MINIMUM LEARNING MATERIAL(MLM) THE SOLID...

MINIMUM LEARNING MATERIAL(MLM)

THE SOLID STATE

TERMS EXPLANATIONS

Amorphous and Crystalline

Solids

Amorphous- short range order, Irregular

arrangement,isotropic,MP not sharp.

eg-glass

Crystalline Solids- long range order, regular

arrangement,anisotropic,MP sharp. Eg : NaCl

Unit Cell Smallest portion of space lattice when repeated generate the

entire lattice.

Molecular solids Constituent particles are molecules Eg Ar, CCl4, H2O (ice)

Covalent or Network solid Constituent particles are held together by co-valent bonds. Eg

SiO2,diamond

No of lattice points per unit

cell

Simple cubic -8, BCC- 9, FCC – 14 , End-Centred- 10

No of atoms per unit cell (z ) Simple cubic -1, BCC- 2, FCC – 4 , End-Centred- 2

Coordination Number FCC- 6:6 BCC- 8:8

Calculation of number of voids Let the number of close packed spheres be N, then:

The number of octahedral voids generated = N

The number of tetrahedral voids generated = 2N

Relation between r and a Simple Cubic 4r=2a , FCC 4r =√2 a BCC 4r =√3 a

Packing Efficiency Simple Cubic52.4% , BCC 68% , FCC 74%

Calculations Involving Unit Cell

Dimensions

d = density, Z = no. of atoms in one unit cell.

M=molar mass (g/mol) a = edge length in cm ,

NA = 6.023× 1023

Stoichiometric Defects (i) Vacancy Defect, (ii) Interstitial Defect, (iii) Frenkel

Defect, (iv) Schottky Defect

Impurity Defects If molten NaCl containing a little amount of SrCl2 is

crystallised, some of the sites of Na+ ions are occupied by

Sr2+ .

Non-Stoichiometric Defects (i) Metal Excess Defect due to anionic vacancies, Metal excess

defect due to the presence of extra cations, (ii) Metal

DAV CENTENARY PUBLIC SCHOOL, PASCHIM ENCLAVE, NEW DELHI - 87

Deficiency Defect,

Zinc oxide is white in colour at

room temperature. On heating

it turns yellow.

On heating it loses oxygen, The excess Zn2+ ions move to

interstitial sites and the electrons to neighbouring interstitial

sites

Frenkel Defect: Cation is dislocated to an interstitial site. It does not change

the density of the solid. Frenkel defect is shown by ionic

substance in which there is a large difference in the size of

ions. Eg ZnS, AgCl, AgBr and AgI due to small size of Zn2+ and

Ag+ ions.

Schottky Defect A vacancy defect. The number of missing cations and anions

are equal. Schotky defect is shown by ionic substance in which

there is a small difference in the size of ions.Density

decreases. Eg NaCl, KCl, CsCl and AgBr.

Metal excess defect due to

anionic vacancies (F-centres )

When NaCl heated in an atmosphere of Na vapour, the Na

atoms deposit on the surface of the crystal. The Cl– ions

diffuse to the surface of the crystal and combine with Na

atoms to give NaCl. The

released electrons diffuse into the crystal and occupy anionic

sites . The anionic sites occupied by unpaired electrons are

called F-centres .They impart yellow colour to the crystals of

NaCl. Similarly, excess of lithium makes LiCl crystals pink and

excess of potassium makes KCl crystals violet (or lilac).

Doping The conductivity of intrinsic semiconductors is increased by

adding an appropriate amount of suitable impurity. This

process is called doping.

n / p -type semiconductors n- type : Si (14th Gp) + As, Sb or Bi (15th Gp)

p-type: Si(14th Gp) + B ,Ga , In or Tl(13th Gp)

13 –15 compounds

12–16 compounds

13 – 15 compounds: InSb, AlP and GaAs.

12 – 16 compounds :ZnS, CdS, CdSe and HgTe

Paramagnetic substances Weakly attracted by a magnetic field. Atoms of these

substances have unpaired electrons. Examples: O2, Cu2+, Fe3+,

Cr3+

Diamagnetic substances Weakly repelled by a magnetic field. Atoms of these

substances have paired electrons.

Eg: H2O, NaCl and C6H6

Ferromagnetism: Strongly attracted by external magnetic field.Domains are

oriented in same direction resulting in combined magnetic

moment. Eg iron, cobalt, nickel, gadolinium and CrO2.They are

attracted very strongly by a magnetic field.


Antiferromagnetism Domains oppositely oriented and cancel out magnetic moment.

Eg MnO.

Ferrimagnetism: Domains aligned in parallel and anti-parallel directions in

unequal numbers.Lose ferrimagnetism on heating and become

paramagnetic. Example: Fe3O4 (magnetite)

Conductors solids with conductivities ranging from 104 to 107 S m–1.

valence band is only partially filled or it overlapsa vacant

conduction band of slightly higher energy. electrons can easily

flow under the influence of electric field

Insulators conductivities ranging from 10–20 to 10–10 S m–1. the gap

between the valence band and conduction band is large. Due to

this the valence electrons cannot jump to the conduction band

and conduct electricity.

Semiconductors intermediate conductivities ranging from 10–6 to 104 S m–1. the

gap between the valence band and nearest conduction band is

small. On applying the electric field, some electrons can jump

to the conduction band and provide low conductivity.

SOLUTIONS

Solid Solutions Gas in solid Solution of hydrogen in palladium

Liquid in solid Amalgam of mercury with sodium

Temp. Vs Conc.

Mass %, ppm, mole fraction and molality are independent of

temperature,

whereas molarity depends on temperature. This is because volume

depends on temperature.


Henry‘s law.

Mole fraction of gas in the solution is proportional to the partial

pressure of the gas

over the solution.

The partial pressure of the gas in vapour phase (p) is proportional to the

mole fraction

of the gas (x) in the solution.

p = KH . x

KH = Henry‘s law constant

( greater the KH value, lower the solubility.)

Different gases have different KH values at the same

temperature

Application of

Henry‘s law.

1.To increase the solubility of CO2 in soft drinks, the bottle is sealed

under high pressure.

2. To avoid bends, the tanks used by scuba divers are filled with air

diluted with helium.

Temp and Solubilty

of gas

Solubility of gas increases with decrease of temperature. It is due to

this reason that aquatic species are more comfortable in cold waters

rather than in warm waters.

Raoult‘s law for

volatile liquids

For a solution of volatile liquids,

The partial vapour pressure of each component in the solution is directly

proportional to its mole fraction. PA α xA pA = pA0 xA

Where pA0 is the vapour pressure of pure component A at the same

temperature.

Similarly, for component B

PB = pB0 xB

where pB0 represents the vapour pressure of the pure component B.

Ideal Solutions

The Ideal solutions

1.Obey Raoult‘s law over the entire range of concentration. 2.

3.

Example : Solution of n-hexane and n-heptane,

Non-ideal Solutions Positive deviation : A-B interactions are weaker than those between


A-A or B-B,

Eg - Mixtures of ethanol and acetone

Negative deviations : Forces between A-A and B-B are weaker than

those between A-B

Eg- mixture of phenol + aniline.

Mixture of chloroform +acetone

Azeotropes

Mixtures have same composition in liquid and vapour phase and boil at a

constant temp.

minimum boiling azeotrope(positive deviation)

eg- 95% aq ethanol

maximum boiling azeotrope(negative deviation)

eg- 68% aq nitric acid

Colligative properties

Properties that depend upon the number of particles of solute and not

on the nature of solute. e.g. Elavation of boiling point, depression of

fruzing point.

Relative Lowering of

Vapour Pressure

= XB

P0A = vapour pressure of solvent.

PA = vapour pressure of solution.

= lowering of vapour pressure

XB = mole fraction of solute.

Boiling Point and

Freezing Point

Boiling point of a liquid is the temperatures at which the vapour pressure

of the liquid becomes equal to the atmospheric pressure.

Freezing point is the temperature at which the solid and the liquid forms

of the substance have the same vapour pressure.

Elevation of Boiling

Point

ΔTb = elevation in boiling point,

Kb = molal elevation constant

Molal elevation constant is the elevation in boiling point when one mole of

solute is dissolved in one kilogram of the solvent.

MB = molecular weight of solute,

WB = weight of solute


WA = weight of solvent

m = molality

Depression of

Freezing Point

ΔTf = depression in freezing point,

Kf = molal depression constant

Molal depression constant or molal cryoscopic constant for the solvent

Kf may be defined as the depression in freezing point of a solution when

one mole of a solute is dissolved in 1 kilogram of the solvent.

MB = molecular weight of solute,

WB = weight of solute

WA = weight of solvent

m = molality

Osmosis Solvent flows through the semi permeable membrane from pure solvent

to the solution.

Osmotic pressure The extra pressure applied on the solution that just stops the flow of

solvent is called osmotic pressure of the solution

Osmotic pressure(π)

RT

R = gas constant

T = Temperature in Kelvin

RT , RT

where C is concentration of solution.

Isotonic solutions Two solutions having same osmotic pressure

Hypertonic Higher osmotic pressure than a particular soln

Hypotonic Lower osmotic pressure than a particular soln

Reverse Osmosis

The direction of osmosis can be reversed if a pressure larger than the

osmotic pressure is applied to the solution side. That is, now the pure

solvent flows out of the solution.

Application : Desalination of sea water

Abnormal Colligative

Properties.

Sometimes while measuring colligative properties abnormal results are

obtained due to the following reasons :

In case of association two or more solute molecules associate to form a


bigger molecule. The number of effective molecules in the solution,

therefore decreases.

Consequently, the value of the collgative property (relative lowering of

vapour pressure, elevation of boiling point, depression of freezing point,

osmotic pressure) is observed to be less than that calculated on the

basis of unassociated molecules.

In case of dissociation of the solute in the solution, the number of

effective solute particles increases. In such cases the value of the

observed collgative property will be greater than that calculated on the

basis of undissociated solute particles.

van‘t Hoff factor

( i )

ratio of normal molar mass to experimentally determined molar mass or

as the ratio of observed colligative property to the calculated colligative

property.

Value of van‘t Hoff

factor(i) for

Strong electrolytes

NaCl, KCl = 2 ; BaCl2 ,CaCl2 = 3 ; Na3PO4 = 4 ; Al2(SO4)3 , K4[Fe(CN)6]

= 5

CH3COOH ( in benzene) = ½

Value of van‘t Hoff

factor(i) for

weak electrolytes

Weak electrolytes dissociation

α = i-1/n-1

association

α = i-1/i/n-1

ELECTROCHEMISTRY

Cell Notation :Zn(s) | Zn2+ (aq) || Cu2+(aq)| Cu(s) ;

To remember LOAN : Left / Oxidation /Anode / Negative


An Ox and a Red Cat ( anode oxidation, cathode

reduction)

Nernst Eqn

EMn+/M = Eo Mn+ /M – log

Zn(s) | Zn2+ (aq) || Cu2+(aq)| Cu(s)

Zn + Cu2+ → Zn2+ + Cu

Ecell = Eo cell – log

Store CuSO4 in Zn

pot?

No. because the following reaction takes place

Zn(s) + CuSO4(aq) → ZnSO4 (aq) + Cu(s) ;

To find out whether the reaction takes place or not, find out the E0 value

of the cell

Zn/Zn2+//Cu2+/Cu

E0 = 0.34(red) – (-)0.76(Ox) = 1.1V

E0 value is positive, ie cell is feasible, reaction can take place.

Electrochemical

Series

Strength of

oxidising agent and

reducing agent.

F2(g) + 2e– → 2F– ; E0 = 2.87

F2(g) is the strongest oxidizing element

Li+ + e– → Li(s) ; E0 = –3.05

Li is the strongest reducing element


Relation between

E0 / Kc & ΔG

ΔrG = – nFE(cell)

ΔrG0 = –nFEo (cell)

ΔrG0 = –2.303RT log K.

R, G, κ, G*

R = Resistance, ρ = resistivity l = length a = area of cross section.

G = conductance

κ = conductivity

Units:- ρ(resistivity) = ohm metre ( m) OR ohm centimetre ( cm)

1 m = 100 cm or 1 cm = 0.01 m

G*= cell constant

Molar Conductivity

(Λm)

Molar conductivity = Λm =


Kohlrausch

law

Limiting molar conductivity of an electrolyte can be represented as the

sum of the individual contributions of the anion and cation of the

electrolyte.

α & Ka Where α is degree of dissociation.

Faraday‘s 1st Laws

The amount of chemical reaction which occurs at any electrode during

electrolysis is proportional to the quantity of electricity passed through

the electrolyte (solution or melt).

Faraday‘s 2nd Laws

The amounts of different substances liberated by the same quantity of

electricity passing through the electrolytic solution are proportional to

their chemical equivalent weights (Atomic Mass of Metal ÷ Number of

electrons required to reduce the cation).

Electrolysis of

NaCl (molten) Cathode : Na+(l) + e– → Na(s)

Anode : Cl–→ ½Cl2+e–

NaCl (aq) Cathode : H2O (l ) + e– → ½H2(g) + OH–

Anode : Cl–→ ½Cl2+e–

H2SO4(dil) Cathode : H+ + e- ½ H2

Anode: 2H2O(l )→ O2(g) + 4H+(aq) + 4e–

H2SO4(conc) Cathode : H+ + e- ½ H2

Anode: 2SO4 2– (aq) → S2O8

2– (aq) + 2e–

AgNO3(aq)-

Ag electrodes

Cathode : Ag+(aq) + e- Ag(s)

Anode: Ag(s) Ag+(aq) + e-

AgNO3(aq)-

Pt electrodes

Cathode : Ag+(aq) + e- Ag(s)

Anode: 2H2O(l )→ O2(g) + 4H+(aq) + 4e–


CuCl(aq)-

Pt electrodes

Cathode : Cu+(aq) + e- Cu(s)

Anode: 2H2O(l )→ O2(g) + 4H+(aq) + 4e–

Pri & Sec Batteries

In the primary batteries, the reaction occurs only once, and cannot be

reused.

Dry Cell

Anode(Zn) : Zn(s) → Zn2+ + 2e–

Cathode(Graphite) : MnO2 + NH4+ + e– → MnO(OH) + NH3 The emf =

1.5 V.

Mercury Cell

Anode(Zn-Hg) : Zn(Hg) + 2OH– → ZnO(s) + H2O + 2e–

Cathode(HgO-C) : HgO + H2O + 2e– → Hg(l ) + 2OH–

EMF= 1.35 V and remains constant during its life as the overall reaction

does not involve any ion in solution whose concentration can change during

its life time.

Lead Storage

Battery

Anode : Pb Cathode : Pb packed with PbO2 Electrolyte : 38%

H2SO4

Anode: Pb(s) + SO4 2–(aq) → PbSO4(s) + 2e–

Cathode: PbO2(s) + SO4 2–(aq) + 4H+(aq) + 2e– → PbSO4 (s) + 2H2O (l )

On charging the battery the reaction is reversed

Fuel Cells

Anode&Cathode : Porous C Electrolyte : Aq NaOH

Cathode: O2(g) + 2H2O(l ) + 4e– 4OH–(aq)

Anode: 2H2 (g) + 4OH–(aq) 4H2O(l) + 4e–

The cell runs continuously as long as the reactants are supplied. Efficiency

of about 70 %. Pollution free. The water vapours produced during the

reaction were condensed and added to the

drinking water supply for the astronauts (Apollo space programme)

Corrosion of Iron

(Rusting)

Oxidation: Fe (s)→ Fe2+ (aq) +2e–

Reduction: O2 (g) + 4H+(aq) +4e– → 2H2O(l)

Atomospheric oxidation :

2Fe2+(aq) + 2H2O(l) + ½O2(g) → Fe2O3(s) + 4H+(aq)

Prevention of

Corrosion

By covering the surface with paint or by some chemicals (e.g. bisphenol). /

Cover the surface by other metals (Sn, Zn, etc.) that are inert or react to

save the object. An electrochemical method (sacrificial electrode like Mg,

Zn, etc.) which corrodes itself but saves the object.

CHEMICAL KINETICS

Rate of a Chemical

Reaction

Change in concentration of a reactant or product in unit time. Unit

=> mol L-1s–1


For RP

Average &

Instantaneous Rate

For 2HI(g) → H2(g) +

I2(g)

Rate of reaction =

Rate law / Rate

equation /

Rate expression

Rate law is the expression in which reaction rate is given in terms of

molar concentration of reactants with each term raised to some

power, which may

or may not be same as the stoichiometric coefficient of the reacting

species in a balanced chemical equation.

For

aA + bB → cC + Dd

The rate expression for this reaction is :

Rate α [A]x [B]y => Rate = k [A]x [B]y

k is the Rate const. (x , y from expts)

Order of a Reaction

(Has to get from expt.

Not from balanced

equation.)

The sum of powers of the concentration of the reactants in the rate

law expression is called the order of that chemical reaction.

If Rate = k [A]x [B]y ( x and y found out experimentally)

Order = x + y

Elementary reactions The reactions taking place in one step are called Elementary

Reactions.

Unit of rate constant

For n th order reaction : mol1-n Ln-1 s-1

Molecularity of a

Reaction

The number of reacting species taking part in an elementary reaction,

which must collide simultaneously in order to bring about a chemical

reaction is called molecularity of a reaction (values are limited from

1 to 3)

Rate Determining Step The overall rate of the reaction is controlled by the slowest step in a

reaction called the rate determining step

Half-life (t1/2) The half-life of a reaction is the time in which the concentration of a


reactant is reduced to one half of its initial concentration. It is

represented as t1/2.

Zero Order Reactions

(Integrated Rate

Equation)

First Order Reactions

(Integrated Rate

Equation)

Dependence of t1/2 on

[R]0

For zero order : t1/2 α [R]0.

For first order reaction t1/2 is independent of [R]0

Example of Zero Order

Reactions

The decomposition of gaseous ammonia on a hot platinum surface is a

zero order reaction at high pressure.

Rate = k [NH3]0 = k

Radioactive decay All natural and artificial radioactive decay of unstable nuclei take

place by first order kinetics

Pseudo First Order

Reaction

When one of two reactants is present in large excess. During the

hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water The

concentration of water does not get altered much during the course

of the reaction and the reaction behaves as first order reaction.

Such reactions are called pseudo first order reactions.

Activation Energy Activation energy is given by the energy difference between

activated complex and the reactant molecules.

Arrhenius equation

k = A e -Ea /RT (exponential form)

ln k = – Ea/RT + ln A (logarithmic form)

A is the Arrhenius factor or the frequency factor or pre-

exponential factor

e -Ea /RT The factor e -Ea /RT corresponds to the fraction of molecules that

have kinetic energy greater than Ea

If k1 and k2 are rate

constants at

T1 and T2

Action of catalyst Catalyst provides an alternate pathway or reaction mechanism by

reducing the activation energy between reactants and products and


hence lowering the potential energy barrier.

Collision frequency (Z)

The number of collisions per second per unit volume of the reaction

mixture.

For A + B → Products Rate = P.ZAB e−Ea/ RT

P is probability or steric factor ZAB is collision frequency of

reactants, A and B.

SURFACE CHEMISTRY

Adsorption Accumulation of molecular species at the surface rather

than in the bulk of a solid or liquid.

Adsorbate & Adsorbent. The substance adsorbed is called adsorbate. And the

surface provider is called adsorbent.

Desorption. The process of removing an adsorbed substance from a

surface.

Adsorption and Absorption In adsorption, the substance is concentrated only at the

surface while in absorption, the substance is uniformly

distributed throughout the bulk of the solid.

Sorption Both adsorption and absorption can take place

simultaneously.

Enthalpy of adsorption Adsorption is exothermic ΔH is -ve.

There is decrease in entropy, ΔS is -ve.

For a process to be spontaneous, ΔG must be negative,

so ΔH should have sufficiently high-ve value.

Surface area & Adsorption Adsorption increases with the increase of surface area

of the adsorbent. Thus, finely

divided metals and porous substances having large

surface areas are good adsorbents.

Physisorption van der Waals‘ forces / not specific /reversible /easily

liquefiable gases adsorbed readily./ Enthalpy of

adsorption is low/ Low temperature is favourable /

Multimolecular layers.

Chemisorption chemical bond /highly specific / irreversible/Enthalpy

of adsorption high / High temperature is favourable /

unimolecular layer.

Adsorption Isotherm Variation of amount of gas adsorbed by the adsorbent

with pressure at constant temperature can be

expressed by means of a curve termed as adsorption

isotherm.

Freundlich isotherm X / m = k.P1/n (n > 1)


log x/m = log k + 1/n log P

Applications of Adsorption High vacuum/masks/Control of humidity/ Removal of

colour/ catalysis/ curing diseases/

Homogeneous catalysis Reactants and the catalyst are in the same phase.eg

Lead Chamber Process of manufacture os sulphuric acid.

Heterogeneous catalysis Reactants and the catalyst are in different phases.eg

contact process of manufacture of sulphuric acid.

Activity of a solid catalyst Depends upon strength of chemisorption Reactants must

get adsorbed reasonably strongly.

Selectivity of a catalyst The selectivity of a catalyst is its ability to direct a

reaction to yield a particular product.

Promoters and poisons Promoters are substances that enhance the activity of a

catalyst while poisons decrease the activity of a

catalyst. For example, in Haber‘s process for

manufacture of ammonia, molybdenum acts as a

promoterfor iron which is used as a catalyst.

Shape- Selective Catalysis by

Zeolites

Depends upon the pore structure of the catalyst and the

size of the reactant and product molecules. Example:-

ZSM-5 (a zeolite) converts alcohols directly into

gasoline (petrol).

Characteristics of enzyme catalysis Highly efficient/ Highly specific / active under optimum

temperature& pH/ Increasing activity in presence of

activators and co-enzymes/ Influence of inhibitors and

poisons

Catalysts used in industrial process. Haber‘s process (NH3) Finely divided Fe, Mo -

promoter; Ostwald‘s process(HNO3) Pt

Contact process(H2SO4) Pt or vanadium pentoxide


(V2O5);

Colloids A colloid is a heterogeneous system in which one

substance is

dispersed (dispersed phase) as very fine particles in

another substance

called dispersion medium.

Diameters of the colloid particles is between 1 and

1000 nm.

Lyophilic colloids Liquid-loving/ can be reconstituted/ reversible/easily

prepared/stable/ eg- gum, starch.

Lyophobic colloids Liquid-hating/ not be reconstituted/ irreversible/

prepared by special methods/ precipitated on addition

of electrolytes/heating/shaking/ not stable./ eg –metal

sols , sulphide sols.

Multimolecular colloids Large number of atoms or molecules of a substance

aggregate eg- gold sol , sulphur sol.

Macromolecular colloids Macromolecules in suitable solvents. Eg-starch, cellulose,

proteins and enzymes; polythene.

Associated colloids (Micelles) Substances at low conc behave as strong electrolytes,

but at high conc exhibit colloidal behaviour due to the

formation of aggregates(micelles.) eg- soaps and

detergents.

Kraft Temperature (Tk) Formation of micelles takes place only above a particular

temperature.

CMC Formation of micelles takes place only above a particular

conc.- critical micelle concentration.

For soaps, the CMC is 10-3mol/l to 10-4 mol/l

Cleansing action of soaps Soap molecules form micelle around the oil droplet in

such a way that hydrophobic part of the stearate ions is

in the oil droplet and hydrophilic part projects out of

the grease droplet like the bristles . Since the polar

groups can interact with water, the oil droplet

surrounded by stearate ions is now pulled in water and

removed from the dirty surface.

Electrical disintegration or

Bredig‘sArc method

This process involves dispersion as well as condensation.

Colloidal sols of metals such as gold, silver, platinum,

etc., can be prepared.

Electric arc is struck between

electrodes of the metal immersed in the dispersion

medium


The intense heat produced vapourises the metal,

which then condenses to form particles of colloidal size.

Peptization Process of converting a ppt into colloid by shaking it

with dispersion medium in the presence of a small

amount of electrolyte. The electrolyte used for this

purpose is called peptizing agent.

Dialysis: Removal of dissolved substance from colloidal by

diffusion through a suitable membrane.

Tyndall effect Colloidal particles scatter light in all directions.

Scattering of light illuminates the path of beam of the

light.

Conditions for Tyndall effect (i) Diameter of particles is not much smaller than

wavelength of the light used;

(ii) Refractive indices of the phase and medium

differ greatly in magnitude

Charge on colloidal particles +ve charged sols Hydrated metallic oxides e.g.,

Al2O3.xH2O, CrO3.xH2O andFe2O3.xH2O,

etc.,/methylene blue sol. /Haemoglobin (blood)

-ve charged sols Metals,/ sulphides e.g., As2S3,

Sb2S3, CdS sols. / eosin, congo red / starch, gum,

gelatin, clay,charcoal.

Origin of charge on colloidal particles Preferential adsorption of ions . The sol particles

acquire positive or negative charge by preferential

adsorption of +ve or –ve ions. When two or more ions are

present in the dispersion medium, preferential

adsorption of the ion common to the colloidal particle

usually takes place.

When silver nitrate solution is added to potassium iodide

solution,the precipitated silver iodide adsorbs iodide

ions from the dispersion medium and negatively charged

colloidal solution results. AgI/I– , Fe2O3.xH2O/Fe3+

However,when KI solution is added to AgNO3 solution,

positively charged sol results due to adsorption of Ag+

ions from dispersion medium. AgI/Ag+

Zeta potential The combination of the two layers of opposite charges

around the colloidal particle is called Helmholtz

electrical double layer.

Potential difference between the fixed layer and the

diffused layer is called Zeta Potential.

Electrophoresis The movement of colloidal particles under an electric


potential is called electrophoresis.

Electroosmosis If electrophoresis is prevented , it is observed that the

dispersion medium begins to move in an electric field.

Coagulation(ppt) The stability of the lyophobic sols is due to the

presence of charge on colloidal particles. If, somehow,

the charge is removed, the particles will come nearer to

each other to form aggregates (or coagulate) and settle

down under the force of gravity.

The process of settling of colloidal particles is called

coagulation or precipitation of the sol.

Reason of Coagulation Electrophoresis/mix oppositely charged

sols/boiling/persistent dialysis/addition of electrolytes.

Hardy-Schulze rule The greater the valence of the flocculating ion added,

the greater is its power to cause precipitation.

For coagulation of a negative sol, : Al3+>Ba2+>Na+

For coagulation of a negative sol :[Fe(CN)6]4– > PO4

3– >

SO4 2– > Cl–

Coagulating value Min. conc of an electrolyte in millimoles/L required to

cause precipitation of a sol in 2 hours

Emulsions Liquid-liquid colloida (i) Oil dispersed in water (O/W

type) - milk and vanishing cream .

(ii) Water dispersed in oil (W/O type).- butter and

cream.

Demulsification Emulsions can be broken into constituent liquids by

heating, freezing, centrifuging, etc.

Artificial Rain By throwing electrified sand , spraying a sol of opposite

charge to the clouds from aeroplane

Formation of Delta River water is a colloidal solution of clay. Sea water

contains a number of electrolytes. When river water

meets the sea water, the electrolytes present in sea

water coagulate the colloidal solution of clay resulting in

its deposition.

Electrical precipitation of

smoke(Cottrellppter)

The smoke, before it comes out from the chimney, is led

through a chamber containing plates having a charge

opposite to that carried by smoke particles. The

particles on coming in contact with these plates lose

their charge and get precipitated.


GENERAL PRINCIPLES AND PROCESS OF EXTRACTION OF ELEMENTS

Mineral It is the combined state in which a metal occurs naturally in the crust of the

earth.

Ore The mineral from which a metal can be profitably and conveniently extracted

is called ore.i.e. All ores are minerals but all minerals are not ores.

Gangue / Matrix: Earthly or undesired materials present in mineral are called gangue

Metallurgy:

The entire scientific and technological process used for isolation of the

metal from its ores is known as metallurgy.

Process of

Extraction

Ore

Pow

dered O

re

Concentrated O

re

Metal

oxid

e

Crude m

etal

Pure

metal

Crushing and grinding

-Hydraulic washing-Magnetic separation-Froth Floatation

-Leaching

-Calcination

-Roasting

-Reduction with ' C '-Reduction with ' CO '-Reduction with ' Al '

-Electolitic Reduction

-Distillation

- Liquation

- Electrolysis- Zone refining- Vapour phase refining

- Chromatographic method

Concentration of ore

Conversion to oxide Reduction Refining

Hydraulic

washing:

In this method, the light (low specific gravity) earthy impurities are removed

from the heavier metallic ore particles by washing with water. It is

therefore, used for the concentration of heavier oxide ores, like haematite

(Fe2O3 ) tinstone (SnO2 ) and gold (Au).

Magnetic

separation

This is based on difference in magnetic properties of the ore component.

Froth floatation

method

It is based on the difference in the wetting qualities of the gangue and

sulfide ore particles with water and oil. Where as the ore particles are

wetted by oil, the gangue particles are wetted by water.

Collectors enhance non-wet ability of the mineral particles. e.g. Pine oil, fatty acid,

xanthates, etc

Froth

stabilizers

stabilizes the forth e.g. Cresol, aniline, etc.

Depressants are reagents used to suppress or depress the formation of forth.

For example, ZnS and PbS can be separated from each other in the forth

floatation process by the addition of small amounts of NaCN which act as

depresent.

NaCN selectively prevents ZnS from coming to the froth by forming soluble

complex But allows PbS to come with the froth.

Leaching Leaching of alumina from bauxite bauxite, usually contains SiO2, iron oxides and titanium oxide (TiO2) as


impurities. Concentration

is carried out by digesting the powdered ore with a concentrated

solution of NaOH Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2Na[Al(OH)4](aq)

2Na[Al(OH)4](aq) + CO2(g) Al2O3.xH2O(s) + 2NaHCO3 (aq)

Al2O3.xH2O(s) Al2O3(s) + xH2O(g)

Leaching of Silver and Gold Leaching done with a dilute solution of NaCN or KCN in the presence of

air (for O2) from which the metal is obtained later by replacement.

4M(s) + 8CN-(aq)+ 2H2O(aq) + O2(g) 4[M(CN)2]-(aq) + 4OH-(aq) (M= Ag or

Au)

Calcination and

Roasting.

Hydro

metallurgy

Calcinaton involves heating when the volatile matter escapes leaving behind

the metal oxide.

In roasting, the ore is heated in a regular supply of air in a furnace at a

temperature below the melting point of the metal.

If the ore contains iron, it is mixed with silica before heating.Iron oxide

‗slags of ‘* as iron silicate and

copper is produced in the form of copper matte which contains Cu2S and

FeS.

flux ‘ is added which combines with ‗ gangue ‗ to form ‗ slag ‗.

Slag separates more easily from the ore than the gangue.

Copper is extracted by hydrometallurgy from low grade ores. It is leached

out using acid or bacteria. The solution containing Cu2+ is treated with scrap


iron or H2

To displace Zn from solution of Zn2+ ions, we need more reactive elements

such as Al, Mg, Ca and K.

But all these metals react with water forming their corresponding ions with

the evolution of H2 gas.

Thus Al, Mg, etc. cannot be used to displace Zn from solution of Zn2+ ions.

Thus, copper can be extracted by hydrometallurgy but not zinc.

Electro

Metallurgy

[For highly

reactive metals]

Highly reactive metals can be isolated from their ores by the process of

electrometallurgy.

The strongest possible reducing agent is an electron; any ionic material can

be reduced to their respective metal by electrolysis.

Pyro

metallurgy [For

moderately

reactive metals]

The process of converting metal oxide in to metal upon strong heating with a

suitable reducing agent is known as pyro metallurgy.

Thermite Reaction:- Reduction of metal oxide with Aluminium .

Cr2O3 + 2Al Al2O3 + 2Cr

Smelting:- Reduction of metal oxide with coke.

Thermo

Metallurgy

[For least

reactive metals]

The metals can be isolated by simply heating the oxide of the metal.

2HgO 2Hg + O2

Ag2O 2Ag + ½ O2

Thermo

dynamic

Principles of

Metallurgy

When the value of ∆G is negative only then the reaction will proceed. If ∆S

is positive, on increasing the temperature (T), the value of T∆S would

increase (∆H < T∆S) and then ∆G will become –ve.

If reactants and products of two reactions are put together (coupling)in a

system and the net ∆G of the two possible reactions is –ve, the overall

reaction will occur. The graphical representation of Gibbs energy was first

used by Ellingham.

This provides a sound basis for considering the choice of reducing agent in

the reduction of oxides. This is known as Ellingham Diagram.

Extraction of

iron from its

oxides

Oxide ores of iron, after concentration through calcination/roasting are

mixed with limestone and coke and fed into a Blast furnace from its top.

Here, the oxide is reduced to the metal.


It can be seen as a couple of two simpler reactions.

At temperatures above 1073K , coke will be reducing the FeO and will itself

be oxidised to CO.

Methods of

refining.

(a) Distillation (b) Liquation

(c) Electrolysis (d) Zone refining

(e) Vapour phase refining (f ) Chromatographic methods

Distillation This method is employed for purification of volatile metals ( Zn, Hg, Cd ) by

heating the ore followed by condensation

Liquation : This method is used when the M.P of the metals are lower than those of the

impurities.

e.g. low melting metal – Sn

Electrolytic

refining

( Zn, Cu )

Anode : Impure metal ( Cu ) Cathode: Pure metal ( Cu )

Electrolyte : Soluble salt of same metal

( acidified CuSO4 solution )

The more basic metal remains in the solution and the less basic ones go to

the anode mud.

Copper is refined using an electrolytic method

Impurities from the blister copper deposit as anode mud which contains

antimony, selenium, tellurium, silver, gold and platinum.

Zone refining Principle-The impurities are more soluble in the melt than in the solid state

of the metal.

A circular mobile heater is fixed at one end of a rod of the impure metal.

The molten zone moves along with the heater which is moved forward. As the

heater moves forward, the pure metal crystallises out of the melt and the

impurities pass on into the adjacent molten zone.

At one end, impurities get concentrated. This end is cut off. This method is

very useful for producing semiconductor and other metals of very high

purity, e.g., germanium, silicon, boron,gallium and indium.

Vapour phase refining

Principle:-The metal is converted into its volatile compound and collected

elsewhere. It is then decomposed to give pure metal.

Eg


Mond Process for Refining Nickel: In this process, nickel is heated in a

stream of carbon monoxide forming a volatile complex, nickel

tetracarbonyl:

van Arkel Method for Refining Zirconium or Titanium The crude metal is heated in an evacuated vessel with iodine. The metal

iodide being more covalent, volatilises.

The metal iodide is decomposed on a tungsten filament, electrically heated

to about 1800K

Chromato graphic methods

This method is based on the principle that different components of a

mixture are differently adsorbed on an adsorbent. The mixture is put in a

liquid or gaseous medium which is moved through the adsorbent.

Aluminium In the metallurgy of aluminium, purified Al2O3 is mixed with Na3AlF6 or CaF2

which lowers the melting point of the mix and brings conductivity.

The fused matrix is electrolysed. Steel cathode and graphite anode are used.

The graphite anode is useful here for reduction to the metal.

The oxygen liberated at anode reacts with the carbon of anode producing CO

and CO2.


Reactions taking

place in Blast

Furnace.

How is wrought

iron or malleable

iron prepared?

Wrought iron or malleable iron is the purest form of commercial iron and is

prepared from cast iron by oxidising impurities in a reverberatory furnace

lined with haematite.

Extraction of

copper from

cuprous oxide

[copper(I)

oxide]

Blister Copper

The sulphide ores are roasted/smelted to give oxides:

The ore is heated in a reverberatory furnace after mixing with silica. In the

furnace, iron oxide ‗slags of‘ as iron silicate and copper is produced in the

form of copper matte. This contains Cu2S and FeS.

The solidified copper obtained has blistered appearance due to the

evolution of SO2 and so it is called blister copper.

Extraction of

zinc from zinc

oxide

The reduction of zinc oxide is done using coke.


THE P-BLOCK ELEMENTS

Structures of Oxides of

Nitrogen

Phosphorus White phosphorus Red phosphorus


Oxoacids of Phosphorus

The structures of some

oxoacids of P

Rhombic sulphur (αsulphur)

Monoclinic sulphur

(βsulphur)


Oxoacids of

Sulphur

Oxoacids of

Halogens

Structures of oxoacids of

Halogens.

BrF3

Xenon-oxygen compounds

Why is Bi (V) a stronger

oxidant than Sb (V)?

Because Bi (III) is more stable than Sb (III) due to inert pair

effect.

Why is red phosphorus less

reactive than white

phosphorus?

This is due to polymeric structure of red phosphorus or angular

strain in P4 molecule of white phosphorus where the angle is only

600


The maximum number of

covalent bonds formed by

nitrogen is 4. Why

Nitrogen has three unpaired electrons and one lone pair of

electrons; therefore, it can form three covalent bonds and one

co-ordinate bond.

Give reasons for the least

reactivity of nitrogen

molecule?

Due to presence of a triple bond between the two N-atoms, the

bond dissociation enthalpy (941.4 KJ mol-1) is very high. Hence N2

is least reactive.

Though nitrogen exhibits +5

oxidation state, it does not

form pentahalide. Give

reason.

Nitrogen does not have d orbitals to expand its covalence beyond

four. That is why it does not form pentahalide

In trimethyl amine, the

nitrogen has a pyramidal

geometry whereas in

trisilylamine N (SiH3)3 ,it

has a planar geometry.

(CH3)3 N is pyramidal due to sp3 hybridization and has a lone pair

of electrons. (SiH3)3 N has sp2 hybridization because lone pair of

nitrogen is donated to vacant d-orbitalof Si.

Why is NH3 a good

complexing agent?

NH3 is a good complexing agent because nitrogen has a lone pair

of electrons which it can donate to form co-ordinate bond.

On being slowly passed

through water, PH3 forms

bubbles but NH3 dissolves.

Why is it so?

N-H bond is more polar than P-H bond. Hence, NH3 forms

hydrogen bonds with H2 Omolecules and hence dissolves in it

whereas PH3 does not dissolve and forms bubbles.

Why does NO2 dimerise? NO2 contains odd number of valence electrons. On dimerisation,

it is converted to a stable N2O4 molecule with even number of

electrons.

Why does PCI3 fume in

moisture?

PCI3 hydrolysis in the presene of moisture giving fumes of HCl.

Why is nitrous acid oxidant

as well as reductant?

The oxidation state of N in nitrous acid (H-O-N) is +3 whick lies

in between its lowest oxidation state of -3 and highest oxidation

state of +5. Since the oxidation state of N in (HNO2) can be

decreased from +3 to any lower value, therefore, it acts as an

oxidizing agent.

Further since the oxidation state of N in HNO2 can be

increased from +3 to +4 or +5, therefore, it acts as a reducing

agent. Thus, nitrous acid acts both as an oxidant as well as a

reductant.

Why are the Group 16 Chalcogens means ore forming. The elements of Group 16 are


elements called chalcogens? called chalcogens because many metals are found as oxides and

sulphides and a few as selenides and tellurides

Oxygen molecule has

formula O2 while sulphur S8-

Why?

Oxygen atom being small in size form multiple bond while S atom

being large in size form single bond with other S atom, the

puckered ring structure S8 is

Sulphur disappears when

boiled with sodium sulphite

solution. Why?

When sodium sulphite is heated with sulphur, we get sodium

thiosulphate which is soluble in water that is why sulphur

disappears.

Na2SO3 +S heat Na2S2O3

Why is I2 more soluble in KI

than in water?

It is due to formation of soluble complex KI3.

I2+KI KI3.

Write an example of a

neutral molecule which is

iso-electronic with CIO-.

OF2 and CIF are iso-electronic with CIO-.

Why is F2O referred to as a

fluoride but CI2O is an

oxide?

F2 Ois called oxygen fluoride because fluorine is more

electronegative than oxygen whereas CI2O is called chlorine

oxide because oxygen is more electronegative than chlorine

Noble gases have largest

radii. Explain.

In noble gases, we can measure only van der Waals‘ radii which

are larger than covalent radii

How does xenon atom form

compounds with fluorine

even though the xenon atom

has a closed shell

configuration?

This is because 1, 2 or 3 electrons from the 5p-orbitals can be

excited to empty 5d-orbitals thus making 2, 4 or 6 half-filled

orbitals available for bond formation

What prompted Bartlett to

prepare first noble gas

compounds.

Barlett observed ionization of energy of O2 is 1180 kJ mol-1

whereas that of Xe is 1170 kJ mol-1.

He could prepare O2 pt F6 which prompted him to prepare

first noble gas compound Xe+ [Pt F6].

Noble gases have very low

boiling points. Why?

Noble gases being monoatomic have no interatomic forces except

weak dispersion forces and therefore, they are liquefied at very

low temperatures. Hence, they have low boiling points.

acidity of oxo acid of

chlorine is

HOCI<HOCIO<HOCIO2<HO

CIO3

Higher the oxidation state of chlorine in oxo acid, stronger the

acid.

Chlorine water has both Chlorine water produces nascent oxygen which is responsible for


oxidizing and bleaching

properties

bleaching action and oxidation.

C12 + H2O 2HC1+ [O]

H3PO2 and H3PO3 act as as

good reducing agents while

H3PO4 does not.

Both H3 PO2 and H3PO3 have P-H bonds, so they act as reducing

agents, but H3PO4, has no P-H bonds, so it cannot act as a

reducing agent.

ICI is more reactive than I2 Because ICI has less bond dissociation enthalpy than I2.

NH3 is a stronger base than

PH3

This is because the lone pair of electrons on N atom in NH3 is

directed and not diffused as it is in PH3 due to larger size of

phosphorus and hence more available for donation.

Sulphur has a greater

tendency for catenation

than oxygen.

Sulphur has a greater tendency for catenation than oxygen

because W-S bond is stronger than O-O bond due to less

interelectronic repulsions.

Bond dissociation energy of

F2 is less than that of CI2

This is due to relatively large electron- electron repulsion among

the lone pairs in F2 molecule where they are much closer to each

other than in case of CI2.

SF4 is easily hydrolysed

whereas SF6 is not easily

hydrolysed.

S atom in SF4 is not sterically protected as it is surrounded by

only four F atoms, so attack of H2O molecules can take place

easily and hence hydrolysis takes place easily. In contract, in SF6,

S is sterically protected by six F atoms. Therefore does not allow

H2O molecules to attack S atoms. As a result of this, SF6 does

not undergo hydrolysis.

XeF2 has a straight linear

structure and not a bent

angular structure

In Xe F2, Xe is sp3d hybridized having 2 bond pair and 3 lone pair

of electrons. The presence of 3 lone of electrons in Xe F2 at

equidistance to have minimum repulsion is responsible for its

linear structure.

NCl3 gets readily

hydrolysed while NF3 does

not.

In NCI3, CI has vacant d-orbitals to accept the lone pair of

electrons donated by O-atom of H2O molecule but in NF3, F does

not have d-orbitals

Elemental nitrogen exists as

a diatomic molecule whereas

elemental phosphorus is a

tetraatomic molecule.

Nitrogen because of its small size forms pπ- pπ multiple bonds

with other nitrogen atoms and thus it exists as a diatomic

molecule (N=N or N2 ). Phosphorus, on the other hand, because of

its large size usually does not form pπ- pπ multiple bonds with

other phosphorus atoms but instead forms single bonds.

Consequently, it exists as a stable tetraatomic (P4)


Molecule.

PH3 has lower boiling point

than NH3.

The electronegativity of N (3.0) is much higher than that of

P(2.1). So, NH3 undergoes extensive intermolecular H-bonding and

hence it exists as an associated molecule. To break these H-

bonds, a large amount of energy is needed. On the other hand,

PH3 does not undergo H-bonding and thus exists as discrete

molecules. Therefore, the boiling point of PH3 is much

lower than that of NH3.

PF5 is known but NF5 is not

known

P has vacant 3d-orbitals in its valence shell while N does not have.

As a result, P can form additional bonds to give PF5 while N

cannot extend its covalency beyond three and hence it forms only

NF3 but not NF5.

Unlike phosphorus, nitrogen

shows little tendency for

catenation

Catenation (i.e., linking of atoms .of the same kind with one

another) is related to the atom-atom bond energy. Greater the

atom-atom bond energy, greater is the catenation. Because of low

N—N bond energy (1638 kJ mol-1) nitrogen shows little tendency

for catenation; P—P bond energy (2O1.6 kJ mol-1 ) is quite high,

hence, it shows more tendency for catenation than nitrogen.

Bismuth is a strong

oxidising agent in the

pentavalent state.

As the inert pair effect is very prominent in Bi, its + 5 oxidation

state is less stable than its + 3 oxidation state. In other words,

bismuth in the pentavalent state can easily accept two electrons

and thus gets reduced to trivalent bismuth.

Bi5+2e- Bi3+

Thus, it acts as a strong oxidising agent

When NaBr is heated with

conc. H2SO4, Br2 is

produced but when NaC1 is

heated with conc. H2SO4

HCl is, produced.

When NaBr is heated with conc. H2SO4, HBr is first produced

which being a reducing agent reduces H2SO4 to SO2 while HBr

itself gets oxidised to Br2.

NaBr + H2 SO4 NaHSO4+ HBr

2 HBr+ H2SO4 2H2O+ SO2 +Br2,

As a result, only Br2 is produced.

Similarly, NaCl reacts with conc. H2SO4 to form HCI but since

HCldoes not act as a reducing agent, it does‘ not get oxidised to

Cl2.

NaCl + H2SO4 NaHSO4 + HCI

HC1 + H2SO4 No action

As a result, only HCI is evolved

Oxygen generally exhibits

an oxidation state of — 2

The electronic configuration of oxygen is ls2 2s2 2px2 2py

1, 2pz1,

i.e., it has two half-filled orbitals and there is no d-orbital


only whereas other

members of its family show

oxidation states of +2, +4

and +6 as well

available for excitation of electrons. Further, it is the most

electronegative element of its family. Hence, it shows oxidation

state of -2 only. Other elements like sulphur have d-orbitals

available for excitation, thereby giving four and six half-filled

orbitals. Moreover, they can combine with more electronegative

elements. Hence, they show oxidation states of +2, +4 and +6 also

H2S acts only as a reducing

agent but SO2 acts both as

a reducing agent as well as

an oxidising agent.

The minimum oxidation number (O.N.) of S is — 2 while its

maximum O.N. is + 6. In SO2, the O.N. is +4, hence, it can not only

increase its O.N. by losing electrons but also reduce its O.N. by

gaining electrons. Thus, it acts both as a reducing agent as well as

an oxidising agent. In contrast, in. H2S, S has an O.N. of — 2.

Thus, it can only increase its U.N by losing electrons and hence

acts only as a reducing agent.

SF6 is known but SH6 is not

known

Fluorine being the strongest oxidising agent oxidises sulphur to

its maximum oxidation state of + 6 and thus forms SF6. In

contrast hydrogen being a very weak oxidising agent cannot

oxidise S to its maximum oxidation state of + 6 and hence does

not form SH6

Compounds of fluorine with

oxygen are called fluorides

of oxygen and not the

oxides of fluorine

This is because fluorine is more electronegative than oxygen.

SCI6 is not known but SF6

is known

Fluorine is a much stronger oxidising agent than chlorine,

therefore, it can easily oxidize sulphur to its maximum oxidation

state of + 6 and hence forms SF6. Chlorine being a weaker

oxidising agent can oxidise sulphur at the maximum to its + 4

oxidation state and hence can form SCI4 but not SCI6.

Sulphur exhibits greater

tendency for catenation

than selenium

As we move from S to Se, the atomic size increases and hence

the strength of E—E bond decreases. Thus, S—S bond is much

stronger than Se—Se bond consequently, S shows greater

tendency for catenation than selenium

Both NO and CIO2, are odd

electron species but NO

dimerises while CIO2 does

not

In NO, the odd electron on N is attracted b only one O-atom but

in ClO2, the odd electron on CI is attracted by two O-atoms.

Thus, the odd electron on N in NC) is localised while the odd

electron on CI in CIO2 is delocalised. Consequently, NO has a

tendency to dimerise but CIO2 does not.

Bleaching of flowers by

chlorine is permanent while

Cl2 bleaches coloured material by oxidation:

Cl2 +H2O 2HCl+O


that by sulphur dioxide is

temporary.

Coloured material + [O] Colourless

and hence bleaching is permanent. In contrast, SO2 bleaches

coloured material by reduction and hence bleaching is temporary

since when the bleached colourless material is exposed to air, it

gets oxidised and the colour is restored,

SO2+2H2O H2SO4+2H

Coloured material + H Colourless material

Coloured material. oxidation

Pentahalides more covalent

than trihalides

High +ve oxidation state, more polarizing power, so covalent

How does ammonia react

with a solution of Cu2+?

Cu2+ (blue) + 4 NH3(aq) [Cu(NH3)4]2+(aq) (deep blue)

What is the covalence of

nitrogen in N2O5 ?

4

Prove that PH3 is basic in

nature?

Reacts with acids like HI . PH3+HI→PH4I

Bond angle in PH4+ is higher

than that in PH3. Why?

in PH3 there is lone pair-bond pair repulsion

white P is heated with conc

NaOH solution in CO2 ?

P4+3NaOH+3H2O PH3+3NaH2PO2 sodium hypophosphite

Are all the five bonds in

PCl5 molecule equivalent?

Two axial bonds are longer than equatorial bonds due to repulsion

What happens when PCl5 is

heated?

PCl5 → PCl3 + Cl2

Hydrolytic reaction of PCl5

in heavy water.

PCl5 + D2O → POCl3 + 2DCl

What is the basicity of

H3PO4?

3

What happens when H3PO3

is heated?

4H3PO3 → 3H3PO4 + PH3 (disproportionation)

The HNH angle is higher

than HPH, HAsH angles.

sp3 hybridisation in NH3

Why does R3P = O exist but

R3N = O does not

Nitrogen cannot form dπ –pπ bond

Why NH3 is basic while BiH3

is only feebly basic.

Small size and hence high electron density on N

Differences between white

P and red P

White P: white waxy/poisonous/ soluble in CS2/ reactive

Red P : grey /nonpoisonous/insoluble in CS2/less reactive

N show catenation N-N single bond is weaker than P-P single bond.


properties less than

phosphorus?

Can PCl5 act as an oxidising

& reducing agent? Justify.

P is in highest oxidation state +5. So it is only oxidizing agent

Elements of Gr- 16 show

lower value of first IE

compared to the

corresponding periods of

group 15. Why?

Due to extra stable half-filled p orbitals electronic

configurations of Group 15 elements

H2S is less acidic than

H2Te. Why?

Due to the decrease in bond (Te–H) dissociation enthalpy

Order of thermal stability

of the hydrides of Group-16

H2O>H2S>H2Se>H2Te>H2Po

Why is H2O a liquid and H2S

a gas ?

Due to intermolecular H-bonding in H2O

Why does O3 act as a

powerful oxidising agent

Due to the ease with which it liberates atoms of nascent oxygen

How is O3 estimated

quantitatively

When ozone reacts with an excess of potassium iodide solution.

iodine is liberated which can be titrated against a standard

solution of sodium thiosulphate

Which form of sulphur

shows paramagnetic

behaviour ?

In vapour state sulphur partly exists as S2 molecule which has

two unpaired electrons in the antibonding π* orbitals like O2 and,

hence, exhibits paramagnetism.

What happens when sulphur

dioxide is passed through an

aqueous solution of Fe(III)

salt?

2Fe3+ + SO2 + H2O 2Fe2+ + SO42- + 4H+

Comment on the nature of

two S–O bonds formed in

SO2

Both the S–O bonds are covalent and have equal strength

How is the presence of SO2

detected ?

It decolourises acidified potassium permanganate(VII) solution

Why is Ka2 << Ka1 for H2SO4

in water ?

H2SO4 is largely dissociated into H+ and HSO4-

Why is dioxygen a gas but

sulphur a solid?

O can form pi bond,exists as discrete O2 molecule, so it is gas.

Knowing the electron gain

enthalpy values for O → O–

and O → O2– as –141 and

702 kJ mol–1 respectively,

Oxides having O2– have high negative lattice enthalpy.


how can you account for the

formation of a large number

of oxides having O2– species

and not O–?

Which aerosols deplete

ozone?

Freons

How is SO2 an air pollutant? It causes acid rain

Halogens have maximum

negative electron gain

enthalpy

Halogens have the smallest size in their respective periods

Although electron gain

enthalpy of fluorine is less

negative as compared to

chlorine, fluorine is a

stronger oxidising agent

than chlorine. Why?

It is due to

(i) low enthalpy of dissociation of F-F bond

(ii) high hydration enthalpy of F–

Fluorine exhibits only –1

oxidation state whereas

other halogens exhibit + 1, +

3, + 5 and + 7 oxidation

states also. Explain.

Fluorine is the most electronegative element and cannot exhibit

any positive oxidation state. Other halogens have d orbitals and

therefore, can expand their octets and show + 1, + 3, + 5 and + 7

Considering the parameters

such as bond dissociation

enthalpy, electron gain

enthalpy and hydration

enthalpy, compare the

oxidising power of F2 and

Cl2.

Electrode potential of F2 is higher than Cl2 so F2 is stronger

oxidizing agent than Cl2

Two examples to show the

anomalous behaviour of F

F forms only one oxo acid , HF is liquid but others are gas

Write the balanced

chemical equation for the

reaction of Cl2 with hot and

concentrated NaOH. Is this

reaction a

disproportionation reaction?

Justify.

3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O

Yes, chlorine from zero oxidation state is changed to –1 and +5

oxidation states.

Name two poisonous gases

prepared from chlorine gas.

Phosgene gas (COCl2) , mustard gas


When HCl reacts with finely

powdered iron, it forms

ferrous chloride and not

ferric chloride. Why?

Its reaction with iron produces H2. 2 Fe+2HCl→FeCl +H2

Liberation of H2 prevents the formation of ferric chloride.

Why are halogens strong

oxidising agents?

Due to low bond dissociation energy, high electronegativity and

large negative electron gain enthalpy

Explain why fluorine forms

only one oxoacid, HOF.

Due to high electronegativity and absence of d-orbitals

Oxygen forms hydrogen

bonding while chlorine does

not.

Due to large size of Cl

Write two uses of ClO2. It is powerful oxidizing agent and chlorinationg agent

Why are halogens coloured? Their molecules absorb light in visible region

What happens when NaCl is

heated with sulphuric acid

in the presence of MnO2.

4NaCl +MnO2 + 4H2SO4 MnCl2 + 4NaHSO4 + Cl2 + H2O

Noble gases have very low

boiling points. Why?

Noble gases being monoatomic have no interatomic forces except

weak dispersion

Does the hydrolysis of XeF6

lead to a redox reaction?

No, the products of hydrolysis are XeOF4 and XeO2F2 where the

oxidation states of all the elements remain the same

Why is helium used in diving

apparatus?

Due to its low solubility in blood

Why has it been difficult to

study the chemistry of

radon?

Due to its very short half life

List the uses of neon and

argon gases.

Neon- in discharge tube. Ar – to make an inert atmosphere

F2, Cl2, Br2, I2 - increasing

bond dissociation enthalpy.

I2< F2 <Br2 <Cl2,

HF, HCl, HBr, HI -

increasing acid strength.

HF<HCl<HBr<HI

NH3, PH3, AsH3, SbH3, BiH3

– increasing base strength.

BiH3 < SbH3< AsH3,< PH3,< NH3,

Which one of the following

does not exist?

(i) XeOF4 (ii) NeF2 (iii) XeF2

(iv) XeF6

NeF2


The d & f BLOCK ELEMENTS

Scandium (Z = 21) is a

transition element but Zinc

(Z = 30) is not?

Incompletely filled 3d orbitals in case of scandium

Zn & Zn2+ have completely filled d orbitals (3d10)

Of the d4 species, Cr 2+ is

strongly reducing while

manganese (III) is strongly

oxidising

E° value for Cr 3+ /Cr2+ is negative (— 0.41 V) while as E° value

for Mn3+ /Mn 2+ is positive (+ 1.57 V). Thus, Cr2+ ions can easily

undergo oxidation to give Cr3+ ions and, therefore, act as strong

reducing agent. On the other hand, Mn 2+ can easily undergo

reduction to give Mn2+ and hence act as oxidising agent.

The d1 configuration is very

unstable in ions

The ions with d1 configuration have the tendency to lose the

only electron present in d-subshell to acquire stable d°

configuration. Therefore, they are unstable and undergo

oxidation or disproportionation

Disproportion reactions? Disproportion reactions are those reactions in which the same

substance undergoes oxidation as well as reduction. In

disproportionation reaction, oxidation number of an element

increases as well decreases to form two different products. For

example,

VI VII IV

3MnO42- +4H+ 2MnO4

- +MnO2 +2H2O

A transition metal exhibits

higher oxidation states in

oxides and fluorides.

A transition metal exhibits higher oxidation states in oxides and

fluorides because oxygen and fluorine are highly electronegative

elements, small in size (and strongest oxidising agents).

For example, osmium shows an oxidation states of +6 in OsF6

and vanadium shows an oxidation states of +5 in VO5

Transition elements show

variable oxidation states.

Transition element which

does not exhibit variable

oxidation state

Transition elements show variable oxidation states because

electrons in ns and (n —1) d-orbitals are available for bond

formation.

Sc

First ionization energies of

5d elements are higher than

those of 3d and 4d elements

Because of weak shielding (or screening) effect of 4f electrons,

the effective nuclear charge acting on the valence electrons in

5d elements is quite high. Hence, the first ionization energies of

5d elements are higher than those of 3d and 4d elements.

Transition elements show

paramagnetic behavior.

Substances containing unpaired electrons are said to be

paramagnetic. Transition elements contain unpaired electrons in


their (n —1) d-orbitals. Hence, they are paramagnetic

copper (At no. 29) considered

a transition metal?

copper in oxidation state +2, i.e., Cu2+ has incompletely filled d-

subshell (3d9).

Oxidising power:VO2+ <

Cr2O72– < MnO4

–

This is due to the increasing stability of the lower species

Account for the irregular

variation of IE

Due to varying degree of stability of different 3d-

configurations

The E0 (M2+/M) values are

not regular

Due to the irregular variation of IE , sublimation enthalpies

E0 of Mn3+/Mn2+ is more

positive than that for

Cr3+/Cr2+ or Fe3+/Fe2+?

Explain.

Much larger third IE of Mn (where the required change

is d5 to d4) is mainly responsible for this.

Highest oxidation state of a

metal exhibited in its oxide

or fluoride

Small size and high electronegativity O or F can oxidise the

metal to its highest oxidation state.

Which is a stronger reducing

agent Cr2+ or Fe2+ and why ?

Cr2+ is stronger reducing agent than Fe2+

Reason: d4 → d3 occurs in case of Cr2+ to Cr3+, d3 is stable

But d6 → d5 occurs in case of Fe2+ to Fe3+

Magnetic moment of a

divalent ion in aqueous

solution if its atomic number

is 25.

d5 Configuration (five unpaired electrons). The magnetic

moment, μ is μ = √5(5+2) = 5.92BM

Calculate the ‗spin only‘

magnetic moment of M2+ (aq)

ion (Z = 27).

M2+ (aq) ion (Z = 27) d7 3 unpaired electrons

μ = √3(3+2) = 3.87BM

What is meant by

disproportionation‘ of an

oxidation state? Give an

example.

When a particular oxidation state becomes less stable relative

to other oxidation states, one lower, one higher, it is said to

undergo disproportionation. For example, Mn (VI) becomes

unstable relative to Mn(VII) and Mn (IV) in acidic solution.

3 MnO4 2– + 4 H+ → 2 MnO4

- + MnO2 + 2H2O

Explain why Cu+ ion is not

stable in aqueous solutions?

Cu+ in aqueous solution undergoes disproportionation,

2Cu+(aq) → Cu2+(aq) + Cu(s) The E0 is favourable for this

Lanthanoids to exhibit +4

oxidation state.

Cerium (Z = 58)

Actinoid contraction is

greater than

lanthanoid contraction. Why?

5f electrons themselves provide poor shielding from element to

element in the actinoid series.

Mn2+ more stable than Fe2+ It is because Mn2+ has 3d5 configuration which is stability.

Stable oxidation state for : 3d3 (V): (+2), +3, +4, and +5 ; 3d5 (Cr): +3, +4, +6


3d3, 3d5, 3d8 and 3d4?

3d5 (Mn): +2, +4, +6, +7 ; 3d8 (Co): +2, +3 (in complexes)

3d4 There is no d4 configuration in the ground state.

Oxometal anions in which the

metal exhibits the oxidation

state equal to its group

number.

Vanadate VO3−, Chromate Cr2O4

− , Permanganate MnO4−

What is lanthanoid

contraction? What are the

consequences of lanthanoid

contraction?

Gradual decrease of atomic radii form La to Lu

Consequences : similar properties of 2nd and 3rd rows transition

elements

In what way is the electronic

configuration of the

transition elements different

from that of the non

transition elements?

In transition elements the last electron goes into penultimate

shell.

Oxidation states of the

lanthanoids?

+3 is the common oxidation No, Other oxidation states +2 and

+4

Transition metals are

paramagnetic

For having unpaired electrons

Enthalpies of atomisation of

the transition metals are

high.

Due to strong metallic bonding

Transition metal form

coloured compounds

Due to incompletely filled d-orbital there is d-d electron

transition

Transition metals and their

many compounds act as good

catalyst.

Because they can adopt variable oxidation states to form

different intermediate

Which of the following will be

coloured

Ti3+,V3+,Cu+,Sc3+,Mn2+, Fe3+ &

Co2+.

Except Sc3+, all others will be coloured due to incompletely filled

3d-orbitals, will give rise to d-d transitions.

Of the d4 species, Cr2+ is

reducing while Mn(III) is

strongly oxidising.

Cr2+ is reducing as it change from d4 to d3, the latter is more

stable Mn(III) to Mn(II) is from 3d4 to 3d5 again 3d5 is stable

Co(II) is stable in aq sol but

in the presence of

Due to CFSE, which more than compensates the 3rd IE.

(CFSE=crystal field splitting energy)


complexing reagents it easily

oxidised.

The d1 configuration is

unstable in ions.

The hydration or lattice energy more than compensates the

ionisation enthalpy involved in removing electron from d1.

Which transition metal has

+1 oxidation state most

frequently and why?

Cu, because with +1 oxidation state an stable configuration, 3d10

Calculate the number of

unpaired electrons in the

following gaseous ions: Mn3+,

Cr3+, V3+ and Ti3+. Which one

of these is the most stable in

aqueous solution?

Unpaired electrons Mn3+ = 4, Cr3+ = 3, V3+ = 2, Ti3+ = 1. Most

stable Cr3+

The lowest oxide of

transition metal is basic, the

highest is

amphoteric/acidic.

In lowest oxidation state ionic bond is formed but in highest

oxidation state covalent bond is formed

A transition metal exhibits

highest oxidation state in

oxides

and fluorides.

O and F are strong oxidizing agents and can excite electrons

Which is the last element in

the series of the actinoids?

Write the electronic

configuration of this element.

Comment on the possible

oxidation state of this

element.

Lawrencium(103)

5f 146d17s2

+3

Preparation of KMnO4

Preparation of K2Cr2O7 prepared from chromite ore (FeCr2O4)


The chromates and

dichromates are

interconvertible in aqueous

solution depending upon pH

of the solution.

Sodium and potassium

dichromates are strong

oxidising agents.

Acidified potassium

dichromate will oxidise

iodides to iodine,

sulphides to sulphur, tin(II)

to tin(IV) and iron(II) salts

to iron(III).

Oxidising reactions of

KMnO4

In acid solutions:


BIOMOLECULES

KEY POINTS EXPLANATIONS

Preparation of

Glucose

Monosaccharides Cannot be hydrolyzed further . eg- glucose, fructose, ribose

Disaccharides Sucrose (α-D- glucose + β-D-fructose) , Maltose(α-D- glucose + α-D-

glucose)

Lactose(β-D-galactose + β-D-glucose )

Polysaccharides Starch (two components—Amylose and Amylopectin) polymer of α-D- glucose

Amylose Water soluble , 15-20% of starch., unbranched chain , C1– C4

glycosidic linkage.

Amylopectin Water insoluble , 80-85% of starch., branched chain polymer, C1–C4 &

C1–C6 glycosidic linkage

Cellulose Straight chain polysaccharide of β -D-glucose units/ joined by C1-

C4glycosidic linkage (β-link), not digestible by human / constituent of cell

wall of plant cells

Glycogen Highly branched polymer of α-D- glucose .found in liver, muscles and brain.

reducing sugars Aldehydic/ ketonic groups free so reduce Fehling‘s/ Tollens solution and. Eg-

maltose and lactose

Non reducing

sugars

Aldehydic/ ketonic groups are bonded so can not reduce Fehling‘s solution

and Tollens‘ reagent. Eg- Sucrose

Anomers. The two cyclic hemiacetal forms of glucose differ only in the configuration

of the hydroxyl group at C1, called anomeric carbon Such isomers, i.e., α –

form and β -form, are called anomers.

Invert sugar Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose

and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is

more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory.

Thus, hydrolysis of sucrose brings about a change in the sign of rotation,

from dextro (+) to laevo (–) and the product is named as invert sugar

Glycosidic

linkage

Linkage between two mono saccharide

Peptide linkage amide formed between –COOH group and –NH2 group


Importance of

Carbo -

Hydrates.

Major portion of our food. / used as storage molecules as starch in plants

and glycogen in animals/.

Cell wall of bacteria and plants is made up of cellulose./wood and cloth are

cellulose /provide raw materials for many important industries like textiles,

paper, lacquers and breweries.

essential amino

acids

which cannot be synthesised in the body and must be obtained through diet,

eg- Valine, Leucine

Nonessential

amino acids

which can be synthesised in the body, eg - Glycine, Alanine

zwitter ion. In aqueous solution, amino acids exist as a dipolar ion known as zwitter ion.

peptide linkage peptide linkage is an amide formed between –COOH group and –NH2 group of

two successive amino acids in peptide chain.

10- str. of

proteins:

sequence of amino acids that is said to be the primary structure of protein

20- str. of

proteins:

secondary structure of protein refers to the shape in which a long

polypeptide chain can exist.

They are found to exist in two types of structures viz. α -helix and β -

pleated sheet structure.

Tertiary

structure of

proteins:

further folding of the secondary structure. It gives rise to two major

molecular shapes viz. fibrous and globular.

Fibrous proteins Polypeptide chains run parallel, held together by hydrogen and disulphide

bonds, fibre– like structure. Water insoluble . Eg- are keratin(in hair, wool,

silk) and myosin (present in muscles).

Globular

proteins

chains of polypeptides coil around to give a spherical shape. water soluble.

Eg-Insulin and albumins

Stab.forces 2°&

3°

hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces

of attraction.

Denaturation of

Proteins

When a protein is subjected to physical change like change in temperature

or chemical change like change in pH, the hydrogen bonds are disturbed. Due

to this, globules unfold and helix get uncoiled and protein loses its biological

activity. This is called denaturation of protein.

(During denaturation 2° and 3° structures are destroyed but 1º structure


remains intact.)

eg- The coagulation of egg white on boiling, curdling of milk

Fat soluble vit These are vitamins A, D, E and K. They are stored in liver and adipose (fat

storing) tissues

Water soluble

vit

B , C . These vitamins must be supplied regularly in diet because they are

readily excreted in urine

Vitamins –

sources-

Deficiency

diseases

Vit- A (Fish liver oil, carrots)- Night blindness / Vitamin B1

(Yeast, milk,)- Beri beri

Vit-B2 (Milk, eggwhite)- Cheilosis

Vit- B6 (Yeast, milk,)- Convulsions

Vit- B12 (Meat, fish,)- anaemia

Vit C(Citrus fruits)- Scurvy, Vit D(Exposure to sunlight, fish and egg

yolk)- Rickets, osteomalacia

Vit E(wheat oil, sunflower oil)- fragility of RBCs / Vit K(leafy

vegetables)- Increased blood clotting time

DNA pentose sugar (D-2-deoxyribose) + phosphoric acid + nitrogenous bases

( A , G , C, T )

Function- reserve of genetic information, protein synthesis(message)

RNA pentose sugar (ribose) + phosphoric acid + nitrogenous bases

(A , G , C, U )

Proteins are synthesised by various RNA molecules in the cell but the

message for the synthesis of a particular protein is present in DNA.

Nucleoside /

tides

Nucleoside sugar + base Nucleotides sugar + base + phosphate

Phosphodiester

link

Linkage between two nucleotides in polynucleotides

Functions of

Nucleic Acids

DNA reserve genetic information, maintain the identity of different species

, is capable of self duplication during cell division, synthesizes protein in the

cell.

How to prove

that –All the six

carbon atoms of

Glucose are

linked in a

straight chain.

How to prove

the presence of

a carbonyl group

(>C = 0) in

glucose.


Carbonyl group

is present as an

aldehydic group

in Glucose

There are five –

OH groups in

Glucose.

glucose pentaacetate

Presence

of one primary

alcoholic (–OH)

group in glucose.

Meaning of D(+)-

glucose

The letters ‗D‘ or ‗L‘ before the name of any compound indicate the

relative configuration of a particular stereoisomer in

relation with a particular isomer of glyceraldehydes.

‗(+)‘ represents dextrorotatory nature.

zwitter ion. In aqueous solution, the carboxyl group can lose a proton and amino group can

accept a proton, giving rise to a dipolar ion known as zwitter ion.


POLYMERS

POLYMER MONOMER(Name & Structure) USES

Addition or Chain Growth Polymer

Polythene Ethene CH2=CH2 Insulator, ,Packing material,

Teflon

(Poly

tetrfluoroethene

Tetrfluoroethene CF2=CF2

Lubricant, Insulator and making non-

stick cooking ware.

Poly acrylonitrile Acrylonitrile CH2=CH-CN For synthetic fibres and synthetic

wool.

Buna S Buta-1,3-diene + Styrene

CH2=CH-CH=CH2 C6H5CH=CH2

Automobile tyres and Foot wears

Buna N

Buta-1,3-diene +

Acrylonitrile

CH2=CH-CH=CH2 CH2=CH-CN

Oil seals, Tank lining

Natural Rubber 2-Methylbuta-1,3-diene (Isoprene) Used for tyres after vulcanization

Synthetic

Rubber(Neoprene)

2-Chlorobuta-1,3-diene

(Chloroprene)

Insulator, making conveyor belts and

printing rollers

Polypropene Propene CH3-CH=CH2 Ropes, toys, pipes and fibres

Polystyrene Styrene C6H5CH=CH2 Insulator, Wrapping material,toys,

Radio and television cabinets.

Polyvinyl chloride

(PVC) Vinyl Chloride CH2=CH-Cl

Rain coats, Hand bags, Vinyl flooring

and water pipe.

Condensation or Step Growth Polymers

Terylene(Dacron)

Ethane-1,2-diol + Benzene-1,4-

dicarboxylic acid

Used for making fibres, safety belts,

tents

Nylon 66

Hexamethylene diamine + Adipic

acid

NH2(CH2)6NH2

HOOC(CH2)4COOH

For making brushes,paratutes and

ropes

Nylon 6 Caprolactum Tyres cords,fabrics and ropes

Bakelite Phenol + Methanal Combs,electrical switches,handles of

utensiles and computer discs

Melamine

formaldehyde

Melamine + Methanal Unbreakable crockery

PHBV

biodegradable

3-Hydroxybutanoic acid +

3-Hydroxypentanoic acid

Specialty packaging, orthopedic

devices, In controlled drug release

Nylon 2 – Nylon 6

biodegradable

Glycine + Amino caproic

acid


H2N-CH2-COOH H2N (CH2)5-

COOH

Urea-

formaldehyde

resin

Urea + Formaldehyde

Unbreakable cups , laminated sheet

Glyptal

Ethane1,2-diol + Benzene-1,2-

dicarboxylic acid

Binding material

in preparation of mixed plastics and

paints

Semi-synthetic

poly Cellulose Acetate (Rayon)

Thermoplastic

polymers

Linear or slightly branched / capable of repeatedly softening on heating

and hardening on cooling. Example : Polythene, Polystyrene, Polyvinyls, etc.

Thermosetting

polymers

Cross linked or heavily branched molecules, / on heating undergo extensive

cross linking in moulds and again become infusible. These cannot be reused.

Examples : Bakelite, Urea-formaldelyde resins.

Homo-polymer &

Co-polymer

Homo-polymer Polymer of a single monomeric species. Example :

Polythene , PVC

Co-polymer Polymer of more than one monomer .Example : Nylon66,

Bakelite

Initiators Benzoyl Peroxide [ C6H5CO-O-O-CO-C6H5 ] (in free radical addition

polymerization)

Vulcanisation of

Rubber.

Natural rubber is soft at high temp and brittle at low temp and absorbs

water. To improve these physical properties, it is heated with sulphur and

an appropriate additive at a temperature range between 373 K to 415 K. On

vulcanisation, sulphur forms cross links at the reactive sites of double

bonds and thus the rubber gets stiffened.

CHEMISTRY IN EVERYDAY LIFE

THERAPEUTIC ACTION OF DIFFERENT DRUGS

Drugs Action Example

Analgesics Relieving pain Aspirin Analgin, Seridon, Anacine,

Analgesics

Narcotic

Reduce tension and pain.

produce unconsciouness.

Opium, Heroin , Pethidine , Codeine,

Morphine

Antibiotics

Produced by micro – organism,

that can inhibit the growth or

destroy other micro-organism.

Ampicillin and Amoxycillin are

synthetic modifications of

Penicillin G(Narrow Spectrum)

Streptomycin, Ampicillin , Amoxycillin

Chloramphenicol Vancomycin, ofloxacin ,


penicillins. These have broad

spectrum.

Antiseptics

Prevent the growth of micro-

organism or kill them but not

harmful to the living tissues.

Dettol, Bithional(in soap)

Tincture iodine, 0.2% phenol,

(Dettol is mixture of chloroxylenol and

terpineol)

Disinfectants

Kills micro-organisms, not safe

for living tissues. It is used

for toilets, instruments.

1% phenol,

chlorine (Cl2) ,

Sulphurdioxide ( SO2)

Antacids

Reduce or neutralise the

acidity. Mg(OH)2

MgCO3

AlPO4

Al(OH)3 gelNaHCO3

Antihistamines

Reduce release of acid.

It is also used to treat allergy

Cimetidine(Tegamet), Ranitidine (Zantac),

Brompheniramine ( Dimetapp)

Terfenadine ( Seldane)

Tranquilizers

Reduce the mental anxiety,

stress, (sleeping pill)

Valium, Serotonin, Veronal,

Equanil,Amytal,Nembutal,Luminal, Seconal

Antipyretics Reduce body temperature Aspirin, Paracetamol, Analgin, Phenacetin.

Antifertility

drugs

These are the steroids used to

control the pregnancy

Norethindrone, Ethynylestradiol

(novestrol )

CHEMICALS IN FOOD

Sweetening

Agent

Saccharine, Aspartame(for cold foods) Alitame

Sucrolose(stable at cooking temp)

Food

Preservative

Salt, sugar, veg. oils, sodium benzoate

CLEANSING AGENTS

Soap Na / K –salt of long chain fatty

acids

Not work in hard water becoz with Ca and

Mg salt soap produce insoluble scum

Anionic

detergents

Sodium laurylsulphate Used in household work / in tooth paste

Cationic

detergents

Cetyltrimethyl ammonium

bromide

Hair conditioner / germicidal properties

Non ionic

detergents

Ester of stearic acid and

polyethylene glycol

Liquid dishwashing

Detergents with highly branched hydrocarbon parts are non biodegradable and hence water

pollutants so branching is minimized which are degradable and pollution is prevented.


NAME REACTIONS

1. Finkelstein CH3Br + NaI CH3-I + NaBr

2. Swarts CH3Br + AgF CH3F + AgBr

3.

Friedel-Crafts

Alkylation + CH3 Cl

Anhydrous AlCl3

CH3

4.

Friedel-Crafts

Acylation

COCH3

CH3COCl

Anhydrous AlCl3

5. Wurtz CH3 Cl + CH3Cl CH3 CH3 + NaCl

2Na

6.

Fittig

Cl

+

Cl

2Na

Dry ether+ NaCl

7.

Wurtz-Fittig

Cl

+ CH3Cl2Na

Dry ether+ NaClCH3

8.

Kolbe

OH

Na OH

ONa

i) CO2

ii) H+

OH

COOH

9.

Reimer-Tiemann

OHONa

CHOCH3Cl + Na OH

H+

OH

CHO

10. Williamson CH3-Br + CH3-ONa CH3-O- CH3 + NaBr

11. Stephen CH3 CN+ SnCl

2 + HCl CH3 CH NH

H3O+

CH3 CHO

12.

Etard

CH3

CrO2Cl

2

H3O+

CHO


13.

Gatterman – Koch

CHO

CO / HCl

Anhydrous AlCl3

14.

Rosenmund

reduction CH3

CCl

O H2

Pd / BaSO4CH3

CH

O

15.

Clemmensen

Reduction CH3

CCH3

OZn - Hg

Conc. HClCH3 CH2 CH3

16.

Wolff-Kishner

reduction CH3

CCH3

O

CH3 CH2 CH3

i) NH2-NH2

ii) KOH / Ethylene glycol /

17. Tollens‘ test

R-CHO + 2 [Ag(NH3)2]+ + 3 OH- R-COO- + 2Ag + 2H2O + 4

NH3

18. Fehling‘s test R-CHO + 2 Cu2+ + 5 OH- R-COO- + Cu2O + 3H2O

19.

Iodoform CH3

CCH3

OI2 / NaOH

OR, NaOICHI

3 + CH

3COONa

20.

Aldol condensation CH3 CHO

dil NaOHCH3 CH CH2

OH

CHO

CH3 CH CH CHO2

21. Cannizzaro HCHO + HCHO

Conc. NaOHHCOONa + CH3 OH

22.

Hell-Volhard-

Zelinsky (HVZ) CH3 COOH

i) Cl2 / Red Phosphorus

ii) H2OCH2 COOH

Cl

23.

Hoffmann bromamide degradation

CH3 C NH2

OBr2

NaOHCH3 NH2

24. Carbylamine

R-NH2 + CHCl3 + 3 KOH

R-NC + 3 KCl + 3 H2O

25.

Diazo

NH2

NaNO2 + dil HCl

273 - 278 K

N2

+Cl

-

26.

Sandmeyer.

N2

+Cl

-

CuCl / HCl

Cl

+ N2


27.

Gatterman

N2

+Cl

-Cl

+ N2

Cu / HCl

28.

Coupling N2

+Cl

-

+ OHH

OH-

N N OH

DISTINGUISH BY A SINGLE CHEMICAL TEST

1. All aldehydes ( R-CHO) give Tollens‘ Test and produce silver mirror.

RCHO + 2 [Ag(NH3)2]+ + 3 OH- RCOO- + 2 Ag + 2H2O + 4 NH3

Tollens‘ Reagent silver ppt

Note: HCOOH(methanoic acid ) also gives this test, ketones(RCOR) do not give this test

2. All aldehydes (R-CHO) and ketones(RCOR) give 2,4-DNP test

RCOR + 2,4-DNP Orange ppt

R-CHO + 2,4-DNP Orange ppt

3. Aldehydes and ketones having CH3CO- (keto methyl) group give Iodoform Test. Alcohols

having CH3CH- group also give Iodoform Test.

|

OH

CH3CHO + 3I2 + 4 NaOH CHI3 + HCOONa + 3 NaI + 3H2O

Yellow ppt

The following compounds give Iodoform Test: ethanol (C2H5OH), propan-2-ol

(CH3CH(OH)CH3), ethanal(CH3CHO), propanone(CH3COCH3), butanone ( CH3COCH2CH3) ,

pentan-2-one (CH3COCH2 CH2CH3) , acetophenone ( PhCOCH3 )

4. All carboxylic acids ( R-COOH) give Bicarbonate Test

RCOOH + NaHCO3 RCOONa + CO2 + H2O

effervescence

5. Phenol gives FeCl3 Test

C6H5OH + FeCl3 (C6H5O)3Fe + 3 HCl

(neutral) (violet color)


6. All primary amines (R/Ar -NH2) give Carbyl Amine Test

R-NH2 + CHCl3 + KOH(alc) R-NC + KCl + H2O

offensive smell

7. Aniline gives Azo Dye Test ( Only for aromatic amines)

C6H5NH2 + NaNO2 + HCl C6H5N2+Cl- ; then add β-naphthol orange dye

8. All alcohols (ROH) give Na-metal test

R-OH + Na R-ONa + H2

bubbles

9. For esters (RCOOR) : Hydrolyses first. Then see the products ( acid & alcohol) and give

a test to identify

them

10. All alkenes (C=C) and alkynes (C≡C) decolorizes Br2 – water from red to colorless

11. Lucas Test to distinguish primary, secondary and tertiary alcohols

Lucas reagent: ZnCl2/HCl

30-alcohol + Lucas reagent immediate turbidity

20-alcohol + Lucas reagent turbidity after sometime

10-alcohol + Lucas reagent no turbidity

KEY FOR CONVERSIONS

Sl

No

Reagent Group Out Group In Remark

1 KMnO4 / H+ -CH2OH -COOH Strong Oxidation (20

alc ketone)

2 LiAlH4 -COOH -CH2OH Strong Reduction

(ketone 20 alc)

3 Cu / 573 K or

CrO3

-CH2OH -CHO Dehydrogenation

4 PCl5 or SOCl2 -OH -Cl

5 Cl2 / Δ or Cl2 / UV -H -Cl Free radical substitution

6 Aq NaOH / KOH -X -OH Nucleophilic substitution

7 KCN -X -CN Step Up

8 AgCN -X -NC

9 Alcoholic KOH -HX = Dehydrohalogenation


(Stzf)

10 Mg / dry ether Mg R-X R-MgX

11 HBr >=< H, Br Merkovnikov

12 H2 / Pd-BaSO4 -COCl -CHO Rosenmund Reduction

13 Zn-Hg / HCl >C=O -CH2- Clemmension Reduction

14 NH3 / Δ -COOH -CONH2 -COOH + NH3 -

COONH4

15 Br2 / NaOH or NaOBr -CONH2 -NH2 Step Down ( Hoffmann)

16 HNO2 or NaNO2/HCl -NH2 -OH HONO

17 CHCl3 / alc KOH -NH2 -NC Carbyl amine

18 P2O5 -CONH2 -CN Dehydration

19 H3O+ -CN -COOH Hydrolysis

20 OH- -CN -CONH2

21 LiAlH4 -CN -CH2NH2 Reduction

22 Red P / Cl2 α-H of acid -Cl HVZ Reaction

In benzene ring

23 Fe / X2 /dark -H -X Halogination

24 CH3Cl / AlCl3(anhyd) -H -CH3 Friedel Craft alkylation

25 CH3COCl / AlCl3(anhyd) -H -COCH3 Friedel Craft acylation

26 Conc.HNO3/con.H2SO4 -H -NO2 Nitration

27 Conc H2SO4 -H -SO3H Sulphonation

28 KMnO4 / H+ -R -COOH Oxidation

29 CrO2Cl2 / H+ -CH3 -CHO Mild oxidation(Etard

Reaction)

30 Sn / HCl or Fe/HCl -NO2 -NH2 Reduction

31 NaOH / 623K / 300 atm -Cl -OH

32 Zn dust / Δ -OH -H

33 NaNO2 / dil HCl / 273-

278 K

-NH2 -N2+Cl- Diazo reaction

34 CuCl / HCl or Cu/HCl -N2+Cl- -Cl Sanmeyer or

Gattermann

35 CuBr / HBr or Cu/HBr -N2+Cl- -Br Sanmeyer or

Gattermann

36 CuCN / KCN -N2+Cl- -CN Sanmeyer

37 KI -N2+Cl- -I

38 HBF4 / Δ -N2+Cl- -F

39 H3PO2 or CH3CH2OH -N2+Cl- -H

40 H2O / 283 K -N2+Cl- -OH


41 HBF4/ NaNO2, Cu / Δ -N2+Cl- -NO2

42 C6H5-OH -N2+Cl- -N=N-C6H5-OH Coupling ( p-hydroxy)

43 C6H5-NH2 -N2+Cl- -N=N-C6H5-NH2 Coupling ( p-amino)

Reactions of Grignard Reagent

Grignard reagent + Any one below + H2O Product

R-MgX

H2O or ROH or RNH2 R-H

H-CHO R-CH2-OH (10 alc)

R-CHO R-CH(OH)-R (20 alc)

R-CO-R R2C(OH)-R (30 alc)

CO2 R-COOH

R-CN R-CO-R

HCOOR Aldehyde

RCOOR Ketone

NB: i) During reaction generally changes take place in the functional group only so see the

functional group very carefully.

ii) Remember structural formula of all the common organic compounds ( with their IUPAC

and common names)

iii) Wurtz Reaction and Aldol Condensation are not included in the table although they are

very important for conversions so study them .

iv) By taking examples practice all the above cases ( from 1 to 43 and Grignards reaction.)

vi) Start practicing NOW !

Directional Properties of groups in benzene ring for electrophilic substitution

Ortho-para directing group: -R , -OH, -NH2, -X, -OR, -NHR, -NR2, -NHCOCH3, -CH2Cl, -

SH, - Ph

Meta-directing group: -NO2 , -CHO , -COOH , COOR , -CN , -SO3H , -COCH3 , -CCl3 , -

NH3+ ,


ELECTRON DISPLACEMENT EFFECTS

+ I : O- , COO- , (CH3)3C , (CH3)2CH , CH3CH2 , CH3 (electron donating)

- I : NR3+ , SR2

+ , NH3+, NO2 , SO2R , CN , COOH , F , Cl , Br , I , OR , OH, NH2 (withdrawing)

+ R ( + M ) : OH , NH2 , OR , NHR , X (electron

donating)

- R ( - M ) : NO2 , CN , CHO , COOH , COCH3 (electron

withdrawing)

DIRECTIVE INFLUENCE OF SUBSTITUENTS IN BENZENE RING

(for electrophilic substitution reactions)

EFFECT OF THE GROUP DIRECTING ACTIVATING / DEACTIVATING

+ I Ortho /

Para

Activating

+ I , + R Ortho /

Para

Activating

- I < + R Ortho / Activating


Para

- I > + R Ortho /

Para

Deactivating

- I Meta Deactivating

- I , - R Meta Deactivating

Example: - I > + R : - X , - CH=CH2 , -CH=CH-COOH , -CH2-Cl

These groups are deactivating but exceptionally o / p directing due to +E effect by the

attacking reagents electron density increases at ortho and para position .

If two groups are present initially

1. When both the groups present in benzene ring are o/p directing than the order of

influence

O- > NH2 > NR2 > OH > OCH3 > NHCOCH3 > CH3 > X

2. When both the groups present in benzene ring are meta directing than the order of

influence :

(CH3)3N+ > NO2 > CN > SO3H > CHO > COCH3 > COOH

3. When one group is o/p and another is m – directing than o/p directing group takes

priority


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