I. Multiple-choice Questions (Type-I)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 – Redox Reactions

by

NCERT ExemplarSolutions-Redox Reactions for Class 11 Chemistry Chapter 8

I. Multiple-choice Questions (Type-I)

1. Which of the following is not an example of redox reaction?

(i). CuO + H2→ Cu + H2O

(ii) Fe2O3 + 3CO → 2Fe + 3CO2

(iii) 2K + F2→ 2KF

(iv) BaCl2 + H2SO4→ BaSO4 + 2HCl

Solution:

Option (iv) is the answer.

2. The more positive the value of Eᶱ, the greater the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below, find out which of the following is the strongest oxidising agent. Eᶱ Values : Fe3+/Fe2+ = + 0.77; I2(s)/I– = + 0.54;

Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80V (i) Fe3+

(ii) I2(s)

(iii) Cu2+

(iv) Ag

Solution:

Option (iv) is the answer.

3. Eᶱvalues of some redox couples is given below. Based on these values

choose the correct option.

Eᶱ values : Br2/Br–= + 1.90; Ag+/Ag(s) = + 0.80

Cu2+/Cu(s) = + 0.34; I2(s)/I– = + 0.54

(i) Cu will reduce Br–

(ii) Cu will reduce Ag

(iii) Cu will reduce I–

(iv) Cu will reduce Br2

Solution:

Option (iv) is the answer.

4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

E°Values: Fe3+/ Fe2+ = +0.77; I2/I- = +0.54;

Cu2+/ Cu = -0.34; Ag+ /Ag = + 0.80 V

(i) Fe3+ and I-

(ii) Ag+ and Cu

(iii) Fe3+ and Cu

(iv) Ag and Fe3+

Solution:

Option (iv) is the answer.

5. Thiosulphate reacts differently with iodine and bromine in the reactions given

below:

2S2O32- + I2 → S4O62- + 2I–

S2O32-+ 2Br2+ 5H2O → 2SO42- + 2Br–+ 10 H+

Which of the following statements justifies the above dual behaviour of

thiosulphate?

(i) Bromine is a stronger oxidant than iodine.

(ii) Bromine is a weaker oxidant than iodine.

(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine

in these reactions.

(iv) Bromine undergoes oxidation and iodine undergoes a reduction in these

reactions.

Solution:

Option (i) is the answer.

6. The oxidation number of an element in a compound is evaluated on the basis

of certain rules. Which of the following rules is not correct in this respect?

(i) The oxidation number of hydrogen is always +1.

(ii) The algebraic sum of all the oxidation numbers in a compound is zero.

(iii) An element in the free or the uncombined state bears oxidation

number zero.

(iv) In all its compounds, the oxidation number of fluorine is – 1.

Solution:

Option (i) is the answer.

7. In which of the following compounds an element exhibits two different

oxidation states.

(i) NH2OH

(ii) NH4NO3

(iii) N2H4

(iv) N3H

Solution:

Option (ii) is the answer.

8. Which of the following arrangements represents an increasing oxidation number of the central atom?

(i) CrO2- , CIO-3, CrO2-4 , MnO-4

(ii) CIO-3, CrO2-4 , MnO-4 , CrO-2

(iii) CrO2+4 , MnO-4 , CrO-2 , CIO3-

(iv) CrO24-, MnO4- , CrO2- , CIO3-

Solution:

Option (i) is the answer.

9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations will the element exhibit the largest oxidation number?

(i) 3d1 4s2

(ii) 3d3 4s2

(iii) 3d5 4s1

(iv) 3d5 4s2

Solution:

Option (iv) is the answer.

10. Identify disproportionation reaction

(i) CH4 + 2O2→ CO2 + 2H2O

(ii) CH4 + 4Cl2→ CCl4 + 4HCl

(iii) 2F2 + 2OH-→ 2F- + OF2 + H2O

(iv) 2NO2 + 2OH–→ NO2- + NO3- + H2O

Solution:

Option (iv) is the answer.

11. Which of the following elements does not show disproportionation tendency?

(i) Cl

(ii) Br

(iii) F

(iv) ISolution:

Option (iii) is the answer.

II. Multiple Choice Questions (Type-II)

In the following questions, two or more options may be correct.

12. Which of the following statement(s) is/are not true about the following

decomposition reaction.

2KClO3 → 2KCl + 3O2

(i) Potassium is undergoing oxidation

(ii) Chlorine is undergoing oxidation

(iii) Oxygen is reduced

(iv) None of the species is undergoing oxidation or reduction

Solution:

Option (i) and (iv) are the answers.

13. identify the correct statement (s) with the following reaction:

Zn + 2HCl → ZnCl2 + H2

(i) Zinc is acting as an oxidant

(ii) Chlorine is acting as a reductant

(iii) Hydrogen ion is acting as an oxidant

(iv) Zinc is acting as a reductant

Solution:

Option (iii) and (iv) are the answers.

14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds?

(i) 3s1

(ii) 3d1 4s2

(iii) 3d2 4s2

(iv) 3s2 3p3

Solution:

Option (iii) and (iv0 are the answers.

15. Identify the correct statements with reference to the given reaction

P4 + 3OH–+ 3H2O → PH3+ 3H2PO2–

(i) Phosphorus is undergoing reduction only.

(ii) Phosphorus is undergoing oxidation only.

(iii) Phosphorus is undergoing oxidation as well as reduction.

(iv) Hydrogen is undergoing neither oxidation nor reduction.

Solution:

Option (iii) and (iv) are the answers.

16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?

(i) Al/Al2+ -E\degree = –1.66

(ii) Fe/Fe2+ -E\degree = – 0.44

(iii) Cu/Cu2+ -E\degree= + 0.34

(iv) F2 (g)/2F- (aq) -E\degree= + 2.87

Solution:

Option (i) and (ii) are the answers.

III. Short Answer Type

17. The reaction

Cl2 (g) + 2OH- (aq) →ClO- (aq) + Cl- (aq) + H2O (l)

represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

Solution:

Hypochlorite ion is the species that bleaches the substance due to its oxidizing action.

18. MnO42- undergoes disproportionation reaction in acidic medium but MnO4–

does not. Give reason.

Solution:

In – MnO4, Mn is in the highest oxidation state i.e. +7. Therefore, it does

not undergo disproportionation. MnO42– undergoes disproportionation

as follows :

3MnO42– + 4H+ → 2MnO4– + MnO2+ 2H2O

19. PbO and PbO2 react with HCl according to the following chemical equations:

2PbO + 4HCl → 2PbCl2 + 2H2O

PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O

Why do these compounds differ in their reactivity?

Solution:

In reaction (i), none of the atoms changes. Therefore, it is not a redox reaction. It is an acid-base reaction because PbO is a basic oxide which reacts with HCl acid.

Reaction (ii) is a redox reaction in which PbO2 gets reduced and acts as an oxidizing agent.

20. Nitric acid is an oxidising agent and reacts with PbO, but it does not react with PbO2. Explain why?

Solution:

Nitric acid is an oxidizing agent and reacts with PbO to give a simple acid-base reaction without any change in oxidation state. In PbO2, Pb is in +4 oxidation state and cannot be oxidized further; hence no reaction takes place between PbO2 and HNO3.

21. Write a balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO4-) reacts with sulphur dioxide gas in acidic medium to produce Mn2+ and hydrogen sulphate ion. (Balance by ion-electron method)

(ii) The reaction of liquid hydrazine (N2H4) with chlorate ion (ClO-3) in basic medium produces nitric oxide gas and chloride ion in the gaseous state. (Balance by oxidation number method)

(iii) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO2-) and oxygen gas.

(Balance by ion-electron method)

Solution:

(i) 2MnO4- + 5SO2+ 2H2O + H+ → 2Mn2+ + 5HSO4-

(ii) N2H4 + ClO3- → Cl- + NO2 + 2H2O

(iii) Cl2O7 + 4H2O2 → 2ClO2- + 3H2O + 4O2+ 2H+

22. Calculate the oxidation number of phosphorus in the following species.

(a) HPO2-3 and PO42-

Solution:

HPO2-3 x= +3

PO42- x=+5

23. . Calculate the oxidation number of phosphorus in the following species.

(a) Na2S2O3

(b) Na2S4O6

(c) Na2SO3

(d) Na2SO4

Solution:

Na2S2O3 x= +2

Na2S4O6 x= +5

Na2SO3 x= +4

Na2SO4 x= +6

24. Balance the following equations by the oxidation number method.

(i) Fe2+ + H+ + Cr2 O2-7→ Cr3+ + Fe3+ + H2O

(ii) I2+ N – O3 → NO2+ I – O3

(iii) I2 + S22– O3 → I– + S42– O

(iv) MnO2 + C2 O2-4→ Mn2+ + CO2

Solution:

(i) 6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

(ii) I2 + 10NO3- + 8H+ → 2IO3- + 10NO2 +4H2O

(iii) 2S2O3 2- + I2 → S4O62- + 2l-

(iv) C2O402- +MnO2 + 4H+ → 2CO2 +MN2+ +2H2O

25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

(i) 3HCI(aq) + HNO3 (aq) → CI2(g) + NOCI (g) + 2H2O(l)

(ii) HgCI2 (aq) + 2KI (aq) → HgI2 (s) + 2KCI (aq)

(iii) Fe2O3 (s) + 3CO(g) → 2Fe (s) + 3CO2(g)

(iv) PCI3 (I) + 3H2O (I) → 3HCI (aq) + H3 Po3 (aq)

(v) 4NH3 + 3O2 (g) → 2N2 (g) + 6H2 O (g)

Solution:

(i) (iii) and (iv) are redox reactions

In (i) Reducing agent: HCl

Oxidizing agent: HNO3

In (iii) Oxidising agent: Fe2O3

Reducing agent: CO

In (iv) Oxidising agent: O2

Reducing agent: NH3

26. Balance the following ionic equations

(i) Cr2O72- + H+ + I-→ Cr3+ + I2 + H2O

(ii) Cr2O2-7 + Fe2+ + H+→ Cr3+ + Fe3+ + H2O

(iii) MnO-4 + SO2-3 + H+→ Mn2+ + SO42- + H2O

(iv) MnO4- + H+ + Br-→ Mn2+ + Br2 + H2O

Solution:

(i) 6I + Cr2O72- + 14H+ → 2Cr3+ + 3I2 + 7H2O

(ii) Cr2O2-7 + 6Fe2+ + 14H+→ 2Cr3+ + 6Fe3+ + 7H2O

(iii) 2MnO-4 + 5SO2-3 + 6H+→ 2Mn2+ + 5SO42- + 3H2O

(iv) 2MnO4- + 16H+ + 10Br-→ 2Mn2+ + 5Br2 + 8H2O

IV. Matching Type

27. Match Column I with Column II for the oxidation states of the central atoms.

Column I Column II (i) Cr2O72– (a) + 3 (ii) MnO4– (b) + 4 (iii) VO3- (c) + 5 (iv) FeF63– (d) + 6 (e) + 7Column II (a) + 3 (b) + 4 (c) + 5 (d) + 6 (e) + 7

Solution:

(i) is d

(ii) is e

(iii) is c

(iv) a

28. Match the items in Column I with relevant items in Column II.

Column I (i) Ions having positive charge (ii) The sum of oxidation number of all atoms in a neutral molecule (iii) The oxidation number of hydrogen ion (H+) (iv) The oxidation number of fluorine in NaF (v) Ions having negative chargeColumn II (a) +7 (b) –1 (c) +1 (d) 0 (e) Cation (f) Anion

Solution:

(i) is e

(ii) is d

(iii) is c

(iv) is b

(v) is f

V. Assertion and Reason Type

In the following questions, a statement of assertion (A) followed by a statement

of the reason (R) is given. Choose the correct option out of the choices given

below each question.

29. Assertion (A): Among halogens, fluorine is the best oxidant.

Reason (R): Fluorine is the most electronegative atom.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (ii) is correct.

30. Assertion (A): In the reaction between potassium permanganate and

potassium iodide, permanganate ions act as an oxidising agent.

Reason (R): Oxidation state of manganese changes from +2 to +7 during

the reaction.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (iii) is correct.

31. Assertion (A): The decomposition of hydrogen peroxide to form water and

oxygen is an example of a disproportionation reaction.

Reason (R): The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O2 and –2 oxidation state in H2O.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (i) is correct.

32. In the following questions, a statement of assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Redox couple is the combination of the oxidised and reduced form of a substance involved in an oxidation or reduction half cell.

Reason (R) : In the representation E° Fe3+/fe2+and E°cu2+/cu , Fe3+ / Fe2+ and Cu2+ are redox couples.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false

Solution;

Option (ii) is correct.

VI. Long Answer Type

33. Explain redox reactions based on electron transfer. Give suitable examples.

Solution:

In a redox reaction if one species loses electrons it’s considered to be undergoing oxidation reaction and acts as oxidizing agent or oxidant, and for species which accept electrons is said to undergo reduction and behave as reductant.

For example, Zinc and HCl reaction

Zn + 2HCl → ZnCl2 + H2

zinc loses electrons to the electronegative atom Cl with the reaction for oxidation and reduction as follows:

Oxidation: Zn→Zn2+ + 2e-

Reduction: 2H+ + 2e- →H2

Thus the transfer of electrons causes the redox reaction to occur.

34. Based on standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E\degree value).

(i) Cu + Zn2+→ Cu2+ + Zn

(ii) Mg + Fe2+→ Mg2+ + Fe

(iii) Br2 + 2Cl-→ Cl2 + 2Br-

(iv) Fe + Cd2+→ Cd + Fe2+

Solution:

(i)Cu + Zn2+→ Cu2+ + Zn

Here Cu undergoes oxidation so it acts as anode and Zn acts as the cathode. So from the table

For cathode E°cathode = -0.76 V

For anode E°anode = 0.52 V

E°cell = -0.24V

As the EMF of the cell is negative the given reaction will not occur spontaneously if they were to form a cell placed as electrodes.

(ii) Mg + Fe2+→ Mg2+ + Fe

Similarly, we can say that Mg undergoes oxidation and Fe undergoes reduction.

E°cathode = -0.44 V

E°anode = -2.36 V

E°cell = +1.92V

Positive EMF implies that the reaction will give out energy and attain stability, thus it will occur spontaneously. So the given redox reaction will occur.

(iii) Br2 + 2Cl-→ Cl2 + 2Br-

Here Br undergoes reduction thus acting as cathode and Cl acting as the anode.

For cathode E°cathode = 1.09 V

For anode E°anode = 1.36 V

E°cell = -0.25

The negative potential prevents easy reaction, so the redox reaction will not occur.

(iv) Fe + Cd2+→ Cd + Fe2+

Fe is the cathode and Cd is the anode

For cathode E°cathode = -0.44 V

For anode E°anode = -0.40 V

E°cell = -0.04v

The negative potential prevents easy reaction, so the redox reaction will not occur.

35. Why does fluorine not show a disproportionation reaction?

Solution:

Fluorine has the highest electronegativity in the entire periodic table with EN value of 3.98; this means that out of 4 bonded electrons 3.98 fractions of it is shared with fluorine. Thus the ability to attract an electron from other elements is more pronounced than the atom with which it is bonded. And as fluorine has the highest reduction potential (E°cell =2.87) in the spectrochemical series, it cannot undergo oxidation itself. Thus cannot display disproportionation reactions.

36. Write redox couples involved in the reactions (i) to (iv) given in question 34.

Solution:

Redox couple is a reducing element along with its oxidizing form. So, for the given example,

(i) Cu2+ /Cu and Zn2+/Zn

(ii) Mg+/Mg and Fe2+/Fe

(iii) Br-/Br and Cl-/Cl

(iv) Fe2+ /Fe and Cd2+/Cd

37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.

NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2, ClO2

Solution:

NaClO4 x= +7

NaClO3, x= +5

NaClO, x=+1

KClO2, x= +3

Cl2O7, x= +7

ClO3, x= +6

Cl2O, x=+1

NaCl, x=-1

Cl2, x=0

ClO2, x= +4

Ascending order of compounds w.r.t their oxidation number is:

NaCl (-1), Cl2(0), Cl2O(+1), KClO2(+3), ClO2(+4), NaClO3(+5), ClO3(+6), Cl2O7=NaClO4(+7).

38. Which method can be used to find out the strength of reductant/oxidant in a solution? Explain with an example.

Strength of a reductant (reducing agent) or oxidant (oxidising agent) can be found out by measuring the relative electrode potential when it’s connected in a solution using a cell.

For example, Fe3+/Fe is the element we want to test with the Standard Hydrogen electrode (SHE). The half-cell reaction for Fe and H are given below.

H+ + e- → H2 E° = 0.0V

Fe3+ + e- → Fe2+ E° = 0.77V

When any element needs to be evaluated, it is placed as an electrode with SHE. The amount of emf it generates in the cell can be considered as the potential of the element.

E°cell = 0-0.77

E°cell = 0.77

The above-assumed configuration of Fe being anode can be reversed, and hence strength Fe as a reductant can be established. Hence the strength of an oxidant can be determined.

Topics of Class 11 Chemistry Chapter 8 Redox Reactions

  1. Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
  2. Redox Reactions in Terms of Electron Transfer Reactions
    1. Competitive Electron Transfer Reactions
  3. Oxidation Number
    1. Types of Redox Reactions
    2. Balancing of Redox Reactions
    3. Redox Reactions as the Basis for Titrations
    4. Limitations of Concept of Oxidation Number
  4. Redox Reactions and Electrode Processes.

The topics covered in the Class 11 syllabus are the basics of the topics to be taught in Class 12.  So, students must study all the topics of Class 11 thoroughly to score well in the Class 12 board examinations as well  Students are advised to go through this NCERT Exemplars to determine their strengths and weaknesses and plan their studies accordingly. Along with NCERT questions, students should try to solve the previous years’ questions and sample papers to get acquainted with the latest exam pattern and marking scheme.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 11 Chemistry Chapter 8

Q1

Will I be able to get my doubts cleared using the NCERT Exemplar Solutions for Class 11 Chemistry Chapter 8?

Students who are not able to get their doubts cleared during class hours can make use of the NCERT Exemplar Solutions at ANAND CLASSES (A School Of Competitions). The subject experts framed the solutions with utmost care with the main objective of helping students with their annual exam preparation. To score good marks in Chemistry, learning the concepts on a daily basis is very important. The concepts are explained in an interactive manner to help students grasp them and perform well in the annual exams.

Q2

What are the topics covered under Chapter 8 of NCERT Exemplar Solutions for Class 11 Chemistry?

Topics of Class 11 Chemistry Chapter 8 Redox Reactions are listed below:
1. Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
2. Redox Reactions in Terms of Electron Transfer Reactions
3. Competitive Electron Transfer Reactions
4. Oxidation Number
5. Types of Redox Reactions
6. Balancing of Redox Reactions
7. Redox Reactions as the Basis for Titrations
8. Limitations of the Concept of Oxidation Number
9. Redox Reactions and Electrode Processes.
To score full marks in this chapter, students are advised to solve the textbook questions using the NCERT Exemplar solutions available at ANAND CLASSES (A School Of Competitions). The solutions of this chapter are framed by expert faculty having vast experience in the relevant subject. They created answers which are concept focused rather than question focused on building a strong foundation of basic concepts which are important from the exam perspective.

Q3

Why should I refer to the NCERT Exemplar Solutions for Class 11 Chemistry Chapter 8 from ANAND CLASSES (A School Of Competitions)?

You should refer to the NCERT Exemplar Solutions for Class 11 Chemistry Chapter 8 from ANAND CLASSES (A School Of Competitions) to understand the concepts easily. The solutions are framed by the teachers based on the understanding abilities of students. Pictorial representation of certain topics improves the visual learning skills of students, which will help them to grasp the concepts easily. NCERT Exemplar Solutions are created in a stepwise manner in order to meet the CBSE standards. Every year, the solutions are updated based on the changes in the CBSE syllabus.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

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CBSE Class 11 Chemistry Syllabus

CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.

Unit-wise CBSE Class 11 Syllabus for Chemistry

Below is a list of detailed information on each unit for Class 11 Students.

UNIT I – Some Basic Concepts of Chemistry

General Introduction: Importance and scope of Chemistry.

Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements,
atoms and molecules.

Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

UNIT II – Structure of Atom

Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

UNIT III – Classification of Elements and Periodicity in Properties

Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

UNIT IV – Chemical Bonding and Molecular Structure

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

UNIT V – Chemical Thermodynamics

Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.
Third law of thermodynamics (brief introduction).

UNIT VI – Equilibrium

Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,
ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).

UNIT VII – Redox Reactions

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

UNIT IX – Hydrocarbons

Classification of Hydrocarbons
Aliphatic Hydrocarbons:
Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.
Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons:

Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.

CBSE Class 11 Chemistry Practical Syllabus

The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.

The table below consists of evaluation scheme of practical exams.

Evaluation SchemeMarks
Volumetric Analysis08
Salt Analysis08
Content Based Experiment06
Project Work04
Class record and viva04
Total30

CBSE Syllabus for Class 11 Chemistry Practical

Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.

A. Basic Laboratory Techniques
1. Cutting glass tube and glass rod
2. Bending a glass tube
3. Drawing out a glass jet
4. Boring a cork

B. Characterization and Purification of Chemical Substances
1. Determination of melting point of an organic compound.
2. Determination of boiling point of an organic compound.
3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.

C. Experiments based on pH

1. Any one of the following experiments:

  • Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
  • Comparing the pH of solutions of strong and weak acids of same concentration.
  • Study the pH change in the titration of a strong base using universal indicator.

2. Study the pH change by common-ion in case of weak acids and weak bases.

D. Chemical Equilibrium
One of the following experiments:

1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions.
2. Study the shift in equilibrium between [Co(H2O)6] 2+ and chloride ions by changing the concentration of either of the ions.

E. Quantitative Estimation
i. Using a mechanical balance/electronic balance.
ii. Preparation of standard solution of Oxalic acid.
iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid.
iv. Preparation of standard solution of Sodium carbonate.
v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.

F. Qualitative Analysis
1) Determination of one anion and one cation in a given salt
Cations‐ Pb2+, Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2+, Zn2+, Co2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4 +
Anions – (CO3)2‐ , S2‐, NO2‐ , SO32‐, SO2‐ , NO ‐ , Cl‐ , Br‐, I‐, PO43‐ , C2O2‐ ,CH3COO‐
(Note: Insoluble salts excluded)

2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.

G) PROJECTS
Scientific investigations involving laboratory testing and collecting information from other sources.

A few suggested projects are as follows:

  • Checking the bacterial contamination in drinking water by testing sulphide ion
  • Study of the methods of purification of water.
  • Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional
    variation in drinking water and study of causes of presence of these ions above permissible
    limit (if any).
  • Investigation of the foaming capacity of different washing soaps and the effect of addition of
    Sodium carbonate on it.
  • Study the acidity of different samples of tea leaves.
  • Determination of the rate of evaporation of different liquids Study the effect of acids and
    bases on the tensile strength of fibres.
  • Study of acidity of fruit and vegetable juices.

Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

Practical Examination for Visually Impaired Students of Class 11

Below is a list of practicals for the visually impaired students.

A. List of apparatus for identification for assessment in practicals (All experiments)
Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand,
dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp
stand, dropper, wash bottle
• Odour detection in qualitative analysis
• Procedure/Setup of the apparatus

B. List of Experiments A. Characterization and Purification of Chemical Substances
1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid
B. Experiments based on pH
1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied
concentrations of acids, bases and salts using pH paper
2. Comparing the pH of solutions of strong and weak acids of same concentration.

C. Chemical Equilibrium
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing
the concentration of eitherions.
2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the
concentration of either of the ions.

D. Quantitative estimation
1. Preparation of standard solution of oxalic acid.
2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard
solution of oxalic acid.

E. Qualitative Analysis
1. Determination of one anion and one cation in a given salt
2. Cations – NH+4
Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO-
(Note: insoluble salts excluded)
3. Detection of Nitrogen in the given organic compound.
4. Detection of Halogen in the given organic compound.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

Q1

How many units are in the CBSE Class 11 Chemistry Syllabus?

There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).

Q2

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.

Q3

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.

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