NCERT Solution - Redox Reactions Class 11 Notes | EduRev

Class 11 : NCERT Solution - Redox Reactions Class 11 Notes | EduRev

 Page 1


  CHAPTER-8 NCERT SOLUTIONS 
 
Question 8.1: 
Assign oxidation numbers to the underlined elements in each of the following species: 
(a) NaH
2
PO
4
 (b) NaHSO
4
 (c) H
4
P
2
O
7
 (d) K
2
MnO
4
 
(e) CaO
2
 (f) NaBH
4
 (g) H
2
S
2
O
7
 (h) KAl(SO
4
)
2
.12 H
2
O 
 
Answer: 
(a)  
Let the oxidation number of P be x. 
We know that, 
Oxidation number of Na = +1 
Oxidation number of H = +1 
Oxidation number of O = –2 
 
Then, we have 
 
Hence, the oxidation number of P is +5. 
(b)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(c)  
 
Then, we have 
 
Hence, the oxidation number of P is + 5. 
(d)  
 
Then, we have 
 
Hence, the oxidation number of Mn is + 6. 
Page 2


  CHAPTER-8 NCERT SOLUTIONS 
 
Question 8.1: 
Assign oxidation numbers to the underlined elements in each of the following species: 
(a) NaH
2
PO
4
 (b) NaHSO
4
 (c) H
4
P
2
O
7
 (d) K
2
MnO
4
 
(e) CaO
2
 (f) NaBH
4
 (g) H
2
S
2
O
7
 (h) KAl(SO
4
)
2
.12 H
2
O 
 
Answer: 
(a)  
Let the oxidation number of P be x. 
We know that, 
Oxidation number of Na = +1 
Oxidation number of H = +1 
Oxidation number of O = –2 
 
Then, we have 
 
Hence, the oxidation number of P is +5. 
(b)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(c)  
 
Then, we have 
 
Hence, the oxidation number of P is + 5. 
(d)  
 
Then, we have 
 
Hence, the oxidation number of Mn is + 6. 
(e)  
 
Then, we have 
 
Hence, the oxidation number of O is – 1. 
(f)  
 
Then, we have 
 
Hence, the oxidation number of B is + 3. 
(g)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(h)  
 
Then, we have 
 
Or, 
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation 
numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the 
water molecule, we have 
 
Hence, the oxidation number of S is + 6. 
 
Page 3


  CHAPTER-8 NCERT SOLUTIONS 
 
Question 8.1: 
Assign oxidation numbers to the underlined elements in each of the following species: 
(a) NaH
2
PO
4
 (b) NaHSO
4
 (c) H
4
P
2
O
7
 (d) K
2
MnO
4
 
(e) CaO
2
 (f) NaBH
4
 (g) H
2
S
2
O
7
 (h) KAl(SO
4
)
2
.12 H
2
O 
 
Answer: 
(a)  
Let the oxidation number of P be x. 
We know that, 
Oxidation number of Na = +1 
Oxidation number of H = +1 
Oxidation number of O = –2 
 
Then, we have 
 
Hence, the oxidation number of P is +5. 
(b)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(c)  
 
Then, we have 
 
Hence, the oxidation number of P is + 5. 
(d)  
 
Then, we have 
 
Hence, the oxidation number of Mn is + 6. 
(e)  
 
Then, we have 
 
Hence, the oxidation number of O is – 1. 
(f)  
 
Then, we have 
 
Hence, the oxidation number of B is + 3. 
(g)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(h)  
 
Then, we have 
 
Or, 
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation 
numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the 
water molecule, we have 
 
Hence, the oxidation number of S is + 6. 
 
Question 8.2: 
What are the oxidation numbers of the underlined elements in each of the following and how do 
you rationalise your results? 
(a) KI
3
 (b) H
2
S
4
O
6
 (c) Fe
3
O
4
 (d) CH
3
CH
2
OH (e) CH
3
COOH 
 
Answer: 
(a) KI
3
 
In KI
3
, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . 
However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI
3
 to 
find the oxidation states. 
In a KI
3
 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule. 
 
Hence, in a KI
3
 molecule, the O.N. of the two I atoms forming the I
2
 molecule is 0, whereas the 
O.N. of the I atom forming the coordinate bond is –1. 
(b) H
2
S
4
O
6
 
 
However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the 
molecule. 
 
The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0. 
(c)  
On taking the O.N. of O as –2, the O.N. of Fe is found to be . However, O.N. cannot be 
fractional. 
Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the 
O.N. of +3. 
 
(d)  
 
Page 4


  CHAPTER-8 NCERT SOLUTIONS 
 
Question 8.1: 
Assign oxidation numbers to the underlined elements in each of the following species: 
(a) NaH
2
PO
4
 (b) NaHSO
4
 (c) H
4
P
2
O
7
 (d) K
2
MnO
4
 
(e) CaO
2
 (f) NaBH
4
 (g) H
2
S
2
O
7
 (h) KAl(SO
4
)
2
.12 H
2
O 
 
Answer: 
(a)  
Let the oxidation number of P be x. 
We know that, 
Oxidation number of Na = +1 
Oxidation number of H = +1 
Oxidation number of O = –2 
 
Then, we have 
 
Hence, the oxidation number of P is +5. 
(b)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(c)  
 
Then, we have 
 
Hence, the oxidation number of P is + 5. 
(d)  
 
Then, we have 
 
Hence, the oxidation number of Mn is + 6. 
(e)  
 
Then, we have 
 
Hence, the oxidation number of O is – 1. 
(f)  
 
Then, we have 
 
Hence, the oxidation number of B is + 3. 
(g)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(h)  
 
Then, we have 
 
Or, 
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation 
numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the 
water molecule, we have 
 
Hence, the oxidation number of S is + 6. 
 
Question 8.2: 
What are the oxidation numbers of the underlined elements in each of the following and how do 
you rationalise your results? 
(a) KI
3
 (b) H
2
S
4
O
6
 (c) Fe
3
O
4
 (d) CH
3
CH
2
OH (e) CH
3
COOH 
 
Answer: 
(a) KI
3
 
In KI
3
, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . 
However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI
3
 to 
find the oxidation states. 
In a KI
3
 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule. 
 
Hence, in a KI
3
 molecule, the O.N. of the two I atoms forming the I
2
 molecule is 0, whereas the 
O.N. of the I atom forming the coordinate bond is –1. 
(b) H
2
S
4
O
6
 
 
However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the 
molecule. 
 
The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0. 
(c)  
On taking the O.N. of O as –2, the O.N. of Fe is found to be . However, O.N. cannot be 
fractional. 
Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the 
O.N. of +3. 
 
(d)  
 
Hence, the O.N. of C is –2. 
(e)  
 
However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in 
different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits 
the oxidation states of +2 and –2 in CH
3
COOH. 
 
 
Question 8.3: 
Justify that the following reactions are redox reactions: 
(a) CuO(s) + H
2
(g) ? Cu(s) + H
2
O(g) 
(b) Fe
2
O
3
(s) + 3CO(g) ? 2Fe(s) + 3CO
2
(g) 
(c) 4BCl
3
(g) + 3LiAlH
4
(s) ? 2B
2
H
6
(g) + 3LiCl(s) + 3 AlCl
3
 (s) 
(d) 2K(s) + F
2
(g) ? 2K+F– (s) 
(e) 4 NH
3
(g) + 5 O
2
(g) ? 4NO(g) + 6H
2
O(g) 
 
Answer: 
(a)  
Let us write the oxidation number of each element involved in the given reaction as: 
 
Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to 
Cu. Also, the oxidation number of H increases from 0 in H
2
 to +1 in H
2
O i.e., H
2
 is oxidized to 
H
2
O. Hence, this reaction is a redox reaction. 
(b)  
Let us write the oxidation number of each element in the given reaction as: 
 
Here, the oxidation number of Fe decreases from +3 in Fe
2
O
3
 to 0 in Fe i.e., Fe
2
O
3
 is reduced to 
Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO
2
 i.e., CO 
is oxidized to CO
2
. Hence, the given reaction is a redox reaction. 
(c)  
The oxidation number of each element in the given reaction can be represented as: 
 
In this reaction, the oxidation number of B decreases from +3 in BCl
3
 to –3 in B
2
H
6
. i.e., BCl
3
 is 
reduced to B
2
H
6
. Also, the oxidation number of H increases from –1 in LiAlH
4
 to +1 in B
2
H
6
 i.e., 
LiAlH
4
 is oxidized to B
2
H
6
. Hence, the given reaction is a redox reaction. 
(d)  
The oxidation number of each element in the given reaction can be represented as: 
Page 5


  CHAPTER-8 NCERT SOLUTIONS 
 
Question 8.1: 
Assign oxidation numbers to the underlined elements in each of the following species: 
(a) NaH
2
PO
4
 (b) NaHSO
4
 (c) H
4
P
2
O
7
 (d) K
2
MnO
4
 
(e) CaO
2
 (f) NaBH
4
 (g) H
2
S
2
O
7
 (h) KAl(SO
4
)
2
.12 H
2
O 
 
Answer: 
(a)  
Let the oxidation number of P be x. 
We know that, 
Oxidation number of Na = +1 
Oxidation number of H = +1 
Oxidation number of O = –2 
 
Then, we have 
 
Hence, the oxidation number of P is +5. 
(b)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(c)  
 
Then, we have 
 
Hence, the oxidation number of P is + 5. 
(d)  
 
Then, we have 
 
Hence, the oxidation number of Mn is + 6. 
(e)  
 
Then, we have 
 
Hence, the oxidation number of O is – 1. 
(f)  
 
Then, we have 
 
Hence, the oxidation number of B is + 3. 
(g)  
 
Then, we have 
 
Hence, the oxidation number of S is + 6. 
(h)  
 
Then, we have 
 
Or, 
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation 
numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the 
water molecule, we have 
 
Hence, the oxidation number of S is + 6. 
 
Question 8.2: 
What are the oxidation numbers of the underlined elements in each of the following and how do 
you rationalise your results? 
(a) KI
3
 (b) H
2
S
4
O
6
 (c) Fe
3
O
4
 (d) CH
3
CH
2
OH (e) CH
3
COOH 
 
Answer: 
(a) KI
3
 
In KI
3
, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . 
However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI
3
 to 
find the oxidation states. 
In a KI
3
 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule. 
 
Hence, in a KI
3
 molecule, the O.N. of the two I atoms forming the I
2
 molecule is 0, whereas the 
O.N. of the I atom forming the coordinate bond is –1. 
(b) H
2
S
4
O
6
 
 
However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the 
molecule. 
 
The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0. 
(c)  
On taking the O.N. of O as –2, the O.N. of Fe is found to be . However, O.N. cannot be 
fractional. 
Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the 
O.N. of +3. 
 
(d)  
 
Hence, the O.N. of C is –2. 
(e)  
 
However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in 
different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits 
the oxidation states of +2 and –2 in CH
3
COOH. 
 
 
Question 8.3: 
Justify that the following reactions are redox reactions: 
(a) CuO(s) + H
2
(g) ? Cu(s) + H
2
O(g) 
(b) Fe
2
O
3
(s) + 3CO(g) ? 2Fe(s) + 3CO
2
(g) 
(c) 4BCl
3
(g) + 3LiAlH
4
(s) ? 2B
2
H
6
(g) + 3LiCl(s) + 3 AlCl
3
 (s) 
(d) 2K(s) + F
2
(g) ? 2K+F– (s) 
(e) 4 NH
3
(g) + 5 O
2
(g) ? 4NO(g) + 6H
2
O(g) 
 
Answer: 
(a)  
Let us write the oxidation number of each element involved in the given reaction as: 
 
Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to 
Cu. Also, the oxidation number of H increases from 0 in H
2
 to +1 in H
2
O i.e., H
2
 is oxidized to 
H
2
O. Hence, this reaction is a redox reaction. 
(b)  
Let us write the oxidation number of each element in the given reaction as: 
 
Here, the oxidation number of Fe decreases from +3 in Fe
2
O
3
 to 0 in Fe i.e., Fe
2
O
3
 is reduced to 
Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO
2
 i.e., CO 
is oxidized to CO
2
. Hence, the given reaction is a redox reaction. 
(c)  
The oxidation number of each element in the given reaction can be represented as: 
 
In this reaction, the oxidation number of B decreases from +3 in BCl
3
 to –3 in B
2
H
6
. i.e., BCl
3
 is 
reduced to B
2
H
6
. Also, the oxidation number of H increases from –1 in LiAlH
4
 to +1 in B
2
H
6
 i.e., 
LiAlH
4
 is oxidized to B
2
H
6
. Hence, the given reaction is a redox reaction. 
(d)  
The oxidation number of each element in the given reaction can be represented as: 
 
In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to 
KF. On the other hand, the oxidation number of F decreases from 0 in F
2
 to –1 in KF i.e., F
2
 is 
reduced to KF. 
Hence, the above reaction is a redox reaction. 
(e)  
The oxidation number of each element in the given reaction can be represented as: 
 
Here, the oxidation number of N increases from –3 in NH
3
 to +2 in NO. On the other hand, the 
oxidation number of O
2
 decreases from 0 in O
2
 to –2 in NO and H
2
O i.e., O
2
 is reduced. Hence, 
the given reaction is a redox reaction. 
 
Question 8.4: 
Fluorine reacts with ice and results in the change: 
H
2
O(s) + F
2
(g) ? HF(g) + HOF(g) 
Justify that this reaction is a redox reaction. 
 
Answer: 
Let us write the oxidation number of each atom involved in the given reaction above its symbol 
as: 
 
Here, we have observed that the oxidation number of F increases from 0 in F
2
 to +1 in HOF. 
Also, the oxidation number decreases from 0 in F
2 
to –1 in HF. Thus, in the above reaction, F is 
both oxidized and reduced. Hence, the given reaction is a redox reaction. 
 
Question 8.5: 
Calculate the oxidation number of sulphur, chromium and nitrogen in H
2
SO
5
, –. 
Suggest structure of these compounds. Count for the fallacy. 
 
Answer: 
(i)  
 
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S 
cannot be more than +6. 
The structure of H
2
SO
5
 is shown as follows: 
 
 
Therefore, the O.N. of S is +6. 
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