends on the nuclear charge (number of protons).
As the nuclear charge increases,the attraction between the nucleus and the valence electrons increases,making it harder to remove an electron.
The nuclear charges are: $S^{2-} (Z=16)$,$Cl^{-} (Z=17)$,$Ar (Z=18)$,and $K^{+} (Z=19)$.
Since $S^{2-}$ has the lowest nuclear charge $(Z=16)$,it has the weakest hold on its electrons,making it the easiest to lose an electron.

(B) The stability of an ion depends on the energy required to reach that oxidation state. $A$ lower sum of ionization energies $(IE)$ indicates that the corresponding ion is more stable.
For $P^{2+}$ and $Q^{2+}$: The sum of $IE_1 + IE_2$ is $2.45 \ kJ/mol$ for $P$ and $2.85 \ kJ/mol$ for $Q$. Since $2.45 < 2.85$,$P^{2+}$ is more stable than $Q^{2+}$. Thus,statement $(A)$ is correct.
For $P^{4+}$ and $Q^{4+}$: The sum of $IE_1 + IE_2 + IE_3 + IE_4$ is $(2.45 + 8.82) = 11.27 \ kJ/mol$ for $P$ and $(2.85 + 6.11) = 8.96 \ kJ/mol$ for $Q$. Since $8.96 < 11.27$,$Q^{4+}$ is more stable than $P^{4+}$.
Therefore,statement $(B)$ ($P^{4+}$ is more stable than $Q^{4+}$) is incorrect,and statement $(C)$ ($P^{4+}$ is less stable than $Q^{4+}$) is correct.
The question asks for the incorrect statement$(s)$. Only $(B)$ is incorrect.

(C) Alkali metals have the lowest first ionization energy $(I.E.)$ in their respective periods.
In the given graph,the points $S$ and $Q$ represent alkali metals because they show the minimum values of $I.E.$
Since the graph plots $I.E.$ against atomic number,the point $S$ appears before $Q$ on the atomic number axis.
Therefore,$S$ represents the alkali metal with the least atomic number among the given options.

(B) Metallic character is inversely proportional to the Ionization Potential $(IP)$.
Elements with lower $IP$ values lose electrons more easily,exhibiting higher metallic character.
Comparing the given values: $P = 17 \ eV$,$Q = 2 \ eV$,$R = 10 \ eV$,$S = 13 \ eV$.
The element $Q$ has the lowest $IP$ value $(2 \ eV)$,therefore it possesses the highest metallic character.

(B) The first ionization energy $(IE_1)$ of $Mg$ $([Ne] 3s^2)$ is higher than $Al$ $([Ne] 3s^2 3p^1)$ because $Mg$ has a stable fully-filled $s$-orbital. Thus,$(Al)_{IE_1} < (Mg)_{IE_1}$ is correct.
For $IE_2$,$Na$ $([Ne] 3s^1)$ loses its second electron from the stable $2p^6$ configuration,while $Mg$ $([Ne] 3s^2)$ loses its second electron from the $3s^1$ orbital. Thus,$(Na)_{IE_2} > (Mg)_{IE_2}$. Therefore,the relation $(Mg)_{IE_2} > (Na)_{IE_2}$ is incorrect.
$(Na)_{IE_1} < (Mg)_{IE_1}$ is correct due to the stable configuration of $Mg$.
$(Mg)_{IE_3} > (Al)_{IE_3}$ is correct because $Mg^{2+}$ has a stable noble gas configuration,making the removal of the third electron very difficult compared to $Al^{2+}$.

(B) Generally,the first ionisation energy increases as we move across a period in the periodic table.
Rare gases (noble gases) possess a stable electronic configuration $(ns^2 np^6)$,which makes it very difficult to remove an electron from their valence shell.
Therefore,they have the highest first ionisation energy in their respective periods.
Consequently,the peaks in the plot of first ionisation energy versus atomic number are occupied by rare gases.

(D) . $Se$ $([Ar] 3d^{10} 4s^2 4p^4)$ and $As$ $([Ar] 3d^{10} 4s^2 4p^3)$. After removing one electron,$Se^+$ has $4p^3$ and $As^+$ has $4p^2$. Removing the second electron from $Se^+$ is harder due to the stable half-filled $p$-orbital,so this statement is correct.
$B$. $C^{2+}$ has $1s^2 2s^2$ configuration,while $N^{2+}$ has $1s^2 2s^2 2p^1$. Removing an electron from the stable $2s^2$ shell of $C^{2+}$ requires more energy than from the $2p^1$ orbital of $N^{2+}$. This statement is correct.
$C$. $F^{2+}$ has $1s^2 2s^2 2p^3$ and $O^{2+}$ has $1s^2 2s^2 2p^2$. Removing the third electron from $F^{2+}$ (half-filled $p$-orbital) is harder than from $O^{2+}$. This statement is correct.
$D$. Noble gases have the highest $I.E.$ in their respective periods,not halogens. Therefore,this statement is incorrect.

(C) The first ionisation energy $(IE_1)$ is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Alkali metals have the lowest first ionisation energy in their respective periods because they have a stable noble gas core plus one valence electron in the $ns^1$ orbital.
Comparing the options:
$A$: $1s^2, 2s^2, 2p^5$ (Fluorine,a halogen,has high $IE_1$)
$B$: $1s^2, 2s^2, 2p^6, 3s^2, 3p^1$ (Aluminium,group $13$)
$C$: $1s^2, 2s^2, 2p^6, 3s^1$ (Sodium,an alkali metal,group $1$)
$D$: $1s^2, 2s^2, 2p^6$ (Neon,a noble gas,has very high $IE_1$)
Therefore,the configuration $1s^2, 2s^2, 2p^6, 3s^1$ corresponds to an alkali metal,which has the minimum first ionisation energy among the given choices.

(C) The ionization energy $(IE)$ is the energy required to remove an electron from a gaseous atom or ion.
Successive ionization energies increase significantly because it becomes harder to remove an electron from an increasingly positive ion.
Option $A$ and $B$ represent the first ionization energy $(IE_1)$ of $Ba$ and $Ca$ respectively.
Option $C$ represents the second ionization energy $(IE_2)$ of $Ca$,which is significantly higher than $IE_1$.
Option $D$ represents the total energy for the removal of two electrons $(IE_1 + IE_2)$ from $Mg$.
Comparing the processes,the removal of the second electron from a cation (as in option $C$) or the cumulative energy for two electrons (as in option $D$) involves higher energy values than a single first ionization.
However,the second ionization energy of $Ca$ $(IE_2)$ is a specific high-energy step compared to the first ionization of larger atoms like $Ba$ or $Ca$. Given the standard comparison of $IE$ values,the process $Ca^+(g) \to Ca^{2+}(g) + e^-$ requires the most energy among the single-step processes listed.

(A) The first ionization energy of elements in a group decreases as we move down the group due to an increase in atomic size and shielding effect.
For the alkaline earth metals (Group $2$),the order of elements from top to bottom is $Be, Mg, Ca, Sr, Ba, Ra$.
Therefore,the decreasing order of first ionization energy is $Mg > Ca > Sr > Ba$.
Thus,the correct option is $A$.

(A) The first ionization energy of alkaline earth metals $(Group \ 2)$ is higher than that of alkali metals $(Group \ 1)$ because alkaline earth metals have a smaller atomic size and a higher nuclear charge compared to alkali metals in the same period.
Due to the higher nuclear charge,the valence electrons are more strongly attracted by the nucleus,making it more difficult to remove an electron,thus resulting in higher ionization energy.