Sample Solution Paper 2 - Chemistry, Class 11 NEET Notes | EduRev

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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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