Classification of Elements and Periodicity in Properties
89574
The second ionization enthalpies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in order
1 $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$
2 Li $>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
3 Be $>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 B $>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The ionization enthalpy depend on the following factors: (i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital. (ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{\text {nd }}$ I.E. all elements have $\mathrm{M}^{+}$ configuration
| | ${c} { Total no. of } | |---|---| | { electrons }$ | ${c} { Outermost } | | { configuration }$ | |$^{+}(=3)$ | 2 | $1 ^2$ (fully filled) | |$^{+}(=4)$ | 3 | ${l}2 s^1 { (has more size) } | | { than carbon) }$ | |$^{+}(=5)$ | 4 | $2 ^2$ (fully filled) | |$C^{+}(Z=6)$ | 5 | $2 s^2 2 p^1$ (Smaller size) | | Hence, correct, order for $2^{\text {nd }}$ I.E. is $\mathrm{Li}^{+}>\mathrm{B}^{+}>\mathrm{C}^{+}>\mathrm{Be}^{+}$ or $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}^{+}$
CG PET -2017
Classification of Elements and Periodicity in Properties
89582
Identify the elements $X$ and $Y$ using the ionisation energy values given below. {lll} | | ${l} { Ionisation } | |---|---| | { energy (Ist) }$ | ${l} / | | { (IInd) }$ | |$X$ | 495 | 4563 | |$$ | 731 | 1450 | |
The electronic configuration of $\mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of $\mathrm{Na}^{+}$. The electronic configuration of $\mathrm{Mg}=[\mathrm{Ne}] 3 \mathrm{~s}^2$ first and second ionisation energy is low.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89585
Choose the correct option regarding the following statements : Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen. Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.
1 Statement - 1 is correct. Statement -2 is wrong
2 Both statement 1 and 2 are wrong
3 Both statement 1 and 2 are correct
4 Statement -1 is wrong. Statement -2 is correct.
Explanation:
Electronic configuration of ${ }_7 \mathrm{~N}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3$ Electronic configuration of ${ }_8 \mathrm{O}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4$ $$ \begin{array}{|l|l|l|} \hline \uparrow \downarrow & \uparrow & \uparrow \\ \hline \end{array} $$ $\mathrm{N}$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming. Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 \mathrm{p}$ - orbital, oxygen acquires stable configuration i.e. $2 \mathrm{p}^3$.On the other hand, in case nitrogen it is not easy to remove one of the three $2 \mathrm{p}$ electronic due to its stable configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89586
Which order among the following is incorrect?
Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is- $\begin{aligned} & \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C} \\ & 520 \quad 801 \quad 899 \quad 1086 \mathrm{~kJ} / \text { Mole }\left(\Delta \mathrm{H}_1\right) \\ & \end{aligned}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Classification of Elements and Periodicity in Properties
89574
The second ionization enthalpies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in order
1 $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$
2 Li $>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
3 Be $>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 B $>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The ionization enthalpy depend on the following factors: (i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital. (ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{\text {nd }}$ I.E. all elements have $\mathrm{M}^{+}$ configuration
| | ${c} { Total no. of } | |---|---| | { electrons }$ | ${c} { Outermost } | | { configuration }$ | |$^{+}(=3)$ | 2 | $1 ^2$ (fully filled) | |$^{+}(=4)$ | 3 | ${l}2 s^1 { (has more size) } | | { than carbon) }$ | |$^{+}(=5)$ | 4 | $2 ^2$ (fully filled) | |$C^{+}(Z=6)$ | 5 | $2 s^2 2 p^1$ (Smaller size) | | Hence, correct, order for $2^{\text {nd }}$ I.E. is $\mathrm{Li}^{+}>\mathrm{B}^{+}>\mathrm{C}^{+}>\mathrm{Be}^{+}$ or $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}^{+}$
CG PET -2017
Classification of Elements and Periodicity in Properties
89582
Identify the elements $X$ and $Y$ using the ionisation energy values given below. {lll} | | ${l} { Ionisation } | |---|---| | { energy (Ist) }$ | ${l} / | | { (IInd) }$ | |$X$ | 495 | 4563 | |$$ | 731 | 1450 | |
The electronic configuration of $\mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of $\mathrm{Na}^{+}$. The electronic configuration of $\mathrm{Mg}=[\mathrm{Ne}] 3 \mathrm{~s}^2$ first and second ionisation energy is low.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89585
Choose the correct option regarding the following statements : Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen. Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.
1 Statement - 1 is correct. Statement -2 is wrong
2 Both statement 1 and 2 are wrong
3 Both statement 1 and 2 are correct
4 Statement -1 is wrong. Statement -2 is correct.
Explanation:
Electronic configuration of ${ }_7 \mathrm{~N}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3$ Electronic configuration of ${ }_8 \mathrm{O}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4$ $$ \begin{array}{|l|l|l|} \hline \uparrow \downarrow & \uparrow & \uparrow \\ \hline \end{array} $$ $\mathrm{N}$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming. Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 \mathrm{p}$ - orbital, oxygen acquires stable configuration i.e. $2 \mathrm{p}^3$.On the other hand, in case nitrogen it is not easy to remove one of the three $2 \mathrm{p}$ electronic due to its stable configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89586
Which order among the following is incorrect?
Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is- $\begin{aligned} & \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C} \\ & 520 \quad 801 \quad 899 \quad 1086 \mathrm{~kJ} / \text { Mole }\left(\Delta \mathrm{H}_1\right) \\ & \end{aligned}$
Classification of Elements and Periodicity in Properties
89574
The second ionization enthalpies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in order
1 $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$
2 Li $>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
3 Be $>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 B $>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The ionization enthalpy depend on the following factors: (i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital. (ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{\text {nd }}$ I.E. all elements have $\mathrm{M}^{+}$ configuration
| | ${c} { Total no. of } | |---|---| | { electrons }$ | ${c} { Outermost } | | { configuration }$ | |$^{+}(=3)$ | 2 | $1 ^2$ (fully filled) | |$^{+}(=4)$ | 3 | ${l}2 s^1 { (has more size) } | | { than carbon) }$ | |$^{+}(=5)$ | 4 | $2 ^2$ (fully filled) | |$C^{+}(Z=6)$ | 5 | $2 s^2 2 p^1$ (Smaller size) | | Hence, correct, order for $2^{\text {nd }}$ I.E. is $\mathrm{Li}^{+}>\mathrm{B}^{+}>\mathrm{C}^{+}>\mathrm{Be}^{+}$ or $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}^{+}$
CG PET -2017
Classification of Elements and Periodicity in Properties
89582
Identify the elements $X$ and $Y$ using the ionisation energy values given below. {lll} | | ${l} { Ionisation } | |---|---| | { energy (Ist) }$ | ${l} / | | { (IInd) }$ | |$X$ | 495 | 4563 | |$$ | 731 | 1450 | |
The electronic configuration of $\mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of $\mathrm{Na}^{+}$. The electronic configuration of $\mathrm{Mg}=[\mathrm{Ne}] 3 \mathrm{~s}^2$ first and second ionisation energy is low.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89585
Choose the correct option regarding the following statements : Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen. Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.
1 Statement - 1 is correct. Statement -2 is wrong
2 Both statement 1 and 2 are wrong
3 Both statement 1 and 2 are correct
4 Statement -1 is wrong. Statement -2 is correct.
Explanation:
Electronic configuration of ${ }_7 \mathrm{~N}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3$ Electronic configuration of ${ }_8 \mathrm{O}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4$ $$ \begin{array}{|l|l|l|} \hline \uparrow \downarrow & \uparrow & \uparrow \\ \hline \end{array} $$ $\mathrm{N}$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming. Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 \mathrm{p}$ - orbital, oxygen acquires stable configuration i.e. $2 \mathrm{p}^3$.On the other hand, in case nitrogen it is not easy to remove one of the three $2 \mathrm{p}$ electronic due to its stable configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89586
Which order among the following is incorrect?
Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is- $\begin{aligned} & \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C} \\ & 520 \quad 801 \quad 899 \quad 1086 \mathrm{~kJ} / \text { Mole }\left(\Delta \mathrm{H}_1\right) \\ & \end{aligned}$
Classification of Elements and Periodicity in Properties
89574
The second ionization enthalpies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $C$ are in order
1 $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$
2 Li $>\mathrm{C}>\mathrm{B}>\mathrm{Be}$
3 Be $>\mathrm{C}>\mathrm{B}>\mathrm{Li}$
4 B $>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$
Explanation:
The ionization enthalpy depend on the following factors: (i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital. (ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{\text {nd }}$ I.E. all elements have $\mathrm{M}^{+}$ configuration
| | ${c} { Total no. of } | |---|---| | { electrons }$ | ${c} { Outermost } | | { configuration }$ | |$^{+}(=3)$ | 2 | $1 ^2$ (fully filled) | |$^{+}(=4)$ | 3 | ${l}2 s^1 { (has more size) } | | { than carbon) }$ | |$^{+}(=5)$ | 4 | $2 ^2$ (fully filled) | |$C^{+}(Z=6)$ | 5 | $2 s^2 2 p^1$ (Smaller size) | | Hence, correct, order for $2^{\text {nd }}$ I.E. is $\mathrm{Li}^{+}>\mathrm{B}^{+}>\mathrm{C}^{+}>\mathrm{Be}^{+}$ or $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}^{+}$
CG PET -2017
Classification of Elements and Periodicity in Properties
89582
Identify the elements $X$ and $Y$ using the ionisation energy values given below. {lll} | | ${l} { Ionisation } | |---|---| | { energy (Ist) }$ | ${l} / | | { (IInd) }$ | |$X$ | 495 | 4563 | |$$ | 731 | 1450 | |
The electronic configuration of $\mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of $\mathrm{Na}^{+}$. The electronic configuration of $\mathrm{Mg}=[\mathrm{Ne}] 3 \mathrm{~s}^2$ first and second ionisation energy is low.
(JEE Main 2021
Classification of Elements and Periodicity in Properties
89585
Choose the correct option regarding the following statements : Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen. Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.
1 Statement - 1 is correct. Statement -2 is wrong
2 Both statement 1 and 2 are wrong
3 Both statement 1 and 2 are correct
4 Statement -1 is wrong. Statement -2 is correct.
Explanation:
Electronic configuration of ${ }_7 \mathrm{~N}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3$ Electronic configuration of ${ }_8 \mathrm{O}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4$ $$ \begin{array}{|l|l|l|} \hline \uparrow \downarrow & \uparrow & \uparrow \\ \hline \end{array} $$ $\mathrm{N}$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming. Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 \mathrm{p}$ - orbital, oxygen acquires stable configuration i.e. $2 \mathrm{p}^3$.On the other hand, in case nitrogen it is not easy to remove one of the three $2 \mathrm{p}$ electronic due to its stable configuration.
Shift-II
Classification of Elements and Periodicity in Properties
89586
Which order among the following is incorrect?
Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is- $\begin{aligned} & \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C} \\ & 520 \quad 801 \quad 899 \quad 1086 \mathrm{~kJ} / \text { Mole }\left(\Delta \mathrm{H}_1\right) \\ & \end{aligned}$
