Oxidation Number and State-Definition, Calculation, Examples, Problems

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

The oxidation number is also referred to as the oxidation state. However, sometimes these terms can have a different meaning depending on whether we are considering the electronegativity of the atoms or not. The oxidation number term is used frequently in coordination chemistry.

In general, the oxidation state or number helps us describe the transfer of electrons. However, students have to note that it is different from a formal charge which determines the arrangement of atoms. The oxidation number/state is also used to determine the changes that occur in redox reactions. Besides, it is quite similar to valence electrons.

What Is the Oxidation Number?

The oxidation number of an atom is defined as the charge that an atom appears to have on forming ionic bonds with other heteroatoms. An atom having higher electronegativity (even if it forms a covalent bond) is assigned a negative oxidation state.

The definition assigns an oxidation state to an atom on conditions that the atom –

i) Bonds with heteroatoms.

ii) Always form ionic bonding by either gaining or losing electrons, irrespective of the actual nature of bonding.

Since an atom can have multiple valence electrons and form multiple bonds, all of them will be assumed to be ionic and assigned an oxidation state equal to the number of electrons involved in the bonding. So, oxidation number or state is a hypothetical case of the assumption of atoms forming an ionic bond.

In the complex cation tetroxoplatinum (PtO4)2+, platinum possesses an oxidation state of 10. Ten is the maximum oxidation state exhibited by any atom. It appears to have lost ten electrons to form the ion. But, the ionization energy required for removing an electron from charges positively species increases heavily.

The larger the charge, difficult it is to remove an electron, and so, the higher the ionization energy. So, the removal of ten electrons is highly hypothetical. Similarly, the addition of electrons also becomes difficult with increasing negative charge.

Oxidation states larger than three, whether positive or negative, are practically impossible. In spite of the assumption, it helps in understanding the changes accompanying the atom undergoing a chemical change.

How to Find the Oxidation Number of an Atom?

The oxidation number or state of an atom/ion is the number of electrons an atom/ion that the molecule has either gained or lost compared to the neutral atom. Electropositive metal atoms of groups 1, 2 and 3 lose a specific number of electrons and always have constant positive oxidation numbers.

In molecules, more electronegative atoms gain electrons from less electronegative atoms and have negative oxidation states. The numerical value of the oxidation state is equal to the number of electrons lost or gained.

The oxidation number or oxidation state of an atom or ion in a molecule/ion is assigned by:

i) Summing up the constant oxidation state of other atoms/molecules/ions that are bonded to it

ii) Equating the total oxidation state of a molecule or ion to the total charge of the molecule or ion

Atoms/ Molecules and Ions That Have Constant Oxidation State (Number)

a) The net charge on neutral atoms or molecules is zero. So, their overall oxidation state is zero.

For example, the oxidation state of elemental atoms such as sodium, magnesium, and iron is zero. Similarly, the net oxidation state of neutral molecules such as oxygen, chlorine, water, ammonia, methane, and potassium permanganate is zero.

The oxidation state of atoms in homo-polar molecules is zero. The oxidation number of an atom in an oxygen molecule is zero.

b) The oxidation state of charged ions is equal to the net charge of the ion. So,

  1. The oxidation number of all alkali metal ions is always = +1
  2. The oxidation number of all alkaline earth metal ions is always = +2
  3. The oxidation number of all boron family metal ions is always = +3
  4. The oxidation number of hydrogen in proton (H+) is +1, and in hydride is -1.
  5. The oxidation number of oxygen in oxide ion(O2-) is -2, and in peroxide ion(O-O2-) is -1.

Calculation of Oxidation Number of an Atom in a Molecule/Ion

Oxidation number of potassium permanganate (KMnO4) = Sum of oxidation number of (K + Mn + 4O) = 0

Oxidation number of permanganate ion (MnO4) = Sum of oxidation number of ( Mn + 4O)= -1

Calculation of Oxidation Number of Atoms Occurring Only Once in a Molecule

Examples 1: Oxidation state of chlorine in KCl

KCl is neutral, and so net charge = 0

Oxidation state of KCl = Oxidation state of potassium + oxidation state of chlorine = 0.

Oxidation state of potassium = +1

Oxidation states → +1 + x = 0: x = -1

Atoms in the species → K Cl

Oxidation state of chlorine in KCl = -1

Example 2: Oxidation number of manganese in permanganate ion MnO4

The charge on the permanganate ion is -1

Oxidation state of permanganate ion = Oxidation state of manganese + 4 oxidation state of oxygen = -1.

Oxidation state of oxygen = -2

Oxidation states → x + (4*-2) = -1: x = +7

Atoms in the species → Mn 4O

Oxidation state of manganese = +7

Example 3: Oxidation number of a metal ion in a complex.

i) Ni(CO) 4.

The total charge of the complex is zero. CO is a neutral molecule.

Oxidation states → x + (4*0) = 0: x = 0

Atoms in the species → Ni 4 CO

Nickel is also in a zero oxidation state.

ii) [CoCl2(NH3)4]Cl.

The complex can be written in the ionic forms as [CoCl2(NH3)4]+Cl.

Metal is in a cationic complex with a unitary positive charge. Ammonia is a neutral ligand, and chlorine has a unit negative charge.

Oxidation number of [CoCl2(NH3)4]+ = Oxidation number of (Co + 2Cl + 4×0) = +1.

Oxidation states → x + (2*-1) + 4*0 = +1: x = +3

Atoms in the species → Co 2Cl 4 NH3

Oxidation number of cobalt in the complex = +3

Calculation of Oxidation Number of Atoms Occurring More Than Once in a Molecule and Having Identical Bonding

Atoms occurring more than once in a molecule may be bonded in an identical way or not. If they are identically bonded, then there is no difference between them, and all the atoms will have the same oxidation numbers. The oxidation state of such an atom in a molecule can be calculated by the normal method.

The average oxidation number will be the same as calculated individually and as a whole number.

Example 1: The number of atoms of chlorine is two in the molecules Cl2O, Cl2O5 and Cl2O7. But, the environment of both atoms of chlorine is the same, as shown by their structures. The oxidation number of the atoms calculated either individually or from the whole molecule is the same.

i) Cl2O:

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

Cl2O is neutral, and so net charge = 0.

Net oxidation state of Cl2O = 2 x Oxidation state of chlorine + 1x Oxidation state of oxygen = 0.

Oxidation state of oxygen = -2.

Oxidation states → 2 x + (-2) = 0: x = +1

Atoms in the species → 2Cl O

Oxidation state of chlorine in Cl2O= 2/2 = +1

ii) Cl2O5:

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

Cl2O5 is neutral, and so net charge = 0

Oxidation state of Cl2O5 = 2 x Oxidation state of chlorine + 5 x oxidation state of oxygen = 0.

Therefore,

Oxidation state of oxygen = -2.

Oxidation states → 2x + (5*-2) = 0: x = +5

Atoms in the species → 2Cl 5O

Oxidation state of chlorine in Cl2O5 = 10/2 = +5

iii) Cl2O7:

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

Cl2O7 is neutral, and so net charge = 0

Oxidation state of Cl2O7 = 2 x Oxidation state of chlorine + 7 x oxidation state of oxygen = 0.

⸪ Oxidation state of oxygen = -2.

Oxidation states → 2x + (7*-2) = 0: x = +7

Atoms in the species → 2Cl 7O

Oxidation state of chlorine in Cl2O = 14/2 = +7

Note: Except the atoms/molecules/ions mentioned, as having a constant oxidation state, the oxidation state of other atoms/molecules and ions will vary depending on the molecule they are present.

In the given examples, the oxidation state of chlorine is not constant, but variable (+1, +5 and +7)

Example 2: Oxidation state of chromium in dichromate anion.

Dichromate ion is Cr2O72-.

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

The charge on the ion is -2.

Oxidation state of dichromate ion = 2 x Oxidation state of chromium + 7 x oxidation state of oxygen = -2.

Oxidation state of oxygen = -2.

Oxidation states → 2x + (7*-2) = -2: x = +6

Atoms in the species → 2Cr 7O

Oxidation state of chromium= 12 / 2 = 6

Calculation of Oxidation Number of Atoms Occurring More Than Once in a Molecule and Having a Difference in Bonding

Atoms having different bond structures will have different oxidation states. Hence, their oxidation state has to be individually determined from their molecular structure. The average oxidation state can be calculated by assuming them to be equal. In such a case, the average oxidation could be fractional rather than a whole integer.

Example 1: Cl2O4

i) The average oxidation state of chlorine

Cl2O4 is neutral, and so net charge = 0

Oxidation state of Cl2O4 = 2 x Oxidation state of chlorine + 4 x oxidation state of oxygen = 0. ⸪

Oxidation state of oxygen = -2.

Oxidation states → 2x + (4*-2) = 0: x = +4

Atoms in the species → 2Cl 4O

Oxidation state of chlorine in Cl2O5 = 8/2 = +4

ii) Structure of Cl2O4

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

Individual oxidation state of oxygen ‘a’ is +7

Individual oxidation state of oxygen ‘b’ is +1

None of the oxygen has a +4 oxidation state.

Example 2:  Cyclopentadienyl anion C5H5 

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

i) Average oxidation number of carbon

Five carbon atoms share the five electrons from five hydrogen atoms and an additional electron of the negative charge by resonance. So, six electrons are shared by five-carbon.

The average oxidation state of each carbon = 6/5 = fraction

ii) Without resonance, four carbon has a -1 oxidation state, and one carbon has a -2 oxidation state.

Oxidation Number of Atoms in a Diatomic Molecule

A diatomic molecule can be either homo or heteronuclear.

i) Homonuclear diatomic molecule

The oxidation number concept is applicable only to heteroatoms forming a molecule. Hence, in a homonuclear diatomic molecule, the oxidation number of the atoms is zero. Also, the oxidation number of hydrogen or oxygen, nitrogen, and chlorine in respective molecules is zero.

ii) Heteronuclear diatomic molecule

In hetero diatomic molecules, all bonds formed between the atoms are considered as ionic.

More electronegative atoms are assumed to take away the bonding electrons from the less electronegative atom. So, the electronegative atom will have a negative oxidation state, and the magnitude is equal to the number of electrons taken by it.

The less electronegative atom is supposed to have lost its electron to the more electronegative atom. So, the less electronegative atom will have a positive oxidation state equal to the number of electrons lost by it.

Example 1: HCl

Chlorine is more electronegative than hydrogen. So, chlorine is assumed to take away the electron from hydrogen. Chlorine, which receives one electron, has an oxidation number of -1, while hydrogen losing one electron, has an oxidation state of +1.

Example 2: H2O

Oxygen is more electronegative than hydrogen. So, the oxygen atom receives one electron each from the two-hydrogen atom and will have an oxidation number of -2. Both hydrogens losing one electron each will have an oxidation number of +1 each.

Fractional Oxidation States

The oxidation state is the number of electrons assumed to have either been lost or taken by heteroatoms during their bonding. Since the numbers of electrons are whole numbers, the oxidation number of individual atoms also has to be a whole integer.

But, there are molecules that contain an atom more than once and each bonded differently. Such atoms shall have different oxidation state at different positions and hence has to be, calculated individually, taking into consideration the atoms it bonds.

Calculation of the oxidation state of the atom using the normal method assumes all the same atoms as equal and will give only an average of the different oxidation states of the same atom in the molecule. This average oxidation state is mostly a fraction instead of a whole number.

So, the fractional oxidation state is always an average oxidation number of the same atoms in a molecule and does not reflect the true state of the oxidation state of atoms.

Example 1: Superoxide -KO2

Potassium ion has an oxidation number of +1. Potassium superoxide molecule being neutral, the oxidation state of two oxygen atoms together is -1.

So, the average oxidation number of oxygen in superoxide is -1/2.

The structure of superoxide ion is given below:

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

As per the structure, one oxygen atom has a zero oxidation state. The second oxygen atom is negatively charged and has a -1 oxidation state. So, the true oxidation state of oxygen atoms is not minus half each but 0 and -1.

Example 2: Fe3O4

Considering the oxidation state of oxygen as -2, the average oxidation state of iron atoms will be +8/3.

But the molecule is a mixture of two compounds of FeO and Fe2O3.

In FeO and Fe2O3, iron is in +2 and +3 oxidation states. So, in Fe3O4, one iron has +2, and the other iron has +3 oxidation states.

Average oxidation state is

\(\begin{array}{l}= +\frac{2+3+3}{3} = +\frac{8}{3}\end{array} \)

Example 3: Tetra-thionate ion

The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

Tetrathionate ion has four sulphur atoms bonded to oxygen as in the structure. Out of the four sulphur atoms, the two-terminal sulphur atoms are connected to three oxygen heteroatoms and one homo sulphur atom. Each terminal sulphur atom forms five bonds with oxygen heteroatoms, so the oxidation state will be +5. The bridging sulphur atoms being homo-nuclear, have zero oxidation state.

The total oxidation of the entire four sulphur atoms is ten.

So, the average oxidation state of sulphur = 10/4 = 2.5

Oxidation and Reduction, Redox Reactions

Atoms and molecules react to form products. The reactions are classified into many types based on the nature of change in the reactants to form products. Whatever may be the reaction types, reactant and product atoms/ions in the reaction may either have the same or a different number of valence electrons.

Reactions, where the number of valence electrons in the reactant atom/ion is different from the product side, are called reduction-oxidation or simply redox reactions. Atom/ion might have either lost or gained electrons during the reaction, and accordingly, atom/ion is said to be either oxidized or reduced.

Atoms/ions in the reactions are represented by their atomic symbol with a superscript. The superscript represents the difference in the number of electrons of the atom/ion compared to the neutral atom. The superscript also has a positive sign if the electron is lost and a negative sign if the electron is gained compared to the neutral atom.

The superscript, along with the sign, is called the ‘oxidation state’ of the atom. The atom may have different oxidation states depending upon the number of electrons either gained or lost. Neutral atoms have zero oxidation state.

ReactionReaction TypeThe Oxidation State of the Atom in the Reactant SideThe Oxidation State of the Atom in the Product Side
M0 → M+ + eOxidation reaction0+1
M+ + e → M0Reduction+10
M2+ → M3+ + eOxidation reaction+2+3
M3+ + e→ M2+Reduction+3+2

In redox reactions:

In redox reactions, atoms or ions either lose or gain electrons and have different oxidation states before and after the reaction.

  • The oxidation number can be positive or zero, or negative
  • The oxidation number has to be an integer, as the number of electrons can only be an integer.
  • The oxidation number cannot be fractional
  • The oxidation number is the same as the oxidation state.

Frequently Asked Questions on Oxidation Number

Q1

What is the oxidation number of alkali and alkaline earth metals?

The oxidation number of alkali and alkaline earth metals are +1 and +2, respectively.

Q2

What is the oxidation number of oxide ions?

The oxidation number of oxide ions is -2.

Q3

What is the oxidation number of hydrogen in HCl?

The oxidation number of hydrogen in HCl is +1.

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.

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.