Classification of Elements and Periodicity: JEE Mains Previous Year Questions (2021 - 2025)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Classification of Elements and Periodicity  
 
(January 2026) 
 
Q1: Consider the elements N, P , O, S, Cl and F . The number of valence electrons present in the 
elements with most and least metallic character from the above list is respectively. 
A: 7 and 5 
B: 6 and 7 
C: 5 and 7 
D: 5 and 6 
Answer: C 
Explanation: 
Metallic character increases down a group and decreases from left to right across a period 
(NCERT trend). 
Given elements: 
Period 2: N, O, F 
Period 3: P , S, Cl  
1) Most metallic element 
In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to 
right: 
P > S > Cl 
So, P is the most metallic among the given elements. 
Valence electrons in (Group 15) = 5. 
2) Least metallic element 
Least metallic means most non-metallic, which is highest towards the top-right of the periodic 
table. Among the given, is the most non-metallic. 
Valence electrons in (Group 17) = 7. 
Final Answer 
Number of valence electrons (most metallic, least metallic) = 5, 7. 
Option C: 5 and 7 
 
Q2: In period 4 of the periodic table, the elements with highest and lowest atomic radii are 
respectively. 
A: Rb & Br 
B: Na & Cl 
C: K & Br 
D: K & Se 
Answer: C 
Explanation: 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Classification of Elements and Periodicity  
 
(January 2026) 
 
Q1: Consider the elements N, P , O, S, Cl and F . The number of valence electrons present in the 
elements with most and least metallic character from the above list is respectively. 
A: 7 and 5 
B: 6 and 7 
C: 5 and 7 
D: 5 and 6 
Answer: C 
Explanation: 
Metallic character increases down a group and decreases from left to right across a period 
(NCERT trend). 
Given elements: 
Period 2: N, O, F 
Period 3: P , S, Cl  
1) Most metallic element 
In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to 
right: 
P > S > Cl 
So, P is the most metallic among the given elements. 
Valence electrons in (Group 15) = 5. 
2) Least metallic element 
Least metallic means most non-metallic, which is highest towards the top-right of the periodic 
table. Among the given, is the most non-metallic. 
Valence electrons in (Group 17) = 7. 
Final Answer 
Number of valence electrons (most metallic, least metallic) = 5, 7. 
Option C: 5 and 7 
 
Q2: In period 4 of the periodic table, the elements with highest and lowest atomic radii are 
respectively. 
A: Rb & Br 
B: Na & Cl 
C: K & Br 
D: K & Se 
Answer: C 
Explanation: 
In period 4 (from k to kr), the atomic radius decreases as we move from left to right because: 
Nuclear charge increases (Z increases), 
Electrons are added in the same principal shell (n = 4), 
So the effective nuclear charge on valence electrons increases and pulls them closer. 
Hence: 
The largest atomic radius is for the ?rst element of the period: K. 
The smallest atomic radius (as per NCERT trend, minimum at halogen) is for the halogen in that 
period: Br. 
So, the elements with highest and lowest atomic radii respectively are: K & Br  
Correct option: C 
 
Q3: The correct order of C, N, O and F in terms of second ionisation potential is 
A: C < N < F < O 
B: C < F < N < O 
C: F < N < C < O 
D: C < O < N < F 
Answer: A 
Explanation: 
To compare second ionization potential con?guration of mono-cation is observed 
 
 
Q4: Given below are two statements: 
Statement I: K > Mg > Al > B is the correct order in terms of metallic character. 
Statement II: Atomic radius is always greater than the ionic radius for any element. 
In the light of the above statements, choose the correct answer from the options given below 
A: Both Statement I and Statement II are false 
B: Statement I is false but Statement II is true 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: C 
Explanation: 
Metallic character means the tendency of an element to lose electrons and form a positive ion. 
In general (as per NCERT trends), metallic character is more in s-block elements than in p-block 
elements. 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Classification of Elements and Periodicity  
 
(January 2026) 
 
Q1: Consider the elements N, P , O, S, Cl and F . The number of valence electrons present in the 
elements with most and least metallic character from the above list is respectively. 
A: 7 and 5 
B: 6 and 7 
C: 5 and 7 
D: 5 and 6 
Answer: C 
Explanation: 
Metallic character increases down a group and decreases from left to right across a period 
(NCERT trend). 
Given elements: 
Period 2: N, O, F 
Period 3: P , S, Cl  
1) Most metallic element 
In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to 
right: 
P > S > Cl 
So, P is the most metallic among the given elements. 
Valence electrons in (Group 15) = 5. 
2) Least metallic element 
Least metallic means most non-metallic, which is highest towards the top-right of the periodic 
table. Among the given, is the most non-metallic. 
Valence electrons in (Group 17) = 7. 
Final Answer 
Number of valence electrons (most metallic, least metallic) = 5, 7. 
Option C: 5 and 7 
 
Q2: In period 4 of the periodic table, the elements with highest and lowest atomic radii are 
respectively. 
A: Rb & Br 
B: Na & Cl 
C: K & Br 
D: K & Se 
Answer: C 
Explanation: 
In period 4 (from k to kr), the atomic radius decreases as we move from left to right because: 
Nuclear charge increases (Z increases), 
Electrons are added in the same principal shell (n = 4), 
So the effective nuclear charge on valence electrons increases and pulls them closer. 
Hence: 
The largest atomic radius is for the ?rst element of the period: K. 
The smallest atomic radius (as per NCERT trend, minimum at halogen) is for the halogen in that 
period: Br. 
So, the elements with highest and lowest atomic radii respectively are: K & Br  
Correct option: C 
 
Q3: The correct order of C, N, O and F in terms of second ionisation potential is 
A: C < N < F < O 
B: C < F < N < O 
C: F < N < C < O 
D: C < O < N < F 
Answer: A 
Explanation: 
To compare second ionization potential con?guration of mono-cation is observed 
 
 
Q4: Given below are two statements: 
Statement I: K > Mg > Al > B is the correct order in terms of metallic character. 
Statement II: Atomic radius is always greater than the ionic radius for any element. 
In the light of the above statements, choose the correct answer from the options given below 
A: Both Statement I and Statement II are false 
B: Statement I is false but Statement II is true 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: C 
Explanation: 
Metallic character means the tendency of an element to lose electrons and form a positive ion. 
In general (as per NCERT trends), metallic character is more in s-block elements than in p-block 
elements. 
So, elements like K and Mg (s-block) show more metallic character than AI and B (p-block). 
Now, compare atomic radius and ionic radius: if an atom forms a cation, its ionic radius 
becomes smaller than the atomic radius. If an atom forms an anion, its ionic radius becomes 
larger than the atomic radius. 
Therefore, ionic radius is not always smaller than atomic radius. Speci?cally, anionic radius is 
greater than atomic radius, but cationic radius is always less than atomic radius for any 
element. 
 
Q5: 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 : 
A: Statement I is true but Statement II is false 
B: Statement I is false but Statement II is true 
C: Both Statement I and Statement II are true 
D: Both Statement I and Statement II are false 
Answer: C 
Explanation: 
Statement I: 
Na: 1s² 2s² 2p6 3s¹ 
After ?rst ionisation, Na? becomes 1s² 2s² 2p6 (noble gas con?guration). 
So, the second ionisation enthalpy of Na means removing an electron from a stable noble gas 
core, which needs very high energy. 
Mg: 1s² 2s² 2p6 3s² 
After ?rst ionisation, Mg? becomes 1s² 2s² 2p6 3s¹. 
So, the second ionisation enthalpy of Mg removes the remaining valence 3s electron, which 
needs less energy than removing from a noble gas core. 
Hence, 
IE2(Na) > IE2(Mg) 
So, Statement I is true. 
Statement II: 
O²? has 8 + 2 = 10 electrons, and F? has 9 + 1 = 10 electrons. 
So, O²? and F? are isoelectronic (same number of electrons). 
In an isoelectronic series, ionic radius decreases as nuclear charge (Z) increases, because 
higher Z pulls electrons more strongly. 
Here, Z(O) = 8 and Z(F) = 9. 
So, F? (higher Z) will be smaller than O²?. 
Therefore, r(O²?) > r(F?). 
So, Statement II is true. 
Correct option: Option C 
Both Statement I and Statement II are true. 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Classification of Elements and Periodicity  
 
(January 2026) 
 
Q1: Consider the elements N, P , O, S, Cl and F . The number of valence electrons present in the 
elements with most and least metallic character from the above list is respectively. 
A: 7 and 5 
B: 6 and 7 
C: 5 and 7 
D: 5 and 6 
Answer: C 
Explanation: 
Metallic character increases down a group and decreases from left to right across a period 
(NCERT trend). 
Given elements: 
Period 2: N, O, F 
Period 3: P , S, Cl  
1) Most metallic element 
In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to 
right: 
P > S > Cl 
So, P is the most metallic among the given elements. 
Valence electrons in (Group 15) = 5. 
2) Least metallic element 
Least metallic means most non-metallic, which is highest towards the top-right of the periodic 
table. Among the given, is the most non-metallic. 
Valence electrons in (Group 17) = 7. 
Final Answer 
Number of valence electrons (most metallic, least metallic) = 5, 7. 
Option C: 5 and 7 
 
Q2: In period 4 of the periodic table, the elements with highest and lowest atomic radii are 
respectively. 
A: Rb & Br 
B: Na & Cl 
C: K & Br 
D: K & Se 
Answer: C 
Explanation: 
In period 4 (from k to kr), the atomic radius decreases as we move from left to right because: 
Nuclear charge increases (Z increases), 
Electrons are added in the same principal shell (n = 4), 
So the effective nuclear charge on valence electrons increases and pulls them closer. 
Hence: 
The largest atomic radius is for the ?rst element of the period: K. 
The smallest atomic radius (as per NCERT trend, minimum at halogen) is for the halogen in that 
period: Br. 
So, the elements with highest and lowest atomic radii respectively are: K & Br  
Correct option: C 
 
Q3: The correct order of C, N, O and F in terms of second ionisation potential is 
A: C < N < F < O 
B: C < F < N < O 
C: F < N < C < O 
D: C < O < N < F 
Answer: A 
Explanation: 
To compare second ionization potential con?guration of mono-cation is observed 
 
 
Q4: Given below are two statements: 
Statement I: K > Mg > Al > B is the correct order in terms of metallic character. 
Statement II: Atomic radius is always greater than the ionic radius for any element. 
In the light of the above statements, choose the correct answer from the options given below 
A: Both Statement I and Statement II are false 
B: Statement I is false but Statement II is true 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: C 
Explanation: 
Metallic character means the tendency of an element to lose electrons and form a positive ion. 
In general (as per NCERT trends), metallic character is more in s-block elements than in p-block 
elements. 
So, elements like K and Mg (s-block) show more metallic character than AI and B (p-block). 
Now, compare atomic radius and ionic radius: if an atom forms a cation, its ionic radius 
becomes smaller than the atomic radius. If an atom forms an anion, its ionic radius becomes 
larger than the atomic radius. 
Therefore, ionic radius is not always smaller than atomic radius. Speci?cally, anionic radius is 
greater than atomic radius, but cationic radius is always less than atomic radius for any 
element. 
 
Q5: 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 : 
A: Statement I is true but Statement II is false 
B: Statement I is false but Statement II is true 
C: Both Statement I and Statement II are true 
D: Both Statement I and Statement II are false 
Answer: C 
Explanation: 
Statement I: 
Na: 1s² 2s² 2p6 3s¹ 
After ?rst ionisation, Na? becomes 1s² 2s² 2p6 (noble gas con?guration). 
So, the second ionisation enthalpy of Na means removing an electron from a stable noble gas 
core, which needs very high energy. 
Mg: 1s² 2s² 2p6 3s² 
After ?rst ionisation, Mg? becomes 1s² 2s² 2p6 3s¹. 
So, the second ionisation enthalpy of Mg removes the remaining valence 3s electron, which 
needs less energy than removing from a noble gas core. 
Hence, 
IE2(Na) > IE2(Mg) 
So, Statement I is true. 
Statement II: 
O²? has 8 + 2 = 10 electrons, and F? has 9 + 1 = 10 electrons. 
So, O²? and F? are isoelectronic (same number of electrons). 
In an isoelectronic series, ionic radius decreases as nuclear charge (Z) increases, because 
higher Z pulls electrons more strongly. 
Here, Z(O) = 8 and Z(F) = 9. 
So, F? (higher Z) will be smaller than O²?. 
Therefore, r(O²?) > r(F?). 
So, Statement II is true. 
Correct option: Option C 
Both Statement I and Statement II are true. 
 
Q6: Elements X and Y belong to Group 15. The difference between the electronegativity values 
of 'X' and phosphorus is higher than that of the difference between phosphorus and 'Y'. 'X' & 'Y' 
are respectively 
A: As & Bi 
B: Bi & N 
C: N & As 
D: As & Sb 
Answer: C 
Explanation: 
As per the NCERT textbook, electronegativity is the ability of an atom to attract a shared pair of 
electrons. For Group 15, the electronegativity decreases as we go down the group. 
Let's look at their approximate electronegativity values (on the Pauling scale): 
EN(N) = 3.0 
EN(P) = 2.1 
EN(As) = 2.0 
EN(Sb) = 1.9 
EN(Bi) = 1.9 
Notice a very important point: The drop in electronegativity from Nitrogen (N) to Phosphorus (P) 
is very large (3.0 – 2.1 = 0.9). However, for the elements below Phosphorus, the values decrease 
very slowly. 
The question states: "The difference between the electronegativity values of 'X' and phosphorus 
is higher than that of the difference between phosphorus and 'Y'." 
Let's write this mathematically. Let EN stand for electronegativity. The absolute difference is 
what matters. 
|EN(X) – EN(P)| > |EN(P) – EN(Y)| 
This means we are looking for an element 'X' whose electronegativity is very different from 
Phosphorus, and an element 'Y' whose electronegativity is very close to Phosphorus. 
Now, let's check each option using the electronegativity values. 
Option A: X = As & Y = Bi 
Difference 1: |EN(X) – EN(P)| = |EN(As) – EN(P)| = |2.0 – 2.1| = 0.1 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(Bi)| = |2.1 – 1.9| = 0.2 
Is 0.1 > 0.2? No. So, this option is incorrect. 
Option B: X = Bi & Y = N 
Difference 1: |EN(X) – EN(P)| = |EN(Bi) – EN(P)| = |1.9 – 2.1| = 0.2 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(N)| = |2.1 – 3.0| = 0.9 
Is 0.2 > 0.9? No. So, this option is incorrect. 
Option C: X = N & Y = As 
Difference 1: |EN(X) – EN(P)| = |EN(N) – EN(P)| = |3.0 – 2.1| = 0.9 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(As)| = |2.1 – 2.0| = 0.1 
Is 0.9 > 0.1? Yes. This matches the condition given in the question. 
Option D: X = As & Y = Sb 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Classification of Elements and Periodicity  
 
(January 2026) 
 
Q1: Consider the elements N, P , O, S, Cl and F . The number of valence electrons present in the 
elements with most and least metallic character from the above list is respectively. 
A: 7 and 5 
B: 6 and 7 
C: 5 and 7 
D: 5 and 6 
Answer: C 
Explanation: 
Metallic character increases down a group and decreases from left to right across a period 
(NCERT trend). 
Given elements: 
Period 2: N, O, F 
Period 3: P , S, Cl  
1) Most metallic element 
In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to 
right: 
P > S > Cl 
So, P is the most metallic among the given elements. 
Valence electrons in (Group 15) = 5. 
2) Least metallic element 
Least metallic means most non-metallic, which is highest towards the top-right of the periodic 
table. Among the given, is the most non-metallic. 
Valence electrons in (Group 17) = 7. 
Final Answer 
Number of valence electrons (most metallic, least metallic) = 5, 7. 
Option C: 5 and 7 
 
Q2: In period 4 of the periodic table, the elements with highest and lowest atomic radii are 
respectively. 
A: Rb & Br 
B: Na & Cl 
C: K & Br 
D: K & Se 
Answer: C 
Explanation: 
In period 4 (from k to kr), the atomic radius decreases as we move from left to right because: 
Nuclear charge increases (Z increases), 
Electrons are added in the same principal shell (n = 4), 
So the effective nuclear charge on valence electrons increases and pulls them closer. 
Hence: 
The largest atomic radius is for the ?rst element of the period: K. 
The smallest atomic radius (as per NCERT trend, minimum at halogen) is for the halogen in that 
period: Br. 
So, the elements with highest and lowest atomic radii respectively are: K & Br  
Correct option: C 
 
Q3: The correct order of C, N, O and F in terms of second ionisation potential is 
A: C < N < F < O 
B: C < F < N < O 
C: F < N < C < O 
D: C < O < N < F 
Answer: A 
Explanation: 
To compare second ionization potential con?guration of mono-cation is observed 
 
 
Q4: Given below are two statements: 
Statement I: K > Mg > Al > B is the correct order in terms of metallic character. 
Statement II: Atomic radius is always greater than the ionic radius for any element. 
In the light of the above statements, choose the correct answer from the options given below 
A: Both Statement I and Statement II are false 
B: Statement I is false but Statement II is true 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: C 
Explanation: 
Metallic character means the tendency of an element to lose electrons and form a positive ion. 
In general (as per NCERT trends), metallic character is more in s-block elements than in p-block 
elements. 
So, elements like K and Mg (s-block) show more metallic character than AI and B (p-block). 
Now, compare atomic radius and ionic radius: if an atom forms a cation, its ionic radius 
becomes smaller than the atomic radius. If an atom forms an anion, its ionic radius becomes 
larger than the atomic radius. 
Therefore, ionic radius is not always smaller than atomic radius. Speci?cally, anionic radius is 
greater than atomic radius, but cationic radius is always less than atomic radius for any 
element. 
 
Q5: 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 : 
A: Statement I is true but Statement II is false 
B: Statement I is false but Statement II is true 
C: Both Statement I and Statement II are true 
D: Both Statement I and Statement II are false 
Answer: C 
Explanation: 
Statement I: 
Na: 1s² 2s² 2p6 3s¹ 
After ?rst ionisation, Na? becomes 1s² 2s² 2p6 (noble gas con?guration). 
So, the second ionisation enthalpy of Na means removing an electron from a stable noble gas 
core, which needs very high energy. 
Mg: 1s² 2s² 2p6 3s² 
After ?rst ionisation, Mg? becomes 1s² 2s² 2p6 3s¹. 
So, the second ionisation enthalpy of Mg removes the remaining valence 3s electron, which 
needs less energy than removing from a noble gas core. 
Hence, 
IE2(Na) > IE2(Mg) 
So, Statement I is true. 
Statement II: 
O²? has 8 + 2 = 10 electrons, and F? has 9 + 1 = 10 electrons. 
So, O²? and F? are isoelectronic (same number of electrons). 
In an isoelectronic series, ionic radius decreases as nuclear charge (Z) increases, because 
higher Z pulls electrons more strongly. 
Here, Z(O) = 8 and Z(F) = 9. 
So, F? (higher Z) will be smaller than O²?. 
Therefore, r(O²?) > r(F?). 
So, Statement II is true. 
Correct option: Option C 
Both Statement I and Statement II are true. 
 
Q6: Elements X and Y belong to Group 15. The difference between the electronegativity values 
of 'X' and phosphorus is higher than that of the difference between phosphorus and 'Y'. 'X' & 'Y' 
are respectively 
A: As & Bi 
B: Bi & N 
C: N & As 
D: As & Sb 
Answer: C 
Explanation: 
As per the NCERT textbook, electronegativity is the ability of an atom to attract a shared pair of 
electrons. For Group 15, the electronegativity decreases as we go down the group. 
Let's look at their approximate electronegativity values (on the Pauling scale): 
EN(N) = 3.0 
EN(P) = 2.1 
EN(As) = 2.0 
EN(Sb) = 1.9 
EN(Bi) = 1.9 
Notice a very important point: The drop in electronegativity from Nitrogen (N) to Phosphorus (P) 
is very large (3.0 – 2.1 = 0.9). However, for the elements below Phosphorus, the values decrease 
very slowly. 
The question states: "The difference between the electronegativity values of 'X' and phosphorus 
is higher than that of the difference between phosphorus and 'Y'." 
Let's write this mathematically. Let EN stand for electronegativity. The absolute difference is 
what matters. 
|EN(X) – EN(P)| > |EN(P) – EN(Y)| 
This means we are looking for an element 'X' whose electronegativity is very different from 
Phosphorus, and an element 'Y' whose electronegativity is very close to Phosphorus. 
Now, let's check each option using the electronegativity values. 
Option A: X = As & Y = Bi 
Difference 1: |EN(X) – EN(P)| = |EN(As) – EN(P)| = |2.0 – 2.1| = 0.1 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(Bi)| = |2.1 – 1.9| = 0.2 
Is 0.1 > 0.2? No. So, this option is incorrect. 
Option B: X = Bi & Y = N 
Difference 1: |EN(X) – EN(P)| = |EN(Bi) – EN(P)| = |1.9 – 2.1| = 0.2 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(N)| = |2.1 – 3.0| = 0.9 
Is 0.2 > 0.9? No. So, this option is incorrect. 
Option C: X = N & Y = As 
Difference 1: |EN(X) – EN(P)| = |EN(N) – EN(P)| = |3.0 – 2.1| = 0.9 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(As)| = |2.1 – 2.0| = 0.1 
Is 0.9 > 0.1? Yes. This matches the condition given in the question. 
Option D: X = As & Y = Sb 
Difference 1: |EN(X) – EN(P)| = |EN(As) – EN(P)| = |2.0 – 2.1| = 0.1 
Difference 2: |EN(P) – EN(Y)| = |EN(P) – EN(Sb)| = |2.1 – 1.9| = 0.2 
Is 0.1 > 0.2? No. So, this option is incorrect. 
Therefore, X is Nitrogen (N) and Y is Arsenic (As). 
 
Q7: The correct trend in the ?rst ionization enthalpies of the elements in the 3
rd
 period of 
periodic table is : 
A: Si < S < Al < P < Cl 
B: Al < Si < S < P < Cl 
C: S < Si < Al < P < Cl 
D: Al < S < P < Si < Cl 
Answer: B 
Explanation: 
In general on moving from left to right in a period ionization energy increases as Z
eff
  increases. 
 
 
Q8: Given below are two statements : 
Statement I : C < O < N < F   the correct order in terms of ?rst ionization enthalpy values. 
Statement II : S > Se > Te > Po > O is the correct order in terms of the magnitude of electron 
gain enthalpy values. 
In the light of the above statements, choose the correct answer from the options given below : 
A: Both Statement I and Statement II are false 
B: Statement I is false but Statement II is true 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Statement I (First ionization enthalpy) 
Across a period, ?rst ionization enthalpy generally increases due to increasing nuclear charge 
and decreasing atomic size. 
But there is an important NCERT exception: 
Nitrogen (N) has a half-?lled con?guration 2p³, which is extra stable. 
Oxygen (O) has con?guration 2p4, where one p-orbital contains a pair of electrons, causing extra 
electron–electron repulsion, so it is easier to remove one electron. 
So, 
IE1(O) < IE1(N)