Class 11 Chemistry Notes GOC - Intermediates of Organic Compounds

Intermediates of Organic Compounds

Electronic Displacement Effect:

It is an effect which exists due to the displacement of electron in specie.

These effects can affect stability of compounds and its properties like acidic nature and basic nature. It can also affect the strength of acidic and basic nature of any compound.

Electronic Displacement Effect can be basically divided into two parts:-

1. Permanent effect
2. Temporary effect

Permanent effect includes:

1. Inductive effect
2. Mesmeric (resonance) effect
3. Hyper conjugation whereas

Temporary effect:

1. Electromeric effect
2. Inductomeric effect

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Class 11 Chemistry Notes General Organic Chemistry

General Organic Chemistry is like soul for preparation of organic chemistry.
First we will see how bond breaks and what the different reasons behind bond breaking.
There are basically two types of bond fission:

1. Homolytic Bond Fission
2. Heterolytic Bond Fission

Homolytic Bond Fission:
Homolysis: The bond cleavage in which each bonded atom gets their own contribution

• Cleavage takes place due to HELP (H = Heat, E = Electricity, L = Light, P = Peroxide)
• Favored when Electro negativity difference is zero.
• Cleavage takes place in non polar solvent.

Heterolytic Bond Fission:

• It is formed when the electronegativity difference between the bonded atoms is high.
• Formation is favored by polar solvent

• Positive charge of the solvent attracts the Negative pole of compound and the
• Negative pole of the solvent attracts Positive, and the bond breaks.

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Class 11 Chemistry Notes Stocihiometry - Important Question Answers

Ex. A volume of 12.53 ml of 0.051 M SeO₂ reacts exactly with 25.5 ml of 0.1 M CrSO₄ which is oxidized to Cr³⁺ To what oxidation state was the selenium converted b the reaction.
Sol.

Ex. In basic solution CrO₄^2- ion oxidizes S₂O₃^2- ion to form Cr(OH)₄^- and SO₄^2- ions respectively. How many ml of 0.154 Na₂CrO₄ solutions are required just to react with 40 ml of 0.246 M Na₂S₂ O₃ solutions.
Sol.

Ex. What mass of N₂H₄ can be oxidized to N₂ by 24 gm of K₂CrO₄ which is reduced to
Sol.

Ex. The ion Aⁿ⁺ is oxidized to AO₃^- by MnO₄^- changing to Mn²⁺ in acid medium. Given that mole of Aⁿ⁺ required mole of MnO₄^-. What is the value of n.
Sol.

Ex. 25 ml of 0.17 M HSO₃^- in strongly acidic solution required the addition of 16.9 ml of 0.01 M MnO₄^- for its complete oxidation to SO₄^2-. In neutral solution it requires 28.6 ml. Assign oxidation no. of ‘Mn’ in each of the products.
Sol.

Ex. A given amount of Fe²⁺ is oxidized by x mol of MnO₄^- in acidic medium. Calculate mol of Cr₂O₇^2- required same amount of Fe²⁺ in acidic medium
Sol.

Ex. Calculate mol of (a) MnO₄^- and (b) Cr₂O₇^2- to oxidize 1 mol of FeC₂O₄ (ferrous oxalate) in acidic medium.
Sol.

Ex. A mixture of Na₂C₂O₄ and KHC₂O₄. H₂C₂O₄ required equal volume of 0.1 M KMnO₄ and 0.1 M NaOH separately. What is the molar ratio of Na₂C₂O₄, KHC2O4 and H₂C₂O₄ in the mixture?
Sol.

Ex. 5.5 gm of a mixture of FeSO₄. 7H₂O and Fe₂ (SO₄)₃. 9H₂O requires 5.4 ml of 0.1 N KMnO₄ solutions for complete oxidation. Calculate the no. of gm moles of hydrated ferric sulphate in the mixture?
Sol.

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Class 11 Chemistry Notes Stocihiometry - Glossary

Aliquot. A portion of the whole, usually a simple fraction. A portion of a sample withdraw from a volumetric flask with a pipet is called as aliquot.
Analytical concentration, The total number of moles per litre of a solute regardless of any reactions that might occur when the solute dissolves. Used synonymously with formality.
Equivalent. The amount of a substance which furnishes or reacts with 1 of H⁺ (acid-base), 1 mol of electrons (redox), or 1 mol of a univalent cation (precipitation and complex formation).
Equivalent weight. The weight in grams of one equivalent of a substance.
Equivalence point. The point in a titration where the number of equivalents of titrant is the same as the number of equivalents of analytic.
End point.The point in a titration where an indicator changes color.
Formula weight. The number of formula weights of all the atoms in the chemical formula of a substance.
Formality. The number formula weights of solute per liter of solution; synonymous with analytical concentration.
Indicator. A chemical substance which exhibits different colors in the presence of excess analyte or titrant.
Normality. The number of equivalents of solute per litre of solution.
Primary standard. A substance available in a pure form or state of known purity which is used in standardizing a solution.
Standardization. The process by which the concentration of a solution is accurately acertained.
Standard solution. A solution whose concentration has been accurately determined.
Titrant. The reagent (a standard solution) which is added from a buret to react with the analyte.

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Class 11 Chemistry Notes Stocihiometry - The Atlas

THE ATLAS : A GOOD WAY TO REVISE CHAPTER

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Class 11 Chemistry Notes Stocihiometry -Double indicator acid-base titration

Double indicator acid-base titration:

In the acid-base titration the equivalence point is known with the help of indicator which changes its color at the end point. In the titration of polyacidic base or polybasic acid there are more than one end point. One indicator is not able to give color change at every end point. So to find out each end point we have to use more than one indicator. For example in the titration of Na₂CO₃ against HCl there are two end points.

Na₂CO₃ + HCl → NaHCO₃ + NaCl

NaHCO₃ + HCl → H₂CO₃ + NaCl

When we use phenolphthalein in the above titration it changes its color at first end point when NaHCO₃ is formed and with it we cannot know second end point. Similarly with methyl orange it changes its color at second end point only and we cannot know first end point. It is because all indicator changes color on the basis of pH of medium. So in titration of against acid phenolphthalein cannot be used.


A sample contains NaOH, Na₂CO₃& NaHCO₃. This solution is titrated with HCl or H₂SO₄

(i) When phenolphthalein (HPh) is used as indicator the reaction are given below.

Na₂CO₃ + HCl → NaHCO₃ + NaCl

This is the half neutralization of Na₂CO₃

NaHCO₃ + HCl → X

NaOH + HCl → NaCl + H₂O

Hence

Meq. Of HCl = Meq. Of NaOH + (1/2) Meq. Of Na₂CO₃

(ii) When methyl orange (MeOH) is used as indicator

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

This is the complete neutralization

NaHCO₃ + HCl → NaCl + H₂O + CO₂

NaOH + HCl → NaCl + H₂O

in this case

Meq. Of HCl = Meq. Of NaOH + Meq. Of Na₂CO₃ + Meq. Of NaHCO₃

(iii) If MeOH is added after the first end point obtained from HPh

NaHCO₃ + NaHCO₃ + HCl → NaCl + H₂O + CO₂

Produced original

Meq. Of HCl = Meq. Of NaHCO₃ (produced) + Meq. Of NaHCO₃ (original)

Note:
When we carry out dilution of solution, Meq eq. Milli mole or mole of substance does not change because they represent amount of substance, however molar concentration may change.
Ex.1 A solution contains a mixture of Na₂CO₃ and NaOH. Using phenolphthalein as indicator 25 ml. of mixture required 19.5 ml of 0.995 N HCl for the end point. With methyl orange. 25 ml of solution required 25 ml. of the same. HCl for the end point. Calculate grams per liter of each substance in the mixture.
Sol.

Let a & b are the Meq. Of Na₂CO₃ and NaOH respectively

When phenolphthalein in is used as indicator

(1/2) Meq. Of Na₂CO₃ + Meq. of NaOH = Meq. of HCl

(a/2) + b = 19.5 × 0.995

(a/2) + b = 19.4 …(1)

When MeOH is used as indicator

Meq. Of Na₂CO₃ + Meq. Of NaOH = Meq. Of HCl

a + b = 25 × 0.995

a + b = 24.875 …(2)

From (1) & (2)

(a/2) = 5.475 Þ a = 10.95

B = 24.875 – 10.95 = 13.925

Wt of Na₂CO₃/lit = (10.95/25) × 10^-3 × (106/3) × 1000 = 23.2 gm/lit

Wt of NaOH/lit = b × 10^-3 × 84/25 × 1000 = 22.28 gm/lit Ans.

Ex.2 2.5 g of a mixture containing CaCO₃, Ca(HCO₃)₂ and NaCl was dissolved in 100 mL water and its 10 mL portion required 10 mL 0.05 M H₂SO₄ solution to reach the phenolphthalein end point. An another 10 mL portion of the same stock solution required 32.35 mL of the same acid solution to reach the methyl orange end point. Determine mass percentage of CaCO₃ and Ca (HCO₃)₃in the original mixture.
Sol.

Let a and b are the Meq. of CaCO₃ and Ca(HCO₃)₂ respectively.

When HPh is used in indicater then only CaCO₃ will react with H₂SO₄

Meq. of H₂SO₄ = (1/2) Meq. of CaCO₃

Meq. of H₂SO₄ = ((10 × 0.05 × 100 × 2)/10) = 10

Meq. of CaCO₃ = 20

When MeOH is used as indicator

Then Meq. or CaCO₃ + Meq. of Ca(HCO₃)₂ = Meq. of H₂SO₄

a + b = ((32.35 × 0.05 × 2)/10) × 100

a + b = 32.35

b = 12.5

wt. of CaCO₃ = a × 10^-3 × 100/2

= 1 gm

wt. of Ca(HCO₃)₂ = b × 10^-3 × 162/2 = 1 gm

% of CaCO₃ = (1/2.5) × 100 = 40%

% of Ca(HCO₃)₂ = (1/2.5) × 100 = 40%

Ex.3. 6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO₂ evolved is absorbed completely in one liter of an aqueous solution of NaOH. A 10 mL portion of this solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. Another 10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl orange end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid and strength of original NaOH solution
Sol.

Let 100 ml of solution contains a Meq. of NaOH & b Meq. of Na₂CO₃

When HPh is used as indicator then

Meq. of NaOH + 1/2 Meq. of Na2CO3=Meq. of HCl

Ex.4 A solution contains Na₂CO₃ and NaHCO₃. 10 mL of the solution requires 2.5 mL of for neutralisation using phenolphthalein as an indicator. Methyl organge is added when a further 2.5 mL of 0.2 M H₂SO₄ was required. Calculate the amount of Na₂CO₃ and NaHCO₃ in one litre of the solution.
Sol.

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Class 11 Chemistry Notes Stocihiometry -Calculation of Chlorine from a Sample of Bleaching Powder

The weight % of available Cl₂ from the given sample of bleaching powder on reaction with dil acids or CO₂ is called available chlorine.
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂

CaOCl₂ + 2CH₃COOH → Ca (CH₃COO) ₂ + H₂O + Cl₂

CaCOl₂ + CO₂ → CaCO₃ + Cl₂
Method of determination:
Bleaching powder + CH₃COOH + KI (starch/Hypo)end point

Meq. Of I₂ = Meq. Of Cl₂ = Meq. Of Hypo solution

% of Cl₂ = (3.55 × x × V (mL)/W (g))

Where x=Molarity of hypo solution

V =mL of hypo solution used in titration.
Ex.1. 10 gm sample of bleaching powder was dissolved into water to make the solution on liter. To this solution 35 mL of 1.0 M Mohr salt solution was added containing enough H₂SO₄. After the reaction was complete, the excess Mohr salt required 30 mL of 0.1 M KMnO₄ for oxidation. The % of available Cl₂ approximately is (mol wt=71)
Sol.2. Meq. Of Mohr’s salt =35×1×1=35

Meq. Of KMnO₄=Meq. Of excess Mohr salt=30×0.1×5=15

Meq. Of Mohr salt reacted with bleaching powder =35-15=20

Meq. Of Cl2=20

Wt. of Cl₂ = 20 × 10^-3 × (71/2) = 0.71 gm

% of Cl₂ = (0.71/10) × 100 = 7.1% Ans.

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Class 11 Chemistry Notes Stocihiometry - Equivalent Concept and Titration

Back titration
Back titration is used in volumetric analysis to find out excess of reagent added by titrating it with suitable reagent. It is also used to find out percentage purity of sample. For example in acid-base titration supposes we have added excess base in acid mixture. To find excess base we can titrate the solution with another acid of known strength.
Ex.2 20 g. of a sample of Ba(OH)₂ is dissolved in 50 ml. of 0.1 N HCl solution. The excess of HCl was titrated with 0.1 N NaOH. The volume of NaOH used was 20 cc. Calculate the percentage of Ba(OH)₂ in the sample.
Sol. Milli eq. of HCl initially =50×0.1=5
Milli eq. of NaOH consumed = Milli eq. of HCl in excess =20×0.1=2
∴ Milli eq. of HCl consumed = Milli eq. of Ba (OH)₂=5-2=3
∴ eq. of Ba(OH)₂ = 3 / 1000 = 3 × 10^-3
Mass of Ba (OH)₂ = 3 × 10^-3 × (171/2) = 0.2565 g
% Ba (OH)₂=(0.2565/20)×100=1.28% Ans.
Ex.3 4.0 g of monobasic, saturated carboxylic acid is dissolved in 100 mL water and its 10 mL portion required 8.0 mL 0.27 M NaOH to reach the equivalence point. In an another experiment, 5.0 g of the same acid is burnt completely and CO₂ produced is absorbed completely in 500 mL of a 2.0 N NaOH solution. A 10 mL portion of the resulting solution is treated with excess of BaCl₂ to precipitate all carbonate and finally titrated with 0.5 N H₂SO₄ solution. Determine the volume of the acid solution that would be required to make this solution neutral.
Sol. Meq. Of NaOH=8×0.27=Meq. Of acid for 10 ml of acid solution
Meq. Of acid for 100 ml of solution
= (8 × 0.27 × 1000)/10 = (4/M) × 10³
M (acid) = 4/(8 × 0.27 × 10) × 1000 = 185.2
Formula of acid. = C_nH_2nO₂
Þ M = 14n + 32 = 185.2
Þ n = 11
Now 5g acid will produce (5/185.2) × 11 = (55/185.2) mol CO₂ after complete combustion.
2NaOH + CO₂ → Na₂CO₃ + H₂O
Total mass of NaOH available = 500 × 2 × 10^-3 = 1.0 mole
Moles of NaOH left unreacted = 1-(2 × 55/185.2) = (76/185.2) in 500 mL
Þ Molarity of NaOH after precipitation of Na₂CO₃ = 0.812
Therefore, 0.812 × 10 = 0.5 × V
Þ V = 16.24 mL Ans.
Precipitation titration:
In ionic reaction we can know strength of unknown solution of salt by titrating it against a reagent with which it can form precipitate. For example NaCl strength can be known by titrating it against AgNO₃ solution with which it forms white ppt. of AgCl.
So Meq. Of NaCl at equivalence point =Meq. Of AgNO₃ used= Meq. Of AgCl formed
Ex.4 A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a standardized acid as follows:
Co (NH₃) _x Cl₃ (aq) + HCl → NH⁺₄(aq) + Co³⁺ (aq) + Cl^-(aq)
A 1.58 g complex required 23.63 mL 1.5 M HCl to reach the equivalence point. Determine formula. If the reaction mixture at equivalence point is titrated with excess of AgNO₃ solution, what mass of AgCl will precipitate out?
Sol. The balanced chemical reaction is:
Co(NH₃)_x Cl₃ + xHCl → xNH⁺₄ + Co³⁺ + (x + 3) Cl^-

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Class 11 Chemistry Notes Stocihiometry -Disproportionate Reactions and Law of Chemical equivalence

Disproportionate Reactions:
A redox reaction in which a same element present in a particular compound in a definite oxidation state is oxidized as well as reduced simultaneously is a disproportionate reaction.

n factor of overall reaction = n₁ × n₂ / n₁ + n₂

Ex.1

MnO₄^- + 8H⁺ + 5e → Mn⁺² + 4H₂O

n factor of KMnO₄ = 5

EX.2

Find the n factor of Na₂S₂O₃ in the following reactions.

(a) Na₂S₂O₃ + I₂ → NaI + Na₂S₄O₆

Sodium thio sodium tetra

Sulphate thionate

(b) Na₂S₂O₃ + I₂ → NaI + Na₂SO₄

Sol.

(a) Change in oxidation number of sulphur=|+2-2.5|=0.5

Change in oxidation number per molecule =0.5×2

∴ n factor of Na₂S₂O₃ = 1

(b) Change in oxidation number of sulphur=|2-6|=4

Change in oxidation number per molecule =4×2

∴ n factor of Na₂S₂O₃ = 8

=> Consider the salt A_x B_y to undergo a reaction so that the element A undergoes a change in O.S. but is present in more than one product with the same oxidation state i.e. A^+c_xB_y → A^+f_aB_y + A^+f_gH

(the superscripts denote the oxidation state of the respective elements).

∴ n=|xc-xf|

=> Salts that react in such a way that more than one type of atom in the salt undergoes oxidation state Change.

A^+c_xB_y → A^+f_aE + J_xB^-1

In this case both A and B are changing their O.SO. ‘s and both of them are either getting oxidized or reduced. In such a case n factor of the compound is the sum of the individual n factors of A and B. i.e.|xc-xf| + |-xC- (-yi)|. Then n factor of A can be understood which is |xc-xf|. The n factor of B is |-xc-(-yi)|

Law of Chemical equivalence:
It states that in any chemical reaction the equivalents of all the reactants and products must be same.

2A+3B→4C

Eq. of A=Equivalents of ‘B’=Equivalents of ‘C’
Equivalents of ‘A’ = (Weight of ‘A’)/Equivalent weight of ‘A’
Or Equivalents of ‘A’ = no. of moles of ‘A’ x n-factor

A + B + C → product

If A & B do not react with each other but C react with both A & B then

Gm eq. of A + Gm eq. of B= gm eq. of C

Or

Meq. Of A + Meq. Of B = gm Meq. Of C

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Class 11 Chemistry Notes Stocihiometry - N factor calculation

N factor here we mean a conversion factor by which we divide molar mass of substance to get equivalent mass and it depends on nature of substance which varies from one condition to another condition. We can divide n-factor calculation is two category.
In case of non-redox reaction.

(a) n factor of acid =Basicity of the acid

Basicity: Number of replaceable H⁺ ion.

Ex.1

n factor of HCl = 1

n factor of CH₃COOH=1

n factor of H₂SO₄=2

(b)n factor of base= acidity of the base

Acidity: Number of replaceable OH^- ion.

Ex.2

n factor of NaOH=1

n factor of Ca(OH)₂=2

n factor of Al(OH)₃=3

n factor of B(OH)₃=1 (because it is a mono basic acid)

(c) n factor Salt: Total number of positive or negative charge.

Ex.3

n factor of NaCl=1

n factor of Na₂SO₄=2

n factor of K₂SO₄. Al₂ (SO₄)₃. 24H₂O = 8

Ex.4

Find the n factor of H₃PO₄ in the following reaction.

H₃PO₄ + Ca (OH)₃ → CaHPO₄ + 2H₂O

Sol.

Basicity of H₃PO₄ in the above reaction is 2

∴ the n factor of H₃PO₄ is 2

In case of redox reaction

(a) From oxidation number

n factor of oxidizing or reducing agent=change in oxidation number per molecule.

⇒ consider a salt A_xB_y in which the O.S. of A is +c. It changes to a compound A_dE in which the O.S. of A is +f. Here we are assuming that B does not undergo any change in O.S. A^+c_xB_y → A^+f_dE (Obviously A_xB_y must have reacted with some other substance to produce the product A_dE. That means other substance has the atom E in it.) The ‘n’ factor is = |xc-xf|.

Ex.5

Find the n factor of KMnO₄ in different medium.

Sol.

(i) In acidic medium

KMnO₄→Mn⁺²

Change in oxidation number of Mn=+7-2=5

∴ the n factor of KMnO₄=5

(ii) Basic medium

KMnO₄→K₂MnO₄

N factor of KMnO₄ = +7 – 6 = 1

(iii) Neutral medium

KMnO₄ → MnO₂

N factor of MnO₂ = +7 – 4 = 3

(b)From ion electron method:

N factor = total number of electrons transferred per mole of the reactant

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Class 11 Chemistry Notes Stocihiometry - Balancing of Equations

Balancing of Equations:

1. Ion-electron method
2. Oxidation number method
   
   [Concept involved that in any chemical reaction e^- cannot be produced so no. of e^- s in O.H. & R.H. should be same]

Note : To predict the product of reaction remember:

• Free halogen on reduction gives halide ion (F₂ → F^-)
• Alkali metals on oxidation give metallic ion with +1 oxidation state.
• In alkaline medium KMnO₄ is reduced to K₂MnO₄ but in strong alkaline or neutral medium it reduces of Mn⁺⁴
• In acid solution KMnO₄ is reduced to Mn²⁺
• H₂O₂ on reduction gives water and on oxidation gives oxygen.
• Dichromate ion in acid solution is reduced to Cr³⁺.

Nature of Oxides Based on Oxidation Number:

Lowest oxidation state → Basic (MnO)

Intermediate oxidation state → Amphoteric (Mn₃O₄, MnO₂)

Highest oxidation state → Acidic (Mn₂O₇)

Rules For Assigning Oxidation Number:

1. Oxidation number of free elements or atoms is zero.
2. Oxidation number of allotropes is zero.
3. Oxidation number of atoms in homo-nuclear molecules is zero.
4. Oxidation number of mono-atomic ions is equal to the algebraic charge on them.
5. Oxidation number of F in compounds is – 1
6. Oxidation number of H in its compounds is +1, except in metal hydrides where it is-1
7. Oxidation number of O is-2in its compound, but in and in peroxides it is 1 and -0.5 in KO₂
8. Oxidation number of alkali metals in their compounds + 1.
9. Oxidation number of alkaline earth metals in their compounds is + 2.
10. The sum of oxidation number of all the atoms in a molecule should be zero and in an ion equal to its charge.

Average Oxidation Number : Find Oxidation Number of Fe in Fe₃O₄.

Fe₃O₄ is FeO. Fe₂O₃

O.N. of Fe in FeO is + 2 ; O.N. of Fe in Fe₂O₃ is + 3.

Therefore average O.N. of three Fe atoms (+2 + 2x (+3))/3 = +(8/3)

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Class 11 Chemistry Notes Stocihiometry - Oxidation and Reduction

Oxidation Number:
It is the charge (real or imaginary) which an atom appears to have when it is in combination. It may be a whole no. or fractional. An element may have different values of oxidation number depending. It depends on nature of compound in which it is present. There are some operational rules to determine oxidation number.
Oxidation:
Addition of oxygen, removal of hydrogen, addition of electro-negative element, removal of electropositive element, loss of electrons, increase in oxidation number (de – electro nation).
Reduction:
Removal of oxygen, addition of hydrogen, removal of electronegative element, addition of electro positive element, gain of electrons, decrease inoxidation number (electro nation).
Redox Reactions:
A reaction in which oxidation & reduction occur simultaneously.
Oxidising Agents:
(oxidants, oxidizers). They oxidise others, themselves are reduced & gain electrons. Eq. O2, O3, HNO3, MnO2, H2O2, halogens, KMnO4, K2Cr2O7, KIO3, CIO3, FeCl3, NaOCl hydrogen ions.

[Atoms are present in their higher oxidation state.
Reducing Agents:
H₂, molecular form is weak reducing agent but Nascent hydrogen is powerful. C, CO, H₂S, SO₂, SnCl₂, sodium thio Sulphate, Al, Na, CaH₂, NaBH₄, LiAlH₄. They reduce others, themselves get oxidised & lose electrons. Also called reductants or reducers.

[Atoms are present in their lower oxidation state.]
Both oxidizing & Reducing Agents:

SO₂, H₂O₂, O₃, NO₂, etc.

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CBSE class 11 chemistry notes

Chemistry Notes Some Basic Concepts of Chemistry

CBSE class 11 Chemistry Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book.

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Chemistry Notes Structure of Atom
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Chemistry Notes Classification of Elements
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Chemistry Notes Chemical Bonding and Molecular Structure
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Chemistry Notes States of Matter
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Chemistry Notes Thermodynamics
Chemistry Notes Thermodynamics CBSE class 11 Chemistry Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book.

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Chemistry Notes Equilibrium
Chemistry Notes Equilibrium CBSE class 11 Chemistry Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book.

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Chemistry Notes Redox Reaction 01
Chemistry Notes Redox Reaction 01 CBSE class 11 Chemistry Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book.

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Chemistry Notes Redox Reaction 02
Chemistry Notes Redox Reaction 02 CBSE class 11 Chemistry Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book.

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Chemistry Notes Hydrogen
Chemistry Notes Hydrogen CBSE class 11 Chemistry Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book.
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CBSE Class 11 physics notes

Key Notes Physical World and Measurement

Key Notes Physical World and Measurement CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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 Key Notes Kinematics

Key Notes Kinematics CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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Key Notes Law of Motion

Key Notes Law of Motion CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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Key Notes Work Energy Power

Key Notes Work Energy Power CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools.

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Key Notes Motion of System of Particles and Rigid Body

Key Notes Motion of System of Particles and Rigid Body CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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Key Notes Gravitation

Key Notes Gravitation CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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Key Notes Properties of Bulk Matter

Key Notes Properties of Bulk Matter CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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 Key Notes Thermodynamics

Key Notes Thermodynamics CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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 Key Notes Behaviour of Perfect Gas and Kinetic Theory of Gases

Key Notes Behaviour of Perfect Gas and Kinetic Theory of Gases CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools.

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Key Notes Oscillations and Waves

Key Notes Oscillations and Waves CBSE class 11 Physics Key notes and summary of the chapter with examples. These are very useful summary notes with neatly explained examples for best revision of the book. CBSE Key notes prepared by KVS and Delhi Schools. 

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