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Page 1
CBSE XI | Chemistry
Sample Paper – 2 Solution
CBSE
Class XI Chemistry
Sample Paper – 2 Solution
Section A
1. 2,2 Dimehylpropane< 2-methylbutane < Pentane.
2. ClF3: T- shape
BF3: Trigonal planar
OR
The electron pairs involved in the bond formation are known as bond pairs or shared
pairs.
3. It is due to delocalization of ?- electrons in benzene it is highly stable.
4. London smog consists of H2SO4 deposited on the particulates suspended in the
atmosphere.
Gases responsible for green house effect are CO2, methane, nitrous oxide, chlorofluro
hydrocarbons and ozone.
5. Alkali metals due to low ionization energy absorbs energy from visible region to radiate
complementary colour.
Section B
6. Iso electronic species are those species (atoms/ions) which have same number of
electrons.
The iso electronic species for F
-
is Na
+
and for Ar is K
+
.
7. Metallic character increases down the group and decreases across the period as we
move from left to right. Hence the increasing order of metallic character is:
P<Si<Be<Mg<Na.
Page 2
CBSE XI | Chemistry
Sample Paper – 2 Solution
CBSE
Class XI Chemistry
Sample Paper – 2 Solution
Section A
1. 2,2 Dimehylpropane< 2-methylbutane < Pentane.
2. ClF3: T- shape
BF3: Trigonal planar
OR
The electron pairs involved in the bond formation are known as bond pairs or shared
pairs.
3. It is due to delocalization of ?- electrons in benzene it is highly stable.
4. London smog consists of H2SO4 deposited on the particulates suspended in the
atmosphere.
Gases responsible for green house effect are CO2, methane, nitrous oxide, chlorofluro
hydrocarbons and ozone.
5. Alkali metals due to low ionization energy absorbs energy from visible region to radiate
complementary colour.
Section B
6. Iso electronic species are those species (atoms/ions) which have same number of
electrons.
The iso electronic species for F
-
is Na
+
and for Ar is K
+
.
7. Metallic character increases down the group and decreases across the period as we
move from left to right. Hence the increasing order of metallic character is:
P<Si<Be<Mg<Na.
CBSE XI | Chemistry
Sample Paper – 2 Solution
9. Given:
-9
-7
8
8
-7
Wavelength of the radiation = 580 nm = 580 × 10 m
= 5.8 × 10 m
Velocity of radiation, c = 3 × 10 m / s
c = ?
3 × 10 m / s
=
5.8 × 10 m
=
14 -1
-
-7
6 -1
5.17 × 10 s
1
Wave number ? =
?
1
=
5.8 × 10 m
= 1.72 × 10 m
10. Root mean square speed is given as:
r.m.s
3RT
u=
M
Here,
T =273 + 27 =300 K
M = 16 g mol-1
R= 8.314 x 10
7
7
r.m.s
2 -1
-1
3x8.314x10 x300
u=
16
=683.9 x10 cmsec
=683.9msec
OR
The given equation is;
3MnO 2 + 4Al ? 3Mn + 2Al 2O 3
Change in oxidation numbers:
Mn: 4 to 0, Al: 0 to 2 and O: -2 to 2
Thus MnO2 is reduced and Al is oxidized.
Page 3
CBSE XI | Chemistry
Sample Paper – 2 Solution
CBSE
Class XI Chemistry
Sample Paper – 2 Solution
Section A
1. 2,2 Dimehylpropane< 2-methylbutane < Pentane.
2. ClF3: T- shape
BF3: Trigonal planar
OR
The electron pairs involved in the bond formation are known as bond pairs or shared
pairs.
3. It is due to delocalization of ?- electrons in benzene it is highly stable.
4. London smog consists of H2SO4 deposited on the particulates suspended in the
atmosphere.
Gases responsible for green house effect are CO2, methane, nitrous oxide, chlorofluro
hydrocarbons and ozone.
5. Alkali metals due to low ionization energy absorbs energy from visible region to radiate
complementary colour.
Section B
6. Iso electronic species are those species (atoms/ions) which have same number of
electrons.
The iso electronic species for F
-
is Na
+
and for Ar is K
+
.
7. Metallic character increases down the group and decreases across the period as we
move from left to right. Hence the increasing order of metallic character is:
P<Si<Be<Mg<Na.
CBSE XI | Chemistry
Sample Paper – 2 Solution
9. Given:
-9
-7
8
8
-7
Wavelength of the radiation = 580 nm = 580 × 10 m
= 5.8 × 10 m
Velocity of radiation, c = 3 × 10 m / s
c = ?
3 × 10 m / s
=
5.8 × 10 m
=
14 -1
-
-7
6 -1
5.17 × 10 s
1
Wave number ? =
?
1
=
5.8 × 10 m
= 1.72 × 10 m
10. Root mean square speed is given as:
r.m.s
3RT
u=
M
Here,
T =273 + 27 =300 K
M = 16 g mol-1
R= 8.314 x 10
7
7
r.m.s
2 -1
-1
3x8.314x10 x300
u=
16
=683.9 x10 cmsec
=683.9msec
OR
The given equation is;
3MnO 2 + 4Al ? 3Mn + 2Al 2O 3
Change in oxidation numbers:
Mn: 4 to 0, Al: 0 to 2 and O: -2 to 2
Thus MnO2 is reduced and Al is oxidized.
CBSE XI | Chemistry
Sample Paper – 2 Solution
11.
For K2MnO4, let the oxidation number of Mn be y
Oxidation Number of each Oxygen atom = -2
Oxidation Number of each K atom = +1
In a molecule, sum oxidation number of various atoms must be equal to zero
0= 2+ y + 4(-2) = y-6
y-6 = 0
y = 6
?
?
For HNO3, let the oxidation number of N be y
Oxidation Number of each O atom = -2ss
Oxidation Number of each H atom = +1
In a molecule, sum oxidation number of various atoms must be equal to zero.
12. The balanced chemical equation is
22
2CO O 2CO
2mol 1mol
2x22.4L 22.4L
? ? ? ?
Volume of oxygen required to convert 2 x 22.4 L of CO at N.T.P. = 22.4 L
Volume of oxygen required to convert 5.2 L of CO at N.T.P. = xL
x
22.4
5.2 2.6
2 22.4
?
OR
1 mole of
12
C atoms = 6.022×10
23
atoms = 12 g
12
1
23
23
2
2
3
atomsof C havemass12g
12
1 atom of C
6.022 10
6.022 10
would have mass g
1.99 10 g
?
?
??
?
?
?
Page 4
CBSE XI | Chemistry
Sample Paper – 2 Solution
CBSE
Class XI Chemistry
Sample Paper – 2 Solution
Section A
1. 2,2 Dimehylpropane< 2-methylbutane < Pentane.
2. ClF3: T- shape
BF3: Trigonal planar
OR
The electron pairs involved in the bond formation are known as bond pairs or shared
pairs.
3. It is due to delocalization of ?- electrons in benzene it is highly stable.
4. London smog consists of H2SO4 deposited on the particulates suspended in the
atmosphere.
Gases responsible for green house effect are CO2, methane, nitrous oxide, chlorofluro
hydrocarbons and ozone.
5. Alkali metals due to low ionization energy absorbs energy from visible region to radiate
complementary colour.
Section B
6. Iso electronic species are those species (atoms/ions) which have same number of
electrons.
The iso electronic species for F
-
is Na
+
and for Ar is K
+
.
7. Metallic character increases down the group and decreases across the period as we
move from left to right. Hence the increasing order of metallic character is:
P<Si<Be<Mg<Na.
CBSE XI | Chemistry
Sample Paper – 2 Solution
9. Given:
-9
-7
8
8
-7
Wavelength of the radiation = 580 nm = 580 × 10 m
= 5.8 × 10 m
Velocity of radiation, c = 3 × 10 m / s
c = ?
3 × 10 m / s
=
5.8 × 10 m
=
14 -1
-
-7
6 -1
5.17 × 10 s
1
Wave number ? =
?
1
=
5.8 × 10 m
= 1.72 × 10 m
10. Root mean square speed is given as:
r.m.s
3RT
u=
M
Here,
T =273 + 27 =300 K
M = 16 g mol-1
R= 8.314 x 10
7
7
r.m.s
2 -1
-1
3x8.314x10 x300
u=
16
=683.9 x10 cmsec
=683.9msec
OR
The given equation is;
3MnO 2 + 4Al ? 3Mn + 2Al 2O 3
Change in oxidation numbers:
Mn: 4 to 0, Al: 0 to 2 and O: -2 to 2
Thus MnO2 is reduced and Al is oxidized.
CBSE XI | Chemistry
Sample Paper – 2 Solution
11.
For K2MnO4, let the oxidation number of Mn be y
Oxidation Number of each Oxygen atom = -2
Oxidation Number of each K atom = +1
In a molecule, sum oxidation number of various atoms must be equal to zero
0= 2+ y + 4(-2) = y-6
y-6 = 0
y = 6
?
?
For HNO3, let the oxidation number of N be y
Oxidation Number of each O atom = -2ss
Oxidation Number of each H atom = +1
In a molecule, sum oxidation number of various atoms must be equal to zero.
12. The balanced chemical equation is
22
2CO O 2CO
2mol 1mol
2x22.4L 22.4L
? ? ? ?
Volume of oxygen required to convert 2 x 22.4 L of CO at N.T.P. = 22.4 L
Volume of oxygen required to convert 5.2 L of CO at N.T.P. = xL
x
22.4
5.2 2.6
2 22.4
?
OR
1 mole of
12
C atoms = 6.022×10
23
atoms = 12 g
12
1
23
23
2
2
3
atomsof C havemass12g
12
1 atom of C
6.022 10
6.022 10
would have mass g
1.99 10 g
?
?
??
?
?
?
CBSE XI | Chemistry
Sample Paper – 2 Solution
13. Configuration (b) is correct.
According to Hund’s rule of maximum multiplicity, pairing of electrons in the orbitals of
a particular subshell does not take place until all the orbitals of the subshell are singly
occupied. Since in the configuration (a) two electrons are present in 2px and no electron
is present in 2pz, it is incorrect as per Hund’s Rule.
OR
When n = 5, l = 0, 1, 2, 3, 4. The order in which the energy of the available orbital’s 4d, 5s
and 5p increases is 5s < 4d < 5p. The total number of orbital’s available is 9. The maximum
number of electrons that can be accommodated is 18; and therefore18 elements are there
in the 5th period.
14.
(a) NH3: sp
3
(b)C2H2 : sp
Dipole moment of CCl4 molecule is zero. Dipole moment is a vector quantity. In
symmetrical molecule dipoles of individual bonds cancel each other giving resultant
dipole moment as zero.
OR
In the formation of PCl5, one s, three p and one d orbitals are involved in hybridization
and give sp
3
d hybrid state.
The ground state and excited state outer electronic configuration of phosphorus (15) are
as:
(a) P (ground state)
3s 3p 3d
Ground state
2p 3s 3p 3d
Excited State
PCl5
Cl Cl Cl Cl Cl
sp
3
d hybrid orbitals filled by electron pairs donated by five Cl atoms.
??
?
? ?
?
?
? ?
?
??
??
?? ??
? ?
Page 5
CBSE XI | Chemistry
Sample Paper – 2 Solution
CBSE
Class XI Chemistry
Sample Paper – 2 Solution
Section A
1. 2,2 Dimehylpropane< 2-methylbutane < Pentane.
2. ClF3: T- shape
BF3: Trigonal planar
OR
The electron pairs involved in the bond formation are known as bond pairs or shared
pairs.
3. It is due to delocalization of ?- electrons in benzene it is highly stable.
4. London smog consists of H2SO4 deposited on the particulates suspended in the
atmosphere.
Gases responsible for green house effect are CO2, methane, nitrous oxide, chlorofluro
hydrocarbons and ozone.
5. Alkali metals due to low ionization energy absorbs energy from visible region to radiate
complementary colour.
Section B
6. Iso electronic species are those species (atoms/ions) which have same number of
electrons.
The iso electronic species for F
-
is Na
+
and for Ar is K
+
.
7. Metallic character increases down the group and decreases across the period as we
move from left to right. Hence the increasing order of metallic character is:
P<Si<Be<Mg<Na.
CBSE XI | Chemistry
Sample Paper – 2 Solution
9. Given:
-9
-7
8
8
-7
Wavelength of the radiation = 580 nm = 580 × 10 m
= 5.8 × 10 m
Velocity of radiation, c = 3 × 10 m / s
c = ?
3 × 10 m / s
=
5.8 × 10 m
=
14 -1
-
-7
6 -1
5.17 × 10 s
1
Wave number ? =
?
1
=
5.8 × 10 m
= 1.72 × 10 m
10. Root mean square speed is given as:
r.m.s
3RT
u=
M
Here,
T =273 + 27 =300 K
M = 16 g mol-1
R= 8.314 x 10
7
7
r.m.s
2 -1
-1
3x8.314x10 x300
u=
16
=683.9 x10 cmsec
=683.9msec
OR
The given equation is;
3MnO 2 + 4Al ? 3Mn + 2Al 2O 3
Change in oxidation numbers:
Mn: 4 to 0, Al: 0 to 2 and O: -2 to 2
Thus MnO2 is reduced and Al is oxidized.
CBSE XI | Chemistry
Sample Paper – 2 Solution
11.
For K2MnO4, let the oxidation number of Mn be y
Oxidation Number of each Oxygen atom = -2
Oxidation Number of each K atom = +1
In a molecule, sum oxidation number of various atoms must be equal to zero
0= 2+ y + 4(-2) = y-6
y-6 = 0
y = 6
?
?
For HNO3, let the oxidation number of N be y
Oxidation Number of each O atom = -2ss
Oxidation Number of each H atom = +1
In a molecule, sum oxidation number of various atoms must be equal to zero.
12. The balanced chemical equation is
22
2CO O 2CO
2mol 1mol
2x22.4L 22.4L
? ? ? ?
Volume of oxygen required to convert 2 x 22.4 L of CO at N.T.P. = 22.4 L
Volume of oxygen required to convert 5.2 L of CO at N.T.P. = xL
x
22.4
5.2 2.6
2 22.4
?
OR
1 mole of
12
C atoms = 6.022×10
23
atoms = 12 g
12
1
23
23
2
2
3
atomsof C havemass12g
12
1 atom of C
6.022 10
6.022 10
would have mass g
1.99 10 g
?
?
??
?
?
?
CBSE XI | Chemistry
Sample Paper – 2 Solution
13. Configuration (b) is correct.
According to Hund’s rule of maximum multiplicity, pairing of electrons in the orbitals of
a particular subshell does not take place until all the orbitals of the subshell are singly
occupied. Since in the configuration (a) two electrons are present in 2px and no electron
is present in 2pz, it is incorrect as per Hund’s Rule.
OR
When n = 5, l = 0, 1, 2, 3, 4. The order in which the energy of the available orbital’s 4d, 5s
and 5p increases is 5s < 4d < 5p. The total number of orbital’s available is 9. The maximum
number of electrons that can be accommodated is 18; and therefore18 elements are there
in the 5th period.
14.
(a) NH3: sp
3
(b)C2H2 : sp
Dipole moment of CCl4 molecule is zero. Dipole moment is a vector quantity. In
symmetrical molecule dipoles of individual bonds cancel each other giving resultant
dipole moment as zero.
OR
In the formation of PCl5, one s, three p and one d orbitals are involved in hybridization
and give sp
3
d hybrid state.
The ground state and excited state outer electronic configuration of phosphorus (15) are
as:
(a) P (ground state)
3s 3p 3d
Ground state
2p 3s 3p 3d
Excited State
PCl5
Cl Cl Cl Cl Cl
sp
3
d hybrid orbitals filled by electron pairs donated by five Cl atoms.
??
?
? ?
?
?
? ?
?
??
??
?? ??
? ?
CBSE XI | Chemistry
Sample Paper – 2 Solution
The longer nature of axial bonds is due to stronger repulsive interactions experienced by
the axial bond pairs from equatorial bond pairs.
15.
(i) 1 mole of N2 is 28 g react with 3 mol of H2 which is 3 g of H2
28 g react of N2 with 3 g of H2
22
22
2 3 3
23
2000 6
2000 g of N react withH
28
428.6g
N is the limiting agent while H is the excess reagent.
28g of N gives 2mol NH 34g NH
34
2000g of N will produce NH 2000 g
28
2428.57 g
?
??
?
?
? ? ?
?
(ii) H2 will left unreacted.
(iii) Mass left unreacted = 1000 g – 428.6 g
= 2428.57 g
16.
(a) Intensive properties: The properties which depends only on the nature of the
substance and not on the amount of the substance are called intensive properties
Example: Density
Adiabatic process: A process in which no heat flows between the system and the
surroundings is called an adiabatic process i.e. q= 0.
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FAQs on Sample Solution Paper 2: Chemistry, Class 11 - Sample Papers for Class 11 Medical and Non-Medical - JEE
| 1. What is the atomic number of an element? |  |
Ans. The atomic number of an element is the number of protons found in the nucleus of an atom of that element. It is a unique number for each element and determines its position in the periodic table.
| 2. What is the difference between an element and a compound? | |
Ans. An element is a pure substance that cannot be broken down into simpler substances by ordinary chemical means. It consists of atoms of the same type. On the other hand, a compound is a substance composed of two or more elements chemically combined in a fixed ratio. It can be broken down into its constituent elements by chemical reactions.
| 3. How is the periodic table organized? | |
Ans. The periodic table is organized based on the atomic number of elements. Elements are arranged in order of increasing atomic number from left to right and top to bottom. The elements in the same group or column have similar chemical properties, while elements in the same period or row have similar energy levels.
| 4. What is the difference between an exothermic and an endothermic reaction? | |
Ans. An exothermic reaction is a chemical reaction that releases energy in the form of heat to the surroundings. It is characterized by a rise in temperature. On the other hand, an endothermic reaction is a chemical reaction that absorbs energy from the surroundings to proceed. It is characterized by a decrease in temperature.
| 5. What is the difference between a physical change and a chemical change? | |
Ans. A physical change is a change in the physical properties of a substance without forming a new substance. Examples include changes in state, shape, or size. In contrast, a chemical change is a change in the chemical composition of a substance, resulting in the formation of one or more new substances. Examples include combustion, rusting, and digestion.
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