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 7

Q70 jee_main_2024_27_jan_morning S-block and F-block General Features

Given below are two statements: Statement (I): The 4f and 5f series of elements are placed separately in the Periodic table to preserve the principle of classification. Statement (II) : S-block elements can be found in pure form in nature. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Solution

Core Logic

Statement I is true: The lanthanides ( $4f$ $4\text{f}$) and actinides ( $5f$ $5\text{f}$) are separated at the bottom of the periodic table to avoid distortion of the periodic trends layout and preserve clean structural classification groups.

Statement II is false: s-block elements (alkali and alkaline earth metals) possess exceptionally low ionization energies, making them highly reactive; thus, they are always encountered as compounds rather than in native pure forms in nature.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q61 jee_main_2024_29_jan_morning Ionization Enthalpy

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: The first ionisation enthalpy decreases across a period. Reason R: The increasing nuclear charge outweighs the shielding across the period. In the light of the above statements, choose the most appropriate from the options given below:

A. $Both A and R are true and R is the correct explanation of A$
B. $A is true but R is false$ $\text{A is true but R is false}$
C. $A is false but R is true$ $\text{A is false but R is true}$
D. $Both A and R are true but R is NOT the correct explanation of A$

Solution

Core Logic

Assertion A states that the first ionisation enthalpy decreases across a period. This is false, because first ionisation energy generally increases along a period from left to right.

Reason R states that the increasing nuclear charge outweighs the shielding across the period. This is true, and it is the exact reason why the atomic radius decreases and ionisation enthalpy increases across a period.

Step 1: Final Conclusion

Since the assertion is incorrect and the reason is a factually correct statement, A is false but R is true.

Pattern Recognition

Recall the horizontal periodic trends: across a period, effective nuclear charge ( $Zeff$ $Z_{\text{eff}}$) dominates, pulling valence electrons tighter, increasing ionisation energy.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q73 jee_main_2024_30_january_evening Periodic Trends in Chemical Properties

Given below are two statements: Statement I: Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18. Statement II: The nature of oxides formed by group 1 element is basic while that of group 17 elements is acidic. In the light above statements, choose the most appropriate from the questions given below:

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

Solution

Core Logic

Statement I: Along a period, the chemical reactivity of elements first decreases and then increases. Group 1 elements (alkali metals) are highly reactive, Group 17 elements (halogens) are also highly reactive, but the elements in the middle (like carbon) are less reactive. Moreover, Group 18 elements (noble gases) are mostly inert. Therefore, statement I is false.

Statement II: Group 1 elements are metals and form basic oxides (e.g., $Na₂O$ $Na_2O$). Group 17 elements are non-metals and form acidic oxides (e.g., $Cl₂O₇$ $Cl_2O_7$). Therefore, statement II is true.

Pattern Recognition

Reactivity is highest at the extreme left (Group 1) and extreme right (Group 17) of the periodic table, excluding noble gases. Metallic character = basic oxides; Non-metallic character = acidic oxides.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q84 jee_main_2024_30_jan_morning IUPAC Nomenclature for Elements > 100

If IUPAC name of an element is "Unununnium" then the element belongs to nth group of periodic table. The value of n is

Numerical Answer. Answer: 11 to 11

Solution

Core Logic

Decode the IUPAC name "Unununnium" into its atomic number. Un = 1 Un = 1 Un = 1 Suffix = ium So, the atomic number is 111.

Step 1: Finding group number

For d-block elements with Z > 100, the group number is the last two digits of the atomic number. Since Z = 111, the element belongs to group 11 (the Copper group).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q jee_main_2024_31_jan_evening Periodic Trends in Properties

Consider the following elements.

Which of the following is/are true about

A^

B^

C^

and

D^

? A. Order of atomic radii:

B^ < A^ < D^ < C^

B. Order of metallic character:

B^ < A^ < D^ < C^

C. Size of the element:

D^ < C^ < B^ < A^

D. Order of ionic radii:

B⁺ < A⁺ < D⁺ < C⁺

Choose the correct answer from the options given below:

A. $A only$ $\text{ A only}$
B. $A, B and D only$ $\text{A, B and D only}$
C. $A and B only$ $\text{A and B only}$
D. $B, C and D only$ $\text{B, C and D only}$

Solution

Core Logic

From the grid layout, $A^$ $A^\prime$ and $B^$ $B^\prime$ are in the same period (left to right), while $C^$ $C^\prime$ and $D^$ $D^\prime$ are below them in the next period. In general, moving along a period from left to right, size decreases and metallic character decreases. Moving down a group, size increases and metallic character increases.

Atomic Radii / Size: $B^ < A^$ $B^\prime < A^\prime$ (across period) $D^ < C^$ $D^\prime < C^\prime$ (across period) $A^ < C^$ $A^\prime < C^\prime$ and $B^ < D^$ $B^\prime < D^\prime$ (down the group) Combining them: $B^ < A^ < D^ < C^$ . (Statement A is correct, C is incorrect).

Metallic Character: $B^ < A^$ $B^\prime < A^\prime$ and $D^ < C^$ $D^\prime < C^\prime$ Overall trend follows size closely for these blocks: $B^ < A^ < D^ < C^$ . (Statement B is correct).

Ionic Radii ( $+$ $+$ ions): Follows the same basic trend as atomic radii: $B⁺ < A⁺ < D⁺ < C⁺$ . (Statement D is correct).

Step 1: Final Selection

Statements A, B, and D are correct. Statement C contradicts A, so it is incorrect. Option 2 is the right choice.

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 7

Solution

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Finding group number

Chapter Mix

Solution

Core Logic

Step 1: Final Selection

Chapter Mix

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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 7

Solution

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Finding group number

Chapter Mix

Solution

Core Logic

Step 1: Final Selection

Chapter Mix

More Classification of Elements and Periodicity in Properties Questions — jee_main_2026_24_january_evening

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