JEE Main Previous Year Questions (2025): d – and f – Block Elements

 Page 1


JEE Main Previous Year Questions 
(2025): d – and f – Block Elements 
Q1: Niobium (???? ) and ruthenium (???? ) have " ?? " and " ?? " number of electrons in 
their respective ???? orbitals. The value of ?? + ?? is ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 11 
Solution: 
We need to determine the number of electrons each element has in its 4d orbitals and then add 
them together. 
1. Niobium ( Nb, Z = 41 ) 
The ground-state electronic configuration of niobium is: 
[Kr]4?? 4
5?? 1
. 
Hence, Niobium has 4 electrons in its 4d orbitals. 
2. Ruthenium ( Ru, Z = 44 ) 
The ground-state electronic configuration of ruthenium is: 
[Kr]4?? 7
5?? 1
. 
Hence, Ruthenium has 7 electrons in its 4d orbitals. 
3. Sum of 4d Electrons 
?? = 4 ( for Nb), ?? = 7 ( for Ru). 
Therefore, 
?? + ?? = 4 + 7 = 11. 
Ans: 11 
Q2: The molar mass of the water insoluble product formed from the fusion of 
chromite ore (???????? ?? ?? ?? ) with ????
?? ????
?? in presence of ?? ?? is ____ ?? ?????? -?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 160 
Solution: 
FeC
2
O
4
+ Na
2
CO
3
?
O
2
 
? 
The chemical reaction can be written as 
The insoluble product is Fe
2
O
3
. 
Molar mass of Fe
2
O
3
 
Fe (Atomic mass) ? 55.845amu 
Molar mass = 55.845 g mol
-1
 
O (Atomic mass) ? 15.999amu 
Page 2


JEE Main Previous Year Questions 
(2025): d – and f – Block Elements 
Q1: Niobium (???? ) and ruthenium (???? ) have " ?? " and " ?? " number of electrons in 
their respective ???? orbitals. The value of ?? + ?? is ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 11 
Solution: 
We need to determine the number of electrons each element has in its 4d orbitals and then add 
them together. 
1. Niobium ( Nb, Z = 41 ) 
The ground-state electronic configuration of niobium is: 
[Kr]4?? 4
5?? 1
. 
Hence, Niobium has 4 electrons in its 4d orbitals. 
2. Ruthenium ( Ru, Z = 44 ) 
The ground-state electronic configuration of ruthenium is: 
[Kr]4?? 7
5?? 1
. 
Hence, Ruthenium has 7 electrons in its 4d orbitals. 
3. Sum of 4d Electrons 
?? = 4 ( for Nb), ?? = 7 ( for Ru). 
Therefore, 
?? + ?? = 4 + 7 = 11. 
Ans: 11 
Q2: The molar mass of the water insoluble product formed from the fusion of 
chromite ore (???????? ?? ?? ?? ) with ????
?? ????
?? in presence of ?? ?? is ____ ?? ?????? -?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 160 
Solution: 
FeC
2
O
4
+ Na
2
CO
3
?
O
2
 
? 
The chemical reaction can be written as 
The insoluble product is Fe
2
O
3
. 
Molar mass of Fe
2
O
3
 
Fe (Atomic mass) ? 55.845amu 
Molar mass = 55.845 g mol
-1
 
O (Atomic mass) ? 15.999amu 
Molar mass = 15.999 g mol
-1
 
Molar mass of ?? ?? 2
?? 3
= 55.845 g mol
-1
× 2 + 15.999 g mol
-1
× 3 
= (111.69 + 47.997)gmol
-1
 
= 159.687 g mol
-1
˜ 160 g mol
-1
 
Q3: Consider the following reactions 
?? + ???????? Little 
 amount 
+ ?? ?? ????
?? ? ?????? ?? ????
?? + Side Products 
?????? ?? ????
?? (vapour) 
+ ???????? ? ?? + ???????? + ?? ?? ?? 
?? + ?? +
? ?? + ?? ?? ?? 
The number of terminal ' ?? ' present in the compound ' C ' is ____ 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 6 
Solution: 
Cr
2
O
7
2-
+ NaCl + H
2
SO
4
? CrO
2
Cl
2
 
CrO
2
Cl
2
 (Vapour) +NaOH? 
Na
2
CrO
4
+ NaCl + H
2
O 
Na
2
CrO
4
+ H
?
? Na
2
Cr
2
O
7
+ H
2
O 
(C) 
Na
2
Cr
2
O
7
? 2Na
?
+ Cr
2
O
7
2-
 
 
No of terminal "O" = 6 
Q4: Among, ???? , ???? , ???? and Cu , identify the element with highest enthalpy of 
atomisation. The spin only magnetic moment value of that element in its +2 oxidation 
state is ____ BM (in nearest integer). 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 4 
Solution: 
 Sc Mn Co Cu 
Enthalpy of 
Atomisation (kJ/mole) 
326 281 425 339 
 
Page 3


JEE Main Previous Year Questions 
(2025): d – and f – Block Elements 
Q1: Niobium (???? ) and ruthenium (???? ) have " ?? " and " ?? " number of electrons in 
their respective ???? orbitals. The value of ?? + ?? is ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 11 
Solution: 
We need to determine the number of electrons each element has in its 4d orbitals and then add 
them together. 
1. Niobium ( Nb, Z = 41 ) 
The ground-state electronic configuration of niobium is: 
[Kr]4?? 4
5?? 1
. 
Hence, Niobium has 4 electrons in its 4d orbitals. 
2. Ruthenium ( Ru, Z = 44 ) 
The ground-state electronic configuration of ruthenium is: 
[Kr]4?? 7
5?? 1
. 
Hence, Ruthenium has 7 electrons in its 4d orbitals. 
3. Sum of 4d Electrons 
?? = 4 ( for Nb), ?? = 7 ( for Ru). 
Therefore, 
?? + ?? = 4 + 7 = 11. 
Ans: 11 
Q2: The molar mass of the water insoluble product formed from the fusion of 
chromite ore (???????? ?? ?? ?? ) with ????
?? ????
?? in presence of ?? ?? is ____ ?? ?????? -?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 160 
Solution: 
FeC
2
O
4
+ Na
2
CO
3
?
O
2
 
? 
The chemical reaction can be written as 
The insoluble product is Fe
2
O
3
. 
Molar mass of Fe
2
O
3
 
Fe (Atomic mass) ? 55.845amu 
Molar mass = 55.845 g mol
-1
 
O (Atomic mass) ? 15.999amu 
Molar mass = 15.999 g mol
-1
 
Molar mass of ?? ?? 2
?? 3
= 55.845 g mol
-1
× 2 + 15.999 g mol
-1
× 3 
= (111.69 + 47.997)gmol
-1
 
= 159.687 g mol
-1
˜ 160 g mol
-1
 
Q3: Consider the following reactions 
?? + ???????? Little 
 amount 
+ ?? ?? ????
?? ? ?????? ?? ????
?? + Side Products 
?????? ?? ????
?? (vapour) 
+ ???????? ? ?? + ???????? + ?? ?? ?? 
?? + ?? +
? ?? + ?? ?? ?? 
The number of terminal ' ?? ' present in the compound ' C ' is ____ 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 6 
Solution: 
Cr
2
O
7
2-
+ NaCl + H
2
SO
4
? CrO
2
Cl
2
 
CrO
2
Cl
2
 (Vapour) +NaOH? 
Na
2
CrO
4
+ NaCl + H
2
O 
Na
2
CrO
4
+ H
?
? Na
2
Cr
2
O
7
+ H
2
O 
(C) 
Na
2
Cr
2
O
7
? 2Na
?
+ Cr
2
O
7
2-
 
 
No of terminal "O" = 6 
Q4: Among, ???? , ???? , ???? and Cu , identify the element with highest enthalpy of 
atomisation. The spin only magnetic moment value of that element in its +2 oxidation 
state is ____ BM (in nearest integer). 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 4 
Solution: 
 Sc Mn Co Cu 
Enthalpy of 
Atomisation (kJ/mole) 
326 281 425 339 
 
Highest Co 
Co
+2
= (Ar)3 d
7
 
 
 
?? = 3
?? = v15 = 3.87
 
Nearest integer = 4 
Q5: Lanthanoid ions with ?? ?? ?? configuration are : 
(A) ????
?? +
 
(B) ????
?? +
 
(C) ????
?? +
 
(D) ????
?? +
 
(E) ????
?? +
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (B) and (E) only 
B. (B) and (C) only 
C. (A) and (D) only 
D. (A) and (B) only 
Ans: D 
Solution: 
Let's analyze the electron configurations of the given lanthanide ions. 
Europium (Eu) 
The neutral atom of Eu (atomic number 63) typically has the configuration: 
[Xe]4?? 7
6?? 2
. 
For Eu
2+
, two electrons are removed, usually from the 6?? orbital, resulting in: 
[Xe]4?? 7
. 
Thus, Eu
2+
 has a 4?? 7
 configuration. 
For Eu
3+
, three electrons are removed (the two 6?? electrons and one 4?? electron), giving: 
[Xe]4?? 6
. 
Therefore, Eu
3+
 does not have a 4?? 7
 configuration. 
Gadolinium (Gd) 
The neutral atom of Gd (atomic number 64) is usually represented as: 
[Xe]4?? 7
5?? 1
6?? 2
. 
For Gd
3+
, three electrons are removed (typically the 5?? 1
 and the two 6?? electrons), resulting in: 
[Xe]4?? 7
. 
Page 4


JEE Main Previous Year Questions 
(2025): d – and f – Block Elements 
Q1: Niobium (???? ) and ruthenium (???? ) have " ?? " and " ?? " number of electrons in 
their respective ???? orbitals. The value of ?? + ?? is ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 11 
Solution: 
We need to determine the number of electrons each element has in its 4d orbitals and then add 
them together. 
1. Niobium ( Nb, Z = 41 ) 
The ground-state electronic configuration of niobium is: 
[Kr]4?? 4
5?? 1
. 
Hence, Niobium has 4 electrons in its 4d orbitals. 
2. Ruthenium ( Ru, Z = 44 ) 
The ground-state electronic configuration of ruthenium is: 
[Kr]4?? 7
5?? 1
. 
Hence, Ruthenium has 7 electrons in its 4d orbitals. 
3. Sum of 4d Electrons 
?? = 4 ( for Nb), ?? = 7 ( for Ru). 
Therefore, 
?? + ?? = 4 + 7 = 11. 
Ans: 11 
Q2: The molar mass of the water insoluble product formed from the fusion of 
chromite ore (???????? ?? ?? ?? ) with ????
?? ????
?? in presence of ?? ?? is ____ ?? ?????? -?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 160 
Solution: 
FeC
2
O
4
+ Na
2
CO
3
?
O
2
 
? 
The chemical reaction can be written as 
The insoluble product is Fe
2
O
3
. 
Molar mass of Fe
2
O
3
 
Fe (Atomic mass) ? 55.845amu 
Molar mass = 55.845 g mol
-1
 
O (Atomic mass) ? 15.999amu 
Molar mass = 15.999 g mol
-1
 
Molar mass of ?? ?? 2
?? 3
= 55.845 g mol
-1
× 2 + 15.999 g mol
-1
× 3 
= (111.69 + 47.997)gmol
-1
 
= 159.687 g mol
-1
˜ 160 g mol
-1
 
Q3: Consider the following reactions 
?? + ???????? Little 
 amount 
+ ?? ?? ????
?? ? ?????? ?? ????
?? + Side Products 
?????? ?? ????
?? (vapour) 
+ ???????? ? ?? + ???????? + ?? ?? ?? 
?? + ?? +
? ?? + ?? ?? ?? 
The number of terminal ' ?? ' present in the compound ' C ' is ____ 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 6 
Solution: 
Cr
2
O
7
2-
+ NaCl + H
2
SO
4
? CrO
2
Cl
2
 
CrO
2
Cl
2
 (Vapour) +NaOH? 
Na
2
CrO
4
+ NaCl + H
2
O 
Na
2
CrO
4
+ H
?
? Na
2
Cr
2
O
7
+ H
2
O 
(C) 
Na
2
Cr
2
O
7
? 2Na
?
+ Cr
2
O
7
2-
 
 
No of terminal "O" = 6 
Q4: Among, ???? , ???? , ???? and Cu , identify the element with highest enthalpy of 
atomisation. The spin only magnetic moment value of that element in its +2 oxidation 
state is ____ BM (in nearest integer). 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 4 
Solution: 
 Sc Mn Co Cu 
Enthalpy of 
Atomisation (kJ/mole) 
326 281 425 339 
 
Highest Co 
Co
+2
= (Ar)3 d
7
 
 
 
?? = 3
?? = v15 = 3.87
 
Nearest integer = 4 
Q5: Lanthanoid ions with ?? ?? ?? configuration are : 
(A) ????
?? +
 
(B) ????
?? +
 
(C) ????
?? +
 
(D) ????
?? +
 
(E) ????
?? +
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (B) and (E) only 
B. (B) and (C) only 
C. (A) and (D) only 
D. (A) and (B) only 
Ans: D 
Solution: 
Let's analyze the electron configurations of the given lanthanide ions. 
Europium (Eu) 
The neutral atom of Eu (atomic number 63) typically has the configuration: 
[Xe]4?? 7
6?? 2
. 
For Eu
2+
, two electrons are removed, usually from the 6?? orbital, resulting in: 
[Xe]4?? 7
. 
Thus, Eu
2+
 has a 4?? 7
 configuration. 
For Eu
3+
, three electrons are removed (the two 6?? electrons and one 4?? electron), giving: 
[Xe]4?? 6
. 
Therefore, Eu
3+
 does not have a 4?? 7
 configuration. 
Gadolinium (Gd) 
The neutral atom of Gd (atomic number 64) is usually represented as: 
[Xe]4?? 7
5?? 1
6?? 2
. 
For Gd
3+
, three electrons are removed (typically the 5?? 1
 and the two 6?? electrons), resulting in: 
[Xe]4?? 7
. 
So, Gd
3+
 has a 4?? 7
 configuration. 
Terbium (Tb) 
The neutral atom of Tb (atomic number 65) typically has the configuration: 
[Xe]4?? 9
6?? 2
. 
For Tb
3+
, removal of three electrons gives: 
[Xe]4?? 8
. 
Hence, Tb
3+
 does not have a 4?? 7
 configuration. 
Samarium (Sm) 
The neutral atom of Sm (atomic number 62) generally has: 
[Xe]4?? 6
6?? 2
. 
For Sm
2+
, after removal of two electrons (likely from the 6?? orbital), the configuration is: 
[Xe]4?? 6
. 
Therefore, Sm
2+
 does not have a 4?? 7
 configuration. 
Thus, the ions with a 4?? 7
 configuration are: 
(A) Eu
2+
 
(B) Gd
3+
 
The correct answer is: Option D - (A) and (B) only. 
Q6: The correct set of ions (aqueous solution) with same colour from the following is: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. V
2+
, Cr
3+
, Mn
3+
 
B. Ti
4+
, V
4+
, Mn
2+
 
C. Sc
3+
, Ti
3+
, Cr
2+
 
D. Zn
2+
, V
3+
, Fe
3+
 
Ans: A 
Solution: 
(1) V
2+
 (Violet), Cr
3+
 (Violet), Mn
3+
 (Violet) 
(2) Zn
2+
 (Colourless), V
3+
 (Green), Fe
3+
 (Yellow) 
(3) Ti
4+
 (Colourless), V
4+
 (Blue), Mn
2+
 (Pink) 
(4) Sc
3+
 (Colourless), Ti
3+
 (Purple), Cr
2+
 (Blue) 
Q7: Consider the following reactions 
?? ?? ????
?? ?? ?? ?
?????? 
-?? ?? ?? 
[?? ] ?
?? ?? ????
?? 
-?? ?? ?? 
[ ?? ] + ?? ?? ????
?? 
The products [?? ] and [?? ], respectively are : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. K
2
Cr(OH )
6
 and Cr
2
O
3
 
B. K
2
CrO
4
 and Cr
2
O
3
 
C. K
2
CrO
4
 and K
2
Cr
2
O
7
 
D. K
2
CrO
4
 and CrO 
Page 5


JEE Main Previous Year Questions 
(2025): d – and f – Block Elements 
Q1: Niobium (???? ) and ruthenium (???? ) have " ?? " and " ?? " number of electrons in 
their respective ???? orbitals. The value of ?? + ?? is ____ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 11 
Solution: 
We need to determine the number of electrons each element has in its 4d orbitals and then add 
them together. 
1. Niobium ( Nb, Z = 41 ) 
The ground-state electronic configuration of niobium is: 
[Kr]4?? 4
5?? 1
. 
Hence, Niobium has 4 electrons in its 4d orbitals. 
2. Ruthenium ( Ru, Z = 44 ) 
The ground-state electronic configuration of ruthenium is: 
[Kr]4?? 7
5?? 1
. 
Hence, Ruthenium has 7 electrons in its 4d orbitals. 
3. Sum of 4d Electrons 
?? = 4 ( for Nb), ?? = 7 ( for Ru). 
Therefore, 
?? + ?? = 4 + 7 = 11. 
Ans: 11 
Q2: The molar mass of the water insoluble product formed from the fusion of 
chromite ore (???????? ?? ?? ?? ) with ????
?? ????
?? in presence of ?? ?? is ____ ?? ?????? -?? . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 160 
Solution: 
FeC
2
O
4
+ Na
2
CO
3
?
O
2
 
? 
The chemical reaction can be written as 
The insoluble product is Fe
2
O
3
. 
Molar mass of Fe
2
O
3
 
Fe (Atomic mass) ? 55.845amu 
Molar mass = 55.845 g mol
-1
 
O (Atomic mass) ? 15.999amu 
Molar mass = 15.999 g mol
-1
 
Molar mass of ?? ?? 2
?? 3
= 55.845 g mol
-1
× 2 + 15.999 g mol
-1
× 3 
= (111.69 + 47.997)gmol
-1
 
= 159.687 g mol
-1
˜ 160 g mol
-1
 
Q3: Consider the following reactions 
?? + ???????? Little 
 amount 
+ ?? ?? ????
?? ? ?????? ?? ????
?? + Side Products 
?????? ?? ????
?? (vapour) 
+ ???????? ? ?? + ???????? + ?? ?? ?? 
?? + ?? +
? ?? + ?? ?? ?? 
The number of terminal ' ?? ' present in the compound ' C ' is ____ 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 6 
Solution: 
Cr
2
O
7
2-
+ NaCl + H
2
SO
4
? CrO
2
Cl
2
 
CrO
2
Cl
2
 (Vapour) +NaOH? 
Na
2
CrO
4
+ NaCl + H
2
O 
Na
2
CrO
4
+ H
?
? Na
2
Cr
2
O
7
+ H
2
O 
(C) 
Na
2
Cr
2
O
7
? 2Na
?
+ Cr
2
O
7
2-
 
 
No of terminal "O" = 6 
Q4: Among, ???? , ???? , ???? and Cu , identify the element with highest enthalpy of 
atomisation. The spin only magnetic moment value of that element in its +2 oxidation 
state is ____ BM (in nearest integer). 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 4 
Solution: 
 Sc Mn Co Cu 
Enthalpy of 
Atomisation (kJ/mole) 
326 281 425 339 
 
Highest Co 
Co
+2
= (Ar)3 d
7
 
 
 
?? = 3
?? = v15 = 3.87
 
Nearest integer = 4 
Q5: Lanthanoid ions with ?? ?? ?? configuration are : 
(A) ????
?? +
 
(B) ????
?? +
 
(C) ????
?? +
 
(D) ????
?? +
 
(E) ????
?? +
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (B) and (E) only 
B. (B) and (C) only 
C. (A) and (D) only 
D. (A) and (B) only 
Ans: D 
Solution: 
Let's analyze the electron configurations of the given lanthanide ions. 
Europium (Eu) 
The neutral atom of Eu (atomic number 63) typically has the configuration: 
[Xe]4?? 7
6?? 2
. 
For Eu
2+
, two electrons are removed, usually from the 6?? orbital, resulting in: 
[Xe]4?? 7
. 
Thus, Eu
2+
 has a 4?? 7
 configuration. 
For Eu
3+
, three electrons are removed (the two 6?? electrons and one 4?? electron), giving: 
[Xe]4?? 6
. 
Therefore, Eu
3+
 does not have a 4?? 7
 configuration. 
Gadolinium (Gd) 
The neutral atom of Gd (atomic number 64) is usually represented as: 
[Xe]4?? 7
5?? 1
6?? 2
. 
For Gd
3+
, three electrons are removed (typically the 5?? 1
 and the two 6?? electrons), resulting in: 
[Xe]4?? 7
. 
So, Gd
3+
 has a 4?? 7
 configuration. 
Terbium (Tb) 
The neutral atom of Tb (atomic number 65) typically has the configuration: 
[Xe]4?? 9
6?? 2
. 
For Tb
3+
, removal of three electrons gives: 
[Xe]4?? 8
. 
Hence, Tb
3+
 does not have a 4?? 7
 configuration. 
Samarium (Sm) 
The neutral atom of Sm (atomic number 62) generally has: 
[Xe]4?? 6
6?? 2
. 
For Sm
2+
, after removal of two electrons (likely from the 6?? orbital), the configuration is: 
[Xe]4?? 6
. 
Therefore, Sm
2+
 does not have a 4?? 7
 configuration. 
Thus, the ions with a 4?? 7
 configuration are: 
(A) Eu
2+
 
(B) Gd
3+
 
The correct answer is: Option D - (A) and (B) only. 
Q6: The correct set of ions (aqueous solution) with same colour from the following is: 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. V
2+
, Cr
3+
, Mn
3+
 
B. Ti
4+
, V
4+
, Mn
2+
 
C. Sc
3+
, Ti
3+
, Cr
2+
 
D. Zn
2+
, V
3+
, Fe
3+
 
Ans: A 
Solution: 
(1) V
2+
 (Violet), Cr
3+
 (Violet), Mn
3+
 (Violet) 
(2) Zn
2+
 (Colourless), V
3+
 (Green), Fe
3+
 (Yellow) 
(3) Ti
4+
 (Colourless), V
4+
 (Blue), Mn
2+
 (Pink) 
(4) Sc
3+
 (Colourless), Ti
3+
 (Purple), Cr
2+
 (Blue) 
Q7: Consider the following reactions 
?? ?? ????
?? ?? ?? ?
?????? 
-?? ?? ?? 
[?? ] ?
?? ?? ????
?? 
-?? ?? ?? 
[ ?? ] + ?? ?? ????
?? 
The products [?? ] and [?? ], respectively are : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. K
2
Cr(OH )
6
 and Cr
2
O
3
 
B. K
2
CrO
4
 and Cr
2
O
3
 
C. K
2
CrO
4
 and K
2
Cr
2
O
7
 
D. K
2
CrO
4
 and CrO 
Ans: C 
Solution: 
 
Q8: Match List - I with List - II 
 List - I  List - II 
(A) Bronze (I) Cu, Ni 
(B) Brass (II) Fe, Cr, Ni, C 
(C) UK silver coin (III) Cu, Zn 
(D) Stainless steel (IV) Cu, Sn 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) 
B. (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 
C. (A)-(III), (B)-(I), (C)-(IV), (D)-(II) 
D. (A)-(III), (B)-(IV), (C)-(II), (D)-(I) 
Ans: B 
Solution: 
To match the items from List I to List II, we need to understand the composition of each 
material: 
Bronze is an alloy composed of copper (Cu) and tin(Sn). 
Brass is an alloy made of copper ( Cu ) and zinc ( Zn ). 
The UK silver coin primarily consists of copper ( Cu ) and nickel ( Ni ). 
Stainless steel is an alloy that includes iron ( Fe ), chromium ( Cr ), nickel ( Ni ), and carbon ( C ). 
Based on these definitions, the correct matches are: 
(A) Bronze corresponds to (IV) Cu, Sn 
(B) Brass corresponds to (III) Cu, Zn 
(C) UK silver coin corresponds to (I) Cu, Ni 
(D) Stainless steel corresponds to (II) Fe, Cr, Ni, C 
These pairings illustrate the composition of each material and ensure accurate alignment 
between List I and List II. 
Q9: Which of the following ions is the strongest oxidizing agent? (Atomic Number of 
???? = ???? , ???? = ???? , ???? = ???? , ???? = ???? )