Classification of Elements and Periodicity in Properties
89602
The first ionisation energy (in $\mathrm{kJ} / \mathrm{mol}$ ) of $\mathrm{Na}$, $\mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ respectively, are:
1 $496,577,737,786$
2 $786,737,577,496$
3 $496,577,786,737$
4 $496,737,577,786$
Explanation:
Ionisaiton energy is directly proportional to the effective nuclear charge (Zeff). As the size of atom increase Zeff also increase i.e. ionisaiton energy also increase. Ionisation energy $\propto$ Zeff Ionisation energy is the amount of energy required to remove the outer shell electron of an atom. Thus, the correct order of ionization energy are- $$ \mathrm{Na}=496, \mathrm{mg}=737, \mathrm{Al}=577, \mathrm{Si}=786 $$
(JEE Main 2020
Classification of Elements and Periodicity in Properties
89596
Identify the elements $X$ and $Y$ using the ionisation energy values given below: {lll} | {l}{ Ionization energy $( / )$} | |---| |$$ | $1^{ {st }}$ | $2^{ {nd }}$ | |$$ | 495 | 4563 | | 731 | 1450
As we know, first ionization energy is always less than second ionization energy. After removal of one electron from outer orbit, it's attains the stable noble gas configuration i.e. $$ \mathrm{Na}^{+} \rightarrow[\mathrm{Ne}], 3 \mathrm{~s}^0 $$ due to which ionization energy of $\mathrm{Na}^{+}$is very low but second ionization energy is very high. $\mathrm{Mg} \rightarrow[\mathrm{Ne}] 3 \mathrm{~s}^2$, The first and second ionization enthalpy of $\mathrm{Mg}$ is low but the third ionization enthalpy is very high due to noble gas configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89597
Assertion (A): The first ionization energy of Be is greater than that of $B$. Reason (R): 2p orbital has lower energy than 2 s orbital.
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $A$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $A$
3 $\mathrm{A}$ is true $\mathrm{R}$ is false
4 $\mathrm{A}$ is false, $\mathrm{R}$ is true
Explanation:
Electronic configuration of- Ionisation energy is defined by the atom which required minimum energy to remove the one electron in outer shell. So, Be has stable subshell to remove one electron is maximum in comparison to B. Now, according to Aufbau principle. $\begin{array}{lllll}1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^6 & 3 \mathrm{~s}^2 & 3 \mathrm{p}^6\end{array} \ldots \ldots \ldots$. So, the energy of $2 s$ is lower than $2 p$. So, the option (c) is correct.
AP EAPCET 23-08-2021 Shift-I
Classification of Elements and Periodicity in Properties
89598
The second ionisation energies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in the order
1 $\mathrm{Li}>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
2 Li $>$ B $>$ C $>$ Be
3 $\mathrm{Be}>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 $\mathrm{B}>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The second ionisation energy is the energy required to remove an electron from a $1^{+}$cation in the gaseous state. $$ \mathrm{X}^{+}(\mathrm{g}) \rightarrow \mathrm{X}^{2+}(\mathrm{g})+\mathrm{e}^{-} $$ Just like the first ionisation energy, the second ionisation energy is affected by size, effective nuclear charge, and electron configuration. Li has highest $\mathrm{IE}_2$ than because the second electron remove from stable noble gas configuration and $\mathrm{B}$ has higher $\mathrm{IE}_2$ than $\mathrm{C}$ due to the extra stability of the $2 \mathrm{~s}^2$ subshell in the $\mathrm{B}^{+}$ion. Therefore, the order is $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$ Hence, the correct option is (b).
Shift-I
Classification of Elements and Periodicity in Properties
89603
Assertion: Ionisation enthalpy of beryllium is higher than that of boron. Reason: Across the period, from left to right, ionisation enthalpy decreases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to periodic trends, boron should have a higher ionization energy but because its valence shell $2 \mathrm{p}^1$ is Shielded by the $2 \mathrm{~s}$ electron, less energy is required to remove the $2 \mathrm{p}$ electrons from a boron atom than is required to remove the $2 \mathrm{~s}$ electron from a beryllium atom.
Classification of Elements and Periodicity in Properties
89602
The first ionisation energy (in $\mathrm{kJ} / \mathrm{mol}$ ) of $\mathrm{Na}$, $\mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ respectively, are:
1 $496,577,737,786$
2 $786,737,577,496$
3 $496,577,786,737$
4 $496,737,577,786$
Explanation:
Ionisaiton energy is directly proportional to the effective nuclear charge (Zeff). As the size of atom increase Zeff also increase i.e. ionisaiton energy also increase. Ionisation energy $\propto$ Zeff Ionisation energy is the amount of energy required to remove the outer shell electron of an atom. Thus, the correct order of ionization energy are- $$ \mathrm{Na}=496, \mathrm{mg}=737, \mathrm{Al}=577, \mathrm{Si}=786 $$
(JEE Main 2020
Classification of Elements and Periodicity in Properties
89596
Identify the elements $X$ and $Y$ using the ionisation energy values given below: {lll} | {l}{ Ionization energy $( / )$} | |---| |$$ | $1^{ {st }}$ | $2^{ {nd }}$ | |$$ | 495 | 4563 | | 731 | 1450
As we know, first ionization energy is always less than second ionization energy. After removal of one electron from outer orbit, it's attains the stable noble gas configuration i.e. $$ \mathrm{Na}^{+} \rightarrow[\mathrm{Ne}], 3 \mathrm{~s}^0 $$ due to which ionization energy of $\mathrm{Na}^{+}$is very low but second ionization energy is very high. $\mathrm{Mg} \rightarrow[\mathrm{Ne}] 3 \mathrm{~s}^2$, The first and second ionization enthalpy of $\mathrm{Mg}$ is low but the third ionization enthalpy is very high due to noble gas configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89597
Assertion (A): The first ionization energy of Be is greater than that of $B$. Reason (R): 2p orbital has lower energy than 2 s orbital.
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $A$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $A$
3 $\mathrm{A}$ is true $\mathrm{R}$ is false
4 $\mathrm{A}$ is false, $\mathrm{R}$ is true
Explanation:
Electronic configuration of- Ionisation energy is defined by the atom which required minimum energy to remove the one electron in outer shell. So, Be has stable subshell to remove one electron is maximum in comparison to B. Now, according to Aufbau principle. $\begin{array}{lllll}1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^6 & 3 \mathrm{~s}^2 & 3 \mathrm{p}^6\end{array} \ldots \ldots \ldots$. So, the energy of $2 s$ is lower than $2 p$. So, the option (c) is correct.
AP EAPCET 23-08-2021 Shift-I
Classification of Elements and Periodicity in Properties
89598
The second ionisation energies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in the order
1 $\mathrm{Li}>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
2 Li $>$ B $>$ C $>$ Be
3 $\mathrm{Be}>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 $\mathrm{B}>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The second ionisation energy is the energy required to remove an electron from a $1^{+}$cation in the gaseous state. $$ \mathrm{X}^{+}(\mathrm{g}) \rightarrow \mathrm{X}^{2+}(\mathrm{g})+\mathrm{e}^{-} $$ Just like the first ionisation energy, the second ionisation energy is affected by size, effective nuclear charge, and electron configuration. Li has highest $\mathrm{IE}_2$ than because the second electron remove from stable noble gas configuration and $\mathrm{B}$ has higher $\mathrm{IE}_2$ than $\mathrm{C}$ due to the extra stability of the $2 \mathrm{~s}^2$ subshell in the $\mathrm{B}^{+}$ion. Therefore, the order is $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$ Hence, the correct option is (b).
Shift-I
Classification of Elements and Periodicity in Properties
89603
Assertion: Ionisation enthalpy of beryllium is higher than that of boron. Reason: Across the period, from left to right, ionisation enthalpy decreases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to periodic trends, boron should have a higher ionization energy but because its valence shell $2 \mathrm{p}^1$ is Shielded by the $2 \mathrm{~s}$ electron, less energy is required to remove the $2 \mathrm{p}$ electrons from a boron atom than is required to remove the $2 \mathrm{~s}$ electron from a beryllium atom.
Classification of Elements and Periodicity in Properties
89602
The first ionisation energy (in $\mathrm{kJ} / \mathrm{mol}$ ) of $\mathrm{Na}$, $\mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ respectively, are:
1 $496,577,737,786$
2 $786,737,577,496$
3 $496,577,786,737$
4 $496,737,577,786$
Explanation:
Ionisaiton energy is directly proportional to the effective nuclear charge (Zeff). As the size of atom increase Zeff also increase i.e. ionisaiton energy also increase. Ionisation energy $\propto$ Zeff Ionisation energy is the amount of energy required to remove the outer shell electron of an atom. Thus, the correct order of ionization energy are- $$ \mathrm{Na}=496, \mathrm{mg}=737, \mathrm{Al}=577, \mathrm{Si}=786 $$
(JEE Main 2020
Classification of Elements and Periodicity in Properties
89596
Identify the elements $X$ and $Y$ using the ionisation energy values given below: {lll} | {l}{ Ionization energy $( / )$} | |---| |$$ | $1^{ {st }}$ | $2^{ {nd }}$ | |$$ | 495 | 4563 | | 731 | 1450
As we know, first ionization energy is always less than second ionization energy. After removal of one electron from outer orbit, it's attains the stable noble gas configuration i.e. $$ \mathrm{Na}^{+} \rightarrow[\mathrm{Ne}], 3 \mathrm{~s}^0 $$ due to which ionization energy of $\mathrm{Na}^{+}$is very low but second ionization energy is very high. $\mathrm{Mg} \rightarrow[\mathrm{Ne}] 3 \mathrm{~s}^2$, The first and second ionization enthalpy of $\mathrm{Mg}$ is low but the third ionization enthalpy is very high due to noble gas configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89597
Assertion (A): The first ionization energy of Be is greater than that of $B$. Reason (R): 2p orbital has lower energy than 2 s orbital.
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $A$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $A$
3 $\mathrm{A}$ is true $\mathrm{R}$ is false
4 $\mathrm{A}$ is false, $\mathrm{R}$ is true
Explanation:
Electronic configuration of- Ionisation energy is defined by the atom which required minimum energy to remove the one electron in outer shell. So, Be has stable subshell to remove one electron is maximum in comparison to B. Now, according to Aufbau principle. $\begin{array}{lllll}1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^6 & 3 \mathrm{~s}^2 & 3 \mathrm{p}^6\end{array} \ldots \ldots \ldots$. So, the energy of $2 s$ is lower than $2 p$. So, the option (c) is correct.
AP EAPCET 23-08-2021 Shift-I
Classification of Elements and Periodicity in Properties
89598
The second ionisation energies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in the order
1 $\mathrm{Li}>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
2 Li $>$ B $>$ C $>$ Be
3 $\mathrm{Be}>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 $\mathrm{B}>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The second ionisation energy is the energy required to remove an electron from a $1^{+}$cation in the gaseous state. $$ \mathrm{X}^{+}(\mathrm{g}) \rightarrow \mathrm{X}^{2+}(\mathrm{g})+\mathrm{e}^{-} $$ Just like the first ionisation energy, the second ionisation energy is affected by size, effective nuclear charge, and electron configuration. Li has highest $\mathrm{IE}_2$ than because the second electron remove from stable noble gas configuration and $\mathrm{B}$ has higher $\mathrm{IE}_2$ than $\mathrm{C}$ due to the extra stability of the $2 \mathrm{~s}^2$ subshell in the $\mathrm{B}^{+}$ion. Therefore, the order is $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$ Hence, the correct option is (b).
Shift-I
Classification of Elements and Periodicity in Properties
89603
Assertion: Ionisation enthalpy of beryllium is higher than that of boron. Reason: Across the period, from left to right, ionisation enthalpy decreases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to periodic trends, boron should have a higher ionization energy but because its valence shell $2 \mathrm{p}^1$ is Shielded by the $2 \mathrm{~s}$ electron, less energy is required to remove the $2 \mathrm{p}$ electrons from a boron atom than is required to remove the $2 \mathrm{~s}$ electron from a beryllium atom.
Classification of Elements and Periodicity in Properties
89602
The first ionisation energy (in $\mathrm{kJ} / \mathrm{mol}$ ) of $\mathrm{Na}$, $\mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ respectively, are:
1 $496,577,737,786$
2 $786,737,577,496$
3 $496,577,786,737$
4 $496,737,577,786$
Explanation:
Ionisaiton energy is directly proportional to the effective nuclear charge (Zeff). As the size of atom increase Zeff also increase i.e. ionisaiton energy also increase. Ionisation energy $\propto$ Zeff Ionisation energy is the amount of energy required to remove the outer shell electron of an atom. Thus, the correct order of ionization energy are- $$ \mathrm{Na}=496, \mathrm{mg}=737, \mathrm{Al}=577, \mathrm{Si}=786 $$
(JEE Main 2020
Classification of Elements and Periodicity in Properties
89596
Identify the elements $X$ and $Y$ using the ionisation energy values given below: {lll} | {l}{ Ionization energy $( / )$} | |---| |$$ | $1^{ {st }}$ | $2^{ {nd }}$ | |$$ | 495 | 4563 | | 731 | 1450
As we know, first ionization energy is always less than second ionization energy. After removal of one electron from outer orbit, it's attains the stable noble gas configuration i.e. $$ \mathrm{Na}^{+} \rightarrow[\mathrm{Ne}], 3 \mathrm{~s}^0 $$ due to which ionization energy of $\mathrm{Na}^{+}$is very low but second ionization energy is very high. $\mathrm{Mg} \rightarrow[\mathrm{Ne}] 3 \mathrm{~s}^2$, The first and second ionization enthalpy of $\mathrm{Mg}$ is low but the third ionization enthalpy is very high due to noble gas configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89597
Assertion (A): The first ionization energy of Be is greater than that of $B$. Reason (R): 2p orbital has lower energy than 2 s orbital.
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $A$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $A$
3 $\mathrm{A}$ is true $\mathrm{R}$ is false
4 $\mathrm{A}$ is false, $\mathrm{R}$ is true
Explanation:
Electronic configuration of- Ionisation energy is defined by the atom which required minimum energy to remove the one electron in outer shell. So, Be has stable subshell to remove one electron is maximum in comparison to B. Now, according to Aufbau principle. $\begin{array}{lllll}1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^6 & 3 \mathrm{~s}^2 & 3 \mathrm{p}^6\end{array} \ldots \ldots \ldots$. So, the energy of $2 s$ is lower than $2 p$. So, the option (c) is correct.
AP EAPCET 23-08-2021 Shift-I
Classification of Elements and Periodicity in Properties
89598
The second ionisation energies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in the order
1 $\mathrm{Li}>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
2 Li $>$ B $>$ C $>$ Be
3 $\mathrm{Be}>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 $\mathrm{B}>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The second ionisation energy is the energy required to remove an electron from a $1^{+}$cation in the gaseous state. $$ \mathrm{X}^{+}(\mathrm{g}) \rightarrow \mathrm{X}^{2+}(\mathrm{g})+\mathrm{e}^{-} $$ Just like the first ionisation energy, the second ionisation energy is affected by size, effective nuclear charge, and electron configuration. Li has highest $\mathrm{IE}_2$ than because the second electron remove from stable noble gas configuration and $\mathrm{B}$ has higher $\mathrm{IE}_2$ than $\mathrm{C}$ due to the extra stability of the $2 \mathrm{~s}^2$ subshell in the $\mathrm{B}^{+}$ion. Therefore, the order is $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$ Hence, the correct option is (b).
Shift-I
Classification of Elements and Periodicity in Properties
89603
Assertion: Ionisation enthalpy of beryllium is higher than that of boron. Reason: Across the period, from left to right, ionisation enthalpy decreases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to periodic trends, boron should have a higher ionization energy but because its valence shell $2 \mathrm{p}^1$ is Shielded by the $2 \mathrm{~s}$ electron, less energy is required to remove the $2 \mathrm{p}$ electrons from a boron atom than is required to remove the $2 \mathrm{~s}$ electron from a beryllium atom.
Classification of Elements and Periodicity in Properties
89602
The first ionisation energy (in $\mathrm{kJ} / \mathrm{mol}$ ) of $\mathrm{Na}$, $\mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ respectively, are:
1 $496,577,737,786$
2 $786,737,577,496$
3 $496,577,786,737$
4 $496,737,577,786$
Explanation:
Ionisaiton energy is directly proportional to the effective nuclear charge (Zeff). As the size of atom increase Zeff also increase i.e. ionisaiton energy also increase. Ionisation energy $\propto$ Zeff Ionisation energy is the amount of energy required to remove the outer shell electron of an atom. Thus, the correct order of ionization energy are- $$ \mathrm{Na}=496, \mathrm{mg}=737, \mathrm{Al}=577, \mathrm{Si}=786 $$
(JEE Main 2020
Classification of Elements and Periodicity in Properties
89596
Identify the elements $X$ and $Y$ using the ionisation energy values given below: {lll} | {l}{ Ionization energy $( / )$} | |---| |$$ | $1^{ {st }}$ | $2^{ {nd }}$ | |$$ | 495 | 4563 | | 731 | 1450
As we know, first ionization energy is always less than second ionization energy. After removal of one electron from outer orbit, it's attains the stable noble gas configuration i.e. $$ \mathrm{Na}^{+} \rightarrow[\mathrm{Ne}], 3 \mathrm{~s}^0 $$ due to which ionization energy of $\mathrm{Na}^{+}$is very low but second ionization energy is very high. $\mathrm{Mg} \rightarrow[\mathrm{Ne}] 3 \mathrm{~s}^2$, The first and second ionization enthalpy of $\mathrm{Mg}$ is low but the third ionization enthalpy is very high due to noble gas configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89597
Assertion (A): The first ionization energy of Be is greater than that of $B$. Reason (R): 2p orbital has lower energy than 2 s orbital.
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $A$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $A$
3 $\mathrm{A}$ is true $\mathrm{R}$ is false
4 $\mathrm{A}$ is false, $\mathrm{R}$ is true
Explanation:
Electronic configuration of- Ionisation energy is defined by the atom which required minimum energy to remove the one electron in outer shell. So, Be has stable subshell to remove one electron is maximum in comparison to B. Now, according to Aufbau principle. $\begin{array}{lllll}1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^6 & 3 \mathrm{~s}^2 & 3 \mathrm{p}^6\end{array} \ldots \ldots \ldots$. So, the energy of $2 s$ is lower than $2 p$. So, the option (c) is correct.
AP EAPCET 23-08-2021 Shift-I
Classification of Elements and Periodicity in Properties
89598
The second ionisation energies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in the order
1 $\mathrm{Li}>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
2 Li $>$ B $>$ C $>$ Be
3 $\mathrm{Be}>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 $\mathrm{B}>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The second ionisation energy is the energy required to remove an electron from a $1^{+}$cation in the gaseous state. $$ \mathrm{X}^{+}(\mathrm{g}) \rightarrow \mathrm{X}^{2+}(\mathrm{g})+\mathrm{e}^{-} $$ Just like the first ionisation energy, the second ionisation energy is affected by size, effective nuclear charge, and electron configuration. Li has highest $\mathrm{IE}_2$ than because the second electron remove from stable noble gas configuration and $\mathrm{B}$ has higher $\mathrm{IE}_2$ than $\mathrm{C}$ due to the extra stability of the $2 \mathrm{~s}^2$ subshell in the $\mathrm{B}^{+}$ion. Therefore, the order is $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$ Hence, the correct option is (b).
Shift-I
Classification of Elements and Periodicity in Properties
89603
Assertion: Ionisation enthalpy of beryllium is higher than that of boron. Reason: Across the period, from left to right, ionisation enthalpy decreases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to periodic trends, boron should have a higher ionization energy but because its valence shell $2 \mathrm{p}^1$ is Shielded by the $2 \mathrm{~s}$ electron, less energy is required to remove the $2 \mathrm{p}$ electrons from a boron atom than is required to remove the $2 \mathrm{~s}$ electron from a beryllium atom.
