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CBSE XI | Chemistry
Sample Paper – 4 Solution
CBSE
Class XI Chemistry
Sample Paper – 4 Solution
Section A
1. Organic compound is fused with sodium metal to convert N, S, P and halogens present in
organic compound to their corresponding sodium salts.
OR
CH3CH2OCH2CH3 and CH3OCH2CH2CH3 are metamers.
2. sp
3
d
hybrid orbitals -Trigonal bipyramidal
sp
3
d
2
hybrid orbitals-Octahedral
OR
Increasing bonds order of ionic character:
Br-H< C-H< F-H< Na-I< Li-Cl< K-F
3. It is due to the delocalization of ? electrons in benzene it is highly stable.
4. Mixing of soil or rock particles in water is called siltation.
5. Because alkali and alkaline earth metals are themselves stronger reducing agents than
the majority of other reducing agents.
6. Atomic radii decrease across a period. Cations is smaller than their parent atoms. Among
isoelectronic species, the one with the larger positive nuclear charge will have a smaller
radius. Hence the largest species is Mg; the smallest one is Al3+.
7.
(a) All of them are isoelectronic in nature and have 10 electrons each.
(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic
radius.
Al
3+
< Mg
2+
< Na
+
< F
–
< O
2-
< N
3-
8.
According to Rydberg equation,
Page 2
CBSE XI | Chemistry
Sample Paper – 4 Solution
CBSE
Class XI Chemistry
Sample Paper – 4 Solution
Section A
1. Organic compound is fused with sodium metal to convert N, S, P and halogens present in
organic compound to their corresponding sodium salts.
OR
CH3CH2OCH2CH3 and CH3OCH2CH2CH3 are metamers.
2. sp
3
d
hybrid orbitals -Trigonal bipyramidal
sp
3
d
2
hybrid orbitals-Octahedral
OR
Increasing bonds order of ionic character:
Br-H< C-H< F-H< Na-I< Li-Cl< K-F
3. It is due to the delocalization of ? electrons in benzene it is highly stable.
4. Mixing of soil or rock particles in water is called siltation.
5. Because alkali and alkaline earth metals are themselves stronger reducing agents than
the majority of other reducing agents.
6. Atomic radii decrease across a period. Cations is smaller than their parent atoms. Among
isoelectronic species, the one with the larger positive nuclear charge will have a smaller
radius. Hence the largest species is Mg; the smallest one is Al3+.
7.
(a) All of them are isoelectronic in nature and have 10 electrons each.
(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic
radius.
Al
3+
< Mg
2+
< Na
+
< F
–
< O
2-
< N
3-
8.
According to Rydberg equation,
CBSE XI | Chemistry
Sample Paper – 4 Solution
12
22
R = 109,677 which is the Rydberg's constant
1
v =
Wavelength
For first line in Balmer series, n = 2, n = 3
Given wavelength of 1st spectral line = 6561 Å
1 1 1 5
Therefore, = R =R
6561 36
23
?? ??
?
? ??
?? ??
12
22
(i)
For second line in Balmer series, n = 2, n =4
1 1 1 3
Therefore, = R = R (ii)
Wavelength 16
24
Dividing eq. (i) by (ii), we get:
Wavelength 5 16
6561 36 3
?
?? ??
?
?? ??
?? ??
?
?
?
Wavelength=4860 Å ?
9.
1
1
2
12
12
12
2
1
2
V = 2 L
T = (23.4 + 273) K
= 296.4 K
T = (26.1 + 273) K
= 299.1 K
From Charles law
VV
TT
VT
V
T
2L 299.1 K
V
296.4 K
2 L 1.009
2.018 L
?
??
?
??
??
?
Page 3
CBSE XI | Chemistry
Sample Paper – 4 Solution
CBSE
Class XI Chemistry
Sample Paper – 4 Solution
Section A
1. Organic compound is fused with sodium metal to convert N, S, P and halogens present in
organic compound to their corresponding sodium salts.
OR
CH3CH2OCH2CH3 and CH3OCH2CH2CH3 are metamers.
2. sp
3
d
hybrid orbitals -Trigonal bipyramidal
sp
3
d
2
hybrid orbitals-Octahedral
OR
Increasing bonds order of ionic character:
Br-H< C-H< F-H< Na-I< Li-Cl< K-F
3. It is due to the delocalization of ? electrons in benzene it is highly stable.
4. Mixing of soil or rock particles in water is called siltation.
5. Because alkali and alkaline earth metals are themselves stronger reducing agents than
the majority of other reducing agents.
6. Atomic radii decrease across a period. Cations is smaller than their parent atoms. Among
isoelectronic species, the one with the larger positive nuclear charge will have a smaller
radius. Hence the largest species is Mg; the smallest one is Al3+.
7.
(a) All of them are isoelectronic in nature and have 10 electrons each.
(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic
radius.
Al
3+
< Mg
2+
< Na
+
< F
–
< O
2-
< N
3-
8.
According to Rydberg equation,
CBSE XI | Chemistry
Sample Paper – 4 Solution
12
22
R = 109,677 which is the Rydberg's constant
1
v =
Wavelength
For first line in Balmer series, n = 2, n = 3
Given wavelength of 1st spectral line = 6561 Å
1 1 1 5
Therefore, = R =R
6561 36
23
?? ??
?
? ??
?? ??
12
22
(i)
For second line in Balmer series, n = 2, n =4
1 1 1 3
Therefore, = R = R (ii)
Wavelength 16
24
Dividing eq. (i) by (ii), we get:
Wavelength 5 16
6561 36 3
?
?? ??
?
?? ??
?? ??
?
?
?
Wavelength=4860 Å ?
9.
1
1
2
12
12
12
2
1
2
V = 2 L
T = (23.4 + 273) K
= 296.4 K
T = (26.1 + 273) K
= 299.1 K
From Charles law
VV
TT
VT
V
T
2L 299.1 K
V
296.4 K
2 L 1.009
2.018 L
?
??
?
??
??
?
CBSE XI | Chemistry
Sample Paper – 4 Solution
10.
- - 2-
3 3 4
a) First, we will write down the oxidation number of each atom
22
Cr(OH) + IO I + CrO
3 5 1 6
b) Write separately oxidation &
??
?
? ? ? ?
2- -
34
- - -
3
2
reduction half reactions
Oxidation half reaction:
Cr(OH) CrO 3e
36
Reduction half reaction:
IO + 6e I
5 1
c) Balance O atoms by adding H O molecules to the side deficient in
??
??
?
??
2- -
3 2 4
- - -
32
'O'
atoms and then balancing H atoms
Cr(OH) + H O CrO 3e
IO + 6e I + 3H O
??
?
2
-
d) Balance H atoms. Since the medium is alkaline, therefore H O molecules
are added to the side deficient in H atoms and equal no. of OH
ions to the other side :
Cr(O
- 2- -
3 4 2
- 2- -
3 2 4 2
- - - -
32
- - -
3 2 2
H) + 5OH CrO 3e +4H O
( Cr(OH) + H O 5OH CrO 3e +5H O)
IO + 6e 3H O I + 6OH
( IO + 6e 6H O I + 3H O
??
? ? ?
??
??
-
- 2- -
3 4 2
+ 6OH
e) Equalise the electrons lost and gained by multiplying the oxidation half
reaction with 2.
2Cr(OH) + 10OH 2CrO 6e +8H O
Adding the oxidation half reaction and reduction half reaction we
??
)
- 2- -
3 4 2
- - - -
32
get
2Cr(OH) + 10OH 2CrO 6e +8H O
IO + 6e 3H O I + 6OH
??
??
_____________________________________
- - 2- -
3 3 4 2
2Cr(OH) IO 4OH 2CrO I +5H O ? ? ? ?
____________
Page 4
CBSE XI | Chemistry
Sample Paper – 4 Solution
CBSE
Class XI Chemistry
Sample Paper – 4 Solution
Section A
1. Organic compound is fused with sodium metal to convert N, S, P and halogens present in
organic compound to their corresponding sodium salts.
OR
CH3CH2OCH2CH3 and CH3OCH2CH2CH3 are metamers.
2. sp
3
d
hybrid orbitals -Trigonal bipyramidal
sp
3
d
2
hybrid orbitals-Octahedral
OR
Increasing bonds order of ionic character:
Br-H< C-H< F-H< Na-I< Li-Cl< K-F
3. It is due to the delocalization of ? electrons in benzene it is highly stable.
4. Mixing of soil or rock particles in water is called siltation.
5. Because alkali and alkaline earth metals are themselves stronger reducing agents than
the majority of other reducing agents.
6. Atomic radii decrease across a period. Cations is smaller than their parent atoms. Among
isoelectronic species, the one with the larger positive nuclear charge will have a smaller
radius. Hence the largest species is Mg; the smallest one is Al3+.
7.
(a) All of them are isoelectronic in nature and have 10 electrons each.
(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic
radius.
Al
3+
< Mg
2+
< Na
+
< F
–
< O
2-
< N
3-
8.
According to Rydberg equation,
CBSE XI | Chemistry
Sample Paper – 4 Solution
12
22
R = 109,677 which is the Rydberg's constant
1
v =
Wavelength
For first line in Balmer series, n = 2, n = 3
Given wavelength of 1st spectral line = 6561 Å
1 1 1 5
Therefore, = R =R
6561 36
23
?? ??
?
? ??
?? ??
12
22
(i)
For second line in Balmer series, n = 2, n =4
1 1 1 3
Therefore, = R = R (ii)
Wavelength 16
24
Dividing eq. (i) by (ii), we get:
Wavelength 5 16
6561 36 3
?
?? ??
?
?? ??
?? ??
?
?
?
Wavelength=4860 Å ?
9.
1
1
2
12
12
12
2
1
2
V = 2 L
T = (23.4 + 273) K
= 296.4 K
T = (26.1 + 273) K
= 299.1 K
From Charles law
VV
TT
VT
V
T
2L 299.1 K
V
296.4 K
2 L 1.009
2.018 L
?
??
?
??
??
?
CBSE XI | Chemistry
Sample Paper – 4 Solution
10.
- - 2-
3 3 4
a) First, we will write down the oxidation number of each atom
22
Cr(OH) + IO I + CrO
3 5 1 6
b) Write separately oxidation &
??
?
? ? ? ?
2- -
34
- - -
3
2
reduction half reactions
Oxidation half reaction:
Cr(OH) CrO 3e
36
Reduction half reaction:
IO + 6e I
5 1
c) Balance O atoms by adding H O molecules to the side deficient in
??
??
?
??
2- -
3 2 4
- - -
32
'O'
atoms and then balancing H atoms
Cr(OH) + H O CrO 3e
IO + 6e I + 3H O
??
?
2
-
d) Balance H atoms. Since the medium is alkaline, therefore H O molecules
are added to the side deficient in H atoms and equal no. of OH
ions to the other side :
Cr(O
- 2- -
3 4 2
- 2- -
3 2 4 2
- - - -
32
- - -
3 2 2
H) + 5OH CrO 3e +4H O
( Cr(OH) + H O 5OH CrO 3e +5H O)
IO + 6e 3H O I + 6OH
( IO + 6e 6H O I + 3H O
??
? ? ?
??
??
-
- 2- -
3 4 2
+ 6OH
e) Equalise the electrons lost and gained by multiplying the oxidation half
reaction with 2.
2Cr(OH) + 10OH 2CrO 6e +8H O
Adding the oxidation half reaction and reduction half reaction we
??
)
- 2- -
3 4 2
- - - -
32
get
2Cr(OH) + 10OH 2CrO 6e +8H O
IO + 6e 3H O I + 6OH
??
??
_____________________________________
- - 2- -
3 3 4 2
2Cr(OH) IO 4OH 2CrO I +5H O ? ? ? ?
____________
CBSE XI | Chemistry
Sample Paper – 4 Solution
11.
(a) Anhydrous AlCl3 is covalent but hydrated AlCl3 is electrovalent because when it is
dissolved in water the high heat of hydration is sufficient to break the covalent bond
of AlCl3 into Al
3+
and Cl
-
ions.
(b) Boric acid behaves as Lewis acid by accepting a pair of electron from OH
-
ion (in
water).
-
3 4 3
B(OH) + 2H-O-H [B(OH) ] + H O
?
?
12. 1 molar solution contains 1mole of solute in 1 L of solution while
1 molal solution contains 1 mole of solute in 1000g of solvent.
Considering density of water as almost 1g/mL, then 1mole of solute is present in
1000mL of water in 1molal solution while 1mole of it is present in less than 1000mL of
water in 1 molar solution
(1000mL solution in molar solution = Volume of solute + Volume of solvent).
Thus 1M solution is more concentrated than 1m solution.
OR
Given:
Molarity of solution = 0.5 M
Volume of solution = 250 cm
3
0.5 M NaCl solution contains 0.5 mole of NaCl in 1 litre of solution.
Number of moles of NaCl in 250 cm
3
0.500
4
0.125mol
?
?
Molar mass of NaCl = 58.44 g
Mass of 0.125 mol of NaCl = 58.44 ×0.125
= 7.305 g of NaCl
Page 5
CBSE XI | Chemistry
Sample Paper – 4 Solution
CBSE
Class XI Chemistry
Sample Paper – 4 Solution
Section A
1. Organic compound is fused with sodium metal to convert N, S, P and halogens present in
organic compound to their corresponding sodium salts.
OR
CH3CH2OCH2CH3 and CH3OCH2CH2CH3 are metamers.
2. sp
3
d
hybrid orbitals -Trigonal bipyramidal
sp
3
d
2
hybrid orbitals-Octahedral
OR
Increasing bonds order of ionic character:
Br-H< C-H< F-H< Na-I< Li-Cl< K-F
3. It is due to the delocalization of ? electrons in benzene it is highly stable.
4. Mixing of soil or rock particles in water is called siltation.
5. Because alkali and alkaline earth metals are themselves stronger reducing agents than
the majority of other reducing agents.
6. Atomic radii decrease across a period. Cations is smaller than their parent atoms. Among
isoelectronic species, the one with the larger positive nuclear charge will have a smaller
radius. Hence the largest species is Mg; the smallest one is Al3+.
7.
(a) All of them are isoelectronic in nature and have 10 electrons each.
(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic
radius.
Al
3+
< Mg
2+
< Na
+
< F
–
< O
2-
< N
3-
8.
According to Rydberg equation,
CBSE XI | Chemistry
Sample Paper – 4 Solution
12
22
R = 109,677 which is the Rydberg's constant
1
v =
Wavelength
For first line in Balmer series, n = 2, n = 3
Given wavelength of 1st spectral line = 6561 Å
1 1 1 5
Therefore, = R =R
6561 36
23
?? ??
?
? ??
?? ??
12
22
(i)
For second line in Balmer series, n = 2, n =4
1 1 1 3
Therefore, = R = R (ii)
Wavelength 16
24
Dividing eq. (i) by (ii), we get:
Wavelength 5 16
6561 36 3
?
?? ??
?
?? ??
?? ??
?
?
?
Wavelength=4860 Å ?
9.
1
1
2
12
12
12
2
1
2
V = 2 L
T = (23.4 + 273) K
= 296.4 K
T = (26.1 + 273) K
= 299.1 K
From Charles law
VV
TT
VT
V
T
2L 299.1 K
V
296.4 K
2 L 1.009
2.018 L
?
??
?
??
??
?
CBSE XI | Chemistry
Sample Paper – 4 Solution
10.
- - 2-
3 3 4
a) First, we will write down the oxidation number of each atom
22
Cr(OH) + IO I + CrO
3 5 1 6
b) Write separately oxidation &
??
?
? ? ? ?
2- -
34
- - -
3
2
reduction half reactions
Oxidation half reaction:
Cr(OH) CrO 3e
36
Reduction half reaction:
IO + 6e I
5 1
c) Balance O atoms by adding H O molecules to the side deficient in
??
??
?
??
2- -
3 2 4
- - -
32
'O'
atoms and then balancing H atoms
Cr(OH) + H O CrO 3e
IO + 6e I + 3H O
??
?
2
-
d) Balance H atoms. Since the medium is alkaline, therefore H O molecules
are added to the side deficient in H atoms and equal no. of OH
ions to the other side :
Cr(O
- 2- -
3 4 2
- 2- -
3 2 4 2
- - - -
32
- - -
3 2 2
H) + 5OH CrO 3e +4H O
( Cr(OH) + H O 5OH CrO 3e +5H O)
IO + 6e 3H O I + 6OH
( IO + 6e 6H O I + 3H O
??
? ? ?
??
??
-
- 2- -
3 4 2
+ 6OH
e) Equalise the electrons lost and gained by multiplying the oxidation half
reaction with 2.
2Cr(OH) + 10OH 2CrO 6e +8H O
Adding the oxidation half reaction and reduction half reaction we
??
)
- 2- -
3 4 2
- - - -
32
get
2Cr(OH) + 10OH 2CrO 6e +8H O
IO + 6e 3H O I + 6OH
??
??
_____________________________________
- - 2- -
3 3 4 2
2Cr(OH) IO 4OH 2CrO I +5H O ? ? ? ?
____________
CBSE XI | Chemistry
Sample Paper – 4 Solution
11.
(a) Anhydrous AlCl3 is covalent but hydrated AlCl3 is electrovalent because when it is
dissolved in water the high heat of hydration is sufficient to break the covalent bond
of AlCl3 into Al
3+
and Cl
-
ions.
(b) Boric acid behaves as Lewis acid by accepting a pair of electron from OH
-
ion (in
water).
-
3 4 3
B(OH) + 2H-O-H [B(OH) ] + H O
?
?
12. 1 molar solution contains 1mole of solute in 1 L of solution while
1 molal solution contains 1 mole of solute in 1000g of solvent.
Considering density of water as almost 1g/mL, then 1mole of solute is present in
1000mL of water in 1molal solution while 1mole of it is present in less than 1000mL of
water in 1 molar solution
(1000mL solution in molar solution = Volume of solute + Volume of solvent).
Thus 1M solution is more concentrated than 1m solution.
OR
Given:
Molarity of solution = 0.5 M
Volume of solution = 250 cm
3
0.5 M NaCl solution contains 0.5 mole of NaCl in 1 litre of solution.
Number of moles of NaCl in 250 cm
3
0.500
4
0.125mol
?
?
Molar mass of NaCl = 58.44 g
Mass of 0.125 mol of NaCl = 58.44 ×0.125
= 7.305 g of NaCl
CBSE XI | Chemistry
Sample Paper – 4 Solution
Section C
13. Given:
12
7
34
8
? 150pm
150 10 m
v 1.5 10 m / s
h 6.626 10 Js
c 3.8 10 m / s
?
?
?
??
??
??
??
Energy of the incident photon
? ? ? ?
34 8
12
16
2
2
31 7
17
16
hc
?
6.626 10 3.8 10
150 10 m
13.25 10 J
Energy of the ejected electron
1
mv
2
1
9.11 10 1.5 10
2
20.49 10
2
1.024 10 J
?
?
?
?
?
?
?
? ? ?
?
?
??
?
? ? ?
?
?
??
The energy with which the electron is bound to the nucleus =13.25×10
-16
J? 1.024×10
-16
J
= 12.226×10
-16
J
The energy with which the electron is bound to the nucleus is 12.226×10
-16
J.
OR
(a) Orbitals which possess equal energies are called degenerate orbitals.
For example: 2px, 2py and 2pz orbital which are oriented alon X, Y and Z axes are
degenerate. Similarly, five d-orbitals dxy, dyz, dzx,
2 2 2
x y z
d and d
?
are degenerate orbitals.
(b) The angular momentum of an electron in a given stationary state can be expressed as
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FAQs on Sample Solution Paper 4: Chemistry, Class 11 - Chemistry for Grade 11
| 1. What is the importance of chemistry in Class 11? |  |
Ans. Chemistry is an important subject in Class 11 as it forms the foundation for higher level studies in science and technology. It helps in understanding the composition, properties, and behavior of matter, which is essential for various fields such as medicine, engineering, and environmental science.
| 2. What are the main topics covered in the Class 11 Chemistry syllabus? | |
Ans. The Class 11 Chemistry syllabus covers various topics such as the structure of atoms, chemical bonding and molecular structure, states of matter, thermodynamics, equilibrium, redox reactions, and basic concepts of organic chemistry.
| 3. What are the benefits of studying chemistry in Class 11? | |
Ans. Studying chemistry in Class 11 provides several benefits. It helps in developing problem-solving skills, critical thinking, and logical reasoning. It also enhances analytical skills, which are useful in various fields. Additionally, it lays the foundation for higher studies in chemistry and related disciplines.
| 4. How can I improve my understanding of chemistry in Class 11? | |
Ans. To improve understanding of chemistry in Class 11, it is important to have a strong conceptual understanding. Regular practice of numerical problems, solving sample papers, and referring to additional study materials can help in enhancing understanding. Seeking clarification from teachers and participating in group discussions can also be beneficial.
| 5. What are some important practical applications of chemistry studied in Class 11? | |
Ans. The knowledge of chemistry gained in Class 11 has practical applications in various fields. It is used in industries for manufacturing chemicals and pharmaceuticals. It is also applied in environmental science to understand and mitigate pollution. Additionally, chemistry plays a crucial role in medicine and healthcare, as well as in the development of new materials and technologies.
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Sample Solution Paper 4: Chemistry, Class 11 Grade 11 Questions
The "Sample Solution Paper 4: Chemistry, Class 11 Grade 11 Questions" guide is a valuable resource for all aspiring students preparing for the
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The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
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The content of Sample Solution Paper 4: Chemistry, Class 11 is prepared as per the latest Grade 11 syllabus.
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