Ionisation energy Questions in English

(B) In general,from left to right in a period,the first ionization enthalpy increases due to an increase in effective nuclear charge.
However,due to the extra stability of half-filled $(p^3)$ and fully-filled $(s^2)$ electronic configurations,the ionization enthalpy of certain elements is higher than their neighbors.
The electronic configurations are: $Mg$ $([Ne] 3s^2)$,$Al$ $([Ne] 3s^2 3p^1)$,$Si$ $([Ne] 3s^2 3p^2)$,$P$ $([Ne] 3s^2 3p^3)$,and $S$ $([Ne] 3s^2 3p^4)$.
Due to the stable $s^2$ configuration of $Mg$,its ionization enthalpy is higher than $Al$.
Due to the stable half-filled $p^3$ configuration of $P$,its ionization enthalpy is higher than $S$.
Thus,the correct increasing order is $Al < Mg < Si < S < P$.

(A) The electronic configurations correspond to the following elements:
$(a)$ $1s^{2} 2s^{2} \rightarrow Be$ (Beryllium)
$(b)$ $1s^{2} 2s^{2} 2p^{4} \rightarrow O$ (Oxygen)
$(c)$ $1s^{2} 2s^{2} 2p^{3} \rightarrow N$ (Nitrogen)
$(d)$ $1s^{2} 2s^{2} 2p^{1} \rightarrow B$ (Boron)
Ionization enthalpy ($\Delta_{i}H$) values are influenced by electronic stability:
$Be$ $(2s^{2})$ has a fully filled orbital,making its $IE$ higher than $B$ $(2s^{2} 2p^{1})$.
$N$ $(2p^{3})$ has a half-filled $p$-orbital,making its $IE$ higher than $O$ $(2p^{4})$.
The values are:
$N = 1402 \ kJ\ mol^{-1}$
$O = 1314 \ kJ\ mol^{-1}$
$Be = 899 \ kJ\ mol^{-1}$
$B = 801 \ kJ\ mol^{-1}$
Matching:
$(a) \rightarrow (ii)$
$(b) \rightarrow (iii)$
$(c) \rightarrow (iv)$
$(d) \rightarrow (i)$
Therefore,the correct option is $(a)$ $\rightarrow (ii), (b)$ $\rightarrow (iii), (c)$ $\rightarrow (iv), (d)$ $\rightarrow (i)$.

(B) The electronic configurations of the elements are:
$Mg (Z=12): [Ne] 3s^2$
$Al (Z=13): [Ne] 3s^2 3p^1$
$P (Z=15): [Ne] 3s^2 3p^3$
$S (Z=16): [Ne] 3s^2 3p^4$
$1$. $Mg$ has a fully filled $3s$ orbital,making it more stable than $Al$,so $I.E._{Mg} > I.E._{Al}$.
$2$. $P$ has a half-filled $3p$ subshell $(3p^3)$,which is highly stable,making its $I.E.$ higher than that of $S$ $(3p^4)$.
$3$. Across a period,$I.E.$ generally increases. Combining these factors,the order is $Al < Mg < S < P$.

(A) The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$,which has a stable half-filled $p$-orbital.
The electronic configuration of $O$ $(Z=8)$ is $1s^2 2s^2 2p^4$.
Due to the half-filled stability of the $2p$ subshell in nitrogen,it requires more energy to remove an electron compared to oxygen.
In oxygen,the $2p^4$ configuration means that two electrons must occupy the same $2p$ orbital,leading to increased inter-electronic repulsion,which makes it easier to remove an electron.
Thus,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation for $A$.

(D) The electronic configurations are: $Be (2s^2)$,$B (2s^2 2p^1)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$.
Due to the stable fully-filled $2s$ orbital,$Be$ has a higher ionization enthalpy than $B$.
Due to the stable half-filled $2p$ orbital,$N$ has a higher ionization enthalpy than $O$.
The overall order of first ionization enthalpy is $B < Be < O < N$.

(B) The elements are identified as:
$A: 3s^{2} \rightarrow Mg$
$B: 3s^{2} 3p^{1} \rightarrow Al$
$C: 3s^{2} 3p^{3} \rightarrow P$
$D: 3s^{2} 3p^{4} \rightarrow S$
Ionization enthalpy generally increases across a period from left to right.
However,there are exceptions due to stable electronic configurations:
$1$. $Mg$ $(3s^{2})$ has a fully filled $s$-orbital,making it more stable than $Al$ $(3s^{2} 3p^{1})$.
$2$. $P$ $(3s^{2} 3p^{3})$ has a half-filled $p$-orbital,making it more stable than $S$ $(3s^{2} 3p^{4})$.
Comparing the values:
$Al < Mg < S < P$
Substituting the labels:
$B < A < D < C$

(D) $1$. Ionization enthalpy decreases down a group. Thus,for alkali metals: $Li > Na > K$. Statement $A$ is correct.
$2$. In noble gases,ionization enthalpy decreases down the group. $Xe$ is below $He, Ne, Ar, Kr$,so it has a lower ionization enthalpy than them. However,$Rn$ is below $Xe$,so $Rn$ has the lowest. Thus,$Xe$ is not the lowest. Statement $B$ is correct.
$3$. Atomic number $37$ is $Rb$ (Group $1$) and atomic number $38$ is $Sr$ (Group $2$). Across a period,ionization enthalpy increases. Thus,$Sr > Rb$. Statement $C$ is correct.
$4$. Atomic number $30$ is $Zn$ $([Ar] 3d^{10} 4s^2)$ and $Ga$ is $([Ar] 3d^{10} 4s^2 4p^1)$. Due to the stable $d^{10}$ configuration and higher effective nuclear charge,$Zn$ has a higher ionization enthalpy than $Ga$. Thus,the statement that $Ga$ is higher than $Zn$ is incorrect.

(C) The general trend for the first ionization enthalpy across the third period is: $Na < Al < Mg < Si$.
Given values are:
$IE(Na) = 496 \ kJ \ mol^{-1}$
$IE(Mg) = 737 \ kJ \ mol^{-1}$
$IE(Si) = 786 \ kJ \ mol^{-1}$
Based on the periodic trend,the ionization enthalpy of $Al$ should be greater than $Na$ but less than $Mg$.
Therefore,$496 < IE(Al) < 737$.
Among the given options,only $577 \ kJ \ mol^{-1}$ falls within this range.
Thus,the correct option is $(C)$.

(A) The difference between the $1^{st}$ and $2^{nd}$ ionisation energies is largest when the removal of the second electron requires breaking a stable,fully-filled shell or subshell configuration.

Electronic configuration	Element/Group
$1s^{2} 2s^{2} 2p^{6}$	Neon (Group $18$)
$1s^{2} 2s^{2} 2p^{6} 3s^{1}$	Sodium (Group $1$)
$1s^{2} 2s^{2} 2p^{6} 3s^{2}$	Magnesium (Group $2$)
$1s^{2} 2s^{2} 2p^{1}$	Boron (Group $13$)

For $1s^{2} 2s^{2} 2p^{6}$ (Neon),the $1^{st}$ ionisation energy is very high due to its stable noble gas configuration. However,the $2^{nd}$ ionisation energy is significantly higher because it involves removing an electron from a stable $2p^{6}$ shell,which is extremely difficult. Thus,the jump between $IE_1$ and $IE_2$ is largest for this configuration.

(C) The ionisation potential $(IP)$ generally decreases as we move down a group in the periodic table due to the increase in atomic size and shielding effect.
$Na$ and $K$ both belong to Group $1$ (alkali metals).
Since $K$ is below $Na$ in the periodic table,the $IP$ of $K$ must be lower than the $IP$ of $Na$.
The $IP$ of $Na$ is $5.14 \ eV$.
Among the given options,only $4.3 \ eV$ is less than $5.14 \ eV$.
Therefore,the correct option is $C$.

(B) $Na$ is an alkali metal and has the lowest ionisation energy due to its large atomic size.
$B$,$N$,and $O$ are non-metals belonging to the same period ($2^{nd}$ period).
Generally,ionisation energy increases from left to right across a period due to an increase in effective nuclear charge.
However,$N$ $(1s^2 2s^2 2p^3)$ has a stable half-filled $p$-orbital configuration,which makes its ionisation energy higher than that of $O$ $(1s^2 2s^2 2p^4)$.
Therefore,the correct order is $Na < B < O < N$.

(A) The elements $O$,$F$,and $Ne$ belong to the same period,i.e.,the $2nd$ period of the periodic table.
As we move from left to right across a period,the atomic radius decreases and the effective nuclear charge increases,which leads to an increase in the first ionisation enthalpy.
The order of increasing ionisation enthalpy for these elements is $O < F < Ne$.
Given values are $1314 \, kJ \, mol^{-1}$ for $O$,$1680 \, kJ \, mol^{-1}$ for $F$,and $2080 \, kJ \, mol^{-1}$ for $Ne$.
Therefore,the correct sequence is $O, F$ and $Ne$.

(B) The first ionization enthalpy values are as follows:
$B$: $801 \ kJ/mol$
$Al$: $577 \ kJ/mol$
$Ga$: $579 \ kJ/mol$
Statement $I$: The decrease from $B$ to $Al$ is $801 - 577 = 224 \ kJ/mol$,while the change from $Al$ to $Ga$ is $579 - 577 = 2 \ kJ/mol$ (an increase). Thus,the decrease from $B$ to $Al$ is indeed much larger than the change from $Al$ to $Ga$. Statement $I$ is correct.
Statement $II$: The electronic configuration of $Ga$ $(Z=31)$ is $[Ar] \ 3d^{10} \ 4s^2 \ 4p^1$. The $3d$ orbitals are completely filled. Statement $II$ is correct.

(A) For element $X$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $3069 - 419 = 2650 \ kJ \ mol^{-1}$,which is very large. This indicates that the removal of the second electron requires breaking a stable noble gas configuration,characteristic of an alkali metal.
For element $Y$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $1145 - 590 = 555 \ kJ \ mol^{-1}$,which is relatively small. This indicates that the second electron is removed from the same valence shell,characteristic of an alkaline earth metal.
Therefore,$X$ is an alkali metal and $Y$ is an alkaline earth metal.

(C) The electronic configurations of the elements are as follows:
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
$F (Z=9): 1s^2 2s^2 2p^5$
Across a period,ionization enthalpy generally increases due to increasing effective nuclear charge.
However,nitrogen $(N)$ has a stable half-filled $2p^3$ configuration,which makes its ionization enthalpy higher than that of oxygen $(O)$.
Therefore,the correct order is $C < O < N < F$.

(D) The first ionization energy $(IE_1)$ is the energy required to remove the first electron from a neutral atom $(Mg \rightarrow Mg^+ + e^-)$.
The second ionization energy $(IE_2)$ is the energy required to remove the second electron from the unipositive ion $(Mg^+ \rightarrow Mg^{2+} + e^-)$.
Since $Mg^{2+}$ is a smaller ion with a higher positive charge compared to $Mg^+$,the electrostatic attraction between the nucleus and the remaining electrons is much stronger,making it significantly harder to remove the second electron.
Therefore,$IE_2 > IE_1$,which means the energy required to form $Mg^{2+}$ from $Mg$ (total energy = $IE_1 + IE_2$) is higher than that required to produce $Mg^+$ $(IE_1)$.
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(C) The first ionization enthalpy $(I.E._1)$ generally 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 (fully filled or half-filled orbitals).
For the given elements in the second period,the order is:
$Li (520 \ kJ/mol) < B (801 \ kJ/mol) < Be (899 \ kJ/mol) < C (1086 \ kJ/mol) < O (1314 \ kJ/mol) < N (1402 \ kJ/mol) < F (1681 \ kJ/mol)$.
$Be$ has a higher $I.E._1$ than $B$ due to its stable $2s^2$ configuration.
$N$ has a higher $I.E._1$ than $O$ due to its stable half-filled $2p^3$ configuration.
Therefore,the correct order is $Li < B < Be < C < O < N < F$.

(C) The first ionisation enthalpy generally increases across a period from left to right because the atomic radius decreases and the effective nuclear charge increases.
Therefore,Assertion $A$ is false.
Reason $R$ states that the increasing nuclear charge outweighs the shielding effect across the period,which is a correct statement explaining why ionisation energy increases.
Thus,$A$ is false but $R$ is true.

(C) The first ionization enthalpy $(IE_1)$ generally increases across a period from left to right.
For the given elements ($Al$,$Si$,$C$,$N$),they belong to the $2^{nd}$ and $3^{rd}$ periods.
$Al$ and $Si$ are in the $3^{rd}$ period,while $C$ and $N$ are in the $2^{nd}$ period.
Elements in the $2^{nd}$ period have higher ionization enthalpies than those in the $3^{rd}$ period due to smaller atomic size.
Comparing $C$ and $N$ in the $2^{nd}$ period,$N$ has a higher $IE_1$ than $C$ because of its stable half-filled $p$-orbital configuration $(2s^2 2p^3)$.
Therefore,the order of $IE_1$ is $Al < Si < C < N$.

(B) The first ionization enthalpy $(IE_1)$ generally 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.
For the given elements in the second period,the order is:
$B (2s^2 2p^1) < Be (2s^2) < C (2s^2 2p^2) < O (2s^2 2p^4) < N (2s^2 2p^3) < F (2s^2 2p^5)$.
Note that $Be$ has a higher $IE_1$ than $B$ due to a fully filled $s$-orbital,and $N$ has a higher $IE_1$ than $O$ due to a half-filled $p$-orbital.
Mapping the elements to their labels: $A=O, B=N, C=Be, D=F, E=B$.
The order is $B < Be < O < N < F$,which corresponds to $E < C < A < B < D$.

(D) $(i)$ The first ionization enthalpy $(IE_1)$ decreases down the group from $B$ to $Al$ due to an increase in atomic size.
$(ii)$ From $Al$ to $Ga$,the $IE_1$ increases due to the poor shielding effect of $d$-electrons (scandide contraction).
$(iii)$ From $Ga$ to $Tl$,the $IE_1$ increases due to the poor shielding effect of $f$-electrons (lanthanide contraction).
$(iv)$ Combining these trends,the correct order is $Tl > Ga > Al$.

(A) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.
However,there are exceptions due to electronic configuration stability:
$1$. $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ orbital,making it more stable than $B$ $(1s^2 2s^2 2p^1)$,so $Be > B$.
$2$. $N$ $(1s^2 2s^2 2p^3)$ has a half-filled $2p$ subshell,which is more stable than $C$ $(1s^2 2s^2 2p^2)$,so $N > C$.
The correct increasing order is $Li < B < Be < C < N$.

Element	First ionization enthalpy $(\Delta_i H / kJ \ mol^{-1})$
$Li$	$520$
$Be$	$899$
$B$	$801$
$C$	$1086$
$N$	$1402$

(C) To determine the value of $n$,we analyze the jump in ionization enthalpies for each element:
$1$. For atomic number $(n+4)$,the jump occurs between $I_2$ and $I_3$ ($1451$ to $7733$),indicating it is an alkaline earth metal (Group $2$).
$2$. For atomic number $(n+3)$,the jump occurs between $I_1$ and $I_2$ ($496$ to $4562$),indicating it is an alkali metal (Group $1$).
$3$. Since $(n+3)$ is an alkali metal and $(n+4)$ is an alkaline earth metal,$(n+3)$ corresponds to $Z=11$ (Sodium) and $(n+4)$ corresponds to $Z=12$ (Magnesium).
$4$. If $n+3 = 11$,then $n = 8$.
$5$. Checking the sequence: $n=8$ (Oxygen),$n+1=9$ (Fluorine),$n+2=10$ (Neon),$n+3=11$ (Sodium). The ionization energies provided for $n+1$ $(1681)$ and $n+2$ $(2081)$ are consistent with Fluorine and Neon respectively. Thus,$n=8$.

(A) The first ionization energy $(IE_1)$ trend for group $14$ elements is $C > Si > Ge > Sn < Pb$.
Statement $I$: The $IE_1$ of $Pb$ $(715 \ kJ/mol)$ is greater than that of $Sn$ $(708 \ kJ/mol)$ due to the poor shielding effect of $4f$ and $5d$ electrons (lanthanoid contraction),which increases the effective nuclear charge. Thus,Statement $I$ is true.
Statement $II$: The $IE_1$ of $Ge$ $(761 \ kJ/mol)$ is less than that of $Si$ $(786 \ kJ/mol)$ because $Si$ is smaller than $Ge$. Thus,Statement $II$ is false.
Therefore,Statement $I$ is true but Statement $II$ is false.

(B) The successive ionization energies are $IE_1 = 800 \ kJ/mol$,$IE_2 = 2427 \ kJ/mol$,$IE_3 = 3658 \ kJ/mol$,$IE_4 = 25024 \ kJ/mol$,and $IE_5 = 32824 \ kJ/mol$.
There is a large jump in energy between the $3^{rd}$ and $4^{th}$ ionization energy $(IE_4 - IE_3 = 25024 - 3658 = 21366 \ kJ/mol)$.
This indicates that the $4^{th}$ electron is being removed from a stable noble gas core.
Therefore,the element has $3$ valence electrons,which places it in Group $13$.

(A) Statement $(I)$ is correct because,in the periodic table,the first ionisation enthalpy generally increases from left to right across a period due to the increase in effective nuclear charge and decrease in atomic size. Thus,group $14$ elements have higher first ionisation enthalpy than group $13$ elements.
Statement $(II)$ is incorrect because group $14$ elements (like carbon and silicon) form strong covalent bonds in their elemental state,leading to much higher melting and boiling points compared to the corresponding group $13$ elements.

(D) The first ionisation enthalpy $(IE_1)$ of Group $13$ elements does not decrease regularly down the group due to the poor shielding effect of $d$ and $f$ orbitals.
The observed order of $IE_1$ for Group $13$ elements is: $B > Tl > Ga > Al > In$.
Therefore,Boron $(B)$ has the highest $IE_1$ and Indium $(In)$ has the lowest $IE_1$.

(C) The given atomic numbers correspond to the following elements:
$32 \Rightarrow Ge$ (Germanium)
$35 \Rightarrow Br$ (Bromine)
$87 \Rightarrow Fr$ (Francium)
$19 \Rightarrow K$ (Potassium)
Ionisation enthalpy generally decreases down a group and increases across a period.
$Fr$ is an alkali metal located in the $7^{th}$ period and $1^{st}$ group.
Due to its large atomic size and the shielding effect of inner shells,$Fr$ has the lowest $1^{st}$ ionisation enthalpy among the given elements.
Therefore,the correct answer is $87$.

(C) The successive ionization energies are $10, 16, 25, 400, 500 \ eV \ mol^{-1}$.
There is a very large jump in ionization energy between the $3^{rd}$ $(25 \ eV \ mol^{-1})$ and $4^{th}$ $(400 \ eV \ mol^{-1})$ ionization energies.
This indicates that the $4^{th}$ electron is being removed from a stable noble gas core.
Therefore,the metal $M$ has $3$ valence electrons and exhibits a stable oxidation state of $+3$.
The formula of its halide (where $X$ is a halogen with $-1$ charge) is $MX_3$.

(D) The $I.P.$ (Ionization Potential) depends on the stability of the electronic configuration and the shielding effect.
Comparing the given configurations:
$A$: $3p^6$ (Noble gas,very stable,high $I.P.$)
$B$: $3p^5$ (High $I.P.$ due to high effective nuclear charge)
$C$: $3s^2$ (Fully filled $s$-orbital,stable,high $I.P.$)
$D$: $3s^2 3p^1$ (The electron is being removed from the $3p$ orbital,which is shielded by the $3s^2$ electrons and is further from the nucleus compared to the $3s$ orbital).
Therefore,the configuration $1s^2 2s^2 2p^6 3s^2 3p^1$ has the least $I.P.$ value.

(A) To identify the element,we look for a large jump in the successive ionization potential values,which indicates the removal of an electron from a stable noble gas core.
$1$. The given values are: $(IP)_1 = 7.1 \ eV$,$(IP)_2 = 10.3 \ eV$,$(IP)_3 = 36.2 \ eV$,and $(IP)_4 = 39.1 \ eV$.
$2$. The jump from $(IP)_2$ to $(IP)_3$ is very large $(36.2 - 10.3 = 25.9 \ eV)$,suggesting that the element has $2$ valence electrons.
$3$. An element with $2$ valence electrons belongs to Group $2$ (alkaline earth metals).
$4$. Among the given options,$Mg$ (Magnesium) is in Group $2$ with the electronic configuration $[Ne] 3s^2$. Therefore,it has $2$ valence electrons.

(D) The electronic configuration of Nitrogen $(N)$ is $1s^2 2s^2 2p^3$.
The electronic configuration of Oxygen $(O)$ is $1s^2 2s^2 2p^4$.
Nitrogen has a half-filled $p$-subshell $(2p^3)$,which provides extra stability to the atom.
Due to this extra stability,more energy is required to remove an electron from Nitrogen compared to Oxygen,which has a partially filled $p$-subshell $(2p^4)$.

(C) The ionisation energy $(IE)$ is inversely proportional to the size of the atom or ion $(IE \propto \frac{1}{\text{size}})$.
For neutral atoms,$O$ has a smaller size than $S$,so $IE(O) > IE(S)$.
For anions,$O^{-}$ has a smaller size than $S^{-}$,so $IE(O^{-}) > IE(S^{-})$.
Comparing neutral atoms and anions,neutral atoms have a much higher $IE$ than their corresponding anions because removing an electron from a negatively charged species is easier than from a neutral one.
Thus,the correct order is $O > S > O^{-} > S^{-}$.

(D) The transition $M_{(g)}^{2+} \longrightarrow M_{(g)}^{3+}$ involves the maximum amount of energy.
As the positive charge on the ion increases,the effective nuclear charge per electron increases.
This results in a stronger electrostatic attraction between the nucleus and the remaining electrons.
Therefore,the energy required to remove the $3^{rd}$ electron (third ionization energy) is significantly higher than the energy required for the removal of the first or second electron.

(A) The ionization enthalpy $(I.E.)$ is the energy required to remove an electron from an isolated gaseous atom.
In the periodic table,as we move down a group,the atomic size increases and the valence electrons are further from the nucleus,which leads to a decrease in $I.E.$
All the given elements $(Li, K, Rb, Cs)$ belong to Group $1$ (Alkali metals).
Since $Li$ is at the top of the group,it has the smallest atomic size and therefore the highest $I.E.$ among the given options.

(A) The ionization enthalpy $(I.E.)$ decreases down the group due to an increase in atomic size.
Therefore,the $I.E.$ of $Te$ is higher than that of $Po$.
Across a period,$I.E.$ increases with an increase in atomic number.
Comparing the given elements,$Po$ (Polonium) is in Group $16$ and Period $6$,$Te$ (Tellurium) is in Group $16$ and Period $5$,$Br$ (Bromine) is in Group $17$ and Period $4$,and $Kr$ (Krypton) is in Group $18$ and Period $4$.
Since $I.E.$ increases from left to right across a period and decreases down a group,$Po$ has the lowest $I.E.$ among the given elements.

(A) The ionization enthalpy $(IE)$ is highest for noble gases due to their stable electronic configuration. Therefore,the $IE$ of $Ar$ and $He$ is higher than that of the halogens $Cl$ and $I$.
Down a group,$IE$ decreases due to an increase in atomic size.
Comparing $He$ and $Ar$,$He$ has a smaller atomic size,so $IE$ of $He > Ar$.
Across a period,$IE$ increases with an increase in atomic number.
Comparing $Cl$ and $Ar$,$Ar$ has a higher $IE$ than $Cl$.
Thus,the overall order of ionization enthalpy is $I < Cl < Ar < He$.
Therefore,$He$ has the highest ionization enthalpy.

(B) The ionisation enthalpy decreases down a group due to an increase in atomic size.
Therefore,the ionisation enthalpy of $Ne$ is greater than that of $Ar$.
Across a period,ionisation enthalpy generally increases with an increase in atomic number.
Comparing the elements in the same period ($3^{rd}$ period),the order is $S < Cl < Ar$.
Combining these trends,the correct decreasing order of ionisation enthalpy is:
$Ne > Ar > Cl > S$.

(A) The first ionisation enthalpy generally increases as we move from left to right across a period due to an increase in effective nuclear charge.
However,$Be$ $(1s^2 2s^2)$ has a fully-filled $2s$ subshell,which is more stable than the $2p^1$ configuration of $B$ $(1s^2 2s^2 2p^1)$.
Therefore,$Be$ has a higher first ionisation enthalpy than $B$.
The correct order is $Li < B < Be < C$.

(B) The third ionisation enthalpy is highest in $alkaline \ earth \ metals$ because they achieve a stable noble gas configuration after the removal of two electrons.
Further removal of an electron from the completely filled orbital requires a very high ionisation enthalpy.

(A) The first ionization enthalpy $(\Delta H_{IE})$ generally increases across a period from left to right and decreases down a group.
Comparing the elements $C$ (Group $14$,Period $2$),$N$ (Group $15$,Period $2$),$Si$ (Group $14$,Period $3$),and $P$ (Group $15$,Period $3$):
$1$. Elements in Period $2$ $(C, N)$ have higher ionization enthalpy than those in Period $3$ $(Si, P)$ due to smaller atomic size.
$2$. Within the same period,Group $15$ elements have higher ionization enthalpy than Group $14$ elements due to the stable half-filled $p$-orbital configuration $(ns^2 np^3)$.
$3$. Combining these trends,the order is $Si < P < C < N$.

(C) The jump between $IE_3$ and $IE_4$ is very large ($2750$ to $11580 \ kJ \cdot mol^{-1}$),indicating that the fourth electron is removed from a stable inner shell (noble gas configuration).
This means the element has $3$ valence electrons.
Among the given options,Aluminum $(Al)$ has $3$ valence electrons $([Ne] 3s^2 3p^1)$.

(B) On moving from left to right in a period,ionisation energy generally increases due to an increase in effective nuclear charge.
Thus,the expected order of ionisation energy for $C$,$N$,$O$,and $F$ is $C < N < O < F$.
However,the ionisation energy of $N$ $(1s^2 2s^2 2p^3)$ is greater than that of $O$ $(1s^2 2s^2 2p^4)$.
This is because $N$ has a stable half-filled $p$-orbital configuration,which requires more energy to remove an electron.
Therefore,the correct order is $C < O < N < F$.

(A) Elements $X, Y$ and $Z$ with atomic numbers $19, 37$ and $55$ are $K, Rb$ and $Cs$ respectively.
These elements belong to group-$I$ of the periodic table.
On moving down the group,the atomic size increases,which leads to a decrease in the ionisation potential.
Therefore,the order of ionisation potential is $X > Y > Z$.
Thus,$Y$ has an ionisation potential between those of $X$ and $Z$.

(B) The first ionisation energy generally increases across a period due to an increase in effective nuclear charge.
Exceptions occur when a more stable electronic configuration (like fully filled or half-filled orbitals) is present in the preceding element.
$Be$ $(2s^2)$ has a higher ionisation energy than $B$ $(2s^2 2p^1)$.
$Mg$ $(3s^2)$ has a higher ionisation energy than $Al$ $(3s^2 3p^1)$.
$N$ $(2s^2 2p^3)$ has a higher ionisation energy than $O$ $(2s^2 2p^4)$.
$Na$ $(3s^1)$ has a lower ionisation energy than $Mg$ $(3s^2)$,which follows the general trend.
Therefore,$Na$ and $Mg$ is not an exception.

(B) The first ionization energy generally increases across a period from left to right due to an increase in effective nuclear charge.
Exceptions occur when a more stable electronic configuration (like fully filled or half-filled orbitals) is present in an element to the left of another.
For $N$ $(2s^2 2p^3)$ and $O$ $(2s^2 2p^4)$,$N$ has a higher $IE$ than $O$ due to its stable half-filled $p$-orbital.
For $Be$ $(2s^2)$ and $B$ $(2s^2 2p^1)$,$Be$ has a higher $IE$ than $B$ due to its stable fully filled $s$-orbital.
For $Mg$ $(3s^2)$ and $Al$ $(3s^2 3p^1)$,$Mg$ has a higher $IE$ than $Al$ due to its stable fully filled $s$-orbital.
For $Na$ $(3s^1)$ and $Mg$ $(3s^2)$,the trend follows the general rule where $IE$ of $Mg > IE$ of $Na$,so this is not an exception.

(B) The first ionisation potential generally increases in a period from left to right and decreases in a group from top to bottom.
Comparing the elements: $K$ (Group $1$,Period $4$),$Na$ (Group $1$,Period $3$),and $Be$ (Group $2$,Period $2$).
Since $K$ and $Na$ are in the same group,$IE_1$ of $K < Na$.
$Be$ is in Period $2$ and Group $2$,which has a higher $IE_1$ than elements in Group $1$ of the same or lower periods.
Thus,the correct order of first ionisation potential is $K < Na < Be$.

(D) The electronic configurations are:
$A$: $1s^2 2s^2 2p^6 3s^1$ (Sodium,$Na$)
$B$: $1s^2 2s^2 2p^6 3s^2 3p^1$ (Aluminium,$Al$)
$C$: $1s^2 2s^2 2p^6 3s^2$ (Magnesium,$Mg$)
$D$: $1s^2 2s^2 2p^6 3s^2 3p^2$ (Silicon,$Si$)
Ionization enthalpy generally increases across a period from left to right.
The elements belong to the $3^{rd}$ period in the order $Na < Mg < Al < Si$.
However,$Mg$ $(3s^2)$ has a fully filled orbital,making it more stable than $Al$ $(3s^2 3p^1)$. Thus,$IE_1$ of $Mg > Al$.
The order is $Na < Al < Mg < Si$,which corresponds to $A < B < C < D$.
Therefore,the correct order of first ionization enthalpy is $D > C > B > A$.

(C) The first ionisation enthalpy values for $Li$,$Be$,and $C$ are $520$,$899$,and $1086 \ kJ \ mol^{-1}$ respectively.
Ionisation enthalpy generally increases from left to right across a period due to an increase in effective nuclear charge and a decrease in atomic size.
The electronic configurations are:
$Li: [He] \ 2s^1$
$Be: [He] \ 2s^2$
$B: [He] \ 2s^2 \ 2p^1$
$C: [He] \ 2s^2 \ 2p^2$
Due to the stable fully-filled $2s$ orbital in $Be$,its ionisation enthalpy is higher than that of $B$,which has a single electron in the $2p$ orbital.
Therefore,the trend in ionisation enthalpy for these elements is $C > Be > B > Li$.
Substituting the given values: $1086 > 899 > B > 520$.
The value for $B$ must be between $520$ and $899 \ kJ \ mol^{-1}$.
Among the given options,$801 \ kJ \ mol^{-1}$ is the only value that fits this range.
Thus,the correct option is $(C)$.