Given below are two statements : Statement I : The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. Statement II : The ionic radius of O2- is larger than that of F-. In the light of the above statements, choose the correct answer from the options given below.

Statement I: Let's analyze the electronic configurations. Na (Z=11): 1s2 2s2 2p6 3s1 Na^+ is 1s2 2s2 2p6 (Stable Neon noble gas core). Mg (Z=12): 1s2 2s2 2p6 3s2 Mg^+ is 1s2 2s2 2p6 3s1. Removing a second electron from Na^+ (IE2) involves disrupting a highly stable, fully-filled 2p6 shell, requiring massive energy. Removing a second electron from Mg^+ (IE2) just removes the 3s1 electron. Thus, IE2 of Na > IE2 of Mg. Statement I is true. Statement II: Both O2- and F- are isoelectronic species, possessing 10 electrons (1s2 2s2 2p6). However, the nuclear charge (number of protons, Z) is different. O2- has 8 protons pulling 10 electrons. F- has 9 protons pulling 10 electrons. Since F- has a higher effective nuclear charge (Zeff), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of O2- > F-. Statement II is true. For isoelectronic species, more negative charge always equals a larger ionic radius (lower Z/e ratio implies less nuclear pull per electron). Class 11 Chemistry: Classification of Elements and Periodicity

Given below are two statements : Statement I : The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. Statement II : The ionic radius of O2- is larger than that of F-. In the light of the above statements, choose the correct answer from the options given below.

Statement I: Let's analyze the electronic configurations. Na (Z=11): 1s2 2s2 2p6 3s1 Na^+ is 1s2 2s2 2p6 (Stable Neon noble gas core). Mg (Z=12): 1s2 2s2 2p6 3s2 Mg^+ is 1s2 2s2 2p6 3s1. Removing a second electron from Na^+ (IE2) involves disrupting a highly stable, fully-filled 2p6 shell, requiring massive energy. Removing a second electron from Mg^+ (IE2) just removes the 3s1 electron. Thus, IE2 of Na > IE2 of Mg. Statement I is true. Statement II: Both O2- and F- are isoelectronic species, possessing 10 electrons (1s2 2s2 2p6). However, the nuclear charge (number of protons, Z) is different. O2- has 8 protons pulling 10 electrons. F- has 9 protons pulling 10 electrons. Since F- has a higher effective nuclear charge (Zeff), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of O2- > F-. Statement II is true. For isoelectronic species, more negative charge always equals a larger ionic radius (lower Z/e ratio implies less nuclear pull per electron). Class 11 Chemistry: Classification of Elements and Periodicity

Given below are two statements : Statement I : The second ionisation enthalpy of Na is larger than the corresponding…

Solution & Explanation

Related Formula

IE₂ requires breaking stable noble gas configurations if M^+ is isoelectronic with a noble gas.

Core Logic

Statement I: Let's analyze the electronic configurations. $Na$ $Na$ ( $Z=11$ $Z=11$): $1s² 2s² 2p⁶ 3s¹ Na^+$ $1s^2 2s^2 2p^6 3s^1 \implies Na^+$ is $1s² 2s² 2p⁶$ $1s^2 2s^2 2p^6$ (Stable Neon noble gas core). $Mg$ $Mg$ ( $Z=12$ $Z=12$): $1s² 2s² 2p⁶ 3s² Mg^+$ $1s^2 2s^2 2p^6 3s^2 \implies Mg^+$ is $1s² 2s² 2p⁶ 3s¹$ $1s^2 2s^2 2p^6 3s^1$. Removing a second electron from $Na^+$ $Na^+$ ( $IE₂$ $IE_2$) involves disrupting a highly stable, fully-filled $2p⁶$ $2p^6$ shell, requiring massive energy. Removing a second electron from $Mg^+$ $Mg^+$ ( $IE₂$ $IE_2$) just removes the $3s¹$ $3s^1$ electron. Thus, $IE₂$ $IE_2$ of Na > $IE₂$ $IE_2$ of Mg. Statement I is true.

Statement II: Both $O²⁻$ $O^{2-}$ and $F⁻$ $F^{-}$ are isoelectronic species, possessing 10 electrons ( $1s² 2s² 2p⁶$ $1s^2 2s^2 2p^6$). However, the nuclear charge (number of protons, Z) is different. $O²⁻$ $O^{2-}$ has 8 protons pulling 10 electrons. $F⁻$ $F^{-}$ has 9 protons pulling 10 electrons. Since $F⁻$ $F^{-}$ has a higher effective nuclear charge ( $Zeff$ $Z_{\text{eff}}$), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of $O²⁻$ $O^{2-}$ > $F⁻$ $F^{-}$. Statement II is true.

Pattern Recognition

For isoelectronic species, more negative charge always equals a larger ionic radius (lower $Z/e$ $Z/e$ ratio implies less nuclear pull per electron).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity

Ionization Enthalpy and Ionic Radius solution diagram 1 for Q59 - JEE Main 2026 Evening — Relevant data for evaluating Statement I and II regarding atomic properties.

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Q70 jee_main_2024_31_jan_morning Electron Gain Enthalpy

The correct sequence of electron gain enthalpy of the elements listed below is A. Ar B. Br C. F D. S Choose the most appropriate from the options given below:

A. $C > B > D > A$ $C > B > D > A$
B. $A > D > B > C$ $A > D > B > C$
C. $A > D > C > B$ $A > D > C > B$
D. $D > C > B > A$ $D > C > B > A$

Solution

Core Logic

Electron gain enthalpy ( $ΔegH$ $\Delta_{eg}H$) values are considered with their signs when ordering from largest to smallest algebraically.

Element	$ΔegH (kJ/mol)$ $\Delta_{eg}H \text{ (kJ/mol)}$
F	-333
S	-200
Br	-325
Ar	+96

Arranging these algebraically from highest to lowest: $+96 > -200 > -325 > -333$ $+96 > -200 > -325 > -333$ Thus, $Ar > S > Br > F$ $Ar > S > Br > F$, which corresponds to $A > D > B > C$ $A > D > B > C$.

Pattern Recognition

Noble gases always have positive electron gain enthalpies (highest algebraically). Halogens have highly negative values (lowest algebraically). Fluorine is highly negative, Bromine is slightly less negative. Sulfur sits in between.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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More Classification of Elements and Periodicity Previous-Year Questions — Page 8

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