JEE Main Previous Year Questions (2025): Classification of Elements and Periodicity in Properties

 Page 1


JEE Main Previous Year Questions 
(2025): Classification of Elements 
and Periodicity in Properties 
Q1: Which of the following electronegativity order is incorrect? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. S<Cl<O<F 
B. Al<Si<C<N 
C. Al<Mg<B<N 
D. Mg<Be<B<N 
Ans: C 
Solution: 
Let's compare the known electronegativities (Pauling scale) of the elements in each option: 
?? ~3.98 
O ~3.44 
?? ~3.04 
Cl~3.16 
?? ~2.55 
B ~2.04 
S ~2.58 (sometimes listed as 2.5-2.58) 
???? ~1.90 
AI~1.61 
???? ~1.57 
Mg ~ 1.31 
Now, check each statement: 
Option A: S<Cl<O<F 
S(2.58)<Cl(3.16)<O(3.44)<F(3.98) 
This order is correct. 
Option B: Al<Si<C<N 
Al(1.61)<Si(1.90)<C(2.55)<N(3.04) 
This order is correct. 
Option C: Al<Mg<B<N 
Actual electronegativities are Al (1.61) and Mg (1.31). 
The statement says "Al <Mg ," which would mean 1.61<1.31, which is wrong. 
Therefore, Option C is incorrect. 
Option D: Mg<Be<B<N 
Mg(1.31)<Be(1.57)<B(2.04)<N(3.04) 
This order is correct. 
Ans: Option C is the incorrect order. 
Page 2


JEE Main Previous Year Questions 
(2025): Classification of Elements 
and Periodicity in Properties 
Q1: Which of the following electronegativity order is incorrect? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. S<Cl<O<F 
B. Al<Si<C<N 
C. Al<Mg<B<N 
D. Mg<Be<B<N 
Ans: C 
Solution: 
Let's compare the known electronegativities (Pauling scale) of the elements in each option: 
?? ~3.98 
O ~3.44 
?? ~3.04 
Cl~3.16 
?? ~2.55 
B ~2.04 
S ~2.58 (sometimes listed as 2.5-2.58) 
???? ~1.90 
AI~1.61 
???? ~1.57 
Mg ~ 1.31 
Now, check each statement: 
Option A: S<Cl<O<F 
S(2.58)<Cl(3.16)<O(3.44)<F(3.98) 
This order is correct. 
Option B: Al<Si<C<N 
Al(1.61)<Si(1.90)<C(2.55)<N(3.04) 
This order is correct. 
Option C: Al<Mg<B<N 
Actual electronegativities are Al (1.61) and Mg (1.31). 
The statement says "Al <Mg ," which would mean 1.61<1.31, which is wrong. 
Therefore, Option C is incorrect. 
Option D: Mg<Be<B<N 
Mg(1.31)<Be(1.57)<B(2.04)<N(3.04) 
This order is correct. 
Ans: Option C is the incorrect order. 
Q2: Match List-I with List-II. 
 
 
 List - I  List - II 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 (I) Ionisation Enthalpy 
(B) B<C<O<N (II) Metallic character 
(C) B<Al<Mg<K (III) Electronegativity 
(D) Si<P<S<Cl (IV) Ionic radii 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (A)-(IV), (B)-(I), (C)- (II), (D)-(III) 
B. (A)-(III), (B)-(IV), (C)- (II), (D)-(I) 
C. (A)-(II), (B)-(III), (C)- (IV), (D)-(I) 
D. (A)-(IV), (B)-(I), (C)- (III), (D)-(II) 
Ans: A 
Solution: 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 
All four ions ( Al
3+
,Mg
2+
,Na
+
,F
-
) are isoelectronic (each has 10 electrons). For isoelectronic 
species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest 
ion here is Al
3+
(Z=13) and the largest is F
-
(Z=9). Thus, this ordering is one of increasing 
ionic radius. 
(?? )? (IV) Ionic radii 
(B) B < C < O < N 
Check first ionization enthalpies ( IE
1
 ): 
B:˜801\, kJ/mol 
C:˜1086\, kJ/mol 
O:˜1314\, kJ/mol 
N:˜1402\, kJ/mol 
Hence, the order of increasing IE
1
 is 
?? <?? <?? <?? . 
This matches the given sequence exactly. 
(?? )? (I) Ionisation enthalpy 
(C) B<Al<Mg<K 
Consider metallic character (the tendency to lose electrons easily, show metallic properties). 
Across a period (left to right), metallic character decreases; down a group, it increases. 
B (metalloid) has the least metallic character here. 
Al (group 13 metal) is more metallic than B. 
Page 3


JEE Main Previous Year Questions 
(2025): Classification of Elements 
and Periodicity in Properties 
Q1: Which of the following electronegativity order is incorrect? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. S<Cl<O<F 
B. Al<Si<C<N 
C. Al<Mg<B<N 
D. Mg<Be<B<N 
Ans: C 
Solution: 
Let's compare the known electronegativities (Pauling scale) of the elements in each option: 
?? ~3.98 
O ~3.44 
?? ~3.04 
Cl~3.16 
?? ~2.55 
B ~2.04 
S ~2.58 (sometimes listed as 2.5-2.58) 
???? ~1.90 
AI~1.61 
???? ~1.57 
Mg ~ 1.31 
Now, check each statement: 
Option A: S<Cl<O<F 
S(2.58)<Cl(3.16)<O(3.44)<F(3.98) 
This order is correct. 
Option B: Al<Si<C<N 
Al(1.61)<Si(1.90)<C(2.55)<N(3.04) 
This order is correct. 
Option C: Al<Mg<B<N 
Actual electronegativities are Al (1.61) and Mg (1.31). 
The statement says "Al <Mg ," which would mean 1.61<1.31, which is wrong. 
Therefore, Option C is incorrect. 
Option D: Mg<Be<B<N 
Mg(1.31)<Be(1.57)<B(2.04)<N(3.04) 
This order is correct. 
Ans: Option C is the incorrect order. 
Q2: Match List-I with List-II. 
 
 
 List - I  List - II 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 (I) Ionisation Enthalpy 
(B) B<C<O<N (II) Metallic character 
(C) B<Al<Mg<K (III) Electronegativity 
(D) Si<P<S<Cl (IV) Ionic radii 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (A)-(IV), (B)-(I), (C)- (II), (D)-(III) 
B. (A)-(III), (B)-(IV), (C)- (II), (D)-(I) 
C. (A)-(II), (B)-(III), (C)- (IV), (D)-(I) 
D. (A)-(IV), (B)-(I), (C)- (III), (D)-(II) 
Ans: A 
Solution: 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 
All four ions ( Al
3+
,Mg
2+
,Na
+
,F
-
) are isoelectronic (each has 10 electrons). For isoelectronic 
species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest 
ion here is Al
3+
(Z=13) and the largest is F
-
(Z=9). Thus, this ordering is one of increasing 
ionic radius. 
(?? )? (IV) Ionic radii 
(B) B < C < O < N 
Check first ionization enthalpies ( IE
1
 ): 
B:˜801\, kJ/mol 
C:˜1086\, kJ/mol 
O:˜1314\, kJ/mol 
N:˜1402\, kJ/mol 
Hence, the order of increasing IE
1
 is 
?? <?? <?? <?? . 
This matches the given sequence exactly. 
(?? )? (I) Ionisation enthalpy 
(C) B<Al<Mg<K 
Consider metallic character (the tendency to lose electrons easily, show metallic properties). 
Across a period (left to right), metallic character decreases; down a group, it increases. 
B (metalloid) has the least metallic character here. 
Al (group 13 metal) is more metallic than B. 
Mg (group 2 metal) is typically more metallic than Al. 
?? (group 1 metal) is the most metallic among these. 
Thus, B<Al<Mg<K is an order of increasing metallic character. 
(?? )? (II) Metallic character 
(D) Si<P<S<Cl 
Check electronegativities: 
Si:˜1.90 
P:˜2.19 
S:˜2.58 
Cl:˜3.16 
They increase in the order 
Si<P<S<Cl, 
which matches the given sequence for increasing electronegativity. 
(?? )? (III) Electronegativity 
Final Matching 
(?? )?(???? ),(?? )?(?? ),(?? )?(???? ),(?? )?(?????? ). 
Looking at the choices given: 
Option A: (?? )-(???? ),(?? )-(?? ),(?? )-(???? ),(?? )-(?????? ) 
This is exactly what we found. 
Ans: Option A 
Q3: Given below are two statements : 
Statement (I): An element in the extreme left of the periodic table forms acidic oxides. 
Statement (II): Acid is formed during the reaction between water and oxide of a 
reactive element present in the extreme right of the periodic table. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are false 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: D 
Solution: 
First, let us re-state the two statements clearly: 
Statement (I): An element in the extreme left of the periodic table forms acidic oxides. 
Statement (II): Acid is formed during the reaction between water and oxide of a reactive 
element present in the extreme right of the periodic table. 
Analyzing Statement (I) 
Page 4


JEE Main Previous Year Questions 
(2025): Classification of Elements 
and Periodicity in Properties 
Q1: Which of the following electronegativity order is incorrect? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. S<Cl<O<F 
B. Al<Si<C<N 
C. Al<Mg<B<N 
D. Mg<Be<B<N 
Ans: C 
Solution: 
Let's compare the known electronegativities (Pauling scale) of the elements in each option: 
?? ~3.98 
O ~3.44 
?? ~3.04 
Cl~3.16 
?? ~2.55 
B ~2.04 
S ~2.58 (sometimes listed as 2.5-2.58) 
???? ~1.90 
AI~1.61 
???? ~1.57 
Mg ~ 1.31 
Now, check each statement: 
Option A: S<Cl<O<F 
S(2.58)<Cl(3.16)<O(3.44)<F(3.98) 
This order is correct. 
Option B: Al<Si<C<N 
Al(1.61)<Si(1.90)<C(2.55)<N(3.04) 
This order is correct. 
Option C: Al<Mg<B<N 
Actual electronegativities are Al (1.61) and Mg (1.31). 
The statement says "Al <Mg ," which would mean 1.61<1.31, which is wrong. 
Therefore, Option C is incorrect. 
Option D: Mg<Be<B<N 
Mg(1.31)<Be(1.57)<B(2.04)<N(3.04) 
This order is correct. 
Ans: Option C is the incorrect order. 
Q2: Match List-I with List-II. 
 
 
 List - I  List - II 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 (I) Ionisation Enthalpy 
(B) B<C<O<N (II) Metallic character 
(C) B<Al<Mg<K (III) Electronegativity 
(D) Si<P<S<Cl (IV) Ionic radii 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (A)-(IV), (B)-(I), (C)- (II), (D)-(III) 
B. (A)-(III), (B)-(IV), (C)- (II), (D)-(I) 
C. (A)-(II), (B)-(III), (C)- (IV), (D)-(I) 
D. (A)-(IV), (B)-(I), (C)- (III), (D)-(II) 
Ans: A 
Solution: 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 
All four ions ( Al
3+
,Mg
2+
,Na
+
,F
-
) are isoelectronic (each has 10 electrons). For isoelectronic 
species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest 
ion here is Al
3+
(Z=13) and the largest is F
-
(Z=9). Thus, this ordering is one of increasing 
ionic radius. 
(?? )? (IV) Ionic radii 
(B) B < C < O < N 
Check first ionization enthalpies ( IE
1
 ): 
B:˜801\, kJ/mol 
C:˜1086\, kJ/mol 
O:˜1314\, kJ/mol 
N:˜1402\, kJ/mol 
Hence, the order of increasing IE
1
 is 
?? <?? <?? <?? . 
This matches the given sequence exactly. 
(?? )? (I) Ionisation enthalpy 
(C) B<Al<Mg<K 
Consider metallic character (the tendency to lose electrons easily, show metallic properties). 
Across a period (left to right), metallic character decreases; down a group, it increases. 
B (metalloid) has the least metallic character here. 
Al (group 13 metal) is more metallic than B. 
Mg (group 2 metal) is typically more metallic than Al. 
?? (group 1 metal) is the most metallic among these. 
Thus, B<Al<Mg<K is an order of increasing metallic character. 
(?? )? (II) Metallic character 
(D) Si<P<S<Cl 
Check electronegativities: 
Si:˜1.90 
P:˜2.19 
S:˜2.58 
Cl:˜3.16 
They increase in the order 
Si<P<S<Cl, 
which matches the given sequence for increasing electronegativity. 
(?? )? (III) Electronegativity 
Final Matching 
(?? )?(???? ),(?? )?(?? ),(?? )?(???? ),(?? )?(?????? ). 
Looking at the choices given: 
Option A: (?? )-(???? ),(?? )-(?? ),(?? )-(???? ),(?? )-(?????? ) 
This is exactly what we found. 
Ans: Option A 
Q3: Given below are two statements : 
Statement (I): An element in the extreme left of the periodic table forms acidic oxides. 
Statement (II): Acid is formed during the reaction between water and oxide of a 
reactive element present in the extreme right of the periodic table. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are false 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: D 
Solution: 
First, let us re-state the two statements clearly: 
Statement (I): An element in the extreme left of the periodic table forms acidic oxides. 
Statement (II): Acid is formed during the reaction between water and oxide of a reactive 
element present in the extreme right of the periodic table. 
Analyzing Statement (I) 
The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline 
earth metals (Group 2). 
These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 
elements) oxides, not acidic oxides. 
For example, ????
?? ?? ,?? ?? ?? ,?????? ,?????? , etc., all from basic solutions (e.g., Na
2
O+H
2
O?2NaOH 
). 
Hence, Statement (I)-that an element in the extreme left forms acidic oxides-is false. 
Analyzing Statement (II) 
The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and 
noble gases in Group 18). 
Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic. 
When these oxides dissolve in water, they typically form acids. 
Example: SO
3
+H
2
O?H
2
SO
4
 (sulfuric acid) 
Example: P
2
O
5
+3H
2
O?2H
3
PO
4
 (phosphoric acid) 
Example: Cl
2
O
7
+H
2
O?2HClO
4
 (perchloric acid) 
Thus, Statement (II)-that acid is formed when water reacts with an oxide of a reactive element 
in the extreme right-is true. 
Conclusion 
Statement (I) is false. 
Statement (II) is true. 
Therefore, the correct choice is: 
(D) Statement (I) is false but Statement (II) is true. 
Q4: The element that does not belong to the same period of the remaining elements 
(modern periodic table) is: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. Platinum 
B. Osmium 
C. Iridium 
D. Palladium 
Ans: D 
Solution: 
Let's analyze the periods for each element: 
Platinum (Pt) has an atomic number of 78 , which places it in period 6 . 
Osmium (Os) has an atomic number of 76, also in period 6. 
Iridium (Ir) has an atomic number of 77 , which is in period 6 as well. 
Palladium (Pd) has an atomic number of 46, placing it in period 5. 
Since Palladium is in period 5 while the other elements are in period 6 , the element that does 
not belong to the same period is: 
Option D: Palladium. 
Page 5


JEE Main Previous Year Questions 
(2025): Classification of Elements 
and Periodicity in Properties 
Q1: Which of the following electronegativity order is incorrect? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. S<Cl<O<F 
B. Al<Si<C<N 
C. Al<Mg<B<N 
D. Mg<Be<B<N 
Ans: C 
Solution: 
Let's compare the known electronegativities (Pauling scale) of the elements in each option: 
?? ~3.98 
O ~3.44 
?? ~3.04 
Cl~3.16 
?? ~2.55 
B ~2.04 
S ~2.58 (sometimes listed as 2.5-2.58) 
???? ~1.90 
AI~1.61 
???? ~1.57 
Mg ~ 1.31 
Now, check each statement: 
Option A: S<Cl<O<F 
S(2.58)<Cl(3.16)<O(3.44)<F(3.98) 
This order is correct. 
Option B: Al<Si<C<N 
Al(1.61)<Si(1.90)<C(2.55)<N(3.04) 
This order is correct. 
Option C: Al<Mg<B<N 
Actual electronegativities are Al (1.61) and Mg (1.31). 
The statement says "Al <Mg ," which would mean 1.61<1.31, which is wrong. 
Therefore, Option C is incorrect. 
Option D: Mg<Be<B<N 
Mg(1.31)<Be(1.57)<B(2.04)<N(3.04) 
This order is correct. 
Ans: Option C is the incorrect order. 
Q2: Match List-I with List-II. 
 
 
 List - I  List - II 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 (I) Ionisation Enthalpy 
(B) B<C<O<N (II) Metallic character 
(C) B<Al<Mg<K (III) Electronegativity 
(D) Si<P<S<Cl (IV) Ionic radii 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (A)-(IV), (B)-(I), (C)- (II), (D)-(III) 
B. (A)-(III), (B)-(IV), (C)- (II), (D)-(I) 
C. (A)-(II), (B)-(III), (C)- (IV), (D)-(I) 
D. (A)-(IV), (B)-(I), (C)- (III), (D)-(II) 
Ans: A 
Solution: 
(A) Al
3+
<Mg
2+
<Na
+
<F
-
 
All four ions ( Al
3+
,Mg
2+
,Na
+
,F
-
) are isoelectronic (each has 10 electrons). For isoelectronic 
species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest 
ion here is Al
3+
(Z=13) and the largest is F
-
(Z=9). Thus, this ordering is one of increasing 
ionic radius. 
(?? )? (IV) Ionic radii 
(B) B < C < O < N 
Check first ionization enthalpies ( IE
1
 ): 
B:˜801\, kJ/mol 
C:˜1086\, kJ/mol 
O:˜1314\, kJ/mol 
N:˜1402\, kJ/mol 
Hence, the order of increasing IE
1
 is 
?? <?? <?? <?? . 
This matches the given sequence exactly. 
(?? )? (I) Ionisation enthalpy 
(C) B<Al<Mg<K 
Consider metallic character (the tendency to lose electrons easily, show metallic properties). 
Across a period (left to right), metallic character decreases; down a group, it increases. 
B (metalloid) has the least metallic character here. 
Al (group 13 metal) is more metallic than B. 
Mg (group 2 metal) is typically more metallic than Al. 
?? (group 1 metal) is the most metallic among these. 
Thus, B<Al<Mg<K is an order of increasing metallic character. 
(?? )? (II) Metallic character 
(D) Si<P<S<Cl 
Check electronegativities: 
Si:˜1.90 
P:˜2.19 
S:˜2.58 
Cl:˜3.16 
They increase in the order 
Si<P<S<Cl, 
which matches the given sequence for increasing electronegativity. 
(?? )? (III) Electronegativity 
Final Matching 
(?? )?(???? ),(?? )?(?? ),(?? )?(???? ),(?? )?(?????? ). 
Looking at the choices given: 
Option A: (?? )-(???? ),(?? )-(?? ),(?? )-(???? ),(?? )-(?????? ) 
This is exactly what we found. 
Ans: Option A 
Q3: Given below are two statements : 
Statement (I): An element in the extreme left of the periodic table forms acidic oxides. 
Statement (II): Acid is formed during the reaction between water and oxide of a 
reactive element present in the extreme right of the periodic table. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are false 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: D 
Solution: 
First, let us re-state the two statements clearly: 
Statement (I): An element in the extreme left of the periodic table forms acidic oxides. 
Statement (II): Acid is formed during the reaction between water and oxide of a reactive 
element present in the extreme right of the periodic table. 
Analyzing Statement (I) 
The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline 
earth metals (Group 2). 
These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 
elements) oxides, not acidic oxides. 
For example, ????
?? ?? ,?? ?? ?? ,?????? ,?????? , etc., all from basic solutions (e.g., Na
2
O+H
2
O?2NaOH 
). 
Hence, Statement (I)-that an element in the extreme left forms acidic oxides-is false. 
Analyzing Statement (II) 
The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and 
noble gases in Group 18). 
Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic. 
When these oxides dissolve in water, they typically form acids. 
Example: SO
3
+H
2
O?H
2
SO
4
 (sulfuric acid) 
Example: P
2
O
5
+3H
2
O?2H
3
PO
4
 (phosphoric acid) 
Example: Cl
2
O
7
+H
2
O?2HClO
4
 (perchloric acid) 
Thus, Statement (II)-that acid is formed when water reacts with an oxide of a reactive element 
in the extreme right-is true. 
Conclusion 
Statement (I) is false. 
Statement (II) is true. 
Therefore, the correct choice is: 
(D) Statement (I) is false but Statement (II) is true. 
Q4: The element that does not belong to the same period of the remaining elements 
(modern periodic table) is: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. Platinum 
B. Osmium 
C. Iridium 
D. Palladium 
Ans: D 
Solution: 
Let's analyze the periods for each element: 
Platinum (Pt) has an atomic number of 78 , which places it in period 6 . 
Osmium (Os) has an atomic number of 76, also in period 6. 
Iridium (Ir) has an atomic number of 77 , which is in period 6 as well. 
Palladium (Pd) has an atomic number of 46, placing it in period 5. 
Since Palladium is in period 5 while the other elements are in period 6 , the element that does 
not belong to the same period is: 
Option D: Palladium. 
Q5: Given below are the atomic numbers of some group 14 elements. The atomic 
number of the element with lowest melting point is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 14 
B. 50 
C. 6 
D. 82 
Ans: B 
Solution: 
Order of M.P. of group 14: C>Si>Ge>Pb>Sn 
Element M.P. ( 
°
C) 
?? =6=C 3730 
?? =14=Si 1410 
?? =32=Ge 937 
?? =50=Sn 232 
?? =82=Pb 327 
 
Q6: Which of the following statements are NOT true about the periodic table? 
A. The properties of elements are function of atomic weights. 
B. The properties of elements are function of atomic numbers. 
C. Elements having similar outer electronic configurations are arranged in same 
period. 
D. An element's location reflects the quantum numbers of the last filled orbital. 
E. The number of elements in a period is same as the number of atomic orbitals 
available in energy level that is being filled. 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. A, C and E Only 
B. A and E Only 
C. B, C and E Only 
D. D and E Only 
Ans: A 
Solution: