JEE Advanced Previous Year Questions (2021 - 2024): p-block Elements | Chemistry for JEE Main & Advanced PDF Download

 Page 1


  
Q1: The species formed on fl uorination of phosphorus pentachloride in a polar organic solvent are:
(a) [PF 4] ?[PF 6] ? and [PCl 4] ?[PF 6] ?
(b) [PCl 4] ?[PCl 4F 2]? and [PCl 4] ?[PF 6]?
(c) PF 3 and PCl 3
(d) PF 5 and PCl 3   [JEE Advanced 2024 Paper 2]
Ans: (b)
When PCl 5 is fl uorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic
species can be understood as follows:
PCl 5 in a polar organic solvent can dissociate into ionic forms, such as [PCl 4]? and [PCl 6]?.
Upon fl uorination, PCl 5 can be converted into [PF 6]? and [PCl 4] ?.
Further fl uorination can lead to the formation of mixed ionic compounds where PCl 5 can form isomers like
[PCl 4]?[PCl 4F2]?.
The two ionic isomers formed are:
[PCl 4]?[PCl 4F2]?, which are colorless crystals.
[PCl 4]?[PF 6] ?, which are white crystals.
Hence, the ionic isomers formed when PCl 5 is fl uorinated in a polar organic solvent are:
Option B: [PCl 4] ?[PCl 4F 2] ? and [PCl 4] ?[PF 6]?
Q2: At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid
(HNO 2) gives the species:
(a) H 3O ?, NO 3?, and NO
(b) H 3O ?, NO 3?, and NO 2
(c) H 3O?, NO ?, and NO 2
(d) H 3O ?, NO 3?, and N 2O    [JEE Advanced 2024 Paper 1]
Ans: (a)
T o determine the products of the disproportionation of nitrous acid (HNO 2) at room temperature, let's
Page 2


  
Q1: The species formed on fl uorination of phosphorus pentachloride in a polar organic solvent are:
(a) [PF 4] ?[PF 6] ? and [PCl 4] ?[PF 6] ?
(b) [PCl 4] ?[PCl 4F 2]? and [PCl 4] ?[PF 6]?
(c) PF 3 and PCl 3
(d) PF 5 and PCl 3   [JEE Advanced 2024 Paper 2]
Ans: (b)
When PCl 5 is fl uorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic
species can be understood as follows:
PCl 5 in a polar organic solvent can dissociate into ionic forms, such as [PCl 4]? and [PCl 6]?.
Upon fl uorination, PCl 5 can be converted into [PF 6]? and [PCl 4] ?.
Further fl uorination can lead to the formation of mixed ionic compounds where PCl 5 can form isomers like
[PCl 4]?[PCl 4F2]?.
The two ionic isomers formed are:
[PCl 4]?[PCl 4F2]?, which are colorless crystals.
[PCl 4]?[PF 6] ?, which are white crystals.
Hence, the ionic isomers formed when PCl 5 is fl uorinated in a polar organic solvent are:
Option B: [PCl 4] ?[PCl 4F 2] ? and [PCl 4] ?[PF 6]?
Q2: At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid
(HNO 2) gives the species:
(a) H 3O ?, NO 3?, and NO
(b) H 3O ?, NO 3?, and NO 2
(c) H 3O?, NO ?, and NO 2
(d) H 3O ?, NO 3?, and N 2O    [JEE Advanced 2024 Paper 1]
Ans: (a)
T o determine the products of the disproportionation of nitrous acid (HNO 2) at room temperature, let's
analyze the chemical behavior of HNO 2 in aqueous solution.
Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced,
yielding two different products. In the case of nitrous acid (HNO 2), the reaction can be represented as
follows:
3HNO 2 ? HNO 3 + H 2O + 2NO
Breaking this down, we see that nitrous acid disproportionates into nitric acid (HNO 3), water (H 2O), and
nitrogen monoxide (NO). Additionally, in an aqueous solution, nitric acid dissociates to form H 3O ?
(hydronium ion) and NO 3? (nitrate ion).
Therefore, the species formed from this disproportionation reaction are H 3O ?, NO 3?, and NO.
From the given options, the correct answer is:
Option A
H 3O ?, NO 3?, and NO
Q3: The compound(s) having peroxide linkage is(are):
(a) H 2S 2O7
(b) H 2S2O 8
(c) H 2S 2O5
(d) H 2SO 5  [JEE Advanced 2024 Paper 2]
Ans: (b), (d)
Page 3


  
Q1: The species formed on fl uorination of phosphorus pentachloride in a polar organic solvent are:
(a) [PF 4] ?[PF 6] ? and [PCl 4] ?[PF 6] ?
(b) [PCl 4] ?[PCl 4F 2]? and [PCl 4] ?[PF 6]?
(c) PF 3 and PCl 3
(d) PF 5 and PCl 3   [JEE Advanced 2024 Paper 2]
Ans: (b)
When PCl 5 is fl uorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic
species can be understood as follows:
PCl 5 in a polar organic solvent can dissociate into ionic forms, such as [PCl 4]? and [PCl 6]?.
Upon fl uorination, PCl 5 can be converted into [PF 6]? and [PCl 4] ?.
Further fl uorination can lead to the formation of mixed ionic compounds where PCl 5 can form isomers like
[PCl 4]?[PCl 4F2]?.
The two ionic isomers formed are:
[PCl 4]?[PCl 4F2]?, which are colorless crystals.
[PCl 4]?[PF 6] ?, which are white crystals.
Hence, the ionic isomers formed when PCl 5 is fl uorinated in a polar organic solvent are:
Option B: [PCl 4] ?[PCl 4F 2] ? and [PCl 4] ?[PF 6]?
Q2: At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid
(HNO 2) gives the species:
(a) H 3O ?, NO 3?, and NO
(b) H 3O ?, NO 3?, and NO 2
(c) H 3O?, NO ?, and NO 2
(d) H 3O ?, NO 3?, and N 2O    [JEE Advanced 2024 Paper 1]
Ans: (a)
T o determine the products of the disproportionation of nitrous acid (HNO 2) at room temperature, let's
analyze the chemical behavior of HNO 2 in aqueous solution.
Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced,
yielding two different products. In the case of nitrous acid (HNO 2), the reaction can be represented as
follows:
3HNO 2 ? HNO 3 + H 2O + 2NO
Breaking this down, we see that nitrous acid disproportionates into nitric acid (HNO 3), water (H 2O), and
nitrogen monoxide (NO). Additionally, in an aqueous solution, nitric acid dissociates to form H 3O ?
(hydronium ion) and NO 3? (nitrate ion).
Therefore, the species formed from this disproportionation reaction are H 3O ?, NO 3?, and NO.
From the given options, the correct answer is:
Option A
H 3O ?, NO 3?, and NO
Q3: The compound(s) having peroxide linkage is(are):
(a) H 2S 2O7
(b) H 2S2O 8
(c) H 2S 2O5
(d) H 2SO 5  [JEE Advanced 2024 Paper 2]
Ans: (b), (d)
Page 4


  
Q1: The species formed on fl uorination of phosphorus pentachloride in a polar organic solvent are:
(a) [PF 4] ?[PF 6] ? and [PCl 4] ?[PF 6] ?
(b) [PCl 4] ?[PCl 4F 2]? and [PCl 4] ?[PF 6]?
(c) PF 3 and PCl 3
(d) PF 5 and PCl 3   [JEE Advanced 2024 Paper 2]
Ans: (b)
When PCl 5 is fl uorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic
species can be understood as follows:
PCl 5 in a polar organic solvent can dissociate into ionic forms, such as [PCl 4]? and [PCl 6]?.
Upon fl uorination, PCl 5 can be converted into [PF 6]? and [PCl 4] ?.
Further fl uorination can lead to the formation of mixed ionic compounds where PCl 5 can form isomers like
[PCl 4]?[PCl 4F2]?.
The two ionic isomers formed are:
[PCl 4]?[PCl 4F2]?, which are colorless crystals.
[PCl 4]?[PF 6] ?, which are white crystals.
Hence, the ionic isomers formed when PCl 5 is fl uorinated in a polar organic solvent are:
Option B: [PCl 4] ?[PCl 4F 2] ? and [PCl 4] ?[PF 6]?
Q2: At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid
(HNO 2) gives the species:
(a) H 3O ?, NO 3?, and NO
(b) H 3O ?, NO 3?, and NO 2
(c) H 3O?, NO ?, and NO 2
(d) H 3O ?, NO 3?, and N 2O    [JEE Advanced 2024 Paper 1]
Ans: (a)
T o determine the products of the disproportionation of nitrous acid (HNO 2) at room temperature, let's
analyze the chemical behavior of HNO 2 in aqueous solution.
Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced,
yielding two different products. In the case of nitrous acid (HNO 2), the reaction can be represented as
follows:
3HNO 2 ? HNO 3 + H 2O + 2NO
Breaking this down, we see that nitrous acid disproportionates into nitric acid (HNO 3), water (H 2O), and
nitrogen monoxide (NO). Additionally, in an aqueous solution, nitric acid dissociates to form H 3O ?
(hydronium ion) and NO 3? (nitrate ion).
Therefore, the species formed from this disproportionation reaction are H 3O ?, NO 3?, and NO.
From the given options, the correct answer is:
Option A
H 3O ?, NO 3?, and NO
Q3: The compound(s) having peroxide linkage is(are):
(a) H 2S 2O7
(b) H 2S2O 8
(c) H 2S 2O5
(d) H 2SO 5  [JEE Advanced 2024 Paper 2]
Ans: (b), (d)
 
 
Single Correct Type 
JEE Advanced 2023 Paper 1 Online 
Q1. MCQ (Single Correct Answer) +?? - ?? 
Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their 
products given in List-II and choose the correct option. 
 
List - I List - II 
(P) ?? ?? ?? ?? + ?? ?? ?? ?? ? (1) ?? (?? )(?????? ?? )????
?? 
(Q) ?? ?? + ?????????? + ?? ?? ?? ?? ? (2) ?? ?? ????
?? 
(R) ?????? ?? + ????
?? ???????? ? (3) ????
?? 
(S) ?? ?? ????
?? + ?? ?? ?? ?? + ?? ???????? ?? ? (4) ???????? ?? 
 (5) ?? ?? ????
?? 
 
A. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
B. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
C. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
D. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
Ans: (D) 
P
2
O
3
+ 3H
2
O ? H
3
PO
3
, P ? 2 
P
4
+ 3NaOH + 3H
2
O ? PH
3
+ NaH
2
PO
2
, Q ? 3 
PCl
5
+ CH
3
COOH ? CH
3
COCl + POCl
3
, R ? 4 
H
3
PO
2
+ 2H
2
O + 4AgNO
3
? H
3
PO
4
+ Ag + HNO
3
,  S ? 5 
 
Page 5


  
Q1: The species formed on fl uorination of phosphorus pentachloride in a polar organic solvent are:
(a) [PF 4] ?[PF 6] ? and [PCl 4] ?[PF 6] ?
(b) [PCl 4] ?[PCl 4F 2]? and [PCl 4] ?[PF 6]?
(c) PF 3 and PCl 3
(d) PF 5 and PCl 3   [JEE Advanced 2024 Paper 2]
Ans: (b)
When PCl 5 is fl uorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic
species can be understood as follows:
PCl 5 in a polar organic solvent can dissociate into ionic forms, such as [PCl 4]? and [PCl 6]?.
Upon fl uorination, PCl 5 can be converted into [PF 6]? and [PCl 4] ?.
Further fl uorination can lead to the formation of mixed ionic compounds where PCl 5 can form isomers like
[PCl 4]?[PCl 4F2]?.
The two ionic isomers formed are:
[PCl 4]?[PCl 4F2]?, which are colorless crystals.
[PCl 4]?[PF 6] ?, which are white crystals.
Hence, the ionic isomers formed when PCl 5 is fl uorinated in a polar organic solvent are:
Option B: [PCl 4] ?[PCl 4F 2] ? and [PCl 4] ?[PF 6]?
Q2: At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid
(HNO 2) gives the species:
(a) H 3O ?, NO 3?, and NO
(b) H 3O ?, NO 3?, and NO 2
(c) H 3O?, NO ?, and NO 2
(d) H 3O ?, NO 3?, and N 2O    [JEE Advanced 2024 Paper 1]
Ans: (a)
T o determine the products of the disproportionation of nitrous acid (HNO 2) at room temperature, let's
analyze the chemical behavior of HNO 2 in aqueous solution.
Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced,
yielding two different products. In the case of nitrous acid (HNO 2), the reaction can be represented as
follows:
3HNO 2 ? HNO 3 + H 2O + 2NO
Breaking this down, we see that nitrous acid disproportionates into nitric acid (HNO 3), water (H 2O), and
nitrogen monoxide (NO). Additionally, in an aqueous solution, nitric acid dissociates to form H 3O ?
(hydronium ion) and NO 3? (nitrate ion).
Therefore, the species formed from this disproportionation reaction are H 3O ?, NO 3?, and NO.
From the given options, the correct answer is:
Option A
H 3O ?, NO 3?, and NO
Q3: The compound(s) having peroxide linkage is(are):
(a) H 2S 2O7
(b) H 2S2O 8
(c) H 2S 2O5
(d) H 2SO 5  [JEE Advanced 2024 Paper 2]
Ans: (b), (d)
 
 
Single Correct Type 
JEE Advanced 2023 Paper 1 Online 
Q1. MCQ (Single Correct Answer) +?? - ?? 
Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their 
products given in List-II and choose the correct option. 
 
List - I List - II 
(P) ?? ?? ?? ?? + ?? ?? ?? ?? ? (1) ?? (?? )(?????? ?? )????
?? 
(Q) ?? ?? + ?????????? + ?? ?? ?? ?? ? (2) ?? ?? ????
?? 
(R) ?????? ?? + ????
?? ???????? ? (3) ????
?? 
(S) ?? ?? ????
?? + ?? ?? ?? ?? + ?? ???????? ?? ? (4) ???????? ?? 
 (5) ?? ?? ????
?? 
 
A. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
B. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
C. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
D. ?? ? ?? ; ?? ? ?? ; ?? ? ?? ; ?? ? ?? 
Ans: (D) 
P
2
O
3
+ 3H
2
O ? H
3
PO
3
, P ? 2 
P
4
+ 3NaOH + 3H
2
O ? PH
3
+ NaH
2
PO
2
, Q ? 3 
PCl
5
+ CH
3
COOH ? CH
3
COCl + POCl
3
, R ? 4 
H
3
PO
2
+ 2H
2
O + 4AgNO
3
? H
3
PO
4
+ Ag + HNO
3
,  S ? 5 
 
2022 
Numerical Type  
JEE Advanced 2022 Paper 1 Online 
Q1. Dissolving ?? . ???? ?? of white phosphorous in boiling ???????? solution in an inert atmosphere 
gives a gas ?? . The amount of ???????? ?? (in ?? ) required to completely consume the gas ?? is 
[Given: Atomic mass of ?? = ?? , ?? = ???? , ???? = ???? , ?? = ???? , ?? = ???? , ???? = ???? ] 
Ans: 2.37 to 2.41 
 
P
4
1.24 g
 or 
0.01 mole 
+ 3NaOH + 3H
2
O ? PH
3
+ 3NaH
2
PO
2
 
As NaOH is present in excess. So, amount of phosphine formed is 0.01 mole (as P
4
 is limiting) 
2PH
3
0.01 mole 
+ 3CuSO
4
? Cu
3
P
2
+ 3H
2
SO
4
 
Amount of CuSO
4
 required =
3×0.01
2
 mole 
Mass of CuSO
4
(in g) required =
0.03
2
× (63 + 32 + 16 × 4) 
=
0.03
2
× 159 = 2.38 g, 
 
JEE Advanced 2022 Paper 2 Online 
Q2.The reaction of ???????? ?? with ?????? gives a paramagnetic gas, which upon reaction with ?? ?? 
produces 
(A) ????
?? ?? 
(B) ?????? ?? 
(C) ????
?? ?? ?? 
(D) ????
?? ?? ?? 
Ans: (C) 
HClO
3
 reacts with HCl according to the following equation, 
2HClO
3
+ 2HCl ? 2ClO
2
+ Cl
2
+ 2H
2
O 
ClO
2
 molecule is paramagnetic, as it contains odd number of electrons. 
2ClO
2
+ 2O
3
? Cl
2
O
6
+ 2O
2
 
 
JEE Advanced 2022 Paper 2 Online 
Q3. The reaction of ???? (????
?? )
?? and ???????? in water produces a precipitate that dissolves upon the 
addition of ?????? of appropriate concentration. The dissolution of the precipitate is due to the 
formation of 
(A) ???????? ?? 
(B) ?? ?????? ?? 
(C) [???????? ?? ]
?? -
 
(D) [???????? ?? ]
?? -