JEE Main Previous Year Questions (2025): Redox Reactions

 Page 1


JEE Main Previous Year Questions 
(2025): Redox Reactions 
Q1: The species which does not undergo disproportionation reaction is: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. ClO
4
-
 
B. ClO
3
-
 
C. ClO
-
 
D. ClO
2
-
 
Ans: A 
Solution: 
Oxidation States: 
In ClO
4
-
(perchlorate), chlorine is in the +7 oxidation state. 
In ClO
3
-
(chlorate), chlorine is in the +5 oxidation state. 
In ClO
2
-
(chlorite), chlorine is in the +3 oxidation state. 
In ClO
-
(hypochlorite), chlorine is in the +1 oxidation state. 
Disproportionation Reaction: 
A disproportionation reaction is one in which a species simultaneously undergoes oxidation and 
reduction. For this to occur, the element must be in an intermediate oxidation state such that it 
can be oxidized to a higher state and reduced to a lower one. 
In ClO
-
and ClO
2
-
, chlorine is in lower oxidation states ( +1 and +3 , respectively), making them 
susceptible to disproportionation. For example, hypochlorite can disproportionate in basic 
solution as follows: 
3ClO
-
? 2Cl
-
+ ClO
3
-
 
In ClO
3
-
, chlorine is in an intermediate oxidation state ( +5 ) that, under certain conditions, can 
undergo disproportionation. 
However, in ClO
4
-
, chlorine is in its highest possible oxidation state ( +7 ) and cannot be oxidized 
further. Since disproportionation requires one part of the species to be oxidized and the other 
reduced, ClO
4
-
is thermodynamically stable and does not undergo disproportionation. 
ClO
4
-
(perchlorate ion) does not undergo disproportionation. 
Q2: Which of the following oxidation reactions are carried out by both ?? ?? ????
?? ?? ?? and 
???????? ?? in acidic medium? 
A. ?? -
? ?? ?? 
B. ?? ?? -
? ?? 
C. ????
?? +
? ????
?? +
 
D. ?? -
? ????
?? -
 
E. ?? ?? ?? ?? 
?? -
? ????
?? ?? -
 
Choose the correct answer from the options given below : 
Page 2


JEE Main Previous Year Questions 
(2025): Redox Reactions 
Q1: The species which does not undergo disproportionation reaction is: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. ClO
4
-
 
B. ClO
3
-
 
C. ClO
-
 
D. ClO
2
-
 
Ans: A 
Solution: 
Oxidation States: 
In ClO
4
-
(perchlorate), chlorine is in the +7 oxidation state. 
In ClO
3
-
(chlorate), chlorine is in the +5 oxidation state. 
In ClO
2
-
(chlorite), chlorine is in the +3 oxidation state. 
In ClO
-
(hypochlorite), chlorine is in the +1 oxidation state. 
Disproportionation Reaction: 
A disproportionation reaction is one in which a species simultaneously undergoes oxidation and 
reduction. For this to occur, the element must be in an intermediate oxidation state such that it 
can be oxidized to a higher state and reduced to a lower one. 
In ClO
-
and ClO
2
-
, chlorine is in lower oxidation states ( +1 and +3 , respectively), making them 
susceptible to disproportionation. For example, hypochlorite can disproportionate in basic 
solution as follows: 
3ClO
-
? 2Cl
-
+ ClO
3
-
 
In ClO
3
-
, chlorine is in an intermediate oxidation state ( +5 ) that, under certain conditions, can 
undergo disproportionation. 
However, in ClO
4
-
, chlorine is in its highest possible oxidation state ( +7 ) and cannot be oxidized 
further. Since disproportionation requires one part of the species to be oxidized and the other 
reduced, ClO
4
-
is thermodynamically stable and does not undergo disproportionation. 
ClO
4
-
(perchlorate ion) does not undergo disproportionation. 
Q2: Which of the following oxidation reactions are carried out by both ?? ?? ????
?? ?? ?? and 
???????? ?? in acidic medium? 
A. ?? -
? ?? ?? 
B. ?? ?? -
? ?? 
C. ????
?? +
? ????
?? +
 
D. ?? -
? ????
?? -
 
E. ?? ?? ?? ?? 
?? -
? ????
?? ?? -
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A, B and C Only 
B. C, D and E Only 
C. B, C and D Only 
D. A, D and E Only 
Ans: A 
Solution: 
I
-
?
H
+
 
I
2
 I
-
?
OH
-
 
IO
3
-
 
S
-2
?
H
+
 
S S
2
O
3
2-
?
OH
-
 
SO
4
2-
 
Fe
+2
? Fe
+3
 
S
2
O
3
2-
?
H
+
 
S ? +SO
4
2-
 
Q3: Match the List - I with List - II 
 List - I (Redox Reaction)  List - II (Type of Redox 
Reaction) 
(A) 
????
?? ( ?? )
+ ?? ?? ?? ( ?? )
?
?? 
????
?? ( ?? )
 +?? ?? ?? ?? (?? )
 
(I) Disproportionation reaction 
(B) 
?? ?????? (?? )
?
?? 
?? ????
(?? )
+ ?? ?? ( ?? )
 
(II) Combination reaction 
(C) 
?? ?? ?? ?? ( ?? )
+ ?? ????
(?? )
?
?? 
?? ?? (?? )
 +?? ?????? (?? )
 
(III) Decomposition reaction 
(D) 
?? ?? ?? ?? ?? (???? )
?
?? 
?? ?? ?? ?? (?? )
+
?? ?? ( ?? )
 
(IV) Displacement reaction 
 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A-III, B-IV, C-I, D-II 
B. A-IV, B-I, C-II, D-III 
C. A-II, B-III, C-I, D-IV 
D. A-II, B-III, C-IV, D-I 
Ans: D 
Solution: 
(A) Combustion of hydrocarbon 
(B) Decomposition into gaseous product. 
(C) Displacement of ' ?? ' by ' Ca ' atom. 
(D) Disproportionation of H
2
O
2
-1
 into O
-2
 and O
°
 oxidation states. 
Page 3


JEE Main Previous Year Questions 
(2025): Redox Reactions 
Q1: The species which does not undergo disproportionation reaction is: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. ClO
4
-
 
B. ClO
3
-
 
C. ClO
-
 
D. ClO
2
-
 
Ans: A 
Solution: 
Oxidation States: 
In ClO
4
-
(perchlorate), chlorine is in the +7 oxidation state. 
In ClO
3
-
(chlorate), chlorine is in the +5 oxidation state. 
In ClO
2
-
(chlorite), chlorine is in the +3 oxidation state. 
In ClO
-
(hypochlorite), chlorine is in the +1 oxidation state. 
Disproportionation Reaction: 
A disproportionation reaction is one in which a species simultaneously undergoes oxidation and 
reduction. For this to occur, the element must be in an intermediate oxidation state such that it 
can be oxidized to a higher state and reduced to a lower one. 
In ClO
-
and ClO
2
-
, chlorine is in lower oxidation states ( +1 and +3 , respectively), making them 
susceptible to disproportionation. For example, hypochlorite can disproportionate in basic 
solution as follows: 
3ClO
-
? 2Cl
-
+ ClO
3
-
 
In ClO
3
-
, chlorine is in an intermediate oxidation state ( +5 ) that, under certain conditions, can 
undergo disproportionation. 
However, in ClO
4
-
, chlorine is in its highest possible oxidation state ( +7 ) and cannot be oxidized 
further. Since disproportionation requires one part of the species to be oxidized and the other 
reduced, ClO
4
-
is thermodynamically stable and does not undergo disproportionation. 
ClO
4
-
(perchlorate ion) does not undergo disproportionation. 
Q2: Which of the following oxidation reactions are carried out by both ?? ?? ????
?? ?? ?? and 
???????? ?? in acidic medium? 
A. ?? -
? ?? ?? 
B. ?? ?? -
? ?? 
C. ????
?? +
? ????
?? +
 
D. ?? -
? ????
?? -
 
E. ?? ?? ?? ?? 
?? -
? ????
?? ?? -
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A, B and C Only 
B. C, D and E Only 
C. B, C and D Only 
D. A, D and E Only 
Ans: A 
Solution: 
I
-
?
H
+
 
I
2
 I
-
?
OH
-
 
IO
3
-
 
S
-2
?
H
+
 
S S
2
O
3
2-
?
OH
-
 
SO
4
2-
 
Fe
+2
? Fe
+3
 
S
2
O
3
2-
?
H
+
 
S ? +SO
4
2-
 
Q3: Match the List - I with List - II 
 List - I (Redox Reaction)  List - II (Type of Redox 
Reaction) 
(A) 
????
?? ( ?? )
+ ?? ?? ?? ( ?? )
?
?? 
????
?? ( ?? )
 +?? ?? ?? ?? (?? )
 
(I) Disproportionation reaction 
(B) 
?? ?????? (?? )
?
?? 
?? ????
(?? )
+ ?? ?? ( ?? )
 
(II) Combination reaction 
(C) 
?? ?? ?? ?? ( ?? )
+ ?? ????
(?? )
?
?? 
?? ?? (?? )
 +?? ?????? (?? )
 
(III) Decomposition reaction 
(D) 
?? ?? ?? ?? ?? (???? )
?
?? 
?? ?? ?? ?? (?? )
+
?? ?? ( ?? )
 
(IV) Displacement reaction 
 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A-III, B-IV, C-I, D-II 
B. A-IV, B-I, C-II, D-III 
C. A-II, B-III, C-I, D-IV 
D. A-II, B-III, C-IV, D-I 
Ans: D 
Solution: 
(A) Combustion of hydrocarbon 
(B) Decomposition into gaseous product. 
(C) Displacement of ' ?? ' by ' Ca ' atom. 
(D) Disproportionation of H
2
O
2
-1
 into O
-2
 and O
°
 oxidation states. 
Q4: Given below are two statements: 
Statement I: In the oxalic acid vs ???????? ?? (in the presence of dil ?? ?? ????
?? ) titration the solution 
needs to be heated initially to ????
°
?? , but no heating is required in Ferrous ammonium sulphate 
(FAS) vs ???????? ?? titration (in the presence of dil ?? ?? ????
?? ) 
Statement II: In oxalic acid vs ???????? ?? titration, the initial formation of ???????? ?? takes place at 
high temperature, which then acts as catalyst for further reaction. In the case of FAS vs 
???????? ?? , heating oxidizes ????
?? +
 into ?? ?? ?? +
 by oxygen of air and error may be introduced in the 
experiment. 
In the light of the above statements, choose the correct answer from the options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Statement I is true but Statement II is false 
C. Both Statement I and Statement II are true 
D. Both Statement I and Statement II are false 
Ans: C 
Solution: 
For the titration: Oxalic acid v/sKMnOKM
4
 
2MnO
4
-
+ 5(COO )
2
2-
+ 16H
+
?
10CO
2
+ 2Mn
2+
+ 8H
2
O
 
This reaction is slow at room temperature, but becomes fast at 60
°
C. Manganese(II) ions 
catalyse the reaction; thus, the reaction is autocatalytic; once manganese(II) ions are formed, it 
becomes faster and faster. 
The titration of FASv/sKMnO
4
 do not require heating because at higher temperature the 
oxidation of Fe
+2
 to Fe
+3
 by atmospheric O
2
 will be prominent. 
Q5: ?? . ?? ?? solution of KI reacts with excess of ?? ?? ????
?? and ?? ????
?? solutions. According 
to equation ?? ?? -
+ ????
?? -
+ ?? ?? +
? ?? ?? ?? + ?? ?? ?? ?? 
Identify the correct statements : 
(A) ?????? ???? of KI solution reacts with ?? . ?????? ?????? of ?????? ?? 
(B) ?????? ???? ???? ???? ???????????????? ???????????? ???????? ?? . ?????? ?????? of ?? ?? ????
?? 
(C) 0.5 L of KI solution produced 0.005 mol of ?? ?? 
(D) Equivalent weight of ?????? ?? is equal to ( 
 Molecular weight 
?? ) 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. (C) and (D) only 
B. (A) and (B) only 
Page 4


JEE Main Previous Year Questions 
(2025): Redox Reactions 
Q1: The species which does not undergo disproportionation reaction is: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. ClO
4
-
 
B. ClO
3
-
 
C. ClO
-
 
D. ClO
2
-
 
Ans: A 
Solution: 
Oxidation States: 
In ClO
4
-
(perchlorate), chlorine is in the +7 oxidation state. 
In ClO
3
-
(chlorate), chlorine is in the +5 oxidation state. 
In ClO
2
-
(chlorite), chlorine is in the +3 oxidation state. 
In ClO
-
(hypochlorite), chlorine is in the +1 oxidation state. 
Disproportionation Reaction: 
A disproportionation reaction is one in which a species simultaneously undergoes oxidation and 
reduction. For this to occur, the element must be in an intermediate oxidation state such that it 
can be oxidized to a higher state and reduced to a lower one. 
In ClO
-
and ClO
2
-
, chlorine is in lower oxidation states ( +1 and +3 , respectively), making them 
susceptible to disproportionation. For example, hypochlorite can disproportionate in basic 
solution as follows: 
3ClO
-
? 2Cl
-
+ ClO
3
-
 
In ClO
3
-
, chlorine is in an intermediate oxidation state ( +5 ) that, under certain conditions, can 
undergo disproportionation. 
However, in ClO
4
-
, chlorine is in its highest possible oxidation state ( +7 ) and cannot be oxidized 
further. Since disproportionation requires one part of the species to be oxidized and the other 
reduced, ClO
4
-
is thermodynamically stable and does not undergo disproportionation. 
ClO
4
-
(perchlorate ion) does not undergo disproportionation. 
Q2: Which of the following oxidation reactions are carried out by both ?? ?? ????
?? ?? ?? and 
???????? ?? in acidic medium? 
A. ?? -
? ?? ?? 
B. ?? ?? -
? ?? 
C. ????
?? +
? ????
?? +
 
D. ?? -
? ????
?? -
 
E. ?? ?? ?? ?? 
?? -
? ????
?? ?? -
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A, B and C Only 
B. C, D and E Only 
C. B, C and D Only 
D. A, D and E Only 
Ans: A 
Solution: 
I
-
?
H
+
 
I
2
 I
-
?
OH
-
 
IO
3
-
 
S
-2
?
H
+
 
S S
2
O
3
2-
?
OH
-
 
SO
4
2-
 
Fe
+2
? Fe
+3
 
S
2
O
3
2-
?
H
+
 
S ? +SO
4
2-
 
Q3: Match the List - I with List - II 
 List - I (Redox Reaction)  List - II (Type of Redox 
Reaction) 
(A) 
????
?? ( ?? )
+ ?? ?? ?? ( ?? )
?
?? 
????
?? ( ?? )
 +?? ?? ?? ?? (?? )
 
(I) Disproportionation reaction 
(B) 
?? ?????? (?? )
?
?? 
?? ????
(?? )
+ ?? ?? ( ?? )
 
(II) Combination reaction 
(C) 
?? ?? ?? ?? ( ?? )
+ ?? ????
(?? )
?
?? 
?? ?? (?? )
 +?? ?????? (?? )
 
(III) Decomposition reaction 
(D) 
?? ?? ?? ?? ?? (???? )
?
?? 
?? ?? ?? ?? (?? )
+
?? ?? ( ?? )
 
(IV) Displacement reaction 
 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A-III, B-IV, C-I, D-II 
B. A-IV, B-I, C-II, D-III 
C. A-II, B-III, C-I, D-IV 
D. A-II, B-III, C-IV, D-I 
Ans: D 
Solution: 
(A) Combustion of hydrocarbon 
(B) Decomposition into gaseous product. 
(C) Displacement of ' ?? ' by ' Ca ' atom. 
(D) Disproportionation of H
2
O
2
-1
 into O
-2
 and O
°
 oxidation states. 
Q4: Given below are two statements: 
Statement I: In the oxalic acid vs ???????? ?? (in the presence of dil ?? ?? ????
?? ) titration the solution 
needs to be heated initially to ????
°
?? , but no heating is required in Ferrous ammonium sulphate 
(FAS) vs ???????? ?? titration (in the presence of dil ?? ?? ????
?? ) 
Statement II: In oxalic acid vs ???????? ?? titration, the initial formation of ???????? ?? takes place at 
high temperature, which then acts as catalyst for further reaction. In the case of FAS vs 
???????? ?? , heating oxidizes ????
?? +
 into ?? ?? ?? +
 by oxygen of air and error may be introduced in the 
experiment. 
In the light of the above statements, choose the correct answer from the options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Statement I is true but Statement II is false 
C. Both Statement I and Statement II are true 
D. Both Statement I and Statement II are false 
Ans: C 
Solution: 
For the titration: Oxalic acid v/sKMnOKM
4
 
2MnO
4
-
+ 5(COO )
2
2-
+ 16H
+
?
10CO
2
+ 2Mn
2+
+ 8H
2
O
 
This reaction is slow at room temperature, but becomes fast at 60
°
C. Manganese(II) ions 
catalyse the reaction; thus, the reaction is autocatalytic; once manganese(II) ions are formed, it 
becomes faster and faster. 
The titration of FASv/sKMnO
4
 do not require heating because at higher temperature the 
oxidation of Fe
+2
 to Fe
+3
 by atmospheric O
2
 will be prominent. 
Q5: ?? . ?? ?? solution of KI reacts with excess of ?? ?? ????
?? and ?? ????
?? solutions. According 
to equation ?? ?? -
+ ????
?? -
+ ?? ?? +
? ?? ?? ?? + ?? ?? ?? ?? 
Identify the correct statements : 
(A) ?????? ???? of KI solution reacts with ?? . ?????? ?????? of ?????? ?? 
(B) ?????? ???? ???? ???? ???????????????? ???????????? ???????? ?? . ?????? ?????? of ?? ?? ????
?? 
(C) 0.5 L of KI solution produced 0.005 mol of ?? ?? 
(D) Equivalent weight of ?????? ?? is equal to ( 
 Molecular weight 
?? ) 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. (C) and (D) only 
B. (A) and (B) only 
C. (B) and (C) only 
D. (A) and (D) only 
Ans: D 
Solution: 
Molarity of KI = 0.1M 
5I
-
+ IO
3
-
+ 6H
+
? 3I
2
+ 3H
2
O (balanced chemical equation) 
5KI + KIO
3
+ 3H
2
SO
4
? 3I
2
+ 3 K
2
SO
4
+ 3H
2
O 
(A) Volume given (KI) = 200 mL= 0.2 L 
Molarity (KI) = 0.1M
or molL
-1 
The molarity formula is used here to calculate the number of moles of K.I. 
Molarity, ?? =
 number of moles 
 volume in ?? 
So, number of moles = Molarity × Volume in L 
For the given volume of KI, 
number of moles = 0.1 mol L
-1
× 0.2 L 
= 0.02 mol 
The stoichiometric ratio between KI and KIO
3
 is 
???? : ???? ?? 3
 
?? -
: ?? ?? 3
-
 
5: 1 
1:
1
5
 
For one mol ?? -
,
1
5
mol ?? ?? 3
-
is used. 
So, for 0.2 mol I
-
, 
Moles of ?? ?? 3
-
= 0.02 ×
1
5
= 0.004 mol 
The statement (A) is correct. 
(B) Volume given (KI) = 200 mL= 0.2 L 
Molarity (KI) = 0.1M 
number of moles of KI(I
-
) =Molarity ×Volume (L) 
= 0.1 mol ?? -1
× 0.2?? = 0.02 mol 
The stoichiometric ratio between KE and H
2
HO
4
 is 
KI: H
2
HO
4
 
5 : 3 
1:
3
5
{3 moles H
2
SO
4
 can give 6H
+
} 
?? -
,
3
5
 
For one mol ?? -
,
3
5
mol H
2
SO
4
 is used. 
So, for 0.02 mol I
-
, 
moles of H
2
SO
4
= 0.02 ×
3
5
= 0.012 mol 
This statement is not correct. 
(C) Volume gien (KI) = 0.5 L 
Page 5


JEE Main Previous Year Questions 
(2025): Redox Reactions 
Q1: The species which does not undergo disproportionation reaction is: 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. ClO
4
-
 
B. ClO
3
-
 
C. ClO
-
 
D. ClO
2
-
 
Ans: A 
Solution: 
Oxidation States: 
In ClO
4
-
(perchlorate), chlorine is in the +7 oxidation state. 
In ClO
3
-
(chlorate), chlorine is in the +5 oxidation state. 
In ClO
2
-
(chlorite), chlorine is in the +3 oxidation state. 
In ClO
-
(hypochlorite), chlorine is in the +1 oxidation state. 
Disproportionation Reaction: 
A disproportionation reaction is one in which a species simultaneously undergoes oxidation and 
reduction. For this to occur, the element must be in an intermediate oxidation state such that it 
can be oxidized to a higher state and reduced to a lower one. 
In ClO
-
and ClO
2
-
, chlorine is in lower oxidation states ( +1 and +3 , respectively), making them 
susceptible to disproportionation. For example, hypochlorite can disproportionate in basic 
solution as follows: 
3ClO
-
? 2Cl
-
+ ClO
3
-
 
In ClO
3
-
, chlorine is in an intermediate oxidation state ( +5 ) that, under certain conditions, can 
undergo disproportionation. 
However, in ClO
4
-
, chlorine is in its highest possible oxidation state ( +7 ) and cannot be oxidized 
further. Since disproportionation requires one part of the species to be oxidized and the other 
reduced, ClO
4
-
is thermodynamically stable and does not undergo disproportionation. 
ClO
4
-
(perchlorate ion) does not undergo disproportionation. 
Q2: Which of the following oxidation reactions are carried out by both ?? ?? ????
?? ?? ?? and 
???????? ?? in acidic medium? 
A. ?? -
? ?? ?? 
B. ?? ?? -
? ?? 
C. ????
?? +
? ????
?? +
 
D. ?? -
? ????
?? -
 
E. ?? ?? ?? ?? 
?? -
? ????
?? ?? -
 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A, B and C Only 
B. C, D and E Only 
C. B, C and D Only 
D. A, D and E Only 
Ans: A 
Solution: 
I
-
?
H
+
 
I
2
 I
-
?
OH
-
 
IO
3
-
 
S
-2
?
H
+
 
S S
2
O
3
2-
?
OH
-
 
SO
4
2-
 
Fe
+2
? Fe
+3
 
S
2
O
3
2-
?
H
+
 
S ? +SO
4
2-
 
Q3: Match the List - I with List - II 
 List - I (Redox Reaction)  List - II (Type of Redox 
Reaction) 
(A) 
????
?? ( ?? )
+ ?? ?? ?? ( ?? )
?
?? 
????
?? ( ?? )
 +?? ?? ?? ?? (?? )
 
(I) Disproportionation reaction 
(B) 
?? ?????? (?? )
?
?? 
?? ????
(?? )
+ ?? ?? ( ?? )
 
(II) Combination reaction 
(C) 
?? ?? ?? ?? ( ?? )
+ ?? ????
(?? )
?
?? 
?? ?? (?? )
 +?? ?????? (?? )
 
(III) Decomposition reaction 
(D) 
?? ?? ?? ?? ?? (???? )
?
?? 
?? ?? ?? ?? (?? )
+
?? ?? ( ?? )
 
(IV) Displacement reaction 
 
Choose the correct answer from the options given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. A-III, B-IV, C-I, D-II 
B. A-IV, B-I, C-II, D-III 
C. A-II, B-III, C-I, D-IV 
D. A-II, B-III, C-IV, D-I 
Ans: D 
Solution: 
(A) Combustion of hydrocarbon 
(B) Decomposition into gaseous product. 
(C) Displacement of ' ?? ' by ' Ca ' atom. 
(D) Disproportionation of H
2
O
2
-1
 into O
-2
 and O
°
 oxidation states. 
Q4: Given below are two statements: 
Statement I: In the oxalic acid vs ???????? ?? (in the presence of dil ?? ?? ????
?? ) titration the solution 
needs to be heated initially to ????
°
?? , but no heating is required in Ferrous ammonium sulphate 
(FAS) vs ???????? ?? titration (in the presence of dil ?? ?? ????
?? ) 
Statement II: In oxalic acid vs ???????? ?? titration, the initial formation of ???????? ?? takes place at 
high temperature, which then acts as catalyst for further reaction. In the case of FAS vs 
???????? ?? , heating oxidizes ????
?? +
 into ?? ?? ?? +
 by oxygen of air and error may be introduced in the 
experiment. 
In the light of the above statements, choose the correct answer from the options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Statement I is true but Statement II is false 
C. Both Statement I and Statement II are true 
D. Both Statement I and Statement II are false 
Ans: C 
Solution: 
For the titration: Oxalic acid v/sKMnOKM
4
 
2MnO
4
-
+ 5(COO )
2
2-
+ 16H
+
?
10CO
2
+ 2Mn
2+
+ 8H
2
O
 
This reaction is slow at room temperature, but becomes fast at 60
°
C. Manganese(II) ions 
catalyse the reaction; thus, the reaction is autocatalytic; once manganese(II) ions are formed, it 
becomes faster and faster. 
The titration of FASv/sKMnO
4
 do not require heating because at higher temperature the 
oxidation of Fe
+2
 to Fe
+3
 by atmospheric O
2
 will be prominent. 
Q5: ?? . ?? ?? solution of KI reacts with excess of ?? ?? ????
?? and ?? ????
?? solutions. According 
to equation ?? ?? -
+ ????
?? -
+ ?? ?? +
? ?? ?? ?? + ?? ?? ?? ?? 
Identify the correct statements : 
(A) ?????? ???? of KI solution reacts with ?? . ?????? ?????? of ?????? ?? 
(B) ?????? ???? ???? ???? ???????????????? ???????????? ???????? ?? . ?????? ?????? of ?? ?? ????
?? 
(C) 0.5 L of KI solution produced 0.005 mol of ?? ?? 
(D) Equivalent weight of ?????? ?? is equal to ( 
 Molecular weight 
?? ) 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. (C) and (D) only 
B. (A) and (B) only 
C. (B) and (C) only 
D. (A) and (D) only 
Ans: D 
Solution: 
Molarity of KI = 0.1M 
5I
-
+ IO
3
-
+ 6H
+
? 3I
2
+ 3H
2
O (balanced chemical equation) 
5KI + KIO
3
+ 3H
2
SO
4
? 3I
2
+ 3 K
2
SO
4
+ 3H
2
O 
(A) Volume given (KI) = 200 mL= 0.2 L 
Molarity (KI) = 0.1M
or molL
-1 
The molarity formula is used here to calculate the number of moles of K.I. 
Molarity, ?? =
 number of moles 
 volume in ?? 
So, number of moles = Molarity × Volume in L 
For the given volume of KI, 
number of moles = 0.1 mol L
-1
× 0.2 L 
= 0.02 mol 
The stoichiometric ratio between KI and KIO
3
 is 
???? : ???? ?? 3
 
?? -
: ?? ?? 3
-
 
5: 1 
1:
1
5
 
For one mol ?? -
,
1
5
mol ?? ?? 3
-
is used. 
So, for 0.2 mol I
-
, 
Moles of ?? ?? 3
-
= 0.02 ×
1
5
= 0.004 mol 
The statement (A) is correct. 
(B) Volume given (KI) = 200 mL= 0.2 L 
Molarity (KI) = 0.1M 
number of moles of KI(I
-
) =Molarity ×Volume (L) 
= 0.1 mol ?? -1
× 0.2?? = 0.02 mol 
The stoichiometric ratio between KE and H
2
HO
4
 is 
KI: H
2
HO
4
 
5 : 3 
1:
3
5
{3 moles H
2
SO
4
 can give 6H
+
} 
?? -
,
3
5
 
For one mol ?? -
,
3
5
mol H
2
SO
4
 is used. 
So, for 0.02 mol I
-
, 
moles of H
2
SO
4
= 0.02 ×
3
5
= 0.012 mol 
This statement is not correct. 
(C) Volume gien (KI) = 0.5 L 
Molarity (KI) = 0.1M 
Number of moles of KI = Molarity × Volume 
= 0.1 mol ?? -1
× 0.5?? = 0.05 mol 
The stoichiometric ratio between I
-
and I
2
 is 
???? : ?? 2
 
?? -
: ?? 2
 
5 : 3 
1:
3
5
 
For 1 mol ?? -
,
3
5
 mol I
2
 is produced. 
So, for 0.05 mol , moles of ?? 2
= 0.05 ×
3
5
= 0.03 mol 
This statement is not correct. 
(D) Equivalent weight of KIO
3
=
 Molecular weight 
5
 
This statement is correct. 
Equivalent weight =
 Molecular weight 
 Valency factor ? number of electrons 
 
Here, valency factor = 5 
Each KIO
3
 molecule gains 5 electrons in this vacation. 
KIO
3
 acts as the ordinating agent, gains electrons from the iodide ion in KI . 
When KIO
3
 is redued to I
2
 each molecule gains 5 electrons. So, the valency factor is 5 and 
equivalent weight is lower than molecular weight. 
This statement is correct. 
Correct statements are A and D. 
Q6: Some ????
?? gas was kept in a sealed container at a pressure of 1 atm and at 273 K . 
This entire amount of ????
?? gas was later passed through an aqueous solution of 
???? (???? )
?? . The excess unreacted ???? (???? )
?? was later neutralized with 0.1 M of 40 mL 
HCl . If the volume of the sealed container of ????
?? was ?? , then ?? is ____ ????
?? (nearest 
integer). [Given : The entire amount of ????
?? ( ?? ) reacted with exactly half the initial 
amount of ???? (???? )
?? present in the aqueous solution.] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 45 
Solution: 
Let moles of CO
2
= n 
moles of Ca(OH )
2
 
total initially = 2n 
excess Ca(OH )
2
= n 
gm equivalent of Ca(OH )
2
= gm equivalent of HCl 
n × 2 = 0.1 ×
40
1000
× 1 
n = 2 × 10
-3