Ionisation energy Questions in English

(B) The correct answer is $(B)$.
$1$. The first ionisation potential $(IE_1)$ of $Al$ $([Ne] 3s^2 3p^1)$ is less than $Mg$ $([Ne] 3s^2)$ because $Mg$ has a stable fully-filled $s$-orbital.
$2$. The second ionisation potential $(IE_2)$ of $Na$ $([Ne] 3s^1 \rightarrow [Ne])$ is much higher than that of $Mg$ $([Ne] 3s^2 \rightarrow [Ne] 3s^1)$ because the second electron in $Na$ is removed from a stable noble gas core $([Ne])$.
$3$. $IE_1$ of $Na$ is less than $Mg$ due to higher effective nuclear charge and stable configuration of $Mg$.
$4$. The third ionisation potential $(IE_3)$ of $Mg$ is higher than $Al$ because the third electron in $Mg$ is removed from a stable noble gas core $([Ne])$,whereas in $Al$ it is removed from the $3s^2$ subshell.

(C) The second ionisation energy is defined as the energy required to remove one mole of electrons from one mole of monovalent gaseous cations of an element to form divalent gaseous cations.
The process is represented as: $M^+(g) \rightarrow M^{2+}(g) + e^-$.

(D) Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.
The process can be represented as: $M(g) + \text{Energy} \rightarrow M^+(g) + e^-$.
Therefore,option $D$ is the correct definition.

(A) The first ionisation energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
As we move from alkali metals (Group $1$) to alkaline earth metals (Group $2$) in the same period,the nuclear charge increases.
An increase in nuclear charge leads to a stronger electrostatic attraction between the nucleus and the valence electrons.
Consequently,more energy is required to remove an electron from alkaline earth metals compared to alkali metals.
Therefore,the first ionisation energies of alkaline earth metals are higher due to the increase in nuclear charge.

(D) . Alkali metals have the lowest ionization potential in their respective periods because they have the largest atomic size and a single valence electron,making it easiest to remove the electron.

(A) The electronic configuration of Nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which contains a half-filled $p$-orbital.
This half-filled configuration provides extra stability to the atom.
Due to this increased stability,more energy is required to remove an electron from the $2p$-orbital of Nitrogen compared to Oxygen $(1s^2 2s^2 2p^4)$,which has a less stable configuration.

(B) The ionization energy $(IE)$ of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state.

(C) The electronic configuration of $Be$ $(Z=4)$ is $1s^2 2s^2$,which is a fully filled $s$-orbital.
The electronic configuration of $B$ $(Z=5)$ is $1s^2 2s^2 2p^1$.
Removing an electron from $Be$ requires breaking a stable,fully filled $2s$ subshell,which requires more energy.
In contrast,the electron in $B$ is removed from the $2p$ subshell,which is easier to remove as it is shielded by the $2s$ electrons and is at a slightly higher energy level.

(A) $E_{1}$ represents the first ionization energy $(I.E._{1})$,which is the energy required to remove the first electron from a neutral atom.
$E_{2}$ represents the second ionization energy $(I.E._{2})$,which is the energy required to remove the second electron from a unipositive ion $(A^{+})$.
Since the $A^{+}$ ion has a higher effective nuclear charge and a smaller size compared to the neutral atom $A$,it holds the remaining electrons more tightly.
Therefore,more energy is required to remove the second electron than the first,leading to the relationship $E_{1} < E_{2}$.

(B) The energy required to remove an electron is known as the ionization energy.
Among the given configurations,$1s^2 2s^2 2p^3$ represents a nitrogen atom $(N)$.
This configuration has a half-filled $p$-orbital $(p^3)$,which provides extra stability to the atom.
Due to this high stability,more energy is required to remove an electron from this configuration compared to the others.
Therefore,the correct option is $B$.

(A) The first ionization potential $(IE_1)$ 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 $Na$ $([Ne] 3s^1)$,$Mg$ $([Ne] 3s^2)$,$Al$ $([Ne] 3s^2 3p^1)$,and $Si$ $([Ne] 3s^2 3p^2)$:
$1$. $Mg$ has a fully filled $3s$ orbital,which is more stable than the $3p^1$ configuration of $Al$,making $IE_1$ of $Mg > Al$.
$2$. The general trend is $Na < Mg > Al < Si$.
Thus,the correct order is $Na < Mg > Al < Si$.

(C) The energy required to remove an electron from the outermost orbit of an isolated gaseous atom is called Ionisation Energy $(I.E.)$.
Carbon has an atomic number of $6$,with an electronic configuration of $1s^2 2s^2 2p^2$.
It has $4$ valence electrons in its outermost shell $(n=2)$.
Therefore,carbon can lose these $4$ electrons one by one,resulting in $4$ successive ionisation energies.

(A) The electronic configurations of the given elements are:
$Mg (Z=12): [Ne] 3s^2$
$Al (Z=13): [Ne] 3s^2 3p^1$
$Si (Z=14): [Ne] 3s^2 3p^2$
$P (Z=15): [Ne] 3s^2 3p^3$
Among these,$P$ has a stable half-filled $p$-orbital configuration $(3p^3)$,which requires more energy to remove an electron compared to the others. Therefore,$P$ has the highest $IE$.

(D) The electronic configuration of $Be$ $(Z=4)$ is $1s^2 2s^2$.
The electronic configuration of $B$ $(Z=5)$ is $1s^2 2s^2 2p^1$.
$Be$ has a fully filled $2s$ orbital,which is more stable than the $2p$ orbital of $B$.
Therefore,more energy is required to remove an electron from $Be$ compared to $B$.
The first ionization potential of $Be$ is approximately $9.32 \ eV$ and that of $B$ is $8.29 \ eV$.
Thus,the correct option is $D$.

(B) The correct answer is $(B)$.
Ionisation energy increases as we remove electrons from more stable configurations or from ions with higher positive charge.
In option $(B)$,$K^{+}$ has a stable noble gas configuration $([Ar])$. Removing an electron from a stable noble gas configuration requires a very high amount of energy compared to the other options,where electrons are being removed from less stable or valence shells.

(D) The first ionization potential increases across a period from left to right and decreases down a group.
The elements $Na$,$Mg$,and $Al$ belong to the $3^{rd}$ period,while $K$ belongs to the $4^{th}$ period.
Among $Na$,$Mg$,and $Al$,the order of ionization potential is $Na < Al < Mg$.
$Mg$ has a higher ionization potential than $Al$ due to its stable fully-filled $3s^2$ electronic configuration.
Therefore,$Mg$ has the maximum ionization potential among the given options.

(C) The successive ionization energies are $IE_1 = 191 \ kcal$,$IE_2 = 578 \ kcal$,$IE_3 = 872 \ kcal$,and $IE_4 = 5962 \ kcal$.
There is a very large jump in energy between the third and fourth ionization energy $(IE_4 - IE_3 = 5962 - 872 = 5090 \ kcal)$.
This indicates that after removing $3$ electrons,the fourth electron is being removed from a stable,noble gas-like inner shell configuration.
Therefore,the element has $3$ valence electrons.

(B) Ionization potential decreases down a group due to an increase in atomic size and shielding effect.
$Li$ and $Cs$ both belong to Group $1$ (Alkali metals).
Since $Cs$ has a much larger atomic size compared to $Li$,the valence electron in $Cs$ is further from the nucleus and experiences a weaker electrostatic force of attraction.
Therefore,$Cs$ has the least ionization potential among the given options.

(C) The ionization potential $(I.P.)$ generally decreases down a group and increases across a period.
$Li$ (Lithium) belongs to Group $1$ (Alkali metals) and has a valence electronic configuration of $2s^1$.
Due to its large atomic size and the presence of a single electron in the outermost shell,it is very easy to remove the valence electron.
Comparing the given elements: $He$ (Noble gas,very high $I.P.$),$H$ (Non-metal,high $I.P.$),$Fe$ (Transition metal),and $Li$ (Alkali metal).
$Li$ has the lowest ionization potential among the given options.

(B) . Increases from left to right. As we move across a period,the atomic radius decreases due to an increase in effective nuclear charge,which makes it harder to remove an electron,thus increasing the ionization energy.

(A) Ionization energy is highest for noble gases because they possess a stable electronic configuration $(ns^2 np^6)$,which makes it extremely difficult to remove an electron.

(C) The electronic configuration of the given elements corresponds to $Al$ $(3p^1)$,$Si$ $(3p^2)$,$P$ $(3p^3)$,and $Ge$ $(4p^2)$.
Ionization energy generally increases across a period.
However,the $p^3$ configuration represents a half-filled stable subshell.
Due to the extra stability of the half-filled $p$-orbital,the energy required to remove an electron from $[Ne] \, 3s^2 \, 3p^3$ is the highest among the given options.

(B) The ionisation potential generally increases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
All the given elements ($N$,$O$,$F$,$Ne$) belong to the $2^{nd}$ period.
However,we must consider the electronic configurations:
$N (1s^2 2s^2 2p^3)$ has a stable half-filled $p$-orbital.
$O (1s^2 2s^2 2p^4)$ has a lower ionisation potential than $N$ due to inter-electronic repulsion.
$F (1s^2 2s^2 2p^5)$ has a higher ionisation potential than $O$.
$Ne (1s^2 2s^2 2p^6)$ has the highest ionisation potential due to a stable noble gas configuration.
Comparing these,$O$ has the lowest ionisation potential among the given options.

(A) The first ionisation potential generally increases across a period from left to right due to an increase in effective nuclear charge.
All these elements belong to the $2^{nd}$ period.
The order of first ionisation potential is $B < C < O < N$.
$N$ has a higher value than $O$ due to its stable half-filled $2p^3$ electronic configuration.
Therefore,$B$ (Boron) has the lowest first ionisation potential among the given elements.

(C) The first ionization energy increases across a period and decreases down a group.
Comparing the elements $N$,$O$,$P$,and $S$:
$N$ $(2s^2 2p^3)$ has a stable half-filled $p$-orbital configuration,which makes it harder to remove an electron compared to $O$ $(2s^2 2p^4)$.
$P$ $(3s^2 3p^3)$ also has a stable half-filled configuration but is in a higher period than $N$,so its ionization energy is lower than that of $N$.
Therefore,$N$ has the highest first ionization energy among the given options.

(C) The second ionization potential is greater than the first ionization potential.
After the removal of the first electron $(e^-)$,the remaining electrons are more strongly attracted by the nucleus due to an increase in the effective nuclear charge $(Z_{eff})$ and a decrease in electron-electron repulsion.
Therefore,more energy is required to remove the second electron.

(C) The peaks in the plot of first ionization energy versus atomic number correspond to the noble gases (rare gases).
This is because noble gases possess a stable,fully-filled valence shell electron configuration $(ns^2 np^6)$,which requires a very high amount of energy to remove an electron $(e^-)$.

(D) The first ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Noble gases,such as $Kr$ (Krypton),have a stable electronic configuration with a completely filled valence shell $(ns^2 np^6)$.
Due to this stable configuration,they have the highest ionization energy in their respective periods.
Therefore,$Kr$ has the highest first ionization energy among the given options.

(B) The first ionisation potential $(I.P.)$ is maximum for hydrogen among the given options due to its extremely small atomic size and high effective nuclear charge relative to its single electron.

(C) The increasing order of ionization enthalpy is $As < S < P$.
On moving from left to right in a period,the ionization energy generally increases.
On moving down the group,the ionization energy decreases.
$As$ is in the $4^{th}$ period,while $S$ and $P$ are in the $3^{rd}$ period,so $As$ has the lowest ionization energy.
Between $S$ $(3p^4)$ and $P$ $(3p^3)$,$P$ has a stable half-filled $3p$ orbital,making it harder to remove an electron compared to $S$.
Thus,the correct order is $As < S < P$.

(D) The ionisation potential generally decreases down a group and increases across a period from left to right.
$B$ (Boron) is a non-metal in period $2$,which has a high ionisation potential.
$K$ (Potassium) and $Cs$ (Caesium) are alkali metals in group $1$,where $Cs$ is below $K$,so $Cs$ has a lower ionisation potential than $K$.
$U$ (Uranium) is an actinoid with a relatively low ionisation potential compared to non-metals like $B$.
Comparing these,the correct order of increasing ionisation potential is $Cs < U < K < B$.

(C) The ionization potential increases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
Among the given elements ($Li$,$B$,$F$,$Ne$),$Ne$ is a noble gas located at the end of the period.
$Ne$ has a stable,fully-filled electronic configuration $(1s^2 2s^2 2p^6)$,which requires a very high amount of energy to remove an electron.
Therefore,$Ne$ has the highest ionization potential.

(B) The first ionisation potential $(I.P.)$ generally decreases as we move down a group due to the increase in atomic size and shielding effect.
In Group $2$ $(Be, Mg, Ca)$,the order of $I.P.$ is $Be > Mg > Ca$.
In Group $1$ $(Li, Na, K)$,the order is $Li > Na > K$.
In Group $14$ $(C, Si, Ge)$,the order is $C > Si > Ge$.
In Group $13$ $(B, Al, Ga)$,the order is $B > Al > Ga$.
Therefore,the correct set is $Be > Mg > Ca$.

(A) The first ionisation potential $(I.P.)$ generally increases from left to right across a period due to an increase in effective nuclear charge.
For the elements $B$ (Boron),$C$ (Carbon),and $N$ (Nitrogen),which belong to the $2^{nd}$ period,the order of increasing first $I.P.$ is $B < C < N$.
Therefore,the correct sequence is $B < C < N$.

(B) Ionization potential is the amount of energy required to remove an outermost valence electron from an isolated gaseous atom.
In a period,ionization potential generally increases from left to right due to an increase in effective nuclear charge.
$Ne$ is a noble gas with a completely filled octet,giving it the highest ionization potential among the given elements.
Comparing $P$ $(3s^2 3p^3)$ and $S$ $(3s^2 3p^4)$: $P$ has a stable half-filled $p$-orbital,making its ionization potential higher than $S$.
Comparing $Mg$ $(3s^2)$ and $Al$ $(3s^2 3p^1)$: $Mg$ has a stable fully-filled $s$-orbital,making its ionization potential higher than $Al$.
Thus,the correct decreasing order is: $Ne > Cl > P > S > Mg > Al$.

(B) The electronic configurations are: $C (2s^2 2p^2)$,$N (2s^2 2p^3)$,and $O (2s^2 2p^4)$.
Nitrogen has a stable half-filled $2p$ subshell,which results in a higher ionization potential compared to both Carbon and Oxygen.
Oxygen has a lower ionization potential than Nitrogen due to inter-electronic repulsion in the $2p^4$ configuration.
Thus,the order of first ionization potential is $C < O < N$ or $N > O > C$.
Comparing this with the given options,option $B$ $(C < N > O)$ correctly represents the relative order where $N$ is the highest.

(A) The ionization potential $(I.P.)$ is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Among the given elements,Lithium $(Li)$ is an alkali metal with a large atomic size and low effective nuclear charge,making it easiest to remove its valence electron.
The $I.P.$ values are approximately: $Li \approx 5.4 \ eV$,$He \approx 24.6 \ eV$,$N \approx 14.5 \ eV$,and $Zn \approx 9.4 \ eV$.
Therefore,$Li$ has the least ionization potential.

(B) The first ionisation energy $(I.E.)$ generally increases as we move from left to right across a period in the periodic table due to an increase in effective nuclear charge and a decrease in atomic size.
$Li$ $(Z=3)$ and $Be$ $(Z=4)$ belong to the same period ($2^{nd}$ period).
Since $Be$ is to the right of $Li$,the first ionisation energy of $Li$ is less than that of $Be$.

(A) The correct answer is $(A)$.
$He$ (Helium) has the highest ionisation energy among the given elements.
This is because $He$ has a stable,fully-filled $1s^2$ electronic configuration,which makes it extremely difficult to remove an electron from its valence shell.

(A) The penetration power of orbitals follows the order $s > p > d > f$.
$s$-orbitals are closest to the nucleus and experience the greatest effective nuclear charge,making them more strongly held by the nucleus.
Therefore,the energy required to remove an electron from an $s$-orbital is the highest among the given options.

(D) The first ionisation energy depends on the electronic configuration and effective nuclear charge.
Electronic configurations are: $Na (Z=11): [Ne] 3s^1$,$Mg (Z=12): [Ne] 3s^2$,$Al (Z=13): [Ne] 3s^2 3p^1$.
$Mg$ has a fully filled $3s$ orbital,which is more stable and requires more energy to remove an electron compared to $Al$ and $Na$.
$Al$ has a higher effective nuclear charge than $Na$,making its first ionisation energy higher than $Na$.
Thus,the decreasing order is $Mg > Al > Na$.

(A) As we move across the period,nuclear charge increases and atomic size decreases,hence ionisation enthalpy generally increases.
For $Al$ $([Ne] 3s^2 3p^1)$,the electron is removed from a partially filled $3p$ orbital,whereas for $Mg$ $([Ne] 3s^2)$,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.
Therefore,the ionisation enthalpy of $Mg$ is greater than that of $Al$.
The correct order of first ionisation enthalpies is: $Na < Mg > Al < Si$.

(A) All the given ions $(O^{2-}, F^{-}, Na^{+}, Mg^{2+})$ are isoelectronic,having $10$ electrons each.
Ionisation potential is the energy required to remove an electron from an atom or ion.
For isoelectronic species,the ionisation potential increases as the nuclear charge (number of protons) increases.
$O^{2-}$ has the lowest nuclear charge $(Z = 8)$ among the given ions.
Therefore,it is easiest to remove an electron from $O^{2-}$,making it the species with the lowest ionisation potential.

(A) The ionisation energy of elements in a group decreases as we move down the group due to an increase in atomic size and the shielding effect.
As the atomic size increases,the outermost electron is further from the nucleus,resulting in a weaker electrostatic attraction between the nucleus and the valence electron.
For group $I$ (alkali metals),the order of ionisation energy is $Li > Na > K > Rb > Cs$.
Therefore,the correct decreasing order is $Li > Na > K > Cs$.

(D) The electronic configuration of $Na$ $(Z=11)$ is $[Ne] 3s^1$. After losing one electron,it achieves the stable noble gas configuration of $Ne$ $(1s^2 2s^2 2p^6)$.
Removing the second electron from this stable configuration requires a very high amount of energy.
The electronic configuration of $Mg$ $(Z=12)$ is $[Ne] 3s^2$. After losing one electron,it becomes $[Ne] 3s^1$,which is relatively easier to ionize further compared to the stable $Ne$ core of $Na^+$.
Therefore,the second ionization potential of sodium is significantly higher than that of magnesium,i.e.,$II_{Na} > II_{Mg}$.

(A) The first ionisation energy depends on the electronic configuration and effective nuclear charge.
$Be$ $(1s^2 2s^2)$ has a fully filled $s$-orbital,making it more stable than $B$ $(1s^2 2s^2 2p^1)$.
$N$ $(1s^2 2s^2 2p^3)$ has a half-filled $p$-orbital,which is more stable than $O$ $(1s^2 2s^2 2p^4)$.
Comparing the groups and periods,the correct order of first ionisation energy is $N > O > Be > B$.

(D) The correct option is $D$.
In the given transitions,the energy required for the removal of an electron is known as the ionization energy.
As the positive charge on the ion increases,the effective nuclear charge experienced by the remaining electrons increases.
This leads to a stronger electrostatic attraction between the nucleus and the valence electrons.
Therefore,the energy required to remove an electron from $M^{2+}_{(g)}$ to form $M^{3+}_{(g)}$ is significantly higher than the energy required for the previous ionization steps.

(D) The ionization potential is the energy required to remove an electron from a gaseous species.
Comparing the species:
$O_2^-$ has an extra electron compared to $O_2$,making it more unstable and easier to lose an electron.
$O$ is an atom,while the others are molecules.
Among the given species,$O_2^-$ has the highest electron density and the least effective nuclear charge per electron,making it the easiest to ionize.
Therefore,$O_2^-$ has the lowest ionization potential.

(A) The ionization energy $(I.E.)$ generally increases from left to right across a period in the periodic table due to an increase in effective nuclear charge and a decrease in atomic radius.
All given elements ($Ge$,$As$,$Se$,$Br$) belong to the $4^{th}$ period.
Their positions from left to right are: $Ge$ ($Group$ $14$) < $As$ ($Group$ $15$) < $Se$ ($Group$ $16$) < $Br$ ($Group$ $17$).
Therefore,$Ge$ has the lowest ionization energy among the given options.

(B) The electronic configuration of $Mg$ $(Z=12)$ is $[Ne] 3s^2$,which is a fully filled stable orbital configuration.
The electronic configuration of $Al$ $(Z=13)$ is $[Ne] 3s^2 3p^1$.
Due to the extra stability of the fully filled $3s$ subshell in $Mg$,it requires more energy to remove an electron compared to $Al$.
Therefore,the first $I.P.$ of $Mg$ is more than $Al$.