Oxidation Number | Definition, How To Find, Rules, Examples

Oxidation number is defined as the total number of electrons that an atom either gains or loses to form a chemical bond with another atom. 

Let’s learn about oxidation number in detail, including its rules and steps to calculate it with the help of examples.

Oxidation Number Definition

Oxidation Number is defined as the charge, whether positive or negative gained by an atom of an element when they form bonds with atoms of other elements either by sharing or transferring electrons.

  • The charge that appears as an oxidation number is the number of electrons lost, gained or shared by the atom during the formation of a bond with atoms of another element in a molecule.
  • It can be positive, negative, zero or fraction depending upon the situation.
  • The oxidation number of an atom is often referred to as its oxidation state.
  • It is generally assigned to atoms in the reaction in which they undergo oxidation and reduction. Such a reaction in which atoms either get oxidised or get reduced is called a redox reaction. The word redox is an acronym of two words reduction and oxidation. 

Note – In assigning oxidation numbers, we simplify the molecular complexity by assuming complete electron transfer, treating bonds as fully ionic. For example, in CO, carbon’s oxidation state is simplified by assuming complete electron loss to oxygen.

List of Oxidation Numbers

ElementOxidation Number
Hydrogen (H)+1
Oxygen (O)-2
Fluorine (F)-1
Chlorine (Cl)-1
Bromine (Br)-1
Iodine (I)-1
Group 1 metals (e.g., Li, Na, K)+1
Group 2 metals (e.g., Mg, Ca, Ba)+2
Group 13 metals (e.g., Al)+3
Group 14 elements (e.g., C in CH4)-4
Group 15 elements (e.g., N in NH3)-3
Group 16 elements (e.g., S in H2S)-2
Group 17 elements (e.g., Cl in HCl)-1

Oxidation number of elements from 1 to 30

H+1
He0
Li+1
Be+2
B+3
C-4/+4
N-3/+3
O-2
F-1
Ne0
Na+1
Mg+2
Al+3
Si-4/+4
P-3/+3/+5
S-2/+4/+6
Cl-1
Ar0
K+1
Ca+2
Sc+3
Ti+2/+3/+4
V+2/+3/+4/+5
Cr+2/+3/+6
Mn+2/+3/+4/+6/+7
Fe+2/+3/+4/+6
Co+2/+3
Ni+2/+3/+4
Cu+1/+2
Zn+2
  

How to Find Oxidation Number

The oxidation number of an atom is the number of electrons lost or gained to form bonds with heteroatoms in a particular molecule. The molecule can be neutral (its overall charge is zero) or may be charged (it has some overall positive or negative charge).

To find the oxidation number of an atom we need to follow the following steps:

Step 1: Assume the oxidation number of the atom to be X which you need to calculate.

Step 2: Mention the oxidation state of other bonded atoms and multiply it with the number of such atoms present in one molecule.

Step 3: Write the oxidation number of all the atoms in the molecule in a linear sum format and equate it to the overall charge of the molecule.

Step 4: Solve for X.

We can learn it with the help of examples below.

Oxidation Number of Sulphur in H2SO4

Oxidation number is defined as the total number of electrons that an atom either gains or loses to form a chemical bond with another atom. 

Oxidation Number of Sulphur in H2SO4

Solution:

Step 1: Assume the oxidation number of sulphur to be x

Step 2: The oxidation number for Hydrogen is +1 and for O is -2.

Step 3: Since the overall charge on the molecule is 0, therefore 2(+1) + X + 4(-2) = 0

Step 4: 2 + X – 8 = 0 ⇒ X – 6 = 0 ⇒ X = +6

Hence, the oxidation number of Sulphur in H2SO4 is +6

Oxidation Number of Chromium in Cr2O72-

Oxidation number is defined as the total number of electrons that an atom either gains or loses to form a chemical bond with another atom. 

Oxidation Number of Chromium in Cr2O72-

Solution:

Step 1: Assume the oxidation number for Chromium be X

Step 2: The oxidation number for oxygen is -2

Step 3: Since the molecule has an overall charge of -2 the equation can be written as 2X + 7(-2)=-2

Step 4: 2X – 14 = -2 ⇒ 2X = +12 ⇒ X = +6

Hence the oxidation number of chromium in Cr2O72- is +6.

It should be noted that in one molecule there can be different oxidation numbers for the same atom. This can be understood from the following example

Multiple Oxidation Numbers of an Atom in a Molecule

Now will learn how to calculate multiple oxidation numbers of an atom in a molecule discussed below:

Oxidation Number of Nitrogen in Ammonium Nitrate i.e. NH4NO3

Solution:

Step 1: In this case, we will split the molecule into two ions Ammonium ion (NH4+) and Nitrate ion (NO3)

Step 2: We will find the oxidation number of Nitrogen in each ion. Assume the oxidation number for Nitrogen to be X in each case

Step 3: For the Ammonium ion, since the molecule has an overall +1 charge, hence equation can be written as X + 4(+1) = +1

Step 4: Solving the equation X + 4 = 1 ⇒ X = -3

Step 5: In the case of nitrate ion, the oxidation number for oxygen is -2 and the molecule has an overall charge of -1.

Step 6: Hence equation, in this case, will be X + 3(-2) = -1

Step 7: Solving the equation we get X – 6 = -1 ⇒ X = +5

Hence the oxidation number of Nitrogen in ammonium nitrate is -3 and +5

In the above examples, we saw that some atoms have fixed oxidation numbers. To understand why some atoms have fixed atomic numbers we need to look at the following mentioned rules of oxidation number.

Oxidation Number Rules

These are the rules to assign constant oxidation numbers to atoms of an element:

Rule 1: The total of all the atoms in a molecule’s oxidation numbers equals zero.

e.g. In KMnO4, the oxidation number of K is +1, the oxidation number of Mn is +7 and the oxidation number of oxygen is −2.

Rule 2: An atom’s oxidation number is always 0 in its most basic form.

e.g. The oxidation number of H, O, N, P, S, Se, Cu, Ag in their element forms is H2, O2, N2, P4, S8, Se8, Cu, Ag respectively, is zero.

Rule 3: Alkali metals (Li, Na, K, Rb, Cs) have an oxidation number of +1 in their compounds.

e.g. In NaCl, the oxidation number of Na is +1.

Rule 4: Alkaline earth metals (Be, Mg, Ca, Sr, Ba) usually have a +2 oxidation number in their compounds.

e.g. The oxidation number of Mg in MgO is +2.

Rule 5: Except in metal hydrides, the oxidation number of H in its compound is always +1.

e.g. In HCl, the oxidation number of H is +1 and in NaH (Sodium hydride), the oxidation number of H is −1.

Rule 6: The fluorine oxidation number is 1 in all its compounds.

e.g. In NaF, the oxidation number of F is −1.

Rule 7: The charge of an ionic compound is equal to the sum of all oxidation states of all the atoms in the compound. 

e.g. In SO42- the oxidation number of Sulphur is +6. The oxidation number of oxygen is −2. Hence the sum is +6+4(-2) = 6-8 = -2.

Rule 8: Except for oxygen and fluorine, the maximum oxidation number of any element is equal to its group number.

e.g. The oxidation number of sulphur in H2S2O8, K2S2O8, S2O–28 and H2SO5 is +6 due to the presence of a peroxide bond.

Rule 9: It’s possible that not all atoms of the same element have the same oxidation number in certain compounds. We get the average result when we compute the oxidation number for that element in such components.

e.g. One sulphur atom in Na2S2O3 has an oxidation number of +6, whereas the other sulphur atoms have an oxidation number of -2. As a result, the average sulphur oxidation state number in Na2S2O3 is +2.

Oxidation number is defined as the total number of electrons that an atom either gains or loses to form a chemical bond with another atom. 

Rule 10: Carbon in organic molecules can have any oxidation number between -4 and +4.

e.g. in HCHO, the oxidation number of carbon is zero.

Rule 11: The common oxidation number of an element is equivalent to its group number from IA to IV A. The formula (Group number –8) gives the common oxidation number of any element from V A to VIII-A.

e.g. 

  • Oxidation number of I A group elements = +1.
  • Oxidation number of II A group elements = +2.
  • Oxidation number of III A group elements = +3.
  • Oxidation number of IV A group elements = +4.
  • Oxidation number of V A group elements = –3.
  • Oxidation number of VI A group elements = –2.
  • Oxidation number of VII A group elements = –1.
  • Oxidation number of VIII A group elements = 0.

Rule 12: C, N, P, and S have oxidation numbers of 4,–3,–3 and –2 in all carbides, nitrides, phosphides, and sulphides, respectively.

e.g. In Mg3N2, the oxidation number of Nitrogen is −3.

Rule 13: In all Metal carbonyls, the oxidation number of metals is zero. 

e.g. In Ni(CO)4, the oxidation number of Ni is zero.

Rule 14: Except for peroxide, superoxides, oxyfluorides, and ozonides, oxygen has an oxidation number of -2 in most of its oxides.

e.g. In Na2O, the oxidation number of O is −2. 

Exceptional Cases for Oxidation Number of Oxygen

There are some exceptional cases for the oxidation number of Oxygen, which are

  • Peroxides: The oxidation number of oxygen in peroxides is 1.  Examples, are H2O2, and Na2O2.
  • Oxidation number of oxygen in fluorine compounds is +2. Examples are F2O or OF2, etc.
  • Superoxides: The oxidation number of oxygen in superoxides is –1/2.
  • Ozonides: Each oxygen atom in an ozonide has an oxidation number of –1/3.

Fixed Oxidation Levels in Atoms, Molecules, and Ions

In neutral atoms and molecules, the overall oxidation state is zero. For example, elements like sodium, magnesium, and iron have an oxidation state of zero. The same goes for neutral molecules like oxygen, chlorine, water, ammonia, methane, and potassium permanganate.

In homo-polar molecules, each atom’s oxidation state is also zero. For instance, in an oxygen molecule, the oxidation state of each oxygen atom is zero.

On the other hand, the oxidation state of charged ions equals the net charge of the ion. This means:

  • All alkali metal ions have an oxidation state of +1.
  • All alkaline earth metal ions have an oxidation state of +2.
  • All boron family metal ions have an oxidation state of +3.
  • Hydrogen has an oxidation state of +1 in a proton (H+) and -1 in a hydride.
  • Oxygen has an oxidation state of -2 in an oxide ion (O2-) and -1 in a peroxide ion (O-O2-).”

Oxidation Number Examples

As of now, we know the basic rules to calculate the Oxidation Number, hence we will find the Oxidation Number of some common compounds which we come across frequently in chemistry.

We will find the Oxidation Number of the following compounds:

Oxidation Number of Manganese in KMnO4

Solution:

 Let the oxidation number of Manganese in KMnO4 be equal to X

The oxidation number of Potassium = +1

The oxidation number of oxygen is =  –2

⇒1 + (X) + 4×(–2)=0

⇒1 + X – 8 = 0

⇒X -7 = 0

⇒X = +7

The oxidation number of Manganese in KMnO4 is +7.

Oxidation Number of Carbon in Na2CO3

Solution:

Oxidation number of oxygen is = –2

Oxidation number of Sodium = +1

Let the oxidation number of carbon be X

⇒2(+1) + X + 3(–2) = 0

⇒2 + X – 6 = 0

⇒X – 4 = 0

⇒X = +4

Therefore X = +4

Oxidation Number of Carbon in Na2CO3 is +4 

Oxidation Number of Nitrogen in Ammonium Nitrite

Solution:

Ammonium Nitrite is an ionic compound containing NH4+ and NO2 ions.

Oxidation number of nitrogen in NH4+

Let oxidation number of N atom is x

⇒ x + 4 = +1

⇒ x = 1 – 4 = -3

Oxidation number of nitrogen in NO2

Let oxidation number of Nitrogen atom be Y

Y + 2(-2) = -1

⇒ Y – 4 = -1

⇒ Y = -1 + 4 = +3

Thus, one atom of Nitrogen in Ammonium Nitrite is in the -3 oxidation state, while the other nitrogen atom is in the +3 oxidation state.

Oxidation Number of Carbon in CN

Solution:

Let Oxidation Number of C atom in CN be x, Oxidation Number for N atom is -3 and the overall charge on the molecule is -1. 

⇒ x – 3 = -1

⇒ x = -1 + 3 = +2

Hence, Carbon atom has oxidation number of +2 in CN

Oxidation Number of N in NH3

Solution:

Let Oxidation Number of N in NH3 be x and we know that Oxidation Number of H is +1.

⇒ x + 3(+1) = 0

⇒ x + 3 = 0

⇒ x = -3

Hence, the Oxidation Number of N atom in NH3 is -3.

Difference between Oxidation and Reduction

Oxidation: The loss of electrons or a rise in the positive oxidation state or decrease in the negative oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as oxidation.

Reduction: The gain of electrons or a drop in the positive oxidation state or increase in the negative oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as reduction . It is referred toas reduction in the oxidation state.

AspectOxidationReduction
DefinitionLoss of electronsGain of electrons
Oxidation stateIncreaseDecrease
Electron transferElectrons are lost or removed from a speciesElectrons are gained or added to a species
ReactivitySpecies acts as an electron donorSpecies acts as an electron acceptor
ExamplesRusting of iron, combustion reactionsReduction of metal ions, hydrogenation reactions

Related:

Oxidation Number- FAQs

What is Oxidation Number?

The charge that appears on an atom of an element when it bonds with atoms of other elements in a molecule is called Oxidation Number.

What is the Common Oxidation Number of Inert Gases?

Inert gases have an oxidation number of zero.

What are the Rules for Calculating Oxidation Numbers?

The rules for oxidation numbers are:

  1. The total of all the atoms in a molecule’s oxidation numbers equals zero.
  2. An atom’s oxidation number in its most basic form is always zero.
  3. In their compounds, the oxidation number of alkali metals (Li, Na ,K ,Rb ,Cs) is always +1.
  4. In their compounds, the oxidation number of alkaline earth metals (Be ,Mg ,Ca ,Sr ,Ba) is always +2.
  5. Except in metal hydrides, the oxidation number of H in its compound is always +1.
  6. Fluorine has an oxidation number of 1 in all of its compounds.
  7. Except for peroxide, super oxides, oxyfluoride, and ozonides, the oxidation number of oxygen in most of its oxides is -2.
  8. Carbon in organic molecules can have any oxidation number between -4 and +4.

What is the Oxidation Number of Mercury(Hg) in Amalgam?

Mercury amalgam has a zero-oxidation number. Each element in an alloy or amalgam has an oxidation number of zero.

What is the Oxidation Number of Oxygen in O3 and in MgO?

  • Oxidation number of O in O3 = 0
  • Oxidation number of O in MgO = –2

What are the Oxidation Numbers of Oxide and Hydride ions?

The oxidation number for oxide ion is -2 and that of hydride ion is -1.

What is the Oxidation Number of Sulphur in S8?

S8 is a polyatomic molecule, it is in its most basic form. As a result, sulphur’s oxidation number in this molecule is zero.

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.