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Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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Atoms, molecules and chemical arithmatic

Chemistry is basically an experimental science. In it we study physical and chemical properties of substance and measure
it upto possibility. The results of measurement can we reported in two steps : (a) Arithmetic number, (b) Unit of measurement.

Every experimental measurement vary slightly from one another and involves some error or uncertainty depending upon
the skill of person making the measurements and measuring instrument. The closeness of the set of values obtained from
identical measurement called precision and a related term, refers to the closeness of a single measurement to its true value
called accuracy.

Significant figures.

In the measured value of a physical quantity, the digits about the correctness of which we are surplus the last
digit which is doubtful, are called the significant figures. Number of significant figures in a physical quantity depends
upon the least count of the instrument used for its measurement.

(1) Common rules for counting significant figures : Following are some of the common rules for
counting significant figures in a given expression:

Rule 1. All non zero digits are significant.
Example : x = 1234 has four significant figures. Again x = 189 has only three significant figures.
Rule 2. All zeros occurring between two non zero digits are significant.
Example : x = 1007 has four significant figures. Again x = 1.0809 has five significant figures.
Rule 3. In a number less than one, all zeros to the right of decimal point and to the left of a non zero digit are
not significant.
Example : x = 0.0084 has only two significant digits. Again, x = 1.0084 has five significant figures. This is on
account of rule 2.
Rule 4. All zeros on the right of the last non zero digit in the decimal part are significant.
Example : x = 0.00800 has three significant figures 8, 0, 0. The zeros before 8 are not significant again 1.00
has three significant figures.
Rule 5. All zeros on the right of the non zero digit are not significant.
Example : x = 1000 has only one significant figure. Again x = 378000 has three significant figures.
Rule 6. All zeros on the right of the last non zero digit become significant, when they come from a measurement.
Example : Suppose distance between two stations is measured to be 3050 m. It has four significant figures.
The same distance can be expressed as 3.050 km or 3.050 ×105 cm . In all these expressions, number of significant
figures continues to be four. Thus we conclude that change in the units of measurement of a quantity does not
change the number of significant figures. By changing the position of the decimal point, the number of significant
digits in the results does not change. Larger the number of significant figures obtained in a measurement, greater is
the accuracy of the measurement. The reverse is also true.
(2) Rounding off : While rounding off measurements, we use the following rules by convention:

Rule 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged.

Example : x = 7.82 is rounded off to 7.8, again x = 3.94 is rounded off to 3.9.

Rule 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one.

Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8.

Rule 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one.

Example : x = 16.351 is rounded off to 16.4, again x = 6.758 is rounded off to 6.8.


Rule 4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
Example : x = 3.250 becomes 3.2 on rounding off, again x = 12.650 becomes 12.6 on rounding off.
Rule 5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd.
Example : x = 3.750 is rounded off to 3.8. again x = 16.150 is rounded off to 16.2.
(3) Significant figure in calculation
(i) Addition and subtraction : In addition and subtraction the following points should be remembered :
(a) Every quantity should be changed into same unit.
(b) If a quantity is expressed in the power of 10, then all the quantities should be changed into power of 10.
(c) The result obtained after addition or subtraction, the number of figure should be equal to that of least, after
decimal point.
(ii) Multiplication and division
(a) The number of significant figures will be same if any number is multiplied by a constant.
(b) The product or division of two significant figures, will contain the significant figures equal to that of least.
Units for measurement.

The chosen standard of measurement of a quantity which has essentially the same nature as that of the
quantity is called the unit of the quantity. Following are the important types of system for unit,

(1) C.G.S. System : Length (centimetre), Mass (gram), Time (second)

(2) M.K.S. System : Length (metre), Mass (kilogram), Time (second)

(3) F.P.S. System : Length (foot), Mass (pound), Time (second)

(4) S.I. System : The 11th general conference of weights and measures (October 1960) adopted
International system of units, popularly known as the SI units. The SI has seven basic units
from which all other units are derived called derived units. The standard prefixes which
helps to reduce the basic units are now widely used.

Dimensional analysis : The seven basic quantities lead to a number of derived quantities such as pressure,
volume, force, density, speed etc. The units for such quantities can be obtained by defining the derived quantity
in terms of the base quantities using the base units. For example, speed (velocity) is expressed in distance/time.

So the unit is m / s or ms −1 . The unit of force (mass × acceleration) is kg ms −2 and the unit for acceleration is

ms −2 .

Seven basic SI units

Length Mass Time Temperature Electric Luminous Amount of
Current Intensity substance

metre (m) Kilogram (kg) Second (s) Kelvin (K) Ampere (A) Candela (Cd) Mole (mol)


Derived Units

Physical quantity Unit Symbol
Area square metre m2
Volume cubic metre m3
Velocity metre per second ms–1
Acceleration metre per second square ms–2
Density kilogram per cubic metre
Molar mass kilogram per mole kg m–3
Molar volume cubic metre per mole kg mol–1
Molar concentration mole per cubic metre m3 mol–1
Force newton (N) mol m–3
Pressure pascal (Pa) kg m s–2
Energy work joule (J)
N m–2
Standard prefixes use to reduce the basic units kg m2 s–2, Nm

Multiple Prefix Symbol Submultiple Prefix Symbol
1024 yotta Y 10–1 deci d
1021 zetta Z 10–2 centi c
1018 exa E 10–3 milli m
1015 peta P 10–6 micro
1012 tera T 10–9 nano µ
109 giga G 10–12 pico n
106 mega M 10–15 femto p
103 kilo k 10–18 atto f
102 hecto h 10–21 zeto a
101 deca da 10–24 yocto z
y

Conversion factors

1 m = 39.37 inch 1 cal = 4.184 J 1 e.s.u. = 3.3356 × 10–10 C 1 mole of a gas = 22.4 L at STP
1 dyne = 10–5 N
1 inch = 2.54 cm 1 eV = 1.602 × 10–19 J 1 atm = 101325 Pa 1 mole a substance = N0 molecules
1 litre = 1000 mL 1 eV/atom =96.5 kJ mol–1 1 bar = 1 × 105 N m–2 1 g atom = N0 atoms
1 gallon (US) = 3.79 L 1 amu = 931.5016 MeV 1 litre atm = 101.3 J t (oF) = 9/5 t (oC) + 32
1 lb = 453.59237 g 1 kilo watt hour = 3600 kJ 1 year = 3.1536 × 107 s 1 g cm–3 = 1000 kg m–3
1 newton =1 kg m s–2 1 horse power = 746 watt 1 debye (D) = 1 × 10 –18 esu cm 1Å = 10–10 m
1 J = 1 Nm =1 kg m2 s–2 1 joule = 107 erg 1nm = 10–9 m


Matter and its classification.

Matter is the physical material of the universe which occupies space and has mass e.g., water, sugar, metals,
plants etc. Matter can be classified as,

MATTER
Everything that has mass and occupies space

Physical classification Chemical classification

Solids Liquids Gases MIXTURES PURE SUBSTANCES

• Variable composition • Fixed composition
• Components retain their characteristic properties • Cannot be separated into simpler substances by
• May be separated into pure components by physical
physical methods
methods
• Properties do not vary
• Mixtures of different composition may have widely
different properties

Homogeneous Hetrogeneous Elements Compounds

• Have same composition • Do not have same • Can not be decom- • Can be decomposed into
composition throughout posed into simpler simpler substances by
throughout substances by chemical chemical changes, always
• Components are distin- changes at constant composition
• Components are guishable for example
carbon and sulphur
indistin-guishable for mixture (gun powder)

example : a gaseous

mixture or a liquid

solution

Inorganic Organic

Metals Metalloids Non-metals

Separation of mixtures or purification of an impure substance.

Each component of a mixture retains its own properties and thus a mixture can be separated into its
components by taking advantages of the difference in their physical and chemical properties. The different methods
which are employed to separate the constituents from a mixture to purify an impure sample of a substance are,

(1) Sedimentation : It is also called gravity separation. It is used for a mixture in which one component is a
liquid and the other is insoluble solid heavier than the liquid. Example : Sand dispersed in water.

(2) Filtration : It is used for a mixture containing two components one of which is soluble in a particular
solvent and the other is not. Example : (i) A mixture of salt and paper using water as solvent (ii) A mixture of sand
and sulphur using carbon disulphide as solvent. (iii) A mixture of glass powder and sugar, using water as a solvent
in which sugar dissolves but glass does not. (iv) A mixture of sand and sulphur, using carbon disulphide as the
solvent in which sulphur dissolves but sand does not.

(3) Sublimation : It is used for a mixture containing a solid component, which sublimes on heating from
non-volatile solids. Example : A mixture of sand + naphthalene or powdered glass + NH4 Cl / camphor / iodine
etc. can be separated by the method of sublimation because substances like naphthalene, NH4 Cl , iodine, camphor
etc. form sublimates whereas sand, glass etc. do not.


(4) Evaporation : It is used for a mixture in which one component is a non–volatile soluble salt and other is a
liquid. Example : Sodium chloride dissolved in sea–water.

(5) Crystallization : It is a most common method for a mixture containing solid components and based
upon the differences in the solubilities of the components of the mixture into a solvent. For separation, a suitable
solvent is first selected. It is of two types :

(i) Simple crystallization : It is applicable when there is a large difference in the solubilities of different
components of a mixture.

(ii) Fractional crystallization : It is applicable when there is a small difference in the solubilities of different
components of a mixture in the same solvent. Example : K 2 Cr2O7 and KCl . Here K 2Cr2O7 is less soluble in

water and hence crystallizes first. A series of repeated crystallization separates the two components in pure form.

(6) Distillation : It is the most important method for purifying the liquids. It involves the conversion of a
liquid to its vapours on heating (vaporisation) and then cooling the vapours back into the liquid (condensation). It
can be used to separate, (i) A solution of a solid in a liquid. e.g., aqueous copper sulphate solution. (ii) A solution
of two liquids whose boiling points are different. Several methods of distillation are employed.

(i) Simple distillation : It is used only for such liquids which boil without decomposition at atmospheric
pressure and contain non–volatile impurities. Example : (a) Pure water from saline water. (b) Benzene from
toluene.

(ii) Fractional distillation : It is used for the separation and purification of a mixture of two or more
miscible liquids having different boiling points. The liquid having low boiling point vaporises first, gets condensed
and is collected in the receiver. The temperature is then raised to the boiling point of second so that the second
liquid vaporises and is collected in other receiver. If two liquids present in a mixture have their boiling points closer
to each other, a fractionating column is used. Example : (a) Crude petroleum is separated into many useful
products such as lubricating oil, diesel oil, kerosene and petrol by fractional distillation. (b) A mixture of acetone and
methyl alcohol.

(iii) Vacuum distillation or distillation under reduced pressure : It is used for such liquids which
decompose on heating to their boiling points. At reduced pressure, the boiling point of liquid is also reduced.

Example : Glycerol is distilled under pressure as it decomposes on heating to its boiling point.

(iv) Steam distillation : It is used for liquids which are partially miscible with water, volatile in steam. e.g.,
aniline, oils etc. are purified by steam distillation. The principle involved is of reduced pressure distillation. If Pw and
Pl are vapour pressures of water and liquid at distillation temperature, then Pw + Pl = P = 76 cm i.e., atmospheric
pressure. Thus, a liquid boils at relatively low temperature than its boiling point in presence of steam.

Example : Some solids like naphthalene, o-nitrophenol which are steam volatile can be purified.
Nitrobenzene, chlorobenzene, essential oils are also extracted or separated by this process

(7) Solvent extraction : It is used for the separation of a compound from its solution by shaking with a
suitable solvent. The extraction follows Nernst distribution law. The solvent used must be insoluble with other phase
in which compound is present as well as the compound should be more soluble in solvent. The extraction becomes
more efficient if the given extracting liquid is used for more number of extractions with smaller amounts than using it
once with all extracting liquid. Example : (i) Aqueous solution of benzoic acid by using benzene. (ii) Aqueous
solution of Iodine by using chloroform or carbon tetrachloride. (iii) Flavour of tea from the tea leaves by boiling with
water.


(8) Magnetic separation : It is used for a mixture in which one component is magnetic while the other is
non–magnetic. Example : iron ore from the non–magnetic impurities.

(9) Chromatography : It is based on the differences in the rates at which different components of a mixture
are absorbed on a suitable solvent. It is used in separation, isolation, purification and identification of a substance. It
was proposed by a Russian botanist Tswett.

(10) Atmolysis : It is used for separation of the mixture of gases or vapours. It is based on the difference in
their rates of diffusion through a porous substance. Example : H 2 , SO2 , CH 4 and O2 , U 235 & U 238 in the form of
their hexa–fluorides.

(11) Electrophoresis : It is based upon the differences in the electrical mobility of the different components
of the mixture.

(12) Ultracentrifugation : It is based upon the difference in sedimentation velocity of the components in a
centrifugal field.

Laws of chemical combination.

Various chemical reactions take place according to the certain laws, known as the Laws of chemical
combination. These are as follows,

(1) Law of conservation of mass : It was proposed by Lavoisier and verified by Landolt. According to this
law, Matter is neither created nor destroyed in the course of chemical reaction though it may change from one form
to other. The total mass of materials after a chemical reaction is same as the total mass before reaction.

Example : A reaction between AgNO3 solution and KI solution.

AgNO3(aq) + KI(aq) → AgI + NaNO3(aq) (yellow ppt.)

Mass of AgNO3(aq) + Mass of KI(aq) = Mass of the ppt. of AgI + Mass of NaNO3(aq)

According to the modified statement of the law, The total sum of mass and energy of the system remains
constant.

(2) Law of constant or definite proportion : It was proposed by Proust. According to this law, A pure
chemical compound always contains the same elements combined together in the fixed ratio of their weights
whatever its methods of preparation may be.

Example : CO2 can be formed by either of the following processes:

(i) By heating CaCO3 : Ca CO3 ∆ → Ca O + CO2

(ii) By heating NaHCO3 : 2 NaHCO3 ∆ → Na2 CO3 + H 2O + CO2

CO2 is collected separately as a product of each reaction and the analysis of CO2 of each collection reveals
that it has the combination ratio of carbon and oxygen as 12 : 32 by weight.

(3) Law of multiple proportion : It was proposed by Dalton and verified by Berzelius. According to this
law, When two elements A and B combine to form more than one chemical compounds then different weights of A,
which combine with a fixed weight of B, are in proportion of simple whole numbers.


Example : Nitrogen forms as many as five stable oxides. The analysis of these oxides (N 2O, NO, N 2O3 , N 2O4 and
N 2O5 ) reveals that for 28 gm. nitrogen, the weight of oxygen that combines is in the ratio 16 : 32 : 48 : 64 :
80 i.e., 1 : 2 : 3 : 4 : 5 in N 2O, NO, N 2O3 , N 2O4 and N 2O5 respectively.

(4) Law of equivalent proportion or law of reciprocal proportion : It was proposed by Ritcher.
According to this law, The weights of the two or more elements which separately react with same weight of a third
element are also the weights of these elements which react with each other or in simple multiple of them.

Example : Formation of H 2 S, H 2O and SO2 can be done as follows, H2S H (2 g)
S H2O
(i) Hydrogen combines with sulphur forming hydrogen sulphide; 2gm. of
hydrogen reacts with 32gm of sulphur. (ii) Hydrogen combines oxygen forming (32 g) SO2 O
water; 2 gm. of hydrogen reacts with 16 gm. of oxygen. (iii) Sulphur combines with
oxygen forming sulphur dioxide; 32 gm. of sulphur reacts with 32 gm. of oxygen (16 g)
i.e., in the ratio 32 : 32. This ratio is double of the ratio weights of these elements
which combine with 2 gm. of hydrogen. i.e., 32/16 : 32/32 = 2 : 1

Law of Reciprocal proportion can be used to obtain equivalent weights of elements. As elements always
combine in terms of their equivalent weights.

(5) Gay-Lussac’s Law: It was proposed by Gay–Lussac and is applicable only for gases. According to this

law, When gases combine, they do so in volumes, which bear a simple ratio to each other and also to the product

formed provided all gases are measured under similar conditions. The Gay-Lussac’s law, was based on

experimental observation.

Example : (i) Reaction between hydrogen and oxygen. H 2(g) + 1 O2(g) → H 2O (v)
2

One volume of H2 reacts with half volume of O2 to give one volume of H 2 O .

(ii) Reaction between nitrogen and hydrogen. N 2(g) + 3 H 2(g) → 2 NH 3(g)

One volume of N2 reacts with three volumes of H 2 to give two volumes of NH3 .

Important hypothesis.

(1) Atomic hypothesis : Keeping in view various law of chemical combinations, a theoretical proof for the
validity of different laws was given by John Dalton in the form of hypothesis called Dalton's atomic hypothesis.
Postulates of Dalton's hypothesis is as followes,

(i) Each element is composed of extremely small particles called atoms which can take part in chemical
combination.

(ii) All atoms of a given element are identical i.e., atoms of a particular element are all alike but differ from
atoms of other element.

(iii) Atoms of different elements possess different properties (including different masses).
(iv) Atoms are indestructible i.e., atoms are neither created nor destroyed in chemical reactions.
(v) Atoms of elements take part to form molecules i.e., compounds are formed when atoms of more than one
element combine.
(vi) In a given compound, the relative number and kinds of atoms are constant.


(2) Modern atomic hypothesis : The main modifications made in Dalton’s hypothesis as a result of new
discoveries about atoms are,

(i) Atom is no longer considered to be indivisible.

(ii) Atoms of the same element may have different atomic weights. e.g., isotopes of oxygen O16 , O17 , and O18 .

(iii) Atoms of different element may have same atomic weights. e.g., isobars Ca 40 and Ar 40 .

(iv) Atom is no longer indestructible. In many nuclear reactions, a certain mass of the nucleus is converted into
energy in the form of α, β and γ rays.

(v) Atoms may not always combine in simple whole number ratios. e.g., in sucrose (C12 H 22O11 ) , the elements
carbon, hydrogen and oxygen are present in the ratio of 12 : 22 : 11 and the ratio is not a simple whole number
ratio.

(3) Berzelius hypothesis : “Equal volumes of all gases contain equal number of atoms under same
conditions of temperature and pressure”. When applied to law of combining volumes, this hypothesis predicts that
atoms are divisible and hence it is contrary to Dalton's hypothesis.

(4) Avogadro’s hypothesis : “Equal volumes of all gases under similar conditions of temperature and
pressure contain equal number of molecules.” Avogadro hypothesis has been found to explain as follows :

(i) Provides a method to determine the atomic weight of gaseous elements.

(ii) Provides a relationship between vapour density (V.D.) and molecular masses of substances.

Vapour density = Volume of definite amount of a gas
Volume of same amount of hydrogen

or Vapour denstiy = Mass of 'n' molecule of a gas
Mass of 'n' molecule of hydrogen

Vapour density = Mass of 1 molecule of a gas
Mass of 1 molecule of hydrogen

or Vapour density = Molecular mass or Molecular mass = 2 × vapour density
2

(iii) It helps in the determination of mass of fixed volume of a particular gas.

Vapour density = Mass of 1 molecule of gas = Mass of 1 ml of gas = Mass of 1 ml of gas
Mass of 1 molecule of H2 Mass of 1 ml of H2 0.0000897

( 1 ml H 2 = 0.0000897 gm.) at NTP

∴ Mass of 1 ml gas = V.D. × 0.0000897 gm.

(iv) It also helps in the determination of molar volume at N.T.P.

 V.D. × 0.0000897 gm. gas has volume = 1 ml

∴ 2× V.D.(i.e., molecular mass) gm. has volume = 1 × 2 × V.D. ml = 22400 ml
V.D. × 0.0000897

∴ Molar mass of a gas or its 1 mole occupies 22.4 L volume at S.T.P.


Note :  22.4 litres of any gas at S.T.P. weigh equal to the molecular weight of the gas expressed in

grams. This is called Gram-molecular volume (G.M.V.) law.

(v) It helps in determination of molecular formulae of gases and is very useful in gas analysis. By knowing the
molecular volumes of reactants and products of reaction, molecular composition can be determined easily.

Atomic, Molecular and Equivalent masses.

(1) Atomic Mass : It is the average relative mass of atom of element as compared with an atom of carbon –
12 isotope taken as 12.

Atomic mass = Average mass of an atom
1/12 × Mass of an atom of C12

Average atomic mass : If an elements exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio m :

n, then average atomic mass = (m × a) + (n × b) . Since the atomic mass is a ratio, it has no units and is expressed in amu,
m + n

1 amu = 1.66 × 10 −24 g. One atomic mass unit (amu) is equal to 1 th of the mass of an atom of carbon-12 isotope.
12

Gram atomic mass (GAM) : Atomic mass of an element expressed in grams is called Gram atomic mass or

gram atom or mole atom.

(i) Number of gram atoms or mole atoms = Mass of an element
GAM

(ii) Mass of an element in gm. = Number of gm. atom × GAM

(iii) Number of atoms in 1 GAM = 6.02 × 10 23

∴ Number of atoms in a given substance = No. of GAM × 6.02 × 10 23 = Mass × 6.02 × 1023
GAM

(iv) Number of atoms in 1 gm of element = 6.02×10 23
Atomic mass

(v) Mass of one atom of the element (in gm.) = GAM
6.02 ×10 23

Methods of determination of atomic mass
(i) Dulong and Pettit's method : According to Dulong and Pettit's law

Atomic mass × Specific heat = 6.4 (approx.)

Atomic mass (approx. ) = 6.4
Specific heat (in cals.)

This law is applicable to solid elements only except Be, B, C and Si.
Atomic mass = Equivalent mass × Valency

Valency = Approximate atomic mass
Equivalent mass

(ii) Vapour density method : It is suitable for elements whose chlorides are volatile.

Valency of the element = Molecular mass of chloride = 2 × Vapour density of chloride
Equivalent mass of chloride Equivalent mass of metal + 35.5


Atomic mass = Equivalent mass of metal × Valency

(iii) Specific heat method : It is suitable only for gases. The two types of specific heats of gases are CP (at
constant pressure) and Cv (at constant volume). Their ratio is known as γ whose value is constant (1.66 for

monoatomic, 1.40 for diatomic and 1.33 for triatomic gases).

Atomic mass of a gaseous element = Molecular mass
Atomaticity

(iv) Volatile chloride method : Different steps are given below

Step I. The element (M) whose atomic mass is to be determined is converted into its volatile chloride whose
vapour density is determined by Victor Meyer method.

Thus, Molecular mass of the chloride = 2 × V. D.
Step II. Equivalent mass of the element (M) of valency X is determined as usual.

Atomic mass of the element = Equivalent mass of the element × X = Z × X
Step III. The formula of the volatile chloride is derived as below,

M X Cl 1

M Cl X

Step IV. Molecular mass of the chloride = (Z × X) + (35.5 × X) = X(Z + 35.5)

From (I) and (IV) X (Z + 35.5) = 2× Vapour density or X= 2 × V.D.
Z + 35.5

Thus, Atomic mass of the element = Z × X

(v) Isomorphism method : It is based on law of isomorphism which states that compounds having identical
crystal structure have similar constitution and chemical formulae.

Example : K 2 SO4 , K 2CrO 4 and K 2 SeO4 (valency of S, Cr, Se = 6),
ZnSO4 .7 H 2O, MgSO4 .7 H 2O, FeSO4 .7 H 2O (valency of Zn, Mg, Fe = 2).

Applications of isomorphism

• The valencies of two elements forming isomorphism salts are essentially same. Thus if valency of one of the
elements is known that of other will be the same.

• Masses of two elements, that combine with the same mass of other elements in their respective isomorphous
compounds, are in the ratio of their atomic masses i.e.,

Mass of element A that combines with a certain mass of other elements Atomic mass of A
Mass of element B that combines with the same mass of other elements = Atomic mass of B

• By knowing the percentage of two elements of their isomorphous compound and atomic mass of one
element, atomic mass of other element can be calculated.

(2) Molecular mass : Molecular mass of a molecule, of an element or a compound may be defined as a

number which indicates how many times heavier is a molecule of that element or compound as compared with

1 of the mass of an atom of carbon–12. Molecular mass is a ratio and hence has no units. It is expressed in a.m.u.
12


Molecular mass = Mass of one molecule of the substance
1 / 12 × Mass of one atom of C - 12

Actual mass of one molecule = Molecular mass ×1.66 ×10−24 gm.

Molecular mass of a substances is the additive property and can be calculated by adding the atomic masses
present in one molecule.

Gram molecular mass (GMM) and Gram molar volume : Molecular mass of an element or compound
when expressed in gm. is called its gram molecular mass, gram molecule or mole molecule.

Number of gm molecules or mole molecules = Mass of substances
GMM

Mass of substances in gm = Number of gm. molecules × GMM

Volume occupied by one mole of any gas at STP is called Gram molar volume. The value of gram molar
volume is 22.4 litres. Volume of 1 mole of any gas at STP = 22.4 litres

Expression for mass and density

Mass of 11.2L of any gas at STP = V.D. of that gas in gm.

Density of a gas at NTP = Mol. mass in gm.
22400 ml

Important generalisations
Number of atoms in a substance = Number of GMM × 6.02 × 1023 × Atomaticity
Number of electrons in given substance = Number of GMM × 6.02 × 1023 × Number of electrons
Methods of determination of molecular mass : following methods are used to determine molecular mass
(i) Diffusion method (For gases) : The ratio of rates of diffusion of two gases is inversely proportional to the
square root of their molecular masses.

r1 = M2
r2 M1

(ii) Vapour density method (For gases only) : Mass of a fixed volume of the vapour is compared with the
mass of the same volume of hydrogen under same conditions. The ratio of these masses is called Vapour density
or Relative density.

Molecular mass = 2 × Vapour desity

Note :  Vapour density of a substance increases due to molecular association (e.g., CH 3COOH, HF)

and decreases due to dissociation (e.g., NH 4 Cl, PCl5 , etc.)

(iii) Victor Meyer method (For volatile liquids or solids) : It is based on Dalton's law of partial pressure and
Avogadro's hypothesis (gram molar volume).

22400 ml of vapours of a substance = Molecular mass of that substance

(iv) Colligative property method (For non-volatile solids) : Discussed in colligative properties of solutions.

Average atomic mass and molecular mass


A (Average atomic mass) = ∑ Ai Xi ; M (Average molecular mass) = ∑ MiXi
∑ X total ∑ X total

Where A1, A2 , A3 ,.... are atomic mass of species 1, 2, 3,.... etc. with % ratio as X1, X 2 , X3 ......... etc. Similar terms
are for molecular masses.

(3) Equivalent mass : The number of parts by mass of a substance that combines with or displaces 1.008
parts by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine is called its equivalent mass (EM). On the
other hand quantity of a substance in grams numerically equal to its equivalent mass is called its gram equivalent
mass (GEM) or gram equivalent.

Number of GEM = Mass of the substance in grams
GEM of the substance

Expressions for equivalent mass (EM)

(i) EM of an element = Atomic mass
Valency

(ii) EM of an acid = Molecular mass
Basicity

(Basicity of acid is the number of replaceable hydrogen atoms in one molecule of the acid).

(iii) EM of a base = Molecular mass
Acidity

(Acidity of a base is the number of replaceable– OH groups in one molecule of the base).

(iv) EM of a salt = Formula mass
Total positive or negative charge

(v) EM of a reducing agent = Formula mass
Number of electrons lost per molecule or Total change in oxidation number

Example : Eq. mass of oxalic acid (C2 H 2O4 )

C2O4 2− → 2 CO2 + 2 e −

(O.N. of C = + 3) (O.N. of C = + 4)

∴ Number of electrons lost = 2
or Change in O.N. per C atom (from + 3 to + 4) = 1

∴ Total change in O.N. =1 × 2 = 2

∴ Equivalent mass of C2 H 2O4 = 90 = 45
2

(vi) EM of an oxidising agent = Number of Formula mass Total change in O.N.
electrons gained per molecule or

Equivalent mass of common oxidising agent changes with the medium of the reaction.
Example : (a) Equivalent mass of KMnO4 in acidic medium


MnO4 − + 8 H + + 5 e − → Mn2 + + 4 H 2O + 5 e −
Change in O.N. of Mn from + 7 to + 2 = + 5

∴ Equivalent mass of KMnO4 158 = 31.60
=5

(b) Equivalent mass of KMnO4 in neutral medium

MnO4 − + 4 H + + 3 e − → MnO2 + 2 H 2O or KMn O4 → Mn O2

∴ Equivalent mass of KMnO4 = 158 = 52.67
3

(c) Equivalent mass of KMnO4 in basic medium

MnO4 − + e − → MnO42− or KMnO4 → K 2 MnO4

Equivalent mass of KMnO4 = 158 = 158
1

(vii) EM of radicals = Formula mass of radical
Number of units of charge

Example : Equivalent mass of SO 2− = 96 = 48
4 2

Methods of determination of equivalent mass

(i) Hydrogen displacement method : The mass of metal which displaces 11200 ml of hydrogen at NTP

from an acid, alkali or alcohol is the equivalent mass of the metal.

(a) Equivalent mass of metal = Mass of metal ×1.008 = W × 1.008g
Mass of H 2 displaced M

(b) Equivalent mass of metal Mass of metal ×11200 = W ×11200
= Vol. (ml) of H 2 displaced at STP V

This method is useful for metals which can displace hydrogen from acids or can combine with hydrogen

(Mg, Zn, Na, Ca etc.)

(ii) Oxide formation method : The mass of the element which combines with 8 grams of oxygen is the

equivalent mass of the element.

(a) Equivalent mass of metal = Mass of metal × 8
Mass of oxygen

(b) Equivalent mass of metal = Vol. Mass of metal in ml × 5600
of O2 at S.T.P.

(iii) Chloride formation method : The mass of an element which reacts with 35.5 gm. of chlorine is the

equivalent mass of that element.

(a) Equivalent mass of metal = Mass of metal × 35.5
Mass of chlorine

(b) Equivalent mass of metal = Vol. Mass of metal STP × 11200
of Cl 2 (in ml.) at


(iv) Neutralisation method : (For acids and bases).

Equivalent mass of acid (or base) = W , Where W = Mass of acid or base in gm., V = Vol. of base or
V×N

acid in litre required for neutralisation and N is Normality of base or acid

(v) Metal displacement method : It is based on the fact that one gm. equivalent of a more electropositive
metal displaces one gm equivalent of a less electropositive metal from its salt solution.

Mass of metal added = Eq. mass of metal added ; W1 = E1
Mass of metal displaced Eq. mass of metal displaced W2 E2

(vi) Electrolytic method : The quantity of substance that reacts at electrode when 1 faraday of electricity is
passed is equal to its gram equivalent mass.

Gram equivalent mass = Electrochemical equivalent × 96500

The ratio of masses of two metals deposited by the same quantity of electricity will be in the ratio of their

equivalent masses. W1 = E1
W2 E2

(vii) Double decomposition method : AB + CD → AD ↓ + CB

Mass of compound AB = Eq. mass of A + Eq. mass of B
Mass of compound AD Eq. mass of A + Eq. mass of D

or Mass of salt taken (W1 ) ) = Eq. mass of salt (E1 ) )
Mass of ppt. obtained (W2 Eq. mass of salt in ppt. (E
2

(viii) Conversion method : When one compound of a metal is converted to another compound of the same
metal, then

Mass of compound I (W1 ) = E + Eq. mass of radical I (E = Eq. mass of the metal)
Mass of compound II (W2 ) E + Eq. mass of radical II

(ix) Volatile chloride method

Valency of metal = 2× V.D. of Chloride = 2× V.D. ∴ E = 2 × V.D. of Chloride − 35.5
Eq. mass of metal chloride E + 35.5 Valency

(x) Silver salt method (For organic acids)

Equivalent Mass of acid = 108 × Mass of silver salt − 107
Mass of Ag metal

Molecular mass of acid = Equivalent mass of acid × Basicity

Example: 1 Oxygen contains 90% O16 and 10% O18 . Its atomic mass is [KCET 1998]

Solution: (b) (a) 17.4 (b) 16.2 (c) 16.5 (d) 17
Example: 2
Average atomic mass of oxygen 90 × 16 + 10 × 18 = 16.20
= 100

The total number of electrons present in 1.6 gm. of methane is [IIT 1976; Roorkee 1985; CPMT 1987, 92]


(a) 6.02 × 1023 (b) 6.02 × 10 22 (c) 6.02 × 1021 (d) 4.02 × 1020

Solution: (a) Number of GMM of methane in 1.6 gm. of methane

Example: 3 = Mass of methane = 1.6 = 0.1
Solution: (b) Mol. mass of methane 16

Example: 4 Number of electrons in 1 molecule of methane (CH4) = 6 + 4 = 10
Solution: (a)
Example:5 Hence, Number of electrons in 1.6 gm. of methane
Solution: (a)
Example: 6 = Number of GMM × 6.02 × 1023 × No. of electrons
Solution: (c)
= 0.1 × 6.02 × 1023 × 10 = 6.02 × 1023

KClO3 on heating decomposes to KCl and O2 . The volume of O2 at STP liberated by 0.1 mole KClO3 is [BIT 1991]

(a) 4.36 L (b) 3.36 L (c) 2.36 L (d) none of these

On heating KClO3 dissociates as:

2 KClO3 ∆ → 2 KCl + 3 O2

2 moles 3 × 22.4 L at STP

 2 moles of KClO3 on heating produces = 67. 2 L of O2 at STP

 0.1 mole of KClO3 on heating produces = 67.2 × 0.1 L = 3.36 L of O2 at STP
2

At S.T.P. the density of CCl4 vapour in g/L will be nearest to [CBSE PMT 1988]

(a) 6.84 (b) 3.42 (c) 10.26 (d) 4.57

1 mole of CCl4 vapour = 12 + 4 × 35.5 = 154 gm = 22.4 L at S.T.P.

∴ Density = 154 gmL−1 = 6.875 gmL−1
22.4

4.4 g of an unknown gas occupies 2.24 litres of volume at NTP. The gas may be [MP PMT 1995]

(a) Carbon dioxide (b) Carbon monoxide (c) Oxygen (d) Sulphur dioxide

Mass of 2.24 litres gas = 4.4 gm.

 Mass of 22.4 litres gas = 44 gm.

Here, out of four given gases, the molecular mass of only carbon dioxide is 44 gm.

One gm. of a metal carbonate gave 0.56 gm. of its oxide on heating. The equivalent mass of the metal is

(a) 40 (b) 30 (c) 20 (d) 10

Mass of metal carbonate = E + Eq. mass of CO3 2− or 1.0 = E + 30
Mass of metal oxide E + Eq. mass of O2 2− 0.56 E+8

or (E + 8) × 1.0 = (E + 30)(0.56) (Where E is the equivalent mass of the metal)

∴ E = 20

The mole concept.

The mole (abbreviated as mol) is the SI base unit for a amount of a chemical species. It is always associated
with a chemical formula and refers to Avogadro’s number ( 6.022 ×1023 ) of particles represented by the formula. It
is designated as N0 . Thus, 12 molecules of HCl is a dozen, 144 molecules of HCl is a gross and 6.022 ×1023
molecules of HCl is a mole.


1 mole of a substance = 6.022 ×1023 species Grams of A Given
Moles of A
The molar mass of a substance is the mass in grams of 1 mole of Molecules of A Divide by
that substance. molar mass

Mole of a substance = mass in grams Multiply
molar mass by N0

Find

Under STP conditions when temperature is 273K and pressure is 1 Calculations are reversed if operations are reversed
atm, volume of one mole of an ideal gas is 22.4L

Example: 7 The number of gram molecules of oxygen in 6.02 ×1024 CO molecules is [IIT 1990]

(a) 10 gm molecules (b) 5 gm molecules (c) 1 gm molecules (d) 0.5 gm molecules

Solution: (b) 6.02 × 1023 molecules = 1 mole of CO [CBSE PMT 1988]

∴ 6.02 × 1024 CO molecules = 10 moles of CO

= 10 gms of Oxygen atom =5 gm molecules of O2

Example: 8 1 c.c of N 2O at NTP contains :

(a) 1.8 × 10 22 atoms (b) 6.02 × 10 23 molecules
224 22400

(c) 1.32 × 10 23 electrons (d) All the above
224

Solution: (d) 22400 c.c. = 6.02 × 10 23 molecules

1 c.c. of N2O = 6.02 × 10 23 molecules
22400

= 3 × 6.02 × 10 23 atoms (Since N 2O has three atoms)
22400

= 6.02 × 10 23 × 22 electron (Because number of electrons in N2O are 22)
22400

Example: 9 The mass of carbon present in 0.5 mole of K4 [Fe(CN)6 ] is

(a) 1.8 g (b) 18 g (c) 3.6 g (d) 36 g

Solution: (d) 1 mole of K4 [Fe(CN)6 ] = 6 gm atoms of carbon
0.5 mole of K4 [Fe(CN)6 ] = 3 gm atoms of carbon
= 3 × 12 = 36 g

Example: 10 The number of moles of oxygen in one litre of air containing 21% oxygen by volume under standard
Solution: (c) conditions is
[CBSE PMT 1995]

(a) 0.186 mole (b) 0.21 mole (c) 0.0093 mole (d) 2.10 mole

 100 ml of air at STP contains 21 ml of O2.

∴ 1000 ml of air at STP contains 210 ml of O2.


∴ No. of moles of O2 = Vol. of O2 in litres under STP conditions = 210 / 1000 = 21 = 0.0093 mole
22.4 litre 22.4 2240

Example: 11 The number of moles of BaCO3 which contains 1.5 moles of oxygen atoms is [EAMCET 1991]
Solution: (a) (d) 6.02 × 1023
(a) 0.5 (b) 1 (c) 3

 1 mole of BaCO3 contains 3 moles of oxygen atoms.

∴ 1 mole (0.5) of BaCO3 contains 1.5 moles of oxygen atoms.
2

Percentage composition & Molecular formula.

(1) Percentage composition of a compound : Percentage composition of the compound is the relative
mass of each of the constituent element in 100 parts of it. If the molecular mass of a compound is M and B is the
mass of an element in the molecule, then

Percentage of element = Mass of element × 100 = X × 100
Molecular mass M

(2) Empirical formula : The chemical formula that gives the simplest whole number ratio between the atoms
of various elements present in one molecule of the compound is known as empirical formula (simplest formula).

For example, the empirical formula of glucose is CH 2 O which shows that C, H and O are present in the ratio
1 : 2 : 1 in a molecule of glucose.

Empirical formula mass of a compound is equal to the sum of atomic masses of all atoms present in the
empirical formula of the compound.

Calculation of the empirical formula : The empirical formula of a chemical compound can be deduced
by knowledge of the, (i) Percentage composition of different elements. (ii) Atomic masses of the elements.

The following steps are involved in the calculation of the empirical formula,

Step I. Calculate the relative number of atoms or atomic ratio.

Atomic ratio = Percentage of an element
Atomic mass of the same element

Step II. Calculate the simplest atomic ratio.

Step III. Calculate the simple whole number ratio.

Step IV. Write the empirical formula.

(3) Molecular formula : The chemical formula that gives the actual number of atoms of various elements
present in one molecule of the compound. For example, the molecular formula of glucose is C6 H12O6 .

Relation between empirical and molecular formula : The molecular formula of a compound is a simple
whole number multiple of its empirical formula.

Molecular formula = n × Empirical formula ; Where n is any integer such as 1, 2, 3….etc.

The value of ‘n’ can be obtained from the following relation: n = Molecular mass .
Empirical formula mass


The molecular mass of a volatile compound can be determined by Victor Meyer’s method or by employing
the relation, Molecular mass = 2 × Vapour density .

Calculation of the molecular formula : The molecular formula of a compound can be deduced from its :
(i) Empirical formula, (ii) Molecular mass
The determination of molecular formula involves the following steps:

Setp I. Calculation of empirical formula from the percentage composition.
Setp II. Calculation of empirical formula mass.
Setp III. Calculation of the value of ‘n’.
Setp IV. Calculation of molecular formula by multiplying the empirical formula of the compound by ‘n’.

Example:12 The oxide of a metal contains 40% by mass of oxygen. The percentage of chlorine in the chloride of the metal
Solution: (b)
is [BIT Ranchi 1997]

(a) 84.7 (b) 74.7 (c) 64.7 (d) 44.7

% of oxygen = 8 × 100 = 40 or, 40(m + 8)= 800 or, m + 8 = 20 or, m =12
m+8

∴ % of chlorine = 35.5 ×100 = 35.5 × 100 = 74.7 (Where m is the atomic mass of metal)
m + 35.5 12 + 35.5

Example: 13 The empirical formula of an organic compound containing carbon and hydrogen is CH 2 . The mass of one
Solution: (a)
litre of this organic gas is exactly equal to that of one litre of N 2 . Therefore, the molecular formula of the

organic gas is [EAMCET 1985]

(a) C2 H 4 (b) C3 H 6 (c) C6 H12 (d) C4 H 8

Molar mass of 1L of gas = mass of 1L N 2

∴ Molecular masses will be equal i.e., molecular mass of the gas = 28, hence formula is C2 H 4

Example: 14 A sample of pure compound is found to have Na = 0.0887 mole, O = 0.132 mole, C = 2.65 ×1022 atoms.

The empirical formula of the compound is [CPMT 1997]

(a) Na2 CO3 (b) Na3 O2 C5

(c) Na0.0887 00.132 C2.65 ×1022 (d) NaCO

Solution: (a)  6.02 × 10 23 atoms of C = 1 mole of C

∴ 2.65 × 10 22 atoms of C= 1× 2.65 × 1022 mole = 2.65 = 0.044 mole
6.02 × 1023 6.02 ×10

Now,

Element Relative number of moles Simplest ratio of moles

Na 0.0887 0.0887 = 2
0.044

O 0.132 0.132 = 3
0.044

C 0.044 0.044 =1
0.044


Thus, the empirical formula of the compound is Na2CO3 .

Example: 15 An organic compound containing C, H and N gave the following on analysis: C = 40%, H = 13.3% and

N = 46.67%. Its empirical formula would be [CBSE PMT 1999, 2002]

(a) CHN (b) C2H2N (c) CH4 N (d) C2H7 N

Solution: (c) Calculation of empirical formula

Element Symbol Percentage At. mass of Relative number Simplest Simplest
of element elements atomic whole number
of atoms ratio atomic ratio

= Percentage
At. mass

C 40 12 40 = 3.33 3.33 =1 1
12 3.33
Carbon

Hydroge H 13.3 1 13.3 = 13.3 13.3 = 4 4
n 1 3.33

N 46.67 14 46.67 = 3.33 3.33 = 1 1
14 3.3
Nitrogen

Thus, the empirical formula is CH4 N .

Example: 16 An organic substance containing C, H and O gave the following percentage composition :

C = 40.687%, H = 5.085% and O = 54.228%. The vapour density of the compound is 59. The molecular
formula of the compound is

(a) C4 H6O4 (b) C4 H6O2 (c) C4 H4O2 (d) None of these

Solution: (a)

Element Symbol Percentage Atomic Relative number Simplest Simplest
of element mass of atomic ratio whole
element of atoms number

= Percentage atomic ratio
At. mass

Carbon C 40.687 12 40.687 = 3.390 3.390 = 1 2
12 3.389

Hydroge H 5.085 1 5.085 = 5.085 5.085 = 1.5 3
n 1 3.389

Oxygen O 54.228 16 54.228 = 3.389 3.389 = 1 2
16 3.389

∴ Empirical formula is C2H3O2
∴ Empirical formula mass of C2H3O2 = 59

Also, Molecular mass = 2 × Vapour density = 2 × 59 = 118

∴ n = Molecular mass = 118 = 2
Empirical formula mass 59

Now, Molecular formula = n × (Empirical formula) = 2 × (C2H3O2 ) = C4H6O4


∴ Molecular formula is C4 H6O4 .

Chemical equations and its balancing.

(1) Chemical equations : Chemical reactions are represented in a concise way by chemical equations. A

chemical equation represents an actual chemical change taking place in terms of the symbols and the formulae of

the reactants and products. e.g., Methane burns in oxygen to produce carbon dioxide and water. The chemical

reaction can be represented as: CH 4 + 2O2 → CO2 + 2H 2O
Reactants Products

Essentials of a chemical equation

(i) It must represent an actual chemical reaction.

(ii) It must be balanced i.e., the total number of atoms of various substances involved on both sides of the

equation must be equal.

(iii) It should be molecular. The elementary gases like hydrogen, oxygen etc. must be written in the molecular

form as H2, O2 etc.
Information conveyed by a chemical equation : A Chemical equation conveys both qualitative and

quantitative information.

(i) Qualitative information : Qualitatively a chemical equation conveys the names of the reactants and

products taking part in the reaction.

(ii) Quantitative information : Quantitatively a chemical equation conveys the following information :

(a) It conveys the actual number of reactants and product species (atoms or molecules) taking part in the
reaction.

(b) It tells the relative masses of the reactants and products participating in the reaction.

(c) It conveys the relative number of reactant and product moles.

(d) It also conveys the volumes of the gaseous reactants and products if present.

Example : Reaction between CaCO3 and aqueous HCl.

CaCO3 + 2HCl → CaCl 2 + H 2O + CO2

100 gm. 73 gm. 111 gm. 18 gm. 44 gm.
1 mole 2 mole 1 mole 1 mole 1 mole
22.4 litres
at STP

Note :  All chemical equations are written under STP conditions provided no other conditions are

mentioned.

Limitations of a chemical equation and their removal : Although a chemical equation conveys a
number of informations, it suffers from certain limitations or drawbacks. The major limitations and the steps taken
for their removal are given below:

(i) Physical states of the reactants and products : The chemical equation fails to convey the physical states of
the reactants and products. These are specified by the use of letters ‘s’(for solids), ‘l’(for liquids), ‘g’(for gases) and
‘aq’(for aqueous solutions).

Example : CaCO3(s) + 2HCl(aq) → CaCl 2(s) + H 2O(l) + CO2(g)


(ii) Conditions of temperature, pressure and catalyst : These conditions are normally not mentioned in the
equation. These can be expressed on the arrow head which separates the reactants from the products.

Example : N 2( g ) + 3 H2(g) Fe,723K → 2 NH3(g)
600 atm

(iii) Speed of reaction : The speed of a particular reaction whether slow or fast can be mentioned by writing
the word slow or fast on the arrow head.

Example : NO2(g) + F2(g) slow→ NO2 F(g) + F(g)

NO2(g) + F(g) fas→t NO2 F(g)

(iv) Heat change accompanying the reaction : The heat evolved or absorbed in a chemical reaction can be
written on the product sides. The S.I. unit of heat is kJ.

Example : CH 4(g) + 2O2(g) → CO2(g) + 2 H 2O(l) + 393.5 KJ
(Heat is evolved)

H 2(g) + I 2(g) → 2 HI(g) − 53.9 KJ
(Heat is absorbed)

(v) Reversible nature of a reaction : Certain chemical reactions proceed both in the forward and backward
directions. The reversible nature of the reaction can be indicated by two arrows pointing in the opposite direction (⇌).

Example : H 2(g) + Cl 2(g) Forward 2 HCl(g)
Backward

(vi) Formation of precipitate and evolution of a gas : Formation of a precipitate in the chemical reaction can
be indicated by writing the word ppt. or by an arrow pointing downwards.

Ag NO3(aq) + NaCl(aq) → AgCl ↓ + NaNO3 (aq)
(ppt)

The evolution of a gas is expressed by an arrow which points upwards.

Mg(s) + 2 HCl(aq) → Mg Cl 2(aq) + H 2(g) ↑

(2) Balancing of chemical equations : A correct chemical equation must be in accordance with the law of
conservation of mass i.e., the number of atoms of each kind in the reactants must be equal to the number of atoms
of same kind in the products. Balancing of a chemical equation means to equalise the atoms of different elements or
compounds which are involved in it.

Let us consider a chemical reaction which occurs due to passing of steam over red hot iron forming iron oxide
and hydrogen gas. It can be represented as:

Skeleton equation : Fe (s) + H 2O (v) → Fe 3O4 (s) + H 2(g)

Balanced equation : 3 Fe(s) + 4 H 2O(v) → Fe 3O4 (s) + 4 H 2(g)

The balancing of equations is done by the following methods:

(i) Hit and Trial method, (ii) Partial Equation method


(iii) Oxidation-Number method, (iv) Ion-Electron method

The first two methods are discussed here, while the remaining two methods will be taken up for discussion in
redox reactions.

(i) Hit and Trial method : This method involves the following steps:
(a) Write the symbols and formulae of the reactants and products in the form of skeleton equation.
(b) If an elementary gas like H2, O2 or N2 etc. appears on either side of the equation, write the same in the
atomic form.
(c) Select the formula containing maximum number of atoms and start the process of balancing.
(d) In case the above method is not convenient, then start balancing the atoms which appear minimum
number of times.
(e) Balance the atoms of elementary gases in the last.
(f) When the balancing is complete, convert the equation into molecular form.

Let us balance the skeleton equation Mg3 N 2 + H 2O → Mg (OH)2 + NH 3
The balancing is done in the following steps

Step I. Balance the Mg atoms Mg 3 N 2 + H 2O → 3 Mg (OH)2 + NH 3
Step II. Balance the N atoms Mg 3 N 2 + H 2O → 3 Mg(OH)2 + 2 NH 3
Step III. Balance the O atoms Mg 3 N 2 + 6 H 2O → 3 Mg (OH)2 + 2 NH 3

The hydrogen atoms are already balanced. Hence, final balanced equation is

Mg 3 N 2 + 6H 2O → 3 Mg (OH)2 + 2 NH 3

(ii) Partial equation method : Chemical equations which involve a large number of reactants and products
can not be balanced easily by the hit and trial method. In partial equation method, the overall reaction is assumed
to take place into two or more simpler reactions known as partial equations. The balancing of the equation involves
the following steps:

(a) Split the chemical equation into two or more simpler equations or partial equations.
(b) Balance each partial equation separately by hit and trial method.

(c) Multiply the partial equations with suitable coefficient if necessary so as to cancel out the final substances
which do not appear in the final equation.

(d) Finally, add up the partial equations to get the final equation.

Let us balance the skeleton equation NaOH + Cl2 → NaCl + NaClO3 + H 2O
This reaction is supposed to take place in the following steps:

The probable partial equations for the above reaction are:

Na OH + Cl2 → Na Cl + Na ClO + H 2O and Na Cl O → Na Cl O3 + NaCl
Balance the partial chemical equations separately by hit and trial method as

2 Na OH + Cl2 → NaCl + Na ClO + H 2O and 3 Na ClO3 → Na ClO3 + 2 NaCl

Multiply the first partial equation by 3 in order to cancel out NaClO which does not appear in the final
equation. Finally add the two partial equations to get the final equation.


2 NaOH + Cl2 → NaCl + NaClO + H 2O ] × 3

3 NaClO → NaClO3 + 2 NaCl
6 NaOH + 3 Cl2 → NaClO3 + 5 NaCl + 3 H 2O

Chemical Stoichiometry.

Calculation based on chemical equations is known as chemical stoichiometry. Stoichiometry can be broadly
classified into two groups: (1) Gravimetric analysis (Stoichiometry-I), (2) Volumetric analysis (Stoichiometry-II)

(1) Gravimetric analysis (Stoichiometry-I) : With the help of chemical equation, we can calculate the
weights of various substances reacting and weight of substances formed. For example,

MgCO3 → MgO + CO2 ↑

This equation implies :

(i) 1 mol of MgCO3 gives 1 mol of MgO and 1 mol of CO2 .
(ii) 84 g of MgCO3 (Mol. wt. of MgCO3 ) gives 40 g of MgO and 44 g of CO2 .
Hence, chemical equation provide us information regarding :

(i) Molar ratio of reactants and products.

(ii) Mass ratio between reactants and products.

(iii) Volume ratio between gaseous reactant and products.

The calculation based upon chemical equation (Stoichiometry–I) are based upon three types namely :

(a) Mass-mass relationship (b) Mass-volume relationship (c) Volume-volume relationship

(2) Volumetric analysis (Stoichiometry-II) : It is a method which involves quantitative determination of
the amount of any substance present in a solution through volume measurements. For the analysis a standard
solution is required. (A solution which contains a known weight of the solute present in known volume of the
solution is known as standard solution.)

To determine the strength of unknown solution with the help of known (standard) solution is known as
titration. Different types of titrations are possible which are summerised as follows :

(i) Redox titrations : To determine the strength of oxidising agents or reducing agents by titration with the help
of standard solution of reducing agents or oxidising agents.

Examples:
K 2Cr2O7 + 4 H 2SO4 → K 2SO4 + Cr2 (SO4 )3 + 4H 2O + 3[O]

[2FeSO4 + H 2SO4 + O → Fe 2 (SO4 )3 + H 2O] × 3

6FeSO4 + K 2Cr2O7 + 7H 2SO4 → 3Fe(SO4 )3 + K 2SO4 + Cr2 (SO4 )3 7H 2O
2KMnO4 + 3H 2SO4 → K 2SO4 + 2MnSO4 + 3H 2O + 5[O]

[2FeSO4 + H 2SO4 + O → Fe 2 (SO4 )3 + H 2O] × 5

10FeSO4 + 2KMnO4 + 8H 2SO4 → 5Fe 2 (SO4 )3 + K 2SO4 + 2MnSO4 + 8H 2O
Similarly with H 2C2O4
2KMnO4 + 3H 2SO4 + 5H 2C2O4 → K 2SO4 + 2MnSO4 + 8H 2O + 10CO2 etc.


(ii) Acid-base titrations : To determine the strength of acid or base with the help of standard solution of base or

acid.

Example: NaOH + HCl → NaCl + H 2O and NaOH + CH3COOH → CH 3COONa + H 2O etc.
(iii) Iodiometric titrations : To determine the reducing agents with the help of standard iodine solution is known

as iodiometry.

For example: As2O3 + 2I 2 + 2H 2O → As2O3 + 4HI
Reducing agent

Na2S2O3 + I 2 → Na2S4 O6 + 2NaI

(iv) Iodometric titrations : To determine the oxidising agent indirectly by titration of liberated I 2 with the help
of standard hypo solution is known as iodometric titrations.

Examples: Oxidising agents such as KMnO4 , K2Cr2O7 , CuSO4 , ferric salts, etc. are reduced quantitatively
when treated with large excess of KI in acidic or neutral medium and liberate equivalent amount of I 2 .

2CuSO4 + 4KI → Cu2 I 2 + 2K 2SO4 + I 2

Kr2Cr2O7 + 7H 2 SO4 + 6KI → Cr2 (SO4 )3 + 4 K 2SO4 + 7H 2O + 3I 2
This I 2 is estimated with hypo

I 2 + 2Na2S2O3 → Na2S4 O6 + 2NaI

(v) Precipitation titrations : To determine the anions like CN − , AsO 3− , PO43− , X− etc, by precipitating with
3

AgNO3 provides examples of precipitation titrations.

NaCl + AgNO3 → AgCl ↓ + NaNO3 ; KSCN + AgNO3 → AsSCN ↓ +KNO3

End point and equivalence point : The point at which titration is stopped is known as end point, while the

point at which the acid and base (or oxidising and reducing agents) have been added in equivalent quantities is

known as equivalence point. Since the purpose of the indicator is to stop the titration close to the point at which the

reacting substances were added in equivalent quantities, it is important that the equivalent point and the end point

be as close as possible.

Normal solution : A solution containing one gram equivalent weight of the solute dissolved per litre is called

a normal solution; e.g. when 40 g of NaOH are present in one litre of NaOH solution, the solution is known as

normal (N) solution of NaOH. Similarly, a solution containing a fraction of gram equivalent weight of the solute

dissolved per litre is known as subnormal solution. For example, a solution of NaOH containing 20 g (1/2 of g eq.

wt.) of NaOH dissolved per litre is a sub-normal solution. It is written as N/2 or 0.5 N solution.

Formula used in solving numerical problems on volumetric analysis

(1) Strength of solution = Amount of substance in g litre−1

(2) Strength of solution = Amount of substance in g moles litre−1

(3) Strength of solution = Normality × Eq. wt. of the solute

(4) Molarity = Moles of solute
Volume in litre

(5) Number of moles = Wt. in gm = M × V(in l) = Volume in litres at NTP (only for gases)
Mol. wt. 22.4

(6) Number of millimoles = Wt. in gm × 1000 = Molarity × Volume in ml.
mol. wt.


(7) Number of equivalents = Wt. in gm = x × No. of moles × Normality × Volume in litre
Eq. wt.

(8) Number of milliequivalents (meq.) = Wt. in gm × 1000 = normality × Volume in ml.
Eq. wt.

(9) Normality = x × No. of millimoles = x × Molarity = Strength in gm litre −1
Eq. wt.

where x = Mol. wt. , x* = valency or change in oxi. Number.
Eq. wt.

(10) Normality formula, N1V1 = N 2V2

(11) % by weight = Wt. of solvent × 100
Wt. of solution

(12) % by volume = Wt. of solvent × 100
Vol. of solution

(13) % by strength = Vol. of solvent × 100
Vol. of solution

(14) Specific gravity = Wt. of solution = Wt. of 1 ml. of solution
Vol. of solution

(15) Formality = Wt. of ionic solute
Formula Wt. of solute × Vinl

(16) Mol. Wt. = V.D × 2 (For gases only)

Limiting reagent or reactant.

In many situations, an excess of one or more substance is available for chemical reaction. Some of these
excess substances will therefore be left over when the reaction is complete; the reaction stops immediately as soon
as one of the reactant is totally consumed.

The substance that is totally consumed in a reaction is called limiting reagent because it determines or limits the
amount of product. The other reactant present in excess are called as excess reagents.

Let us consider a chemical reaction which is initiated by passing a spark through a reaction vessel containing
10 mole of H2 and 7 mole of O2.

2 H 2 (g) + O2 (g) → 2 H 2O (v)

Moles before reaction 10 7 0

Moles after reaction 0 2 10

The reaction stops only after consumption of 5 moles of O2 as no further amount of H2 is left to react with
unreacted O2. Thus H2 is a limiting reagent in this reaction

Example: 17 When a solution containing 4.77 gm. of NaCl is added to a solution of 5.77 gm. of AgNO3, the weight of
Solution: (c)
precipitated AgCl is [IIT 1978]

(a) 11.70 gm. (b) 9.70 gm. (c) 4.86 gm. (d) 2.86 gm.

AgNO3 + NaCl → AgCl + NaNO3
0
Moles before mixing 5.77 4.77 0
108 + 14 + 48 58.5

= 0.0339 = 0.0815 (Here AgNO3 is limiting reactant, thus)


Moles after mixing 0 0.0815 – 0.0339 0.0339 0.0339

= 0.0476

∴ Moles of AgCl formed = 0.0339

∴ Mass of AgCl formed = Mol. mass × No. of moles = 143.5 × 0.0339 = 4.864 gm.

Example: 18 The volume of oxygen at STP required to completely burn 30 ml of acetylene at STP is [Orissa JEE 1997]
Solution: (b)
(a) 100 ml (b) 75 ml (c) 50 ml (d) 25 ml

The balanced chemical equation for the reaction can be written as:

C2 H 2 + 5 / 2 O2 → 2 CO2 + H 2O

1 Vol. 5 / 2 Vol.

1ml 5 / 2 ml

30 ml 30 × 5 / 2 = 75 ml

Hence, volume of the oxygen at STP required to burn 30 ml of acetylene at STP = 75 ml.

Example: 19 What is the volume (in litres) of oxygen at STP required for complete combustion of 32 g of CH4 [EAMCET 2001]
Solution: (b)
(a) 44.8 (b) 89.6 (c) 22.4 (d) 179.2

According to Avogadro's hypothesis, volume occupied by one mole of any gas at STP is 22.4 litres.

CH 4 (g) + 2 O2(g) → CO2(g) + 2 H 2O (l)

1 mole 2 moles
2 moles 4 moles
2×16 gm. = 32 gm. 4 × 22.4 litres = 89.6 litres

Example: 20 A metal oxide has the formula Z 2O3 . It can be reduced by hydrogen to give free metal and water. 0.1596 g

of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

[CBSE PMT 1989]

(a) 27.9 (b) 159.6 (c) 79.8 (d) 55.8

Solution: (d) Valency of metal in Z 2O3 = 3

Z 2O3 + 3H 2 → 2Z + 3H 2O

0.1596 gm of Z 2O3 react with H 2 = 6mg = 0.006 gm

1 gm of H 2 react with Z 2O3 = 0.1596 = 26.6 gm
0.006

Equivalent wt. of Z 2O3 = 26.6 = equivalent wt. of Z + equivalent wt. of O = E + 8 = 26.6 or E = 18.6

Atomic weight of Z = 18.6 × 3 = 55.8


Structure of atom

John Dalton 1808, believed that matter is made up of extremely minute indivisible particles, called atom
which can takes part in chemical reactions. These can neither be created nor be destroyed. However, modern
researches have conclusively proved that atom is no longer an indivisible particle. Modern structure of atom is
based on Rutherford’s scattering experiment on atoms and on the concepts of quantization of energy.

Composition of atom.

The works of J.J. Thomson and Ernst Rutherford actually laid the foundation of the modern picture of the
atom. It is now believed that the atom consists of several sub-atomic particles like electron, proton, neutron,
positron, neutrino, meson etc. Out of these particles, the electron, proton and the neutron are called fundamental
subatomic particles and others are non-fundamental particles.

Electron (–1eo)

(1) It was discovered by J.J. Thomson (1897) and is negatively charged particle. Electron is a component
particle of cathode rays.

(2) Cathode rays were discovered by William Crooke's & J.J. Thomson (1880) using a cylindrical hard
glass tube fitted with two metallic electrodes. The tube has a side tube with a stop cock. This tube was known as
discharge tube. They passed electricity (10,000V) through a discharge tube at very low pressure (10 −2 to
10 −3 mm Hg) . Blue rays were emerged from the cathode. These rays were termed as Cathode rays.

Cathode rays Gas at low TC Vaccum pump
Cathode pressure Anode

High voltage
–+

Discharge tube experiment for production of cathode rays

(3) Properties of Cathode rays
(i) Cathode rays travel in straight line.
(ii) Cathode rays produce mechanical effect, as they can rotate the wheel placed in their path.
(iii) Cathode rays consist of negatively charged particles known as electron.
(iv) Cathode rays travel with high speed approaching that of light (ranging between 10 −9 to 10 −11 cm/sec)
(v) Cathode rays can cause fluorescence.
(vi) Cathode rays heat the object on which they fall due to transfer of kinetic energy to the object.
(vii) When cathode rays fall on solids such as Cu, X − rays are produced.
(viii) Cathode rays possess ionizing power i.e., they ionize the gas through which they pass.
(ix) The cathode rays produce scintillation the photographic plates.
(x) They can penetrate through thin metallic sheets.


(xi) The nature of these rays does not depend upon the nature of gas or the cathode material used in
discharge tube.

(xii) The e/m (charge to mass ratio) for cathode rays was found to be the same as that for an e −

(−1.76 × 108 coloumb per gm). Thus, the cathode rays are a stream of electrons.

Note :  When the gas pressure in the discharge tube is 1 atmosphere no electric current flows

through the tube. This is because the gases are poor conductor of electricity.

 The television picture tube is a cathode ray tube in which a picture is produced due to fluorescence
on the television screen coated with suitable material. Similarly, fluorescent light tubes are also
cathode rays tubes coated inside with suitable materials which produce visible light on being hit with
cathode rays.

(4) R.S. Mullikan measured the charge on an electron by oil drop experiment. The charge on each electron
is − 1.602 × 10 −19 C.

(5) Name of electron was suggested by J.S. Stoney. The specific charge (e/m) on electron was first
determined by J.J. Thomson.

(6) Rest mass of electron is 9.1 × 10−28 gm = 0.000549amu = 1 / 1837 of the mass of hydrogen atom.

(7) According to Einstein’s theory of relativity, mass of electron in motion is, m′ = Rest mass of electron(m)
[1 − (u / c)2 ]

Where u = velocity of electron, c= velocity of light.

When u=c than mass of moving electron =∞.

(8) Molar mass of electron = Mass of electron × Avogadro number = 5.483 × 10 −4.

(9) 1.1 × 10 27 electrons =1gram.

(10) 1 mole electron = 0.5483 mili gram.

(11) Energy of free electron is≈ 0. The minus sign on the electron in an orbit, represents attraction between
the positively charged nucleus and negatively charged electron.

(12) Electron is universal component of matter and takes part in chemical combinations.

(13) The physical and chemical properties of an element depend upon the distribution of electrons in outer shells.

(14) The radius of electron is 4.28 × 10 −12 cm.

(15) The density of the electron is = 2.17 × 10 −17 g / mL .

Example : 1 The momentum of electron moving with 1/3rd velocity of light is (in g cm sec–1)

(a) 9.69 × 10−8 (b) 8.01× 1010 (c) 9.652 × 10 −18 (d) None

Solution: (c) Momentum of electron, ‘p’ = m′ × u

Where m′ is mass of electron in motion = m ; Also u = c / 3

1 − (u / c)2


∴ Momentum = 9.108 × 10 −28 × 3 × 1010 = 9.108 × 10 −28 × 3 × 1010 = 9.652 × 10 −18 g cm sec −1
3 0.94 × 3
 c  2
 × 
1 − 3 c

Example: 2 An electron has a total energy of 2 MeV. Calculate the effective mass of the electron in kg and its speed.
Assume rest mass of electron 0.511 MeV.

(a) 2.9 ×108 (b) 8.01×108 (c) 9.652 ×108 (d) None

Solution: (a) Mass of electron in motion = 2 amu (1 amu = 931 MeV)
931

= 2 × 1.66 × 10 − 27 kg = 3.56 × 10−30 kg (1 amu = 1.66 × 10−27 kg )
931

Let the speed of the electron be u.

m 0.511 × 1.66 × 10− 27 0.911× 10−30
931
m′ = 1 − (u / c)2 or 3.56 × 10−30 = =
 2  2
1 −  3 u  1 −  3 u 
 × 108  × 108

or 1 −  u  2 = 0.06548 or u2 = 9 × 1016 × 0.93452 or u = 2.9 × 108 m
 3 × 108 

Example: 3 A electron of rest mass 1.67 × 10−27 kg is moving with a velocity of 0.9c (c = velocity of light). Find its mass
and momentum.

(a) 10.34 × 10−19 (b) 8.01× 1010 (c) 9.652 × 10 −18 (d) None

Solution: (a) Mass of a moving object can be calculated using Einsten’s theory of relativity :

m′ = m m = rest mass (given), u = velocity (given), c = velocity of light

1 − (u / c)2

m′ = 1.67 × 10 −27 = 3.83 × 10 −27 kg
2
 0.9c 
1 −  c 

Momentum ' p' = m′ × u

p = 3.83 × 10−27 × 0.9c = 10.34 × 10−19 kg ms−1

Proton (1H1, H+, P)

(1) Proton was discovered by Goldstein and is positively charged particle. It is a component particle of
anode rays.

(2) Goldstein (1886) used perforated cathode in the discharge tube and repeated Thomson's experiment
and observed the formation of anode rays. These rays also termed as positive or canal rays.


Anode rays Cathode rays

TC Vaccum pump

Perforated High voltage
cathode –+

Perforated tube experiment for production of anode rays

(3) Properties of anode rays

(i) Anode rays travel in straight line.

(ii) Anode rays are material particles.
(iii) Anode rays are positively charged.
(iv) Anode rays may get deflected by external magnetic field.
(v) Anode rays also affect the photographic plate.
(vi) The e/m ratio of these rays is smaller than that of electrons.
(vii) Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the tube. It is

maximum when gas present in the tube is hydrogen.
(viii) These rays produce flashes of light on ZnS screen.

(4) Charge on proton = 1.602 × 10 −19 coulombs = 4.80 × 10 −10 e.s.u.

(5) Mass of proton = Mass of hydrogen atom= 1.00728amu = 1.673 × 10 −24 gram = 1837 of the mass of electron.

(6) Molar mass of proton = mass of proton × Avogadro number = 1.008 (approx).

(7) Proton is ionized hydrogen atom (H + ) i.e., hydrogen atom minus electron is proton.

(8) Proton is present in the nucleus of the atom and it's number is equal to the number of electron.

(9) Mass of 1 mole of protons is ≈ 1.007 gram.

(10) Charge on 1 mole of protons is ≈ 96500 coulombs.

(11) The volume of a proton (volume = 4 πr 3 ) is ≈ 1.5× 10 −38 cm3 .
3

(12) Specific charge of a proton is 9.58 × 10 4 Coulomb/gram.


Neutron (on1, N)

(1) Neutron was discovered by James Chadwick (1932) according to the following nuclear reaction,

4 Be 9 + 2 He 4 → 6 C12 + on1 or 5 B11 + 2 He 4 → 7 N 14 + on1

(2) The reason for the late discovery of neutron was its neutral nature.

(3) Neutron is slightly heavier (0.18%) than proton.

(4) Mass of neutron = 1.675 × 10 −24 gram = 1.675 × 10−27 kg = 1.00899 amu ≈ mass of hydrogen atom.

(5) Specific charge of a neutron is zero.

(6) Density = 1.5 × 10−14 gram / c.c.

(7) 1 mole of neutrons is ≈ 1.008 gram.
(8) Neutron is heaviest among all the fundamental particles present in an atom.

(9) Neutron is an unstable particle. It decays as follows :

0 n1 → 1 H1 + −1e0 + 0ν 0
neutron proton electron anti nutrino

(10) Neutron is fundamental particle of all the atomic nucleus, except hydrogen or protium.

Comparison of mass, charge and specific charge of electron, proton and neutron

Name of constant Unit Electron(e–) Proton(p+) Neutron(n)

Mass (m) amu 0.000546 1.00728 1.00899
kg 9.109 × 10–31 1.673 × 10–27 1.675 × 10–24
Relative 1/1837 1 1

Charge(e) Coulomb (C) – 1.602 × 10–19 +1.602 × 10–19 Zero
Specific charge (e/m) esu – 4.8 × 10–10 +4.8 × 10–10 Zero
–1 +1 Zero
Relative 1.76 × 108 9.58 × 104 Zero
C/g

• The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C12 , i.e. 1.660 × 10 −27 kg .
Other non fundamental particles

Particle Symbol Nature Charge esu Mass Discovered by
×10–10 (amu)

Positron e + ,1e 0 , β + + + 4.8029 0.000548 Anderson (1932)
Neutrino ν 0 0 6
– < Pauli (1933) and Fermi (1934)
Anti-proton p− + – 4.8029
– + 4.8029 0.00002 Chamberlain Sugri (1956) and
Positive mu µ + 1.00787 Weighland (1955)
meson Yukawa (1935)
0.1152

Negative mu µ − – 4.8029 0.1152 Anderson (1937)
meson


Positive pi meson π+ + + 4.8029 0.1514 Powell (1947)
π−
Negative pi – – 4.8029 0.1514
π0
meson

Neutral pi meson 0 0 0.1454

Atomic number, Mass number and Atomic species.

(1) Atomic number or Nuclear charge
(i) The number of protons present in the nucleus of the atom is called atomic number (Z).

(ii) It was determined by Moseley as,

ν = a(Z − b) or aZ − ab

where, ν = X − rays frequency ν s −1
Z= atomic number of the metal

a & b are constant. Z

(iii) Atomic number = Number of positive charge on nucleus = Number of protons in nucleus = Number

of electrons in nutral atom.

(iv) Two different elements can never have identical atomic number.

(2) Mass number
(i) The sum of proton and neutrons present in the nucleus is called mass number.

Mass number (A) = Number of protons + Number of neutrons or Atomic number (Z)
or Number of neutrons = A – Z .

(ii) Since mass of a proton or a neutron is not a whole number (on atomic weight scale), weight is not
necessarily a whole number.

(iii) The atom of an element X having mass number (A) and atomic number (Z) may be represented by a

symbol, Element Mass number
Atomic number
ZXA

e.g. 9 F 19 , 8 O16 , 7 N 14 etc.

Note :  A part of an atom up to penultimate shell is a kernel or atomic core.

 Negative ion is formed by gaining electrons and positive ion by the loss of electrons.

 Number of lost or gained electrons in positive or negative ion =Number of protons ± charge on ion.


(3) Different Types of Atomic Species

Atomic species Similarities Differences Examples

Isotopes (i) Atomic No. (Z) (i) Mass No. (A) (i) 1 H, 2 H, 3 H
(Soddy) (ii) No. of protons (ii) No. of neutrons 1 1 1
(iii) No. of electrons (iii) Physical properties
(iv) Electronic (ii) 16 O, 17 O, 18 O
configuration 8 8 8

(iii) 35 Cl, 37 Cl
17 17

(v) Chemical properties

(vi) Position in the
periodic table

Isobars (i) Mass No. (A) (i) Atomic No. (Z) (i) 40 Ar, 40 K, 40 Ca
(ii) No. of nucleons 18 19 20
(ii) No. of protons,
electrons and neutrons (ii) 130 Te, 130 Xe, 130 Ba
52 54 56

(iii)Electronic
configuration

(iv) Chemical properties

(v) Position in the
perodic table.

Isotones No. of neutrons (i) Atomic No. (i) 30 Si, 31 P, 32 S
14 15 16
Isodiaphers (ii) Mass No., protons
Isoelectronic and electrons. (ii) 39 K, 40 Ca
19 20
species (iii) Electronic
configuration (iii) 3 H, 4 He
1 2
(iv) Physical and
chemical properties (iv) 13 C, 14 N
6 7

(v) Position in the
periodic table.

Isotopic No. (i) At No., mass No., (i) 92 U 235 , 90 Th231
(N – Z) or (A – 2Z) electrons, protons, (ii) 19 K 39 , 9 F 19
neutrons.

(ii) Physical and (iii) 29 Cu65 , 24 Cr 55
chemical properties.

(i) No. of electrons At. No., mass No. (i) N 2O, CO2 , CNO− (22e − )
(ii) CO, CN − , N 2 (14e − )
(ii) Electronic (iii) H − , He, Li+ , Be 2+ (2e − )
configuration


Isosters (i) No. of atoms (iv)

(ii) No. of electrons P 3−, S2−, Cl −, Ar, K +and Ca2+(18e−)
(iii) Same physical and
chemical properties. (i) N 2 and CO
(ii) CO2 and N 2O
(iii) HCl and F2
(iv) CaO and MgS

(v) C6 H6 and B3 N 3 H6

Note :  In all the elements, tin has maximum number of stable isotopes (ten).

Average atomic weight/ The average isotopic weight

= % of 1st isotope × relative mass of 1st isotope + % of 2nd isotope × relative mass of 2nd isotope
100

Example : 4 The characteristics X- ray wavelength for the lines of the kα series in elements X and Y are 9.87Å and 2.29Å
respectively. If Moseley’s equation ν = 4.9 × 107 (Z − 0.75) is followed, the atomic numbers of X and Y are

(a) 12, 24 (b) 10, 12 (c) 6, 12 (d) 8, 10

Solution : (a) ν=c
λ

νx = 3 × 10 8 = 5.5132 × 108
9.87 × 10 −10

νy = 3 × 108 = 11.4457 × 108
2.29 × 10 −10

using Moseley’s equation we get

∴ 5.5132 × 108 = 4.9 × 107 (Z x − 0.75) …..(i)
and 11.4457 × 108 = 4.90 × 107 (Zy − 0.75) ….. (ii)
On solving equation (i) and (ii) Z x = 12, Zy = 24.

Example : 5 If the straight line is at an angle 45° with intercept, 1 on ν − axis, calculate frequency ν when atomic

number Z is 50.

(a) 2000 s −1 (b) 2010 s −1 (c) 2401 s −1 (d) None

Solution : (c) ν = tan 45° = 1 = a ν s −1 a= tanθ
ab=1
∴ ν = 50 − 1 = 49 ab=intercept
ν = 2401s −1. θ

Z


Example : 6 What is atomic number Z when ν = 2500 s −1 ?

(a) 50 (b) 40 (c) 51 (d) 53

Solution : (c) ν = 2500 = Z − 1, Z = 51.

Example : 7 Atomic weight of Ne is 20.2. Ne is a mixutre of Ne 20 and Ne 22 . Relative abundance of heavier isotope is
Solution:(d)
(a) 90 (b) 20 (c) 40 (d) 10

Average atomic weight/ The average isotopic weight

= % of 1st isotope × relative mass of 1st isotope + % of 2nd isotope × relative mass of 2nd isotope
100

∴ 20.2 = a × 20 + (100 − a) × 22 ; ∴ a = 90 ; per cent of heavier isotope = 100 − 90 = 10
100

Example : 8 The relative abundance of two isotopes of atomic weight 85 and 87 is 75% and 25% respectively. The
Solution:(b) average atomic weight of element is

(a) 75.5 (b) 85.5 (c) 87.5 (d) 86.0

Average atomic weight/ The average isotopic weight

= % of 1st isotope × relative mass of 1st isotope + % of 2nd isotope × relative mass of 2nd isotope
100

= 85 × 75 + 87 × 25 = 85.5
100

Example : 9 Nitrogen atom has an atomic number of 7 and oxygen has an atomic number of 8. The total number of
Solution : (c) electrons in a nitrate ion is

(a) 30 (b) 35 (c) 32 (d) None

Number of electrons in an element = Its atomic number

So number of electrons in N=7 and number of electrons in O=8.

Formula of nitrate ion is NO −
3

So, in it number of electrons

= 1 × number of electrons of nitrogen +3 × number of electrons of oxygen +1 (due to negative charge)

= 1 × 7 + 3 × 8 + 1 = 32

Example :10 An atom of an element contains 11 electrons. Its nucleus has 13 neutrons. Find out the atomic number and
approximate atomic weight.

(a) 11, 25 (b) 12, 34 (c) 10, 25 (d) 11, 24

Solution : (d) Number of electrons =11

∴ Number of protons = Number of electron =11

Number of neutrons = 13

Atomic number of element = Number of proton = Number of electrons =11

Further, Atomic weight = Number of protons + Number of neutrons =11 + 13=24

Example : 11 How many protons, neutrons and electrons are present in (a) 31 P (b) 40 Ar (c) 108 Ag ?
15 18 47


Solution : The atomic number subscript gives the number of positive nuclear charges or protons. The neutral atom
contains an equal number of negative electrons. The remainder of the mass is supplied by neutrons.

Atom Protons Electrons Neutrons
15 15 31 – 15=16
31 P 18 18 40 – 18=22
15 47 47 108 – 47=61

40 Ar
18

108 Ag
47

Example :12 State the number of protons, neutrons and electrons in C12 and C14 .

Solution : The atomic number of C12 is 6. So in it number of electrons = 6
Number of protons =6; Number of neutrons =12 – 6=6

The atomic number of C14 is 6. So in it number of electrons = 6
Number of protons = 6; Number of neutrons =14 – 6=8

Example :13 Predict the number of electrons, protons and neutrons in the two isotopes of magnesium with atomic number
12 and atomic weights 24 and 26.

Solution : Isotope of the atomic weight 24, i.e. 12 Mg 24 . We know that
Number of protons = Number of electrons =12
Further, Number of neutrons = Atomic weight – Atomic number =24 – 12 =12

Similarly, In isotope of the atomic weight 26, i.e. 12 Mg 26
Number of protons = Number of electrons =12
Number of neutrons = 26 – 12 = 14

Electromagnetic Radiations.

(1) Light and other forms of radiant energy propagate without any medium in the space in the form of waves
are known as electromagnetic radiations. These waves can be produced by a charged body moving in a magnetic
field or a magnet in a electric field. e.g. α − rays, γ − rays, cosmic rays, ordinary light rays etc.

(2) Characteristics : (i) All electromagnetic radiations travel with the velocity of light. (ii) These consist of
electric and magnetic fields components that oscillate in directions perpendicular to each other and perpendicular to
the direction in which the wave is travelling.

(3) A wave is always characterized by the following five characteristics:

(i) Wavelength : The distance between two Crest Wavelength
nearest crests or nearest troughs is called the
wavelength. It is denoted by λ (lambda) and Vibrating
is measured is terms of centimeter(cm), source
angstrom(Å), micron( µ ) or nanometre (nm).
Energy

1Å = 10 −8 cm = 10 −10 m

Trough


1µ = 10−4 cm = 10−6 m

1nm = 10−7 cm = 10−9 m

1cm = 108 Å = 104 µ = 107 nm

(ii) Frequency : It is defined as the number of waves which pass through a point in one second. It is denoted
by the symbol ν (nu) and is expressed in terms of cycles (or waves) per second (cps) or hertz (Hz).

λν = distance travelled in one second = velocity =c

ν=c
λ

(iii) Velocity : It is defined as the distance covered in one second by the wave. It is denoted by the letter ‘c’.
All electromagnetic waves travel with the same velocity, i.e., 3 × 1010 cm / sec .

c = λν = 3 × 1010 cm / sec

Thus, a wave of higher frequency has a shorter wavelength while a wave of lower frequency has a
longer wavelength.

(iv) Wave number : This is the reciprocal of wavelength, i.e., the number of wavelengths per centimetre. It
is denoted by the symbol ν (nu bar). It is expressed in cm−1 or m−1 .

ν =1
λ

(v) Amplitude : It is defined as the height of the crest or depth of the trough of a wave. It is denoted by the

letter ‘A’. It determines the intensity of the radiation.

The arrangement of various types of electromagnetic radiations in the order of their increasing or
decreasing wavelengths or frequencies is known as electromagnetic spectrum.

Name Wavelength (Å) Frequency (Hz) Source

Radio wave 3 × 1014 − 3 × 107 1 × 105 − 1 × 109 Alternating current of high
frequency

Microwave 3 × 107 − 6 × 106 1 × 109 − 5 × 1011 Klystron tube
Infrared (IR) 6 × 106 − 7600 5 × 1011 − 3.95 × 1016 Incandescent objects

Visible 7600 − 3800 3.95 × 1016 − 7.9 × 1014 Electric bulbs, sun rays

Ultraviolet (UV) 3800 − 150 7.9 × 1014 − 2 × 1016 Sun rays, arc lamps with
mercury vapours

X-Rays 150 − 0.1 2 × 1016 − 3 × 1019 Cathode rays striking metal
plate

γ − Rays 0.1 − 0.01 3 × 1019 − 3 × 10 20 Secondary effect of

radioactive decay

Cosmic Rays 0.01- zero 3 × 10 20 − infinity Outer space


Atomic spectrum - Hydrogen spectrum.

Atomic spectrum

(1) Spectrum is the impression produced on a photographic film when the radiation (s) of particular
wavelength (s) is (are) analysed through a prism or diffraction grating. It is of two types, emission and absorption.

(2) Emission spectrum : A substance gets excited on heating at a very high temperature or by giving energy
and radiations are emitted. These radiations when analysed with the help of spectroscope, spectral lines are
obtained. A substance may be excited, by heating at a higher temperature, by passing electric current at a very low
pressure in a discharge tube filled with gas and passing electric current into metallic filament. Emission spectra is of
two types,

(i) Continuous spectrum : When sunlight is passed through a prism, it gets dispersed into continuous
bands of different colours. If the light of an incandescent object resolved through prism or
spectroscope, it also gives continuous spectrum of colours.

(ii) Line spectrum : If the radiations obtained by the excitation of a substance are analysed with help of a
spectroscope a series of thin bright lines of specific colours are obtained. There is dark space in
between two consecutive lines. This type of spectrum is called line spectrum or atomic spectrum..

(3) Absorption spectrum : When the white light of an incandescent substance is passed through any substance,
this substance absorbs the radiations of certain wavelength from the white light. On analysing the transmitted light
we obtain a spectrum in which dark lines of specific wavelengths are observed. These lines constitute the absorption
spectrum. The wavelength of the dark lines correspond to the wavelength of light absorbed.

Hydrogen spectrum

(1) Hydrogen spectrum is an example of line emission spectrum or atomic emission spectrum.
(2) When an electric discharge is passed through hydrogen gas at low pressure, a bluish light is emitted.
(3) This light shows discontinuous line spectrum of several isolated sharp lines through prism.
(4) All these lines of H-spectrum have Lyman, Balmer, Paschen, Barckett, Pfund and Humphrey series. These
spectral series were named by the name of scientist discovered them.
(5) To evaluate wavelength of various H-lines Ritz introduced the following expression,

ν = 1 = ν = R  1 − 1
λ c  n12 
 n22 

Where R is universal constant known as Rydberg’s constant its value is 109, 678 cm−1 .

Thomson's model.

(1) Thomson regarded atom to be composed of positively charged protons and negatively charged electrons.

The two types of particles are equal in number thereby making atom electrically +– + Positively charged
neutral. sphere

(2) He regarded the atom as a positively charged sphere in which negative – +– +– Electron
electrons are uniformly distributed like the seeds in a water melon. + –+

–+

(3) This model failed to explain the line spectrum of an element and the Positive charge spreaded throughout the sphere


scattering experiment of Rutherford.
Rutherford's nuclear model.

(1) Rutherford carried out experiment on the bombardment of thin (10–4 mm) Au foil with high speed
positively charged α − particles emitted from Ra and gave the following observations, based on this experiment :

(i) Most of the α − particles passed without any deflection.

(ii) Some of them were deflected away from their path.

(iii) Only a few (one in about 10,000) were returned back to their original direction of propagation.

(iv) The scattering of α − particles ∝ 1 .
 θ 
sin 4  2 

Scattering of α -particle

θ
b

r0 Nucleus

α-particle
(energy E eV)

(2) From the above observations he concluded that, an atom consists of
(i) Nucleus which is small in size but carries the entire mass i.e. contains all the neutrons and protons.
(ii) Extra nuclear part which contains electrons. This model was similar to the solar system.

Planetry


+ Nucleu

10–15 m

10–10
Size of the nucleus = 1 Fermi =

10–15 m

(3) Properties of the Nucleus

(i) Nucleus is a small, heavy, positively charged portion of the atom and located at the centre of the atom.
(ii) All the positive charge of atom (i.e. protons) are present in nucleus.
(iii) Nucleus contains neutrons and protons, and hence these particles collectively are also referred to as nucleons.
(iv) The size of nucleus is measured in Fermi (1 Fermi = 10–13 cm).

(v) The radius of nucleus is of the order of 1.5 × 10 −13 cm. to 6.5 × 10 −13 cm. i.e. 1.5 to 6.5 Fermi.
Generally the radius of the nucleus ( rn ) is given by the following relation,

rn = ro (= 1.4 × 10−13 cm) × A1/ 3


This exhibited that nucleus is 10−5 times small in size as compared to the total size of atom.

(vi) The Volume of the nucleus is about 10 −39 cm3 and that of atom is 10 −24 cm3 , i.e., volume of the

nucleus is 10 −15 times that of an atom.

(vii) The density of the nucleus is of the order of 1015 g cm−3 or 108 tonnes cm−3 or 1012 kg / cc . If

nucleus is spherical than, mass of the nucleus mass number
volume of the nucleus
Density = = 4
3
6.023 × 10 23 × πr 3

(4) Drawbacks of Rutherford's model

(i) It does not obey the Maxwell theory of electrodynamics, according to it “A small charged particle
moving around an oppositely charged centre continuously loses its energy”. If an electron does so, it
should also continuously lose its energy and should set up spiral motion ultimately failing into the
nucleus.

(ii) It could not explain the line spectra of H − atom and discontinuous spectrum nature.

e–

Unstability of atom

Example:14 Assuming a spherical shape for fluorine nucleus, calculate the radius and the nuclear density of fluorine
Solution : nucleus of mass number 19.

We know that,

r = (1.4 × 10 −13 )A1 / 3 = 1.4 × 10 −13 × 191 / 3 = 3.73 × 10−13 cm (A for F=19)

Volume of a fluorine atom = 4 πr 3 = 4 × 3.14 × (3.73 × 10−13 )3 = 2.18 × 10 −37 cm3
3 3

Mass of single nucleus = Mass of one mol of nucleus = 19 g
Avogadro' s number 6.023 × 10 23

Thus Density of nucleus = Mass of single nucleus 10 1 = 7.616 = 1013 g cm−1
Volume of single nucleus = 6.023 × 10 23 × 2.18 × 10−37

Example: 15 Atomic radius is the order of 10−8 cm, and nuclear radius is the order of 10−13 cm. Calculate what fraction of
atom is occupied by nucleus.

Solution : Volume of nucleus = (4 / 3)pr 3 = (4 / 3)p × (10−13 )3 cm3

Volume of atom = (4 / 3)pr 3 = (4 / 3)p × (10 −8 )3 cm3

∴ Vnucleus 10 −39 = 10 −15 or Vnucleus = 10 −15 × Vatom
Vatom = 10 −24


Planck's Quantum theory and Photoelectric effect.

Planck's Quantum theory

(1) Max Planck (1900) to explain the phenomena of 'Black body radiation' and 'Photoelectric effect' gave
quantum theory. This theory extended by Einstein (1905).

(2) If the substance being heated is a black body (which is a perfect absorber and perfect radiator of energy)
the radiation emitted is called black body radiation.

(3) Main points

(i) The radiant energy which is emitted or absorbed by the black body is not continuous but
discontinuous in the form of small discrete packets of energy, each such packet of energy is called a
'quantum'. In case of light, the quantum of energy is called a 'photon'.

(ii) The energy of each quantum is directly proportional to the frequency (ν ) of the radiation, i.e.
E ∝ ν or E = hν = hc
λ

where, h = Planck's constant = 6.62×10–27 erg. sec. or 6.62 × 10−34 Joules sec .

(iii) The total amount of energy emitted or absorbed by a body will be some whole number quanta.

Hence E = nhν , where n is an integer.

(iv) The greater the frequency (i.e. shorter the wavelength) the greater is the energy of the radiation.

thus, E1 = ν1 = λ2
E2 ν2 λ1

(v) Also E = E1 + E2 , hence, hc hc hc or 1 = 1 + 1 .
=+ λ λ1 λ2

λ λ1 λ2

Example: 16 Suppose 10 −17 J of energy is needed by the interior of human eye to see an object. How many photons of
green light (λ = 550 nm) are needed to generate this minimum amount of energy

(a) 14 (b) 28 (c) 39 (d) 42

Solution : (b) Let the number of photons required =n
Example: 17
n hc = 10 −17 ; n= 10 −17 × λ = 10 −17 × 550 × 10 −9 = 27.6 = 28 photons
λ hc 6.626 × 10−34 × 3 × 108

Assuming that a 25 watt bulb emits monochromatic yellow light of wave length 0.57µ. The rate of emission of

quanta per sec. will be

(a) 5.89 × 1013 sec −1 (b) 7.28 × 1017 sec −1 (c) 5 × 1010 sec −1 (d) 7.18 × 1019 sec −1

Solution: (d) Let n quanta are evolved per sec.

n hc  25 J sec −1 ; n 6.626 × 10 −34 × 3 × 10 8 = 25 ; n = 7.18 × 1019 sec −1
λ  0.57 × 10 −6
=

Photoelectric effect

(1) When radiations with certain minimum frequency (ν 0 ) strike the surface of a metal, the electrons are
ejected from the surface of the metal. This phenomenon is called photoelectric effect and the electrons emitted are
called photo-electrons. The current constituted by photoelectrons is known as photoelectric current.

(2) The electrons are ejected only if the radiation striking the surface of the metal has at least a minimum
frequency (ν 0 ) called Threshold frequency. The minimum potential at which the plate photoelectric current
becomes zero is called stopping potential.


(3)The velocity or kinetic energy of the electron ejected depend upon the frequency of the incident radiation
and is independent of its intensity.

(4) The number of photoelectrons ejected is proportional to the intensity of incident radiation.
(5) Einstein’s photoelectric effect equation : According to Einstein,
Maximum kinetic energy of the ejected electron = absorbed energy – threshold energy

1 mv 2 = hν − hν 0 = hc  1 − 1
2 max  λ 
 λ0 

where, ν 0 and λ0 are threshold frequency and threshold wavelength.

Note :  Nearly all metals emit photoelectrons when exposed to u.v. light. But alkali metals like

lithium, sodium, potassium, rubidium and caesium emit photoelectrons even when exposed to

visible light.

U.V. light Metal Photo electrons
Visible light No photo electrons
Metal other Photo electrons
Visible light than alkali

metals

Alkali
metals

 Caesium (Cs) with lowest ionisation energy among alkali metals is used in photoelectric cell.
Bohr’s atomic model.

(1) This model was based on the quantum theory of radiation and the classical law of physics. It gave new
idea of atomic structure in order to explain the stability of the atom and emission of sharp spectral lines.

(2) Postulates of this theory are :

(i) The atom has a central massive core nucleus where all the protons and neutrons are present. The size
of the nucleus is very small.

(ii) The electron in an atom revolve around the nucleus in certain discrete orbits. Such orbits are known as
stable orbits or non – radiating or stationary orbits.

(iii) The force of attraction between the nucleus and the electron is equal to centrifugal force of the moving
electron.

Force of attraction towards nucleus = centrifugal force

(iv) An electron can move only in those permissive orbits in which the angular momentum (mvr) of the

electron is an integral multiple of h / 2π . Thus, mvr = n h


Where, m = mass of the electron, r = radius of the electronic orbit, v = velocity of the electron in its orbit.


(v) The angular momentum can be h , 2h , 3h ,...... nh . This principal is known as quantization of
2π 2π 2π 2π

angular momentum. In the above equation ‘n’ is any integer which has been called as principal

quantum number. It can have the values n=1,2,3, ------- (from the nucleus). Various energy levels are

designed as K(n=1), L(n=2), M(n=3) ------- etc. Since the electron present in these orbits is associated

with some energy, these orbits are called energy levels.

(vi) The emission or absorption of radiation by the atom takes place when an electron jumps from one
stationary orbit to another.

E1 E1

E1 – E2 = hν E1 – E2 = hν
E2 E2

Emission Absorption

(vii) The radiation is emitted or absorbed as a single quantum (photon) whose energy hν is equal to the
difference in energy ∆E of the electron in the two orbits involved. Thus, hν = ∆E

Where ‘h’ =Planck’s constant, ν = frequency of the radiant energy. Hence the spectrum of the atom
will have certain fixed frequency.

(viii) The lowest energy state (n=1) is called the ground state. When an electron absorbs energy, it gets
excited and jumps to an outer orbit. It has to fall back to a lower orbit with the release of energy.

(3) Advantages of Bohr’s theory

(i) Bohr’s theory satisfactorily explains the spectra of species having one electron, viz. hydrogen atom,
He + , Li2+ etc.

(ii) Calculation of radius of Bohr’s orbit : According to Bohr, radius of orbit in which electron moves is

r =  h2 . n2
 2me  Z
 4π 2 k

[ ] [ ]where, n =Orbit number, m =Mass number 9.1× 10−31kg , e =Charge on the electron 1.6 × 10−19
[ ]Z =Atomic number of element, k = Coulombic constant 9 × 109 Nm2c −2

After putting the values of m,e,k,h, we get.

rn = n2 × 0.529 Å or rn = n2 × 0.529nm
Z Z

(a) For a particular system [e.g., H, He+ or Li+2]

r ∝ n2 [Z = constant]

Thus we have r1 = n12 i.e., r1 : r2 : r3 ........... :: 1 : 4 : 9....... r1 < r2 < r3
r2 n22

(b) For particular orbit of different species

r∝ 1 [Z =constant] Considering A and B species, we have rA = ZB
Z rB ZA


Thus, radius of the first orbit H, He + , Li+2 and Be +3 follows the order: H > He + > Li+2 > Be +3
(iii) Calculation of velocity of electron

Vn = 2πe 2 ZK , Vn =  Ze 2 1/ 2
nh  
 mr 

For H atom, Vn = 2.188 × 108 cm. sec −1
n

(a) For a particular system [H, He+ or Li+2]

V∝ 1 [Z = constant] Thus, we have, V1 = n2
n V2 n1

The order of velocity is V1 > V2 > V3 ......... or V1 : V2 : V3 .......... :: 1 : 1 : 1 ........
2 3

(b) For a particular orbit of different species

V ∝ Z [n =constant] Thus, we have H < He + < Li+2
(c) For H or He+ or Li+2, we have

V1 : V2 = 2 : 1; V1 : V3 = 3 : 1; V1 : V4 = 4 : 1
(iv) Calculation of energy of electron in Bohr’s orbit

Total energy of electron = K.E. + P.E. of electron = kZe 2 kZe 2 = kZe 2
2r −r − 2r

Substituting of r, gives us E = − 2π 2 mZ 2e 4 k 2 Where, n=1, 2, 3………. ∞
n2h2

Putting the value of m, e, k, h,π we get

E = 21.8 × 10 −12 × Z2 erg per atom = −21.8 × 10 −19 × Z2 J per atom(1J = 107 erg)
n2 n2

E = −13.6 × Z2 eV per atom(1eV = 1.6 ×10-19 J) = −313.6 × Z2 kcal. / mole (1 cal = 4.18J)
n2 n2

or − 1312 Z 2 kJmol −1
n2

(a) For a particular system[H, He+ or Li+2]

E ∝ − 1 [Z =constant] Thus, we have E1 = n22
n2 E2 n12

The energy increase as the value of n increases

(b) For a particular orbit of different species

E ∝ −Z 2 [n =constant] Thus, we have EA = Z 2
EB A

Z 2
B

For the system H, He+ , Li+2, Be+3 (n-same) the energy order is H > He + > Li+2 > Be +3

The energy decreases as the value of atomic number Z increases.


When an electron jumps from an outer orbit (higher energy) n2 to an inner orbit (lower
energy) n1, then the energy emitted in form of radiation is given by

∆E = En2 − En1 = 2π 2k 2me 4 Z 2  1 1  ⇒ ∆E = 13.6 Z 2  1 1  eV / atom
h2 n12 − n22 n12 − n22

As we know that E = hν , c = νλ and ν = 1 = ∆E , = 2π 2k 2me 4 Z 2  1 − 1 
λ hc ch3 n12 n22

This can be represented as 1 = ν = RZ 2  1 1  where, R = 2π 2k 2me 4 R is known as Rydberg
λ n12 − n22 ch3

constant. Its value to be used is 109678cm−1.

(4) Quantisation of energy of electron

(i) In ground state : No energy emission. In ground state energy of atom is minimum and for 1st orbit of

H-atom, n=1.

∴E1 = −13.6eV.

(ii) In excited state : Energy levels greater than n1 are excited state. i.e. for H- atom n2 , n3 , n4 are excited
state. For H- atom first excitation state is = n2

(iii) Excitation potential : Energy required to excite electron from ground state to any excited state.

Ground state → Excited state
Ist excitation potential = E2 − E1 = −3.4 + 13.6 = 10.2 eV.
IInd excitation potential = E3 − E1 = −1.5 + 13.6 = 12.1 eV.
(iv) Ionisation energy : The minimum energy required to relieve the electron from the binding of nucleus.

Eionisation = E∞ − En = +13.6 Z2 eV .
eff.

n2

(v) Ionisation potential : Vionisation = Eionisation
e

(vi) Separation energy : Energy required to excite an electron from excited state to infinity.

S.E. = E∞ − Eexcited .

(vii) Binding energy : Energy released in bringing the electron from infinite to any orbit is called its binding
energy (B.E.).

Note :  Principal Quantum Number 'n' = 13.6 .
(B.E.)

(5) Spectral evidence for quantisation (Explanation for hydrogen spectrum on the basisof bohr atomic model)
(i) The light absorbed or emitted as a result of an electron changing orbits produces characteristic
absorption or emission spectra which can be recorded on the photographic plates as a series of lines,
the optical spectrum of hydrogen consists of several series of lines called Lyman, Balmar, Paschen,
Brackett, Pfund and Humphrey. These spectral series were named by the name of scientist who
discovered them.
(ii) To evaluate wavelength of various H-lines Ritz introduced the following expression,

ν = 1 = ν = R  1 − 1
λ c  n12 
 n22 


where, R is = 2π 2me 4 Rydberg's constant
ch 3
=

It's theoritical value = 109,737 cm–1 and It's experimental value = 109,677.581cm−1

This remarkable agreement between the theoretical and experimental value was great achievment of

the Bohr model.

(iii) Although H- atom consists only one electron yet it's spectra consist of many spectral lines as shown in fig.

n=8 Humphrey series
n=7
n=6 Pfund
n=5
n=4 series

Energy level n=3 Brackett
series
n=2
Paschen
n=1 series
Lyman
series Balmer
series

(iv) Comparative study of important spectral series of Hydrogen

S.No. Spectral Lies in the Transition λmax = n12n22 λmin = n12 λ max = n 2
series region n2 > n1 (n22 − n12)R R λ min 2
n1 = 1
(1) Lymen Ultraviolet n2 = 2,3,4....∞ n22 − n12
series region
n1 = 1 and n2 = 2 n1 = 1 and n2 = ∞ 4
(2) Balmer 3
series 4 1
λ max = 3R λ min = R
(3) Paschen
series Visible n1 = 2 n1 = 2 and n2 = 3 n1 = 2 and n2 = ∞
region n2 = 3,4,5....∞
36 4 9
λ max = 5R λ min = R 5

Infra red n1 = 3 n1 = 3 and n2 = 4 n1 = 3 and n2 = ∞ 16
7
region n2 = 4,5,6....∞ 144 9
7R R
λ max = λ min =


(4) Brackett Infra red n1 = 4 n1 = 4 and n2 = 5 n1 = 4 and n2 = ∞ 25
series region n2 = 5,6,7....∞ 9
16 × 25 16
λmax = 9R λ min = R 36
11
(5) Pfund Infra red n1 = 5 n1 = 5 and n2 = 6 n1 = 5 and n2 = ∞
series region n2 = 6,7,8....∞ 49
25 × 36 25 13
λ max = 11R λ min = R

(6) Humphrey Far n1 = 6 n1 = 6 and n2 = 7 n1 = 6 and n2 = ∞
n2 = 7,8....∞
series infrared 36 × 49 36
13R R
region λ max = λ min =

(v) If an electron from nth excited state comes to various energy states, the maximum spectral lines

obtained will be = n(n − 1) . n= principal quantum number.
2

as n=6 than total number of spectral lines = 6(6 − 1) = 30 = 15.
2 2

(vi) Thus, at least for the hydrogen atom, the Bohr theory accurately describes the origin of atomic spectral lines.

(6) Failure of Bohr Model

(i) Bohr theory was very successful in predicting and accounting the energies of line spectra of hydrogen
i.e. one electron system. It could not explain the line spectra of atoms containing more than one
electron.

(ii) This theory could not explain the presence of multiple spectral lines.

(iii) This theory could not explain the splitting of spectral lines in magnetic field (Zeeman effect) and in
electric field (Stark effect). The intensity of these spectral lines was also not explained by the Bohr
atomic model.

(iv) This theory was unable to explain of dual nature of matter as explained on the basis of De broglies
concept.

(v) This theory could not explain uncertainty principle.

(vi) No conclusion was given for the concept of quantisation of energy.

Example: 18 If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be

Solution : (c) (a) 4 r2 (b) 4r2 (c) 9 r3 (d) 9r2
Example: 19 9 4
Solution : (b)
r = n2h2 ∴ r2 22 ∴ r3 = 9 r2
4π 2mZe 2 r3 = 32 4

Number of waves made by a Bohr electron in one complete revolution in 3rd orbit is

(a) 2 (b) 3 (c) 4 (d) 1

Circumference of 3rd orbit = 2πr3
According to Bohr angular momentum of electron in 3rd orbit is

mvr3 = 3 h or h = 2πr3
2π mv 3

by De-Broglie equation, λ = h
mv


∴λ = 2πr3 ∴2πr3 = 3λ
3

i.e. circumference of 3rd orbit is three times the wavelength of electron or number of waves made by Bohr
electron in one complete revolution in 3rd orbit is three.

Example: 20 The degeneracy of the level of hydrogen atom that has energy − R11 is
Solution : (a) 16

(a) 16 (b) 4 (c) 2 (d) 1

En = − RH ∴ − RH = − RH
n2 n2 16

i.e. for 4 th sub-shell

n=4 1 2 3

1=0

m=0 +1 0 +1 +2 –1 0 +1 +2 –3 –2 –1 0 +1 +2 +3
one s three p five d seven f

i.e. 1+3+5+7=16, ∴ degeneracy is 16

Example: 21 The velocity of electron in the ground state hydrogen atom is 2.18 × 108 ms −1. Its velocity in the second orbit
would be

(a) 1.09 × 108 ms −1 (b) 4.38 × 108 ms −1 (c) 5.5 × 105 ms −1 (d) 8.76 × 108 ms −1

Solution : (a) We know that velocity of electron in nth Bohr's orbit is given by

v = 2.18 × 106 Z m/s
n

for H, Z = 1

 v1 = 2.18 × 106 m/s
1

 v2 = 2.18 × 106 m / s = 1.09 × 106 m / s
2

Example: 22 The ionization energy of the ground state hydrogen atom is 2.18 × 10−18 J. The energy of an electron in its
second orbit would be

(a) −1.09 × 10−18 J (b) − 2.18 × 10−18 J (c) − 4.36 × 10−18 J (d) − 5.45 × 10−19 J

Solution : (d) Energy of electron in first Bohr's orbit of H-atom

E = − 2.18 × 10−18 J ( ionization energy of H = 2.18 × 10−18 J )
n2

E2 = − 2.18 × 10−18 J = −5.45 × 10−19 J
22

Example: 23 The wave number of first line of Balmer series of hydrogen atom is 15200 cm−1 . What is the wave number of
first line of Balmer series of Li3+ ion.

(a) 15200cm−1 (b) 6080 cm−1 (c) 76000cm−1 (d) 1,36800 cm−1

Solution : (d) For Li 3+ v = v for H × z 2 =15200 ×9= 1,36800 cm−1

Example: 24 The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530Å. The radius for the first excited
state (n = 2) orbit is (in Å)


(a) 0.13 (b) 1.06 (c) 4.77 (d) 2.12

Solution : (d) The Bohr radius for hydrogen atom (n = 1) = 0.530Å

The radius of first excited state (n = 2) will be = 0.530 × n2 = 0.530 × (2)2 = 2.120Å
Z 1

Example: 25 How many chlorine atoms can you ionize in the process Cl → Cl+ + e−, by the energy liberated from the
following process :

Cl + e − → Cl − for 6 × 10 23 atoms
Given electron affinity of Cl = 3.61eV, and IP of Cl = 17.422 eV

(a) 1.24 × 1023 atoms (b) 9.82 × 10 20 atoms (c) 2.02 × 1015 atoms (d) None of these

Solution : (a) Energy released in conversion of 6 × 10 23 atoms of Cl − ions = 6 × 10 23 × electron affinity

= 6× 10 23 × 3.61 = 2.166 × 10 24 eV.

Let x Cl atoms are converted to Cl + ion
Energy absorbed = x × ionization energy

x × 17.422 = 2.166 × 10 24 ; x = 1.243 × 10 23 atoms

Example: 26 The binding energy of an electron in the ground state of the He atom is equal to 24eV. The energy required to
remove both the electrons from the atom will be

(a) 59eV (b) 81eV (c) 79eV (d) None of these

Solution : (c) Ionization energy of He = Z2 × 13.6 = 22 × 13.6 = 54.4eV
Example: 27 n2 12

Energy required to remove both the electrons

= binding energy + ionization energy

= 24.6 + 54.4 = 79eV

The wave number of the shortest wavelength transition in Balmer series of atomic hydrogen will be

(a) 4215 Å (b) 1437Å (c) 3942Å (d) 3647Å

Solution : (d) 1 = RZ 2  1 − 1  =109678 × 12 ×  1 − 1 
λ shortest n12 n22  22 ∞2 

λ = 3.647 × 10−5 cm = 3647Å

Example: 28 If the speed of electron in the Bohr's first orbit of hydrogen atom is x, the speed of the electron in the third

Solution : (b) Bohr's orbit is

Example: 29 (a) x/9 (b) x/3 (c) 3x (d) 9x
Solution : (b)
Example: 30 According to Bohr's model for hydrogen and hydrogen like atoms the velocity of an electron in an atom is

quantised and is given by v ∝ 2πZe 2 so v ∝ 1 in this cass n = 3
nh n

Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is

(a) n=1 to n=2 (b) n = 3 to n = 8 (c) n = 2 to n = 1 (d) n = 8 to n = 3

Absorption line in the spectra arise when energy is absorbed i.e., electron shifts from lower to higher orbit, out

of a & b, b will have the lowest frequency as this falls in the Paschen series.

The frequency of the line in the emission spectrum of hydrogen when the atoms of the gas contain electrons in

the third energy level are

(a) 1.268 × 1014 Hz and 2.864 × 1016 Hz (b) 3.214 × 1010 Hz and 1.124 × 1012 Hz

(c) 1.806 × 1012 Hz and 6.204 × 1015 Hz (d) 4.568 × 1014 Hz and 2.924 × 1015 Hz

Solution : (d) If an electron is in 3rd orbit, two spectral lines are possible

(a) When it falls from 3rd orbit to 2nd orbit.


In equation ν = 3.289 × 1015 1 − 1
 
 n12 n22 

ν1 = 3.289 × 10 15  1 − 1  = 3.289 × 1015 × 5 = 4.568 × 1414 Hz
 22 32  36

(b) When it falls from 3rd orbit to 1st orbit :

ν2 = 3.289 × 1015 × 1 − 1  = 3.289 × 1015 × 8 = 2.924 × 1015 Hz
1 32  9

Example: 31 If the first ionisation energy of hydrogen is 2.179 × 10−18 J per atom, the second ionisation energy of helium

per atom is

(a) 8.716 × 10−18 J (b) 5.5250 kJ (c) 7.616 × 10−18 J (d) 8.016 × 10−13 J

Solution : (a) For Bohrs systems : energy of the electron ∝ Z2
n2
Example: 32
Solution : (b) Ionisation energy is the difference of energies of an electron (E∞ ), when taken to infinite distance and Er
Example: 33 when present in any Bohr orbit and Eα is taken as zero so ionisation energy becomes equal to the energy of
electron in any Bohr orbit.

EH ∝ Z 2 ; E He ∝ Z 2 or EH = 1 [as Z H = 1, Z He = 2,nH = 1,nHe = 1]
H He E He 2× 2

nH2 nH2 e

or EHe = EH × 4 = 2.179 × 10 −18 × 4 = 8.716 × 10 −18 Joule per atom.

The ionization energy of hydrogen atom is 13.6eV. What will be the ionization energy of He +
(a) 13.6eV (b) 54.4eV (c) 122.4eV (d) Zero

I.E. of He + = 13.6eV × Z 2

13.6eV × 4 = 54.4eV

The ionization energy of He + is 19.6 × 10−18 J atom–1. Calculate the energy of the first stationary state of Li +2

(a) 19.6 × 10−18 J atom-1 (b) 4.41 × 10−18 J atom–1

(c) 19.6 × 10 −19 J atom -1 (d) 4.41 × 10 −17 J atom −1

Solution : (d) I.E. of He + = E × 22(Z for He = 2)

I.E. of Li 2+ = E × 33 (Z for Li=3)

I.E.(He + ) 4 or I.E. (Li 2+ ) = 9 × I.E.(He + ) = 9 × 19.6 × 10 −18 = 4.41 × 10−17 J atom–1
∴ I.E.(Li 2+ ) = 9 4 4

Bohr – Sommerfeld’s model.

(1) In 1915, Sommerfield introduced a new atomic model to explain the fine spectrum of hydrogen atom.

(2) He gave concept that electron revolve round the nucleus in elliptical orbit. Circular orbits are formed in
special conditions only when major axis and minor axis of orbit are equal.

(3) For circular orbit, the angular momentum = nh where n= principal quantum number only one

component i.e. only angle changes.

(4) For elliptical orbit, angular momentum = vector sum of 2 components. In elliptical orbit two components are,

(i) Radial component (along the radius) = nr h


Where, n r = radial quantum number

(ii) Azimuthal component = n φ h


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