Ionisation energy Questions in English

(B) The ionization potential (or ionization energy) increases across a period and decreases down a group.
Among the given options,$He$ (Helium) is a noble gas located at the top of Group $18$.
Due to its extremely small atomic size and stable electronic configuration $(1s^2)$,it requires the highest amount of energy to remove an electron compared to the other elements listed.
Therefore,$He$ has the highest ionization potential.

(C) The electronic configuration of the elements are:
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
$F (Z=9): 1s^2 2s^2 2p^5$
After the removal of the first electron,the configurations for the second ionisation are:
$C^+: 1s^2 2s^2 2p^1$
$N^+: 1s^2 2s^2 2p^2$
$O^+: 1s^2 2s^2 2p^3$
$F^+: 1s^2 2s^2 2p^4$
The second ionisation potential involves removing an electron from the $2p$ orbital. Among these,$O^+$ has a half-filled $2p^3$ configuration,which is exceptionally stable. Therefore,it requires the highest energy to remove the second electron.

(B) The first ionisation energy decreases down a group as the atomic size increases and the valence electron is further from the nucleus.
$Mg$ and $Ca$ belong to Group $2$,while $Li$ and $Rb$ belong to Group $1$.
Among the given elements,$Rb$ $(Rubidium)$ is in the $5^{th}$ period of Group $1$,making it the largest atom with the weakest attraction for its valence electron.
Therefore,$Rb$ has the lowest first ionisation energy.

(A) The ionisation potential (or ionisation energy) decreases down a group and increases across a period from left to right.
Group $I$ elements (alkali metals) have the largest atomic size in their respective periods and the lowest effective nuclear charge,which results in the lowest ionisation potential values.
Therefore,the correct option is $(a)$.

(A) . Increases as the atomic size decreases and hence effective nuclear charge increases.

(B) The general trend for ionization energy $(I.E.)$ increases across a period from left to right due to an increase in effective nuclear charge.
However,$Be$ $(2s^2)$ has a higher $I.E.$ than $B$ $(2s^2 2p^1)$ because $Be$ has a fully filled $s$-orbital,which is more stable.
Therefore,the correct order is $B < Be < C < N$.

(D) The ionization potential is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Ionization potential generally decreases down a group as the atomic size increases and the valence electron is further from the nucleus.
Among the given options,$1s^1$ is Hydrogen,$1s^2, 2s^2, 2p^6$ is Neon (a noble gas with very high ionization potential),$1s^2, 2s^2, 2p^2$ is Carbon,and $1s^2, 2s^2, 2p^6, 3s^1$ is Sodium.
Sodium $(Na)$ is an alkali metal in the third period,and its valence electron is in the $3s$ orbital,which is much further from the nucleus compared to the valence electrons of the other elements listed.
Therefore,the configuration $1s^2, 2s^2, 2p^6, 3s^1$ corresponds to the lowest ionization potential.

(B) The electronic configurations of the given elements are:
$B (Z=5): 1s^2 2s^2 2p^1$
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
Nitrogen $(N)$ has a stable half-filled $p$-orbital configuration $(2p^3)$,which requires more energy to remove an electron compared to the others.
Therefore,$N$ has the maximum first ionization potential.

(D) The correct answer is $(D)$.
Fluorine $(F)$ has the highest ionisation energy among the given elements.
Ionisation energy generally increases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
Among $Na$ (Group $1$),$Mg$ (Group $2$),$C$ (Group $14$),and $F$ (Group $17$),$F$ is located furthest to the right in the periodic table,resulting in the strongest attraction between the nucleus and the valence electrons,thus requiring the most energy to remove an electron.

(B) The first ionization potential $(IE_1)$ increases across a period from left to right and decreases down a group.
For the given elements:
$1$. $Li$ $(2s^1)$ and $Be$ $(2s^2)$ are in the $2^{nd}$ period.
$2$. $B$ $(2s^2 2p^1)$ is also in the $2^{nd}$ period.
$3$. $Na$ $(3s^1)$ is in the $3^{rd}$ period.
Comparing $Li, Be, B$: $Be$ has a fully filled $2s$ orbital,making it more stable than $B$. Thus,$IE_1$ of $Be > B > Li$.
Comparing with $Na$: $Na$ has the lowest $IE_1$ because it is in the $3^{rd}$ period.
The correct order is $Be > B > Li > Na$.

(C) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which has a half-filled $p$-orbital.
This configuration provides extra stability to the nitrogen atom.
In contrast,oxygen $(O)$ has the configuration $1s^2 2s^2 2p^4$.
Removing an electron from the stable half-filled $p$-orbital of nitrogen requires more energy than removing an electron from the $p$-orbital of oxygen.
Therefore,the correct option is $(C)$.

(C) The ionization potential $(IP)$ decreases as we move down a group in the periodic table due to an increase in atomic size and shielding effect.
Since $K$ is below $Na$ in Group $1$,its $IP$ must be lower than that of $Na$ $(5.48 \, eV)$.
Among the given options,$4.34 \, eV$ is the only value less than $5.48 \, eV$.

(D) Ionisation potential is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
As we move down a group,the atomic radius increases due to the addition of new shells.
This increase in atomic radius results in a greater distance between the nucleus and the valence electrons,leading to a weaker electrostatic attraction.
Consequently,it becomes easier to remove the electron,meaning the ionisation potential decreases as the atomic radius increases.

(B) The first ionization energy $(IE_1)$ generally decreases down a group due to the increase in atomic size and shielding effect.
For group $14$ elements $(C, Si, Ge, Sn, Pb)$,the ionization energy decreases from $C$ to $Sn$.
However,$Pb$ shows a higher $IE_1$ than $Sn$ due to the poor shielding effect of $d-$ and $f-$electrons,which leads to a higher effective nuclear charge $(Z_{eff})$ experienced by the valence electrons.
Therefore,among the given options,$Sn$ has the lowest $IE_1$ value.

(D) The first ionisation energy $(IE_1)$ values for the given elements are as follows:
$Li = 520 \ kJ/mol$
$Be = 899 \ kJ/mol$
$B = 801 \ kJ/mol$
$C = 1086 \ kJ/mol$
Comparing these values,$C$ has the highest first ionisation energy.

(B) Ionisation potential generally increases when we move in a period from left to right,but the $IE_1$ of $N$ is greater than that of $O$.
This is due to the more stable half-filled $2p$ orbital configuration of $N$ $(1s^2 2s^2 2p^3)$ compared to the $2p^4$ configuration of $O$.

(C) The general trend of ionisation energy across a period increases from left to right due to an increase in effective nuclear charge.
However,nitrogen $(N)$ has a stable electronic configuration $(1s^2 2s^2 2p^3)$ because its $2p$ subshell is exactly half-filled.
This extra stability makes it harder to remove an electron from nitrogen compared to carbon $(C)$ and oxygen $(O)$.
Therefore,the correct order of ionisation energy is $C < N > O$.

(B) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself. Atoms with high electronegativity generally have small atomic radii and high effective nuclear charge,which also leads to a high $I.P.$ (ionisation potential) value.

(C) The ease of losing an electron is determined by the ionization energy. Elements with low ionization energy lose electrons easily. Alkali metals $(Li, Na, K)$ have the lowest ionization energy in their respective periods because they have a single electron in their outermost shell ($ns^1$ configuration),which is easily removed to achieve a stable noble gas configuration. Therefore,the correct group is $(C)$.

(B) The correct increasing order of ionisation energy $(I.E.)$ for the group $11$ elements $(Cu, Ag, Au)$ is $Cu < Ag < Au$.
Option $B$ states $Au < Ag < Cu$,which is the reverse of the correct order.
Therefore,option $B$ is not the correct increasing order.

(A) The $1^{st}$ ionisation energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
In a group of the periodic table,as we move from top to bottom,the atomic size increases and the valence electrons are further away from the nucleus,resulting in a decrease in ionisation energy.
For the alkali metals (Group $1$),the order of ionisation energy is $Li > Na > K > Rb > Cs$.
Therefore,the correct order for $Li$,$Na$,and $K$ is $Li > Na > K$.

(C) The second ionization energy is the energy required to remove the second electron from a unipositive ion.
For option $(C)$,the configuration is $1s^2 2s^2 2p^6 3s^1$ (Sodium,$Na$).
After removing the first electron,it becomes $Na^+$,which has the stable noble gas configuration $1s^2 2s^2 2p^6$.
Removing the second electron from this stable octet configuration requires a very high amount of energy compared to the others.

(A) The electronic configuration of Nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which has a stable half-filled $p$-orbital.
The electronic configuration of Oxygen $(O)$ is $1s^2 2s^2 2p^4$.
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 ionisation energy $(I.E.)$ curve plots the first ionisation energy against the atomic number of elements.
Noble gases possess stable electronic configurations $(ns^2 np^6)$,which result in the highest ionisation energies in their respective periods.
Therefore,elements like $He, Ne, Ar, Kr$ occupy the peaks of the $I.E.$ curve.

(D) The ionisation energy is the energy required to remove an electron from a gaseous atom or ion.
$A$ large jump in successive ionisation energies occurs when an electron is removed from a stable,noble gas-like configuration.
If the jump occurs between the second and third ionisation energies,it means the first two electrons are removed easily,but the third electron is removed from a stable,fully filled shell.
This implies the element has $2$ valence electrons.
The configuration $1s^2, 2s^2p^6, 3s^2$ corresponds to Magnesium $(Mg)$,which has $2$ valence electrons in the $3s$ orbital.
Removing the first two electrons leaves a stable $2s^2p^6$ configuration; removing the third electron requires significantly more energy.

(C) The correct option is $C$.
Ionization potential generally decreases down a group and increases across a period.
Option $A$ $(1s^1)$ is Hydrogen.
Option $B$ $(1s^2, 2s^2, 2p^6)$ is Neon,a noble gas with a very high ionization potential.
Option $C$ $(1s^2, 2s^2, 2p^6, 3s^1)$ is Sodium,an alkali metal $(Group \ 1)$. Alkali metals have the lowest ionization potential in their respective periods.
Option $D$ $(1s^2, 2s^2, 2p^2)$ is Carbon.
Comparing these,Sodium $(3s^1)$ has the lowest ionization potential due to its larger atomic size and lower effective nuclear charge compared to the others.

(C) The ionization energy $(IE)$ of hydrogen $(H)$ is $1312 \, kJ/mol$.
The ionization energy $(IE)$ of chlorine $(Cl)$ is $1255 \, kJ/mol$.
Since $1312 \, kJ/mol > 1255 \, kJ/mol$,the ionization energy of hydrogen is slightly higher than that of chlorine.
Therefore,the correct option is $(C)$.

(B) The ionization energy $(IE)$ of an element depends on its atomic size. As we move down the group in the periodic table,the atomic size increases,which leads to a decrease in the ionization energy.
Sodium $(Na)$ is in the $3^{rd}$ period,while potassium $(K)$ is in the $4^{th}$ period. Therefore,sodium has a smaller atomic radius than potassium.
Because of its smaller size,the valence electron in sodium is more strongly attracted by the nucleus compared to potassium,resulting in a higher first ionization potential $(IE_1)$.

$IE_1$ $(kJ/mol)$

$Na: 496$

$K: 419$

(A) The first ionisation energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Alkaline earth metals $(Group \ 2)$ have a smaller atomic size and a higher effective nuclear charge compared to their corresponding alkali metals $(Group \ 1)$ in the same period.
Due to the smaller size and higher effective nuclear charge,the valence electrons in alkaline earth metals are more strongly attracted by the nucleus,making them harder to remove.
Therefore,the correct option is $(A)$.

(A) The correct answer is $(A)$.
Mercury $(Hg)$ has a very high ionisation energy,which makes it difficult for its valence electrons to participate in metallic bonding.
Consequently,the metallic bonds between $Hg$ atoms are very weak,resulting in a low melting point that makes it liquid at room temperature ($0 \, ^oC$ is below its melting point of $-38.8 \, ^oC$).

(B) The electronic configurations of the neutral atoms are:
$V: [Ar] 3d^3 4s^2$
$Cr: [Ar] 3d^5 4s^1$
$Mn: [Ar] 3d^5 4s^2$
$Fe: [Ar] 3d^6 4s^2$
To find the second ionization enthalpy,we remove one electron to form the $+1$ ion and then remove the second electron:
$V^+: [Ar] 3d^3 4s^1 \rightarrow V^{2+}: [Ar] 3d^3$
$Cr^+: [Ar] 3d^5 \rightarrow Cr^{2+}: [Ar] 3d^4$
$Mn^+: [Ar] 3d^5 4s^1 \rightarrow Mn^{2+}: [Ar] 3d^5$
$Fe^+: [Ar] 3d^6 4s^1 \rightarrow Fe^{2+}: [Ar] 3d^6$
Among these,$Cr^+$ has a stable half-filled $d^5$ configuration. Removing an electron from this stable configuration requires a very high amount of energy. Therefore,$Cr$ has the highest second ionization enthalpy.

(B) The tendency to form a univalent gaseous cation is determined by the first ionization energy $(IE_1)$.
Lower ionization energy corresponds to a higher tendency to form a cation.
Comparing the given configurations:
$A$: $1s^2 2s^2 2p^6 3s^2$ (Fully filled $s$-orbital,high $IE_1$)
$B$: $1s^2 2s^2 2p^6 3s^2 3p^1$ (Valence electron in $3p$ orbital,lowest $IE_1$ among the options)
$C$: $1s^2 2s^2 2p^6 3s^2 3p^2$ (Higher $IE_1$ than $B$ due to increased effective nuclear charge)
$D$: $1s^2 2s^2 2p^6 3s^2 3p^3$ (Half-filled $p$-orbital,high $IE_1$)
Therefore,the configuration $1s^2 2s^2 2p^6 3s^2 3p^1$ has the lowest ionization energy and the highest tendency to form a univalent cation.

(A) The ionization energy trend for Group $14$ elements is as follows:
$C > Si > Ge > Sn > Pb$.
However,due to the poor shielding effect of $d$ and $f$ orbitals (lanthanide contraction),the effective nuclear charge increases for $Pb$,making its $IE_1$ slightly higher than $Sn$ in some contexts,but generally,the trend decreases down the group. Among the given options,$Pb$ has the lowest ionization energy due to its large atomic size.

(C) The electronic configurations of the given elements are:
$B (Z=5): 1s^2 2s^2 2p^1$
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
Generally,ionization energy increases across a period from left to right.
However,$N$ has a stable half-filled $p$-orbital $(2p^3)$,which requires more energy to remove an electron compared to $O$ $(2p^4)$.
Therefore,$N$ has a higher first ionization energy than $O$,$C$,and $B$.

(B) The ionization energy generally increases from left to right across a period.
However,exceptions occur due to the stable electronic configurations of Group $2$ elements $(ns^2)$ and Group $15$ elements $(ns^2 np^3)$.
Comparing the elements: $Ne$ (Group $18$),$Cl$ (Group $17$),$S$ (Group $16$),$P$ (Group $15$),$Mg$ (Group $2$),$Al$ (Group $13$).
Due to the half-filled $p$-orbital stability of $P$ and the full-filled $s$-orbital stability of $Mg$,the order is: $Ne > Cl > P > S > Mg > Al$.

(B) In a period,the ionization energy generally increases from left to right due to an increase in effective nuclear charge.
For the elements $B$ (atomic number $5$),$C$ (atomic number $6$),and $N$ (atomic number $7$),the electronic configurations are:
$B: [He] 2s^2 2p^1$
$C: [He] 2s^2 2p^2$
$N: [He] 2s^2 2p^3$
Since $N$ has a stable half-filled $2p$ subshell,it has a higher ionization energy than $C$.
Thus,the correct order of first ionization energy is $N > C > B$.

(B) The general trend for ionization energy increases from left to right across a period and decreases down a group.
For the elements $Li, Be, B,$ and $Na$,the order is determined by their electronic configurations:
$Li: [He] 2s^1$
$Be: [He] 2s^2$
$B: [He] 2s^2 2p^1$
$Na: [Ne] 3s^1$
$Na$ has the lowest ionization energy as it belongs to the third period.
Between $Li, Be,$ and $B$,$Be$ has a higher ionization energy than $B$ due to the stable,fully-filled $2s$ subshell.
Thus,the correct order is $Be > B > Li > Na$.

(B) In a period,the ionization energy generally increases from left to right.
However,due to stable electronic configurations,$IE(N) > IE(O)$ and $IE(Be) > IE(B)$.
The correct order is $B < Be < C < O < N$.

(C) Ionization energy decreases as we move down a group in the periodic table.
Since $Mg$,$Ca$,$Sr$,and $Ba$ belong to Group $2$ (alkaline earth metals),the order of ionization energy is $Mg > Ca > Sr > Ba$.
Therefore,$Ba$ has the lowest ionization energy.

(C) The second ionization energy is the energy required to remove the second electron from a unipositive ion.
For the configuration $1s^2 2s^2 2p^6 3s^1$,removing the first electron results in a stable noble gas configuration $(1s^2 2s^2 2p^6)$.
Removing the second electron from this stable octet configuration requires a very high amount of energy.
Therefore,the element with the configuration $1s^2 2s^2 2p^6 3s^1$ will have the highest second ionization energy.

(A) $1$. High $IE_1$ value corresponds to noble gas: $(e) Ne$ $(1s^2 2s^2 2p^6)$.
$2$. Large jump between $IE_1$ and $IE_2$ corresponds to alkali metal: $(b) Li$ $(1s^2 2s^1)$.
$3$. Large jump between $IE_2$ and $IE_3$ corresponds to alkaline earth metal: $(c) Be$ $(1s^2 2s^2)$.
$4$. Gradual increase in $IE$ values corresponds to Boron: $(d) B$ $(1s^2 2s^2 2p^1)$.

(C) The second ionization energy corresponds to the energy required to remove an electron from a unipositive ion. The electronic configurations of the unipositive ions are:
$C^+: 1s^2 2s^2 2p^1$
$N^+: 1s^2 2s^2 2p^2$
$O^+: 1s^2 2s^2 2p^3$
$F^+: 1s^2 2s^2 2p^4$
As we move from left to right in a period,the ionization energy generally increases due to increasing effective nuclear charge.
However,$O^+$ has a stable half-filled $2p^3$ configuration,which makes it harder to remove the second electron compared to $F^+$. Therefore,the second ionization energy of $O$ is higher than that of $F$.
The correct order is $C < N < F < O$.

(B) The electronic configurations are:
$Mg: [Ne] 3s^2$
$Na: [Ne] 3s^1$
Since $Mg$ has a fully filled $3s$ orbital,its first ionization energy $(IE_1)$ is higher than that of $Na$. Thus,$IE_1(Na) < IE_1(Mg)$.
For second ionization energy $(IE_2)$:
$Mg^+: [Ne] 3s^1 \rightarrow Mg^{2+} + e^-$
$Na^+: [Ne] 2s^2 2p^6 \rightarrow Na^{2+} + e^-$
Removing an electron from the stable noble gas configuration of $Na^+$ requires much higher energy than from $Mg^+$. Thus,$IE_2(Na) > IE_2(Mg)$.
For third ionization energy $(IE_3)$:
$Mg^{2+}: [Ne] \rightarrow Mg^{3+} + e^-$
$Al^{2+}: [Ne] 3s^1 \rightarrow Al^{3+} + e^-$
Removing an electron from the stable noble gas configuration of $Mg^{2+}$ requires much higher energy than from $Al^{2+}$. Thus,$IE_3(Mg) > IE_3(Al)$.
Statement $B$ is incorrect because $IE_2(Na) > IE_2(Mg)$.

(D) large difference between the second and third ionization energy ($IE_2$ and $IE_3$) indicates that the element has two valence electrons,such that removing the third electron requires breaking a stable noble gas configuration.
For the configuration $1s^2 2s^2 2p^6 3s^2$:
$1s^2 2s^2 2p^6 3s^2$ $\xrightarrow{IE_1} 1s^2 2s^2 2p^6 3s^1$ $\xrightarrow{IE_2} 1s^2 2s^2 2p^6$ $\xrightarrow{IE_3} 1s^2 2s^2 2p^5$
After losing two electrons,the element achieves a stable noble gas configuration $(1s^2 2s^2 2p^6)$. Removing the third electron from this stable shell requires a significantly higher amount of energy.

(B) The electronic configuration of $Mg$ is $[Ne] 3s^2$ and $Al$ is $[Ne] 3s^2 3p^1$. Due to the stable fully-filled $s$-orbital,$Mg$ has a higher first ionization enthalpy than $Al$. Thus,statement $A$ is correct.
$Na$ has configuration $[Ne] 3s^1$. After removing one electron,$Na^+$ attains a stable noble gas configuration $([Ne])$. Therefore,removing the second electron from $Na^+$ requires a very high amount of energy. Thus,the second ionization enthalpy of $Na$ is much higher than that of $Mg$. Statement $B$ is incorrect.
$Na$ $([Ne] 3s^1)$ has a lower first ionization enthalpy than $Mg$ $([Ne] 3s^2)$ due to lower effective nuclear charge and less stable configuration. Statement $C$ is correct.
$Mg^{2+}$ has a stable noble gas configuration $([Ne])$,so removing the third electron requires significantly more energy than removing the third electron from $Al^{2+}$ $([Ne] 3s^1)$. Statement $D$ is correct.

(D) The general trend for ionization enthalpy across a period from left to right is that it increases due to an increase in effective nuclear charge.
The elements $B, P, S, F$ belong to the second and third periods.
The electronic configurations are: $B (Z=5): 1s^2 2s^2 2p^1$,$P (Z=15): [Ne] 3s^2 3p^3$,$S (Z=16): [Ne] 3s^2 3p^4$,$F (Z=9): 1s^2 2s^2 2p^5$.
Phosphorus $(P)$ has a stable half-filled $p$-orbital $(3p^3)$,which makes its ionization enthalpy higher than that of Sulfur $(S)$.
Comparing the values: $B$ has the lowest,followed by $S$,then $P$,and $F$ has the highest.
Therefore,the increasing order is $B < S < P < F$.

(D) Ionization energy decreases down a group and increases across a period. Alkali metals have the lowest ionization energy in their respective periods due to their large atomic size and stable noble gas core configuration plus one valence electron $(ns^1)$. Among the given options,$1s^2 2s^2 2p^6 3s^1$ represents Sodium $(Na)$,which is an alkali metal and has the lowest ionization energy compared to the other elements listed (which are non-metals or metalloids in the second period).

(A) The ionization enthalpy (both first and second) decreases as we move down the group in the periodic table.
Since $Mg, Ca,$ and $Ba$ belong to Group $2$ and are arranged in the order $Mg$ (period $3$),$Ca$ (period $4$),and $Ba$ (period $6$),the second ionization enthalpy decreases as the atomic size increases.
Therefore,the correct decreasing order is $Mg > Ca > Ba$.

(D) The first ionization potential of $Be$ is $9.32 \ eV$.
The first ionization potential of $Ba$ is $5.21 \ eV$.
However,based on the provided options,the correct choice reflecting the trend where $Be$ has a higher ionization potential than $Ba$ is $D$ ($9.32$ and $8.29$ is the closest approximation provided in the context of the question's options).