Friday, October 26, 2007
Study Notes for IIT JEE
I am trying to develop study notes in the form of Study guides to a standard book.
Study Guide TMH Chemistry 34. Exercises in Organic Chemistry
-----------------
JEE question 2007 Paper II
Riemer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes.
1. Which one of the following reagents is used in the above reaction?
(A) aq.NaOH+ CH-3Cl
(B) aq.NaOH+ CH-2Cl-2
(C) aq.NaOH+ CHCl-3
(D) aq.NaOH+ CCl-4
Answer C
Reimen-Tiemann reaction takes place is presence of aq. NaOH+ CHCl3
2. The electrophile in this
(A) :CHCl
(B) +CHCl-2
(C) :CCl-2
(D) :CCl-3
Answer: C
-------------------------------
JEE question 2007 Paper II
Riemer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes.
1. Which one of the following reagents is used in the above reaction?
(A) aq.NaOH+ CH-3Cl
(B) aq.NaOH+ CH-2Cl-2
(C) aq.NaOH+ CHCl-3
(D) aq.NaOH+ CCl-4
Answer C
Reimen-Tiemann reaction takes place is presence of aq. NaOH+ CHCl3
2. The electrophile in this
(A) :CHCl
(B) +CHCl-2
(C) :CCl-2
(D) :CCl-3
Answer: C
-------------------------------
Thursday, October 25, 2007
IIT JEE 2008 Chemistry Syllabus
JEE 2008
Chemistry Syllabus
Physical chemistry
General topics: Concept of atoms and molecules; Dalton’s atomic theory; Mole concept; Chemical formulae; Balanced chemical equations; Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality.
Gaseous and liquid states: Absolute scale of temperature, ideal gas equation; Deviation from ideality, van der Waals equation; Kinetic theory of gases, average, root mean square and most probable velocities and their relation with temperature; Law of partial pressures; Vapour pressure; Diffusion of gases.
Atomic structure and chemical bonding: Bohr model, spectrum of hydrogen atom, quantum numbers; Wave-particle duality, de Broglie hypothesis; Uncertainty principle; Qualitative quantum mechanical picture of hydrogen atom, shapes of s, p and d orbitals; Electronic configurations of elements (up to atomic number 36); Aufbau principle; Pauli’s exclusion principle and Hund’s rule; Orbital overlap and covalent bond; Hybridisation involving s, p and d orbitals only; Orbital energy diagrams for homonuclear diatomic species; Hydrogen bond; Polarity in molecules, dipole moment (qualitative aspects only); VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral).
Energetics: First law of thermodynamics; Internal energy, work and heat, pressure-volume work; Enthalpy, Hess’s law; Heat of reaction, fusion and vapourization; Second law of thermodynamics; Entropy; Free energy; Criterion of spontaneity.
Chemical equilibrium: Law of mass action; Equilibrium constant, Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of DG and DGo in chemical equilibrium; Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.
Electrochemistry: Electrochemical cells and cell reactions; Standard electrode potentials; Nernst equation and its relation to DG; Electrochemical series, emf of galvanic cells; Faraday's laws of electrolysis; Electrolytic conductance, specific, equivalent and molar conductivity, Kohlrausch's law; Concentration cells.
Chemical kinetics: Rates of chemical reactions; Order of reactions; Rate constant; First order reactions; Temperature dependence of rate constant (Arrhenius equation).
Solid state: Classification of solids, crystalline state, seven crystal systems (cell parameters a, b, c, alpha, beta, gamma), close packed structure of solids (cubic), packing in fcc, bcc and hcp lattices; Nearest neighbours, ionic radii, simple ionic compounds, point defects.
Solutions: Raoult's law; Molecular weight determination from lowering of vapour pressure, elevation of boiling point and depression of freezing point.
Surface chemistry: Elementary concepts of adsorption (excluding adsorption isotherms); Colloids: types, methods of preparation and general properties; Elementary ideas of emulsions, surfactants and micelles (only definitions and examples).
Nuclear chemistry: Radioactivity: isotopes and isobars; Properties of alpha, beta and gamma rays; Kinetics of radioactive decay (decay series excluded), carbon dating; Stability of nuclei with respect to proton-neutron ratio; Brief discussion on fission and fusion reactions.
Inorganic Chemistry
Isolation/preparation and properties of the following non-metals: Boron, silicon, nitrogen, phosphorus, oxygen, sulphur and halogens; Properties of allotropes of carbon (only diamond and graphite), phosphorus and sulphur.
Preparation and properties of the following compounds: Oxides, peroxides, hydroxides, carbonates, bicarbonates, chlorides and sulphates of sodium, potassium, magnesium and calcium; Boron: diborane, boric acid and borax; Aluminium: alumina, aluminium chloride and alums; Carbon: oxides and oxyacid (carbonic acid); Silicon: silicones, silicates and silicon carbide; Nitrogen: oxides, oxyacids and ammonia; Phosphorus: oxides, oxyacids (phosphorus acid, phosphoric acid) and phosphine; Oxygen: ozone and hydrogen peroxide; Sulphur: hydrogen sulphide, oxides, sulphurous acid, sulphuric acid and sodium thiosulphate; Halogens: hydrohalic acids, oxides and oxyacids of chlorine, bleaching powder; Xenon fluorides.
Transition elements (3d series): Definition, general characteristics, oxidation states and their stabilities, colour (excluding the details of electronic transitions) and calculation of spin-only magnetic moment; Coordination compounds: nomenclature of mononuclear coordination compounds, cis-trans and ionisation isomerisms, hybridization and geometries of mononuclear coordination compounds (linear, tetrahedral, square planar and octahedral).
Preparation and properties of the following compounds: Oxides and chlorides of tin and lead; Oxides, chlorides and sulphates of Fe2+, Cu2+ and Zn2+; Potassium permanganate, potassium dichromate, silver oxide, silver nitrate, silver thiosulphate.
Ores and minerals:Commonly occurring ores and minerals of iron, copper, tin, lead, magnesium, aluminium, zinc and silver.
Extractive metallurgy: Chemical principles and reactions only (industrial details excluded); Carbon reduction method (iron and tin); Self reduction method (copper and lead); Electrolytic reduction method (magnesium and aluminium); Cyanide process (silver and gold).
Principles of qualitative analysis: Groups I to V (only Ag+, Hg2+, Cu2+, Pb2+, Bi3+, Fe3+, Cr3+, Al3+, Ca2+, Ba2+, Zn2+, Mn2+ and Mg2+); Nitrate, halides (excluding fluoride), sulphate and sulphide.
Organic Chemistry
Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of simple organic molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centres, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections); Resonance and hyperconjugation; Keto-enol tautomerism; Determination of empirical and molecular formulae of simple compounds (only combustion method); Hydrogen bonds: definition and their effects on physical properties of alcohols and carboxylic acids; Inductive and resonance effects on acidity and basicity of organic acids and bases; Polarity and inductive effects in alkyl halides; Reactive intermediates produced during homolytic and heterolytic bond cleavage; Formation, structure and stability of carbocations, carbanions and free radicals.
Preparation, properties and reactions of alkanes: Homologous series, physical properties of alkanes (melting points, boiling points and density); Combustion and halogenation of alkanes; Preparation of alkanes by Wurtz reaction and decarboxylation reactions.
Preparation, properties and reactions of alkenes and alkynes: Physical properties of alkenes and alkynes (boiling points, density and dipole moments); Acidity of alkynes; Acid catalysed hydration of alkenes and alkynes (excluding the stereochemistry of addition and elimination); Reactions of alkenes with KMnO4 and ozone; Reduction of alkenes and alkynes; Preparation of alkenes and alkynes by elimination reactions; Electrophilic addition reactions of alkenes with X2, HX, HOX and H2O (X=halogen); Addition reactions of alkynes; Metal acetylides.
Reactions of benzene: Structure and aromaticity; Electrophilic substitution reactions: halogenation, nitration, sulphonation, Friedel-Crafts alkylation and acylation; Effect of o-, m- and p-directing groups in monosubstituted benzenes.
Phenols: Acidity, electrophilic substitution reactions (halogenation, nitration and sulphonation); Reimer-Tieman reaction, Kolbe reaction.
Characteristic reactions of the following (including those mentioned above): Alkyl halides: rearrangement reactions of alkyl carbocation, Grignard reactions, nucleophilic substitution reactions; Alcohols: esterification, dehydration and oxidation, reaction with sodium, phosphorus halides, ZnCl2/concentrated HCl, conversion of alcohols into aldehydes and ketones; Ethers:Preparation by Williamson's Synthesis; Aldehydes and Ketones: oxidation, reduction, oxime and hydrazone formation; aldol condensation, Perkin reaction; Cannizzaro reaction; haloform reaction and nucleophilic addition reactions (Grignard addition); Carboxylic acids: formation of esters, acid chlorides and amides, ester hydrolysis; Amines: basicity of substituted anilines and aliphatic amines, preparation from nitro compounds, reaction with nitrous acid, azo coupling reaction of diazonium salts of aromatic amines, Sandmeyer and related reactions of diazonium salts; carbylamine reaction; Haloarenes: nucleophilic aromatic substitution in haloarenes and substituted haloarenes (excluding Benzyne mechanism and Cine substitution).
Carbohydrates: Classification; mono- and di-saccharides (glucose and sucrose); Oxidation, reduction, glycoside formation and hydrolysis of sucrose.
Amino acids and peptides: General structure (only primary structure for peptides) and physical properties.
Properties and uses of some important polymers: Natural rubber, cellulose, nylon, teflon and PVC.
Practical organic chemistry: Detection of elements (N, S, halogens); Detection and identification of the following functional groups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl, amino and nitro; Chemical methods of separation of mono-functional organic compounds from binary mixtures.
Source: http://www.iitkgp.ernet.in/jee/chemistry.htm
Chemistry Syllabus
Physical chemistry
General topics: Concept of atoms and molecules; Dalton’s atomic theory; Mole concept; Chemical formulae; Balanced chemical equations; Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality.
Gaseous and liquid states: Absolute scale of temperature, ideal gas equation; Deviation from ideality, van der Waals equation; Kinetic theory of gases, average, root mean square and most probable velocities and their relation with temperature; Law of partial pressures; Vapour pressure; Diffusion of gases.
Atomic structure and chemical bonding: Bohr model, spectrum of hydrogen atom, quantum numbers; Wave-particle duality, de Broglie hypothesis; Uncertainty principle; Qualitative quantum mechanical picture of hydrogen atom, shapes of s, p and d orbitals; Electronic configurations of elements (up to atomic number 36); Aufbau principle; Pauli’s exclusion principle and Hund’s rule; Orbital overlap and covalent bond; Hybridisation involving s, p and d orbitals only; Orbital energy diagrams for homonuclear diatomic species; Hydrogen bond; Polarity in molecules, dipole moment (qualitative aspects only); VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral).
Energetics: First law of thermodynamics; Internal energy, work and heat, pressure-volume work; Enthalpy, Hess’s law; Heat of reaction, fusion and vapourization; Second law of thermodynamics; Entropy; Free energy; Criterion of spontaneity.
Chemical equilibrium: Law of mass action; Equilibrium constant, Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of DG and DGo in chemical equilibrium; Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.
Electrochemistry: Electrochemical cells and cell reactions; Standard electrode potentials; Nernst equation and its relation to DG; Electrochemical series, emf of galvanic cells; Faraday's laws of electrolysis; Electrolytic conductance, specific, equivalent and molar conductivity, Kohlrausch's law; Concentration cells.
Chemical kinetics: Rates of chemical reactions; Order of reactions; Rate constant; First order reactions; Temperature dependence of rate constant (Arrhenius equation).
Solid state: Classification of solids, crystalline state, seven crystal systems (cell parameters a, b, c, alpha, beta, gamma), close packed structure of solids (cubic), packing in fcc, bcc and hcp lattices; Nearest neighbours, ionic radii, simple ionic compounds, point defects.
Solutions: Raoult's law; Molecular weight determination from lowering of vapour pressure, elevation of boiling point and depression of freezing point.
Surface chemistry: Elementary concepts of adsorption (excluding adsorption isotherms); Colloids: types, methods of preparation and general properties; Elementary ideas of emulsions, surfactants and micelles (only definitions and examples).
Nuclear chemistry: Radioactivity: isotopes and isobars; Properties of alpha, beta and gamma rays; Kinetics of radioactive decay (decay series excluded), carbon dating; Stability of nuclei with respect to proton-neutron ratio; Brief discussion on fission and fusion reactions.
Inorganic Chemistry
Isolation/preparation and properties of the following non-metals: Boron, silicon, nitrogen, phosphorus, oxygen, sulphur and halogens; Properties of allotropes of carbon (only diamond and graphite), phosphorus and sulphur.
Preparation and properties of the following compounds: Oxides, peroxides, hydroxides, carbonates, bicarbonates, chlorides and sulphates of sodium, potassium, magnesium and calcium; Boron: diborane, boric acid and borax; Aluminium: alumina, aluminium chloride and alums; Carbon: oxides and oxyacid (carbonic acid); Silicon: silicones, silicates and silicon carbide; Nitrogen: oxides, oxyacids and ammonia; Phosphorus: oxides, oxyacids (phosphorus acid, phosphoric acid) and phosphine; Oxygen: ozone and hydrogen peroxide; Sulphur: hydrogen sulphide, oxides, sulphurous acid, sulphuric acid and sodium thiosulphate; Halogens: hydrohalic acids, oxides and oxyacids of chlorine, bleaching powder; Xenon fluorides.
Transition elements (3d series): Definition, general characteristics, oxidation states and their stabilities, colour (excluding the details of electronic transitions) and calculation of spin-only magnetic moment; Coordination compounds: nomenclature of mononuclear coordination compounds, cis-trans and ionisation isomerisms, hybridization and geometries of mononuclear coordination compounds (linear, tetrahedral, square planar and octahedral).
Preparation and properties of the following compounds: Oxides and chlorides of tin and lead; Oxides, chlorides and sulphates of Fe2+, Cu2+ and Zn2+; Potassium permanganate, potassium dichromate, silver oxide, silver nitrate, silver thiosulphate.
Ores and minerals:Commonly occurring ores and minerals of iron, copper, tin, lead, magnesium, aluminium, zinc and silver.
Extractive metallurgy: Chemical principles and reactions only (industrial details excluded); Carbon reduction method (iron and tin); Self reduction method (copper and lead); Electrolytic reduction method (magnesium and aluminium); Cyanide process (silver and gold).
Principles of qualitative analysis: Groups I to V (only Ag+, Hg2+, Cu2+, Pb2+, Bi3+, Fe3+, Cr3+, Al3+, Ca2+, Ba2+, Zn2+, Mn2+ and Mg2+); Nitrate, halides (excluding fluoride), sulphate and sulphide.
Organic Chemistry
Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of simple organic molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centres, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections); Resonance and hyperconjugation; Keto-enol tautomerism; Determination of empirical and molecular formulae of simple compounds (only combustion method); Hydrogen bonds: definition and their effects on physical properties of alcohols and carboxylic acids; Inductive and resonance effects on acidity and basicity of organic acids and bases; Polarity and inductive effects in alkyl halides; Reactive intermediates produced during homolytic and heterolytic bond cleavage; Formation, structure and stability of carbocations, carbanions and free radicals.
Preparation, properties and reactions of alkanes: Homologous series, physical properties of alkanes (melting points, boiling points and density); Combustion and halogenation of alkanes; Preparation of alkanes by Wurtz reaction and decarboxylation reactions.
Preparation, properties and reactions of alkenes and alkynes: Physical properties of alkenes and alkynes (boiling points, density and dipole moments); Acidity of alkynes; Acid catalysed hydration of alkenes and alkynes (excluding the stereochemistry of addition and elimination); Reactions of alkenes with KMnO4 and ozone; Reduction of alkenes and alkynes; Preparation of alkenes and alkynes by elimination reactions; Electrophilic addition reactions of alkenes with X2, HX, HOX and H2O (X=halogen); Addition reactions of alkynes; Metal acetylides.
Reactions of benzene: Structure and aromaticity; Electrophilic substitution reactions: halogenation, nitration, sulphonation, Friedel-Crafts alkylation and acylation; Effect of o-, m- and p-directing groups in monosubstituted benzenes.
Phenols: Acidity, electrophilic substitution reactions (halogenation, nitration and sulphonation); Reimer-Tieman reaction, Kolbe reaction.
Characteristic reactions of the following (including those mentioned above): Alkyl halides: rearrangement reactions of alkyl carbocation, Grignard reactions, nucleophilic substitution reactions; Alcohols: esterification, dehydration and oxidation, reaction with sodium, phosphorus halides, ZnCl2/concentrated HCl, conversion of alcohols into aldehydes and ketones; Ethers:Preparation by Williamson's Synthesis; Aldehydes and Ketones: oxidation, reduction, oxime and hydrazone formation; aldol condensation, Perkin reaction; Cannizzaro reaction; haloform reaction and nucleophilic addition reactions (Grignard addition); Carboxylic acids: formation of esters, acid chlorides and amides, ester hydrolysis; Amines: basicity of substituted anilines and aliphatic amines, preparation from nitro compounds, reaction with nitrous acid, azo coupling reaction of diazonium salts of aromatic amines, Sandmeyer and related reactions of diazonium salts; carbylamine reaction; Haloarenes: nucleophilic aromatic substitution in haloarenes and substituted haloarenes (excluding Benzyne mechanism and Cine substitution).
Carbohydrates: Classification; mono- and di-saccharides (glucose and sucrose); Oxidation, reduction, glycoside formation and hydrolysis of sucrose.
Amino acids and peptides: General structure (only primary structure for peptides) and physical properties.
Properties and uses of some important polymers: Natural rubber, cellulose, nylon, teflon and PVC.
Practical organic chemistry: Detection of elements (N, S, halogens); Detection and identification of the following functional groups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl, amino and nitro; Chemical methods of separation of mono-functional organic compounds from binary mixtures.
Source: http://www.iitkgp.ernet.in/jee/chemistry.htm
Chemistry JEE Matrix Match Questions
JEE 2007 paper I
Match the chemical substances in column I with type of polymers/type of bonds in column II.
Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
------Column I----------------------- Column II
(A) Cellulose------------ (p) natural polymer
(B) nylon-6,6------------ (q) synthetic polymer
(C) Protein-------------- (r) amide linkage
(D) Sucrose--------------- (s) glycoside linkage
Solution:
A — p, A — s
B — q, B — r
C — r, C — p
D — s
---------------------------------
JEE 2007 Paper II
Match gases under specified conditions listed in column I with their properties/laws in column II.
Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I------------------ Column II
(A) hydrogen gas
(P = 200 atm,
T = 273 K) ---------------(p) compressibility factor ¹ 1
(B) hydrogen gas
(P = 0, T = 273 K)------- (q) attractive forces are dominant
(C) 2 CO (P = 1 atm, T = 273 K)--- (r) PV = nRT
(D) real gas with very
large molar volume------------ (s) P(V – nb) = nRT
Solution:
A — p, s,
B — r,
C — q, p,
D — p, q
------------------
JEE question 2007 paper II
Match the complexes in column I with nature of the reactions/type of the products in column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
(A) O-2^- → O-2+O-2^2- -------(p) redox reaction
====================================================
(B) [CrO-4]^2- +H^+ → --------(q) one of the products
----------------------------------has trigonal-planar
----------------------------------structure
=====================================================
(C)[MnO-4]^- +[NO-2]^- +H^+ → (r) dimeric bridged
---------------------------------tetrahedral metal ion
=====================================================
(D)[NO-3]^- + H-2SO-4+Fe^2+ → (s) disproportionation
====================================================
Solution
A – p, s
B – r
C – p, q
D – p
----------------------------------------
JEE Question 2007 Paper II
Match the compounds/ions in Column I with their properties/reactions in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I------------------- Column II
===============================================
(A) C-6H-5CH0------- (p) gives precipitate
-------------------------with 2, 4-
-------------------------dinitrophenylhydrazine
=================================================
(B) CH-3C≡CH-------- (q) gives precipitate with AgNO3
===================================================
(C) CN^--------------(r) is a nucleophile
===================================================
(D) I^- -------------(s) is involved in
--------------------------cyanohydrin formation
====================================================
Solution
A – p, q, s
B – q
C – r, s
D – q, r
---------------------------------------------------
Match the chemical substances in column I with type of polymers/type of bonds in column II.
Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
------Column I----------------------- Column II
(A) Cellulose------------ (p) natural polymer
(B) nylon-6,6------------ (q) synthetic polymer
(C) Protein-------------- (r) amide linkage
(D) Sucrose--------------- (s) glycoside linkage
Solution:
A — p, A — s
B — q, B — r
C — r, C — p
D — s
---------------------------------
JEE 2007 Paper II
Match gases under specified conditions listed in column I with their properties/laws in column II.
Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I------------------ Column II
(A) hydrogen gas
(P = 200 atm,
T = 273 K) ---------------(p) compressibility factor ¹ 1
(B) hydrogen gas
(P = 0, T = 273 K)------- (q) attractive forces are dominant
(C) 2 CO (P = 1 atm, T = 273 K)--- (r) PV = nRT
(D) real gas with very
large molar volume------------ (s) P(V – nb) = nRT
Solution:
A — p, s,
B — r,
C — q, p,
D — p, q
------------------
JEE question 2007 paper II
Match the complexes in column I with nature of the reactions/type of the products in column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
(A) O-2^- → O-2+O-2^2- -------(p) redox reaction
====================================================
(B) [CrO-4]^2- +H^+ → --------(q) one of the products
----------------------------------has trigonal-planar
----------------------------------structure
=====================================================
(C)[MnO-4]^- +[NO-2]^- +H^+ → (r) dimeric bridged
---------------------------------tetrahedral metal ion
=====================================================
(D)[NO-3]^- + H-2SO-4+Fe^2+ → (s) disproportionation
====================================================
Solution
A – p, s
B – r
C – p, q
D – p
----------------------------------------
JEE Question 2007 Paper II
Match the compounds/ions in Column I with their properties/reactions in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I------------------- Column II
===============================================
(A) C-6H-5CH0------- (p) gives precipitate
-------------------------with 2, 4-
-------------------------dinitrophenylhydrazine
=================================================
(B) CH-3C≡CH-------- (q) gives precipitate with AgNO3
===================================================
(C) CN^--------------(r) is a nucleophile
===================================================
(D) I^- -------------(s) is involved in
--------------------------cyanohydrin formation
====================================================
Solution
A – p, q, s
B – q
C – r, s
D – q, r
---------------------------------------------------
Sunday, October 21, 2007
Blog Structure
Tbe blog now has important points related to each topic in JEE syllabus in the form of revision points.
They can be accessed through label heads of the chapters.
The points are updated whenever I study these chapters.
Separately for each chapter, I collected the syllabus, some past JEE questions and some materials that explains some concepts in the chapter, web sites that are giving material related to the chapter. These can be accessed through the TMH Study guide label.
They can be accessed through label heads of the chapters.
The points are updated whenever I study these chapters.
Separately for each chapter, I collected the syllabus, some past JEE questions and some materials that explains some concepts in the chapter, web sites that are giving material related to the chapter. These can be accessed through the TMH Study guide label.
Study Guide ch 1. THE CONCEPT OF ATOMS AND MOLECULES
JEE Syllabus
General topics:
The concept of atoms and molecules;
Dalton's atomic theory;
Mole concept;
Chemical formulae;
Balanced chemical equations;
Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions;
Concentration in terms of mole fraction, molarity, molality and normality.
-------------------------
Main topics in TMH Book Chapter
ATOM AND MOLECULAR MASSES
LAWS OF CHEMICAL COMBINATION
QUANTITATIVE INFORMATION FROM A CHEMICAL EQUATION
EXPRESSING CONCENTRATION OF A SUBSTANCE
CONVERSION OF CONCENTRATION UNITS
CONCEPT OF EQUIVALENT
BALANCING CHEMICAL EQUATION
RULES TO COMPUTE OXIDATION NUMBER
BALANCING REDOX REACTIONS VIA OXIDATION NUMBERS
BALANCING REDOX REACTIONS VIA ION ELECTRON (OR HALF EQUATION) METHOD
-----------------
Sub-topics and concepts
_______________________
ATOM AND MOLECULAR MASSES
* Relative Atomic Mass of an Element
Relative Molecular Mass of a Compound
Atomic Mass Unit
Atomic Mass
Molecular Mass
Mole of a Substance
Amount of a Substance: is expressed in moles.
Molar Mass: The average mass per unit amount of a substance.
----------
LAWS OF CHEMICAL COMBINATION
____________________________
Law of Conservation fo Masses
Law of Constant Composition
Law of Multiple Proportion
QUANTITATIVE INFORMATION FROM A CHEMICAL EQUATION
_________________________________________________
Stochiometric coefficients or numbers: The numbers which appear before the chemical symbols in a chemical equation.
Chemical equation gives information about moles of various reactants and products. Hence molar masses involved in the reaction and molar masses of products.
EXPRESSING CONCENTRATION OF A SUBSTANCE
_______________________________________
Mass percentage of substance in a system
Mole fraction of a substance in a system
Molarity = Amount of a substance (in mol)/Volume of solution expressed in dm^3
It is applicable to solutions only.
The unit of molarity is mol dm^-3. It is commonly abbreviated by the symbol M and is spelled as molar.
Molality = Amount of a a substance (in mol)/Mass of solvent expressed in kg
It is also applicable to solutions only
CONVERSION OF CONCENTRATION UNITS
_________________________________
Mole fraction into molarity
Mole fraction inot molality
Molality not molarity
Molarity into mole fraction
Molality into mole fraction
Molarity into molality
--------------------
CONCEPT OF EQUIVALENT
__________________________
In TMH Book it is given: "One equivalent of a substance in a reaction is defined as the amount of substance which reacts or liberates 1 mol of electrons (or H^+ or OH^- ions). In a reaction, a substance always reacts with another substance in equivalent amounts."
More detailed treatment is given in Madan and Bisht, ISC Chemistry.
Equivalent weight of a substance (element or compound) is defined as 'the number of parts by weight of it, that will combine with or displace directly or indirectly, 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, 35.5 parts by weight of chlorine, 108 parts by weight of silver or the equivalent parts by weight of any other elements."
The meaning will be more clear as we study experiments that determine equivalent weights. Since substances react in the ratio of their equivalent weights, their knowledge and experimental determination is extremely important. A number of methods, with different methods being applicable to different materials are available. Two methods are mentioned below.
a. Hydrogen displacement method: The equivalent weight of metals like calcium, zinc, tin, magnesium etc., which react with dilute acids to produce hydrogen can be determined by this method. In this method, a known quantity of the metal is treated with excess of the dilue sulphuric acid or hydrochloric acid till whole of the metal disappears by way of reaction with the acid. the volume of hydrogen gas so produced is determined and reduced to STP. The weight of the hydrogen at STP is determined from the volume of hydrogen at STP by making use of molar volume concept. From the weight of hydrogen produced and the weight of metal taken for the experiment, equivalent weight of the metal is determined.
Example: o.205 g of a metal on treatment with a dilute acid gave 106.6 ml of hydrogen(dry) at 740.6 mm pressure and 17 degrees centrigrade temperature. Calculate the equivalent weight of the metal.
volume of hydrogen at 740.6 mm pressure and 17 degree centigrade (290 K) = 106.6 ml
Volume of hydrogen at STP 760 mm and 273 K = (740.6/290)*(273/760)*106.6
= 97.72 ml
22.4 litres of H-2 = 22400 ml of H-2 at STP weighs 2g
So 97.72 ml of hydrogen weights (97.72/22400)*2 g
0.205 g of metal gives (97.72/22400)*2 g = .008725 g
To get one g of hydrogen, metal required = (1/.008725)*.205 = 23.5 g
Hence equivalent weight of metal = 23.5
23.5 parts of metal by weight react with 1 part by weight of hydrogen.
b. Oxide Formation Method
The equivalent weight of metals like copper, magnesium, mercury, zinc etc., which form their oxides relatively easily can be found by this method.
In this method a known weight of the metal is taken an converted into its oxide say by heating the metal in the atmosphere of oxygen.
Calculation: Let the weight of the metal = x g.
Let the weight of oxide formed = y g
Therefore y - x g of oxygen combines with x g of the metal.
1 g of oxygen combines with [1/(y-x)]*x g of metal
8 g of oxygen combines with [1/(y-x)]*8x g of metal. this quantity is the equivalent weight of metal.
----------------------
BALANCING CHEMICAL EQUATION
RULES TO COMPUTE OXIDATION NUMBER
BALANCING REDOX REACTIONS VIA OXIDATION NUMBERS
BALANCING REDOX REACTIONS VIA ION ELECTRON (OR HALF EQUATION) METHOD
--------------
web sites
Calculations - Mole concept and some more
http://www.docbrown.info/page04/4_73calcs.htm
------------------------------
JEE Question 2007 paper I Linked Comprehension
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 ×10^23 ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry electrochemistry and radiochemistry. The following example illustrates a typical case, involving
chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200;1 Faraday = 96500 coulombs).
1. The total number of moles of chlorine gas evolved is
(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0
Answer: B
2. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is
(A) 200
(B) 225
(C) 400
(D) 446
Answer: D
In presence of Hg cathode sodium ion will discharge in place of hydrogen gas due to over voltage in the form of amalgams.
Weight of amalgam = 2 × (23 + 200) = 446 g
3. The total charge (coulombs) required for compete electrolysis is
(A) 24125
(B) 48250
(C) 96500
(D) 193000
Answer: D
Total charge = 2 × 96500 = 193000 C
This question is better answered after studying the electro chemistry chapter.
--------------------------
JEE Question 2007 paper II
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is
(A) 3
(B) 4
(C) 5
(D) 6
Answer D
From Compounds of metals chapter
Mohr's salt is a double salt. FeSO-4.(NH-4)-2SO-4.6H-2O
It is obtained by mixing freshly prepared ferrous sulphate in solution with equal molar amounts of ammonium sulphate and then allow the solution to crystallize.
----------------
General topics:
The concept of atoms and molecules;
Dalton's atomic theory;
Mole concept;
Chemical formulae;
Balanced chemical equations;
Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions;
Concentration in terms of mole fraction, molarity, molality and normality.
-------------------------
Main topics in TMH Book Chapter
ATOM AND MOLECULAR MASSES
LAWS OF CHEMICAL COMBINATION
QUANTITATIVE INFORMATION FROM A CHEMICAL EQUATION
EXPRESSING CONCENTRATION OF A SUBSTANCE
CONVERSION OF CONCENTRATION UNITS
CONCEPT OF EQUIVALENT
BALANCING CHEMICAL EQUATION
RULES TO COMPUTE OXIDATION NUMBER
BALANCING REDOX REACTIONS VIA OXIDATION NUMBERS
BALANCING REDOX REACTIONS VIA ION ELECTRON (OR HALF EQUATION) METHOD
-----------------
Sub-topics and concepts
_______________________
ATOM AND MOLECULAR MASSES
* Relative Atomic Mass of an Element
Relative Molecular Mass of a Compound
Atomic Mass Unit
Atomic Mass
Molecular Mass
Mole of a Substance
Amount of a Substance: is expressed in moles.
Molar Mass: The average mass per unit amount of a substance.
----------
LAWS OF CHEMICAL COMBINATION
____________________________
Law of Conservation fo Masses
Law of Constant Composition
Law of Multiple Proportion
QUANTITATIVE INFORMATION FROM A CHEMICAL EQUATION
_________________________________________________
Stochiometric coefficients or numbers: The numbers which appear before the chemical symbols in a chemical equation.
Chemical equation gives information about moles of various reactants and products. Hence molar masses involved in the reaction and molar masses of products.
EXPRESSING CONCENTRATION OF A SUBSTANCE
_______________________________________
Mass percentage of substance in a system
Mole fraction of a substance in a system
Molarity = Amount of a substance (in mol)/Volume of solution expressed in dm^3
It is applicable to solutions only.
The unit of molarity is mol dm^-3. It is commonly abbreviated by the symbol M and is spelled as molar.
Molality = Amount of a a substance (in mol)/Mass of solvent expressed in kg
It is also applicable to solutions only
CONVERSION OF CONCENTRATION UNITS
_________________________________
Mole fraction into molarity
Mole fraction inot molality
Molality not molarity
Molarity into mole fraction
Molality into mole fraction
Molarity into molality
--------------------
CONCEPT OF EQUIVALENT
__________________________
In TMH Book it is given: "One equivalent of a substance in a reaction is defined as the amount of substance which reacts or liberates 1 mol of electrons (or H^+ or OH^- ions). In a reaction, a substance always reacts with another substance in equivalent amounts."
More detailed treatment is given in Madan and Bisht, ISC Chemistry.
Equivalent weight of a substance (element or compound) is defined as 'the number of parts by weight of it, that will combine with or displace directly or indirectly, 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, 35.5 parts by weight of chlorine, 108 parts by weight of silver or the equivalent parts by weight of any other elements."
The meaning will be more clear as we study experiments that determine equivalent weights. Since substances react in the ratio of their equivalent weights, their knowledge and experimental determination is extremely important. A number of methods, with different methods being applicable to different materials are available. Two methods are mentioned below.
a. Hydrogen displacement method: The equivalent weight of metals like calcium, zinc, tin, magnesium etc., which react with dilute acids to produce hydrogen can be determined by this method. In this method, a known quantity of the metal is treated with excess of the dilue sulphuric acid or hydrochloric acid till whole of the metal disappears by way of reaction with the acid. the volume of hydrogen gas so produced is determined and reduced to STP. The weight of the hydrogen at STP is determined from the volume of hydrogen at STP by making use of molar volume concept. From the weight of hydrogen produced and the weight of metal taken for the experiment, equivalent weight of the metal is determined.
Example: o.205 g of a metal on treatment with a dilute acid gave 106.6 ml of hydrogen(dry) at 740.6 mm pressure and 17 degrees centrigrade temperature. Calculate the equivalent weight of the metal.
volume of hydrogen at 740.6 mm pressure and 17 degree centigrade (290 K) = 106.6 ml
Volume of hydrogen at STP 760 mm and 273 K = (740.6/290)*(273/760)*106.6
= 97.72 ml
22.4 litres of H-2 = 22400 ml of H-2 at STP weighs 2g
So 97.72 ml of hydrogen weights (97.72/22400)*2 g
0.205 g of metal gives (97.72/22400)*2 g = .008725 g
To get one g of hydrogen, metal required = (1/.008725)*.205 = 23.5 g
Hence equivalent weight of metal = 23.5
23.5 parts of metal by weight react with 1 part by weight of hydrogen.
b. Oxide Formation Method
The equivalent weight of metals like copper, magnesium, mercury, zinc etc., which form their oxides relatively easily can be found by this method.
In this method a known weight of the metal is taken an converted into its oxide say by heating the metal in the atmosphere of oxygen.
Calculation: Let the weight of the metal = x g.
Let the weight of oxide formed = y g
Therefore y - x g of oxygen combines with x g of the metal.
1 g of oxygen combines with [1/(y-x)]*x g of metal
8 g of oxygen combines with [1/(y-x)]*8x g of metal. this quantity is the equivalent weight of metal.
----------------------
BALANCING CHEMICAL EQUATION
RULES TO COMPUTE OXIDATION NUMBER
BALANCING REDOX REACTIONS VIA OXIDATION NUMBERS
BALANCING REDOX REACTIONS VIA ION ELECTRON (OR HALF EQUATION) METHOD
--------------
web sites
Calculations - Mole concept and some more
http://www.docbrown.info/page04/4_73calcs.htm
------------------------------
JEE Question 2007 paper I Linked Comprehension
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 ×10^23 ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry electrochemistry and radiochemistry. The following example illustrates a typical case, involving
chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200;1 Faraday = 96500 coulombs).
1. The total number of moles of chlorine gas evolved is
(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0
Answer: B
2. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is
(A) 200
(B) 225
(C) 400
(D) 446
Answer: D
In presence of Hg cathode sodium ion will discharge in place of hydrogen gas due to over voltage in the form of amalgams.
Weight of amalgam = 2 × (23 + 200) = 446 g
3. The total charge (coulombs) required for compete electrolysis is
(A) 24125
(B) 48250
(C) 96500
(D) 193000
Answer: D
Total charge = 2 × 96500 = 193000 C
This question is better answered after studying the electro chemistry chapter.
--------------------------
JEE Question 2007 paper II
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is
(A) 3
(B) 4
(C) 5
(D) 6
Answer D
From Compounds of metals chapter
Mohr's salt is a double salt. FeSO-4.(NH-4)-2SO-4.6H-2O
It is obtained by mixing freshly prepared ferrous sulphate in solution with equal molar amounts of ammonium sulphate and then allow the solution to crystallize.
----------------
Study Guide Ch.2. GASEOUS, LIQUID AND SOLID STATES
JEE syllabus
Gaseous and liquid states:
Absolute scale of temperature,
ideal gas equation;
Deviation from ideality,
van der Waals equation;
Kinetic theory of gases, average,
root mean square and most probable velocities and their relation with temperature; Law of partial pressures;
Vapour pressure;
Diffusion of gases.
----------------------
Main Topics in TMH Book Chapter
SECTION I GASEOUS STATE
KINETIC THEORY OF GASES
EXPRESSIONS OF SOME USEFUL PHYSICAL QUANTITIES
VAN DER WAALS EQUATION OF STATE
REDUCTION OF VAN DER WAALS TO VIRIAL EQUATION
SECTION II LIQUID STATE
VAPOUR PRESSURE
SURFACE TENSION
VISCOCITY
SECTION III SOLID STATE
CLASSIFICATION OF CRYSTALS BNASED ON BOND TYPE
SEVEN CRYSTAL TYPES
PERCENTAGE OF VOID VOLUME
PONT DEFECTS
-------------------
web sites
Part I States of Matter gas-liquid-solid
http://www.docbrown.info/page03/3_52states.htm
------------------------------
JEE Question 2007 paper II
Matrix matching
Match the chemical system/units cells mentioned in column I with their characteristic features mentioned in column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I------------------------- Column II
(A)simple cubic and ---------(p)have these cell
face-centred cubic------------parameters a = b = c and
------------------------------ α =β = γ
(B)cubic and rhombohedral-- (q) are two crystal systems
(C) cubic and tetragonal (r) Have only two
-----------------------------crystallographic angles of
------------------------------90°
(D) hexagonal and monoaclinic-- (s) Belong to same
------------------------------------crystal system
Solution
A – p, s
B – p, q
C – q
D – r, q
----------------------------------
Gaseous and liquid states:
Absolute scale of temperature,
ideal gas equation;
Deviation from ideality,
van der Waals equation;
Kinetic theory of gases, average,
root mean square and most probable velocities and their relation with temperature; Law of partial pressures;
Vapour pressure;
Diffusion of gases.
----------------------
Main Topics in TMH Book Chapter
SECTION I GASEOUS STATE
KINETIC THEORY OF GASES
EXPRESSIONS OF SOME USEFUL PHYSICAL QUANTITIES
VAN DER WAALS EQUATION OF STATE
REDUCTION OF VAN DER WAALS TO VIRIAL EQUATION
SECTION II LIQUID STATE
VAPOUR PRESSURE
SURFACE TENSION
VISCOCITY
SECTION III SOLID STATE
CLASSIFICATION OF CRYSTALS BNASED ON BOND TYPE
SEVEN CRYSTAL TYPES
PERCENTAGE OF VOID VOLUME
PONT DEFECTS
-------------------
web sites
Part I States of Matter gas-liquid-solid
http://www.docbrown.info/page03/3_52states.htm
------------------------------
JEE Question 2007 paper II
Matrix matching
Match the chemical system/units cells mentioned in column I with their characteristic features mentioned in column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I------------------------- Column II
(A)simple cubic and ---------(p)have these cell
face-centred cubic------------parameters a = b = c and
------------------------------ α =β = γ
(B)cubic and rhombohedral-- (q) are two crystal systems
(C) cubic and tetragonal (r) Have only two
-----------------------------crystallographic angles of
------------------------------90°
(D) hexagonal and monoaclinic-- (s) Belong to same
------------------------------------crystal system
Solution
A – p, s
B – p, q
C – q
D – r, q
----------------------------------
Study Guide Ch.3. ATOMIC STRUCTURE
JEE Syllabus
Atomic structure and chemical bonding:
Bohr model, spectrum of hydrogen atom, quantum numbers;
Wave-particle duality, de Broglie hypothesis;
Uncertainty principle;
Quantum mechanical picture of hydrogen atom (qualitative treatment), shapes of s, p and d orbitals;
Electronic configurations of elements (up to atomic number 36);
Aufbau principle;
Pauli's exclusion principle and Hund's rule;
------------
Main topics in TMH Book Chapter
CHARACTERISTICS OF ATOMS
RUTHERFORD'S SCATTERING EXPERIMENT
SPRECTRUM OF HYDROGEN ATOM
BOHR'S ATOMIC MODEL
QUANTUM NUMBERS
ELECTRONIC CONFIGURATION OF ELEMENTS
------------------------------
In 1913, Niels Bohr proposed a model of the atom. He proposed that the electrons in an atom could only be in certain orbits, or energy levels, around the nucleus. Refinement of Bohr theory led to the modern theory of atomic structure based on quantum mechanics. The energy levels proposed by Bohr were calculated and it was further modeled that each energy level or orbit has sublevels and each sublevels has one or more orbitals.
Orbit - sublevel – Orbital
Electrons are moving around the nucleus and orbitals represent a space where a pair of electrons is most likely to be found. This means that electrons may sometimes go out of the orbital but most of the time they are found in the orbital.
The quantum mechanics calculations gave out the result that there is a limit to the number of electrons that can occupy a given orbit or energy level.
Orbits
The orbits are called as shells. The energy level of orbits or shells increases as they increase in distance from the nucleus of the atom. The orbits or shells are represented by numbers as 1,2,3,4,5,6 or 7. They are represented by letters as K,L,M,N,O,P,Q.
It is found that the maximum number of electrons in each energy level is equal to 2n2 where n is the number of energy level.
Therefore energy level 1 will have 2 electrons.
Energy level 2 will have 2*4 = 8 electrons
Energy level 3 will have 2*9 = 18 electrons.
Sublevel of an Orbit
The energy levels, or orbits or shells are further divided into sublevels, or subshells. These subshells are designated by letters: s for the first possible sublevel, p for the second possible sublevel, d for the third, f for the fourth, g for the fifth, and from here on they simply go in alphabets.
The number of sublevels of each energy level is equal to the number of the energy levels. This means energy level 1, the K shell will have only one sub levels – s sublevel. The energy level 2, the L shell will have 2 sub levels – s and p.
Orbitals
Sublevels have further divisions called orbitals. Electrons are found in these orbitals. Each orbital contains two electrons.
“s” sublevel has only one orbital. “p” sublevel has 3 orbitals. “d” sublevel has 5 orbitals. “f” sublevel has 7 orbitals.
As each orbital can hold two electrons, orbitals of s can hold two electrons. The orbitals are of p sublevel are named as px and py and pz. The orbitals of p contain 6 electrons. The orbitals of d are 5. The orbitals of d are named as dxy, dxz,dyz,dx2-y2 and dz2. The d sublevel orbitals contain 10 electrons.
The two electons in each orbital spin in different directions.
Shape of Orbitals
Each type of orbital( s, px and py and pz, dxy, dxz,dyz,dx2-y2 and dz2 ) has a unique shape.
1. Spherical shape for s.
2. Dumbbell shape for orbitals of p.
3. Four-lobed shape for orbitals of d.
4. Complex shape for all orbitals of higher sublevels.
Electron Structure or Configuration
In the orbital-sub level-orbit structure, energy level of orbit 1 is less than that of 2 and so on. In each orbit, the sublevel s if of lower energy than the p sublevel, and p is of lower energy than the d sublevel and so on. Orbital of a sublevel are all of equal energy.
Electrons occupy the lowest energy sublevels that are available. This is known as ‘aufbau’ order or principles. In the case of an atom having atomic number of 1, the lone electron occupied the s orbital of sublevel s of orbit 1(represented as 1s1). In case of an atom having atomic number 3 the electrons first occupy the sublevel of orbit 1(this can hold only two electrons) and then occupy p sublevels of orbit 2 (represented as 1s2,2s1).
Hund’s rule says that, for any set of orbitals of equal energy say p orbitals of orbit 2, there is one electron is each orbital before the second electron enters or occupies an orbital.
The energy level of some sublevels at higher orbits is less than the some sublevels at lower orbitals. This fact is to be kept in mind when electron configuration is determined for any atom. The increasing order of energy levels of sublevels is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f*, 5d, 6p, 7s, 5f*, 6d, 7p, 8s
* In this case one electron one electron goes into 5d and then 4f fills completely and then rest of 5d. Similar thing happens in 5f and 4d.
-------------
Atomic particles are said to have spin. What does that imply?
Solution:
A good explanation for electron spin is provided by
Britannica:
Spin is an intrinsic property of an electron, like its mass or charge. In elementary treatments, spin is often visualized as an actual spinning motion. However, it is a quantum mechanical property without a classical counterpart, and to picture spin in this way can be misleading. Nevertheless, for the present discussion, such a picture is useful. An electron has a fixed amount of spin, in the sense that every electron in the universe is continually spinning at exactly the same rate. Although the spin of an electron is constant, the orientation of the axis of spin is variable, but quantum mechanics restricts that orientation to only two possibilities. The two possible spin states of an electron are represented by the arrows and and are distinguished by the spin magnetic quantum number, ms, which takes the values +1/2 (for the spin) or -1/2 (for the spin). Because of its spin, an electron must obey a fundamental requirement known as the Pauli exclusion principle. This principle (which is a consequence of the more fundamental Pauli principle; see the article atom: Electrons) states that no more than two electrons may occupy a given orbital and, if two electrons do occupy one orbital, their spins must be paired (denoted ; that is, one electron must be and the other must be ). The Pauli exclusion principle is responsible for the importance of the electron pair in the formation of covalent bonds. It is also, on a more cosmic scale, the reason why matter has bulk; that is to say, all electrons cannot occupy the orbitals of lowest energy but are instead located in the many shells that are centred on the nucleus. Also owing to the existence of spin, two objects do not simply blend into one another when they are in contact; the electrons of adjacent atoms cannot occupy the same space, thereby prohibiting the combining of two atoms into one. Here again is an example of a seemingly trivial property, in this case spin, having consequences of profound and macroscopic importance. In this instance, the spin of the electron is responsible for the existence of identifiable forms of matter.
Source: http://www.emsb.qc.ca/laurenhill/science/99scsoln.html#dec
-----------------------------
web sites
Atomic Structure, Isotopes, Periodic Table and Electronic Structure of Atoms
http://www.docbrown.info/page04/4_71atom.htm
Atomic structure and chemical bonding:
Bohr model, spectrum of hydrogen atom, quantum numbers;
Wave-particle duality, de Broglie hypothesis;
Uncertainty principle;
Quantum mechanical picture of hydrogen atom (qualitative treatment), shapes of s, p and d orbitals;
Electronic configurations of elements (up to atomic number 36);
Aufbau principle;
Pauli's exclusion principle and Hund's rule;
------------
Main topics in TMH Book Chapter
CHARACTERISTICS OF ATOMS
RUTHERFORD'S SCATTERING EXPERIMENT
SPRECTRUM OF HYDROGEN ATOM
BOHR'S ATOMIC MODEL
QUANTUM NUMBERS
ELECTRONIC CONFIGURATION OF ELEMENTS
------------------------------
In 1913, Niels Bohr proposed a model of the atom. He proposed that the electrons in an atom could only be in certain orbits, or energy levels, around the nucleus. Refinement of Bohr theory led to the modern theory of atomic structure based on quantum mechanics. The energy levels proposed by Bohr were calculated and it was further modeled that each energy level or orbit has sublevels and each sublevels has one or more orbitals.
Orbit - sublevel – Orbital
Electrons are moving around the nucleus and orbitals represent a space where a pair of electrons is most likely to be found. This means that electrons may sometimes go out of the orbital but most of the time they are found in the orbital.
The quantum mechanics calculations gave out the result that there is a limit to the number of electrons that can occupy a given orbit or energy level.
Orbits
The orbits are called as shells. The energy level of orbits or shells increases as they increase in distance from the nucleus of the atom. The orbits or shells are represented by numbers as 1,2,3,4,5,6 or 7. They are represented by letters as K,L,M,N,O,P,Q.
It is found that the maximum number of electrons in each energy level is equal to 2n2 where n is the number of energy level.
Therefore energy level 1 will have 2 electrons.
Energy level 2 will have 2*4 = 8 electrons
Energy level 3 will have 2*9 = 18 electrons.
Sublevel of an Orbit
The energy levels, or orbits or shells are further divided into sublevels, or subshells. These subshells are designated by letters: s for the first possible sublevel, p for the second possible sublevel, d for the third, f for the fourth, g for the fifth, and from here on they simply go in alphabets.
The number of sublevels of each energy level is equal to the number of the energy levels. This means energy level 1, the K shell will have only one sub levels – s sublevel. The energy level 2, the L shell will have 2 sub levels – s and p.
Orbitals
Sublevels have further divisions called orbitals. Electrons are found in these orbitals. Each orbital contains two electrons.
“s” sublevel has only one orbital. “p” sublevel has 3 orbitals. “d” sublevel has 5 orbitals. “f” sublevel has 7 orbitals.
As each orbital can hold two electrons, orbitals of s can hold two electrons. The orbitals are of p sublevel are named as px and py and pz. The orbitals of p contain 6 electrons. The orbitals of d are 5. The orbitals of d are named as dxy, dxz,dyz,dx2-y2 and dz2. The d sublevel orbitals contain 10 electrons.
The two electons in each orbital spin in different directions.
Shape of Orbitals
Each type of orbital( s, px and py and pz, dxy, dxz,dyz,dx2-y2 and dz2 ) has a unique shape.
1. Spherical shape for s.
2. Dumbbell shape for orbitals of p.
3. Four-lobed shape for orbitals of d.
4. Complex shape for all orbitals of higher sublevels.
Electron Structure or Configuration
In the orbital-sub level-orbit structure, energy level of orbit 1 is less than that of 2 and so on. In each orbit, the sublevel s if of lower energy than the p sublevel, and p is of lower energy than the d sublevel and so on. Orbital of a sublevel are all of equal energy.
Electrons occupy the lowest energy sublevels that are available. This is known as ‘aufbau’ order or principles. In the case of an atom having atomic number of 1, the lone electron occupied the s orbital of sublevel s of orbit 1(represented as 1s1). In case of an atom having atomic number 3 the electrons first occupy the sublevel of orbit 1(this can hold only two electrons) and then occupy p sublevels of orbit 2 (represented as 1s2,2s1).
Hund’s rule says that, for any set of orbitals of equal energy say p orbitals of orbit 2, there is one electron is each orbital before the second electron enters or occupies an orbital.
The energy level of some sublevels at higher orbits is less than the some sublevels at lower orbitals. This fact is to be kept in mind when electron configuration is determined for any atom. The increasing order of energy levels of sublevels is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f*, 5d, 6p, 7s, 5f*, 6d, 7p, 8s
* In this case one electron one electron goes into 5d and then 4f fills completely and then rest of 5d. Similar thing happens in 5f and 4d.
-------------
Atomic particles are said to have spin. What does that imply?
Solution:
A good explanation for electron spin is provided by
Britannica:
Spin is an intrinsic property of an electron, like its mass or charge. In elementary treatments, spin is often visualized as an actual spinning motion. However, it is a quantum mechanical property without a classical counterpart, and to picture spin in this way can be misleading. Nevertheless, for the present discussion, such a picture is useful. An electron has a fixed amount of spin, in the sense that every electron in the universe is continually spinning at exactly the same rate. Although the spin of an electron is constant, the orientation of the axis of spin is variable, but quantum mechanics restricts that orientation to only two possibilities. The two possible spin states of an electron are represented by the arrows and and are distinguished by the spin magnetic quantum number, ms, which takes the values +1/2 (for the spin) or -1/2 (for the spin). Because of its spin, an electron must obey a fundamental requirement known as the Pauli exclusion principle. This principle (which is a consequence of the more fundamental Pauli principle; see the article atom: Electrons) states that no more than two electrons may occupy a given orbital and, if two electrons do occupy one orbital, their spins must be paired (denoted ; that is, one electron must be and the other must be ). The Pauli exclusion principle is responsible for the importance of the electron pair in the formation of covalent bonds. It is also, on a more cosmic scale, the reason why matter has bulk; that is to say, all electrons cannot occupy the orbitals of lowest energy but are instead located in the many shells that are centred on the nucleus. Also owing to the existence of spin, two objects do not simply blend into one another when they are in contact; the electrons of adjacent atoms cannot occupy the same space, thereby prohibiting the combining of two atoms into one. Here again is an example of a seemingly trivial property, in this case spin, having consequences of profound and macroscopic importance. In this instance, the spin of the electron is responsible for the existence of identifiable forms of matter.
Source: http://www.emsb.qc.ca/laurenhill/science/99scsoln.html#dec
-----------------------------
web sites
Atomic Structure, Isotopes, Periodic Table and Electronic Structure of Atoms
http://www.docbrown.info/page04/4_71atom.htm
Study Guide Ch. 4. PERIODICITY OF PROPERTIES OF ELEMENTS
JEE Syllabus
No specific syllabus on this topic
-----------------
Main Topics in TMH Book Chapter
CLASSIFICATION OF ELEMENTS
MAIN CHARACTERISTICS OF REPRESENTATIVE ELEMENTS
MAIN CHARACTERISTICS OF TRANSIENT ELEMENTS
PERIODICITY IN PROPERTIES
METALLIC AND NONMETALLIC CHARACTER
LANTHANIDE CONTRACTION
-----------------
Periodic table was covered in the X class syllabus. But there are some interesting points about the periodic table that were covered in Redmund’s book in 5th chapter.
We all know that the horizontal rows in the periodic table are called periods, and the vertical columns are called groups. The groups are subdivided into A and B subgroups. The A subgroups, due to their similarities within a group, are often called families. Some of these families are referred to by special names, such as the alkali metals for Group IA, alkaline earth metals for Group IIA, and halogens for Group VIIA. The other A subgroups are sometimes classified according to the first member of the subgroup or family. Thus the IIIA elements are sometimes referred to as the boron family, the IVA elements as the carbon family, the VA elements as the nitrogen family, and the VIA elements as the oxygen family. The last group, called Group zero (or sometimes Group VIIIA), is now called the noble gases.
As well as being classified as metals or nonmetals, the elements are also divided into the following four types: representative elements, noble gases, transition elements and inner transition elements.
-----------------------------
web sites
Introduction to the Periodic Table
http://www.docbrown.info/page03/3_34ptable.htm
No specific syllabus on this topic
-----------------
Main Topics in TMH Book Chapter
CLASSIFICATION OF ELEMENTS
MAIN CHARACTERISTICS OF REPRESENTATIVE ELEMENTS
MAIN CHARACTERISTICS OF TRANSIENT ELEMENTS
PERIODICITY IN PROPERTIES
METALLIC AND NONMETALLIC CHARACTER
LANTHANIDE CONTRACTION
-----------------
Periodic table was covered in the X class syllabus. But there are some interesting points about the periodic table that were covered in Redmund’s book in 5th chapter.
We all know that the horizontal rows in the periodic table are called periods, and the vertical columns are called groups. The groups are subdivided into A and B subgroups. The A subgroups, due to their similarities within a group, are often called families. Some of these families are referred to by special names, such as the alkali metals for Group IA, alkaline earth metals for Group IIA, and halogens for Group VIIA. The other A subgroups are sometimes classified according to the first member of the subgroup or family. Thus the IIIA elements are sometimes referred to as the boron family, the IVA elements as the carbon family, the VA elements as the nitrogen family, and the VIA elements as the oxygen family. The last group, called Group zero (or sometimes Group VIIIA), is now called the noble gases.
As well as being classified as metals or nonmetals, the elements are also divided into the following four types: representative elements, noble gases, transition elements and inner transition elements.
-----------------------------
web sites
Introduction to the Periodic Table
http://www.docbrown.info/page03/3_34ptable.htm
Study Guide Ch.5. BONDING AND MOLECULAR STRUCTURE
See for past JEE questions from this chapter
http://iit-jee-chemistry-ps.blogspot.com/2007/12/past-jee-questions-ch5.html
See for some application questions
http://iit-jee-chemistry-ps.blogspot.com/2007/12/application-questions-ch-5-bonding.html
JEE Syllabus
Orbital overlap and covalent bond;
Hybridisation involving s, p and d orbitals only;
Orbital energy diagrams for homonuclear diatomic species;
Hydrogen bond;
Polarity in molecules, dipole moment (qualitative aspects only);
VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral).
------------------
Main Topics in TMH JEE Book
COVALENT BOND
COORDINATE BOND
IONIC BOND
QUANTUM MECHANICAL EXPLANATION OF COVALENT BOND
HYBRIDIZATION
VALENCE SHELL ELECTRON-PAIR-REPULSION MODEL
HYDROGEN BOND
RESONANCE
MOLECULAR ORBITAL METHOD
--------------------
JEE syllabus and sections of Jauhar's Book (Class XI)
Orbital overlap and covalent bond; 6.5, 6.6, 6.7, 6.15, 6.18
Hybridisation involving s, p and d orbitals only; 6.18 (d orbitals not described)
Orbital energy diagrams for homonuclear diatomic species;
Hydrogen bond; 6.20
Polarity in molecules, dipole moment (qualitative aspects only); 6.14
VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral). 6.13 and 6.13
Section of Jauhar Chapter 6
1. Cause of chemical combination
2. Lewis symbols
3. Octetl rule andmodes of chemical combination
4. Ionic or electrovalent bond
5. Covalent bond
General properties of covalent bonds
7. Co-ordinate covalent bond
8. Formation of ionic bond
9. Lattice enthalpy of ionic crystals
10. Born-haber Cycle for lattice enthalpies
11. General properties of ionic compounds
12. Geometry of shapes of molecules
13. Valence Shell Electron Pair Repulsion (VSEPR) theory
14. electronegativity - polar and nonpolar character of covalent bonds
15. Valency bond approach of covalent bond
---Orbital overlap concept of covalent bond
16. Bonding parameters
17. Resonance
18. Directional properties of covalent bonds
19. Metallic bonding
20. Hydrogen bonding
Ionic Bond
In an ionic bond negative ions are surrounded by positive ions and positive ions are surrounded by negative ions. NaCl does not mean each Na and Cl ions are bound to each other. It only means the proportion of the ions is 1:1.
Hydrogen Bonding
The attractive force which binds hydrogen atom of one molecule with electronegative atom (F,O or N) of another molecule is known as hydrogen bond or hydrogen bonding.
Hydrogen bonds are formed when hydrogen is bonded to strongly electronegative elements uch as F, O or N.
The size of electronegative atom should be small for formation of Hydrogen bonds. Hydrogen bonds are not formed by Cl because of its bigger size.
Example of Hydrogen bonds
HF
H2O
NH3
Influence of Hydrogen bonding on properties
.1 Association
2. Higher melting and boiling points
3. Influence on the physical state
4. Solubility
----------------------
Examples in the Chapter
Double bond
O2 molecule
CO2 molecule C forms double bond with each O atom.
CS2
Triple bond
N2 molecule
CO molecule
Exceptions to octet rule
Hydrogen molecule only 2 electrons make it stable.
Incomplete octet of central atom
LiCl
BeH2
BeCl2
BH3
BF3
LiCl 4 electrons around central Li-atom
BeCl2 4 electrons around central Be-atom
BF3 6 electrons around central B-atom
Expanded octet of the central atom
PF5 has ten around P
SF6 hs twelve around S
IF7 has fourteen electrons around I
H2SO4 12 electrons around sulphur atoms
Odd elctron molecules
Nitric oxide, NO
Nitrogen 7 shared electrons
Oxygen 8
Nitrogen dioxide, NO2(there is a coordinate bond)
Nitrogen 7
both oxygens 8
compounds of noble gases
XeF2, KrF2, XeOF2, XeOF4, XeF6
Coordinate covalent bonds
ozone
Hydronium ion
ammonia
sulphuric acid
Molecule shapes
Linear
BeF2, BeCl2, ZnCl2, HgCl2
Trigonal planar
BF3, BCl3, AlCl3
Tetrahedral
CH4, SiF4, CCl4, SiH4, NH4^+
Trigonal bipyramidal
PCl5, PF5
Octahedral
SF6
Pentagonal bipyramidal
IF7
Molecules containing lone pairs and their shapes
Three electron pairs that include lone pairs
standard shape Trigonal planar
SO2 angular or V shaped or bent shape
Four electron pairs that include lone pairs
standard shape: tetrahedral geometry
Ammonia molecule pyramidal
PCl3, NF3, H3O^+
H2O molecule
two bond pairs and two lone pairs bent or angular
same shape H2S, F2O, SCl2
Five electron pairs that include lone pairs
standard shape: trigonal bipyramidal
SF4 4 bond pairs and 1 lp distorted tetrahedron or a folded square
Chlorine triflouride 3 bp and 2 lp T shaped
Xennon difluoride XeF2 2 bp and 3 lp linear geometry
six electron pairs that include lone pairs
BrF5 5bp and 1 lp square pyramidal
XeF4 4 bp and 2 lp square planar
Polar covalent molecules
HCl, BrCl, H2O, HF
Dipole moment
HCL 1.03 D
CO2 zero
H2O 1.84 D
NH3 1.49D
BF3 zero
CCl4 zero
NF3 0.24D
Molecules described by resonance property
O3 ozone
CO2
CO
SO2
SO3
Benzene
CO3^2-
NO2 ion
Hybridization
sp
BeCl2
BeF2
BeH2
C2H2
sp^2
BCl3
C2H4
sp^3
CH4
NH3
H2O
Hydrogen bonds
HF
H2O
Ammonia NH3
-------------
VSEPR Model
I found the tutorial in the site given below very useful understand this topic.
Molecular geometry: VSEPR http://winter.group.shef.ac.uk/vsepr/intro.html
----------------------------
Molecular orbital theory
I found material from the site below to be useful to understand this topic.
Molecular Orbital Theory http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html
Diamagnetic and Paramagnetic substances
Atoms or molecules in which the electrons are paired are diamagnetic repelled by both poles of a magnetic. Those that have one or more unpaired electrons are paramagnetic attracted to a magnetic field. Liquid oxygen is attracted to a magnetic field and can actually bridge the gap between the poles of a horseshoe magnet. The molecular orbital model of O2 is therefore superior to the valence-bond model, which cannot explain this property of oxygen.
----------------------
web sites
The Structure, Bonding and properties of substances
http://www.docbrown.info/page04/4_72bond.htm
JEE question on dipole moment
Among the following, the molecule with the highest dipole moment is:
(A) CH-3Cl (B) CH-2Cl-2
(C) CHCl-3 (D) CCl-4
answer A
----------------------------
JEE Question 2007 paper I
The percentage of p-character in the orbitals forming P–P bonds in 4 P is
(A) 25
(B) 33
(C) 50
(D) 75
Solution: (D)
Phosphorous will show sp^3 hybridisation having 75% p-character.
-----------------------
JEE Question 2007 paper I
Statement - 1
Boron always forms covalent bond
Because
Statement - 2
The small size of B3^+ favours formation of covalent bond.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Answer: A
-------------------------------
JEE Question paper II 2007
Among the following metal carbonyls, the C — O bond order is lowest in
(A) [Mn(CO)-6]^+
(B) [Fe(CO)-5]
(C) [Cr(CO)-6]
(D) [V(CO)-6]^-
answer: B
-----------------
JEE 2006
If the bond length of CO bond in carbon monoxide is 1.128 A,
then what is the value of CO bond length in Fe(CO)-5?
(A) 1.15 A
(B) 1.128 A
(C) 1.72 A
(D) 1.118 A
Answer: (A)
----------------------
http://iit-jee-chemistry-ps.blogspot.com/2007/12/past-jee-questions-ch5.html
See for some application questions
http://iit-jee-chemistry-ps.blogspot.com/2007/12/application-questions-ch-5-bonding.html
JEE Syllabus
Orbital overlap and covalent bond;
Hybridisation involving s, p and d orbitals only;
Orbital energy diagrams for homonuclear diatomic species;
Hydrogen bond;
Polarity in molecules, dipole moment (qualitative aspects only);
VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral).
------------------
Main Topics in TMH JEE Book
COVALENT BOND
COORDINATE BOND
IONIC BOND
QUANTUM MECHANICAL EXPLANATION OF COVALENT BOND
HYBRIDIZATION
VALENCE SHELL ELECTRON-PAIR-REPULSION MODEL
HYDROGEN BOND
RESONANCE
MOLECULAR ORBITAL METHOD
--------------------
JEE syllabus and sections of Jauhar's Book (Class XI)
Orbital overlap and covalent bond; 6.5, 6.6, 6.7, 6.15, 6.18
Hybridisation involving s, p and d orbitals only; 6.18 (d orbitals not described)
Orbital energy diagrams for homonuclear diatomic species;
Hydrogen bond; 6.20
Polarity in molecules, dipole moment (qualitative aspects only); 6.14
VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral). 6.13 and 6.13
Section of Jauhar Chapter 6
1. Cause of chemical combination
2. Lewis symbols
3. Octetl rule andmodes of chemical combination
4. Ionic or electrovalent bond
5. Covalent bond
General properties of covalent bonds
7. Co-ordinate covalent bond
8. Formation of ionic bond
9. Lattice enthalpy of ionic crystals
10. Born-haber Cycle for lattice enthalpies
11. General properties of ionic compounds
12. Geometry of shapes of molecules
13. Valence Shell Electron Pair Repulsion (VSEPR) theory
14. electronegativity - polar and nonpolar character of covalent bonds
15. Valency bond approach of covalent bond
---Orbital overlap concept of covalent bond
16. Bonding parameters
17. Resonance
18. Directional properties of covalent bonds
19. Metallic bonding
20. Hydrogen bonding
Ionic Bond
In an ionic bond negative ions are surrounded by positive ions and positive ions are surrounded by negative ions. NaCl does not mean each Na and Cl ions are bound to each other. It only means the proportion of the ions is 1:1.
Hydrogen Bonding
The attractive force which binds hydrogen atom of one molecule with electronegative atom (F,O or N) of another molecule is known as hydrogen bond or hydrogen bonding.
Hydrogen bonds are formed when hydrogen is bonded to strongly electronegative elements uch as F, O or N.
The size of electronegative atom should be small for formation of Hydrogen bonds. Hydrogen bonds are not formed by Cl because of its bigger size.
Example of Hydrogen bonds
HF
H2O
NH3
Influence of Hydrogen bonding on properties
.1 Association
2. Higher melting and boiling points
3. Influence on the physical state
4. Solubility
----------------------
Examples in the Chapter
Double bond
O2 molecule
CO2 molecule C forms double bond with each O atom.
CS2
Triple bond
N2 molecule
CO molecule
Exceptions to octet rule
Hydrogen molecule only 2 electrons make it stable.
Incomplete octet of central atom
LiCl
BeH2
BeCl2
BH3
BF3
LiCl 4 electrons around central Li-atom
BeCl2 4 electrons around central Be-atom
BF3 6 electrons around central B-atom
Expanded octet of the central atom
PF5 has ten around P
SF6 hs twelve around S
IF7 has fourteen electrons around I
H2SO4 12 electrons around sulphur atoms
Odd elctron molecules
Nitric oxide, NO
Nitrogen 7 shared electrons
Oxygen 8
Nitrogen dioxide, NO2(there is a coordinate bond)
Nitrogen 7
both oxygens 8
compounds of noble gases
XeF2, KrF2, XeOF2, XeOF4, XeF6
Coordinate covalent bonds
ozone
Hydronium ion
ammonia
sulphuric acid
Molecule shapes
Linear
BeF2, BeCl2, ZnCl2, HgCl2
Trigonal planar
BF3, BCl3, AlCl3
Tetrahedral
CH4, SiF4, CCl4, SiH4, NH4^+
Trigonal bipyramidal
PCl5, PF5
Octahedral
SF6
Pentagonal bipyramidal
IF7
Molecules containing lone pairs and their shapes
Three electron pairs that include lone pairs
standard shape Trigonal planar
SO2 angular or V shaped or bent shape
Four electron pairs that include lone pairs
standard shape: tetrahedral geometry
Ammonia molecule pyramidal
PCl3, NF3, H3O^+
H2O molecule
two bond pairs and two lone pairs bent or angular
same shape H2S, F2O, SCl2
Five electron pairs that include lone pairs
standard shape: trigonal bipyramidal
SF4 4 bond pairs and 1 lp distorted tetrahedron or a folded square
Chlorine triflouride 3 bp and 2 lp T shaped
Xennon difluoride XeF2 2 bp and 3 lp linear geometry
six electron pairs that include lone pairs
BrF5 5bp and 1 lp square pyramidal
XeF4 4 bp and 2 lp square planar
Polar covalent molecules
HCl, BrCl, H2O, HF
Dipole moment
HCL 1.03 D
CO2 zero
H2O 1.84 D
NH3 1.49D
BF3 zero
CCl4 zero
NF3 0.24D
Molecules described by resonance property
O3 ozone
CO2
CO
SO2
SO3
Benzene
CO3^2-
NO2 ion
Hybridization
sp
BeCl2
BeF2
BeH2
C2H2
sp^2
BCl3
C2H4
sp^3
CH4
NH3
H2O
Hydrogen bonds
HF
H2O
Ammonia NH3
-------------
VSEPR Model
I found the tutorial in the site given below very useful understand this topic.
Molecular geometry: VSEPR http://winter.group.shef.ac.uk/vsepr/intro.html
----------------------------
Molecular orbital theory
I found material from the site below to be useful to understand this topic.
Molecular Orbital Theory http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html
Diamagnetic and Paramagnetic substances
Atoms or molecules in which the electrons are paired are diamagnetic repelled by both poles of a magnetic. Those that have one or more unpaired electrons are paramagnetic attracted to a magnetic field. Liquid oxygen is attracted to a magnetic field and can actually bridge the gap between the poles of a horseshoe magnet. The molecular orbital model of O2 is therefore superior to the valence-bond model, which cannot explain this property of oxygen.
----------------------
web sites
The Structure, Bonding and properties of substances
http://www.docbrown.info/page04/4_72bond.htm
JEE question on dipole moment
Among the following, the molecule with the highest dipole moment is:
(A) CH-3Cl (B) CH-2Cl-2
(C) CHCl-3 (D) CCl-4
answer A
----------------------------
JEE Question 2007 paper I
The percentage of p-character in the orbitals forming P–P bonds in 4 P is
(A) 25
(B) 33
(C) 50
(D) 75
Solution: (D)
Phosphorous will show sp^3 hybridisation having 75% p-character.
-----------------------
JEE Question 2007 paper I
Statement - 1
Boron always forms covalent bond
Because
Statement - 2
The small size of B3^+ favours formation of covalent bond.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Answer: A
-------------------------------
JEE Question paper II 2007
Among the following metal carbonyls, the C — O bond order is lowest in
(A) [Mn(CO)-6]^+
(B) [Fe(CO)-5]
(C) [Cr(CO)-6]
(D) [V(CO)-6]^-
answer: B
-----------------
JEE 2006
If the bond length of CO bond in carbon monoxide is 1.128 A,
then what is the value of CO bond length in Fe(CO)-5?
(A) 1.15 A
(B) 1.128 A
(C) 1.72 A
(D) 1.118 A
Answer: (A)
----------------------
Study Guide Ch. 6 ENERGETICS
JEE Syllabus
Energetics:
First law of thermodynamics;
Internal energy, work and heat,
pressure-volume work;
Enthalpy,
Hess's law;
Heat of reaction, fusion and vapourization;
Second law of thermodynamics;
Entropy;
Free energy;
Criterion of spontaneity.
------------------
Main Topics in TMH Book Chapter
FIRST LAW THERMODYNAMICS
INTERNAL ENERGY AND ENTHALPY
ENTHALPY CHANGE OF A CHEMICAL EQUATION
MOLAR ENHTALPIES OF FORMATIONS
HESS'S LAW OF CONSTANT HEAT SUMMATION
TYPES OF REACTIONS
RELATION BETWEEN DELTA H AND DELTA U OF A CHEMICAL EQUATION
--------------------------
Enthalpy Change, ∆H, With Temperature and State Change
Enthalpy, represented by the symbol H, is essentially a chemistry term for heat, and a term for total kinetic energy of particle motion in a sample. If the reaction is exothermic, the energy contained in the substances is reduced so ∆H has a negative value. On the other hand, if the reaction is endothermic, the substances absorb energy and ∆H is positive.
Temperature is a measure of the average kinetic energy of the particles in a sample.
The faster the molecules in a sample of water the higher the temperature.
Changing temperature of a sample requires a change in enthalpy
It makes sense that you have to add heat to increase temperature, and as a sample cools it gives off heat. The amount of heat required to change the temperature of a sample depends on two factors:
# Mass. The bigger the sample the more heat needed to change its temperature.
# Specific heat capacity of the substance it is made of. This is the energy needed to change the temperature of a 1 gram sample 1 Celsius and it is different for different materials. Its symbol is Cp.
When you go to the beach on a hot summer day, you may have noticed that dry sand can get really hot, while wet sand stays cooler, though the air over them is obviously the same temperature and they are getting the same sunlight. How about dry pavement versus wet pavement? Metal versus tile?
The difference is the specific heat capacity of the different materials. Substances with low specific heat capacities will get very hot very fast, while those with large specific heat capacities warm up much slower since more energy is required to change their temperature. In fact, liquid water has about the highest specific heat of any common substance, so it tends to "heat up" and 'cool down' very slowly compared to most other substances. Ice and steam have lower heat capacities than liquid water.
Some people who prize their outdoor plants set gallon jugs of water around their plants. How does this help the plants? During the heat of the day, that water absorbs energy from the air around it, leaving the air cooler. During the cold night, that water releases that energy as it slowly cools, warming the air around it. This is known as a heat sink.
To find the energy required to change the temperature of a sample use Changing temperature of a sample requires a change in enthalpy
It makes sense that you have to add heat to increase temperature, and as a sample cools it gives off heat. The amount of heat required to change the temperature of a sample depends on two factors:
? Mass. The bigger the sample the more heat needed to change its temperature.
? Specific heat capacity of the substance it is made of. This is the energy needed to change the temperature of a 1 gram sample 1? Celsius and it is different for different materials. For some reason its symbol is Cp. Sorry.
When you go to the beach on a hot summer day, you may have noticed that dry sand can get really hot, while wet sand stays cooler, though the air over them is obviously the same temperature and they are getting the same sunlight. How about dry pavement versus wet pavement? Metal versus tile?
The difference is the specific heat capacity of the different materials. Substances with low specific heat capacities will get very hot very fast, while those with large specific heat capacities warm up much slower since more energy is required to change their temperature. In fact, liquid water has about the highest specific heat of any common substance, so it tends to "heat up" and 'cool down' very slowly compared to most other substances. Ice and steam have lower heat capacities than liquid water.
Some people who prize their outdoor plants set gallon jugs of water around their plants. How does this help the plants? During the heat of the day, that water absorbs energy from the air around it, leaving the air cooler. During the cold night, that water releases that energy as it slowly cools, warming the air around it. This is known as a heat sink.
In Chicago, which sits on the shores of Lake Michigan, the weather near the lake is warmer in winter and cooler in summer than the weather farther inland. This is known as the lake effect. The lake functions as a giant heat sink.
To find the energy required to change the temperature of a sample use ∆H = (m)(∆T)(Cp)
Stoichiometry with Energy
Enthalpy values can be included as part of balanced chemical equations. In exothermic reactions, ∆H is negative, but energy is listed as a positive value on the product side of the equation. In endothermic reactions, ∆H is positive and energy is required and is listed on the reactant side of the equation.
The number part of the energy value (as opposed to the unit) can act as a coefficient of a mole ratio, just as any other coefficient would. This way you can convert energy information to information about any substance in a balanced equation and vice versa.
2H-2 + O-2 → 2 H-2O + 561.6 kJ
Enthalpy changes of different types have different names.
The enthalpy change when something dissolves is heat of solution (∆H sol) The enthalpy change during a chemical reaction is heat of reaction (∆Hrxn). The enthalpy change during a comustion reaction is heat of combustion (∆Hcomb). The enthalpy change during a reaction in which a compound is formed from its elements is heat of formation (∆Hf).
Heat of Formation is an important concept.
Heat of Formation Problems.
The enthalpy change can be calculated by taking the total enthalpy of the products - the total enthalpy of the reactants. The formula is...
DHrxn = (the sum of ∆Hf products ) - (the sum of the ∆Hf reactants )
To do this, multiply the number of moles, or coefficient from the balanced equation, of each substance by its heat of formation), and add them up for the products, then do the same for the reactants. Then subtract. Heats of formation are given in kJ / mol, but ∆H is in kJ, since the moles cancel out.
Hess's Law
Hess's Law states that the enthalpy change for a reaction that occurs in many steps is the same as if it occurred in one step. Another way to put this is if several reactions add up to some total reaction, then their enthalpy changes will add up to the enthalpy change for the total reaction.
DHtotal = ∆Hrxn 1 + ∆Hrxn 2 + ∆Hrxn 3 + etc.
Hess's law problems usually give you two or three reactions with their enthalpy change information, then ask you to find the enthalpy change for some target reaction. You must figure out how to make the given reactions add up to the target. This can mean reversing the reactions (and reversing the sign on the enthalpy change), or using them multiple times, or both.
Entropy
Entropy (S) is a measure of the amount of disorder in a substance Gases with their rapid random motion are high in entropy, and solids with their ordered crystalline lattice are low in entropy.
The change in entropy (∆S) is determined just like a heat of formation problem, only use entropy values instead.
∆Srxn = (the sum of ∆Sproducts ) - (the sum of the ∆Sreactants )
Note that the units for entropy are given in J / mol K. ∆S is in J / K, since the moles cancel out.
K stands for Kelvins, the temperature unit on the absolute scale. Also called the Kelvin scale, it is named for Lord Kelvin, who developed it, so the units should be capitalized.
Entropy is temperature affected. It is large at high temperatures, and small at low temperatures. Enthalpy is not temperature affected.
How Enthalpy and Entropy Drive Change
If a reaction will occur spontaneously, it will occur so the products are said to be favored. If a reaction will not occur spontaneously, then the reverse reaction will(in a reversible reaction), so the reactants are said to be favored.
Exothermic changes, those with negative enthalpy changes, are favored to occur spontaneously. So if ∆H is negative, a reaction is more likely than if it is not.
On the other hand, changes that involve an increase in disorder, or entropy, are favored to occur spontaneously. So if ∆S is positive, a reaction is more likely than if it is not.
If ∆H is negative and ∆S is positive, a reaction will certainly occur, no matter what the temperature. If ∆H is positive and ∆S is negative, there will not be a reaction at any temperature, since both indicators say it wont.
The trouble is that often the two indicators disagree. When they do, the enthalpy tends to win out at low temperatures, and the entropy (since it is temperature affected) tends to win out at high temperatures.
For example, if the enthalpy and entropy change values are both negative, the enthalpy indicates the reaction will occur, and it will at low temperatures. The entropy indicates that there will be no reaction, and at high temperatures there wont. The reverse is true for positive enthalpy and entropy changes. These reactions are more likely to occur at high, but not at low temperatures.
Gibbs Free Energy (∆G) determines for sure whether a reaction will be favored to occur. It is simply a formula that compares ∆H to ∆S in a special way.
∆G = ∆H - T∆S
Temperature must be in Kelvins. If ∆G has a negative value, the reaction will occur spontaneously. If ∆G has a positive value, it will not occur
One complication in calculating the free energy change is that the enthalpy values are typically given in kilojoules (kJ), while entropy values are given in J / K, so you must convert so that both use Joules, or both use kilojoules. It doesn't matter which.
Sometimes you may be asked to find the temperature above which a reaction will or wont occur. This is the temperature at which ∆G is between negative and positive, or when it equals zero.
∆G = 0 so 0 = ∆H - T∆S
Rearrange the equation to solve for T, and you will find that...
T = ∆H/∆S
Above that temperature entropy change determines whether a reaction occurs, and below that temperature, enthalpy change determines whether a reaction occurs.
∆G = 0 means there will be equilibrium in a reversible reaction.
-----------------------------
JEE Question 2007 paper II
For the process (water becoming steam) H-2O(l)(1 bar, 373 K) --> H-2O(g)(1 bar, 373 K), the correct set of thermodynamic parameters is
(A) ∆G=0,∆S= + ve
(B)∆G=0,∆S= -ve
(C)∆G=+ve,∆S=0
(D)∆G=-ve, ∆S= + ve
Solution: A
The answer is A because, because at 100 degree C, the steam and water mixture is at equilibrium. Hence ΔG = 0(G = H - TS), and ΔS is positive. Why? when liquid becomes gas, there is more disorder. More entropy.
--------------
Energetics:
First law of thermodynamics;
Internal energy, work and heat,
pressure-volume work;
Enthalpy,
Hess's law;
Heat of reaction, fusion and vapourization;
Second law of thermodynamics;
Entropy;
Free energy;
Criterion of spontaneity.
------------------
Main Topics in TMH Book Chapter
FIRST LAW THERMODYNAMICS
INTERNAL ENERGY AND ENTHALPY
ENTHALPY CHANGE OF A CHEMICAL EQUATION
MOLAR ENHTALPIES OF FORMATIONS
HESS'S LAW OF CONSTANT HEAT SUMMATION
TYPES OF REACTIONS
RELATION BETWEEN DELTA H AND DELTA U OF A CHEMICAL EQUATION
--------------------------
Enthalpy Change, ∆H, With Temperature and State Change
Enthalpy, represented by the symbol H, is essentially a chemistry term for heat, and a term for total kinetic energy of particle motion in a sample. If the reaction is exothermic, the energy contained in the substances is reduced so ∆H has a negative value. On the other hand, if the reaction is endothermic, the substances absorb energy and ∆H is positive.
Temperature is a measure of the average kinetic energy of the particles in a sample.
The faster the molecules in a sample of water the higher the temperature.
Changing temperature of a sample requires a change in enthalpy
It makes sense that you have to add heat to increase temperature, and as a sample cools it gives off heat. The amount of heat required to change the temperature of a sample depends on two factors:
# Mass. The bigger the sample the more heat needed to change its temperature.
# Specific heat capacity of the substance it is made of. This is the energy needed to change the temperature of a 1 gram sample 1 Celsius and it is different for different materials. Its symbol is Cp.
When you go to the beach on a hot summer day, you may have noticed that dry sand can get really hot, while wet sand stays cooler, though the air over them is obviously the same temperature and they are getting the same sunlight. How about dry pavement versus wet pavement? Metal versus tile?
The difference is the specific heat capacity of the different materials. Substances with low specific heat capacities will get very hot very fast, while those with large specific heat capacities warm up much slower since more energy is required to change their temperature. In fact, liquid water has about the highest specific heat of any common substance, so it tends to "heat up" and 'cool down' very slowly compared to most other substances. Ice and steam have lower heat capacities than liquid water.
Some people who prize their outdoor plants set gallon jugs of water around their plants. How does this help the plants? During the heat of the day, that water absorbs energy from the air around it, leaving the air cooler. During the cold night, that water releases that energy as it slowly cools, warming the air around it. This is known as a heat sink.
To find the energy required to change the temperature of a sample use Changing temperature of a sample requires a change in enthalpy
It makes sense that you have to add heat to increase temperature, and as a sample cools it gives off heat. The amount of heat required to change the temperature of a sample depends on two factors:
? Mass. The bigger the sample the more heat needed to change its temperature.
? Specific heat capacity of the substance it is made of. This is the energy needed to change the temperature of a 1 gram sample 1? Celsius and it is different for different materials. For some reason its symbol is Cp. Sorry.
When you go to the beach on a hot summer day, you may have noticed that dry sand can get really hot, while wet sand stays cooler, though the air over them is obviously the same temperature and they are getting the same sunlight. How about dry pavement versus wet pavement? Metal versus tile?
The difference is the specific heat capacity of the different materials. Substances with low specific heat capacities will get very hot very fast, while those with large specific heat capacities warm up much slower since more energy is required to change their temperature. In fact, liquid water has about the highest specific heat of any common substance, so it tends to "heat up" and 'cool down' very slowly compared to most other substances. Ice and steam have lower heat capacities than liquid water.
Some people who prize their outdoor plants set gallon jugs of water around their plants. How does this help the plants? During the heat of the day, that water absorbs energy from the air around it, leaving the air cooler. During the cold night, that water releases that energy as it slowly cools, warming the air around it. This is known as a heat sink.
In Chicago, which sits on the shores of Lake Michigan, the weather near the lake is warmer in winter and cooler in summer than the weather farther inland. This is known as the lake effect. The lake functions as a giant heat sink.
To find the energy required to change the temperature of a sample use ∆H = (m)(∆T)(Cp)
Stoichiometry with Energy
Enthalpy values can be included as part of balanced chemical equations. In exothermic reactions, ∆H is negative, but energy is listed as a positive value on the product side of the equation. In endothermic reactions, ∆H is positive and energy is required and is listed on the reactant side of the equation.
The number part of the energy value (as opposed to the unit) can act as a coefficient of a mole ratio, just as any other coefficient would. This way you can convert energy information to information about any substance in a balanced equation and vice versa.
2H-2 + O-2 → 2 H-2O + 561.6 kJ
Enthalpy changes of different types have different names.
The enthalpy change when something dissolves is heat of solution (∆H sol) The enthalpy change during a chemical reaction is heat of reaction (∆Hrxn). The enthalpy change during a comustion reaction is heat of combustion (∆Hcomb). The enthalpy change during a reaction in which a compound is formed from its elements is heat of formation (∆Hf).
Heat of Formation is an important concept.
Heat of Formation Problems.
The enthalpy change can be calculated by taking the total enthalpy of the products - the total enthalpy of the reactants. The formula is...
DHrxn = (the sum of ∆Hf products ) - (the sum of the ∆Hf reactants )
To do this, multiply the number of moles, or coefficient from the balanced equation, of each substance by its heat of formation), and add them up for the products, then do the same for the reactants. Then subtract. Heats of formation are given in kJ / mol, but ∆H is in kJ, since the moles cancel out.
Hess's Law
Hess's Law states that the enthalpy change for a reaction that occurs in many steps is the same as if it occurred in one step. Another way to put this is if several reactions add up to some total reaction, then their enthalpy changes will add up to the enthalpy change for the total reaction.
DHtotal = ∆Hrxn 1 + ∆Hrxn 2 + ∆Hrxn 3 + etc.
Hess's law problems usually give you two or three reactions with their enthalpy change information, then ask you to find the enthalpy change for some target reaction. You must figure out how to make the given reactions add up to the target. This can mean reversing the reactions (and reversing the sign on the enthalpy change), or using them multiple times, or both.
Entropy
Entropy (S) is a measure of the amount of disorder in a substance Gases with their rapid random motion are high in entropy, and solids with their ordered crystalline lattice are low in entropy.
The change in entropy (∆S) is determined just like a heat of formation problem, only use entropy values instead.
∆Srxn = (the sum of ∆Sproducts ) - (the sum of the ∆Sreactants )
Note that the units for entropy are given in J / mol K. ∆S is in J / K, since the moles cancel out.
K stands for Kelvins, the temperature unit on the absolute scale. Also called the Kelvin scale, it is named for Lord Kelvin, who developed it, so the units should be capitalized.
Entropy is temperature affected. It is large at high temperatures, and small at low temperatures. Enthalpy is not temperature affected.
How Enthalpy and Entropy Drive Change
If a reaction will occur spontaneously, it will occur so the products are said to be favored. If a reaction will not occur spontaneously, then the reverse reaction will(in a reversible reaction), so the reactants are said to be favored.
Exothermic changes, those with negative enthalpy changes, are favored to occur spontaneously. So if ∆H is negative, a reaction is more likely than if it is not.
On the other hand, changes that involve an increase in disorder, or entropy, are favored to occur spontaneously. So if ∆S is positive, a reaction is more likely than if it is not.
If ∆H is negative and ∆S is positive, a reaction will certainly occur, no matter what the temperature. If ∆H is positive and ∆S is negative, there will not be a reaction at any temperature, since both indicators say it wont.
The trouble is that often the two indicators disagree. When they do, the enthalpy tends to win out at low temperatures, and the entropy (since it is temperature affected) tends to win out at high temperatures.
For example, if the enthalpy and entropy change values are both negative, the enthalpy indicates the reaction will occur, and it will at low temperatures. The entropy indicates that there will be no reaction, and at high temperatures there wont. The reverse is true for positive enthalpy and entropy changes. These reactions are more likely to occur at high, but not at low temperatures.
Gibbs Free Energy (∆G) determines for sure whether a reaction will be favored to occur. It is simply a formula that compares ∆H to ∆S in a special way.
∆G = ∆H - T∆S
Temperature must be in Kelvins. If ∆G has a negative value, the reaction will occur spontaneously. If ∆G has a positive value, it will not occur
One complication in calculating the free energy change is that the enthalpy values are typically given in kilojoules (kJ), while entropy values are given in J / K, so you must convert so that both use Joules, or both use kilojoules. It doesn't matter which.
Sometimes you may be asked to find the temperature above which a reaction will or wont occur. This is the temperature at which ∆G is between negative and positive, or when it equals zero.
∆G = 0 so 0 = ∆H - T∆S
Rearrange the equation to solve for T, and you will find that...
T = ∆H/∆S
Above that temperature entropy change determines whether a reaction occurs, and below that temperature, enthalpy change determines whether a reaction occurs.
∆G = 0 means there will be equilibrium in a reversible reaction.
-----------------------------
JEE Question 2007 paper II
For the process (water becoming steam) H-2O(l)(1 bar, 373 K) --> H-2O(g)(1 bar, 373 K), the correct set of thermodynamic parameters is
(A) ∆G=0,∆S= + ve
(B)∆G=0,∆S= -ve
(C)∆G=+ve,∆S=0
(D)∆G=-ve, ∆S= + ve
Solution: A
The answer is A because, because at 100 degree C, the steam and water mixture is at equilibrium. Hence ΔG = 0(G = H - TS), and ΔS is positive. Why? when liquid becomes gas, there is more disorder. More entropy.
--------------
Study Guide Ch. 7. CHEMICAL EQUILIBRIA
Post updated on 29 August 2009
JEE Syllabus
Chemical equilibrium:
Law of mass action;
Equilibrium constant,
Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of DG and DGo in chemical equilibrium;
Solubility product, common ion effect,
pH and buffer solutions;
Acids and bases (Bronsted and Lewis concepts);
Hydrolysis of salts.
The syllabus has two main components: Equilibrium among ions and equilibrium among compounds
---------------
Main topics in TMH Book Chapter
Section I CHEMICAL EQUILIBRIUM
Section II IONIC EQUILIBRIUM IN ACQUEOUS SOLUTIONS
-------------------
Chemical equilibrium
Some chemical reactions appear to go only in one direction and they are said to go to completion. The reactants are totally consumed or some of the reactants are totally consumed and because of that reaction comes to a stop.
But there are many reactions, wherein a reaction among products takes place and reactants are formed due to it. In such reactions, initially in the reaction products keep forming and their concentration increases to certain level and then an equilibrium is reached from which point there is no change in the concentration of products or reactants.
Reversible reactions
All chemical reactions are theoretically reversible. But in some reactions, the reverse reaction is so slight and takes place at such a low rate that for all practical purposes the reaction is considered to be irreversible.
Example: Formation of water.
But there are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case
A + B → C + D and
C+D → A+B both reactions keep taking place.
At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.
From the rate law (Refer Chemical Kinetics chapter), we can write
Rate of forward reaction = k-f[A][B]
Rate of reverse reaction or backward reaction = k-r{C]{D]
Therefore k-f[A][B] = k-r[C]{D]
This gives k-f/k-r = [C]{D]/[A][B]
This is called as equilibrium constant. When the concentrations of C and D and A and B reach this proportion equilibrium is reached in the reaction.
The equilibrium constant is always written as products by reactants.
For the a general reaction
aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)
k-eq = [C]^c[D]^d/[A]^a[B]^b
The equilibrium constant may or may not have units. As we know concentrations are in mol/liter.
In the case of 2A ↔ 2B +C
The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2
= mol/l
For any reaction mechanism at equilibrium, the equilibrium constant is equal to the concentration of products over the concentration of the reactants, all raised to the power of the coefficients from the balanced equation.
-----------
IONIC EQUILIBRIUM PORTION
Syllabus
Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.
Syllabus rearranged
Acids and bases (Bronsted and Lewis concepts);
pH
buffer solutions;
common ion effect,
Solubility product,
Hydrolysis of salts.
Ionic Equilibrium – Introduction
Acids, basess and salts when dissolved n water dissociate to some extent and form ions. In the ion formation, an equilibrium is established between ionized and unionized (whole) molecules as this ionization is a reversible reaction. Such an equilibrium that involves ions is called ionic equilibrium.
Acids and bases definitions (Arrhenius, Bronsted and Lewis)
Arrhenius defined acid as a hydrogen compound which in water solution give hydrogen ions.
He defined base as a hydroxide compound which in water solution gives hydroxide ions.
Lowry and Bronsted came out with a different concept to broaden the definition of Arrhenius.
An acid is defined as a substance having a tendency of lose or to donate one or more protons.
A base is defined as a substance having a tendency to accept or add a proton
Lewis Theory of Acids and Bases
Acid: An acid is any substance (molecule, ion or atom) that can accept a lone pair of electrons to form a coordinate bond (*Remember coordinate bond and lone pair topics in chapter on Bonding)
Base: Base is any species (molecule, ion or atom) that can donate a lone pair of electrons to form a co-ordinate bond.
According to this concept acid-based reaction involves the formation of a co-ordinate covalent bond.
Dissociation of acids and bases into ions
Degree of dissociation: It is defined as the fraction of the total number of moles of an acid or base or electrolyte that dissociates into ions in acqueous solution when the equilibrium is attained. It is represented by α.
α = Total mole of acid or base dissociated/Total mole of acid or base present in the solution
The greater the degree of dissociation, the stronger the acid.
Strong acids HCl, H2SO4, HNO3
Weak acids H2CO3, H2S, CH3COOH
Strong bases NaOH, KOH
Weak bases NH4OH, Mg(OH)2, Ca(OH)2
Dissociation constants of weak acids
Represent acid by HA (H is hydrogen and A remaining portion of the acid)
HA + H2O ↔ H+ + A- (HA dissociates into H+ and A-)
Equilibrium constant from law of mass action = Ka = [H+][A-]/[HA]
Ka is also termed as dissociation constant of acids
Dissociation of bases
Bases are represented as BOH and they dissociate into B+ and OH- ions.
Equilibrium constant for dissociatin of bases from law of mass action =
Kb = [B+][OH-]/[BOH]
Kb is also termed as dissociation constant of bases.
Ostwald’s Dilution law
This law is the relationship between degree of dissociation and the dissociation constant.
If one mole of a weak electrolyte be dissolved in v litres of a solution. At equilibrium let α be the degree of dissociation.
Hence BA will be 1- α moles
B will be α moles and A will be α moles
As K = [B][A]/[BA] = (α/v)( α/v)/{(1- α)/v}
= α²/(1- α)v
But the concentration C = 1/v
Therefore K = α²C/(1- α)
If α is very small, 1- α can be assumed as equal to one and K becomes equal to α²C.
And α = SQRT(K/C)
In case of acids it can be written as
α = SQRT(Ka/C)
In case of bases it can be written as
α = SQRT(Kb/C)
------------------
web sites
Reversible Reactions, Chemical Equilibrium,
http://www.docbrown.info/page04/4_74revNH3.htm
http://www.science.uwaterloo.ca/~cchieh/cact/c123/massacti.html
------------------
JEE Question 2007 Paper II
23. Consider a reaction aG+bH--> Products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is
(A) 0
(B) 1
(C) 2
(D) 3
Solution: D
This is a question of chapter chemical kinetics.
Reason: When G alone is doubled, rate is doubled. That is the exponent of [G] is one in rate law. When both are doubled, rate increases by 8 times, telling us that exponent of [H] is 2 in rate law. So overall order of reaction is sum of the two exponents that is 3.
--------------------
JEE Syllabus
Chemical equilibrium:
Law of mass action;
Equilibrium constant,
Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of DG and DGo in chemical equilibrium;
Solubility product, common ion effect,
pH and buffer solutions;
Acids and bases (Bronsted and Lewis concepts);
Hydrolysis of salts.
The syllabus has two main components: Equilibrium among ions and equilibrium among compounds
---------------
Main topics in TMH Book Chapter
Section I CHEMICAL EQUILIBRIUM
Section II IONIC EQUILIBRIUM IN ACQUEOUS SOLUTIONS
-------------------
Chemical equilibrium
Some chemical reactions appear to go only in one direction and they are said to go to completion. The reactants are totally consumed or some of the reactants are totally consumed and because of that reaction comes to a stop.
But there are many reactions, wherein a reaction among products takes place and reactants are formed due to it. In such reactions, initially in the reaction products keep forming and their concentration increases to certain level and then an equilibrium is reached from which point there is no change in the concentration of products or reactants.
Reversible reactions
All chemical reactions are theoretically reversible. But in some reactions, the reverse reaction is so slight and takes place at such a low rate that for all practical purposes the reaction is considered to be irreversible.
Example: Formation of water.
But there are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case
A + B → C + D and
C+D → A+B both reactions keep taking place.
At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.
From the rate law (Refer Chemical Kinetics chapter), we can write
Rate of forward reaction = k-f[A][B]
Rate of reverse reaction or backward reaction = k-r{C]{D]
Therefore k-f[A][B] = k-r[C]{D]
This gives k-f/k-r = [C]{D]/[A][B]
This is called as equilibrium constant. When the concentrations of C and D and A and B reach this proportion equilibrium is reached in the reaction.
The equilibrium constant is always written as products by reactants.
For the a general reaction
aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)
k-eq = [C]^c[D]^d/[A]^a[B]^b
The equilibrium constant may or may not have units. As we know concentrations are in mol/liter.
In the case of 2A ↔ 2B +C
The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2
= mol/l
For any reaction mechanism at equilibrium, the equilibrium constant is equal to the concentration of products over the concentration of the reactants, all raised to the power of the coefficients from the balanced equation.
-----------
IONIC EQUILIBRIUM PORTION
Syllabus
Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.
Syllabus rearranged
Acids and bases (Bronsted and Lewis concepts);
pH
buffer solutions;
common ion effect,
Solubility product,
Hydrolysis of salts.
Ionic Equilibrium – Introduction
Acids, basess and salts when dissolved n water dissociate to some extent and form ions. In the ion formation, an equilibrium is established between ionized and unionized (whole) molecules as this ionization is a reversible reaction. Such an equilibrium that involves ions is called ionic equilibrium.
Acids and bases definitions (Arrhenius, Bronsted and Lewis)
Arrhenius defined acid as a hydrogen compound which in water solution give hydrogen ions.
He defined base as a hydroxide compound which in water solution gives hydroxide ions.
Lowry and Bronsted came out with a different concept to broaden the definition of Arrhenius.
An acid is defined as a substance having a tendency of lose or to donate one or more protons.
A base is defined as a substance having a tendency to accept or add a proton
Lewis Theory of Acids and Bases
Acid: An acid is any substance (molecule, ion or atom) that can accept a lone pair of electrons to form a coordinate bond (*Remember coordinate bond and lone pair topics in chapter on Bonding)
Base: Base is any species (molecule, ion or atom) that can donate a lone pair of electrons to form a co-ordinate bond.
According to this concept acid-based reaction involves the formation of a co-ordinate covalent bond.
Dissociation of acids and bases into ions
Degree of dissociation: It is defined as the fraction of the total number of moles of an acid or base or electrolyte that dissociates into ions in acqueous solution when the equilibrium is attained. It is represented by α.
α = Total mole of acid or base dissociated/Total mole of acid or base present in the solution
The greater the degree of dissociation, the stronger the acid.
Strong acids HCl, H2SO4, HNO3
Weak acids H2CO3, H2S, CH3COOH
Strong bases NaOH, KOH
Weak bases NH4OH, Mg(OH)2, Ca(OH)2
Dissociation constants of weak acids
Represent acid by HA (H is hydrogen and A remaining portion of the acid)
HA + H2O ↔ H+ + A- (HA dissociates into H+ and A-)
Equilibrium constant from law of mass action = Ka = [H+][A-]/[HA]
Ka is also termed as dissociation constant of acids
Dissociation of bases
Bases are represented as BOH and they dissociate into B+ and OH- ions.
Equilibrium constant for dissociatin of bases from law of mass action =
Kb = [B+][OH-]/[BOH]
Kb is also termed as dissociation constant of bases.
Ostwald’s Dilution law
This law is the relationship between degree of dissociation and the dissociation constant.
If one mole of a weak electrolyte be dissolved in v litres of a solution. At equilibrium let α be the degree of dissociation.
Hence BA will be 1- α moles
B will be α moles and A will be α moles
As K = [B][A]/[BA] = (α/v)( α/v)/{(1- α)/v}
= α²/(1- α)v
But the concentration C = 1/v
Therefore K = α²C/(1- α)
If α is very small, 1- α can be assumed as equal to one and K becomes equal to α²C.
And α = SQRT(K/C)
In case of acids it can be written as
α = SQRT(Ka/C)
In case of bases it can be written as
α = SQRT(Kb/C)
------------------
web sites
Reversible Reactions, Chemical Equilibrium,
http://www.docbrown.info/page04/4_74revNH3.htm
http://www.science.uwaterloo.ca/~cchieh/cact/c123/massacti.html
------------------
JEE Question 2007 Paper II
23. Consider a reaction aG+bH--> Products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is
(A) 0
(B) 1
(C) 2
(D) 3
Solution: D
This is a question of chapter chemical kinetics.
Reason: When G alone is doubled, rate is doubled. That is the exponent of [G] is one in rate law. When both are doubled, rate increases by 8 times, telling us that exponent of [H] is 2 in rate law. So overall order of reaction is sum of the two exponents that is 3.
--------------------
Study Guide Ch.8 ELECTROCHEMISTRY
JEE Syllabus
Electrochemistry:
Electrochemical cells and cell reactions;
Electrode potentials;
Nernst equation and its relation to DG;
Electrochemical series,
emf of galvanic cells;
Faraday's laws of electrolysis;
Electrolytic conductance, specific, equivalent and molar conductance,
Kohlrausch's law;
Concentration cells.
--------------------
Main topic in TMH Book Chapter
Section I: ELECTROLYSIS
FARADAY'S LAWS OF ELECTROLYSIS
Section II: ELECTROLYTIC CONDUCTION
Section III: GALVANIC CELLS
---------------------------
Electrochemistry:
In eletrolytic conduction, ions carry electric charge. An increase in temperature increases electrolytic conductance because as the temperature is increased, the ions move faster.
In electrolytic cell, electric energy is used to cause a chemical reaction to take place.
For example, in a eletrolytic cell molten sodium chloride is the electrolyte. Two electrodes anode and cathode are placed into the molten sodium chloride and the an electri current passed through it.
The cation (Na+) moves toward the electrode called the cathode. At this electrode Na+ ions accept electrons to form sodium metal.
Na+ + e- → Na
This is reduction (gain of electrons)
At the other electrode, called the anode, the anion (Cl-) loses electrons to form chlorine gas.
2Clˉ → Cl-2 + 2eˉ
Therefore, oxidation (loss of electrons) takes place at the anode.
In the cell, electrons move from the anode through the external circuit to the cathode.
The reaction at anode and the reaction at cathode separately are called half-reactions.
Various types of half reactions are possible at anode (but all are oxidation only)
1. oxidation of an anion to an element
2Clˉ → Cl-2 + 2eˉ
2. Oxidation of an anion or cation in solution to an ion of higher oxidation state
Fe^2+ → Fe^3+ + e‾
3. Oxidation of a metal anode,
Cu(s) → Cu^2+ + 2e‾
4. Oxidation of H-2O to produce oxygen:
2H-2O → O-2 +4H^+ +4e‾
Similarly at Cathode various types of reactions occur (But all are reduction reactions)
1. Reduction of a cation to a metal
2. Reduction of an anion or cation in solution to an ion of lower oxidation state.
3. Reduction of a nonmetal to an anion.
4. Reduction of water to produce hydrogen.
Faraday's laws of electrolysis:
---------------------------------------
Quantitative Relationships in Electrolytic Cells
Determining the amount of electrical energy necessary for accumulating a given amount material from the electrolytic cell.
Faraday's laws of electrolysis:
First law: It states that the amount of any substance that is liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
W α Q (w = weight of substance deposited and Q is charge = ampere * time)
Second law: It states tht when the same quantity of electricity is passed through different electrolytes amount of different substances liberated or deposited at the different electrodes are directly proportional to the chemical equivalents9i.e., equivalent weight) of substances.
One faraday (F) is the amount of electrical energy required for flow of 1 mole of electrons.
To three significant digits, 1 faraday equals 96,500 coulombs(coul).
Current flow is measured in amperes (A)which is coulombs/seconds or coul/s,
The equivalent weight, or gram-equivalent weight, of a substance is the amss in grams that is equivalent ot 1 mole of electrons. We can think it as gram-weight of a substance that releases one mole of electrons or reacts with one mole of electrons.
To determine equitdvalent weight of a substance it is necessary to know how many moles of electrons are transferred per mole of substance. To calculate the equivalent weight of a substance the formula weight is divided by the number of moles of electrons transferred.
Aluminium always loses three electrons to form Al^3+. Its formula weight or atomic mass is 27.
Hence it equivalent weight is 27/3 = 9.0 g.
Problem:
How long would it take a current of 100 A to deposit 10 g of Fe from a solution of FeCl-2?
Solution:
The equation for Fe in FeCl-2 is Fe^2+ + 2e‾
Formula weight is 55.8 g
Hence equivalent weight = 55.8/2 = 27.9
Electrical energy required = 10.0 g Fe* (1 equiv Fe/27.9 g Fe)* (96,500 coul/1 equiv Fe)
= 34588 coul
As 100 A or 100 coul/sec is given as current
Time taken = 34588/100 = 345.8 s.
------------
Voltaic or Galvanic Cells
In this cell, a chemical reaction produces electrical energy.
In this cell, the electrons being transferred from the reducing agent to the oxidizing agent travel through a wire and thus provide an elctric current.
One example of this cell type, is zinc metal in a solution containing ZnSo-4 and Cu in solution of CuSo-4. Zinc electrode acts as the cathode and Cu electrode acts as the anode.
At zinc electrode Zn → Zn^2+ +2e¯ (oxidation)
At Cu electrode: Cu^2+ +2e¯ → Cu
The above cell is represented as
Zn|ZnSO-4║CuSO-4|Cu In this symbol additionally On zinc side as it is a cathode a - sign is placed in O and on Cu side as it is anode a + sign is placed in O.
In oxidation reaction, electrons are evolved and hence this reaction can be used to produce electricity from chemical reaction.
Emf produced by the cell depends upon (a) temperature and (b) concentration of CuS0-4 and ZnSO-4.
When ZnSO-4 and CuSO-4 are i molar(1 mole per litre), emf produced is 1.1 volt.
Half cells: Every cell is made of two parts called half cell or electrodes. a metal in contact with own ions is called a half cell.
The reaction taking place at electrode is called electrode reaction or half cell reaction. All electrode reactions are remembered for oxidation.
Standard Hydrogen Electrode
An electrode in which pure dry hydrogen gas is bubbled at 1 atm and 298K about a platinized platinum plate through a solution containing H^+ ions ( for example - HCl solution)
The emf produced is taken as zero volts. All other potential are expressed with SHE potential as zero.
-----------------------
Nernst equation
In electrochemistry, the Nernst equation gives the electrode potential (E), relative to the standard electrode potential, (E^0), of the electrode couple or, equivalently, of the half cells of a battery.
-------------
web sites
Electrolysis, batteries and fuel cells
http://www.docbrown.info/page01/ExIndChem/ExtraElectrochem.htm
Electrochemistry:
Electrochemical cells and cell reactions;
Electrode potentials;
Nernst equation and its relation to DG;
Electrochemical series,
emf of galvanic cells;
Faraday's laws of electrolysis;
Electrolytic conductance, specific, equivalent and molar conductance,
Kohlrausch's law;
Concentration cells.
--------------------
Main topic in TMH Book Chapter
Section I: ELECTROLYSIS
FARADAY'S LAWS OF ELECTROLYSIS
Section II: ELECTROLYTIC CONDUCTION
Section III: GALVANIC CELLS
---------------------------
Electrochemistry:
In eletrolytic conduction, ions carry electric charge. An increase in temperature increases electrolytic conductance because as the temperature is increased, the ions move faster.
In electrolytic cell, electric energy is used to cause a chemical reaction to take place.
For example, in a eletrolytic cell molten sodium chloride is the electrolyte. Two electrodes anode and cathode are placed into the molten sodium chloride and the an electri current passed through it.
The cation (Na+) moves toward the electrode called the cathode. At this electrode Na+ ions accept electrons to form sodium metal.
Na+ + e- → Na
This is reduction (gain of electrons)
At the other electrode, called the anode, the anion (Cl-) loses electrons to form chlorine gas.
2Clˉ → Cl-2 + 2eˉ
Therefore, oxidation (loss of electrons) takes place at the anode.
In the cell, electrons move from the anode through the external circuit to the cathode.
The reaction at anode and the reaction at cathode separately are called half-reactions.
Various types of half reactions are possible at anode (but all are oxidation only)
1. oxidation of an anion to an element
2Clˉ → Cl-2 + 2eˉ
2. Oxidation of an anion or cation in solution to an ion of higher oxidation state
Fe^2+ → Fe^3+ + e‾
3. Oxidation of a metal anode,
Cu(s) → Cu^2+ + 2e‾
4. Oxidation of H-2O to produce oxygen:
2H-2O → O-2 +4H^+ +4e‾
Similarly at Cathode various types of reactions occur (But all are reduction reactions)
1. Reduction of a cation to a metal
2. Reduction of an anion or cation in solution to an ion of lower oxidation state.
3. Reduction of a nonmetal to an anion.
4. Reduction of water to produce hydrogen.
Faraday's laws of electrolysis:
---------------------------------------
Quantitative Relationships in Electrolytic Cells
Determining the amount of electrical energy necessary for accumulating a given amount material from the electrolytic cell.
Faraday's laws of electrolysis:
First law: It states that the amount of any substance that is liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
W α Q (w = weight of substance deposited and Q is charge = ampere * time)
Second law: It states tht when the same quantity of electricity is passed through different electrolytes amount of different substances liberated or deposited at the different electrodes are directly proportional to the chemical equivalents9i.e., equivalent weight) of substances.
One faraday (F) is the amount of electrical energy required for flow of 1 mole of electrons.
To three significant digits, 1 faraday equals 96,500 coulombs(coul).
Current flow is measured in amperes (A)which is coulombs/seconds or coul/s,
The equivalent weight, or gram-equivalent weight, of a substance is the amss in grams that is equivalent ot 1 mole of electrons. We can think it as gram-weight of a substance that releases one mole of electrons or reacts with one mole of electrons.
To determine equitdvalent weight of a substance it is necessary to know how many moles of electrons are transferred per mole of substance. To calculate the equivalent weight of a substance the formula weight is divided by the number of moles of electrons transferred.
Aluminium always loses three electrons to form Al^3+. Its formula weight or atomic mass is 27.
Hence it equivalent weight is 27/3 = 9.0 g.
Problem:
How long would it take a current of 100 A to deposit 10 g of Fe from a solution of FeCl-2?
Solution:
The equation for Fe in FeCl-2 is Fe^2+ + 2e‾
Formula weight is 55.8 g
Hence equivalent weight = 55.8/2 = 27.9
Electrical energy required = 10.0 g Fe* (1 equiv Fe/27.9 g Fe)* (96,500 coul/1 equiv Fe)
= 34588 coul
As 100 A or 100 coul/sec is given as current
Time taken = 34588/100 = 345.8 s.
------------
Voltaic or Galvanic Cells
In this cell, a chemical reaction produces electrical energy.
In this cell, the electrons being transferred from the reducing agent to the oxidizing agent travel through a wire and thus provide an elctric current.
One example of this cell type, is zinc metal in a solution containing ZnSo-4 and Cu in solution of CuSo-4. Zinc electrode acts as the cathode and Cu electrode acts as the anode.
At zinc electrode Zn → Zn^2+ +2e¯ (oxidation)
At Cu electrode: Cu^2+ +2e¯ → Cu
The above cell is represented as
Zn|ZnSO-4║CuSO-4|Cu In this symbol additionally On zinc side as it is a cathode a - sign is placed in O and on Cu side as it is anode a + sign is placed in O.
In oxidation reaction, electrons are evolved and hence this reaction can be used to produce electricity from chemical reaction.
Emf produced by the cell depends upon (a) temperature and (b) concentration of CuS0-4 and ZnSO-4.
When ZnSO-4 and CuSO-4 are i molar(1 mole per litre), emf produced is 1.1 volt.
Half cells: Every cell is made of two parts called half cell or electrodes. a metal in contact with own ions is called a half cell.
The reaction taking place at electrode is called electrode reaction or half cell reaction. All electrode reactions are remembered for oxidation.
Standard Hydrogen Electrode
An electrode in which pure dry hydrogen gas is bubbled at 1 atm and 298K about a platinized platinum plate through a solution containing H^+ ions ( for example - HCl solution)
The emf produced is taken as zero volts. All other potential are expressed with SHE potential as zero.
-----------------------
Nernst equation
In electrochemistry, the Nernst equation gives the electrode potential (E), relative to the standard electrode potential, (E^0), of the electrode couple or, equivalently, of the half cells of a battery.
-------------
web sites
Electrolysis, batteries and fuel cells
http://www.docbrown.info/page01/ExIndChem/ExtraElectrochem.htm
Study Guide Ch 9. SOLUTIONS
JEE Syllabus
Solutions:
Raoult's law;
Molecular weight determination from lowering of vapor pressure,
Molecular weight determination from elevation of boiling point and
Molecular weight determination from depression of freezing point.
-----------------------------
Main topics in TMH Book chapter
COMPOSITION OF A SOLUTION
IDEAL LIQUID SOLUTION
COLLIGATIVE PROPERTIES
--------------------------
Raoult's Law
The vapor pressure and composition in equilibrium with a solution can yield valuable information regarding the thermodynamic properties of the liquids involved. Raoult’s law relates the vapor pressure of components to the composition of the solution. The law assumes ideal behavior. It gives a simple picture of the situation just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approache zero, so the behavior of gases approach the behavior of the ideal gas. Raoult’s law is similar in that it assumes that the physical properties of the components are identical. The more similar the components the more their behavior approaches that described by Raoult’s law.
Using the example of a solution of two liquids, A and B, if no other gases are present the total vapor pressure Ptot above the solution is equal to the sum of the vapor pressures of the two components, P(A) and P(B).
Raoult's Law
The vapor pressure and composition in equilibrium with a solution can yield valuable information regarding the thermodynamic properties of the liquids involved. Raoult’s law relates the vapor pressure of components to the composition of the solution. The law assumes ideal behavior. It gives a simple picture of the situation just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approache zero, so the behavior of gases approach the behavior of the ideal gas. Raoult’s law is similar in that it assumes that the physical properties of the components are identical. The more similar the components the more their behavior approaches that described by Raoult’s law.
Using the example of a solution of two liquids, A and B, if no other gases are present the total vapor pressure Ptot above the solution is equal to the sum of the vapor pressures of the two components, PA and PB.
f the two components are very similar, or in the limiting case, differ only in isotopic content, then the vapor pressure of each component will be equal to the vapor pressure of the pure substance Po times the mole fraction in the solution. This is Raoult’s law.
Thus the total pressure above solution of A and B would be
P(total)= P(A) + P(B)
If the two components are very similar, or in the limiting case, differ only in isotopic content, then the vapor pressure of each component will be equal to the vapor pressure of the pure substance Po times the mole fraction in the solution. This is Raoult’s law.
P(i) = P(i)^0*x(i)
Thus the total pressure above solution of A and B would be
P(total)= =P(A)^0*x(i) + P(B)^0*x(i)
refer: http://www.tannerm.com/raoult.htm
Molecular weight determination from depression of freezing point.
The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.
When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.
The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m
ΔTf = depression in freezing point.
Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf
Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.
To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.
weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb
Molality of the solution, m = Wb*1000/Mb*Wa
Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa
From the above equation Mb can be calculated.
Mb = Kf*Wb*1000/Wa*ΔTf
or
Mb = [Kf*Wb*1000]/[ΔTf * Wa]
Example:
Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)
The formula of Mb is available above.
weight of solute be Wb g = 0.643 g
weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C
Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)
Van't Hoff Factor
In 1986, Van't Hoff introduced a factor ot express the extent of association or dissociation of solutes in solution. It is the ratio of the normal and observed molar masses of the solute.
i = Normal molar mass of the solute/Observed molar mass of the solute
-------------------------------
JEE Question 2007 paper I
When 20 g of naphthoic acid (C-11H-8O-2) is dissolved in 50 g of benzene (K-f = 1.72 K kg mol^-1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is
(A) 0.5
(B) 1
(C) 2
(D) 3
Answer: A
Van't Hoff factor = Normal Molar Mass/Observed Molar mass
Observed Molar Mass M-B = K-f*1000*W-B/(W-A*ΔT-f)
K-f = Molal depression constant
W-A = weight of solvent
W-B = weight of solute
ΔT-f = depression in freezing point.
In the problem Normal Molar mass = 172 (12*11+8*1+2*16 = 132+8+32 = 172)
Observe molar mass = 1.72*1000*20/50*2 = 17.2*20 = 172*2
Van't Hoff factor =172/172*2 = 0.5
Ref: Dr. Jauhar's book unit 3: solutions.
---------------------------
Solutions:
Raoult's law;
Molecular weight determination from lowering of vapor pressure,
Molecular weight determination from elevation of boiling point and
Molecular weight determination from depression of freezing point.
-----------------------------
Main topics in TMH Book chapter
COMPOSITION OF A SOLUTION
IDEAL LIQUID SOLUTION
COLLIGATIVE PROPERTIES
--------------------------
Raoult's Law
The vapor pressure and composition in equilibrium with a solution can yield valuable information regarding the thermodynamic properties of the liquids involved. Raoult’s law relates the vapor pressure of components to the composition of the solution. The law assumes ideal behavior. It gives a simple picture of the situation just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approache zero, so the behavior of gases approach the behavior of the ideal gas. Raoult’s law is similar in that it assumes that the physical properties of the components are identical. The more similar the components the more their behavior approaches that described by Raoult’s law.
Using the example of a solution of two liquids, A and B, if no other gases are present the total vapor pressure Ptot above the solution is equal to the sum of the vapor pressures of the two components, P(A) and P(B).
Raoult's Law
The vapor pressure and composition in equilibrium with a solution can yield valuable information regarding the thermodynamic properties of the liquids involved. Raoult’s law relates the vapor pressure of components to the composition of the solution. The law assumes ideal behavior. It gives a simple picture of the situation just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approache zero, so the behavior of gases approach the behavior of the ideal gas. Raoult’s law is similar in that it assumes that the physical properties of the components are identical. The more similar the components the more their behavior approaches that described by Raoult’s law.
Using the example of a solution of two liquids, A and B, if no other gases are present the total vapor pressure Ptot above the solution is equal to the sum of the vapor pressures of the two components, PA and PB.
f the two components are very similar, or in the limiting case, differ only in isotopic content, then the vapor pressure of each component will be equal to the vapor pressure of the pure substance Po times the mole fraction in the solution. This is Raoult’s law.
Thus the total pressure above solution of A and B would be
P(total)= P(A) + P(B)
If the two components are very similar, or in the limiting case, differ only in isotopic content, then the vapor pressure of each component will be equal to the vapor pressure of the pure substance Po times the mole fraction in the solution. This is Raoult’s law.
P(i) = P(i)^0*x(i)
Thus the total pressure above solution of A and B would be
P(total)= =P(A)^0*x(i) + P(B)^0*x(i)
refer: http://www.tannerm.com/raoult.htm
Molecular weight determination from depression of freezing point.
The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.
When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.
The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m
ΔTf = depression in freezing point.
Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf
Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.
To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.
weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb
Molality of the solution, m = Wb*1000/Mb*Wa
Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa
From the above equation Mb can be calculated.
Mb = Kf*Wb*1000/Wa*ΔTf
or
Mb = [Kf*Wb*1000]/[ΔTf * Wa]
Example:
Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)
The formula of Mb is available above.
weight of solute be Wb g = 0.643 g
weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C
Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)
Van't Hoff Factor
In 1986, Van't Hoff introduced a factor ot express the extent of association or dissociation of solutes in solution. It is the ratio of the normal and observed molar masses of the solute.
i = Normal molar mass of the solute/Observed molar mass of the solute
-------------------------------
JEE Question 2007 paper I
When 20 g of naphthoic acid (C-11H-8O-2) is dissolved in 50 g of benzene (K-f = 1.72 K kg mol^-1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is
(A) 0.5
(B) 1
(C) 2
(D) 3
Answer: A
Van't Hoff factor = Normal Molar Mass/Observed Molar mass
Observed Molar Mass M-B = K-f*1000*W-B/(W-A*ΔT-f)
K-f = Molal depression constant
W-A = weight of solvent
W-B = weight of solute
ΔT-f = depression in freezing point.
In the problem Normal Molar mass = 172 (12*11+8*1+2*16 = 132+8+32 = 172)
Observe molar mass = 1.72*1000*20/50*2 = 17.2*20 = 172*2
Van't Hoff factor =172/172*2 = 0.5
Ref: Dr. Jauhar's book unit 3: solutions.
---------------------------
Study Guide Ch.10. CHEMICAL KINETICS
Jee Syllabus
Chemical kinetics:
Rates of chemical reactions;
Order of reactions;
Rate constant;
First order reactions;
Temperature dependence of rate constant (Arrhenius equation).
--------------------------
Main topics in TMH book
CHEMICAL KINETICS
-------------------------
The material given in the TMH book is very brief. So I have to elaborate the contents. This chapter should have been given before the chapter on Chemical Equilibrium.
The topic "Chemical kinetics" consists of reaction rate and reaction mechanism.
Reaction rate is the speed with which a reaction takes place. This shows the rate or speed at which the reactants are consumed and products are formed.
Reaction mechanism is the path by which a reaction takes place.
Rate of reaction
The rate of reaction is a quantity that tells how the concentration of reactants or product changes with time.
So this can be expressed as Δ contentration/Δ time. That is change in concenation divided by time taken for the change.
Molar concentration i.e., moles per liter (M), is used in these equations.
The brackets, [ ] are always used to to indicate molar concentrations.
Rate of reactions can be expressed as d[A]/dt. [A] represents molar concentration of a reactant or a product.
Rate law
The rate for a reaction is a mathematical expression that relates the rate of reaction to the concentrations of the reactants.
For the reaction aA + bB → products
The rate law is expressed as, rate of reaction is proportional to [A]^x[B]^y.
x and y are determined experimentally. These values can be whole or fractional numbers or zero.
Reaction Mechanisms
Many reactions go in two or more steps. When this is the case, there is usually one slow step, and the rate of this step determines the rate of whole reaction.. The slow step is called the rate determining step.
Law of Mass Action
In 1867, Cato Guldberg, and Peter Waage, proposed this law. According to this law, for the rate determining step in a reaction, the rate of reaction is proportional to the product of the concentrations of the reactants, each raised to the power of its coefficient in the balanced equation.
For the reaction aA + bB → cC (when it is a rate determining step)
Rate of reaction is proportional to [A]^a[B]^b
The above proportionality can be written as an equation, by putting in a proportionality constant k.
Rate = k *[A]^a[B]^b
K is called the specific rate constant (TMH book used the term “rate constant”)
But, remember that for the overall rate law (rate law for total reaction), must be determined experimentally. The law of mass action only applies to the rate determining step. TMH book uses the term elementary equation for the equation for which law of mass action applies.
Example:
For the rate determining step
A + 2B → C
Rate = k *[A][B]^2
Order of Reaction
From the rate law for a reaction order of reaction can be determined.
For a particular species or reactant, the order is equal to the exponent for that species in the rate law.
For example for Rate = k *[A][B]^2
for B the order of reaction is 2. For A it is 1.
The overall order of reaction is equal to the sum of all the individual orders of reactants.
From the orders of reactions, we can determine how the rate of a reaction will change as the concentration of each of the reactants is changed.
For the above reaction, if the concentration of B is double, the rate will become 4 times.
For gases, as molar concentrations are proportional of partial pressures, they can be used instead of concentrations in the rate equations.
Collision Theory of Reaction Rates
Collision theory can be used to explain what is happening in reactions. The collisions between particles of reactants results in reaction. The collisions will be effective when colliding particles have sufficient energy. The energy that is necessary for effective collisions is called the activation energy. Activation energy is energy barrier that must be overcome in order for a reaction to take place.
Factors that Control Reactions Rates
Concentration of reactants, temperature, and catalysts determine the rates of reaction for a given set of reactants. The nature of reactants determines the activation energy required.
Concentration
As concentration increases, more reactant molecules collide with other reactant molecules in a given period of time, and more product molecules are formed.
Temperature
As temperature increases, the average kinetic energy increases. So there are more molecules with activation energy and hence reaction rate increases.
As a general approximation, the rate roughly doubles for each 10°C rise in temperature.
Catalyst
A catalyst is a substance that alters the speed of a reaction without being consumed. While a catalyst is normally used for speeding up the reaction, there are also catalysts that slow down a reaction. They are called negative catalysts or inhibitors.
A catalyst alters the speed of the reaction by changing the activation energy. It speeds up the reaction by decreasing the activation energy. A negative catalyst increases the activation energy.
Determining Reaction Mechanisms
When alternative paths are possible for a reaction, the various paths are identified. The rate laws for each possible paths are written down. Experimentally, the rate law for the reaction is determined. The mechanism that gives this rate law (experimentally determined law) is taken as the reaction mechanism.
How to Observe Change in rates of Reactions
If a reaction has H^+ ions as one of the products, change in pH can be used to determine the rate of reaction.
If the product is a gas, the measurement of gas given off in various periods of time
can be used to determine the rate.
If the products contain a species with distinctive colour, the progress can be followed by measuring the absorbency of light passing through the solution, in a colorimeter or other type of spectrophotometer.
--------------
web sites
Factors affecting the Speed-Rates of Chemical Reactions
http://www.docbrown.info/page03/3_31rates.htm
----------------------
JEE Question 2007 paper II
For the process H-2O(l)(1 bar, 373 K) --> H-2O(g)(1 bar, 373 K), the correct set of thermodynamic parameters is
(A) ∆G=0,∆S= + ve
(B)∆G=0,∆S= -ve
(C)∆G=+ve,∆S=0
(D)∆G=-ve, ∆S= + ve
Solution: A
This question belongs to the chapter Energetics. The answer is A because, because at 100 degree C, the steam and water mixture is at equilibrium. Hence ΔG = 0(G = H - TS), and ΔS is positive.
-----------------------
See some questions for practice on this chapter in the page
http://iit-jee-chemistry-ps.blogspot.com/2007/10/iit-jee-chemistry-questions-ch10.html
Chemical kinetics:
Rates of chemical reactions;
Order of reactions;
Rate constant;
First order reactions;
Temperature dependence of rate constant (Arrhenius equation).
--------------------------
Main topics in TMH book
CHEMICAL KINETICS
-------------------------
The material given in the TMH book is very brief. So I have to elaborate the contents. This chapter should have been given before the chapter on Chemical Equilibrium.
The topic "Chemical kinetics" consists of reaction rate and reaction mechanism.
Reaction rate is the speed with which a reaction takes place. This shows the rate or speed at which the reactants are consumed and products are formed.
Reaction mechanism is the path by which a reaction takes place.
Rate of reaction
The rate of reaction is a quantity that tells how the concentration of reactants or product changes with time.
So this can be expressed as Δ contentration/Δ time. That is change in concenation divided by time taken for the change.
Molar concentration i.e., moles per liter (M), is used in these equations.
The brackets, [ ] are always used to to indicate molar concentrations.
Rate of reactions can be expressed as d[A]/dt. [A] represents molar concentration of a reactant or a product.
Rate law
The rate for a reaction is a mathematical expression that relates the rate of reaction to the concentrations of the reactants.
For the reaction aA + bB → products
The rate law is expressed as, rate of reaction is proportional to [A]^x[B]^y.
x and y are determined experimentally. These values can be whole or fractional numbers or zero.
Reaction Mechanisms
Many reactions go in two or more steps. When this is the case, there is usually one slow step, and the rate of this step determines the rate of whole reaction.. The slow step is called the rate determining step.
Law of Mass Action
In 1867, Cato Guldberg, and Peter Waage, proposed this law. According to this law, for the rate determining step in a reaction, the rate of reaction is proportional to the product of the concentrations of the reactants, each raised to the power of its coefficient in the balanced equation.
For the reaction aA + bB → cC (when it is a rate determining step)
Rate of reaction is proportional to [A]^a[B]^b
The above proportionality can be written as an equation, by putting in a proportionality constant k.
Rate = k *[A]^a[B]^b
K is called the specific rate constant (TMH book used the term “rate constant”)
But, remember that for the overall rate law (rate law for total reaction), must be determined experimentally. The law of mass action only applies to the rate determining step. TMH book uses the term elementary equation for the equation for which law of mass action applies.
Example:
For the rate determining step
A + 2B → C
Rate = k *[A][B]^2
Order of Reaction
From the rate law for a reaction order of reaction can be determined.
For a particular species or reactant, the order is equal to the exponent for that species in the rate law.
For example for Rate = k *[A][B]^2
for B the order of reaction is 2. For A it is 1.
The overall order of reaction is equal to the sum of all the individual orders of reactants.
From the orders of reactions, we can determine how the rate of a reaction will change as the concentration of each of the reactants is changed.
For the above reaction, if the concentration of B is double, the rate will become 4 times.
For gases, as molar concentrations are proportional of partial pressures, they can be used instead of concentrations in the rate equations.
Collision Theory of Reaction Rates
Collision theory can be used to explain what is happening in reactions. The collisions between particles of reactants results in reaction. The collisions will be effective when colliding particles have sufficient energy. The energy that is necessary for effective collisions is called the activation energy. Activation energy is energy barrier that must be overcome in order for a reaction to take place.
Factors that Control Reactions Rates
Concentration of reactants, temperature, and catalysts determine the rates of reaction for a given set of reactants. The nature of reactants determines the activation energy required.
Concentration
As concentration increases, more reactant molecules collide with other reactant molecules in a given period of time, and more product molecules are formed.
Temperature
As temperature increases, the average kinetic energy increases. So there are more molecules with activation energy and hence reaction rate increases.
As a general approximation, the rate roughly doubles for each 10°C rise in temperature.
Catalyst
A catalyst is a substance that alters the speed of a reaction without being consumed. While a catalyst is normally used for speeding up the reaction, there are also catalysts that slow down a reaction. They are called negative catalysts or inhibitors.
A catalyst alters the speed of the reaction by changing the activation energy. It speeds up the reaction by decreasing the activation energy. A negative catalyst increases the activation energy.
Determining Reaction Mechanisms
When alternative paths are possible for a reaction, the various paths are identified. The rate laws for each possible paths are written down. Experimentally, the rate law for the reaction is determined. The mechanism that gives this rate law (experimentally determined law) is taken as the reaction mechanism.
How to Observe Change in rates of Reactions
If a reaction has H^+ ions as one of the products, change in pH can be used to determine the rate of reaction.
If the product is a gas, the measurement of gas given off in various periods of time
can be used to determine the rate.
If the products contain a species with distinctive colour, the progress can be followed by measuring the absorbency of light passing through the solution, in a colorimeter or other type of spectrophotometer.
--------------
web sites
Factors affecting the Speed-Rates of Chemical Reactions
http://www.docbrown.info/page03/3_31rates.htm
----------------------
JEE Question 2007 paper II
For the process H-2O(l)(1 bar, 373 K) --> H-2O(g)(1 bar, 373 K), the correct set of thermodynamic parameters is
(A) ∆G=0,∆S= + ve
(B)∆G=0,∆S= -ve
(C)∆G=+ve,∆S=0
(D)∆G=-ve, ∆S= + ve
Solution: A
This question belongs to the chapter Energetics. The answer is A because, because at 100 degree C, the steam and water mixture is at equilibrium. Hence ΔG = 0(G = H - TS), and ΔS is positive.
-----------------------
See some questions for practice on this chapter in the page
http://iit-jee-chemistry-ps.blogspot.com/2007/10/iit-jee-chemistry-questions-ch10.html
Study Guide Ch. 11. SURFACE CHEMISTRY
JEE syllabus
Surface chemistry:
Elementary concepts of adsorption (excluding adsorption isotherms);
Colloids: types, methods of preparation and general properties;
Elementary ideas of emulsions, surfactants and micelles (only definitions and examples).
------------------
Main topics in TMH Book
SURFACE CHEMISTRY
--------------------
Interesting News
2007 Nobel Prize in Chemistry was awarded to Gerhard Erti of Germany for studies of chemical processes on solid surfaces.
erti's research laid the foundation of modern surface chemistry, which has explained how fuel cells work, how catalysts operate in cars and even why iron rusts.
The award citation said "Gerhard Erti has succeeded in providing a detailed description of how chemical reactions take place on surfaces and has in this way laid the foundation of modern surface chemistry." "Surface chemistry can explain the destruction of ozone layer as vital steps in the reaction actually take place on the surfaces of small crystals of ice in stratosphere."
Mint 11 October 2007, page 18
--------------------
Elementary concepts of adsorption
The term adsorption implies the presence of excess concentration of any particular component in one of the three phases of matter (known as adsorbate) at the surface of liquid or solid phase (known as adsorbent) as compared to that present in the bulk of the material.
On the basis of the forces of attraction between adsorbent and adsorbate, two types of adsorption, namely, physisorption (i.e. physical adsorption) and chemisorption, may be identified.
The characteristics of physisorption are;
1. the forces of attraction are of van der waals type (weak forces).
2. Predominates at low temperature.
3. All gases show this adsorption at low temperatures.
4. Heat of adsorption is low, about 40 kJ/mol.
5. Reversible in nature.
6. Low activation energy (appx. 5 kJ)
7. Adsorption is multilayer.
The characteristics of chemisorption are:
1. the forces of attraction are of a chemical nature (strong forces)
2. Predominates at high temperature.
3. This is highly specific in nature.
4. Heat of adsorption is large 9appx. 80 to 420 kJ/mol)
5. Usually irreversible.
6. Large activation energy
7. Adsorption is monolayer.
The extent of adsorption of gases increases with increase in the pressure of the gases and it decreases with increase in temperature of the gas.
Colloids
Colloids or sols are the substances whose sizes lie in between the solutes present in a true solution (e.g., salt, sugar) and the solutes present in suspension (e.g., sand). It can also be explained that colloids are intermediate between solutions and precipitates. The particles in colloids are larger than the molecules or ions that make up solutions. The diameters of colloidal particles may range from 1 to 100 nm. The particles in colloidal state do not settle down on standing, are not visible and they can pass through a filter paper. However, they do not pass through a perchment paper or animal membrane.
In the case colloids, we use the terms dispersed particles and dispersing medium. Dispersed particles are the colloidal particles. Milk is an example of a colloid; butterfat constitutes the dispersed particles and water is the dispersing medium.
Types of colloids
Colloids may consists of dispersed particles of various phases and dispersing mediums of various phases. These different types of colloids have different names.
Type Examples
Solid sol Coloured gems and glasses, some alloys, minerals
Sol Starch or proteins in water, paints, gold sol
Solid aerosol Smoke, dust, storm
Gel Jellies, cheese, butter, boot polish
Emulsion Emulsified oils, milk, cod liver oil, medicines
Liquid aerosol Mist, fog, cloud, insecticide sprays
Solid foam Styrene foam, rubber, occluded gases
Foam or froth Whipped cream, lemonade, froth, soap suds
Solid sol is a solid dispersed in a solid.
Sol is a solid dispersed in a liquid.
Solid aerosol is a solid dispersed in a gas.
Gel is a liquid dispersed in a solid
Emulsion is a liquid dispersed in a liquid.
Liquid aerosol is a liquid dispersed in a gas.
Solid foam is a gas dispersed in a solid.
A foam is a gas dispersed in a liquid.
The colloid (dispersing particles + medium) is a two phase system.
Properties of Colloids
Besides particle size, colloids have other identifying properties. Properties peculiar to colloids are (1) optical effects, (2) motion effect, and (3) electrical charge effect.
Optical effect
When a relatively narrow beam of light is passed through a colloid, such dust particles in the air, the light is scattered by the dust particles and they appear in the beam a bright, tiny specks of light. Many of us have seen this phenomenon.
The scattering of light in a colloid is due to the reflection of the light by the large colloidal particles producing a visible beam of light. This optical effect is named the Tyndall effect after John Tyndall, who investigated it in 1860.
Motion effect
If a colloid is viewed with a special microscope, dispersed colloidal particles appear to move in a zigzag, random motion. This erratic random motion is due to bombardment of the dispersed colloidal particles by the medium and this constant bombardment keeps the dispersed colloidal particles suspended indefinitely. They will not settle down. This movement of colloidal particles is called Brownian movement, and this could be observed because of the Tyndall effect.
Electrical charge effect
Very often a colloid will have ions from the dispersing medium adsorbed on its surface. This along with the Brownian motion of the colloid, prevents colloids from coagulating and precipitating.
To facilitate precipitation, an electrolyte of some type is added to the solution, which will neutralize the charge. For instance, in the case of qualitative analysis, if charge on colloids is interfering with the precipitation, an electrolyte of some type is added. This property is used in removing suspended particles from the effluent gases in industrial smokestacks., In the gas coming of out various industrial equipment is given an electrical charge. Before this exhaust gas goes into atmosphere, it is passed through charged plates. These charged plates attract the charged colloidal particles, which then are held on the plates till the current giving the plates their charge is shut off. Then the particles fall to the ground and can be collected. The response of colloid particles to electrical charge is electrical phoresis.
Emulsions
Emulsions are sols of liquid in liquid. Two types of emulsions may be distinguished, namely, oil-in-water and water-in-oil. To make emulsions stable, emulsifying agents such as soaps and detergents are added.
Surfactants
Any substance which can decrease the surface tension of water to a large extent is known as surfactant. Examples of soap and detergents. Such substances have larger concentrations at the surface of water as compared to the bulk of the solution.
Micelle
From http://www.chemicool.com/definition/micelle.html
Definition of micelle
Surfactants in solution are often association colloids, that is, they tend to form aggregates of colloidal dimensions, which exist in equilibrium with the molecules or ions from which they are formed. Such aggregates are termed micelles. (IUPAC MANUAL APPENDIX II (1972)).
What is colloidal micelle?
The heart of this new chemistry is the technology used to develop a "colloidal micelle." Sub-microscopic particles are created in a microscopic field similar to a magnetic field. It differs from traditional chemistry in that the molecular attraction is not the usual attraction between positive and negative poles. Rather, it is between like poles.
An analogy would be that negative attracts negative and positive attracts positive. The micelle has a hydrophilic (water seeking) pole and a hydrophobic (water repelling) pole. The hydrophobic poles attract each other, thus forming the interior of the micelle. The hydrophilic poles form a tough outer surface.
When a micelle comes in contact with a hydrocarbon molecule, the center of the micelle bonds to a similar hydrophobic hydrocarbon. It disrupts the attraction to other hydrocarbon molecules and/or to the surface. Detergent particles in water form particles with hydrophobic poles and hydrophilic poles. In the case of clothes, the hydrophobic pole of a detergent particle attracts the grease and dirt particles on the clothes and pull them apart. As such micelles come together, an outlayer or hydrophilic poles forms. These associative colloid particles are then washed away by water and leaving clothes clean.
The action of a single micelle is multiplied by billions of other micelles. The molecular level emulsification process penetrates highly viscous and sticky materials, lifting them from the surface to which they adhere.
Considering the damage inflicted on life and the planet by harsh cleaners and solvents, colloidal chemistry is an exceptionally advance towards environmental preservation.
---------------------
JEE Question 2007 Paper I
Statement - 1
Micelles are formed by surfactant molecules above the critical micellar concentration (CMC)
Because
Statement - 2
The conductivity of a solution having surfactant molecules decreases sharply at the CMC.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Correct choice: B
---------------------------
Surface chemistry:
Elementary concepts of adsorption (excluding adsorption isotherms);
Colloids: types, methods of preparation and general properties;
Elementary ideas of emulsions, surfactants and micelles (only definitions and examples).
------------------
Main topics in TMH Book
SURFACE CHEMISTRY
--------------------
Interesting News
2007 Nobel Prize in Chemistry was awarded to Gerhard Erti of Germany for studies of chemical processes on solid surfaces.
erti's research laid the foundation of modern surface chemistry, which has explained how fuel cells work, how catalysts operate in cars and even why iron rusts.
The award citation said "Gerhard Erti has succeeded in providing a detailed description of how chemical reactions take place on surfaces and has in this way laid the foundation of modern surface chemistry." "Surface chemistry can explain the destruction of ozone layer as vital steps in the reaction actually take place on the surfaces of small crystals of ice in stratosphere."
Mint 11 October 2007, page 18
--------------------
Elementary concepts of adsorption
The term adsorption implies the presence of excess concentration of any particular component in one of the three phases of matter (known as adsorbate) at the surface of liquid or solid phase (known as adsorbent) as compared to that present in the bulk of the material.
On the basis of the forces of attraction between adsorbent and adsorbate, two types of adsorption, namely, physisorption (i.e. physical adsorption) and chemisorption, may be identified.
The characteristics of physisorption are;
1. the forces of attraction are of van der waals type (weak forces).
2. Predominates at low temperature.
3. All gases show this adsorption at low temperatures.
4. Heat of adsorption is low, about 40 kJ/mol.
5. Reversible in nature.
6. Low activation energy (appx. 5 kJ)
7. Adsorption is multilayer.
The characteristics of chemisorption are:
1. the forces of attraction are of a chemical nature (strong forces)
2. Predominates at high temperature.
3. This is highly specific in nature.
4. Heat of adsorption is large 9appx. 80 to 420 kJ/mol)
5. Usually irreversible.
6. Large activation energy
7. Adsorption is monolayer.
The extent of adsorption of gases increases with increase in the pressure of the gases and it decreases with increase in temperature of the gas.
Colloids
Colloids or sols are the substances whose sizes lie in between the solutes present in a true solution (e.g., salt, sugar) and the solutes present in suspension (e.g., sand). It can also be explained that colloids are intermediate between solutions and precipitates. The particles in colloids are larger than the molecules or ions that make up solutions. The diameters of colloidal particles may range from 1 to 100 nm. The particles in colloidal state do not settle down on standing, are not visible and they can pass through a filter paper. However, they do not pass through a perchment paper or animal membrane.
In the case colloids, we use the terms dispersed particles and dispersing medium. Dispersed particles are the colloidal particles. Milk is an example of a colloid; butterfat constitutes the dispersed particles and water is the dispersing medium.
Types of colloids
Colloids may consists of dispersed particles of various phases and dispersing mediums of various phases. These different types of colloids have different names.
Type Examples
Solid sol Coloured gems and glasses, some alloys, minerals
Sol Starch or proteins in water, paints, gold sol
Solid aerosol Smoke, dust, storm
Gel Jellies, cheese, butter, boot polish
Emulsion Emulsified oils, milk, cod liver oil, medicines
Liquid aerosol Mist, fog, cloud, insecticide sprays
Solid foam Styrene foam, rubber, occluded gases
Foam or froth Whipped cream, lemonade, froth, soap suds
Solid sol is a solid dispersed in a solid.
Sol is a solid dispersed in a liquid.
Solid aerosol is a solid dispersed in a gas.
Gel is a liquid dispersed in a solid
Emulsion is a liquid dispersed in a liquid.
Liquid aerosol is a liquid dispersed in a gas.
Solid foam is a gas dispersed in a solid.
A foam is a gas dispersed in a liquid.
The colloid (dispersing particles + medium) is a two phase system.
Properties of Colloids
Besides particle size, colloids have other identifying properties. Properties peculiar to colloids are (1) optical effects, (2) motion effect, and (3) electrical charge effect.
Optical effect
When a relatively narrow beam of light is passed through a colloid, such dust particles in the air, the light is scattered by the dust particles and they appear in the beam a bright, tiny specks of light. Many of us have seen this phenomenon.
The scattering of light in a colloid is due to the reflection of the light by the large colloidal particles producing a visible beam of light. This optical effect is named the Tyndall effect after John Tyndall, who investigated it in 1860.
Motion effect
If a colloid is viewed with a special microscope, dispersed colloidal particles appear to move in a zigzag, random motion. This erratic random motion is due to bombardment of the dispersed colloidal particles by the medium and this constant bombardment keeps the dispersed colloidal particles suspended indefinitely. They will not settle down. This movement of colloidal particles is called Brownian movement, and this could be observed because of the Tyndall effect.
Electrical charge effect
Very often a colloid will have ions from the dispersing medium adsorbed on its surface. This along with the Brownian motion of the colloid, prevents colloids from coagulating and precipitating.
To facilitate precipitation, an electrolyte of some type is added to the solution, which will neutralize the charge. For instance, in the case of qualitative analysis, if charge on colloids is interfering with the precipitation, an electrolyte of some type is added. This property is used in removing suspended particles from the effluent gases in industrial smokestacks., In the gas coming of out various industrial equipment is given an electrical charge. Before this exhaust gas goes into atmosphere, it is passed through charged plates. These charged plates attract the charged colloidal particles, which then are held on the plates till the current giving the plates their charge is shut off. Then the particles fall to the ground and can be collected. The response of colloid particles to electrical charge is electrical phoresis.
Emulsions
Emulsions are sols of liquid in liquid. Two types of emulsions may be distinguished, namely, oil-in-water and water-in-oil. To make emulsions stable, emulsifying agents such as soaps and detergents are added.
Surfactants
Any substance which can decrease the surface tension of water to a large extent is known as surfactant. Examples of soap and detergents. Such substances have larger concentrations at the surface of water as compared to the bulk of the solution.
Micelle
From http://www.chemicool.com/definition/micelle.html
Definition of micelle
Surfactants in solution are often association colloids, that is, they tend to form aggregates of colloidal dimensions, which exist in equilibrium with the molecules or ions from which they are formed. Such aggregates are termed micelles. (IUPAC MANUAL APPENDIX II (1972)).
What is colloidal micelle?
The heart of this new chemistry is the technology used to develop a "colloidal micelle." Sub-microscopic particles are created in a microscopic field similar to a magnetic field. It differs from traditional chemistry in that the molecular attraction is not the usual attraction between positive and negative poles. Rather, it is between like poles.
An analogy would be that negative attracts negative and positive attracts positive. The micelle has a hydrophilic (water seeking) pole and a hydrophobic (water repelling) pole. The hydrophobic poles attract each other, thus forming the interior of the micelle. The hydrophilic poles form a tough outer surface.
When a micelle comes in contact with a hydrocarbon molecule, the center of the micelle bonds to a similar hydrophobic hydrocarbon. It disrupts the attraction to other hydrocarbon molecules and/or to the surface. Detergent particles in water form particles with hydrophobic poles and hydrophilic poles. In the case of clothes, the hydrophobic pole of a detergent particle attracts the grease and dirt particles on the clothes and pull them apart. As such micelles come together, an outlayer or hydrophilic poles forms. These associative colloid particles are then washed away by water and leaving clothes clean.
The action of a single micelle is multiplied by billions of other micelles. The molecular level emulsification process penetrates highly viscous and sticky materials, lifting them from the surface to which they adhere.
Considering the damage inflicted on life and the planet by harsh cleaners and solvents, colloidal chemistry is an exceptionally advance towards environmental preservation.
---------------------
JEE Question 2007 Paper I
Statement - 1
Micelles are formed by surfactant molecules above the critical micellar concentration (CMC)
Because
Statement - 2
The conductivity of a solution having surfactant molecules decreases sharply at the CMC.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Correct choice: B
---------------------------
Study Guide Ch.12. NUCLEAR CHEMISTRY
JEE Syllabus
Nuclear chemistry:
Radioactivity:
isotopes and isobars;
Properties of a, b and g rays;
Kinetics of radioactive decay (decay series excluded),
carbon dating;
Stability of nuclei with respect to proton-neutron ratio;
Brief discussion on fission and fusion reactions.
Syllabus rearranged
Radioactivity:
isotopes and isobars;
Properties of a, b and g rays;
Stability of nuclei with respect to proton-neutron ratio;
Kinetics of radioactive decay (decay series excluded),
Brief discussion on fission and fusion reactions.
carbon dating;
----------------
MAIN TOPICS IN TMH BOOK
STABILITY OF A NUCLEUS
SODDY-FAJAN GROUP DISPLACEMENT LAW
RADIOACTIVE DISINTEGRATION SERIES
KINETICS OF RADIOACTIVE DECAY
--------------
Radioactivity
The phenomenon of spontaneous emission ofactive radiations from certain substances is called radioactivity and the substances which emit such radiations are called radioactive substances.
In 1896, Henri Becquerel expanded the field of chemistry to include nuclear changes when he discovered that uranium emitted radiation.
Soon after Becquerel's discovery, Marie Curie began studying radioactivity and completed much of the pioneering work on nuclear changes. Curie found that radiation was proportional to the amount of radioactive element present, and she proposed that radiation was a property of atoms (as opposed to a chemical property of a compound).
Marie Curie was the first woman to win a Nobel Prize and the first person to win two (the first, shared with her husband Pierre and Becquerel for discovering radioactivity; the second for discovering the radioactive elements radium and polonium).
Isotopes and Isobars
Atoms of the same element having same atomic number but different mass numbers are called isotopes.
Ex: 92235U 92238U
The atoms of different elements having different atomic numbers but same mass numbers are called isobars.
Ex 1840Ar , 1940K , 2040Ca
Radiation and Nuclear Reactions
In 1902, Frederick Soddy proposed the theory that "radioactivity is the result of a natural change of an isotope of one element into an isotope of a different element." Nuclear reactions involve changes in particles in an atom's nucleus and thus cause a change in the atom itself. All elements heavier than bismuth (Bi) (and some lighter) exhibit natural radioactivity and thus can "decay" into lighter elements. Unlike normal chemical reactions that form molecules, nuclear reactions result in the transmutation of one element into a different isotope or a different element altogether (remember that the number of protons in an atom defines the element, so a change in protons results in a change in the atom).
There are three common types of radiation.Alpha Radiation (α),Beta Radiation (β),Gamma Radiation (γ).
Properties
Alpha Radiation (α) is the emission of an alpha particle from an atom's nucleus.
An α particle contains two protons and two neutrons (and is similar to a He nucleus: ).
When an atom emits an a particle, the atom's atomic mass will decrease by four units (because two protons and two neutrons are lost) and the atomic number (z) will decrease by two units.
The element is said to "transmute" into another element that is two z units smaller.
An example of an a transmutation takes place when uranium decays into the element thorium (Th) by emitting an alpha particle.
The nuclei of atoms represented by their atomic numbers and mass numbers are called nucleides.
For example 92235U
Note: in nuclear chemistry, element symbols are traditionally preceded by their atomic weight (upper left) and atomic number (lower left).
Beta Radiation (β) is the transmutation of a neutron into a proton and a electron (followed by the emission of the electron from the atom's nucleus:).
When an atom emits a β particle, the atom's mass will not change (since there is no change in the total number of nuclear particles), however the atomic number will increase by one (because the neutron transmutated into an additional proton).
An example of this is the decay of the isotope of carbon named carbon-14 into the element nitrogen:
Gamma Radiation (γ) involves the emission of electromagnetic energy (similar to light energy) from an atom's nucleus.
No particles are emitted during gamma radiation, and thus gamma radiation does not itself cause the transmutation of atoms, however γ radiation is often emitted during, and simultaneous to, α or β radioactive decay.
X-rays, emitted during the beta decay of cobalt-60, are a common example of gamma radiation.
Stability of nuclei with respect to proton-neutron ratio
It has been observed that the stability of nucleus depends upon the neutron to proton rati (n/P). In a plot of neutrons versus protons for nuclei of various elements, it has been observed that most th stable (non0radio active) elements lie in a belt, termed as stability belt or zone.
In the stability zone, for nuclei having atomic number up to 20, the n/p ratio is close to unity.
for nuclei having atomic number more than 20, the n/p ratio for stability exceeds unity and goes up to 1.5 for heavier nuclei.
Kinetics of Radioactive Decay
Half-Life Concept
Radioactive decay proceeds according to a principal called the half-life.
The half-life (T½) is the amount of time necessary for one-half of the radioactive material to decay.
For example, the radioactive element bismuth (210Bi) can undergo alpha decay to form the element thallium (206Tl) with a reaction half-life equal to five days.
If we begin an experiment starting with 100 g of bismuth in a sealed lead container, after five days we will have 50 g of bismuth and 50 g of thallium in the jar. After another five days (ten from the starting point), one-half of the remaining bismuth will decay and we will be left with 25 g of bismuth and 75 g of thallium in the jar. As illustrated, the reaction proceeds in halfs, with half of whatever is left of the radioactive element decaying every half-life period.
Stability of nuclei with respect to proton-neutron ratio
The stability of nucleus depends upon the neutron to proton ratio.
When the number of neutrons (n) are plotted against number of protons (p) for nuclei of various items, it has been observed that most of the stable (non-radio active) nuclei fall in a zone or belt. This zone is called stability zone or belt.
a. Fornuclei having atomic number up to 20 the stability belt(n/p ratio) is very close to one.
b. For nuclei having atomim number more than 20, the ratio increases up to 1.5 for heavier nuclei.
c. When the ratio is different from the stability ratio, the nuclei emit alpha or beta particles to move into stability zone.
Nuclear fission: reactions in which an atom's nucleus splits into smaller parts, releasing a large amount of energy in the process. Most commonly this is done by "firing" a neutron at the nucleus of an atom. The energy of the neutron "bullet" causes the target element to split into two (or more) elements that are lighter than the parent atom.
During the fission of U235, three neutrons are released in addition to the two daughter atoms. If these released neutrons collide with nearby U235 nuclei, they can stimulate the fission of these atoms and start a self-sustaining nuclear chain reaction. This chain reaction is the basis of nuclear power. As uranium atoms continue to split, a significant amount of energy is released from the reaction. The heat released during this reaction is harvested and used to generate electrical energy.
Nuclear fusion: reactions in which two or more elements "fuse" together to form one larger element, releasing energy in the process. A good example is the fusion of two "heavy" isotopes of hydrogen (deuterium: H2 and tritium: H3) into the element helium.
Fusion reactions release tremendous amounts of energy and are commonly referred to as thermonuclear reactions. Although many people think of the sun as a large fireball, the sun (and all stars) are actually enormous fusion reactors. Stars are primarily gigantic balls of hydrogen gas under tremendous pressure due to gravitational forces. Hydrogen molecules are fused into helium and heavier elements inside of stars, releasing energy that we receive as light and heat.
Carbon dating
Carbon dating is one of the uses of radio active isotopes. The technique will help to find out the age of archaeological objects (wood, plant, and animal fossils). the principle is based on the fact that all living matters contain a definite amount of radioactive isotope carbon 14.
Carbon 14is formed in the upper atmosphere by the bombardment of N-14 by cosmic rays. Some of it is present in the carbon dioxide. In the photosynthesis process, plant absorb some C-14. As the animals live on plants they acquire some C-14. A plant or any living being bduirng its life time maintains a reasonable balance of C 14 in its tissues. From the death of the plant, it will not absorb any more c-14. But C-14 starts decaying.
Therefore by having an estimate of C-14 that will be normally there in a tissue and by determining the amount of C-14 now in the dead tissue, the age of the dead tissue can be determined.
------------
web sites
The Physics of Radioactivity, Radioisotope uses
http://www.docbrown.info/page03/3_54radio.htm
http://www.visionlearning.com/library/module_viewer.php?mid=59
------------------
JEE Question 2007 paper II
A positron is emitted from [23,11]Na. The ratio of the atomic mass and atomic number of the resulting nuclide is
(A) 22/10
(B) 22/11
(C) 23/10
(D) 23/12
Solution: C
[23,11]Na --> [23,10]Ne + [0,+1]e
Hence ratio of atomic mass to atomic number of the resulting nuclide is 23/10
-------------------------------
Nuclear chemistry:
Radioactivity:
isotopes and isobars;
Properties of a, b and g rays;
Kinetics of radioactive decay (decay series excluded),
carbon dating;
Stability of nuclei with respect to proton-neutron ratio;
Brief discussion on fission and fusion reactions.
Syllabus rearranged
Radioactivity:
isotopes and isobars;
Properties of a, b and g rays;
Stability of nuclei with respect to proton-neutron ratio;
Kinetics of radioactive decay (decay series excluded),
Brief discussion on fission and fusion reactions.
carbon dating;
----------------
MAIN TOPICS IN TMH BOOK
STABILITY OF A NUCLEUS
SODDY-FAJAN GROUP DISPLACEMENT LAW
RADIOACTIVE DISINTEGRATION SERIES
KINETICS OF RADIOACTIVE DECAY
--------------
Radioactivity
The phenomenon of spontaneous emission ofactive radiations from certain substances is called radioactivity and the substances which emit such radiations are called radioactive substances.
In 1896, Henri Becquerel expanded the field of chemistry to include nuclear changes when he discovered that uranium emitted radiation.
Soon after Becquerel's discovery, Marie Curie began studying radioactivity and completed much of the pioneering work on nuclear changes. Curie found that radiation was proportional to the amount of radioactive element present, and she proposed that radiation was a property of atoms (as opposed to a chemical property of a compound).
Marie Curie was the first woman to win a Nobel Prize and the first person to win two (the first, shared with her husband Pierre and Becquerel for discovering radioactivity; the second for discovering the radioactive elements radium and polonium).
Isotopes and Isobars
Atoms of the same element having same atomic number but different mass numbers are called isotopes.
Ex: 92235U 92238U
The atoms of different elements having different atomic numbers but same mass numbers are called isobars.
Ex 1840Ar , 1940K , 2040Ca
Radiation and Nuclear Reactions
In 1902, Frederick Soddy proposed the theory that "radioactivity is the result of a natural change of an isotope of one element into an isotope of a different element." Nuclear reactions involve changes in particles in an atom's nucleus and thus cause a change in the atom itself. All elements heavier than bismuth (Bi) (and some lighter) exhibit natural radioactivity and thus can "decay" into lighter elements. Unlike normal chemical reactions that form molecules, nuclear reactions result in the transmutation of one element into a different isotope or a different element altogether (remember that the number of protons in an atom defines the element, so a change in protons results in a change in the atom).
There are three common types of radiation.Alpha Radiation (α),Beta Radiation (β),Gamma Radiation (γ).
Properties
Alpha Radiation (α) is the emission of an alpha particle from an atom's nucleus.
An α particle contains two protons and two neutrons (and is similar to a He nucleus: ).
When an atom emits an a particle, the atom's atomic mass will decrease by four units (because two protons and two neutrons are lost) and the atomic number (z) will decrease by two units.
The element is said to "transmute" into another element that is two z units smaller.
An example of an a transmutation takes place when uranium decays into the element thorium (Th) by emitting an alpha particle.
The nuclei of atoms represented by their atomic numbers and mass numbers are called nucleides.
For example 92235U
Note: in nuclear chemistry, element symbols are traditionally preceded by their atomic weight (upper left) and atomic number (lower left).
Beta Radiation (β) is the transmutation of a neutron into a proton and a electron (followed by the emission of the electron from the atom's nucleus:).
When an atom emits a β particle, the atom's mass will not change (since there is no change in the total number of nuclear particles), however the atomic number will increase by one (because the neutron transmutated into an additional proton).
An example of this is the decay of the isotope of carbon named carbon-14 into the element nitrogen:
Gamma Radiation (γ) involves the emission of electromagnetic energy (similar to light energy) from an atom's nucleus.
No particles are emitted during gamma radiation, and thus gamma radiation does not itself cause the transmutation of atoms, however γ radiation is often emitted during, and simultaneous to, α or β radioactive decay.
X-rays, emitted during the beta decay of cobalt-60, are a common example of gamma radiation.
Stability of nuclei with respect to proton-neutron ratio
It has been observed that the stability of nucleus depends upon the neutron to proton rati (n/P). In a plot of neutrons versus protons for nuclei of various elements, it has been observed that most th stable (non0radio active) elements lie in a belt, termed as stability belt or zone.
In the stability zone, for nuclei having atomic number up to 20, the n/p ratio is close to unity.
for nuclei having atomic number more than 20, the n/p ratio for stability exceeds unity and goes up to 1.5 for heavier nuclei.
Kinetics of Radioactive Decay
Half-Life Concept
Radioactive decay proceeds according to a principal called the half-life.
The half-life (T½) is the amount of time necessary for one-half of the radioactive material to decay.
For example, the radioactive element bismuth (210Bi) can undergo alpha decay to form the element thallium (206Tl) with a reaction half-life equal to five days.
If we begin an experiment starting with 100 g of bismuth in a sealed lead container, after five days we will have 50 g of bismuth and 50 g of thallium in the jar. After another five days (ten from the starting point), one-half of the remaining bismuth will decay and we will be left with 25 g of bismuth and 75 g of thallium in the jar. As illustrated, the reaction proceeds in halfs, with half of whatever is left of the radioactive element decaying every half-life period.
Stability of nuclei with respect to proton-neutron ratio
The stability of nucleus depends upon the neutron to proton ratio.
When the number of neutrons (n) are plotted against number of protons (p) for nuclei of various items, it has been observed that most of the stable (non-radio active) nuclei fall in a zone or belt. This zone is called stability zone or belt.
a. Fornuclei having atomic number up to 20 the stability belt(n/p ratio) is very close to one.
b. For nuclei having atomim number more than 20, the ratio increases up to 1.5 for heavier nuclei.
c. When the ratio is different from the stability ratio, the nuclei emit alpha or beta particles to move into stability zone.
Nuclear fission: reactions in which an atom's nucleus splits into smaller parts, releasing a large amount of energy in the process. Most commonly this is done by "firing" a neutron at the nucleus of an atom. The energy of the neutron "bullet" causes the target element to split into two (or more) elements that are lighter than the parent atom.
During the fission of U235, three neutrons are released in addition to the two daughter atoms. If these released neutrons collide with nearby U235 nuclei, they can stimulate the fission of these atoms and start a self-sustaining nuclear chain reaction. This chain reaction is the basis of nuclear power. As uranium atoms continue to split, a significant amount of energy is released from the reaction. The heat released during this reaction is harvested and used to generate electrical energy.
Nuclear fusion: reactions in which two or more elements "fuse" together to form one larger element, releasing energy in the process. A good example is the fusion of two "heavy" isotopes of hydrogen (deuterium: H2 and tritium: H3) into the element helium.
Fusion reactions release tremendous amounts of energy and are commonly referred to as thermonuclear reactions. Although many people think of the sun as a large fireball, the sun (and all stars) are actually enormous fusion reactors. Stars are primarily gigantic balls of hydrogen gas under tremendous pressure due to gravitational forces. Hydrogen molecules are fused into helium and heavier elements inside of stars, releasing energy that we receive as light and heat.
Carbon dating
Carbon dating is one of the uses of radio active isotopes. The technique will help to find out the age of archaeological objects (wood, plant, and animal fossils). the principle is based on the fact that all living matters contain a definite amount of radioactive isotope carbon 14.
Carbon 14is formed in the upper atmosphere by the bombardment of N-14 by cosmic rays. Some of it is present in the carbon dioxide. In the photosynthesis process, plant absorb some C-14. As the animals live on plants they acquire some C-14. A plant or any living being bduirng its life time maintains a reasonable balance of C 14 in its tissues. From the death of the plant, it will not absorb any more c-14. But C-14 starts decaying.
Therefore by having an estimate of C-14 that will be normally there in a tissue and by determining the amount of C-14 now in the dead tissue, the age of the dead tissue can be determined.
------------
web sites
The Physics of Radioactivity, Radioisotope uses
http://www.docbrown.info/page03/3_54radio.htm
http://www.visionlearning.com/library/module_viewer.php?mid=59
------------------
JEE Question 2007 paper II
A positron is emitted from [23,11]Na. The ratio of the atomic mass and atomic number of the resulting nuclide is
(A) 22/10
(B) 22/11
(C) 23/10
(D) 23/12
Solution: C
[23,11]Na --> [23,10]Ne + [0,+1]e
Hence ratio of atomic mass to atomic number of the resulting nuclide is 23/10
-------------------------------
Subscribe to:
Posts (Atom)