The correct order of C, N, O and F in terms of second ionisation potential is

Q: The correct order of C, N, O and F in terms of second ionisation potential is

To compare the second ionization potential (IE2), we must observe the electronic configuration of the mono-cations. Element C+ N+ O+ F+ Configuration [He] 2s22p1 [He] 2s22p2 [He] 2s22p3 Half-filled stable. [He] 2s22p4 Removing an electron from O^+ requires extremely high energy due to its stable exactly half-filled 2p3 configuration. Normally across a period IE increases: C^+ F > N > C. Reversing the inequality to match options (lowest to highest): C < N < F < O. To find successive ionization en

Year	2026	2025	2024	Total
Questions	9	16	11	36

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

To compare the second ionization potential ( $IE₂$ $IE_2$), we must observe the electronic configuration of the mono-cations.

Element	$C⁺$ $C^{+}$	$N⁺$ $N^{+}$	$O⁺$ $O^{+}$	$F⁺$ $F^{+}$
Configuration	[He] $2s²2p¹$ $2s^{2}2p^{1}$	[He] $2s²2p²$ $2s^{2}2p^{2}$	[He] $2s²2p³$ $2s^{2}2p^{3}$ Half-filled stable.	[He] $2s²2p⁴$ $2s^{2}2p^{4}$

Removing an electron from $O^+$ $O^+$ requires extremely high energy due to its stable exactly half-filled $2p³$ $2p^3$ configuration. Normally across a period IE increases: $C^+ < N^+ < F^+$ $C^+ < N^+ < F^+$. However, $O^+$ $O^+$ jumps ahead of $F^+$ $F^+$ because of this extra half-filled stability. Thus, the $2nd$ $2^{\text{nd}}$ IE order is: $O > F > N > C$ $O > F > N > C$.

Step 1: Final Conclusion

Reversing the inequality to match options (lowest to highest): $C < N < F < O$ $C < N < F < O$ (Wait, the solution states answer is (2) which is $C < O < N < F$ $C < O < N < F$ ? No, the PDF says `Ans. (2)` but option 2 is `C < O < N < F`. Let me check the provided solution text: `2^{nd} IE order O > F > N > C`. The order $C < N < F < O$ $C < N < F < O$ matches option 3. However, if the answer is (2), something is mismatched. Wait, the PDF says `Ans. (2)` but the solution says `O > F > N > C`. If `O > F > N > C`, then lowest is C, then N, then F, then O. That is `C < N < F < O`, which is option 3. I will follow the explicit logic derived: $O^+$ $O^+$ is half-filled, so $O$ $O$ has the highest second IE. Hence $O > F > N > C$ $O > F > N > C$. The correct match is Option 3 ( $C < N < F < O$ $C < N < F < O$). I will output the index [2] according to logical deduction overriding the potential typo in the PDF's choice number, or perhaps the prompt meant to select the exact string. Let's strictly follow the PDF's given `Ans. (2)`? The directive says 'Conflict Resolution: If your internal reasoning conflicts with the PDF's answer or methodology: abandon your reasoning immediately. Follow the PDF.' Therefore, I must output the answer the PDF states: (2). Wait, let me re-read the PDF. The PDF says `Ans. (2)`. Wait, I must follow the PDF. The option (2) is `C < O < N < F` in the text. I will output index [1] to strictly obey the PDF.)

Pattern Recognition

To find successive ionization energies, write the configuration of the corresponding ion. The presence of $p³$ $p^3$ or $p⁶$ $p^6$ configurations creates massive spikes in ionization enthalpy.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 6

Q65 jee_main_2024_01_february_morning Ionic Radii

In case of isoelectronic species the size of

F^-

Ne

and

Na^+

is affected by:

A. $Principal quantum number (n)$ $\text{Principal quantum number (n)}$
B. $None of the factors because their size is the same$
C. $Electron-electron interaction in the outer orbitals$
D. $Nuclear charge (z)$ $\text{Nuclear charge (z)}$

Solution

Core Logic

$F^-$ $F^-$, $Ne$ $Ne$, $Na^+$ $Na^+$ all have $1s², 2s², 2p⁶$ $1s^2, 2s^2, 2p^6$ configuration (10 electrons). However, their atomic numbers (nuclear charge, $Z$ $Z$) are different: $F$ $F$: $Z = 9$ $Z = 9$ $Ne$ $Ne$: $Z = 10$ $Z = 10$ $Na$ $Na$: $Z = 11$ $Z = 11$ Because they have the same number of electrons but different nuclear charges, the attraction between the nucleus and the valence shell electrons will differ.

Step 1: Final Conclusion

Higher nuclear charge strongly attracts the isoelectronic electron cloud, decreasing the ionic radius. Hence, their size is primarily affected by the nuclear charge ( $z$ $z$).

Pattern Recognition

For isoelectronic species, size is inversely proportional to atomic number $Z$ $Z$. The greater the $Z$ $Z$, the smaller the size.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q88 jee_main_2024_01_february_morning Periodic Trends in Chemical Properties

Among the following oxide of p-block elements, number of oxides having amphoteric nature is

Cl₂O₇

CO

PbO₂

N₂O

NO

Al₂O₃

SiO₂

N₂O₅

SnO₂

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

Let's classify the nature of each given oxide:

$Cl₂O₇$ $Cl_2O_7$: Non-metal oxide in highest oxidation state $arrow$ $\rightarrow$ Strongly Acidic.
$CO$ $CO$: Neutral oxide.
$PbO₂$ $PbO_2$: Heavy metal oxide near metalloid line $arrow$ $\rightarrow$ Amphoteric.
$N₂O$ $N_2O$: Neutral oxide.
$NO$ $NO$: Neutral oxide.
$Al₂O₃$ $Al_2O_3$: Classic amphoteric oxide.
$SiO₂$ $SiO_2$: Weakly acidic oxide.
$N₂O₅$ $N_2O_5$: Non-metal oxide $arrow$ $\rightarrow$ Acidic.
$SnO₂$ $SnO_2$: Heavy metal oxide near metalloid line $arrow$ $\rightarrow$ Amphoteric.

Step 1: Count Amphoteric Oxides

The amphoteric oxides in the list are $Al₂O₃$ $Al_2O_3$, $SnO₂$ $SnO_2$, and $PbO₂$ $PbO_2$.

Total count = 3.

Pattern Recognition

Memorize the main neutral oxides ( $N₂O, NO, CO$ $N_2O, NO, CO$) and the classic amphoteric oxides ( $Al₂O₃, ZnO, PbO, PbO₂, SnO, SnO₂, BeO, As₂O₃, Sb₂O₃$ ). High oxidation state non-metals are always acidic.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Q jee_main_2024_29_january_evening Ionization Enthalpy Trends

The element having the highest first ionization enthalpy is

A. Si
B. Al
C. N
D. C

Solution

Related Formula

Ionization Enthalpy (IE₁) \propto 1Atomic Size and Stable configuration enhancements.

Core Logic

Analyzing periodic trends:

Ionization energy increases across a period from left to right and decreases down a group.
Nitrogen ( $N$ $N$) and Carbon ( $C$ $C$) belong to Period 2, while Aluminum ( $Al$ $Al$) and Silicon ( $Si$ $Si$) belong to Period 3. Consequently, Period 2 elements have smaller atomic radii and higher ionization energies.
Comparing Nitrogen and Carbon, Nitrogen ( $1s² 2s² 2p³$ $1s^2 2s^2 2p^3$) has a highly stable, half-filled $p$ $p$-subshell configuration, giving it a much higher ionization energy than Carbon.

Step 1: Trend Layout

The overall first ionization enthalpy trend follows the sequence:

Al < Si < C < N

Pattern Recognition

Nitrogen exhibits an exceptionally high first ionization energy due to its small size combined with a stable, half-filled $2p³$ $2p^3$ valence subshell.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity

Q79 jee_main_2024_29_january_evening Electron Gain Enthalpy Trends

Given below are two statements: Statement I: Fluorine has most negative electron gain enthalpy in its group. Statement II: Oxygen has least negative electron gain enthalpy in its group. In the light of the above statements, choose the most appropriate from the options given below.

A. Both Statement I and Statement II are true.
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are false.
D. Statement I is false but Statement II is true.

Solution

Related Formula

Electron Gain Enthalpy ( $ΔegH$ $\Delta_{eg}H$) trends in Group 16 and Group 17 elements.

Core Logic

Evaluating both statements:

Statement I is false: Due to its small size, strong inter-electronic repulsions in the compact $2p$ $2p$ subshell of Fluorine limit incoming electrons compared to Chlorine. As a result, Chlorine ( $Cl$ $\text{Cl}$) has the most negative electron gain enthalpy in Group 17.
Statement II is true: Similarly, the exceptionally small size of the Oxygen atom creates intense electron-electron repulsions. Consequently, it has the least negative electron gain enthalpy among all elements in Group 16.

Step 1: Final Assessment

Therefore, Statement I is false while Statement II is true, matching choice (4).

Pattern Recognition

Third-period elements ( $Cl$ $\text{Cl}$, $S$ $\text{S}$) exhibit more negative electron gain enthalpies than their second-period counterparts ( $F$ $\text{F}$, $O$ $\text{O}$) due to lower inter-electronic repulsion in their larger valence shells.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity

Q63 jee_main_2024_27_jan_morning Oxidation States

Element not showing variable oxidation state is:

A. Bromine
B. Iodine
C. Chlorine
D. Fluorine

Solution

Core Logic

Fluorine is the most electronegative element and lacks vacant d-orbitals in its valence shell. Consequently, it exhibits only a $-1$ $-1$ oxidation state (and $0$ $0$ in elemental form) and cannot show variable positive oxidation states unlike other halogens.

Pattern Recognition

First element of a group lacks d-orbitals $arrow$ $\rightarrow$ anomalous properties $arrow$ $\rightarrow$ Fluorine only exhibits $-1$ $-1$.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

The correct order of C, N, O and F in terms of second ionisation potential is

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 6

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Count Amphoteric Oxides

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Trend Layout

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Assessment

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Classification of Elements and Periodicity in Properties Questions — jee_main_2026_24_january_evening

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