Sample Solution Paper 4 - Chemistry, Class 11 NEET Notes | EduRev

Chemistry Class 11

NEET : Sample Solution Paper 4 - Chemistry, Class 11 NEET Notes | EduRev

 Page 1


  
 
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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