Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight
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New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (ii) is the correct answer.
The oxidation states are given below-
In NH2OH oxidation of N is -1.
NH4NO3 exists as NH4+.NO3-,
Thus, the oxidation state of N in NH4+ is -3 while in NO3? is +5.
In N2H4 oxidation of N is -2
In N3H oxidation of N is
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (i) is the correct answer
In ionic hydrides hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (i) is the correct answer.
Standard reduction potential of bromine is higher than Iodine, hence Bromine is a stronger oxidant than iodine.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iv) is the correct answer.
E? will be negative for the pair Ag and Fe3+. Hence the reaction is not feasible.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iv) is the correct answer
Reduction potential is defined as the tendency of the specie to get reduced. More positive the value of E® , greater is the tendency of the species to get reduced and stronger is the oxidising agent.
On the basis of the given E® values, the order of getting reduced is:
Br2Ag + I2Cu2+
Hence, Cu has the least tendency to get reduced and will itself gets oxidise and reduce other species as: Br2? , Ag+ and I2.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iv) is the correct answer
The strongest oxidising agent means it has greater tendency to oxidise other species and itself gets easily reduced. So higher the E? values, stronger is the oxidising agent it is. Thus, Ag+ having the highest positive E? value among the given systems, is the strongest oxidising agent.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
option (iv) is the correct answer.
Redox reaction is defined as the simultaneous oxidation and reduction of reacting species. Thus, change in oxidation state will decide whether a reaction is redox or not. Thus, assigning the oxidation states as:
(i) CuO + H2→ Cu + H2O
Here, oxidation of H and reduction of Cu is taking place.
(ii) Fe2O3 + 3CO→ 2Fe + 3CO2
Here, oxidation of C and reduction of Fe is taking place.
(iii) 2K + F2→2KF
Here, the oxidation of K and reduction of F is taking place.
(iv) BaCl2 + H
New answer posted
4 months agoContributor-Level 10
This is a Short answer type question as classified in NCERT Exemplar
(i) The balanced chemical is given as –

(ii) The balanced chemical is given as –

(iii) The balanced chemical is given as –

(iv) The balanced chemical is given as –

New answer posted
4 months agoContributor-Level 10
(i) We can write the given reaction along with their oxidation numbers as-

As, chlorine is oxidised in hydrochloric acid (behaving as reducing agent) -as its oxidation number is increases during the reaction from -1 to 0) and nitric acid is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +5 to +3). Hence, it is the redox reaction.
(ii) We can write the given reaction along with their oxidation numbers as-

Here, oxidation number of none of the atoms change hence it is not a redox reaction.
(iii) We can write
New answer posted
4 months agoContributor-Level 10
This is a Short answer type question as classified in NCERT Exemplar
(i) We can balance the given equation by oxidation number method-

(a)Balance the increase and decrease in O.N.

(b) Balancing H and O atoms by adding H+ and H2O molecules
(ii) We can balance the given equation by oxidation number method-

Total decrease in O.N. = 1
To equilize O.N. multiply NO3-, by 10
I2 + 10 NO3-? 10NO2 + IO3-
Balancing atoms other than O and H
I2 + 10 NO3-? 10NO2 + 2 IO3-
Balancing O and H
I2 + 10 NO3- + 8H+? 10NO2 + 2 IO3- + 4H2O
(iii) We can balance the given equation by oxidation number method

Total increase in O.N. =
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