JEE Advanced Previous Year Questions (2021 - 2024): Solid State | Chemistry for JEE Main & Advanced PDF Download

 Page 1


 
Q1: In a metal decient oxide sample, M ?Y 2O4 (M and Y are metals), M is present in both +2
and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M² ? ions present in
M is 1/3, the value of X is ______.
(a) 0.25
(b) 0.33
(c) 0.67
(d) 0.75    [JEE Advanced 2024 Paper 2]
Ans: (d)
T o nd the value of X in the metal-decient oxide sample M Y O , where M is present in both
+2 and +3 oxidation states and Y is in the +3 oxidation state, and given that the fraction of M
ions present in M is 
1
3
, follow these steps:
Charge Balance Equation:
The total charge contributed by the cations (M and Y) must balance the total negative charge
contributed by the oxide ions O:
x(M + M ) + 2(Y ) = 4(-2)
Oxidation States and Charges:
Let the total number of M ions be x.
Given 
1
3
 of M ions are M , the remaining 
2
3
 are M .
Number of Ions:
Number of M ions = 
x
3
Number of M ions = 
2x
3
T otal Positive Charge:
Charge from M ions = 
x
3
 × 2 = 
2x
3
x 2 4
2+
2+ 3+ 3+
2+ 3+
2+
3+
2+
Page 2


 
Q1: In a metal decient oxide sample, M ?Y 2O4 (M and Y are metals), M is present in both +2
and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M² ? ions present in
M is 1/3, the value of X is ______.
(a) 0.25
(b) 0.33
(c) 0.67
(d) 0.75    [JEE Advanced 2024 Paper 2]
Ans: (d)
T o nd the value of X in the metal-decient oxide sample M Y O , where M is present in both
+2 and +3 oxidation states and Y is in the +3 oxidation state, and given that the fraction of M
ions present in M is 
1
3
, follow these steps:
Charge Balance Equation:
The total charge contributed by the cations (M and Y) must balance the total negative charge
contributed by the oxide ions O:
x(M + M ) + 2(Y ) = 4(-2)
Oxidation States and Charges:
Let the total number of M ions be x.
Given 
1
3
 of M ions are M , the remaining 
2
3
 are M .
Number of Ions:
Number of M ions = 
x
3
Number of M ions = 
2x
3
T otal Positive Charge:
Charge from M ions = 
x
3
 × 2 = 
2x
3
x 2 4
2+
2+ 3+ 3+
2+ 3+
2+
3+
2+
Charge from M ions = 
2x
3
 × 3 = 2x
Charge from Y ions = 2 × 3 = 6
T otal Positive Charge Calculation:
2x
3
 + 2x + 6
T otal Negative Charge:
Charge from 4 oxygen ions (O ):
4 × (-2) = -8
Setting Up the Charge Balance:
2x
3
 + 2x + 6 = 8
Solving for x:
2x
3
 + 2x = 2
2x + 6x
3
 = 2
8x
3
 = 2
8x = 6
x = 
6
8
x = 0.75
Thus, the value of X is 0.75.
Answer: Option D: 0.75
3+
3+
2-
Page 3


 
Q1: In a metal decient oxide sample, M ?Y 2O4 (M and Y are metals), M is present in both +2
and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M² ? ions present in
M is 1/3, the value of X is ______.
(a) 0.25
(b) 0.33
(c) 0.67
(d) 0.75    [JEE Advanced 2024 Paper 2]
Ans: (d)
T o nd the value of X in the metal-decient oxide sample M Y O , where M is present in both
+2 and +3 oxidation states and Y is in the +3 oxidation state, and given that the fraction of M
ions present in M is 
1
3
, follow these steps:
Charge Balance Equation:
The total charge contributed by the cations (M and Y) must balance the total negative charge
contributed by the oxide ions O:
x(M + M ) + 2(Y ) = 4(-2)
Oxidation States and Charges:
Let the total number of M ions be x.
Given 
1
3
 of M ions are M , the remaining 
2
3
 are M .
Number of Ions:
Number of M ions = 
x
3
Number of M ions = 
2x
3
T otal Positive Charge:
Charge from M ions = 
x
3
 × 2 = 
2x
3
x 2 4
2+
2+ 3+ 3+
2+ 3+
2+
3+
2+
Charge from M ions = 
2x
3
 × 3 = 2x
Charge from Y ions = 2 × 3 = 6
T otal Positive Charge Calculation:
2x
3
 + 2x + 6
T otal Negative Charge:
Charge from 4 oxygen ions (O ):
4 × (-2) = -8
Setting Up the Charge Balance:
2x
3
 + 2x + 6 = 8
Solving for x:
2x
3
 + 2x = 2
2x + 6x
3
 = 2
8x
3
 = 2
8x = 6
x = 
6
8
x = 0.75
Thus, the value of X is 0.75.
Answer: Option D: 0.75
3+
3+
2-
 
 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
Page 4


 
Q1: In a metal decient oxide sample, M ?Y 2O4 (M and Y are metals), M is present in both +2
and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M² ? ions present in
M is 1/3, the value of X is ______.
(a) 0.25
(b) 0.33
(c) 0.67
(d) 0.75    [JEE Advanced 2024 Paper 2]
Ans: (d)
T o nd the value of X in the metal-decient oxide sample M Y O , where M is present in both
+2 and +3 oxidation states and Y is in the +3 oxidation state, and given that the fraction of M
ions present in M is 
1
3
, follow these steps:
Charge Balance Equation:
The total charge contributed by the cations (M and Y) must balance the total negative charge
contributed by the oxide ions O:
x(M + M ) + 2(Y ) = 4(-2)
Oxidation States and Charges:
Let the total number of M ions be x.
Given 
1
3
 of M ions are M , the remaining 
2
3
 are M .
Number of Ions:
Number of M ions = 
x
3
Number of M ions = 
2x
3
T otal Positive Charge:
Charge from M ions = 
x
3
 × 2 = 
2x
3
x 2 4
2+
2+ 3+ 3+
2+ 3+
2+
3+
2+
Charge from M ions = 
2x
3
 × 3 = 2x
Charge from Y ions = 2 × 3 = 6
T otal Positive Charge Calculation:
2x
3
 + 2x + 6
T otal Negative Charge:
Charge from 4 oxygen ions (O ):
4 × (-2) = -8
Setting Up the Charge Balance:
2x
3
 + 2x + 6 = 8
Solving for x:
2x
3
 + 2x = 2
2x + 6x
3
 = 2
8x
3
 = 2
8x = 6
x = 
6
8
x = 0.75
Thus, the value of X is 0.75.
Answer: Option D: 0.75
3+
3+
2-
 
 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
?? ?? = 4?? ?? =
32
v3
?? ?? … . (5) (for BCC, using Eq. (1)) 
?? ?? = 2?? ?? = 8?? ?? … . (6)  (for simple cubic, using Eq. (3)) 
 
Comparing the edge lengths : 
From Eq. (4), (5) and (6), we find : 
?? ?? > ?? ?? > ?? ?? 
This means Option B: ?? ?? > ?? ?? is correct. 
 
Packing Efficiencies: 
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows: 
? FCC : 
?? ?? ?????? =
4 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? v2
 
? BCC: 
?? ?? ?????? =
2 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
3?? v3
32
 
? Simple Cubic : 
?? ?? ????
=
1 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? 6
 
Comparing these packing efficiencies, we find: 
?? ?? ?????? > ?? ?? ?????? > ?? ?? ????
 
This means Option A : Packing efficiency of unit cell of ?? > ?? > ?? is correct. 
5 Calculating Densities : 
The densities of x and y are given by the formula ?? =
?? ·?? ?? ?? ·?? 3
, where ?? is the number of 
atoms in the unit cell, ?? is the molar mass of the metal, ?? ?? is Avogadro's number, and ?? 
is the edge length of the unit cell. 
? For ?? (FCC), ?? = 4 : 
?? ?? =
4?? ?? ?? ?? (?? ?? )
3
 
? For y (BCC), ?? = 2 : 
Page 5


 
Q1: In a metal decient oxide sample, M ?Y 2O4 (M and Y are metals), M is present in both +2
and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M² ? ions present in
M is 1/3, the value of X is ______.
(a) 0.25
(b) 0.33
(c) 0.67
(d) 0.75    [JEE Advanced 2024 Paper 2]
Ans: (d)
T o nd the value of X in the metal-decient oxide sample M Y O , where M is present in both
+2 and +3 oxidation states and Y is in the +3 oxidation state, and given that the fraction of M
ions present in M is 
1
3
, follow these steps:
Charge Balance Equation:
The total charge contributed by the cations (M and Y) must balance the total negative charge
contributed by the oxide ions O:
x(M + M ) + 2(Y ) = 4(-2)
Oxidation States and Charges:
Let the total number of M ions be x.
Given 
1
3
 of M ions are M , the remaining 
2
3
 are M .
Number of Ions:
Number of M ions = 
x
3
Number of M ions = 
2x
3
T otal Positive Charge:
Charge from M ions = 
x
3
 × 2 = 
2x
3
x 2 4
2+
2+ 3+ 3+
2+ 3+
2+
3+
2+
Charge from M ions = 
2x
3
 × 3 = 2x
Charge from Y ions = 2 × 3 = 6
T otal Positive Charge Calculation:
2x
3
 + 2x + 6
T otal Negative Charge:
Charge from 4 oxygen ions (O ):
4 × (-2) = -8
Setting Up the Charge Balance:
2x
3
 + 2x + 6 = 8
Solving for x:
2x
3
 + 2x = 2
2x + 6x
3
 = 2
8x
3
 = 2
8x = 6
x = 
6
8
x = 0.75
Thus, the value of X is 0.75.
Answer: Option D: 0.75
3+
3+
2-
 
 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
?? ?? = 4?? ?? =
32
v3
?? ?? … . (5) (for BCC, using Eq. (1)) 
?? ?? = 2?? ?? = 8?? ?? … . (6)  (for simple cubic, using Eq. (3)) 
 
Comparing the edge lengths : 
From Eq. (4), (5) and (6), we find : 
?? ?? > ?? ?? > ?? ?? 
This means Option B: ?? ?? > ?? ?? is correct. 
 
Packing Efficiencies: 
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows: 
? FCC : 
?? ?? ?????? =
4 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? v2
 
? BCC: 
?? ?? ?????? =
2 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
3?? v3
32
 
? Simple Cubic : 
?? ?? ????
=
1 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? 6
 
Comparing these packing efficiencies, we find: 
?? ?? ?????? > ?? ?? ?????? > ?? ?? ????
 
This means Option A : Packing efficiency of unit cell of ?? > ?? > ?? is correct. 
5 Calculating Densities : 
The densities of x and y are given by the formula ?? =
?? ·?? ?? ?? ·?? 3
, where ?? is the number of 
atoms in the unit cell, ?? is the molar mass of the metal, ?? ?? is Avogadro's number, and ?? 
is the edge length of the unit cell. 
? For ?? (FCC), ?? = 4 : 
?? ?? =
4?? ?? ?? ?? (?? ?? )
3
 
? For y (BCC), ?? = 2 : 
?? ?? =
2?? ?? ?? ?? (?? ?? )
3
 
?? ?? =
3
2
?? ?? and ?? ?? = 3?? ?? . This implies that ?? ?? =
?? ?? 2
. Thus, 2?? ?? = ?? ?? , we can write : 
?? ?? ?? ?? =
2?? ?? ?? ?? × (
?? ?? ?? ?? )
3
 
Substituting the given relation 2?? ?? = ?? ?? into Eq. (7), we get : 
?? ?? ?? ?? = 2 × 1 × (
32/3
2v2
)
3
. 
After simplifying Eq. (8), we find ?? ?? > ?? ?? . 
This means Option D : Density of ?? > ?? is correct. 
So, the correct answers are Options A, B, and D.