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Subject : Chemistry NTSE STAGE-I

S. No. Topics Page No.

1. Structure of Atom 1 - 12

2. Acids, Bases and Salts 13 - 23

3. Periodic Table 24-38

4. Mole Concept 39-57

.13RPCCP


1111PAGE # 1

STRUCTURE OF ATOM

DALTON�S ATOMIC THEORY

In 1808 John Dalton proposed atomic theory ofmatter, assuming atoms are ultimate indivisibleparticles of matter based on the law of conservationof mass and law of definite proportion.

The important points of Dalton�s theory are -

(i) Elements consist of small indivisible particlescalled atoms and atoms take part in chemicalreactions.

(ii) Atoms of same element are alike in all respect.

(iii) Atoms of different elements are different in allrespect.

(iv) Atoms cannot be created or destroyed.

(v) Atoms combine in a fixed, small, whole number toform compound atoms called molecules.

Note :The term � Element � was coined by Lavoisier.

(a) Merits :

(i) Dalton�s theory explains the law of conservation of

mass (point iv) and law of constant proportion (point v).

(ii) Atoms of elements take part in chemical reactionthis is true till today.

(iii) Atoms combine in whole numbers to formmolecules (point v).

(b) Demerits :

(i) The atom is no longer supposed to be indivisible.The atom is not a simple particle but a complex one.

(ii) He could not explain that why do atoms of sameelement combined with each other.

(iii) Atoms of the same element may not necessarilybe identical in all aspects.There are a number of elements whose atomspossess different masses. All these atoms of thesame element with same atomic number but differentmass number are called isotopes.e.g.

1H1,

1H2,

1H3 are the three isotopes of hydrogen.

(iv) Atoms of different elements may not necessarilybe different in all aspects. There are a number ofelements whose atoms possess same massnumber. All these elements with different atomicnumber but same mass number are called isobars.e.g.

20Ca40 and

18Ar40 are isobars of each other.

ELECTRON

Electrons are the fundamental particles of allsubstances.

(a) Cathode Rays - Discovery of Electron :

The nature and existence of electron was establishedby experiments on conduction of electricity throughgases.

Note :

In 1859, Julius Plucker started the study of conductionof electricity through gases at low pressure in a

discharge tube.

A number of interesting things happen when a highvoltage (say, 10,000 V) is applied across theelectrodes of the discharge tube, and the pressure ofthe gas inside the tube is lowered.

(i) When the pressure of the gas in the dischargetube is at atmospheric pressure and a high voltageis applied across the electrodes, nothing noticeablehappens. But as we lower the pressure and increasethe voltage, sparking or irregular streaks of light areseen in the tube. This is called positive column.

(ii) As the pressure of gas is reduced further, the lengthof the positive column reduces, a fine glow can beseen at the cathode. The dark space or gap leftbetween the cathode and the positive column iscalled the Faraday�s dark space.


2222PAGE # 2

(iii) When the pressure of gas is reduced to about 1mm of Hg, the cathode glow moves away from thecathode, creating a dark space between cathode andthe cathode glow. This dark space is called Crookesdark space.

(iv) The Crookes dark space expands with further fallin pressure at 0.1 mm of Hg. The positive columngets split into a number of bands called striations.

(v) At pressure 0.01 mm of Hg or less, the striationsmove towards the anode and vanish finally. At thisstage the glass tubes begins to glow at the endopposite to the cathode. This phenomenon is calledfluorescence.

Thus, some sort of invisible rays travel from thenegative electrode to the positive electrode. Since thenegative electrode is called cathode, these rays werecalled cathode rays. The colour of glow depends uponthe nature of the glass used. For soda glass the

fluorescence is of yellowish green colour.

(b) Properties of cathode rays :

(i) Cathode rays travel in a straight line at a highvelocity and generate normally from the surface ofthe cathode. If an opaque object is placed in the pathof cathode rays its shadow falls on opposite side ofthe cathode. It shows that cathode rays travel instraight lines.

Cathode

High voltage Anode

To vacuumpump Shadow

Object

+

�

Note :Cathode rays travel with very high velocities rangingfrom 109 to 1011 cm per second.

(ii) They are a beam of minute material particles havingdefinite mass and velocity. When a light paddle wheelis placed in the path of the cathode rays, the bladesof the paddle wheel begin to rotate. This also provesthat cathode rays have mechanical energy.

(iii) They consist of negatively charged particles. Whenthe cathode rays pass through an electric field, theybent towards the positive plate of the electric field.This indicates that cathode rays are negativelycharged.

(iv) Cathode rays can affect the photographic plate.

(v) The nature of cathode rays is independent of thenature of gas used in discharge tube or material ofcathode.

(vi) Cathode rays are deflected in the magnetic fieldalso.

N

S

+�

High voltage

Deflection of cathode rays in magnetic field

(vii) If cathode rays are focused on a thin metal foil,the metal foil gets heated up to incandescence.

(viii) When cathode rays fall on materials having highatomic mass, new type of penetrating rays of verysmall wavelength are emitted which are called X -rays.

Thus, investigations on cathode rays showed thatthese consisted of negatively charged particles.

Note :

The negatively charged particles of cathode rays werecalled �negatrons� by Thomson. The name negatronwas changed to �electron� by Stoney.

(c) Characteristics of electron :

(i) Electrons are sub - atomic particles whichconstitute cathode rays.

(ii) In 1897, J.J.Thomson determined the charge tomass (e/m) ratio of electron by studying the deflectionsof cathode rays in electric and magnetic fields. Thevalue of e/m has been found to be 1.7588 × 108

coulombs/g. The e/m for electrons from differentgases was found to be the same. This indicates thatatoms of all kinds have the same kind of negativelycharged particles. Thus electrons are the commonconstituents of all atoms.

Note :A cathode ray tube is used to measure the charge tomass ratio of the electrons.

3333PAGE # 3

(iii) Charge on the electron :The charge (e) on an electron was determined byRobert Millikan in 1909. Millikan found the charge onoil drops to be -1.6 × 10-19 C or its multiples. So, thecharge on an electron is to be -1.6 × 10-19 coulombs / unit.

(iv) Mass of an electron :

By Thomson�s experiment e/m = 1.76 × 1011 C/kgBy Millikan�s experiment e = � 1.6 × 10-19 C

So mass of electron (m) = 11

19

101.76

101.6

= 9.1096 × 10-31 kgMass of an electron in amu = 0.000549

(v) Mass of electron in comparison to that ofhydrogen :Mass of hydrogen = 1.008 amu= 1.008 × 1.66 × 10-24 g ( since 1 amu = 1.66 × 10-24 g )= 1.673 × 10-24 g

electron of Massatom hydrogen of Mass

= 28�

24

109.1096

10731.6

= 1837

Note :

Thus, the mass of an electron is 1837

1 times the

mass of a hydrogen atom.

PROTON

(a) Anode Rays (Canal rays) :

It has been established that electron is a negatively

charged particle and present in all the atoms. As anatom is electrically neutral, there must be somepositively charged particles present in the atom toneutralize the negative charges of the electrons. Ithas been confirmed by experiments. ScientistGoldstein in 1886 discovered the existence of a newtype of rays in the discharge tube. He carried out theexperiment in discharge tube containing perforatedcathode. It was observed that when high potentialdifference was applied between the electrodes, notonly cathode rays were produced but also a new typeof rays were produced simultaneously from anode,moving from anode towards cathode and passedthrough the holes of cathode.

Anode rays

Anode Perforated cathode

High voltage source

Fluorescence

Note :Anode rays are called canal rays because they passthrough the canals or holes of the cathode. Theserays are also called anode rays since they originatefrom the anode side. Anode rays are produced from apositively charged electrode, therefore these were

named positive rays by J.J.Thomson.

(b) Characteristics of Anode Rays :

(i) Anode rays travel in straight lines.

(ii) These rays rotate the light paddle wheel placed intheir path. This shows that anode rays are made upof material particles.

(iii) Anode rays are deflected by magnetic or electricfield. In the electric field they get deflected towardsnegatively charged plate. This indicates that theserays are positively charged.

(iv) The anode rays affect photographic plate.

(v) The nature of anode rays depend upon the type ofgas used.

(c) Discovery of Proton :

J.J.Thomson in 1906, found that particles obtained

in the discharge tube containing hydrogen have e/mvalue as 9.579 × 104 coulomb/g. This was themaximum value of e/m observed for any positiveparticle. It was thus assumed that the positiveparticles given by hydrogen represent fundamentalparticle of positive charge. This particle was namedproton.

H e H+ (Proton)

Note :

The name �proton� was given by Rutherford in 1911.

(d) Characteristics of Proton :

(i) A proton is a sub - atomic particle which constituteanode rays produced when hydrogen is taken in thedischarge tube.

(ii) Charge of a proton :

Proton is a positively charged particle. The charge on

a proton is equal but opposite to that on an electron.Thus, the charge on a proton is

+1.602 × 10�19

coulombs/ unit.

(iii) Mass of a proton :

The mass of a proton is equal to the mass of a

hydrogen atom.m

p= 1.0073 amu= 1.673 × 10-24 g= 1.673 × 10-27 kg

4444PAGE # 4

(iv) Mass of proton relative to mass of electron :

electronanofMassprotonaofMass

=g109.1

g101.67328

24

= 1837

Thus, the mass of a proton is 1837 times larger thanthe mass of an electron.

(v) Charge to mass ratio for a proton : The e/m ofparticles constituting the anode rays is different fordifferent gases.

me

of proton = 24

19

011.673

101.602

= 9.579 × 104 C/g

THOMSON MODEL OF AN ATOM

J.J. Thomson (1898) tried to explain the structure ofatom. He proposed that an atom consists of a sphereof positive electricity in which electrons are embeddedlike plum in pudding or seeds evenly distributed inred spongy mass in case of a watermelon. The radiusof the sphere is of the order 10�8 cm.

(a) Merits :

(i) Thomson�s model could explain the electrical

neutrality of an atom.

(ii) Thomson �s model could explain why only

negatively charged particles are emitted when a metalis heated as he considered the positive charge to beimmovable by assuming it to be spread over the totalvolume of the atom.

(iii) He could explain the formation of ions and ioniccompounds.

(b) Demerits :

This model could not satisfy the facts proposed byRutherford through his alpha particle scatteringexperiment and hence was discarded.

RUTHERFORD MODEL OF AN ATOM

(a) Rutherford�s Alpha Particle Scattering

Experiment (1909) :

Ernest Rutherford and his coworkers performednumerous experiments in which - particles emittedfrom a radioactive element such as polonium wereallowed to strike thin sheets of metals such as gold orplatinum.

(i) A beam of -particles (He2+) was obtained by placingpolonium in a lead box and letting the alpha particlescome out of a pinhole in the lead box. This beam of -rays was directed against a thin gold foil (0.0004 cm).A circular screen coated with zinc sulphide was placedon the other side of the foil.

(ii) About 99.0% of the -particles passed undeflectedthrough the gold foil and caused illumination of zincsulphide screen.

(iii) Very few -particles underwent small and largedeflections after passing through the gold foil.(iv) A very few (about 1 in 20,000) were deflectedbackward on their path at an angle of 180º.

Rutherford was able to explain these observationsas follows:

(i) Since a large number of -particles pass throughthe atom undeflected, hence, there must be largeempty space within the atom.

(ii) As some of the -particles got deflected, therefore,there must be something massive and positivelycharged structure present in the atom.

(iii) The number of -particles which get deflected isvery small, therefore, the whole positive charge in theatom is concentrated in a very small space.

(iv) Some of the -particles retracted their path i.e.came almost straight back towards the sources as aresult of their direct collisions with the heavy mass.

Note : - particles are made up of two protons and twoneutrons and are Helium (He) nuclei.

(b) Rutherford Nuclear Model of Atom (1911) :

Rutherford proposed a new picture of the structure of

atom.

Main features of this model are as follows-

(i) The atom of an element consists of a smallpositively charged �Nucleus� which is situated at the

centre of the atom and which carries almost the entiremass of the atom.

(ii) The electrons are distributed in the empty spaceof the atom around the nucleus in different concentriccircular paths (orbits).

5555PAGE # 5

(iii) The number of electrons in the orbits is equal tothe number of positive charges (protons) in thenucleus.

(iv) Volume of nucleus is very small as compared tothe volume of atom.

(v) Most of the space in the atom is empty.

Note :

Rutherford�s model is also called �Planetary model�.

(c) Defects in Rutherford�s Model :

(i) Rutherford did not specify the number of electronsin each orbit.

(ii) According to electromagnetic theory, if a chargedparticle (like electron) is accelerated around anothercharged particle (like protons in nucleus) then therewould be continuous loss of energy due to continuousemission of radiations. This loss of energy wouldslow down the speed of electron and eventually theelectron would fall into the nucleus. But such acollapse does not occur. Rutherford�s model could

not explain this theory.

(iii) If the electron loses energy continuously, theobserved spectrum should be continuous but theactual observed spectrum consists of well definedlines of definite frequencies. Hence the loss of energyis not continuous in an atom.

BOHR MODEL OF AN ATOM (1913)

To overcome the objections to Rutherford�s model

and to explain the hydrogen spectrum, Bohr proposeda quantum mechanical model of the atom.

The important postulates on which Bohr�s model is

based are the following -

(i) The atom has a nucleus where all the protons arepresent. The size of the nucleus is very small. It ispresent at the centre of the atom.

(ii) Each stationary orbit is associated with a definiteamount of energy. The greater is distance of the orbitfrom the nucleus, more shall be the energy associatedwith it. These orbits are also called energy levels andare numbered as 1, 2, 3, 4 ------or K, L, M, N ---- fromnucleus to outwards.

(iii) By the time, the electron remains in any one of theallowed stationary orbits, it does not lose energy.Such a state is called ground or normal state.

(iv) The emission or absorption of energy in the formof radiation can only occur when an electron jumpsfrom one stationary orbit to another.E = E

final - E

initial = h

Where h is Planck�s constant (h = 6.625 × 10�34 Js)Energy is absorbed when the electron jumps fromlower to higher orbit and is emitted when it movesfrom higher to lower orbit.

When the electron moves from inner to outer orbit byabsorbing definite amount of energy, the new state ofthe electron is said to be excited state.

(v) Negatively charged electrons revolves around thenucleus in circular path. The force of attractionbetween the nucleus and the electron is equal tocentrifugal force of the moving electron.Force of attraction towards nucleus = Centrifugal force

(vi) Out of infinite number of possible circular orbitsaround the nucleus, the electron can revolve only inthose orbits whose angular momentum is an integral

multiple of 2

h, i.e. mvr = n

2h

where :m = mass of the electronv = velocity of electronr = radius of the orbit, andn =1,2,3 ---- number of the orbit.The angular momentum can have values such as

2h

, 2h2

, 2h3

, but it cannot have a fractional value.

Thus, the angular momentum is quantized. Thespecified circular orbits (quantized) are calledstationary orbits.

RADII OF VARIOUS ORBITS

Radii of various orbits can be given by formula.

r = 22

22

mkZe4

hn

Note :

Greater is the value of �n� larger is the size of atom.

On the other hand, greater is the value of �Z� smaller

is the size of the atom.

For hydrogen atom, Z = 1; so r = 22

22

mke4

hn

Now putting the values of h, , m, e and k.

r = 219�931�2

234�2

)106.1()109()101.9()14.3(4

)10625.6(n

= 0.529 ×n2 × 10�10 m = 0.529 × n2 Å

= 0.529 × 10�8 × n2 cm

Thus, radius of 1st orbit

= 0.529 × 10�8 × 12 = 0.529 × 10�8 cm

= 0.529 × 10�10 m = 0.529 Å

Radius of 2nd orbit

= 0.529 × 10�8 × 22 = 2.11 × 10�8 cm

= 2.11 × 10�10 m = 2.11 Å

Radius of 3rd orbit

= 0.529 × 10�8 × 32 = 4.76 × 10�8 cm

= 4.76 × 10�10 m = 4.76 Å

6666PAGE # 6

Energy of an electron in Bohr�s orbit can be given

by the formula :

E = 22

4222

hn

meKZ2�

For hydrogen atom, Z = 1So,

22

422

hn

mek2�E

Putting the values of , k, m, e and h.

E = � 234�2

419�31�292

)10625.6(n

)106.1()101.9()109()14.3(2

= � 2

19�

n

1079.21 J per atom

= 2n

6.13� eV per atom (1 J = 6.2419 × 1018 eV)

Note :

The negative sign indicates that the electron is underattraction towards nucleus, i.e. it is bound to thenucleus.

The electron has minimum energy in the first orbitand its energy increases as n increases, i.e., itbecomes less negative. The electron can have amaximum energy value of zero when n = . The zeroenergy means that the electron is no longer bound tothe nucleus , i.e. , it is not under the force of attractiontowards nucleus.

VELOCITY OF AN ELECTRON IN BOHR'S ORBIT

Velocity of an electron in Bohr�s orbit can be given by

the formula :

v = nZ

he2 2

Substituting the values of h, , e.

v = nZ

× 27

210

106.625

)104.8(3.142

v = nZ

× 2.188 × 108 cm/sec ----------- (iii)

v = n

10188.2 8

cm/sec (For hydrogen , Z = 1)

v1 = 2.188 × 108 cm/sec

v2 =

21

× 2.188 × 108 cm/sec = 1.094 × 108 cm/sec

v3 =

31

× 2.188 × 108 cm/sec = 0.7293 × 108 cm/sec

Here v1, v

2 and v

3 are the velocities of electron in first,

second and third Bohr orbit in hydrogen.

NEUTRONS

In 1932, James Chadwick bombarded the elementberyllium with - particles. He observed the emissionof a radiation with the following properties -

(i) The radiation was highly penetrating.

(ii) The radiation remained unaffected in the electricor magnetic field i.e. the radiation was neutral.

(iii) The particle constituting the radiation had the samemass as that of the proton. These neutral particleswere called neutrons.

)(Beryllium

Be94 +

particle)(á

He42

(Carbon)

C126 +

(Neutron)

n10

COMPARATIVE STUDY OF ELECTRON,

PROTON AND NEUTRON

Property Electron Proton Neutron

Symbol e p n

Nature Negatively charged Positively charged Neutral

Relative charge

-1 +1 0

Absolute charge

�1.602 × 10-19 C +1.602 × 10

-19 C 0

Relative mass

1 1

Absolute mass

9.109 × 10-28 g 1.6725 × 10

-24 g 1.6748 × 10-24 g

18371

ATOMIC STRUCTURE

An atom consists of two parts -(a) Nucleus(b) Extra - nuclear region

(a) Nucleus :

Nucleus is situated at the centre of an atom. All theprotons & neutrons are situated in the nucleus,therefore, the entire mass of an atom is almostconcentrated in the nucleus. The overall charge ofnucleus is positive due to the presence of positivelycharged protons (neutrons have no charge). Theprotons & neutrons are collectively called nucleons.

Note :

The radius of the nucleus of an atom is of the order of10�13 cm and its density is of the order of 1014 g/cm3.

(b) Extra Nuclear Region :

In extra nuclear part or in the region outside thenucleus, electrons are present which revolve aroundthe nucleus in orbits of fixed energies. These orbitsare called energy levels. These energy levels aredesignated as K, L, M, N & so on.

(i) The maximum number of electrons that can be

accommodated in a shell is given by the formula

2n2.(n = shell number i.e. 1,2,3 -------)

7777PAGE # 7

Shell n 2n2 max. no.of electronsK 1 2(1)2 2L 2 2(2)2 8M 3 2(3)2 18N 4 2(4)2 32

K L

2 8

M N

18 32+N

ucleus Electron shells

Maximum number ofelectrons which can be accommodated in the various shells

First energy levelSecond energy level

Third energy levelFourth energy level

(ii) Each energy level is further divided into subshellsdesignated as s,p,d,f .1st shell (K) contains 1 subshell

(s)

2nd shell (L) contains

2 subshells (s,p)

3rd shell (M) contains 3 subshells (s,p,d)4th shell (N) contains 4 subshells (s,p,d,f).

(iii) Shells are divided into sub-shells, sub shellsfurther contain orbitals.

(A) An orbital may be defined as

�A region in the three - dimensional space around the

nucleus where the probability of finding the electron ismaximum.�

(B)The maximum capacity of each orbital is that of twoelectrons.

Note :

The maximum number of orbitals that can be presentin a shell is given by the formula n2.

(C) Types of orbitals :

(1) s-orbitals : The s-subshell contains just one orbitalwhich is non-directional & spherically symmetrical inshape. The maximum number of electrons which canbe accommodated in s-orbital is 2.

Z

X

Y

s- orbital(2) p - orbitals : The p-subshell contains threeorbitals which have dumb-bell shape and a directionalcharacter. The three p-orbitals are designated as p

x,

py & p

z which are oriented in the perpendicular axis

(x,y,z). The maximum number of electrons which canbe accommodated in the p subshell is 6 (2 electronsin each of three orbitals).

yy

px

z

x

y

py

z

x

pz

y

z

x

(3) d - orbitals : The d-subshells contains 5 orbitals whichare double dumb-bell in shape. These orbitals aredesignated as d

xz, d

xy, d

yz, 22 yxd

, 2zd . The d-subshell

can accommodate a maximum of 10 electrons.

yx

dxy

z

yx

dxz

z

yx

dyz

z

z

22

z

dz2

x yyx

dx � y

(4) f-orbitals : The f-subshell contains 7 orbitals whichare complex in structure.The f-subshell canaccommodate a maximum of 14 electrons.

Note :

Letters s, p, d & f have originated from the wordssharp, principal, diffused & fundamental respectively.(iv) Differences between orbit and orbital :

S.No. Orbit Orbital

1It is well defined circular path around the nucleus in which the electron revolves.

It is a region in three dimensional space around the nucleus where the probability of finding electron is maximum.

2 It is circular in shape.s,p and d-orbitals are spherical, dumb-bell and double dumb-bell in shape respectively.

3It represents that an electron moves around the nucleus in one plane.

It represents that an electron can move around nucleus along three dimensional space (x,y and z axis).

4

It represents that position as well as momentum of an electron can be known simultaneously with certainty. It is against Heisenberg's uncertainty principle.

It represents that position as well as momentum of an electron cannot be known simultaneously with certainty. It is in accordance with Heisenberg's uncertainty principle.

5The maximum number of electrons in an

orbit is 2n2 where 'n' is the number of the orbit.

The maximum number of electrons in an orbital is two.

8888PAGE # 8

Note :Heisenberg�s uncertainty principle - �It is impossible

to determine exactly and simultaneously both theposition and momentum (or velocity) of an electronor of any other moving particle.�

QUANTUM NUMBERS

To describe the position and energy of electron in anatom, four numbers are required, which are knownas quantum numbers.Four quantum numbers are :(a) Principal quantum number(b) Azimuthal quantum number(c) Magnetic quantum number(d) Spin quantum number

(a) Principal Quantum Number :

(i) It is denoted by �n�.

(ii) It represents the name, size and energy of theorbit or shell to which the electron belongs.

(iii) Higher is the value of �n� , greater is the distance

of the shell from the nucleus.r

1 < r

2 < r

3 < r

4 < r

5 < ----

(iv) Higher is the value of �n�, greater is the magnitude

of energy.E

1 < E

2 < E

3 < E

4 < E

5 ----

(v) Maximum number of electrons in a shell is givenby 2n2.Shell Max. number of electronsFirst (n =1) 2 × 12 = 2Second (n = 2) 2 × 22 = 8Third ( n = 3) 2 × 32 = 18Fourth ( n = 4) 2 × 42 = 32

(vi) Angular momentum can also be calculated usingprincipal quantum number.

2ðnh

mvr

(vii) Value of n is from 1 to

(viii) Every shell is given a specific alphabetic name.First shell (n = 1) is known as K shell.Second shell (n = 2) is known as L shell.Third shell (n = 3) is known as M shell and so on.

Note :Principal quantum number was given by Bohr.

(b) Azimuthal Quantum Number :

(i) It is represented by ��.

Note :

Azimuthal quantum number is also called angularquantum number, subsidiary quantum number or

secondary quantum number.(ii) For a given value of n values of are from 0 to n �1Value of n Values of 1 (1st shell) 02 (2nd shell) 0,13 (3rd shell) 0,1,24 (4th shell) 0,1,2,3

(iii) It represents the sub-shell present in shell. = 0 represents s sub shell. = 1 represents p sub shell. = 2 represents d sub shell. = 3 represents f sub shell.

(iv) Number of sub-shell in a shell = Principal quantumnumber of shell.

(v) Maximum value of is always less than the valueof n. So 1p, 1f, 2d, 2f, 3f subshells are not possible.s will start from 1sp will start from 2pd will start from 3df will start from 4f

(vi) Relative energy of various sub-shell in a shell areas follows -s < p < d < f

(vii) Subshells having equal values but with different nvalues have similar shapes but their sizes increasesas the value of �n� increases. 2s-subshell is greater in

size than 1s- subshell. Similarly 2p, 3p, 4p subshellshave similar shapes but their sizes increase inorder 2p < 3p < 4p.

(viii) Maximum no. of electrons present in a subshell= 2 (2 +1)Subshell Max. electronss ( = 0) 2 (2 × 0 +1) = 2

p ( = 1) 2 (2 × 1 +1) = 6

d ( = 2) 2 (2 × 2 +1) = 10

f ( = 3) 2 (2 × 3 +1) = 14

Note :Azimuthal quantum number was given bySommerfeld.

(c) Magnetic quantum number :

(i) It is denoted by �m�.

(ii) It represents the orbitals present in sub-shell. Anorbital can be defined as :

�Region in the three - dimensional space around the

nucleus where the probability of finding an electron ismaximum�.

(iii) For a given value of , values of m are from �through 0 to +. m0 01 �1, 0, +1

2 �2, �1, 0, +1, +2

3 �3, �2, �1, 0, +1, +2, +3

(iv) Maximum number of orbitals in a sub-shell= (2+1)Sub shell Orbitalss ( = 0) (2 × 0 +1) = 1

p ( = 1) (2 × 1 +1) = 3

d ( = 2) (2 × 2 +1) = 5

f ( = 3) (2 × 3 +1) = 7

(v) Maximum number of orbitals in a shell = n2

Shell Max. orbitalsFirst (n = 1) 12 = 1Second (n = 2) 22 = 4Third (n = 3) 32 = 9Fourth (n = 4) 42 = 16

9999PAGE # 9

(vi) It represents the orientation of orbital in threedimensional space.When l = 0, m = 0, i.e. one value implies that �s�subshell has only one space orientation and hence,it can be arranged in space only in one way along x,yor z axis. Thus, �s� orbital has a symmetrical spherical

shape.

Z

X

Y

s- orbital

When = 1,�m� has three values �1, 0, +1 . It implies

that �p� subshell of any energy shell has three space

orientations, i.e. three orbitals. Each p-orbital hasdumb-bell shape. Each one is disposedsymmetrically along one of the three axis. p orbitalshave directional character.

orbital Pz

Px

Py

m 0 ±1 ±1

yy

px

z

x

y

py

z

x

pz

y

z

x

When = 2 �m� has five values �2, �1, 0, +1, +2. It

implies that d-subshell of any energy shell has fiveorientations, i.e. five orbitals. All the five orbitals arenot identical in shape. Four of the d-orbitals

dxy

, dyz

, dzx

, 22 y�xd contain four lobes while fifth orbital

dz2 consists of only two lobes.

yx

dxy

z

yx

dxz

z

yx

dyz

z

z

22

z

dz2

x yyx

dx � y

There are seven f-orbitals designated as

2yzf , 2xzf , 3zf , ),y�x(x 22f ),y�x(y 22f

),y�x(z 22f and xyzf .

Their shapes are complicated ones.

(vii) Characteristics of orbitals :

(A) All orbitals of a subshell possess same energyi.e., they are degenerate.

(B) All orbitals of the same shell differ in the directionof their space orientation.

(C) Total number of orbitals in a shell is equal to n2.

Note :

Magnetic quantum number was given by Zeeman.

(d) Spin Quantum Number :

(i) It is denoted by �s�.

(ii) It represents the direction of spin of electron aroundits own axis.

(iii) Clockwise spin is represented by +1/2 or andanticlockwise by �1/2 or .

(iv) Maximum two electrons with opposite spin canbe placed in an orbital.

(v) Electrons with same spin are called spin paralleland those with opposite spin are called spin paired.

Note :

Spin quantum number was given by Gold Schmidt.

ELECTRONIC CONFIGURATION OF AN ATOM

(i) The arrangement of the electrons in different shellsis known as the electronic configuration of theelement.

(ii) Each of the orbits can accommodate a fixed numberof electrons. Maximum number of electrons in an orbitis equal to 2n2, where �n� is the number of the orbit.

(iii) Electrons are filled in the increasing order of energy,i.e. K < L < M < N ......

(iv) In the outermost shell of any atom, the maximumpossible number of electrons is 8, except in the firstshell which can have at the most 2 electrons.

Note :If the outermost shell has its full quota of 8 electronsit is said to be an octet. If the first shell has its fullquota of 2 electrons, it is said to be duplet.The pictorial representation of Bohr�s model of

hydrogen, helium, carbon, sodium and calcium atomshaving 1, 2, 6, 11 and 20 electrons respectively areshown in the figure where the centre of the circlerepresents the nucleus.

10101010PAGE # 10

(a) Significance of Electronic Configuration :

The electronic configuration of an atom plays animportant role in determining the chemical behaviourof an element.

(i) When the atoms of an element have completelyfilled outermost shell, the element will be chemicallyunreactive. For example the noble gases (He, Ne, Ar,Kr, Xe and Rn) have completely filled outermost shelli.e. contains 8 electrons (except helium which hastwo valence electrons) in outermost shell.

(ii) When the atom of an element has less than 8electrons in its outermost shell, the element will bereactive.

ORDER OF FILLING OF ELECTRONS IN SUBSHELLS

There are different rules governing the filling ofsubshells. They are described briefly as follows -

(a) Aufbau Principle :

The filling of subshells in atoms is based on theirenergies. Electrons first occupy the subshell withlowest energy and progressively fil l the othersubshells in increasing order of energy.

Note :

The subshell with lowest energy is filled first.The order of energy of different subshells of an atom is -

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p< 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p.

The number present before the subshells like 1,2,3------ represents the number of the shell i.e. n.The order of fi l l ing of different sub-shells isrepresented diagrammatically as follows :

1s

2s 2p

3p 3d

4d

5d 5f

4f

6d

4p

5p

6p

7p

3s

4s

5s

6s

7s

(b) Pauli�s Exclusion Principle :

According to Pauli�s exclusion principle �an orbital

cannot accommodate more than two electrons. If

there are two electrons in an orbital they must have

opposite spins.�

(c) Hund�s Rule of Maximum Multiplicity :

According to this rule :�no electron pairing takes place in the orbitals with

equivalent energy until each orbital in the givensubshell contains one electron & the spins of allunpaired electrons are parallel i.e. in the samedirection�.

Electronic configuration of some elements -

Atom ic num ber

Sym bol of the elem ent

Nam e of the elem ent

Electronic configuration

1 H Hydrogen 1s 1

2 He Helium 1s 2

3 Li Lithium 1s 2, 2s 1

4 Be Beryllium 1s 2, 2s 2

5 B Boron 1s 2, 2s 2 , 2p1

6 C Carbon 1s 2, 2s 2 , 2p2

7 N Nitrogen 1s 2, 2s 2 , 2p3

8 O Oxygen 1s 2, 2s 2 , 2p4

9 F Fluorine 1s 2, 2s 2 , 2p5

10 Ne Neon 1s 2, 2s 2 , 2p6

11 Na Sodium 1s 2, 2s 2 , 2p6 ,3s 1

12 Mg Magnes ium 1s 2, 2s 2 , 2p6 ,3s 2

13 Al Alum inium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p1

14 Si Silicon 1s 2, 2s 2 , 2p6 ,3s 2 ,3p2

15 P Phos phorus 1s 2, 2s 2 , 2p6 ,3s 2 ,3p3

16 S Sulphur 1s 2, 2s 2 , 2p6 ,3s 2 ,3p4

17 Cl Chlorine 1s 2, 2s 2 , 2p6 ,3s 2 ,3p5

18 Ar Argon 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6

19 K Potas s ium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, , 4s 1

20 Ca Calcium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, , 4s 2

21 Sc Scandium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d1, 4s 2

22 Ti Titanium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d2, 4s 2

23 V Vanadium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d3, 4s 2

24 Cr Chrom ium 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d5, 4s 1

25 Mn Manganes e 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d5, 4s 2

26 Fe Iron 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d6, 4s 2

27 Co Cobalt 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d7, 4s 2

28 Ni Nickel 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d8, 4s 2

29 Cu Copper 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d10, 4s 1

30 Zn Zinc 1s 2, 2s 2 , 2p6 ,3s 2 ,3p6, 3d10, 4s 2

VALENCE SHELL AND VALENCE ELECTRONS

The outermost shell of an atom is known as thevalence shell. The electrons present in the valence

shell of an atom are known as valence electrons.

The remainder of the atom i.e. the nucleus and otherelectrons is called the core of the atom. Electronspresent in the core of an atom are known as coreelectrons.e.g.The electronic configuration of the sodium (Na) atomis :-Na (11) K L M

2 8 1

Thus, valence electron in Na atom = 1 and coreelectrons in Na atom = 2 + 8 = 10

(a) Significance of Valence Electrons :

(i) The valence electrons of an atom are responsiblefor, and take part in, chemical changes.

(ii) The valence electrons in an atom determine themode of chemical combination.

11111111PAGE # 11

(iii) The valence electrons determine the combiningcapacity or the valency of the atom. The number ofelectrons in an atom that actually take part in bondformation is known as the valency of the element.e.g. In the carbon atom, there are four valenceelectrons.

C(6) K L2 4

The carbon atom is capable of forming four bonds.Hence, the valency of carbon is four.

(iv) If the outermost shell of an atom is completelyfilled, its valency is zero.The outermost shells of helium, neon, argon, kryptonetc. are completely filled. Hence the valency of theseelements is zero.

(v) Elements having the same number of valenceelectrons in their atoms possess similar chemicalproperties.

e.g. All alkali metals have one valence electron intheir atoms. Hence, their chemical properties aresimilar.

(vi) Elements having different number of valenceelectrons in their atoms show different chemicalproperties.

e.g. Let us consider the electronic configuration ofalkali metals and halogens. Alkali metal atoms havesingle valence electron whereas halogen atoms haveseven valence electrons. It is observed that thechemical properties of the alkali metals are entirelydifferent from those of halogens.(vii) The number of the valence shell in the atom of anelement determines the period number of theelement in the periodic table.e.g. Sodium (Na) :

Valence shell number = 3.period number = 3

Calcium (Ca)Valence shell number = 4period number = 4

(viii) Elements with 1, 2 or 3 valence electrons in theiratoms are metals.

Note :Hydrogen and helium are exceptions to this rule.Hydrogen and helium atoms have 1 and 2 valenceelectrons respectively, but they are non-metals.

(ix) Elements with 4, 5, 6, 7 or 8 valence electrons intheir atoms are non metals.e.g. carbon (C), nitrogen (N) and oxygen (O) are nonmetals.

6C = 2,4

7N = 2,5

8O = 2,6

Note :Whether the atom while combining with other atomscan form ionic or covalent bonds is determined bythe number of valence electrons present in the atom.

VALENCY

Valency of an element is the combining capacity ofthe atoms of the element with atoms of the same ordifferent elements. The combining capacity of theatoms was explained in terms of their tendency toattain a fully filled outermost shell (stable octet orduplet)

Note :The number of electrons gained, lost or contributedfor sharing by an atom of the element gives us directlythe combining capacity or valency of the element.

Valency of an element is determined by the numberof valence electrons in an atom of the element.The valency of an element = number of valenceelectrons (if the number of valence electrons is1 to 4)The valency of an element = 8� number of valence

electrons. (if the number of valence electrons is 4 to 8)

ISOTOPES

The isotopes of an element have the same atomic

number but different atomic masses.

Note :

The term isotope was given by Margaret Todd.

The difference in their masses is due to the presenceof different number of neutrons.

e.g. Isotopes of hydrogen :

Hydrogen isotopes

1. Atomic number 1 1 1

2. No. of protons 1 1 1

3. No. of electrons 1 1 1

4. Mass number 1 2 3

5. No. of neutrons 0 1 2

H11

ot iumPr

H21

Deu te r ium

H31

Tr itium

Isotopes of oxygen :

Oxygen isotopes

1. Atomic number 8 8 8

2. No. of protons 8 8 8

3. No. of electrons 8 8 8

4. Mass number 16 17 18

5. No. of neutrons 8 9 10

O168 O17

8 O188

Note :All the isotopes of an element have identical chemical

properties.

12121212PAGE # 12

(a) Characteristics of Isotopes :

(i) The physical properties of the isotopes of anelement are different. This is due to the fact thatisotopes have different numbers of neutrons in theirnuclei. Hence mass, density and other physicalproperties of the isotopes of an element are different.

(ii) All the isotopes of an element contain the samenumber of electrons. So, they have the sameelectronic configuration with the same number ofvalence electrons. Since the chemical properties ofan element are determined by the number of valenceelectrons in its atom, all the isotopes of an elementhave identical chemical properties.

(b) Reason for the Fractional Atomic Masses

of Elements :

The atomic masses of many elements are in fractionand not whole number. The fractional atomic massesof elements are due to the existence of their isotopeshaving different masses.e.g.The atomic mass of chlorine is 35.5 u. Chlorine has

two isotopes Cl3517 and Cl37

17 with abundance of 75%

and 25% respectively. Thus the average mass of achlorine atom will be 75% of 35 and 25% of 37, whichis 35.5 u.So, Average atomic mass of chlorine

= 35 × 10075

+ 37 × 10025

= 1002625

+ 100925

= 26.25 + 9.25= 35.5 u.Thus, the average atomic mass of chlorine is 35.5 u.

(c) Applications of Radioactive Isotopes :

(i) In agriculture : Certain elements such as boron,cobalt, copper, manganese, zinc and molybdenumare necessary in very minute quantities for plantnutrition. By radioactive isotopes we can identify thepresence and requirement of these element in the

nutrition of plants.

(ii) In industry : Isotopes are used for coating on thearm of clock to see in dark. To identify the cracks inmetal casting.

(iii) In medicine :Thyroid, bone diseases, braintumors and cancer are controlled or destroyed with

the help of radioactive isotope PÉ,Na, 3215

13153

2411 etc.

(iv) Determination of the mechanism of chemicalreaction by replacing an atom or molecule by itsisotopes.

(v) In carbon dating : Will and Libby (1960) developedthe technique of radiocarbon to determine the age ofplant, fossil and archeological sample.

Note :Isotopes (Like Uranium-238) are used in Nuclearreactor to produce energy and power.

ISOBARS

The atoms of different elements with different atomicnumbers, which have same mass number are calledisobars.

e.g. C146 and N14

7 are isobars.

Ar4018 and Ca40

20 are isobars.

Isobars

1. Atomic number 18 20

2. Mass number 40 40

3. No. of electrons 18 20

4. No. of protons 18 20

5. No. of neutrons 22 20

6. Electronic configuration 2, 8, 8 2, 8, 8, 2

Ar4018 Ca40

20

NoteIsobars contain different number of electrons,protons and neutrons.

ISOTONES

Isotones may be defined as the atoms of differentelements containing same number of neutrons.

e.g. C136 and N14

7 (Both contain 7 neutrons)

Si3014 , P31

15 and S3216 (All three contain 16 neutrons)

ISOELECTRONIC

Ion or atom or molecule which have the same numberof electrons are called as isoelectronic species.

e.g. Cl17 Ar18K19

220Ca

No. of electrons 18 18 18 18

PAGE # 13

ACIDS, BASES AND SALTS

ACIDS

Substances with sour taste are regarded as acids.

Lemon juice, vinegar, grape fruit juice and spoilt milk

etc. taste sour since they are acidic. Many substances

can be identified as acids based on their taste but

some of the acids like sulphuric acid have very strong

action on the skin which means that they are corrosive

in nature. In such cases it would be according to

modern definition -

An acid may be defined as a substance which

releases one or more H+ ions in aqueous solution.

Acids are mostly obtained from natural sources.

(a) Classification of Acids :

(i) Classification of acids on the basis of theirSourceOn the basis of their source, acids can be classifiedin two categories :(A) Organic acids (B) Inorganic acids

(A) Organic acidsThe acids which are usually obtained fromorganisms are known as organic acids. Oxalic acid[(COOH)2], acetic acid (CH3COOH) etc. are verycommon examples of organic acids. Some otherorganic acids with their natural sources are given inthe following Table.

Some Organic Acids with Their Natural Sources

S.No. Organic acid Natural sources S.No. Organic acid Natural sources

1 Acetic acid Micro-organism 7 Oleic acid Olive oil

2 Citric acid Citrus fruits (likeorange and lemon)

8 Stearic acid Fats

3 Butyric acid Rancid butter 9 Amino acid Proteins

4 Formic acid Sting of bees and ants 10 Uric acid Urine

5 Lactic acid Sour milk 11 Tartaric acid Tamarind

6 Malic acid Apples 12 Oxalic acid Tomatoes

It may be noted that all organic acids contain carbonas one of their constituting elements. These are weakacids and, therefore, do not ionise completely in theiraqueous solutions. Since these acids do not ionisecompletely in their aqueous solutions, therefore, theirsolutions contains both ions as well asundissociated molecules. For example, formic acid�saqueous solution contains H3O+, HCOO� as well asundissociated HCOOH molecules.

HCOOH + H2O H3O+ + HCOO�

Formic acid Hydronium ion Formate ion

(B) Inorganic Acids. The acids which are usually obtained from mineralsare known as inorganic acids. Since the acids areobtained from minerals, therefore, these acids arealso called mineral acids. Some common examplesof inorganic acids are : Hydrochloric acid (HCl),Sulphuric acid (H2SO4), Nitric acid (HNO3) etc. It may be pointed out that except carbonic acid(H2CO3), these acids do not contain carbon. Acidslike HCl, H2SO4 and HNO3 are strong acids whichionise completely in their aqueous solutions and,therefore, their aqueous solutions do not contain anyundissociated molecules.

(ii) Classification of acids on the basis of their

Basicity :

The basicity of an acid is defined as the number of

hydronium ions [H3O+ (aq.)] that can be produced by

the complete ionisation of one molecule of that acid

in aqueous solution.

For example, basicity of HCl, H2SO4, H3PO4 is 1, 2

and 3 respectively because one molecule of these

acids, on ionisation, produces 1, 2 and 3 hydronium

ions in aqueous solution respectively.

It may be pointed out here that the basicity of an acid

is determined by number of hydronium ions produced

per molecule of an acid on ionisation and not the

number of hydrogen atoms present in one molecule

of an acid. For example, basicity of acetic acid

(CH3COOH) is 1 because one molecule of acetic acid,

on ionisation in aqueous solution, produces one

hydronium ion although one molecule of acetic acid

contains four hydrogen atoms.

CH3COOH + H2O H3O+ + CH3COO�

Acetic acid Hydronium ion Acetate ionOn the basis of basicity, the acids can be classifiedas under :


PAGE # 14

(A) Monobasic Acids :When one molecule of an acid on complete ionisationproduces one hydronium ion (H3O+) in aqueoussolution, the acid is said to be a monobasic acid.

Examples of Monobasic Acids.Some examples of monobasic acids are :(i) Hydrochloric acid (HCl)(ii) Hydrobromic acid (HBr)(iii) Nitric acid (HNO3)(iv) Acetic acid (CH3COOH)(v) Formic acid (HCOOH)

Characteristics of Monobasic Acids.Two important characteristics of monobasic acidsare :

(i) A monobasic acid ionises in one step in aqueoussolution. For example,HCl + H2O H3O+ + Cl�

(Single step ionisation)

(ii) A monobasic acid forms only single salt or a normalsalt. For example,

HCl + NaOH NaCl + H2O Sodium chloride (Normal salt)

(B) Dibasic Acids :When one molecule of an acid on complete ionisationproduces two hydronium ions (H3O+) in aqueoussolution, the acid is said to be a dibasic acid.

Examples of Dibasic Acids :Some examples of dibasic acids are :(i) Sulphuric acid (H2SO4)(ii) Sulphurous acid (H2SO3)(iii) Carbonic acid (H2CO3)(iv) Oxalic acid [(COOH)2](v) Hydrofluoric acid (HF)

Characteristics of Dibasic Acids :Two important characteristics of dibasic acids are :

(i) A dibasic acid ionises in two steps in aqueoussolution. For example, sulphuric acid which is a dibasicacid ionises to produce bisulphate ion (HSO4

�) in thefirst step which further ionises to produce sulphate ion(SO4

2�) in the second step.H2SO4 + H2O H3O+ + HSO4

�

Sulphuric acid Bisulphate ion

HSO4� + H2O H3O+ + SO4

2�

Sulphate ion

\ (ii) Because of the presence of two replaceablehydrogen ions, a dibasic acid forms two series of saltsi.e., an acid salt and a normal salt. For example, H2SO4reacts with NaOH to form NaHSO4 (an acid salt) andNa2SO4 (a normal salt)NaOH + H2SO4 NaHSO4 + H2O

Sodium hydrogensulphate(An acid salt)

2NaOH + H2SO4 Na2SO4 + 2H2OSodium sulphate(Normal salt)

(C) Tribasic Acids :

When one molecule of an acid on complete ionisationproduces three hydronium ions (H3O+) in aqueoussolution, the acid is said to be a tribasic acid.An example of tribasic acids is Phosphoric acid(H3PO4).

(D) Tetrabasic Acids :

When one molecule of an acid on complete ionisationproduces four hydronium ions (H3O+) in aqueoussolution, the acid is said to be a tetrabasic acid.An example of tetrabasic acids is silicic acid (H4SiO4).

(iii) Classification of acids on the basis of theirstrength :

We know that acids ionise in the aqueous solution toproduce hydronium ions. So, the strength of an aciddepends upon the degree of ionisation, usuallydenoted by the letter alpha ().Degree of ionisation of an acid ()

= Number of molecules of the acid undergoing ionisation

100Total number of acid molecules

More the degree of ionisation () of an acid, more strongerit will be. Generally, if the degree of ionisation () for anacid is greater than 30%, it is considered to be a strongacid. If it is less than 30%,it is considered to be a weakacid.On the basis of degree of ionisation, the acids can beclassified as under:

(A) Strong Acids :The acids which undergo almost complete ionisationin a dilute aqueous solution, thereby producing a highconcentration of hydronium ions (H3O

+) are known asstrong acids.

Examples of strong acids :Some examples of strong acids are :(i) Hydrochloric acid (HCl)(ii) Sulphuric acid (H2SO4)(iii) Nitric acid (HNO3)

All these three mineral acids are considered to bestrong acids because they ionise almost completelyin their dilute aqueous solutions.

(B) Weak Acids :The acids which undergo partial or incompleteionisation in a dilute aqueous solution, therebyproducing a low concentration of hydronium ions(H3O

+) are known as weak acids.

Examples of weak acids :Some examples of weak acids are :(i) Acetic acid (CH3COOH)(ii) Formic acid (HCOOH)(iii) Oxalic acid [(COOH)2](iv) Carbonic acid (H2CO3)(v) Sulphurous acid (H2SO3)(vi) Hydrogen sulphide (H2S)(vii) Hydrocyanic acid (HCN)

The aqueous solution of weak acids contain both ionsas well as undissociated molecules.

PAGE # 15

(iv) Classification on the basis of Concentration ofthe Acid :

By the term concentration, we mean the amount ofwater present in the given sample of acid solution inwater.

(A) Concentrated Acid :The sample of an acid which contains very small or noamount of water is called a concentrated acid.

(B) Dilute Acid :The sample of an acid which contains far more amountof water than its own mass is known as a dilute acid

It must be mentioned here that concentration of anacid simply tells the amount of water in the acid. It maynot be confused with strength of an acid, which is ameasure of concentration of hydronium ion it producesin aqueous solution.

A concentrated acid may not necessarily be a strongacid while a dilute acid may not necessarily be a weakacid. A strong acid will remain strong even if it is dilutebecause it produces a large concentration of hydroniumions in aqueous solution. On the other hand, a weakacid will remain weak even when concentratedbecause it will produce lesser concentration ofhydronium ions in aqueous solution.

CHEMICAL FORMULAE, TYPES AND USES OF SOME COMMON ACIDS

Name Type Chemical Formula Where found or used

Carbonic acid Mineral acid H2CO3 In soft drinks and lends fizz.

Nitric acid Mineral acid HNO3

Used in the manufacture of explosives (TNT, Nitroglycerine) and fertilizers (Ammonium nitrate,

Calcium nitrate, Purification of Au, Ag)

Hydrochloric acid Mineral acid HClIn purification of common salt, in textile industry

as bleaching agent, to make aqua regia, in stomach as gastric juice, used in tanning industry

Sulphuric acid Mineral acid H2SO4

Commonly used in car batteries, in the manufacture of fertilizers (Ammonium

sulphate, super phosphate) detergents etc, in paints, plastics, drugs, in manufacture of artificial

silk, in petroleum refining.

Phosphoric acid Mineral acid H3PO4 Used in antirust paints and in fertilizers.

Formic acid Organic acid HCOOHFound in the stings of ants and bees, used in tanning leather, in medicines for treating gout.

Acetic acid Organic acid CH3COOH Found in vinegar, used as solvent in the manufacture of dyes and perfumes.

Lactic acid Organic acid CH3CH(OH)COOH Responsible for souring of milk in curd.

Benzoic acid Organic acid C6H5COOH Used as a food preservative.

Citric acid Organic acid C6H8O7 Present in lemons, oranges and citrus fruits.

Tartaric acid Organic acid C4H6O6 Present in tamarind.

(b) Chemical Properties of Acids :

(i) Action with metals :Dilute acids like dilute HCl and dilute H

2SO

4 react with

certain active metals to evolve hydrogen gas.

2Na(s) + 2HCl (dilute) 2NaCl(aq) + H2(g)

Mg(s) + H2SO

4 (dilute) MgSO

4(aq) + H

2(g)

Metals which can displace hydrogen from dilute acidsare known as active metals. e.g. Na, K, Zn, Fe, Ca, Mgetc.

Zn(s) + H2SO

4 (dilute) ZnSO

4(aq) + H

2(g)

The active metals which lie above hydrogen in theactivity series are electropositive and more reactive innature. Their atoms lose electrons to form positive ionsand these electrons are accepted by H+ ions of theacid. As a result, H

2 is evolved.

e.g.Zn(s) Zn2+ (aq) + 2e�

2H+(aq) + SO4

2� (aq) + 2e� H2(g) + SO

42�(aq)

Zn(s) + 2H+(aq) Zn++(aq) + H2(g)

(ii) Action with metal oxides :Acids react with metal oxides to form salt and water.These reactions are mostly carried out upon heating.e.g.ZnO(s) + 2HCl (aq) ZnCl

2(aq) + H

2O()

MgO(s) + H2SO

4(aq) MgSO

4(aq) + H

2O()

CuO(s) + 2HCl(aq.) CuCl2(aq) + H

2O()

(Black) (Bluish green)

(iii) Action with metal carbonates and metalbicarbonates : Both metal carbonates and bicarbonatesreact with acids to evolve CO

2 gas and form salts.

e.g. CaCO

3(s)+ 2HCl(aq) CaCl

2(aq) + H

2O() + CO

2(g)

Calcium Calcium carbonate chloride

2NaHCO3(s) + H2SO4(aq) Na2SO4(aq) + 2H2O(aq) + 2CO2(g)

Sodium Sodiumbicarbonate sulphate

(iv) Action with bases :Acids react with bases to give salt and water.HCl (aq) + NaOH(aq) NaCl + H

2O

PAGE # 16

BASE

Substances with bitter taste and soapy touch areregarded as bases. Since many bases like sodiumhydroxide and potassium hydroxide have corrosiveaction on the skin and can even harm the body, soaccording to the modern definition -A base may be defined as a substance capable ofreleasing one or more OH¯ ions in aqueous solution.

(a) Alkalies :

Some bases like sodium hydroxide and potassiumhydroxide are water soluble. These are known asalkalies. Therefore water soluble bases are known asalkalies eg. KOH, NaOH.Bases like Cu(OH)

2, Fe(OH)

3 and Al(OH)

3 these are

not alkalies.

A list of a few typical bases along with their chemicalformulae and uses is given below-

Name Commercial Name Chemical Formula

Uses

Sodium hydroxide Caustic soda NaOHIn manufacture of soap, paper, pulp, rayon, refining of petroleum etc.

Potassium hydroxide Caustic potash KOHIn alkaline storage batteries, manufacture of soap, absorbing CO2 gas etc.

Calcium hydroxide Slaked lime Ca(OH)2In manufacture of bleaching powder, softening of hard water etc.

Magnesium hydroxide Milk of magnesia Mg(OH)2 As an antacid to remove acidity from stomach.

Aluminium hydroxide � Al(OH)3 As foaming agent in fire extinguishers.

Ammonium hydroxide � NH4OH In removing grease stains from clothes and in cleaning window panes.

(b) Chemical Properties :

1. Action with metals :Metals like zinc, tin and aluminium react with strongalkalies like NaOH (caustic soda), KOH (causticpotash) to evolve hydrogen gas.

Zn(s) + 2NaOH(aq) Na2ZnO

2(aq) + H

2(g)

Sodium zincate

Sn(s) + 2NaOH(aq) Na2SnO

2(aq) + H

2(g)

Sodium stannite

2Al(s)+ 2NaOH + 2H2O 2NaAlO

2(aq) + 3H

2(g)

Sodium meta aluminate

2. Action with non-metallic oxides :Acids react with metal oxides, but bases react withoxides of non-metals to form salt and water.e.g.2NaOH(aq) + CO

2(g) Na

2CO

3(aq) + H

2O()

Ca(OH)2(s) + SO

2(g) CaSO

3(s) + H

2O()

Ca(OH)2(s) + CO

2(g) CaCO

3(s) + H

2O()

CONDUCTING NATURE OF ACID

AND BASE SOLUTIONS

Acids are the substances which contain one or morehydrogen atoms in their molecules which they canrelease in water as H+ ions. Similarly, bases are thesubstances which contain one or more hydroxyl groupsin their molecules which they can release in water asOH¯ ions . Since the ions are the carrier of charge

therefore, the aqueous solutions of both acids andbases are conductors of electricity.

Experiment :

In a glass beaker, take a dilute solution of hydrochloric

acid (HCl). Fix two small nails of iron in a rubber cork in

the beaker as shown in the figure. Connect the nails to

the terminals of a 6 volt battery through a bulb. Switch

on the current and bulb will start glowing. This shows

that the electric current has passed through the acid

solution. As the current is carried by the movement of

ions, this shows that in solution HCl has ionised to

give H+ and Cl� ions. Current will also be in a position

to pass if the beaker contains in it dilute H2SO

4 (H+ ions

are released in aqueous solution). Similarly, aqueous

solutions containing NaOH or KOH will also be

conducting due to release of OH� ions.

Battery Bulb in circuit

Switch

Dilute HCl

Iron Nails

Rubber cork

Bulb will not glow if glucose (C6H

12O

6) or ethyl alcohol

(C2H

5OH) solution is kept in the beaker. This means

that both of them will not give any ions in solution.

PAGE # 17

Acids BasesSour in taste. Bitter in taste.Change Colours of indicatorse.g. litmus turns from blue tored, phenolphthalein remains colourless.

Change colours of indicators e.g.litmus turns from red to blue phenolphthalein turns fromcolourless to pink.

Show electrolytic conductivityin aqueous solution.

Show electrolytic conductivity inaqueous solution

Acidic properties disappearwhen react with bases (Neutralization)

Basic properties disappear whenreact with acids (Neutralization)

Acids decompose carbonatesalts.

No decomposition of carbonatesalts by bases

ROLE OF WATER IN THE IONISATION

OF ACIDS AND BASES

Substances can act as acids and bases only in thepresence of water (in aqueous solution). In dry statewhich is also called anhydrous state, these characterscannot be shown. Actually, water helps in the ionisationof acid or base by separating the ions. This is alsoknown as dissociation and is explained on the basisof a theory called Arrhenius theory of acids and bases.In the dry state, hydrochloric acid is known as hydrogenchloride gas i.e. HCl(g). It is not in the position to giveany H+ ions. Therefore, the acidic character is notshown. Now, let us pass the gas through water takenin a beaker with the help of glass pipe. H

2O molecules

are of polar nature which means that they have partial

negative charge (�) on oxygen atom and partial positive

charge (+) on hydrogen atoms. They will try to form asort of envelope around the hydrogen atoms as wellas chlorine atoms present in the acid and thus help intheir separation as ions. These ions are said to behydrated ions.

HCl(g) + H2O H

3O+ + Cl�(aq)

(Hydrated ions)

The electrical current is carried through these ions.The same applies to other acids as well as bases.Thus we conclude that -(i) acids can release H+ ions only in aqueous solution.(ii) base can release OH� ions only in aqueous solution.(iii) hydration helps in the release of ions from acidsand bases.

DILUTION OF ACIDS AND BASES

Acids and bases are mostly water soluble and can bediluted by adding the required amount of water. Withthe addition of water the amount of acid or base perunit volume decrease and dilution occurs. The processis generally exothermic in nature. When a concentratedacid like sulphuric acid or nitric acid is to be dilutedwith water, acid should be added dropwise to watertaken in the container with constant stirring.

THEORIES OF ACIDS AND BASES

(a) Arrhenius Theory :

This concept was given in 1884 .

According to this theory all substances which give H+

ions when dissolved in water are called acids, while

those which ionise in water to give �OH ions are calledbases.The main points of this theory are -

(i) An acid or base when dissolved in water, splits intoions. This is known as ionisation.

(ii) Upon dilution, the ions get separated from each other.This is known as dissociation of ions.

(iii) The fraction of the acid or base which dissociatesinto ions is called its degree of dissociation and isdenoted by alpha which can be calculated by thefollowing formula :

= moleculesofno.total

mequilibriuatddissociatemoleculesofNo.

(iv) The degree of dissociation depends upon the natureof acid or base. Strong acids and bases are highlydissociated, while weak acids and bases aredissociated to lesser extent.

(v) The electric current is carried by the movement ofions. Greater the ionic mobility more will be theconductivity of the acid or base.

(vi) The H+ ions do not exist as such and exist incombination with molecules of H

2O as H

3O+ ions

(known as hydronium ion).

H+ + H2O H

3O+

HCl + H2O H

3O+ + Cl�

e.g.HA + H

2O H

3O+ + A¯

Acid

H2SO

4 + 2H

2O 2H

3O+ + SO

4�2

Acid

BOH Water B+ + OH¯

Base

NaOH Water Na+ + OH¯

Base

NH4OH Water NH

4+ + OH¯

Base

(A) Limitations of Arrhenius theory :

� It is applicable only to aqueous solutions. For theacidic or basic properties, the presence of water isabsolutely necessary.

� The concept does not explain the acidic or basicproperties of acids or bases in non - aqueous solvents.

� It fails to explain the basic nature of compounds like

NH3, Na

2CO

3 etc., which do not have �OH in their

molecules to furnish �OH ions.

� It fails to explain the acidic nature of non - proticcompounds like SO

2, P

2O

5, CO

2, NO

2 etc., which do not

have hydrogen in their molecules to furnish H+ ions.

� It fails to explain the acidic nature of certain salts likeAlCl

3 etc., in aqueous solutions.

PAGE # 18

(b) Acid Base Concept of Bronsted and

Lowry :

This theory was given by Bronsted, a Danish chemistand Lowry, an English chemist independently in 1923.According to it, an acid is a substance, molecule or ionwhich has a tendency to release the proton(protogenic) and similarly a base has a tendency toaccept the proton (protophilic).

e.g.

HCl + H2O H

3O+ + Cl�

In this reaction, HCl acts as an acid because it donatesa proton to the water molecule. Water, on the otherhand, behaves as a base by accepting a proton.

Note :Bronsted and Lowry theory is also known as protondonor and proton acceptor theory.Other examples :

(i) CH3COOH + H

2O H

3O+ + CH

3COO�

(ii) NH4

+ + H2O H

3O+ + NH

3

(iii) NH3 + H

2O NH

4+ + OH�

In the reactions (i) and (ii) water is acting as a base,while in reaction (iii) it is acting as an acid.Thus watercan donate as well as accept H+ and hence can act asboth acid and base.

Note :The species like H

2O, NH

3, CH

3COOH which can act

as both acid and base are called amphiprotic.

Moreover according to theory, an acid on losing aproton becomes a base, called conjugate base, whilethe base by accepting proton changes to acid calledconjugate acid.

Here CH3COO� ion is conjugate base of CH

3COOH,

while H3O+ ion is conjugate acid of H

2O.

(i) Merits :

(A) Besides water any other solvent, which has thetendency to accept or lose a proton may decide theacidic or basic behaviour of the dissolved substance.

(B) This theory states that the terms acid and base arecomparative. A substance may act as an acid in onesolvent, while as a base in another solvent.

e.g. Acetic acid acts as an acid in water while as abase in HF.

(ii) Demerits :

(A) Many acid - base reactions proceed without H+

transfer.

e.g. SO2 + SO

3 SO2+ + SO

42-

(c) Lewis theory :

The theory was given by G.N. Lewis in 1938. Accordingto it, an acid is a species which can accept a pair ofelectrons, while the base is one which can donate apair of electrons.

Note :It is also known as electron pair donor and electronpair acceptor theory.

e.g.(i) FeCl

3 and AlCl

3 are Lewis acids, because the central

atoms have only six electrons after sharing and needtwo more electrons.

(ii) NH3 is a Lewis base as it has a pair of electrons

which can be easily donated.Lewis acids :- CH

3+, H+, BF

3, AlCl

3, FeCl

3 etc.

Lewis base :- NH3, H

2O, R�O�R, R � OH, CN¯, OH¯

etc.

(i) Characteristics of species which can act as Lewisacids :

(A) Molecules in which the central atom hasincomplete octet : Lewis acids are electron deficientmolecules such as BF

3, AlCl

3, GaCl

3 etc.

H N + AlCl3 3�� [H N 3 AlCl ]3

(B) Molecules in which the central atom has emptyd-orbitals : The central atom of the halides such asTiCl

4, SnCl

4, PCl

3, PF

5, SF

4, TeCl

4. etc., have vacant d-

orbitals. These can, therefore, accept an electron pairand act as Lewis acids.

(C) Simple cations : All cations are expected to act asLewis acid, since they are electron deficient in nature.

(D) Molecules having a multiple bond betweenatoms of dissimilar electronegativity : Typicalexamples of molecules belonging to this class ofLewis acids are CO

2, SO

2 and SO

3.

(ii) Characteristics of species which can act as Lewisbases :

(A) Neutral species having at least one lone pair ofelectrons : For example, ammonia amines, alcoholsetc, act as Lewis bases as they contain a pair ofelectrons.

(B) Negatively charged species or simple anions :For example chloride (Cl�), cyanide (CN�), hydroxide(OH�) ions etc. act as Lewis bases.

(C) Multiple bonded compounds : The compoundssuch as CO, NO, ethylene, acetylene etc. can act asLewis bases.

Note :It may be noted that all Bronsted bases are also Lewisbases, but all Lewis acids are not Bronsted acids.

(iii) Limitations of Lewis theory :

(A) Lewis theory fails to explain the relative strength ofacids and bases.

PAGE # 19

INDICATORS

, An indicator indicates the nature of a particular solutionwhether acidic, basic or neutral. Apart from this, indicatoralso represents the change in nature of the solutionfrom acidic to basic and vice versa. Indicators arebasically coloured organic substances extracted fromdifferent plants. A few common acid base indicators are-

(a) Litmus :

Litmus is a purple dye which is extracted from �lichen�

a plant belonging to variety Thallophyta. It can also beapplied on paper in the form of strips and is availableas blue and red strips. A blue litmus strip, when dippedin an acid solution acquires red colour. Similarly a redstrip when dipped in a base solution becomes blue.

(b) Phenolphthalein :

It is also an organic dye and acidic in nature. In neutralor acidic solution, it remains colourless while in thebasic solution, the colour of indicator changes to pink.

(c) Methyl Orange :

Methyl orange is an orange or yellow coloured dye andbasic in nature. In the acidic medium the colour ofindicator becomes red and in the basic or neutralmedium, its colour remains unchanged.

(d) Red Cabbage Juice :

It is purple in colour in neutral medium and turns redor pink in the acidic medium. In the basic or alkalinemedium, its colour changes to green.

(e) Turmeric Juice :

It is yellow in colour and remains as such in the neutraland acidic medium. In the basic medium its colourbecomes reddish or deep brown.

Note :Litmus is obtained from LICHEN plant.

NEUTRALISATION

It may be defined as a reaction between acid andbase present in aqueous solution to form salt andwater.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O()

Basically neutralisation is the combination betweenH+ ions of the acid with OH� ions of the base to form H

2O.

e.g.H+(aq) + Cl�(aq) + Na+(aq) + OH�(aq) Na+(aq) + Cl�(aq) + H

2O()

H+(aq) + OH�(aq) H2O()

Neutralisation reaction involving an acid and base isof exothermic nature. Heat is evolved in allneutralisation reactions. If both acid and base arestrong, the value of heat energy evolved remains sameirrespective of their nature.

e.g .

HCl (aq) + NaOH (aq) NaCl (aq) + H2O () + 57.1 KJ

Strong Strongacid base

HNO3 (aq) + KOH (aq) KNO3 (aq) + H2O () + 57.1 KJ

Strong Strong

acid base

Strong acids and strong bases are completely ionisedof their own in the solution. No energy is needed fortheir ionisation. Since the cation of base and anion ofacid on both sides of the equation cancel outcompletely, the heat evolved is given by the followingreaction -

H+ (aq) + OH� (aq) H2O () + 57.1 KJ

APPLICATIONS OF NEUTRALISATION

(i) People particularly of old age suffer from acidityproblems in the stomach which is caused mainly dueto release of excessive gastric juices containing HCl.The acidity is neutralised by antacid tablets whichcontain sodium hydrogen carbonate (baking soda),magnesium hydroxide etc.

(ii) The stings of bees and ants contain formic acid. Itscorrosive and poisonous effect can be neutralised byrubbing soap which contains NaOH (an alkali).

(iii) The stings of wasps contain an alkali and itspoisonous effect can be neutralised by an acid likeacetic acid (present in vinegar).

(iv) Farmers generally neutralize the effect of acidity inthe soil caused by acid rain by adding slaked lime(Calcium hydroxide) to the soil.

According to Arrhenius theory, an acid releases H+ ionin aqueous solution. The concentration of these ionsis expressed by enclosing H+ in square bracket i.e. as[H+]. Thus, greater the [H+] ions, stronger will be theacid. However, according to pH scale, lesser the pHvalue, stronger will be the acid. From the abovediscussion, we can conclude that pH value and H+ ionconcentration are inversely proportional to each other.

The relation between them can also be expressed as-

pH = � log [H+] = log

H

1

So, negative logarithm of hydrogen ion concentrationis known as pH.e.g.Let the [H+] of an acid solution be 10�3 M. Its pH can becalculated as -pH = � log [H+]

= � log [10�3]= (�) (�3) log 10

= 3 ( log 10 = 1)

PAGE # 20

Note :Just as the [H+] of a solution can be expressed in termsof pH value, the [OH�] can be expressed as pOH.

Mathematically , pOH = � log [OH�] = log ]OH[

1�

Moreover, pH + pOH = 14.Thus, if pH value of solution is known, its pOH valuecan be calculated.

Note :There are some solutions which have definite pH i.e.,their pH do not change on dilution or on standing forlong. Such solutions are called buffer solutions.

Ex.1 Calculate the concentration of H3O+ ions and OH� ions

in (a) 0.01 M solution of HCl (b) 0.01 M solution ofNaOH at 298 K, assuming that HCl and NaOH arecompletely ionized under the given conditions.

Sol. (a) HCl(aq) + H2O(l) H3O+(aq) + Cl�(aq)

Since HCl is completely ionized [H3O+] = [HCl] = 0.01 Mor 1 × 10�2 mol L�1

[OH�] = ]OH[

]K[

3

w

= 2�

14�

101

101

1 × 10�12 mol L�1

(b) NaOH(aq) Na+(aq) + OH�(aq)Since NaOH is completely ionized[OH�] = [NaOH] = 0.01 Mor 1 × 10�2 mol L�1

[H3O+] =

]OH[

K�

w = 2�

14�

101

101

= 1× 10�12 mol L�1.

Ex-2 Calculate the concentration of H3O+ ions and pH in

0.005 M solution of Ba(OH)2 at 298 K, assuming that

Ba(OH)2

is completely ionized under the givenconditions.

Sol. Ba(OH)2(aq) Ba2+(aq) + 2OH�(aq)

Since Ba(OH)2 is completely ionized and on ionization

its one molecule gives two OH� ions. [OH�] = 2[Ba(OH)

2 ]

= 2 × 0.005 = 0.01 mol L�1

= 1 × 10�2 mol L�1

[H3O+] = ]OH[

K�

w = 2�

14�

101

101

= 1 × 10�12 mol L�1

pH = 12

SALTS

A substance formed by neutralization of an acidwith a base is called a salt.e.g.

Ca(OH)2(aq)+ H

2SO

4(aq) CaSO

4(aq) + 2H

2O()

Cu(OH)2(aq) + 2HNO

3(aq) Cu(NO

3)

2(aq) + 2H

2O()

NaOH(aq) + HCl(aq) NaCl(aq) + H2O()

(a) Classification of Salts :

(i) It is based on their Mode of Formation :

(A) Normal Salts :

The salts which are obtained by complete replacementof the ionisable hydrogen atoms of an acid by a metallicor an ammonium ion are known as normal salts. Forexample, normal salts NaCl and Na2SO4 are formedby the complete replacement of ionisable hydrogenatoms of HCl and H2SO4 respectively

HCl + NaOH NaCl + H2O Sodium chloride (Normal salt)

H2SO4 + 2NaOH Na2SO4 + 2H2O Sodium sulphate (Normal salt)

Some of the salts are given below in the table.

S.No. Parent Acid Normal Salts

1Hydrochloric acid (HCl)

NaCl, KCl, MgCl2, AlCl3, ZnCl2,

CaCl2 and NH4Cl.

2Nitric acid (HNO3)

NaNO3, KNO3, Mg(NO3)2,Al(NO3)3,

Zn(NO3)2, Ca(NO3)2.

3Sulphuric acid (H2SO4)

Na2SO4, K2SO4, MgSO4, Al2(SO4)3,

ZnSO4, CaSO4.

4Acetic acid (CH3COOH)

CH3COONa, CH3COOK, (CH3COO)2Ca,

(CH3COO)2 Pb, CH3COONH4.

5Carbonic acid (H2CO3)

Na2CO3, K2CO3, MgCO3, ZnCO3,

CaCO3, (NH4)2CO3.

6Sulphurous acid (H2SO3)

Na2SO3, K2SO3, MgSO3, ZnSO3,

CaSO3, (NH4)2 SO3.

7Phosphoric acid (H3PO4)

Na3PO4, K3PO4, Mg3(PO4)2,

Zn3(PO4)2, Ca3(PO4)2, (NH4)3PO4.

Some Examples of Normal Salts with their Parent Acids :

(B) Acid Salts : The salts which are obtained by thepartial replacement of ionisable hydrogen atoms of apolybasic acid by a metal or an ammonium ion areknown as acid salts.

These are usually formed when insufficient amount ofthe base is taken for the neutralisation of the acid. Forexample, when insufficient amount of NaOH is takento neutralise H2SO4, we get an acid salt NaHSO4.H2SO4 + NaOH NaHSO4 + H2O

(Insufficient amount) Sodium hydrogensulphate(Acid salt)

In this case, only one hydrogen atom out of two hasbeen replaced by sodium atom. Since there is onemore hydrogen atom in NaHSO4 which can bereplaced, therefore, it further reacts with anothermolecule of NaOH to produce Na2SO4 which is anormal salt.

NaHSO4 + NaOH Na2SO4 + H2OSodium Sodium

hydrogensulphate sulphate(Acid salt) (Normal salt)

Acid salts ionise in aqueous solution to producehydronium ions (H3O+), therefore, they exhibit all theproperties of acids.

PAGE # 21

Some other examples of acid salts are given in Table.

SOME ACID SALTS WITH THEIR PARENT ACIDS

S.No. Parent Acid Acid Salts

1Sulphuric acid (H2SO4)

NaHSO4, KHSO4, Ca(HSO4)2

2Carbonic acid (H2CO3)

NaHCO3, KHCO3, Ca(HCO3)2, Mg(HCO3)2

3Sulphurous acid (H2SO3)

NaHSO3, KHSO3, Ca(HSO3)2, Mg(HSO3)2

4Phosphoric acid(H3PO4)

NaH2PO4, Na2HPO4, KH2PO4, K2HPO4, Ca(H2PO4)2, CaHPO4

(C) Basic Salts : The salts which are formed by partialreplacement of hydroxyl (�OH) groups of a di-or a

triacidic base by an acid radical are known as basicsalts.

These are usually formed when an insufficient amountof acid is taken for the neutralisation of the base. Forexample, when insufficient amount of HCl is added toLead hydroxide, Basic lead chloride [Pb(OH)Cl] isformed

Pb(OH)2 + HCl Pb (OH)Cl + H2OLead hydroxide Basic Lead chloride(Diacidic base) (Basic salt)

Basic salts, for example, Pb(OH)Cl further reacts withHCl to form normal salts

Pb (OH) Cl + HCl PbCl2 + H2OBasic Lead Chloride Lead chloride(Basic salts) (Normal salt)

Some other important examples of basic salts are :

(i) Basic copper chloride, Cu(OH)Cl.(ii) Basic copper nitrate, Cu(OH)NO3(iii) Basic lead nitrate, Pb(OH)NO3.

(D) Double Salts : The salts which are obtained by thecrystallisation of two simple salts from a mixture oftheir saturated solutions are known as double salts.For example, a double salt potash alum [K2SO4.Al2(SO4)3. 24H2O] is prepared by mixing saturatedsolutions of two simple salts, K2SO4 and Al2(SO4)3 andcrystallization of the mixture.

K2SO4 + Al2(SO4)3+ 24H2O Crystallisation

K2SO4. Al2(SO4)3. 24 H2O

Potassium Potash alumsulphate (Double salt)

Some other examples of double salts are :(i) Mohr�s Salt, FeSO4 .(NH4)2SO4. 6H2O,(ii) Dolomite, CaCO3. MgCO3,(iii) Carnallite, KCl. MgCl2.6H2O

(E) Mixed Salts : The salts which contain more thanone type of acidic or basic radicals are called mixedsalts. For example, Sodium potassium carbonate(NaKCO3) is a mixed salt containing two basic radicalssodium and potassium. Similarly, calcium oxy chloride,

Ca(OCI)Cl is also a mixed salt containing two acidradicals OCI� and Cl�.Some other important examples of mixed salts are :Sodium potassium sulphate (NaKSO4) (containing twobasic radicals), Disodium potassium phosphate(Na2KPO4) (containing two basic radicals).

(ii) Classification of salt solutions based on pHvalues :Salts are formed by the reaction between acidsand bases. Depending upon the nature of the acidsand bases or upon the pH values, the salt solutionsare of three types.

(A) Neutral salt solutions : Salt solutions of strongacids and strong bases are neutral and have pH equalto 7. They do not change the colour of litmus solution.e.g. NaCl, KCl, NaNO

3, Na

2SO

4 etc.

(B) Acidic salt solutions : Salt solutions of strong ac-ids and weak bases are of acidic nature and have pHless than 7. They change the colour of blue litmussolution to red.e.g. (NH

4)

2SO

4, NH

4Cl etc.

In both these salts, the base NH4OH is weak while the

acids H2SO

4 and HCl are strong.

(C) Basic salt solutions : Salt solutions of strong basesand weak acids are of basic nature and have pH morethan 7. They change the colour of red litmus solution toblue.e.g. Na

2CO

3, K

3PO

4 etc.

In both the salts, bases NaOH and KOH are strongwhile the acids H

2CO

3 and H

3PO

4 are weak.

Some Important Chemical Compounds :

Common name : Table SaltChemical name : Sodium chlorideChemical formula : NaCl

Sodium chloride (NaCl) also called common salt ortable salt is the most essential part of our diet. Chemi-cally it is formed by the reaction between solutions ofsodium hydroxide and hydrochloric acid. Sea water isthe major source of sodium chloride where it is presentin dissolved form along with other soluble salts suchas chlorides and sulphates of calcium and magne-sium. It is separated by some suitable methods. De-posits of the salts are found in different parts of theworld and is known as rock salt. When pure, it is awhite crystalline solid, however, it is often brown due tothe presence of impurities.

Uses :

(i) Essential for life : Sodium chloride is quite essen-tial for life. Biologically, it has a number of functions toperform such as in muscle contraction, in conductionof nerve impulse in the nervous system and is alsoconverted in hydrochloric acid which helps in the di-gestion of food in the stomach. When we sweat, thereis loss of sodium chloride along with water. It leads tomuscle cramps. Its loss has to be compensated suit-ably by giving certain salt preparations to the patient.Electrol powder is an important substitute of commonsalt.

PAGE # 22

(ii) Raw material for chemicals : Sodium chloride isalso a very useful raw material for different chemicals.A few out of these are hydrochloric acid (HCl), washingsoda (Na

2CO

3.10H

2O), baking soda (NaHCO

3) etc.

Upon electrolysis of a solution of the common salt(brine), sodium hydroxide, chlorine and hydrogen areobtained.Electrolysis of aqueous solution of NaCl :

2NaCl(aq) isElectrolys 2NaOH(aq) + Cl2(g)+ H

2 (g)

Reaction takes place as follows -

� NaCl Na+ + Cl�

� 2Cl� Cl2 (g) + 2e� (anode reaction)

� 2H2O + 2e� H

2 + 2OH� (cathode reaction)

� Na+ + OH� NaOH

(iii) In leather industries : It is used in leather industryfor the leather tanning.

(iv) In severe cold, rock salt is spread on icy roads tomelt ice.

(v) It is also used as a fertilizer for sugar beet.

CAUSTIC SODA

Chemical name : Sodium hydroxideChemical formula : NaOH

Preparation : Sodium hydroxide is prepared byelectrolysing a concentrated solution of sodium chloride.This process is done in Castner - Kellner cell.

2NaCl 2Na+ + 2Cl�

2Cl� Cl2 + 2e�

2Na+ + 2Hg + 2e� 2NaHg

Sodium amalgam

2NaHg + 2H2O 2NaOH + H

2 + 2Hg

Uses :

(i) It is used in soaps, detergents, paper and silkindustries.(ii) It is used in refining of petroleum.(iii) It is used as a laboratory reagent.(iv) It is used in dye industry.(v) It is used in concentration of bauxite ore.

WASHING SODA

Chemical name : Sodium carbonate decahydrateChemical formula : Na

2CO

3.10H

2O

Recrystallization of sodium carbonate :

Sodium carbonate is recrystallized by dissolving inwater to get washing soda.

(b) Uses :

(i) It is used as cleansing agent for domestic purposes.

(ii) It is used in softening of hard water and controllingthe pH of water.

(iii) It is used in the manufacture of glass.

(iv) Due to its detergent properties, it is used as a con-stituent of several dry soap powders.

(v) It also finds use in photography, textile and paperindustries etc.

(vi) It is used in the manufacture of borax (Na2B

4O

7. 10H

2O).

BAKING SODA

Baking soda is sodium hydrogen carbonate or sodiumbicarbonate (NaHCO

3).

(a) Preparation :

It is obtained as an intermediate product in the prepa-ration of sodium carbonate by Solvay process. In thisprocess, a saturated solution of sodium chloride inwater is saturated with ammonia and then carbon di-oxide gas is passed into the solution. Sodium chlo-ride is converted into sodium bicarbonate which, be-ing less soluble, separates out from the solution.

2NH3 (g) + H

2O () + CO

2 (g) (NH

4)

2CO

3(aq)

(NH4)2CO

3(aq)+2NaCl(aq) Na

2CO

3 (aq) + 2NH

4Cl (aq)

Na2CO

3 (aq) + H

2O () + CO

2 (g) 2NaHCO

3 (s)

(b) Properties :

(i) It is a white, crystalline substance that forms analkaline solution with water. The aqueous solution ofsodium bicarbonate does not change the colour ofmethyl orange but gives pink colour withphenolphthalein.(Phenolphthalein and methyl orange are dyes usedas acid-base indicators.)(ii) When heated above 543 K, it is converted intosodium carbonate.

2NaHCO3 (s)

Na

2CO

3 (s) + CO

2 (g) + H

2O ()

(c) Uses :

(i) It is used in the manufacture of baking powder. Bak-ing powder is a mixture of potassium hydrogentartarate and sodium bicarbonate. During the prepa-ration of bread the evolution of carbon dioxide causesbread to rise (swell).

(ii) It is largely used in the treatment of acid spillageand in medicine as soda bicarb, which acts as anantacid.

(iii) It is an important chemical in the textile, tanning,paper and ceramic industries.

PAGE # 23

(iv) It is also used in a particular type of fire extinguisher.The following diagram shows a fire extinguisher thatuses NaHCO

3 and H

2SO

4 to produce CO

2 gas. The

extinguisher consists of a conical metallic container(A) with a nozzle (Z) at one end. A strong solution ofNaHCO

3 is kept in the container. A glass ampoule (P)

containing H2SO

4 is attached to a knob (K) and placed

inside the NaHCO3 solution. The ampoule can be bro-

ken by hitting the knob. As soon as the acid comes incontact with the NaHCO

3 solution, CO

2 gas is formed.

When enough pressure in built up inside the container,CO

2 gas rushes out through the nozzle (Z). Since CO

2

does not support combustion, a small fire can be putout by pointing the nozzle towards the fire. The gas isproduced according to the following reaction.2NaHCO

3 (aq) + H

2SO

4 (aq) Na

2SO

4 (aq) +

2H2O() + 2CO

2(g)

Knob (K)

(P)

(A)

(Z)

CO2

NaHCO3

H SO2 4��

��

Soda-Acid Fire Extinguisher

BLEACHING POWDER

Bleaching powder is commercially called �chloride of

lime or �chlorinated lime�. It is principally calcium oxy-

chloride having the following formula :

Bleaching powder is prepared by passing chlorine overslaked lime at 313 K.

Ca(OH) (aq) + Cl (g) Ca(OCl)Cl (s) + H O (g)2 2 2

313 K

Slaked lime Bleaching powder

Actually bleaching powder is not a compound but amixture of compounds :CaOCl

2.4H

2O, CaCl

2.Ca(OH)

2.H

2O

(a) Uses :

(i) It is commonly used as a bleaching agent in paperand textile industries.

(ii) It is also used for disinfecting water to make it freefrom germs.

(iii) It is used to prepare chloroform.

(iv) It is also used to make wool shrink-proof.

PLASTER OF PARIS

(a) Preparation :

It is prepared by heating gypsum (CaSO4.2H

2O) at about

373 K in large steel pots with mechanical stirrer, or ina revolving furnace.

373 K2(CaSO .2H O) (CaSO ) .H O + 3H O4 2 4 2 22

Gypsum Plaster of Paris

or CaSO .2H O CaSO . H O + H O 4 2 4 2 2

12

32

The temperature is carefully controlled, as at highertemperature gypsum is fully dehydrated. The proper-ties of dehydrated gypsum are completely different fromthose of Plaster of Paris.

(b) Properties :

(i) Action with water : When it is dissolved in water, it getscrystallized and forms gypsum.

(c) Uses :

When finely powdered Plaster of Paris is mixed withwater and made into a paste, it quickly sets into a hardmass. In the process, its volume also increasesslightly. These properties find a number of uses. Addi-tion of water turns Plaster of Paris back into gypsum.(i) It is used in the laboratories for sealing gaps whereair tight arrangement is required.(ii) It is used for making toys, cosmetics and casts ofstatues.(iii) It is used as a cast for setting broken bones.(iv) It also finds use in making moulds in pottery.(v) It is used for making surfaces smooth and for mak-ing designs on walls and ceilings.

HYDRATED SALTS - SALTS CONTAINING

WATER OF CRYSTALLISATION:

Certain salts contain definite amount of some watermolecules loosely attached to their own molecules.These are known as hydrated salts and are of crystal-line nature. The molecules of water present are knownas �water of crystallisation�.

In coloured crystalline and hydrated salts, themolecules of water of crystallisation also account fortheir characteristic colours. Thus, upon heating of hy-drated salt, its colour changes since molecules ofwater of crystallisation are removed and the salt be-comes anhydrous.For example, take a few crystals of blue vitriol i.e. hy-drated copper sulphate in a dry test tube or boilingtube. Heat the tube from below. The salt will change toa white anhydrous powder and water droplet will ap-pear on the walls of the tube. Cool the tube and add afew drops of water again. The white anhydrous pow-der will again acquire blue colour.

CuSO4. 5H

2O

CuSO4 + 5H

2O

Copper sulphate Copper sulphate

(Hydrated) (Anhydrous)

24242424PAGE # 24

PERIODIC TABLE

DEFINITION

A periodic table may be defined as the table givingthe arrangement of all the known elements accordingto their properties so that elements with similarproperties fall within the same vertical column andelements with dissimilar properties are separated.

EARLY ATTEMPTS TO CLASSIFY ELEMENTS

(a) Metals and Non-Metals :

Among the earlier classifications, Lavoisier classifiedthe elements as metals and non-metals. However,this classification proved to be inadequate. In 1803,John Dalton published a table of relative atomicweights (now called atomic masses). This formed

an important basis of classification of elements.

(b) Dobereiner�s Triads:

(i) In 1817, J.W. Dobereiner a German Chemist gave

this arrangement of elements.

(A) He arranged elements with similar properties inthe groups of three called triads.

(B) According to Dobereiner the atomic mass of thecentral element was merely the arithmetic mean ofatomic masses of the other two elements.

e.g.

Elements of the triad Symbol Atomic mass

Lithium Li 7

Sodium Na 23

Potassium K 39

Atomic mass of sodium =

2

potassiumofmassAtomiclithiumofmasscAtomi

= 2

397 = 23

Some examples of triads are given in the table :

(ii) Limitations of Dobereiner�s Classification :

(A) Atomic mass of the three elements of some triadsare almost same.e.g. Fe, Co, Ni

(B) It was restricted to few elements, thereforediscarded.

(c) Newlands� Law of Octaves :

In 1866, an English chemist, John Newlands,proposed a new system of grouping elements withsimilar properties. He tried to correlate the propertiesof elements with their atomic masses. He arrangedthe then known elements in the order of increasingatomic masses. He started with the element havingthe lowest atomic mass (hydrogen) and ended atthorium which was the 56th element. He observedthat every eighth element had properties similar tothat of the first.

Newlands called this relation as a law of octavesdue to the similarity with the musical scale.

(i) Newlands� arrangement of elements into �Octaves�:

Notes of Music sa (do) re (re) ga (mi) ma (fa) pa (so) dha (la) ni (ti)

H Li Be B C N O

F Na Mg Al Si P S

Cl K Ca Cr Ti Mn Fe

Co and Ni Cu Zn Y In As Se

Br Rb Sr Ce and La Zr � �

Elements

(ii) Limitations of law of octaves : The law of octaves

has the following limitations :

(A) The law of octaves was found to be applicableonly upto calcium. It was not applicable to elementsof higher atomic masses.

(B) Position of hydrogen along with fluorine andchlorine was not justified on the basis of chemicalproperties.

(C) Newlands placed two elements in the same

slot to fit elements in the table. He also placed someunlike elements under the same slot. For example,cobalt and nickel are placed in the same slot and inthe column of fluorine, chlorine and bromine. Butcobalt and nickel have properties quite different fromfluorine, chlorine and bromine. Similarly, iron whichhas resemblances with cobalt and nickel in itsproperties has been placed far away from theseelements.

Thus, it was realized that Newlands� law of octaves

worked well only with lighter elements. Therefore,this classification was rejected.


25252525PAGE # 25

(d) Lother Meyer�s Classification :

In 1869, Lother Meyer studied the physical propertieslike volume, melting point, boiling point etc. of differentelements.He plotted a graph between atomic masses againsttheir respective atomic volumes for a number ofelements. He found the following observations -(i) Elements with similar properties occupiedsimilar positions on the curve.

(ii) Alkali metals (Li, Na, K, Rb, Cs etc.) having largeratomic volumes occupied the crests .

(iii) Transition elements (V, Fe, Co, Cu etc.) occupiedthe troughs.

(iv) The halogens (F, Cl, Br, etc.) occupied theascending portions of the curve before the inertgases.

(v) Alkaline earth metals (Mg, Ca, Sr, Ba etc.)occupied positions at about the mid points ofdescending portions of the curve.

70

60

50

40

30

20

10

20 40 8060 100 1200

Atomic mass

Ato

mic

Vol

ume

(cm

per

mol

e of

ato

ms)

3 �

� �

��

��

�

�

��

� �

��

�

�

��

��

�

�

�

�

��

�Mg

LiNa

K

Ca Br

Rb

Sr Ba

Cs

I

FeV Co

Zn

Cu

Cl

Change of Atomic Volume with Atomic Mass.

140

BeF

Drawback of Lother Meyer�s classification : Thiswas a hypothetical classification and it was difficult toremember the positions of different elements.

(e) Mendeleev�s Periodic Table :

The major credit for a systematic classification ofelements goes to Mendeleev. He tried to group theelements on the basis of some fundamental propertyof the atoms. When Mendeleev started his work, only63 elements were known. He examined therelationship between atomic masses of the elementsand their physical and chemical properties.Among chemical properties, Mendeleev concentratedmainly on the compound formed by elements withoxygen and hydrogen. He selected these twoelements because these are very reactive and formedcompound with most of the elements known at thattime. The formulae of the compounds formed withthese elements (i.e. oxides and hydrides) wereregarded as one of the basic properties of an elementfor its classification.

(i) Mendeleev�s periodic law : This law states that

the physical and chemical properties of the elements

are the periodic function of their atomic masses.

This means that when the elements are arranged in

the order of their increasing atomic masses, the

elements with similar properties recur at regular

intervals. Such orderly recurring properties in a cyclic

fashion are said to be occurring periodically. This is

responsible for the name periodic law or periodic

table.

(ii) Merits of Mendeleev�s periodic table : Mendeleev�s

periodic table was one of the greatest achievements in

the development of chemistry. Some of the important

contributions of his periodic table are given below :

(A) Systematic study of elements : He arranged

known elements in order of their increasing atomic

masses considering the fact that elements with

similar properties should fall in the same vertical

column.

(B) Correction of atomic masses : The Mendeleev�s

periodic table could predict errors in the atomic

masses of elements based on their positions in the

table. Therefore atomic masses of certain elements

were corrected. For example, atomic mass of

beryllium was corrected from 13.5 to 9. Similarly, with

the help of this table, atomic masses of indium, gold,

platinum etc. were corrected.

(C) Mendeleev predicted the properties of those

missing elements from the known properties of the

other elements in the same group. Eka-boron, eka -

aluminium and eka -silicon names were given for

scandium , gallium and germanium (not discovered

at the time of Mendeleev ).

(D) Position of noble gases : Noble gases like helium

(He), neon (Ne) and argon (Ar) were mentioned in

many studies. However, these gases were discovered

very late because they are very inert and are present

in extremely low concentrations. One of the

achievements of Mendeleev�s periodic table was that

when these gases were discovered, they could be

placed in a new group without disturbing the existing

order.

(iii) Limitations of Mendeleev�s periodic table :Inspite of many advantages, the Mendeleev�s periodic

table has certain defects also. Some of these are

given below -

(A) Position of hydrogen : Position of hydrogen in

the periodic table is uncertain. It has been placed in

1A group with alkali metals, but certain properties of

hydrogen resemble those of halogens. So, it may be

placed in the group of halogens as well.

26262626PAGE # 26

(B) Position of isotopes : Isotopes are the atoms ofthe same element having different atomic masses.Therefore, according to Mendeleev�s classification

these should be placed at different places dependingupon their atomic masses. For example, hydrogenisotopes with atomic masses 1, 2 and 3 should beplaced at three places. However, isotopes have notbeen given separate places in the periodic tablebecause of their similar properties.

(C) Anomalous pairs of elements : In certain pairsof elements, the increasing order of atomic masseswas not obeyed. In these, Mendeleev placedelements according to similarities in their propertiesand not in increasing order of their atomic masses.

For example :� The atomic mass of argon is 39.9 and that ofpotassium 39.1. But argon is placed beforepotassium in the periodic table.

� The positions of cobalt and nickel are not in proper

order. Cobalt (at. mass = 58.9) is placed before nickel( at. mass = 58.7).

� Tellurium (at. mass = 127.6) is placed before

iodine (at. mass = 126.9).

(D) Some similar elements are separated, in theperiodic table. For example copper (Cu) and mercury(Hg). On the other hand, some dissimilar elementshave been placed together in the same group.e.g. Copper (Cu), silver (Ag) and gold (Au) have beenplaced in group 1 along with alkali metals. Similarly,manganese (Mn) is placed in the group of halogens.

(E) Cause of periodicity : Mendeleev could notexplain the cause of periodicity among the elements.

51

90Zr =

27272727PAGE # 27

MODERN PERIODIC TABLE

(a) Introduction :

In 1913, an English physicist, Henry Moseley showed

that the physical and chemical properties of the atoms

of the elements are determined by their atomic

number & not by their atomic masses. Consequently,

the periodic law was modified.

(b) Modern Periodic Law (Moseley�s

Periodic Law) :

�Physical and chemical properties of an element are

the periodic function of its atomic number��.The

atomic number gives us the number of protons in

the nucleus of an atom and this number increases

by one in going from one element to the next.

Elements, when arranged in the order of increasing

atomic number Z, lead us to the classification known

as the Modern Periodic Table. Prediction of properties

of elements could be made with more precision when

elements were arranged on the basis of increasing

atomic number.

(c) Periodicity :

The repetition of elements with similar properties after

certain regular intervals, when the elements are

arranged in order of increasing atomic number, is

called periodicity.

(d) Cause of Periodicity :

The periodic repetition of the properties of the

elements is due to the recurrence of similar valence

shell (outermost shell) electronic configuration after

certain regular intervals.

e.g. Alkali metals have similar electronic configuration

(ns1) and therefore, have similar properties.

Alkali Metals

3 Lithium

Sodium

Potassium

Rubidium

Caesium

Francium

11

19

37

55

87

Atomic number Element

2,1

2,8,1

2,8,8,1

2,8,18,8,1

2,8,18,18,8,1

2,8,18,32,18,8,1

Electronic configuration

Li

Na

K

Rb

Cs

Fr

Symbol

(e) Long Form of Periodic Table :

(i) The long form of periodic table is based uponModern periodic law. Long form of periodic table isthe contribution of Range, Werner, Bohr and Bury.

(ii) This table is also referred to as Bohr�s table since

it follows Bohr�s scheme of the arrangement of

elements into four types based on electronicconfiguration of elements.

(iii) Long form of periodic table consists of horizontal

rows (periods) and vertical columns (groups).

(f) Description of Periods :

(i) A horizontal row of periodic table is called a period.

(ii) There are seven periods numbered as 1, 2 , 3 , 4,5, 6 and 7.

(iii ) Each period consists of a series of elementshaving the same outermost shell.

(iv) Each period starts with an alkali metal having

outermost shell electronic configuration ns1.

(v) Each period ends with a noble gas with outermostshell electronic configuration ns2 np6 except heliumhaving outermost electronic configuration 1s2.

(vi) Each period starts with the filling of a new energylevel.

(A) 1st period : This period is called very short periodbecause this period contains only 2 elements H andHe.

(B) 2nd and 3rd periods : These periods are calledshort periods because these periods contain 8elements. 2nd period starts from

3Li to

10Ne and 3rd

period starts from 11

Na to 18

Ar.

(C) 4th and 5th periods : These periods are calledlong periods because these periods contain 18elements. 4th period starts from

19K to

36Kr and 5th

period starts from 37

Rb to 54

Xe.

(D) 6th period : This period is called very long period.This period contains 32 elements. Out of the 32elements 14 elements belong to Lanthanoid series(

58Ce to

71Lu). 6th period starts from

55Cs to

86Rn.

(E) 7th period : This period is called as incompleteperiod. It contains 25 elements. Out of the 25elements 14 elements belong to Actinoid series (

90Th

to 103

Lr). 7th period starts from 87

Fr to 111

Rg.

Note : Modern periodic table consists of seven periodsand eighteen groups.

(1 ) n = 1 st 2 8 8

18 18

32 25

(2 ) n = 2 nd

(3 ) n = 3 rd

(4 ) n = 4 th

(5 ) n = 5 th

(6 ) n = 6 th

(7 ) n = 7 th

Very short period Short period Short period Long period Long period

Very long period Incomplete period

Periods Called as No. of Elements

28282828PAGE # 28

Different elements belonging to a particular periodhave different electronic configurations and havedifferent number of valence electrons. That is whyelements belonging to a particular period havedifferent properties.

(g) Description of Groups :

(i) A vertical column of elements in the periodic tableis called a group.

(ii) There are eighteen groups numbered as 1, 2 , 3,4, 5, ------------ 13, 14, 15, 16,17,18.

(iii) A group consists of a series of elements havingsimilar valence shell electronic configuration andhence exhibit similar properties.

e.g. Li, Na, K belong to the same group and have 1electron in their valence shell.

(iv) The group 18 is also known as zero group becausethe valency of the elements of this group is zero.

Note : The elements of 18th or zero group are callednoble gases.

(v) The elements present in groups 1,2,13 to 17 arecalled normal or representative elements.

(vi) Elements of group 1 and 2 are called alkali metalsand alkaline earth metals respectively.

(vii) Elements present in group 17 are calledhalogens.

Note : Elements present in a period have differentproperties, while elements present in a group havesimilar properties.

Note : Modern periodic table is based on atomic

number, not on atomic mass.

(h) Merits of Modern Periodic Table :

(i) Anomalous pairs : The original periodic law basedon atomic masses is violated in case of four pairs ofelements in order to give them positions on the basisof properties. The elements having higher atomicmasses have been assigned position before theelements having lower atomic masses at four placesas shown below -

Ar K Co Ni Te I Th Pa

At. Mass 40 39 60 59 128 127 232 231

At. No. 18 19 27 28 52 53 90 91

(a) (b) (c) (d)

The discrepancy disappears, if the elements arearranged in order of increasing atomic numbers.

(ii) Position of isotopes : Isotopes are atoms of thesame element having different atomic masses, butsame atomic number. All the isotopes of an elementwill be given different positions, if atomic mass istaken as a basis. This shall disturb the symmetry ofthe table. In modern table, one position is fixed forone atomic number and since all the isotopes of anelement have the same atomic number, these areassigned only one position.

(iii) Elements with similar properties were placedtogether and elements with dissimilar properties wereseparated in modern periodic table.

(iv) Cause of periodicity : Modern periodic tableexplains the cause of periodicity among the elements.

(i) Demerits of Modern Periodic Table :

Following are the demerits of modern periodic table -

(i) Position of hydrogen : Position of hydrogen was

uncertain in the periodic table.

(ii) Position of lanthanides and actinides : Thepositions of lanthanides and actinides were alsouncertain in the periodic table.

CLASSIFICATION OF THE ELEMENTS

It is based on the type of subshells which receivesthe differentiating electron (i.e. last electron).

(a) s- Block Elements :

When last electron enters the s- orbital of theoutermost (nth) shell, the elements of this class arecalled s- block elements.Characteristics :

(i) Group 1 & 2 elements constitute the s - block.

(ii) General electronic configuration is ns1�2 .

(iii) s - block elements lie on the extreme left of theperiodic table.

(iv) This block includes metals only, except H.

Note : The total number of elements in s-block is 13(including hydrogen).

(b) p-Block Elements :

When differentiating electron enters the p - orbital ofthe nth orbit, elements of this class are called p - block

elements.

Characteristics :

(i) Elements of group 13 to 18 constitute the p - block.

(ii) General electronic configuration is ns2np1-6 .

(iii) p - block elements lie on the extreme right of theperiodic table.

(iv) This block includes some metals, all non-metalsand metalloids.

Note : The total number of elements in p-block is 31.

29292929PAGE # 29

(c) d - Block Elements:

When differentiating electron enters the (n�1)d orbital,

then elements of this class are called d - blockelements.

Characteristics :

(i) Elements of group 3 to 12 constitute the d - block.

(ii) General electronic configuration is (n � 1)d1�10 ns0- 2.

(iii) d - block elements lie in the centre of the periodictable.

(iv) All the d - block elements are metals and most ofthem form coloured complexes or ions.

Note : The total number of elements in d-block is 39.

(d) f - block elements :

When last electron enters into f - orbital of (n � 2)th

shell, elements of this class are called f - blockelements.

Characteristics :

(i) All f - block elements belong to 3rd group.

(ii) General electronic configuration is (n � 2)f 1 � 14

(n� 1)d0-1 ns2.(iii) f-block elements are present in two separaterows below the periodic table.

(iv) All f-block elements are metals only.

The elements of f- block have been classified intotwo series :

(A) Lanthanides : 14 elements present after elementlanthanum (

57La) are called lanthanides.1st inner

transition series of metals or 4f - series, contains 14elements i.e.

58Ce to 71Lu .

(B) Actinides : 14 elements present after elementactinium (

89Ac) are called actinides. 2nd inner transition

series of metals or 5f- series, also contains 14 elementsi.e. 90Th to 103Lr.

Note : The total number of elements in f-block is 28.

Division of the periodic table into various blocks :

Different Types of Elements :(i) Noble gases : The elements belonging to group

18 are called noble gases or aerogenous. They are

also known as inert gases because their outermost

orbitals are completely filled. Helium (He) is an

exception. It has only two electrons which are present

in s-orbital.

(ii) Representative elements : Elements in which

atoms have all shells complete except outermost shell

which is incomplete. Except 18th group, all s - block

and p - block elements are collectively called normal

or representative elements.

(iii) Transition elements : Those elements which

have partially filled d - orbitals in neutral state or in

any stable oxidation state are called transition

elements.

Note : Zn, Cd and Hg are d-block elements , but nottransition elements, because they do not containpartially filled d-orbitals.

(iv) Inner transition elements :- They contain threeincomplete outermost shell and were also referredto as rare earth elements, since their oxides wererare in earlier days.

(v) Diagonal relationship : Some elements of 2nd

and 3rd period show diagonal relationship amongthem. They represent the same properties of twoperiods. This relation is known as diagonal relation.

(vi) Transuranium elements : The elements withatomic number greater than 92 (Z > 92) are known astransuranium elements.

30303030PAGE # 30

1.00

791

HH

ydro

gen

6.94

13

LiLi

thiu

m

22.9

9011

Na

So

dium 39

.098

19

KP

ota

ssiu

m

85.4

6837

Rb

Rub

idiu

m13

2.9

155

Cs

Cae

siu

m

(223

)87

Fr

Fra

nciu

m

9.0

122

4

Be

Ber

ylliu

m

24.

30

51

2

Mg

Mag

nesi

um

40.

07

82

0

Ca

Cal

cium

87.

62

38

Sr

Str

ontiu

m1

37.3

35

6

Ba

Bar

ium

(226

)8

8

Ra

Rad

ium

44.

956

21

Sc

Sca

ndiu

m

88.

906

39

YY

ttriu

m

47.

867

22

TiT

itan

ium

50.9

4223

VVa

nad

ium

51.

996

24

Cr

Ch

rom

ium

54.

93

82

5

Mn

Man

gan

ese

91.

224

40

Zr

Zir

con

ium

178

.49

72

Hf

Ha

fniu

m

(261

)1

04

Rf

Rut

herf

ord

ium

92.9

0641

Nb

Nio

bium 18

0.95

73

TaTa

ntal

um

(26

2)10

5

Db

Dub

nium

42

Mo

Mol

ybde

num

183

.84

74

WT

ung

ste

n

(26

6)1

06

Sg

Se

abor

gium

95.

94(9

8)4

3

TcTe

chne

tium

186

.21

75

Re

Rh

eniu

m

(264

)1

07

Bh

Bo

hriu

m

55.8

45

26

Fe

Iron

58.9

3327

Co

Co

balt

58.9

6328

Ni

Nic

kel

63

.54

629

Cu

Cop

per

10

7.8

747

Ag

Silv

er1

96.

97

79

Au

Go

ld (27

2)11

1

Rg

Roe

ntg

eniu

m

(28

1)11

0

Ds

Dar

mst

adtiu

m

(268

)10

9

Mt

Mei

tner

ium

(27

7)

108

Hs

Has

siu

m

190.

2376

Os

Osm

ium

101

.07

44

Ru

Rut

heni

um

102

.91

45

Rh

Rho

dium

192

.22

77

IrIr

idiu

m

195.

0878

Pt

Pla

tinum

106.

4246

Pd

Pa

llad

ium

65.3

930

Zn

Zin

c

112.

4148

Cd

Cad

miu

m

200.

5980

Hg

Me

rcur

y

69.

723

31

Ga

Ga

lliu

m

72.

64

32

Ge

Ge

rma

nium

74.9

2233

As

Ars

enic

78.

963

4

Se

Se

leni

um

79.

904

35

Br

Bro

min

e

83.8

036

Kr

Kry

pto

n

114

.82

49

In Ind

ium

118.

715

0

Sn

Tin

121

.76

51

Sb

Ant

imon

y

127

.60

52

TeTe

lluriu

m

126

.90

53

IIo

dine

131

.29

54

Xe

Xen

on

204

.38

81

TiT

hal

lium

207.

282

Pb

Lead

208.

9883

Bi

Bis

mu

th

(20

9)84

Po

Po

loni

um

(210

)8

5

At

Ast

atin

e

(222

)8

6

Rn

Rad

on

26.9

82

13

Al

Alu

min

ium

10.8

115

B1

2.01

16

CC

arbo

n

14.

007

7

NN

itrog

en

15.9

998

OO

xyge

n

18.9

989

FF

luor

ine

20.1

8010

Ne

Neo

n

39.9

4818

Ar

Arg

on

35

.45

317

Cl

Chl

orin

e

32.

065

16

SS

ulph

ur

30.

974

15

PP

hosp

horu

s

28.0

8614

Si

Sili

con

Bor

on

4.0

026

2

He

Hel

ium

BB

oron

Sym

bol

51

0.81

113

Ato

mic

nu

mbe

r

Ele

me

nt n

ame

Rel

ativ

e a

tom

ic m

ass

IIIA

Gro

up IU

PA

C

IIA

IA

IIIB

IVB

VB

VIB

VIIB

VIII

B

1

2

34

56

78

910

1112

IBIIB

13III

A14

1516

17

18

IVA

VA

VIA

VIIA

VIII

A

1 2 3 4 5 6 7

Per

iod

Gro

up

LA

NTH

AN

OID

ES

(22

7)89

Ac

Act

iniu

m

14

0.1

258

Ce

Cer

ium

23

2.0

490

Th

Tho

rium

140

.91

59

Pr

Pra

seod

ymiu

m

231

.04

91

Pa

Pro

tact

iniu

m

144

.24

60

Nd

Ne

odym

ium

238

.03

92

UU

rani

um

(145

)61

Pm

Pro

met

hiu

m

150.

3662

Sm

Sam

ariu

m

(24

4)

94

Pu

Plu

toni

um

(237

)9

3

Np

Nep

tuni

um

151

.96

63

Eu

Eur

opiu

m

(243

)9

5

Am

Am

eric

ium

15

7.2

564

Gd

Ga

dolin

ium

(247

)96

Cm

Cur

ium

158.

9365

Tb

Terb

ium (247

)9

7

Bk

Ber

keliu

m

162

.50

66

Dy

Dys

pros

ium

(25

1)98

Cf

Ca

lifor

niu

m

164

.93

67

Ho

Hol

miu

m

(25

2)99

Es

Ein

ste

iniu

m

197.

2668

Er

Erb

ium

168

.93

69

Tm

Thu

lium

173

.04

70

Yb

Ytte

rbiu

m

174.

9771

LuLu

tetiu

m

(257

)1

00

Fm

Fer

miu

m

(258

)10

1 Md

Men

dele

viu

m

(259

)10

2 No

No

beliu

m

(262

)1

03

LrLa

wre

nciu

m

138.

9157

LaLa

ntha

num

s�B

lock

Ele

men

ts

Tran

sitio

n M

etal

s (d

�B

lock

Ele

men

ts)

p�B

lock

Ele

men

ts

Inn

er -

Tra

nsi

tio

n M

etal

s (f

-Blo

ck E

lem

ents

)

AC

TIN

OID

ES

PE

RIO

DIC

TA

BL

E O

F T

HE

EL

EM

EN

TS

31313131PAGE # 31

PERIODICITY IN PROPERTIES

(a) Atomic Volume :

Atomic volume increases in a group from top tobottom. The increase is due to the increase in thenumber of energy levels.In a period from left to right, atomic volume variescyclically, i.e. it decreases at first for some elements,becomes minimum in the middle and then increases.The following two factors explain this trend -(i) atomic radii decrease due to increase of nuclearcharge.

(ii) the number of valence electrons increases in aperiod.

As to accommodate all the valence electrons, the volumeincreases. These two factors oppose each other. Theeffect of first factor is more on the left hand side and thatof the second factor is more on the right hand side in theperiodic table . The volumes are in cm3.

Period/Group

1 2 3 4 5 6 7 8

2 Li13

Be5

B5

C5

N14

O11

F15

Ne17

3 Na24

Mg14

Al10

Si12

P17

S16

Cl19

Ar24

4 K46

Ca26

Ga12

Ge13

As16

Se16

Br23

Kr33

Incr

ease

s

Note : The maximum value of atomic volume (87cm3) is observed in the case of francium.

(b) Density :

The density of the elements in solid state variesperiodically with their atomic numbers . At first, thedensity increases gradually in a period and becomesmaximum somewhere for the central members andthen starts decreasing afterwards gradually. The valueof densities in the table are in g/cc.

Period/Group

1 2 3 4 5 6 7

2Li

0.54Be

1.85B

2.35C

2.26N�

O�

F�

3Na

0.97Mg

1.74Al

2.70Si

2.34P

1.82S

2.1Cl�

4K

0.86Ca

1.55Ga

5.90Ge

5.32As

5.77Se

4.19Br

3.19

5Rb

1.53Sr

2.63In

7.31Sn

7.27Sb

6.70Te

6.25I

4.94

6Cs

1.90Ba

3.62Tl

11.85Pb

11.34Bi

9.81Po

9.14At�

Note : In solids, Iridium (Ir) has the highest density(22.61 g/cc) and in liquids, mercury (Hg) has the

highest density (13.6 g/cc).

(c) Melting and Boiling Points :

The melting points of the elements exhibit someperiodicity with rise of atomic number. It is observedthat elements with low values of atomic volumes havehigh melting points, while elements with high valuesof atomic volumes have low melting points. In general,melting points of elements in any period at firstincrease and become maximum somewhere in thecentre and thereafter begin to decrease. Values inthe table are in ºC.

Period/Group

1 2 3 4 5 6 7 8

2Li

181Be

1287B

2180C

3727N

�210

O�219

F�220

Ne�249

3Na98

Mg650

Al660

Si1420

P44

S119

Cl�101

Ar�189

Tungsten has the maximum melting point (3410ºC)

amongst metals and carbon has the maximummelting point (3727ºC) amongst non-metal. Helium

has the minimum melting point (�270ºC) amongst

all elements . The metals Cs (m.p. = 28.5ºC), Ga

(m.p. = 30ºC) and Hg (m.p. = �39ºC) are known in

liquid state at 30ºC.

The boiling points of the elements also show similartrends, however, the regularities are not striking asnoted in the case of melting points.

(d) Atomic Radius :

(i) Covalent radius : It may be defined as one - half ofthe distance between the centres of the nuclei of twosimilar atoms bonded by a single covalent bond.

A B

12 AB = rcovalent

(of element X)

X X

e.g. The internuclear distance between two hydrogenatoms in H

2 molecule is 74 pm. Therefore, the

covalent radius of hydrogen atom is 37 pm.

Note : Covalent radius is generally used fornon - metals.

(ii) Vander Waal�s radius : It may be defined as halfof the internuclear distance between two adjacentatoms of the same element belonging to two nearestneighbouring molecules of the same substance.

E H

X XX X

12

EH = rvander Waals

32323232PAGE # 32

Characteristics :

(A) Covalent radius of the elements is shorter than

its Vander Waal�s radius.

(B) The formation of covalent bond involves

overlapping of atomic orbitals, as a result of this, the

internuclear distance between the covalently bonded

atoms is less than the internuclear distance between

the non bonded atoms.

e.g. Vander Waals radius of helium is 1.20 Å

(iii) Metallic radius (Crystal radius) : Metallic radius

may be defined as half of the internuclear distance

between two adjacent atoms in a metallic lattice.

C

X X

D

12 CD = rcrystal

(of element X)

� The metallic radius of an atom is always larger

than its covalent radius.

Note : The order of different radius is - r Vander Waals

> rMetallic

> rCovalent

(iv) Variation of atomic radii in a period : As we move

from left to right across a period, there is a regular

decrease in atomic radii of the representative

elements. This is due to the fact that number of energy

shells remains the same in a period, but nuclear

charge increases gradually as the atomic number

increases. This increases the force of attraction

towards nucleus which brings contraction in size.

This can also be explained on the basis of

effective nuclear charge which increases gradually

in a period i.e. electron cloud is attracted more strongly

towards nucleus as the effective nuclear charge

becomes more and more as we move in a period.

The increased force of attraction brings contraction

in size.

(v) Variation of atomic radii in a group : Atomic radii

in a group increase as the atomic number increases.

The increase in size is due to extra energy shells

which outweigh the effect of increased nuclear charge.

The following table illustrates the periodicity in atomic

radii (covalent radii) of representative elements. The

radii given in the table are in angstrom (Å).

Periodicity in atomic radii (covalent radii)

Period/Group

1 2 3 4 5 6 7

2 Li1.23

Be0.89

B0.80

C0.77

N0.75

O0.73

F0.72

3 Na1.54

Mg1.36

Al1.20

Si1.17

P1.10

S1.04

Cl0.99

4 K2.03

Ca1.74

Ga1.26

Ge1.22

As1.20

Se1.16

Br1.14

5 Rb2.16

Sr1.91

In1.44

Sn1.41

Sb1.40

Te1.36

I

1.33

6 Cs2.35

Ba1.98

Tl1.48

Pb1.47

Bi1.46

Po1.46

Incr

ease

s

Decreases

1 H0.32

Atoms of zero group elements do not form chemicalbonds among themselves. Hence for them VanderWaals radii are considered.

Element He Ne Ar Kr Xe

Vander Waalsradii (in Aº )

1.20 1.60 1.91 2.00 2.20

The sudden increase in atomic radii in comparisonto the halogens (the elements of 7th group) in caseof inert gases, is due to the fact that, Vander Waalsradii are considered which always possess highervalues than covalent radii.The decrease in the size of transition elements issmall since the differentiating electrons enter intoinner �d� levels. The additional electrons into (n�1)d

levels effectively screen much of increased nuclearcharge on the outer ns electrons and therefore, sizeremains almost constant.

However, in vertical columns of transition elements,there is an increase in size from first member tosecond member as expected, but from secondmember to third member, there is very small changein size and sometimes sizes are same. This is dueto Lanthanide contraction (in the lanthanideelements differentiating electrons enter into 4f-levels).Since these electrons do not effectively screen thevalence electrons from the increased nuclear charge,the size gradually decreases. This decrease is termedlanthanide contraction.

Conclusions(i) The alkali metals which are present at the extremeleft of the periodic table have the largest size in aperiod.

(ii) The halogens which are present at the extremeright of the periodic table have the smallest size.

33333333PAGE # 33

(iii) The size of the atoms of inert gases are, however,larger than those of preceding halogens because ininert gases van der Waals' radii are taken intoconsideration.

(iv) In a group of transition elements, there is anincrease in size from first member to second memberas expected but from second member to thirdmember, there is very small change in size andsometimes sizes are same. This is due to Lanthanidecontraction.

(iv) Ionic radius : It is the distance between the nucleusand outermost shell of an ion or it is the distancebetween the nucleus and the point where the nucleusexerts its influence on the electron cloud.

(A) The radius of the cation is always smaller thanthe atomic radius of its parent atom. This is due tothe fact that nuclear charge in the case of a cation isacting on a lesser number of electrons and pulls themcloser.

(B) The radius of the anion is always larger than theatomic radius of its parent atom. In an anion aselectron or electrons are added to the neutral atom,the nuclear charge acts on more electrons so thateach electron is held less tightly and thereby theelectron cloud expands.

Comparative sizes of atoms and their cations

AtomAtomic radii (crystal, Å)

Corresponding cations Ionic radii (Å)

Li 1.52 Li+ 0.59

Na 1.86 Na+ 0.99

K 2.31 K+ 1.33

Mg 1.60 Mg2+ 0.65

Ba 2.22 Ba2+ 1.35

Al 1.43 Al3+ 0.50Pb 1.75 Pb2+

1.32

Conclusions

� The radius of cation (positive ion) is always smallerthan that of the parent atom.

� The radius of anion (negative ion) is always largerthan that of the parent atom.

� The ionic radii in a particular group increase inmoving from top to bottom.

� In a set of species having the same number ofelectrons (isoelectronic), the size decreases as thecharge on the nucleus increases.

� The size of the cations of the same elementdecreases with the increase of positive charge.

(e) Ionisation Energy (IE) :

Ionisation Energy (IE) of an element is defined as theamount of energy required to remove an electron froman isolated gaseous atom of that element resultingin the formation of a positive ion.

(i) Characteristics :

(A) The energy required to remove the outermostelectron from an atom is called first ionisation energy(IE)

1.

After removal of one electron, the atom changes intomonovalent positive ion.M(g) + IE

1 M+(g) + e�

(B) The minimum amount of energy required toremove an electron from monovalent positive ion ofthe element is known as second ionisation energy(IE)

2.

M+(g) + IE2 M2+(g) + e�

(C) The first, second etc. ionisation energies arecollectively known as successive ionisation energies.M2+(g) + IE

3 M3+(g) + e�

In general (IE)1 < (IE)

2 < (IE)

3 so on, because, as the

number of electrons decreases, the attractionbetween the nucleus and the remaining electronsincreases considerably and hence subsequentionisation energies increase.

(D) Units : Ionisation energy is expressed either interms of electron volts per atom (eV/atom) or Kilojoulesper mole of atoms (KJ mol � 1) or K cal mol � 1.1 eV/atom = 96.49 KJ/mol = 23.06 Kcal/mol = 1.602 × 10�19

J/atom

(ii) Factors influencing ionisation energy :

(A) Size of the atom : Ionisation energy decreases withincrease in atomic size. As the distance between theoutermost electrons and the nucleus increases, the forceof attraction between the valence shell electrons and thenucleus decreases. As a result, outermost electrons areheld less firmly and lesser amount of energy is requiredto knock them out.For example, ionisation energy decreases in a groupfrom top to bottom with increase in atomic size.

(B) Nuclear charge : The ionisation energyincreases with increase in the nuclear charge. Thisis due to the fact that with increase in the nuclearcharge, the electrons of the outermost shell are morefirmly held by the nucleus and thus greater amount ofenergy is required to pull out an electron from the atom.For example, ionisation energy increases as we movefrom left to right along a period due to increase in nuclearcharge.

(C) Shielding effect : The electrons in the inner shellsact as a screen or shield between the nucleus andthe electrons in the outermost shell. This is calledshielding effect or screening effect. Larger the numberof electrons in the inner shells, greater is thescreening effect and smaller the force of attractionand thus ionisation energy decreases.

These electrons shield the outer

electrons from the nucleus

This electron does not feel the full inward pull

of the positive charge of the nucleus

34343434PAGE # 34

(D) Penetration effect of the electrons : Theionisation energy increases as the penetration effectof the electrons increases. It is a well known fact thatthe electrons of the s-orbital have the maximumprobability of being found near the nucleus and thisprobability goes on decreasing in case of p, d and forbitals of the same energy level.

Greater the penetration effect of electrons more firmlythe electrons will be held by the nucleus and thushigher will be the ionisation energy of the atom.

For example, ionisation energy of aluminium iscomparatively less than magnesium as outermostelectron is to be removed from p-orbital (having lesspenetration effect) in aluminium, whereas inmagnesium it will be removed from s-orbital (havinglarger penetration effect) of the same energy level.

Note : With in the same energy level,the penetrationeffect decreases in the order s > p > d > f

(E) Electronic Configuration : If an atom has exactly

half-filled or completely filled orbitals, then such anarrangement has extra stability.The removal of anelectron from such an atom requires more energythan expected. For example,

E1 of

Be > E

1 of B

Be (Z = 4)

)stablemore(orbitalfilledCompletely

s2,s1 22

B (Z = 5)

)stableless(orbitalfilledPartially

p2,s2,s1 122

As noble gases have completely filled electronicconfigurations, they have highest ionisation energiesin their respective periods.

(iii) Variation of ionisation energy in a period :

In general, the value of ionisation energy increaseswith increase in atomic number across a period. Thiscan be explained on the basis of the fact that onmoving across the period from left to right-

(A) nuclear charge increases regularly.

(B) addition of electrons occurs in the same shell.

(C) atomic size decreases.

(iv) Variation of ionisation energy in a group :

In general, the value of ionisation energy decreaseswhile moving from top to bottom in a group.This isbecause -

(A) effective nuclear charge decreases regularly.

(B) addition of electrons occurs in a new shell.

(C) atomic size increases.

(v) Conclusions :

(i) In each period, alkali metals show lowest first

ionisation enthalpy. Caesium has the minimum value.

(ii) In each period, noble gases show highest first

ionisation enthalpy. Helium has the maximum value

of first jonisation enthalpy.

(iii) The representative elements show a large range

of values of first ionisation enthalpies, metals having

low while non-metals have high values.

(iv) Generally. ionisation enthalpies of transition

metals increase slowly as we move from left to right

in a period. The f-block elements also show only a

small variation in the values of first ionisation

enthalpies.

(f) Electron Affinity (EA) :

Electron affinity is defined as the energy released in

the process of adding an electron to a neutral atom in

the gaseous state to form a negative ion.

X(g) + e� X�(g) + Energy (E.A.)

Cl(g) + e� Cl� (g) + 349 KJ/mol

The electron affinity of chlorine is 349 KJ/mol.

The addition of second electron to an anion is

opposed by electrostatic repulsion and hence the

energy has to be supplied for the addition of second

electron.

O(g) + e� O� (g) + Energy (EA -)

O�(g) + e� O2� (g) � Energy (EA-)

(EA -) is exothermic whereas, (EA-) is endothermic.

(i) Units : Kilo joules per mole (KJ/mol) of atoms or

electron volts per atom (eV/atom).

(ii) Factors affecting electron affinity:

(A) Nuclear charge : Greater the magnitude of

nuclear charge greater will be the attraction for the

incoming electron and as a result, larger will be the

value of electron affinity.

Electron affinity Nuclear charge.

(B) Atomic size : Larger the size of an atom is, more

will be the distance between the nucleus and the

incoming electron and smaller will be the value of

electron affinity.

sizeAtomic1

E.A.

(C) Electronic configuration : Stable the electronic

configuration of an atom lesser will be its tendency to

accept the electron and lower will be the value of its

electron affinity.

35353535PAGE # 35

(iii) Variation of electron affinity in a period : On

moving across the period the atomic size decreases

and nuclear charge increases. Both these factors

result into greater attraction for the incoming electron,

therefore electron affinity in general increases in a

period from left to right.

(iv) Variation of electron affinity in a group : On

moving down a group, the atomic size as well as

nuclear charge increase, but the effect of increase in

atomic size is much more pronounced than that of

nuclear charge and thus, the incoming electron feels

less attraction consequently, electron affinity

decreases on going down the group.

(v) Some irregularities observed in general trend:

(A) Halogens have the highest electron affinities in

their respective periods. This is due to the small size

and high effective nuclear charge of halogens.

Halogens have seven electrons in their valence shell.

By accepting one more electron they can attain stable

electronic configuration of the nearest noble gas. Thus

they have maximum tendency to accept an additional

electron.

(B) Due to stable electronic configuration of noble

gases electron affinities are zero.

(C) Be, Mg, N and P also have exceptionally low

values of electron affinities due to their stable

electronic configurations.

Be = 1s2, 2s2 N = 1s2, 2s2, 2p3

Mg = 1s2, 2s2 , 2p6, 3s2 P = 1s2, 2s2, 2p6, 3s2, 3p3

Conclusion

(i) The electron gain enthalpies, in general, become

less negative in going down from top to bottom in a

group. This is due to increase in size on moving down

a group. This factor is predominant in comparison to

other factor, i.e., increase in nuclear charge.

Na K Rb Cs�53 �48 �47 �46 KJ mol

�1

Cl Br I At�349 �325 �295 � 270KJ mol

�1

(ii) The electron gain enthalpies of oxygen and fluorine,

the members of the second period, have less

negative values than the elements sulphur and

chlorine of the third period. This is due to small size

of the atoms of oxygen and fluorine. As a result, there

is a strong inter-electronic repulsion when extra

electron is added to these atoms, i.e., electron density

is high and the addition of electron is not easy. Thus,

the electron gain enthalpies of third period elements,

sulphur and chlorine, have more negative values than

corresponding elements oxygen and fluorine.

O �141 kJ mol�1 F �328 kJ mol

�1

S �200 kJ mol�1 Cl �349 kJ mol

�1

Similar trend is observed in nitrogen & phosphorous

N +31 kJ mol�1

P � 74 kJ mol�1

(iii) In general, electron gain enthalpy becomes more

and more negative from left to right in a period. This

is due to decrease in size and increase in nuclear

charge as the atomic number increases in a period.

Both these factors favour the addition of an extra

electron due to higher force of attraction by the nucleus

for the incoming electron.

(iv) Electron gain enthalpies of some of the members

of alkaline earth metals, noble gases and nitrogen

are positive.

This is because they have stable configurations.

Alkaline earth metals have stable configurations due

to completely filled ns orbital while nitrogen has extra

stability due to half filled p-orbitals (1s2, 2s2, 2p3) i.e.,

symmetrical configuration. These atoms resist the

addition of extra electron as they do not want to disturb

their stability.

Noble gases have ns2np6 configuration, i.e., no place

for incoming electron. In case the extra electron is to

be accommodated, it will occupy its position on a

new principal shell, i.e., it will be weakly attracted

towards nucleus. Such anion will be extremely

unstable. Helium has also stable 1s2 configuration

and cannot accommodate the incoming electron.

(v) Halogens have highest negative electron gain

enthalpies.

Following two factors are responsible for this:

� Small atomic size and high nuclear charge of

halogens in a period.

� Halogens have the general electronic configuration

of ns2 np5, i.e., one electron less than stable noble

gas (ns2 np6) configuration.

Thus, halogens have very strong tendency to accept

an additional electron and their electron gain

enthalpies are, therefore, high.

� Importance of Electron Gain Enthalpy : Certain

properties of the elements can be predicted on the

basis of values of electron gain enthalpies.

36363636PAGE # 36

(i) The elements having high negative values ofelectron gain enthalpy are capable of acceptingelectron easily. They form anions and thus form ionic(electrovalent) compounds. These elements areelectronegative in nature.

(ii) The elements having high negative electron gainenthalpies act as strong oxidising agents, for example,F, CI, Br, O, S, etc.On the basis of the general trend of ionisationenthalpy and electron gain enthalpy, the followingproperties can be predicted:

(i) Metallic nature decreases in a period while non-metallic nature increases. Metallic nature increasesin a group while non-metallic nature decreases. Thearrow () represents a group and () represents aperiod.

Metallic

Increases

DecreasesMetallic(Electro + ve)

Non-metallic

Decreases

IncreasesNon-metallic (Electro-ve)

(ii) Reducing nature decreases in a period whileoxidising nature increases. The reducing natureincreases in a group while oxidising naturedecreases.

Reducing nature

Increases

DecreasesReducingnature ;

Oxidising nature

Decreases

IncreasesOxidising nature

(iii) Stability of metal increases while activity of themetal decreases in a period and in a group stabilitydecreases.

Stability of the metal

Decreases

IncreasesStability of the metal

; Activity of the metal

Increases

DecreasesActivity of the metal

(g) Electronegativity :

Electronegativity is a measure of the tendency of an

element to attract electrons towards itself in a

covalently bonded molecule .

(i) Factors influencing electronegativity :

(A) The magnitude of electronegativity of an element

depends upon its ionisation potential & electron

affinity. Higher ionisation potential & electron affinity

values indicate higher electronegativity value.

(B) With increase in atomic size the distance

between nucleus and valence shell electrons

increases, therefore the force of attraction between

the nucleus and the valence shell electrons

decreases and hence the electronegativity values

also decrease.

(C) In higher oxidation state, the element has higher

magnitude of positive charge. Thus, due to more

positive charge on element, it has higher polarising

power. Thus, with increase in the oxidation state of

element, its electronegativity also increases.

Note :

Polarising power is the power of an ion (cation) to

distort the other ion.

(D) With increase in nuclear charge, force of attraction

between nucleus and the valence shell electrons

increases and, therefore electronegativity value

increases.

(E) The electronegativity of the same element

increases as the s-character in the hybrid orbitals

increases.

Hybrid orbital sp3 sp2 sp

s-character 25% 33% 50%

Electronegativity increases

(ii) Variation of Electronegativity in a group : On

moving down the group atomic number increases,

so nuclear charge also increases. Number of shells

also increases, so atomic radius increases.

Therefore electronegativity decreases on moving

down the group.

(iii) Variation of Electronegativity in a period : While

moving across a period left to right atomic number,

nuclear charge increases & atomic radius decreases.

Therefore electronegativity increases along a period.

37373737PAGE # 37

Differences between Electron gain enthalpy and Electronegativity

S.No. Electron gain enthalpy Electronegativity

1It is the tendency of an isolated atom to attract electron.

It is the tendency of an atom in a combined state, i.e., in a molecule to attract the shared pair of electrons.

2 It can be measured experimentally.It is a relative number and cannot be determined experimentally.

3Its units are electron volts per atom or kilo joules per mole or kilo calories per mole.

It has no units but merely a number.

4It is a constant quantity for a particular element.

It has no units but merely a number.

5Its periodicity is not regular in a period or a group.

Electronegativity of an element is not constant. It depends on a number of factors such as hybridised state. Oxidation state, etc. The periodicity is regular in a period but not so regular in groups.

(h) Nature of Oxides :

In a period, the nature of the oxides varies from basicto acidic. Na

2O MgO AI

2O

3 SiO

2 P

2O

5 Cl

2O

7

Strongly Basic Amphoteric Weakly Stronglybasic acidic acidic

In a group, basic nature increases or acidic naturedecreases. Oxides of the metals are generally basicand oxides of the nonnmetals are acidic. The oxidesof the metalloids are amphoteric. The oxides of AI,Zn, Sn, As and Sb are amphoteric. We can summarisethat as the electronegativity of element increases,acidic character of oxides increases. When anelement forms a number of oxides, the acidic natureincreases as the percentage of oxygen increases.

N O2

NO

N O2 3

NO2

N O2 5

Aci

dic

natu

re

incr

eas

es

Neutral oxide

P O2 3

P O2 4

P O2 5

Aci

dic

natu

re

incr

ease

s

Aci

dic

natu

re

incr

ease

s

MnOMn O2 3

MnO2

MnO3

Mn O2 7

Basic Basic NeutralAcidic Acidic

CO, N2O, NO and H

2O are neutral oxides. The oxides

CO2, N

2O

5, P

2O

3, P

2O

5, SO

2, SO

3, Cl

2O

7, etc., are called

acid anhydrides as these combine with water to formoxy-acids.CO

2 H

2CO

3P

2O

3 H

3PO

3

SO3 H

2SO

4N

2O

5 HNO

3

P2O

5 H

3PO

4Cl

2O

7 HClO

4

N2O

3 HNO

2SO

2 H

2SO

3

(i) Nature of Oxy-Acids :

In a period, the strength of the oxy-acids formed bynon-metals increases from ieft to right.

II Period strength increases

III Period strength

increases

In a group, the strength of the oxy-acids of non-metals

decreases.

V group strength decreases

VII group strength decreases

If a non-metal forms a number of oxy-acids, the

strength increases with the increase of percentage

of oxygen.

Sulphur forms two oxy-acids H2SO

3 and H

2SO

4.H

2SO

4

is stronger acid than H2SO

3.

Chlorine forms a number of oxy-acids:

Greater is the oxidation state of central atom more

will be the acidic strength.

( j) Nature of Hydrides :

The nature of the hydrides changes from basic to

acidic in a period from left to right.

NH3

H2O HF

weak base neutral weak acid

PH3

H2S HCl

very weak weak acid strong acid

base

In a group, the acidic nature of the hydrides of non-

metals increases. The reducing nature also increases

but stability decreases from top to bottom

38383838PAGE # 38

� Anomalous behaviour of the elements of secondperiod :It has been observed that in the case of representativeelements, the first element in each, i.e. lithium in thefirst group, beryllium in the second group and boronto fluorine in the group 13 to 17, differ in many respectfrom the other member of their respective group. Theanomalous behaviour of the first member of eachgroup is attributed to following reasons :

(a) small atomic radius of the atom and ionic radiusof its ion.

(b) high electronegativity

(c) non-availability of d-orbitals in their valence shell.

(d) tendency to form multiple bonds by carbon,nitrogen and oxygen

(e) high charge/radius ratio

PAGE # 39

MOLE CONCEPT

ATOMS

All the matter is made up of atoms. An atom is thesmallest particle of an element that can take part in achemical reaction. Atoms of most of the elementsare very reactive and do not exist in the free state (assingle atom).They exist in combination with the atomsof the same element or another element.Atoms are very, very small in size. The size of an atomis indicated by its radius which is called "atomicradius" (radius of an atom). Atomic radius ismeasured in "nanometres"(nm).1 metre = 109 nanometre or 1nm = 10-9 m.Atoms are so small that we cannot see them underthe most powerful optical microscope.

Note :

Hydrogen atom is the smallest atom of all , having anatomic radius 0.037nm.

(a) Symbols of Elements :

A symbol is a short hand notation of an element whichcan be represented by a sketch or letter etc.Dalton was the first to use symbols to representelements in a short way but Dalton's symbols forelement were difficult to draw and inconvenient touse, so Dalton's symbols are only of historicalimportance. They are not used at all.

It was J.J. Berzelius who proposed the modernsystem of representing en element.The symbol of an element is the "first letter" or the"first letter and another letter" of the English name orthe Latin name of the element.

e.g. The symbol of Hydrogen is H.The symbol of Oxygen is O.There are some elements whose names begin withthe same letter. For example, the names of elementsCarbon, Calcium, Chlorine and Copper all begin withthe letter C. In such cases, one of the elements isgiven a "one letter "symbol but all other elements aregiven a "first letter and another letter" symbol of theEnglish or Latin name of the element. This is to benoted that "another letter" may or may not be the"second letter" of the name. Thus,The symbol of Carbon is C.The symbol of Calcium is Ca.The symbol of Chlorine is Cl.

The symbol of Copper is Cu (from its Latin nameCuprum)It should be noted that in a "two letter" symbol, thefirst letter is the "capital letter" but the second letter isthe small letter

English Name of the Element

Symbol

Hydrogen HHelium HeLithium LiBoron B

Carbon CNitrogen NOxygen OFluorine F

Neon NeMagnesium MgAluminium Al

Silicon SiPhosphorous P

Sulphur SChlorine ClArgon Ar

Calcium Ca

Symbol Derived from English Names

Symbols Derived from Latin Names

English Name of the Element

SymbolLatin Name of the Element

Sodium Na Natrium

Potassium K Kalium

(b) Significance of The Symbol of an

Element :

(i) Symbol represents name of the element.

(ii) Symbol represents one atom of the element.

(iii) Symbol also represents one mole of the element.That is, symbol also represent 6.023 × 1023 atoms ofthe element.

(iv) Symbol represent a definite mass of the elementi.e. atomic mass of the element.

Example :(i) Symbol H represents hydrogen element.

(ii) Symbol H also represents one atom of hydrogenelement.

(iii) Symbol H also represents one mole of hydrogenatom.

(iv) Symbol H also represents one gram hydrogenatom.


PAGE # 40

IONS

An ion is a positively or negatively charged atom orgroup of atoms.Every atom contains equal number of electrons(negatively charged) and protons (positively charged).Both charges balance each other, hence atom iselectrically neutral.

(a) Cation :

If an atom has less electrons than a neutral atom,then it gets positively charged and a positivelycharged ion is known as cation.e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc.A cation bears that much units of positive charge asthere are the number of electrons lost by the neutralatom to form that cation.

e.g. An aluminium atom loses 3 electrons to formaluminium ion, so aluminium ion bears 3 units ofpositive charge and it is represented as Al3+.

(b) Anion :

If an atom has more number of electrons than that ofneutral atom, then it gets negatively charged and anegatively charged ion is known as anion.e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.

An anion bears that much units of negative charge asthere are the number of electrons gained by theneutral atom to form that anion.e.g. A nitrogen atom gains 3 electrons to form nitrideion, so nitride ion bears 3 units of negative chargeand it is represented as N3-.

Note :Size of a cation is always smaller and anion is alwaysgreater than that of the corresponding neutral atom.

(c) Monoatomic ions and polyatomic ions :(i) Monoatomic ions : Those ions which are formedfrom single atoms are called monoatomic ions orsimple ions.e.g. Na+, Mg2+ etc.

(ii) Polyatomic ions : Those ions which are formedfrom group of atoms joined together are calledpolyatomic ions or compound ions.e.g. Ammonium ion (NH

4+) , hydroxide ion (OH�) etc.

which are formed by the joining of two types of atoms,nitrogen and hydrogen in the first case and oxygen andhydrogen in the second.

(d) Valency of ions :

The valency of an ion is same as the charge presenton the ion.If an ion has 1 unit of positive charge, its valency is 1and it is known as a monovalent cation. If an ion has2 units of negative charge, its valency is 2 and it isknown as a divalent anion.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS

Note :Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present inthem.

PAGE # 41

LAWS OF CHEMICAL COMBINATION

The laws of chemical combination are theexperimental laws which led to the idea ofatoms being the smallest unit of matter. The laws ofchemical combination played a significant role in the

development of Dalton�s atomic theory of matter.

There are two important laws of chemical combination.These are:

(i) Law of conservation of mass(ii) Law of constant proportions

(a) Law of Conservation of Mass or Matter :

This law was given by Lavoisier in 1774 . According tothe law of conservation of mass, matter can neitherbe created nor be destroyed in a chemical reaction.

OrThe law of conservation of mass means that in achemical reaction, the total mass of products is equalto the total mass of the reactants. There is no changein mass during a chemical reaction.Suppose we carry out a chemical reaction between Aand B and if the products formed are C and D then,A + B C + D

Suppose 'a' g of A and 'b' g of B react to produce 'c' g ofC and 'd' g of D. Then, according to the law ofconservation of mass, we have,

a + b = c + dExample :When Calcium Carbonate (CaCO

3) is heated, a

chemical reaction takes place to form Calcium Oxide(CaO) and Carbon dioxide (CO

2). It has been found

by experiments that if 100 grams of calcium carbonateis decomposed completely, then 56 grams of CalciumOxide and 44 grams of Carbon dioxide are formed.

Since the total mass of products (100g ) is equal tothe total mass of the reactants (100g), there is nochange in the mass during this chemical reaction.The mass remains same or conserved.

(b) Law of Constant Proportions / Law of

Definite Proportions :

Proust, in 1779, analysed the chemical composition(types of elements present and percentage ofelements present ) of a large number of compoundsand came to the conclusion that the proportion ofeach element in a compound is constant (or fixed).According to the law of constant proportions: Achemical compound always consists of the sameelements combined together in the same proportionby mass.

Note :The chemical composition of a pure substance isnot dependent on the source from which it is obtained.

Example :

Water is a compound of hydrogen and oxygen. It canbe obtained from various sources (like river, sea, welletc.) or even synthesized in the laboratory. Fromwhatever source we may get it, 9 parts by weight ofwater is always found to contain 1 part by weight ofhydrogen and 8 parts by weight of oxygen. Thus, inwater, this proportion of hydrogen and oxygen alwaysremains constant.

Note :The converse of Law of definite proportions that whensame elements combine in the same proportion, thesame compound will be formed, is not always true.

(c) Law of Multiple Proportions :

According to it, when one element combines with theother element to form two or more different compounds,the mass of one element, which combines with aconstant mass of the other, bears a simple ratio toone another.

Simple ratio means the ratio between small naturalnumbers, such as 1 : 1, 1 : 2, 1 : 3e.g.Carbon and oxygen when combine, can form twooxides that are CO (carbon monoxide), CO

2 (carbon

dioxide).In CO,12 g carbon combined with 16 g of oxygen.In CO

2,12 g carbon combined with 32 g of oxygen.

Thus, we can see the mass of oxygen which combinewith a constant mass of carbon (12 g) bear simpleratio of 16 : 32 or 1 : 2

Note :The law of multiple proportion was given by Dalton in1808.

Sample Problem :

1. Carbon is found to form two oxides, which contain42.8% and 27.27% of carbon respectively. Show thatthese figures illustrate the law of multiple proportions.

Sol. % of carbon in first oxide = 42.8% of oxygen in first oxide = 100 - 42.8 = 57.2% of carbon in second oxide = 27.27 % of oxygen in second oxide = 100 - 27.27 = 72.73

For the first oxide -Mass of oxygen in grams that combines with 42.8 gof carbon = 57.2 Mass of oxygen that combines with 1 g of carbon =

1.3442.857.2

g

For the second oxide -Mass of oxygen in grams that combines with 27.27 gof carbon = 72.73 Mass of oxygen that combines with 1 g of carbon =

2.6827.2772.73

g

Ratio between the masses of oxygen that combinewith a fixed mass (1 g) of carbon in the two oxides= 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence,this illustrates the law of multiple proportion.

PAGE # 42

(d) Law of Reciprocal Proportions :

According to this law the ratio of the weights of twoelement A and B which combine separately with afixed weight of the third element C is either the sameor some simple multiple of the ratio of the weights inwhich A and B combine directly with each other.e.g.

The elements C and O combine separately with thethird element H to form CH

4 and H

2O and they combine

directly with each other to form CO2.

H 24

CH4 H O2

C O16

12

12

CO2 32

In CH4, 12 parts by weight of carbon combine with 4

parts by weight of hydrogen. In H2O, 2 parts by weight

of hydrogen combine with 16 parts by weight ofoxygen. Thus the weight of C and O which combinewith fixed weight of hydrogen (say 4 parts by weight)are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8.Now in CO

2, 12 parts by weight of carbon combine

directly with 32 parts by weight of oxygen i.e. theycombine directly in the ratio 12 : 32 or 3 : 8 which isthe same as the first ratio.

Note :

The law of reciprocal proportion was put forward byRitcher in 1794.

Sample Problem :2. Ammonia contains 82.35% of nitrogen and 17.65%

of hydrogen. Water contains 88.90% of oxygen and11.10% of hydrogen. Nitrogen trioxide contains63.15% of oxygen and 36.85% of nitrogen. Show thatthese data illustrate the law of reciprocal proportions.

Sol

In NH3, 17.65 g of H combine with N = 82.35 g

1g of H combine with N = 17.65

82.35 g = 4.67 g

In H2O, 11.10 g of H combine with O = 88.90 g

1 g H combine with O = 11.10

88.90g = 8.01 g

Ratio of the weights of N and O which combinewith fixed weight (=1g) of H= 4.67 : 8.01 = 1 : 1.72

In N2O

3, ratio of weights of N and O which combine

with each other = 36.85 : 63.15 = 1 : 1.71

Thus the two ratio are the same. Hence it illustratesthe law of reciprocal proportions.

(e) Gay Lussac�s Law of Gaseous Volumes :

Gay Lussac found that there exists a definiterelationship among the volumes of the gaseousreactants and their products. In 1808, he put forwarda generalization known as the Gay Lussac�s Law of

combining volumes. This may be stated as follows :

When gases react together, they always do so in

volumes which bear a simple ratio to one another

and to the volumes of the product, if these are also

gases, provided all measurements of volumes are

done under similar conditions of temperature and

pressure.

e.g.Combination between hydrogen and chlorine to form

hydrogen chloride gas. One volume of hydrogen and

one volume of chlorine always combine to form two

volumes of hydrochloric acid gas.

H2 (g) + Cl

2 (g) 2HCl (g)

1vol. 1 vol. 2 vol.

The ratio between the volume of the reactants and

the product in this reaction is simple, i.e., 1 : 1 : 2.

Hence it illustrates the Law of combining volumes.

(f) Avogadro�s Hypothesis :

This states that equal volumes of all gases under

similar conditions of temperature and pressure

contain equal number of molecules.

This hypothesis has been found to explain elegantly

all the gaseous reactions and is now widely

recognized as a law or a principle known as Avogadro�s

Law or Avogadro�s principle.

The reaction between hydrogen and chlorine can be

explained on the basis of Avogadro�s Law as follows :

Hydrogen + Chlorine Hydrogen chloride gas 1 vol. (By experiment)1 vol. 2 vol.

n molecules. n molecules. 2n molecules.(By Avogadro's Law)

21 molecules. molecules. 1 molecules. (By dividing throughout by 2n)

1 Atom 1 Atom 1 Molecule (Applying Avogadro's hypothesis)

21

It implies that one molecule of hydrogen chloride gas

is made up of 1 atom of hydrogen and 1 atom of

chlorine.

(i) Applications of Avogadro�s hypothesis :

(A) In the calculation of atomicity of elementarygases.

e.g.

2 volumes of hydrogen combine with 1 volume of

oxygen to form two volumes of water vapours.Hydrogen + Oxygen Water vapours2 vol. 1 vol. 2 vol.

PAGE # 43

Applying Avogadro�s hypothesis

Hydrogen + Oxygen Water vapours2 n molecules n molecules 2 n molecules

or 1 molecule 2

1 molecule 1 molecule

Thus1 molecule of water contains 2

1 molecule of

oxygen. But 1 molecule of water contains 1 atom of

oxygen. Hence. 2

1 molecule of oxygen = 1 atom of

oxygen or 1 molecules of oxygen = 2 atoms of oxygeni.e. atomicity of oxygen = 2.

(B) To find the relationship between molecular massand vapour density of a gas.

Vapour density (V.D.) = hydrogenofDensitygas ofDensity

=

epressur and temp. same the at hydrogen of volume same the of Mass

gas the of volume certain a of Mass

If n molecules are present in the given volume of a gasand hydrogen under similar conditions of temperatureand pressure.

V.D. = hydrogen of molecules n of Massgas the of molecules n of Mass

= hydrogen of molecule 1 of Massgas the of molecule 1 of Mass

= hydrogen of mass Moleculargas the of mass Molecular

= 2

mass Molecular

(since molecular mass of hydrogen is 2)Hence, Molecular mass = 2 × Vapour density

ATOMIC MASS UNIT

The atomic mass unit (amu) is equal to one-twelfth(1/12) of the mass of an atom of carbon-12.The massof an atom of carbon-12 isotope was given the atomicmass of 12 units, i.e. 12 amu or 12 u.The atomic masses of all other elements are nowexpressed in atomic mass units.

RELATIVE ATOMIC MASS

The atomic mass of an element is a relative quantityand it is the mass of one atom of the element relativeto one -twelfth (1/12) of the mass of one carbon-12atom. Thus, Relative atomic mass

= atom12Coneofmass

12

1

elementtheofatomoneofMass

[1/12 the mass of one C-12 atom = 1 amu, 1 amu =1.66 × 10�24 g = 1.66 × 10�27 kg.]

Note :One amu is also called one dalton (Da).

GRAM-ATOMIC MASS

The atomic mass of an element expressed in grams

is called the Gram Atomic Mass of the element.

The number of gram -atoms

= elementtheofmassAtomicGram

gramsinelementtheofMass

e.g.

Calculate the gram atoms present in (i) 16g of oxygen

and (ii) 64g of sulphur.

(i) The atomic mass of oxygen = 16.

Gram-Atomic Mass of oxygen (O) = 16 g.

No. of Gram-Atoms = 1616

= 1

(ii) The gram-atoms present in 64 grams of sulphur.

= sulphurofMass AtomicGram64

= 32

64= 2

AtomicNumber Element Symbol

Atomicmass

1 Hydrogen H 12 Helium He 43 Lithium Li 74 Beryllium Be 95 Boron B 116 Carbon C 127 Nitrogen N 148 Oxygen O 169 Fluorine F 1910 Neon Ne 2011 Sodium Na 2312 Magnesium Mg 2413 Aluminium Al 2714 Silicon Si 2815 Phosphorus P 3116 Sulphur S 3217 Chlorine Cl 35.518 Argon Ar 4019 Potassium K 3920 Calcium Ca 40

RELATIVE MOLECULAR MASS

The relative molecular mass of a substance is themass of a molecule of the substance as comparedto one-twelfth of the mass of one carbon -12 atomi.e.,Relative molecular mass

= atom12Coneofmass

12

1

substancetheofmoleculeoneofMass

The molecular mass of a molecule, thus, representsthe number of times it is heavier than 1/12 of themass of an atom of carbon-12 isotope.

PAGE # 44

GRAM MOLECULAR MASS

The molecular mass of a substance expressed in

grams is called the Gram Molecular Mass of the

substance . The number of gram molecules

= ancetsubstheofmassmolecularGram

gramsintancesubstheofMass

e.g.

(i) Molecular mass of hydrogen (H2) = 2u.

Gram Molecular Mass of hydrogen (H2) = 2 g .

(ii) Molecular mass of methane (CH4) = 16u

Gram Molecular Mass of methane (CH4) = 16 g.

e.g. the number of gram molecules present in 64 g of

methane (CH4).

= 4CHofmassmolecularGram

64 =

16

64 = 4.

(a) Calculation of Molecular Mass :

The molecular mass of a substance is the sum of

the atomic masses of its constituent atoms present

in a molecule.

Ex.3 Calculate the molecular mass of water.

(Atomic masses : H = 1u, O = 16u).

Sol. The molecular formula of water is H2O.

Molecular mass of water = ( 2 × atomic mass of H)

+ (1 × atomic mass of O)

= 2 × 1 + 1 × 16 = 18

i.e., molecular mass of water = 18 amu.

Ex.4 Find out the molecular mass of sulphuric acid.

(Atomic mass : H = 1u, O = 16u, S = 32u).

Sol. The molecular formula of sulphuric acid is H2SO

4.

Molecular mass of H2SO

4

= (2 × atomic mass of H) + ( 1 × atomic mass of S)

+ ( 4 × atomic mass of O)

= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98

i.e., Molecular mass of H2SO

4= 98 amu.

FORMULA MASS

The term �formula mass� is used for ionic compounds

and others where discrete molecules do not exist,

e.g., sodium chloride, which is best represented as

(Na+Cl�)n, but for reasons of simplicity as NaCl or

Na+Cl�. Here, formula mass means the sum of the

masses of all the species in the formula.

Thus, the formula mass of sodium chloride = (atomic

mass of sodium) + (atomic mass of chlorine)

= 23 + 35.5

= 58.5 amu

EQUIVALENT MASS

(a) Definition :

Equivalent mass of an element is the mass of theelement which combine with or displaces 1.008 partsby mass of hydrogen or 8 parts by mass of oxygen or35.5 parts by mass of chlorine.

(b) Formulae of Equivalent Masses of differentsubstances :(i) Equivalent mass of an element =

element the ofValency element the of wt.Atomic

(ii) Eq. mass an acid = acid the ofBasicity acid the of wt. Mol.

Basicity is the number of replaceable H+ ions fromone molecule of the acid.

(iii) Eq. Mass of a base = base the ofAcidity base the of wt. Mol.

Acidity is the number of replaceable OH� ions fromone molecule of the base

(iv) Eq. mass of a salt

= metal ofvalency atoms metal of Numbersalt the of wt. Mol.

(v) Eq. mass of an ion = ion the on Charge

ion the of wt.Formula

(vi) Eq. mass of an oxidizing/reducing agent

=

substance the ofmolecule

oneby gained or lost electrons of No.

wt At. or wt.Mol

Equivalent weight of some compounds are given inthe table :

S.No. CompoundEquivalent

weight

1 HCl 36.5

2 H2SO4 49

3 HNO3 63

4 45

5 .2H2O63

6 NaOH 40

7 KOH 56

8 CaCO3 50

9 NaCl 58.5

10 Na2CO3 53

COOH

COOH

PAGE # 45

In Latin, mole means heap or collection or pile. Amole of atoms is a collection of atoms whose totalmass is the number of grams equal to the atomicmass in magnitude. Since an equal number of molesof different elements contain an equal number ofatoms, it becomes convenient to express theamounts of the elements in terms of moles. A molerepresents a definite number of particles, viz, atoms,molecules, ions or electrons. This definite number iscalled the Avogadro Number (now called the Avogadroconstant) which is equal to 6.023 × 1023.

A mole is defined as the amount of a substance thatcontains as many atoms, molecules, ions, electronsor other elementary particles as there are atoms inexactly 12 g of carbon -12 (12C).

(a) Moles of Atoms :

(i) 1 mole atoms of any element occupy a mass whichis equal to the Gram Atomic Mass of that element.

e.g. 1 Mole of oxygen atoms weigh equal to GramAtomic Mass of oxygen, i.e. 16 grams.

(ii) The symbol of an element represents 6.023 x 1023

atoms (1 mole of atoms) of that element.

e.g : Symbol N represents 1 mole of nitrogen atomsand 2N represents 2 moles of nitrogen atoms.

(b) Moles of Molecules :

(i) 1 mole molecules of any substance occupy a masswhich is equal to the Gram Molecular Mass of thatsubstance.

e.g. : 1 mole of water (H2O) molecules weigh equal toGram Molecular Mass of water (H2O), i.e. 18 grams.

(ii) The symbol of a compound represents 6.023 x1023 molecules (1 mole of molecules) of thatcompound.

e.g. : Symbol H2O represents 1 mole of watermolecules and 2 H2O represents 2 moles of watermolecules.

Note :

The symbol H2O does not represent 1 mole of H2molecules and 1 mole of O atoms. Instead, itrepresents 2 moles of hydrogen atoms and 1 moleof oxygen atoms.

Note :The SI unit of the amount of a substance is Mole.

(c) Mole in Terms of Volume :

Volume occupied by 1 Gram Molecular Mass or 1mole of a gas under standard conditions oftemperature and pressure (273 K and 1atm.pressure) is called Gram Molecular Volume. Its valueis 22.4 litres for each gas.Volume of 1 mole = 22.4 litre (at STP)

Note :The term mole was introduced by Ostwald in 1896.

SOME IMPORTANT RELATIONS AND FORMULAE

(i) 1 mole of atoms = Gram Atomic mass = mass of6.023 × 1023 atoms

(ii) 1 mole of molecules = Gram Molecular Mass= 6.023 x 1023 molecules(iii) Number of moles of atoms

= elementofMassAtomicGram

gramsinelementofMass

(iv) Number of moles of molecules

= substanceofMassMolecularGram

gramsinsubstanceofMass

(v) Number of moles of molecules

= AN

N

numberAvogadro

elementofmolecules ofNo.

Ex.5 To calculate the number of moles in 16 grams ofSulphur (Atomic mass of Sulphur = 32 u).

Sol. 1 mole of atoms = Gram Atomic Mass.So, 1 mole of Sulphur atoms = Gram Atomic Mass ofSulphur = 32 grams.Now, 32 grams of Sulphur = 1 mole of SulphurSo, 16 grams of Sulphur= (1/32) x 16 = 0.5 molesThus, 16 grams of Sulphur constitute 0.5 mole ofSulphur.

6.023 × 10

(N ) Atoms

23

A

6.023 × 10

(N ) molecules

23

A

1 Mole

1 gram atomof element

1 gram molecule of substance

1 gram formula mass of substance

In terms of particles

In terms of mass

22.4 litre

In term ofvolume

PROBLEMS BASED ON THE MOLE CONCEPT

Ex.6 Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium = 23 u)

Sol. Number of moles

= elementofMassAtomicGram

gramsinelementtheofMass =

23

5.75= 0.25 mole

or,1 mole of sodium atoms = Gram Atomic mass ofsodium = 23g.23 g of sodium = 1 mole of sodium.

5.75 g of sodium = 23

5.75mole of sodium = 0.25 mole

PAGE # 46

Ex.7 What is the mass in grams of a single atom ofchlorine ? (Atomic mass of chlorine = 35.5u)

Sol. Mass of 6.022 × 1023 atoms of Cl = Gram AtomicMass of Cl = 35.5 g.

Mass of 1 atom of Cl =23106.022

g35.5

= 5.9 × 10�23 g.

Ex.8 The density of mercury is 13.6 g cm�3. How manymoles of mercury are there in 1 litre of the metal ?(Atomic mass of Hg = 200 u).

Sol. Mass of mercury (Hg) in grams = Density(g cm�3)× Volume (cm3)= 13.6 g cm�3 × 1000 cm3 = 13600 g.

Number of moles of mercury

= mercuryofMassAtomicGramgramsinmercuryofMass

=200

13600 = 68

Ex.9 The mass of a single atom of an element M is3.15× 10�23 g . What is its atomic mass ? Whatcould the element be ?

Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms= mass of 1 atom × 6.022 × 1023

= (3.15 × 10�23g) × 6.022 × 1023

= 3.15 × 6.022 g = 18.97 g.

Atomic Mass of the element = 18.97uThus, the element is most likely to be fluorine.

Ex.10 An atom of neon has a mass of 3.35 × 10�23 g.How many atoms of neon are there in 20 g of thegas ?

Sol. Number of atoms

= atom1ofMass

massTotal = 23�103.35

02

= 5.97 × 1023

Ex.11 How many grams of sodium will have the samenumber of atoms as atoms present in 6 g ofmagnesium ?(Atomic masses : Na = 23u ; Mg =24u)

Sol. Number of gram -atom of Mg

= MassAtomicGramgramsinMgofMass

= 246

= 4

1

Gram Atoms of sodium should be = 4

1

1 Gram Atom of sodium = 23 g

4

1 gram atoms of sodium = 23 ×

4

1 = 5.75 g

Ex.12 How many moles of Cr are there in 85g of Cr2S

3 ?

(Atomic masses : Cr = 52 u , S =32 u)Sol. Molecular mass of Cr

2S

3 = 2 × 52 + 3 × 32 = 104

+ 96 = 200 u.200g of Cr

2S

3 contains = 104 g of Cr.

85 g of Cr2S

3 contains =

200

85104 g of Cr = 44.2g

Thus, number of moles of Cr = 52

44.2 = 0.85 .

Ex.13 What mass in grams is represented by

(a) 0.40 mol of CO2,

(b) 3.00 mol of NH3,

(c) 5.14 mol of H5IO

6

(Atomic masses : C=12 u, O=16 u, N=14 u,

H=1 u and I = 127 u)

Sol. Weight in grams = number of moles × molecular

mass.

Hence,

(a) mass of CO2 = 0.40 × 44 = 17.6 g

(b) mass of NH3 = 3.00 × 17 = 51.0 g

(c) mass of H5IO

6 = 5.14 × 228 = 1171.92g

Ex.14 Calculate the volume in litres of 20 g of hydrogen

gas at STP.

Sol. Number of moles of hydrogen

= hydrogenofMassMolecularGram

grams in hydrogenofMass =

220

= 10

Volume of hydrogen = number of moles × Gram

Molecular Volume.

= 10 ×22.4 = 224 litres.

Ex.15 The molecular mass of H2SO

4 is 98 amu.

Calculate the number of moles of each elementin 294 g of H

2SO

4.

Sol. Number of moles of H2SO

4 =

98

294= 3 .

The formula H2SO

4 indicates that 1 molecule of

H2SO

4 contains 2 atoms of H, 1 atom of S and 4

atoms of O. Thus, 1 mole of H2SO

4 will contain 2

moles of H,1 mole of S and 4 moles of O atomsTherefore, in 3 moles of H

2SO

4 :

Number of moles of H = 2 × 3 = 6

Number of moles of S = 1 × 3 = 3

Number of moles of O = 4 × 3 = 12

Ex.16 Find the mass of oxygen contained in 1 kg ofpotassium nitrate (KNO

3).

Sol. Since 1 molecule of KNO3 contains 3 atoms of

oxygen, 1 mol of KNO3 contains 3 moles of

oxygen atoms. Moles of oxygen atoms = 3 × moles of KNO

3

= 3 × 101

1000 = 29.7

(Gram Molecular Mass of KNO3 = 101 g)

Mass of oxygen = Number of moles × Atomic

mass= 29.7 × 16 = 475.2 g

Ex.17 You are asked by your teacher to buy 10 moles ofdistilled water from a shop where small bottleseach containing 20 g of such water are available.How many bottles will you buy ?

Sol. Gram Molecular Mass of water (H2O) = 18 g

10 mol of distilled water = 18 × 10 = 180 g.

Because 20 g distilled water is contained in 1bottle,

180 g of distilled water is contained in =20

180

bottles = 9 bottles.

Number of bottles to be bought = 9

PAGE # 47

Ex.18 6.022 × 1023 molecules of oxygen (O2) is equal tohow many moles ?

Sol. No. of moles =

AN

N

moleculesofno.sAvogadro'

oxygenofmoleculesofNo. =

23

23

106.023

106.023

= 1

PERCENTAGE COMPOSITION

The percentage composition of elements in acompound is calculated from the molecular formulaof the compound.The molecular mass of the compound is calculatedfrom the atomic masses of the various elementspresent in the compound. The percentage by massof each element is then computed with the help of thefollowing relations.Percentage mass of the element in the compound

= massMolecular

elementtheofmassTotal× 100

Ex.19 What is the percentage of calcium in calciumcarbonate (CaCO

3) ?

Sol. Molecular mass of CaCO3 = 40 + 12 + 3 × 16

= 100 amu.Mass of calcium in 1 mol of CaCO

3 = 40g.

Percentage of calcium = 100

10040 = 40 %

Ex.20 What is the percentage of sulphur in sulphuricacid (H

2SO

4) ?

Sol. Molecular mass of H2SO

4 = 1 × 2 + 32 + 16 × 4 = 98

amu.

Percentage of sulphur = 98

10032 = 32.65 %

Ex.21. What are the percentage compositions ofhydrogen and oxygen in water (H

2O) ?

(Atomic masses : H = 1 u, O = 16 u)

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.

H2O has two atoms of hydrogen.

So, total mass of hydrogen in H2O = 2 amu.

Percentage of H = 18

1002 = 11.11 %

Similarly,

percentage of oxygen = 18

10016 = 88.88 %

The following steps are involved in determining theempirical formula of a compound :(i) The percentage composition of each element is

divided by its atomic mass. It gives atomic ratio of the

elements present in the compound.

(ii) The atomic ratio of each element is divided by theminimum value of atomic ratio as to get the simplestratio of the atoms of elements present in thecompound.

(iii) If the simplest ratio is fractional, then values ofsimplest ratio of each element is multiplied bysmallest integer to get the simplest whole numberfor each of the element.

(iv) To get the empirical formula, symbols of variouselements present are written side by side with theirrespective whole number ratio as a subscript to thelower right hand corner of the symbol.

(v) The molecular formula of a substance may bedetermined from the empirical formula if the molecularmass of the substance is known. The molecularformula is always a simple multiple of empiricalformula and the value of simple multiple (n) isobtained by dividing molecular mass with empiricalformula mass.

n = MassFormulaEmpiricalMassMolecular

Ex-22.A compound of carbon, hydrogen and nitrogencontains these elements in the ratio of 9:1:3.5 respectively.Calculate the empirical formula. If its molecular mass is108, what is the molecular formula ?

Sol.

ElementMassRatio

AtomicMass

Relative Numberof Atoms

Simplest Ratio

Carbon 9 12 0.75×4 = 3

Hydrogen 1 1 1 × 4 = 4

Nitrogen 3.5 14 0.25 ×4 = 1

0.75129

111

0.25143.5

Empirical ratio = C3H

4N

Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54

n = MassFormulaEmpiricalMassMolecular

= 254

108

Thus, molecular formula of the compound

= (Empirical formula)2

= (C3H

4N)

2

= C6H

8N

2

Ex.23. A compound on analysis, was found to have thefollowing composition :(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen= 69.50%, (iv) Hydrogen = 6.22%. Calculate themolecular formula of the compound assuming thatwhole hydrogen in the compound is present as waterof crystallisation. Molecular mass of the compoundis 322.

Sol.

Element PercentageAtomic mass

Relative Number of atoms

Simplest ratio

Sodium 14.31 23 0.622

Sulphur 9.97 32 0.311

Hydrogen 6.22 1 6.22

Oxygen 69.50 16 4.34

20.311

0.622

10.311

0.311

200.311

6.22

140.311

4.34

23

31.14

1

22.6

16

50.69

32

97.9

PAGE # 48

The empirical formula = Na2SH

20O

14

Empirical formula mass= (2 × 23) + 32 + (20 × 1) + (14 × 16)

= 322Molecular mass = 322Molecular formula = Na

2SH

20O

14

Whole of the hydrogen is present in the form of waterof crystallisation. Thus, 10 water molecules arepresent in the molecule.So, molecular formula = Na

2SO

4. 10H

2O

CONCENTRATION OF SOLUTIONS

(a) Strength in g/L :

The strength of a solution is defined as the amount ofthe solute in grams present in one litre (or dm3) of thesolution, and hence is expressed in g/litre or g/dm3.

Strength in g/L = litreinsolutionofVolume

graminsoluteofWeight

(b) Molarity :

Molarity of a solution is defined as the number ofmoles of the solute dissolved per litre (or dm3) ofsolution. It is denoted by �M�. Mathematically,

M = litreinsolutiontheofVolume

soluteofmolesofNumber

litreinsolutionofVolume

soluteofMassMoleculargram/GraminsoluteofMass

M can be calculated from the strength as given below :

M = solute of mass Molecular

litre per grams in Strength

If �w� gram of the solute is present in V cm3 of a givensolution , then

M = massMolecular

w ×

V

1000

e.g. a solution of sulphuric acid having 4.9 grams of itdissolved in 500 cm3 of solution will have its molarity,

M = massMolecular

w×

V

1000

M = 98

4.9 ×

500

1000 = 0.1

(c) Formality :

In case of ionic compounds like NaCl, Na2CO

3 etc.,

formality is used in place of molarity. The formality ofa solution is defined as the number of gram formulamasses of the solute dissolved per litre of thesolution. It is represented by the symbol �F�. The term

formula mass is used in place of molecular massbecause ionic compounds exist as ions and not asmolecules. Formula mass is the sum of the atomicmasses of the atoms in the formula of the compound.

litreinsolutionofVolume

soluteofMasslagram/FormuinsoluteofMass

(d) Normality :

Normality of a solution is defined as the number ofgram equivalents of the solute dissolved per litre (dm3)of given solution. It is denoted by �N�.Mathematically,

N = litre in solution the of Volume

solute of sequivalent gram of Number

N = litre in solution the of Volume

solute of weight equivalent/ graminsolute of Weight

N can be calculated from the strength as given below :

N = solute of mass Equivalent

litre per grams in Strength=

E

S

If �w� gram of the solute is present in V cm3 of a givensolution.

N = solute the of mass Equivalentw

× V

1000

e.g. A solution of sulphuric acid having 0.49 gram ofit dissolved in 250 cm3 of solution will have itsnormality,

N = solute the of mass Equivalentw

× V

1000

N = 49

0.49×

250

1000 = 0.04

(Eq. mass of sulphuric acid = 49).

Solution Seminormal

Decinormal

Centinormal

Normality101

1001

21

Some Important Formulae :

(i) Milli equivalent of substance = N × V

where , N normality of solutionV Volume of solution in mL

(ii) If weight of substance is given,

milli equivalent (NV) = E1000w

Where, W Weight of substance in gramE Equivalent weight of substance

(iii) S = N × E

S Strength in g/LN Normality of solutionE Equivalent weight

(iv) Calculation of normality of mixture :

Ex.24 100 ml of 10

NHCl is mixed with 50 ml of

5

NH

2SO

4 .

Find out the normality of the mixture.Sol. Milli equivalent of HCl + milli equivalent H

2SO

4

= milli equivalent of mixtureN

1 V

1 + N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

50

51

100101 N

3 × 150

N3 =

150

20 =

15

2= 0.133

PAGE # 49

Ex.25 100 ml of 10

NHCl is mixed with 25 ml of

5

NNaOH.

Find out the normality of the mixture.

Sol. Milli equivalent of HCl � milli equivalent of NaOH

= milli equivalent of mixtureN

1 V

1 � N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

25

51

�100101

= N3 × 125

N3 =

251

Note :

1 milli equivalent of an acid neutralizes 1 milliequivalent of a base.

(e) Molality :

Molality of a solution is defined as the number ofmoles of the solute dissolved in 1000 grams of thesolvent. It is denoted by �m�.Mathematically,

m = gram in solvent the of Weightsolute the of moles of Number

× 1000 �m� can

be calculated from the strength as given below :

m = solute of mass Molecularsolvent of gram 1000 per Strength

If �w� gram of the solute is dissolved in �W� gram of the

solvent then

m = solute the of mass Mol.

w×

W1000

e.g. A solution of anhydrous sodium carbonate(molecular mass = 106) having 1.325 grams of it,dissolved in 250 gram of water will have its molality -

m = 250

1000106

1.325 = 0.05

Note :

Relationship Between Normality and Molarity of aSolution :Normality of an acid = Molarity × Basicity

Normality of base = Molarity × Acidity

Ex.26 Calculate the molarity and normality of a solutioncontaining 0.4 g of NaOH dissolved in 500 cm3 ofsolvent.

Sol. Weight of NaOH dissolved = 0.5 gVolume of the solution = 500 cm3

(i) Calculation of molarity :Molecular weight of NaOH = 23 + 16 + 1 = 40

Molarity =litre in solution of Volume

solute of weightmolecularsolute/ of Weight

= 500/1000

0.5/40 = 0.025

(ii) Calculation of normality :Normality

=litre in solution of Volume

solute of weightequivalentsolute/ of Weight

=500/1000

0.5/40 = 0.025

Ex.27 Find the molarity and molality of a 15% solutionof H

2SO

4 (density of H

2SO

4 solution = 1.02 g/cm3)

(Atomic mass : H = 1u, O = 16u , S = 32 u)Sol. 15% solution of H

2SO

4 means 15g of H

2SO

4 are

present in 100g of the solution i.e.Wt. of H

2SO

4 dissolved = 15 g

Weight of the solution = 100 gDensity of the solution = 1.02 g/cm3 (Given)

Calculation of molality :

Weight of solution = 100 gWeight of H

2SO

4 = 15 g

Wt. of water (solvent) = 100 � 15 = 85 g

Molecular weight of H2SO

4 = 98

15 g H2SO

4 =

98

15= 0.153 moles

Thus ,85 g of the solvent contain 0.153 moles .

1000 g of the solvent contain= 85

0.153× 1000 = 1.8 mole

Hence ,the molality of H2SO

4 solution = 1.8 m

Calculation of molarity :

15 g of H2SO

4 = 0.153 moles

Vol. of solution = solution ofDensity

solution of Wt.

= 1.02

100 = 98.04 cm3

This 98.04 cm3 of solution contain H2SO

4 = 0.153

moles1000 cm3 of solution contain H

2SO

4

= 98.04

0.153 × 1000 = 1.56 moles

Hence the molarity of H2SO

4 solution = 1.56 M

(f) Mole Fraction :

The ratio between the moles of solute or solvent tothe total moles of solution is called mole fraction.

mole fraction of solute = Nn

n

solutionofMoles

soluteofMoles

= W/Mw/m

w/m

Mole fraction of solvent = Nn

N

solutionofMoles

solventofMoles

= W/Mw/m

W/M

where,

n number of moles of solute

N number of moles of solvent

m molecular weight of solute

M molecular weight of solvent

w weight of solute

W weight of solvent

PAGE # 50

Ex.28 Find out the mole fraction of solute in 10% (by weight)urea solution.weight of solute (urea) = 10 gweight of solution = 100 gweight of solvent (water) = 100 � 10 = 90g

mole fraction of solute = solutionofMoles

soluteofMoles =

W/Mw/m

w/m

=

18/9060/10

60/10

= 0.032

Note :Sum of mole fraction of solute and solvent is alwaysequal to one.

STOICHIOMETRY

(a) Quantitative Relations in Chemical

Reactions :

Stoichiometry is the calculation of the quantities ofreactants and products involved in a chemicalreaction.

It is based on the chemical equation and on therelationship between mass and moles.N

2(g) + 3H

2(g) 2NH

3(g)

A chemical equation can be interpreted as follows -1 molecule N

2 + 3 molecules H

2 2 moleculesNH

3(Molecular interpretation)

1 mol N2 + 3 mol H

2 2 mol NH

3

(Molar interpretation)28 g N

2 + 6 g H

2 34 g NH

3

(Mass interpretation)1 volume N

2 + 3 volume H

2 2 volume NH

3

(Volume interpretation)

Thus, calculations based on chemical equations aredivided into four types -

(i) Calculations based on mole-mole relationship.

(ii) Calculations based on mass-mass relationship.

(iii) Calculations based on mass-volume relationship.

(iv) Calculations based on volume -volumerelationship.

(i) Calculations based on mole-mole relationship :In such calculations, number of moles of reactantsare given and those of products are required.Conversely, if number of moles of products are given,then number of moles of reactants are required.

Ex.29 Oxygen is prepared by catalytic decompositionof potassium chlorate (KClO

3). Decomposition

of potassium chlorate gives potassium chloride(KCl) and oxygen (O

2). How many moles and how

many grams of KClO3 are required to produce

2.4 mole O2.

Sol. Decomposition of KClO3 takes place as,

2KClO3(s) 2KCl(s) + 3O

2(g)

2 mole KClO3 3 mole O

2

3 mole O2 formed by 2 mole KClO

3

2.4 mole O2 will be formed by

4.2

32

mole KClO3 = 1.6 mole KClO

3Mass of KClO

3 = Number of moles × molar mass

= 1.6 × 122.5 = 196 g

(ii) Calculations based on mass-mass relationship:In making necessary calculation, following steps arefollowed -

(a) Write down the balanced chemical equation.

(b) Write down theoretical amount of reactants andproducts involved in the reaction.

(c) The unknown amount of substance is calculatedusing unitary method.

Ex.30 Calculate the mass of CaO that can be preparedby heating 200 kg of limestone CaCO

3 which is

95% pure.

Sol. Amount of pure CaCO3 = 200

10095

= 190 kg

= 190000 gCaCO

3(s) CaO(s) + CO

2(g)

1 mole CaCO3 1 mole CaO

100 g CaCO3 56 g CaO

100 g CaCO3 give 56 g CaO

190000 g CaCO3 will give=

10056

× 190000 g CaO

= 106400 g = 106.4 kg

Ex.31 Chlorine is prepared in the laboratory by treatingmanganese dioxide (MnO

2) with aqueous

hydrochloric acid according to the reaction -

MnO2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

How many grams of HCl will react with 5 g MnO2 ?

Sol. 1 mole MnO2 reacts with 4 mole HCl

or 87 g MnO2 react with 146 g HCl

5 g MnO2 will react with =

87146

× 5 g HCl = 8.39 g HCl

Ex.32 How many grams of oxygen are required to burncompletely 570 g of octane ?

Sol. Balanced equation

2C H + 25O 8 18 2 16CO + 18H O2 2

2 mole2 × 114

25 mole25 × 32

First method : For burning 2 × 114 g of the octane,

oxygen required = 25 × 32 g

For burning 1 g of octane, oxygen required =1142

3225

g

Thus, for burning 570 g of octane, oxygen required

= 1142

3225

× 570 g = 2000 g

Mole Method : Number of moles of octane in 570grams

114570

= 5.0

For burning 2.0 moles of octane, oxygen required= 25 mol = 25 × 32 g

For burning 5 moles of octane, oxygen required

= 0.23225

× 5.0 g = 2000 g

Proportion Method : Let x g of oxygen be required forburning 570 g of octane. It is known that 2 × 114 g of

the octane requires 25 × 32 g of oxygen; then, the

proportion.

etanocg1142oxygeng3225

= etanocg570

x

x = 1142

5703225

= 2000 g

PAGE # 51

Ex.33 How many kilograms of pure H2SO

4 could be

obtained from 1 kg of iron pyrites (FeS2) according to

the following reactions ?4FeS

2 + 11O

2 2Fe

2O

3 + 8SO

2

2SO2 + O

2 2SO

3

SO3 + H

2O H

2SO

4

Sol. Final balanced equation,4FeS + 15O + 8H O2 2 2

2Fe O + 8H SO2 3 2 4

8 mole8 × 98 g

4 mole4 × 120 g

4 × 120 g of FeS2 yield H

2SO

4 = 8 × 98 g

1000 g of FeS2 will yield H

2SO

4 =

1204988

× 1000

= 1633.3 g

(iii) Calculations involving mass-volume relationship :In such calculations masses of reactants are givenand volume of the product is required and vice-versa.1 mole of a gas occupies 22.4 litre volume at STP.Mass of a gas can be related to volume according tothe following gas equation -PV = nRT

PV = mw

RT

Ex-34. What volume of NH3 can be obtained from 26.75 g

of NH4Cl at 27ºC and 1 atmosphere pressure.

Sol. The balanced equation is -

NH Cl(s) 4 NH (g) + HCl(g)3

1 mol1 mol53.5 g

53.5 g NH4Cl give 1 mole NH

3

26.75 g NH4Cl will give

5.531

× 26.75 mole NH3

= 0.5 molePV = nRT1 ×V = 0.5 × 0.0821 × 300

V = 12.315 litre

Ex-35 What quantity of copper (II) oxide will react with2.80 litre of hydrogen at STP ?

Sol. CuO + H 2 Cu + H O2

1 mol79.5 g

1 mol22.4 litre at NTP

22.4 litre of hydrogen at STP reduce CuO = 79.5 g2.80 litre of hydrogen at STP will reduce CuO

= 4.225.79

× 2.80 g = 9.93 g

Ex-36 Calculate the volume of carbon dioxide at STPevolved by strong heating of 20 g calcium carbonate.

Sol. The balanced equation is -

CaCO 3 CaO + CO 2

1 mol = 22.4 litre at STP

1 mol 100 g

100 g of CaCO3 evolve carbon dioxide = 22.4 litre

20 g CaCO3 will evolve carbon dioxide

= 100

4.22 × 20 = 4.48 litre

Ex.37 Calculate the volume of hydrogen liberated at 27ºC

and 760 mm pressure by heating 1.2 g of magnesiumwith excess of hydrochloric acid.

Sol. The balanced equation is

Mg + 2HCl MgCl + H2 2

1 mol 24 g

24 g of Mg liberate hydrogen = 1 mole

1.2 g Mg will liberate hydrogen = 0.05 mole

PV = nRT1 × V = 0.05 × 0.0821 × 300

V = 1.2315 litre

(iv) Calculations based on volume volumerelationship :These calculations are based on two laws :(i) Avogadro�s law (ii) Gay-Lussac�s Law

e.g.N (g) + 3H (g)2 2 2NH (g) (Avogadro's law)3

2 mol 2 × 22.4 L

1 mol1 × 22.4 L

3 mol3 × 22.4 L

(under similar conditions of temperature andpressure, equal moles of gases occupy equalvolumes)N (g) + 3H (g)2 2 2NH (g)3

1 vol 3 vol 2 vol(Gay- Lussac's Law)

(under similar conditions of temperature andpressure, ratio of coefficients by mole is equal to ratioof coefficient by volume).

Ex-38 One litre mixture of CO and CO2 is taken. This is

passed through a tube containing red hot charcoal.The volume now becomes 1.6 litre. The volume aremeasured under the same conditions. Find thecomposition of mixture by volume.

Sol. Let there be x mL CO in the mixture , hence, there willbe (1000 � x) mL CO

2. The reaction of CO

2 with red

hot charcoal may be given as -

CO (g) + C(s)2 2CO(g) 2 vol.2(1000 � x)

1 vol.(1000 �x)

Total volume of the gas becomes = x + 2(1000 � x)

x + 2000 � 2x = 1600

x = 400 mL volume of CO = 400 mL and volume of CO

2 = 600 mL

Ex-39 What volume of air containing 21% oxygen by volumeis required to completely burn 1kg of carbon containing100% combustible substance ?

Sol. Combustion of carbon may be given as,

C(s) + O (g) 2 CO (g)2

1 mol12 g

1 mol32 g

12 g carbon requires 1 mole O2 for complete

combustion

1000 g carbon will require 1000121 mole O

2 for

combustion, i.e. , 83.33 mole O2

Volume of O2 at STP = 83.33 × 22.4 litre

= 1866.66 litre 21 litre O

2 is present in 100 litre air

1866.66 litre O2 will be present in

21100

× 1866.66 litre air

= 8888.88 litre or 8.89 × 103 litre

PAGE # 52

Ex-40 An impure sample of calcium carbonate contains80% pure calcium carbonate 25 g of the impuresample reacted with excess of hydrochloric acid.Calculate the volume of carbon dioxide at STPobtained from this sample.

Sol. 100 g of impure calcium carbonate contains = 80 gpure calcium carbonate25 g of impure calcium carbonate sample will contain

= 10080

× 25 = 20 g pure calcium carbonate

The desired equation is -

CaCO + 2HCl 3 CaCl + CO + H O2 2 2

1 mol100 g

22.4 litre at STP

100 g pure CaCO3 liberate = 22.4 litre CO

2.

20 g pure CaCO3 liberate = 20

1004.22

= 4.48 litre CO2

VOLUMETRIC CALCULATIONS

The quantitative analysis in chemistry is primarilycarried out by two methods, viz, volumetric analysisand gravimetric analysis.In the first method the massof a chemical species is measured by measurementof volume, whereas in the second method it is deter-mined by taking the weight.

The strength of a solution in volumetric analysis isgenerally expressed in terms of normality, i.e., num-ber of equivalents per litre but since the volume in thevolumetric analysis is generally taken in millilitres(mL), the normality is expressed by milliequivalentsper millilitre.

USEFUL FORMULAE FOR

VOLUMETRIC CALCULATIONS

(i) milliequivalents = normality × volume in millilitres.

(ii) At the end point of titration, the two titrants, say 1and 2, have the same number of milliequivalents,i.e., N

1V

1 = N

2V

2, volume being in mL.

(iii) No. of equivalents = 1000

.e.m.

(iv) No. of equivalents for a gas =

)STPat.eq1of.vol(volumeequivalentSTPatVolume

(v) Strength in grams per litre = normality × equivalent

weight.

(vi) (a) Normality = molarity × factor relating mol. wt.

and eq. wt.(b) No. of equivalents = no. of moles × factor relat

ing mol. wt. and eq. wt.

Ex.41 Calculate the number of milli equivalent of H2SO

4

present in 10 mL of N/2 H2SO

4 solution.

Sol. Number of m.e. = normality × volume in mL =21

× 10 = 5.

Ex.42 Calculate the number of m.e. and equivalents ofNaOH present in 1 litre of N/10 NaOH solution.

Sol. Number of m.e. = normality × volume in mL

= 101

× 1000 = 100

Number of equivalents = 1000

.e.mof.no =

1000100

= 0.10

Ex.43 Calculate number of m.e. of the acids present in(i) 100 mL of 0.5 M oxalic acid solution.(ii) 50 mL of 0.1 M sulphuric acid solution.

Sol. Normality = molarity × basicity of acid

(i) Normality of oxalic acid = 0.5 × 2 = 1 N

m.e. of oxalic acid = normality × vol. in mL = 1 × 100

= 100.(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N

m.e. of sulphuric acid = 0.2 × 50 = 10

Ex.44 A 100 mL solution of KOH contains 10milliequivalents of KOH. Calculate its strength in nor-mality and grams/litre.

Sol. Normality = mLinvolume.e.mof.no

= 1.010010

strength of the solution = N/10Again, strength in grams/litre = normality × eq. wt.

= 56101 = 5.6 gram/litre.

56

156

acidity.wtmolecular

KOHof.wt.eq

Ex.45 What is strength in gram/litre of a solution of H2SO

4,

12 cc of which neutralises 15 cc of 10N

NaOH

solution ?

Sol. m.e. of NaOH solution = 101

× 15 = 1.5

m.e. of 12 cc of H2SO

4 = 1.5

normality of H2SO

4 =

125.1

Strength in grams/litre = normality × eq. wt.

= 12

5.1 × 49 grams/litre

= 6.125 grams/litre.

49

298

basicitywt.molecular

SOHofwt.eq. 42

Ex.46 What weight of KMnO4 will be required to prepare

250 mL of its 10N

solution if eq. wt. of KMnO4 is 31.6 ?

Sol. Equivalent weight of KMnO4 = 31.6

Normality of solution (N) = 101

Volume of solution (V) = 250 ml

1000NEV

W ; W = 1000

2506.31101 79.0

406.31 g

PAGE # 53

Ex.47 100 mL of 0.6 N H2SO

4 and 200 mL of 0.3 N HCl

were mixed together. What will be the normality of theresulting solution ?

Sol. m.e. of H2SO

4 solution = 0.6 × 100 = 60

m.e. of HCl solution = 0.3 × 200 = 60

m.e. of 300 mL (100 + 200) of acidic mixture= 60 + 60 = 120.

Normality of the resulting solution = .voltotal.e.m

= 300120

= 52

N.

Ex.48 A sample of Na2CO

3. H

2O weighing 0.62 g is added

to 100 mL of 0.1 N H2SO

4. Will the resulting solution

be acidic, basic or neutral ?

Sol. Equivalents of Na2CO

3. H

2O =

6262.0

= 0.01

62

2124

OH.CONaof.wt.eq 232

m.e. of Na2CO

3. H

2O = 0.01 × 1000 = 10

m.e. of H2SO

4 = 0.1 × 100 = 10

Since the m.e. of Na2CO

3. H

2O is equal to that of H

2SO

4,

the resulting solution will be neutral.

(a) Introduction :

Volumetric analysis is a method of quantitativeanalysis. It involves the measurement of the volumeof a known solution required to bring about thecompletion of the reaction with a measured volumeof the unknown solution whose concentration orstrength is to be determined. By knowing the volumeof the known solution, the concentration of the solutionunder investigation can be calculated. Volumetricanalysis is also termed as titrimetric analysis.

(b) Important terms used in volumetric

analysis :

(i) Titration : The process of addition of the knownsolution from the burette to the measured volume ofsolution of the substance to be estimated until thereaction between the two is just complete, is termedas titration. Thus, a titration involves two solutions:

(a) Unknown solution and (b) Known solution or stan-dard solution.

(ii) Titrant : The reagent or substance whose solu-tion is employed to estimate the concentration of un-known solution is termed titrant. There are two typesof reagents or titrants:

(A) Primary titrants : These reagents can beaccurately weighed and their solutions are not to bestandardised before use. Oxalic acid, potassiumdichromate, silver nitrate, copper sulphate, ferrousammonium sulphate, sodium thiosulphates etc., arethe examples of primary titrants.

(B) Secondary titrants : These reagents cannotaccurately weighed and their solutions are to bestandardised before use. Sodium hydroxide,potassium hydroxide, hydrochloric acid, sulphuricacid, iodine, potassium permanganate etc. are theexamples of secondary titrants.

(iii) Standard solution : The solution of exactly knownconcentration of the titrant is called the standardsolution.(iv) Titrate : The solution consisting the substance tobe estimated is termed unknown solution. Thesubstance is termed titrate.

(v) Equivalence point : The point at which the reagent(titrant) and the substance (titrate) under investigationare chemically equivalent is termed equivalence pointor end point.

(vi) Indicator : It is the auxiliary substance used forphysical (visual) detection of the completion of titrationor detection of end point is termed as indicator.Indicators show change in colour or turbidity at thestage of completion of titration.

(c) Concentraion representation of solution

(A) Strength of solution : Grams of solute dissolvedper litre of solution is called strength of solution'

(B) Parts Per Million (ppm) : Grams of solutedissolved per 106 grams of solvent is calledconcentration of solution in the unit of Parts Per Million(ppm). This unit is used to represent hardness ofwater and concentration of very dilute solutions.

(C) Percentage by mass : Grams of solute dissolvedper 100 grams of solution is called percentage bymass.

(D) Percentage by volume : Millilitres of solute per100 mL of solution is called percentage by volume.For example, if 25 mL ethyl alcohol is diluted withwater to make 100 mL solution then the solution thusobtained is 25% ethyl alcohol by volume.

(E) Mass by volume percentage :Grams of solutepresent per 100 mL of solution is called percentagemass by volume.For example, let 25 g glucose is dissolved in water tomake 100 mL solution then the solution is 25% massby volume glucose.

(d) Classification of reactions involved in

volumetric analysis

(A) Neutralisation reactions

The reaction in which acids and bases react to formsalt called neutralisation.

e.g., HCI + NaOH NaCI + H2O

H+(acid)

+ OH�

(base) H

2O (feebly ionised)

The titration based on neutralisation is calledacidimetry or alkalimetry.

PAGE # 54

(B) Oxidation-reduction reactions

The reactions involving simultaneous loss and gainof electrons among the reacting species are calledoxidation reduction or redox reactions, e.g., let usconsider oxidation of ferrous sulphate (Fe2+ ion) bypotassium permanganate (MnO

4� ion) in acidic

medium.

MnO4� + 8H+ + 5e� Mn2+ + 4H

2O

(Gain of electrons or reduction)5 [Fe2+ Fe3+ + e�](Loss of electrons or oxidation)

______________________________________________________________

MnO4� + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

________________________________________________________________In the above reaction, MnO

4� acts as oxidising agent

and Fe2+ acts as reducing agent.

The titrations involving redox reactions are called redoxtitrations. These titrations are also called accordingto the reagent used in the titration, e.g., iodometric,cerimetric, permanganometric and dichromometrictitrations

(C) Precipitation reactions :A chemical reaction in which cations and anionscombine to form a compound of very low solubility (inthe form of residue or precipitate), is calledprecipitation.BaCl

2 + Na

2SO

4 BaSO

4 + 2NaCl

(white precipitate)The titrations involving precipitation reactions arecalled precipitation titrations.

(D) Complex formation reactions :

These are ion combination reactions in which asoluble sl ightly dissociated complex ion orcompound is formed.Complex compounds retain their identity in thesolution and have the properties of the constituentions and molecules.e.g. CuSO

4 + 4NH

3 [Cu(NH

3)

4]SO

4

(complex compound)AgNO

3 + 2KCN K[Ag(CN)

2] + KNO

3

(complex compound)

2CuSO4 + K

4[Fe(CN)

6] Cu

2[Fe(CN)

6] + 2K

2SO

4

(complex compound)The titrations involving complex formation reactionsare called complexometric titrations.

The determination of concentration of bases bytitration with a standard acid is called acidimetry andthe determination of concentration of acid by titrationwith a standard base is called alkalimetry.The substances which give different colours withacids and base are called acid base indicators. Theseindicators are used in the visual detection of theequivalence point in acid-base titrations.

The acid-base indicators are also called pHindicators because their colour change according tothe pH of the solution.

In the selection of indicator for a titration, followingtwo informations are taken into consideration :

(i) pH range of indicator(ii) pH change near the equivalence point in thetitration.

The indicator whose pH range is included in the pHchange of the solution near the equivalence point, istaken as suitable indicator for the titration.

(i) Strong acid-strong base titration : In the titrationof HCl with NaOH, the equivalence point lies in thepH change of 4�10. Thus, methyl orange, methyl red

and phenolphthalein will be suitable indicators.

(ii) Weak acid-strong base titration : In the titrationof CH

3COOH with NaOH the equivalence point lies

between 7.5 and 10. Hence, phenolphthalein (8.3�10) will be the suitable indicator.

(iii) Weak base-strong acid titration : In the titrationof NH

4OH (weak base) against HCl (strong acid) the

pH at equivalence point is about 6.5 and 4. Thus,methyl orange (3.1�4.4) or methyl red (4.2�6.3) will

be suitable indicators.

(iv) Weak acid-weak base titration : In the titration ofa weak acid (CH

3COOH) with weak base (NH

4OH)

the pH at the equivalence point is about 7, i.e., liesbetween 6.5 and 7.5 but no sharp change in pH isobserved in these titrations. Thus, no simple indica-tor can be employed for the detection of the equiva-lence point.

(v) Titration of a salt of a weak acid and a strongbase with strong acid:

H2CO

3 + 2NaOH Na

2CO

3 + 2H

2O

Weak acid Strong base

Na2CO

3 when titrated with HCl, the following two

stages are involved :Na

2CO

3 + HCl NaHCO

3 + NaCl (First stage)

pH = 8.3, near equivalence pointNaHCO

3 + HCl NaCl + H

2CO

3 (Second stage)

pH = 4, near equivalence point

For first stage, phenolphthalein and for second stage,methyl orange will be the suitable indicator.

In this titration the following indicators are mainly used :

(i) Phenolphthalein (weak organic acid) : It showscolour change in the pH range (8 � 10)

(ii) Methyl orange (weak organic base) : It showscolour change in the pH range (3.1 � 4.4). Due to

lower pH range, it indicates complete neutralisationof whole of the base.

PAGE # 55

Let for complete neutralisation of Na2CO

3, NaHCO

3 and NaOH, x,y and z mL of standard HCl are required. The

titration of the mixture may be carried by two methods as summarised below :

Mixture Phenlphthalein Methyl orange Phenolphthalein Methyl orange from from from after first end beginning beginning point

1. NaOH z + (x/2) (x + z) z + x/2 x/2 (for remaining 50% + Na CO2 3

beginning

Na CO )

2. NaOH z + 0 (z + y) z + 0 y (for remaining 100% + NaHCO NaHCO

3. Na CO (x/2) + 0 (x + y) (x/2) + 0 x/2 + y (for remaining 50% + NaHCO of Na CO and 100% NaHCO are indicated)

2 3

3 3

2 3

3 2 3

3

Volume of HClused with

Volume of HCl used

An indicator is a substance which is used to deter-mine the end point in a titration. In acid-base titra-tions organic substances (weak acids or weak bases)are generally used as indicators. They change theircolour within a certain pH range. The colour changeand the pH range of some common indicators aretabulated below:________________________________________Indicator pH range Colour

change________________________________________

Methyl orange 3.2 � 4.5 Red to orange

Methyl red 4.4 � 6.5 Red to yellow

Litmus 5.5 � 7.5 Red to blue

Phenol red 6.8 � 8.4 Yellow to red

Phenolphthalein 8.3 � 10.5 Colourless to pink________________________________________

Theory of acid-base indicators : Two theories havebeen proposed to explain the change of colour ofacid-base indicators with change in pH.

1. Ostwald's theory:According to this theory

(a) The colour change is due to ionisation of the acid-base indicator. The unionised form has differentcolour than the ionised form.

(b) The ionisation of the indicator is largely affected inacids and bases as it is either a weak acid or a weakbase. In case, the indicator is a weak acid, itsionisation is very much low in acids due to commonions while it is fairly ionised in alkalies. Similarly if theindicator is a weak base, its ionisation is large inacids and low in alkalies due to common ions.

Considering two important indicators phenolphtha-lein (a weak acid) and methyl orange (a weak base),Ostwald's theory can be illustrated as follows:

Phenolphthalein: It can be represented as HPh. Itionises in solution to a small extent as:

HPh H+ + Ph�

Colourless Pink

Applying law of mass action,

K = ]HPh[

]Ph][H[

PAGE # 56

The undissociated molecules of phenolphthalein arecolourless while ph� ions are pink in colour. In pres-ence of an acid, the ionisation of HPh is practicallynegligible as the equilibrium shifts to left hand sidedue to high concentration of H+ ions. Thus, the solu-tion would remain colourless. On addition of alkali,hydrogen ions are removed by OH� ions in the form ofwater molecules and the equilibrium shifts to righthand side. Thus, the concentration of ph� ions in-creases in solution and they impart pink colour to thesolution.

Let us derive Hendetson's equation for an indicator

HIn + H2O H

3+O + In�

'Acid form' 'Base form'

Conjugate acid-base pair

KIn =

]HIn[]OH][In[ 3

KIn = Ionization constant of indicator

[H3

+O] = KIn ]In[

]HIn[

pH = �log10

[H3

+O] = �log10

[KIn] � log

10 ]In[

]HIn[

pH = pKIn + log

10

]HIn[]In[

(Henderson's equation for

indicator)At equivalence point ;[In�] = [HIn] and pH = pK

In

Methyl orange : It is a very weak base and can berepresented as MeOH. It is ionised in solution to giveMe+ and OH� ions.

MeOH Me+ +OH�

Orange Red

Applying law of mass action,

K = ]MeOH[

]OH][Me[

In presence of an acid, OH� ions are removed in theform of water molecules and the above equilibriumshifts to right hand side. Thus, sufficient Me+ ions areproduced which impart red colour to the solution. Onaddition of alkali, the concentration of OH� ions in-creases in the solution and the equilibrium shifts toleft hand side, i.e., the ionisation of MeOH is practi-cally negligible. Thus, the solution acquires the colourof unionised methyl orange molecules, i.e. orange.

This theory also explains the reason why phenol-phthalein is not a suitable indicator for titrating a weakbase against strong acid. The OH� ions furnished bya weak base are not sufficient to shift the equilibriumtowards right hand side considerably, i.e., pH is notreached to 8.3. Thus, the solution does not attainpink colour. Similarly, it can be explained why methylorange is not a suitable indicator for the titration ofweak acid with strong base.

SOLUBILITY

The solubility of a solute in a solution is alwaysexpressed with respect to the saturated solution.

(a) Definition :

The maximum amount of the solute which can bedissolved in 100g (0.1kg) of the solvent to form asaturated solution at a given temperature.Suppose w gram of a solute is dissolved in W gramof a solvent to make a saturated solution at a fixedtemperature and pressure. The solubility of the solutewill be given by -

W

w× 100 =

solventtheofMass

solutetheofMass× 100

For example, the solubility of potassium chloride inwater at 20ºC and 1 atm. is 34.7 g per 100g of water.

This means that under normal conditions 100 g ofwater at 20ºC and 1 atm. cannot dissolve more than

34.7g of KCl.

(b) Effect of Temperature and Pressure on

Solubility of a Solids :

The solubility of a substance in liquids generallyincreases with rise in temperature but hardly changeswith the change in pressure. The effect of temperaturedepends upon the heat energy changes whichaccompany the process.

Note :If heat energy is needed or absorbed in the process,it is of endothermic nature. If heat energy is evolvedor released in the process, it is of exothermic nature.

(i) Effect of temperature on endothermic dissolutionprocess : Most of the salts like sodium chloride,potassium chloride, sodium nitrate, ammoniumchloride etc. dissolve in water with the absorption ofheat. In all these salts the solubility increases withrise in temperature. This means that sodium chloridebecomes more soluble in water upon heating.

(ii) Effect of temperature on exothermic dissolutionprocess : Few salts like lithium carbonate, sodiumcarbonate monohydrate, cerium sulphate etc.dissolve in water with the evolution of heat. Thismeans that the process is of exothermic nature. Inthese salts the solubility in water decreases with risein temperature.

Note :

1. While expressing the solubility, the solution mustbe saturated but for expressing concentration (masspercent or volume percent), the solution need not tobe saturated in nature.2. While expressing solubility, mass of solvent isconsidered but for expressing concentration themass or volume of the solution may be taken intoconsideration.

(c) Effect of Temperature on the Solubility

of a Gas

(i) The solubility of a gas in a liquid decreases withthe rise in temperature.

(ii) The solubility of gases in liquids increases onincreasing the pressure and decreases on decreas-ing the pressure.

PAGE # 57

SAMPLE PROBLEMS

Ex.49 12 grams of potassium sulphate dissolves in75 grams of water at 60ºC. What is the solubility

of potassium sulphate in water at that temperature ?

Sol. Solubility = 100solventofmasssoluteofmass

= 75

12×100 = 16 g

Thus, the solubility of potassium sulphate inwater is 16 g at 60ºC.

Ex.50 4 g of a solute are dissolved in 40 g of water toform a saturated solution at 25ºC. Calculate the

solubility of the solute.

Sol. Solubility = solventofMass

soluteofMass× 100

Mass of solute = 4 gMass of solvent = 40 g

Solubility = 404

× 100 = 10 g

Ex.51 (a) What mass of potassium chloride would beneeded to form a saturated solution in 50 g ofwater at 298 K ? Given that solubility of the salt is46g per 100g at this temperature.

(b) What will happen if this solution is cooled ?

Sol. (a) Mass of potassium chloride in 100 g of waterin saturated solution = 46 gMass of potassium chloride in 50 g of water insaturated solution.

= 10046

× (50g) = 23 g

(b) When the solution is cooled, the solubility ofsalt in water will decrease. This means, that uponcooling, it will start separating from the solutionin crystalline form.

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