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TM (Chemistry) Periodic Classification 1 Forum for Competitions SCF 100–10 2, Chho ti Barad ari, P ATIALA–147001 Ph: 0175-501202 9, 5012030, Fax: 0175-5012028 www.lakshyainstitute.com Forum for Competitions Patiala Centre : TEL. 0175-5012029, 5012030 Bathinda Centre : TEL. 0164-5012028, 5012029 Lakshya Forum for Competitions Pvt. Ltd. www.lakshyainstitute.com PERIODIC CLASSIFICATION Lecture – 1 Introduction 1. Dobereiner’s law of triad 2. Newland’s law of octave 3. Lother Meyer’s curve 4. Mendeleev’s periodic t able 5. Modern periodic table Lecture – 2 1. Positi on of el ements in peri od ic ta bl e 2. Screening effect 3. Atomic radius 4. Ionic Radius Lecture – 3 1. Ionization energy 2. Electron affinity Lecture – 4 1. Electron affinity 2. Electronegativity LECTURE - 1 Introduction In the very beginning elements were classified into metals & non metals. After the discovery of very large number of elements the need for their classification in small groups was felt. Attempts were made to classify the elements according to one property or other. After the establishment of Dalton’s atomic theory, the chemists took atomic mass as the fundamental property of the element & tried to seek relationship between the properties of the elements & their atomic masses. 1. DOBEREINER’S LAW OF TRIADS (a) He classified the ele ments in a group of three elements called triads. ( b) Acco rding to Dobereiner  the atomic mass of the central element was nearly the arithmetic mean of atomic masses of other two element. (c) He classifie d some co mmon element s as followi ng :
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PERIODIC CLASSIFICATIONLecture – 1

Introduction

1. Dobereiner’s law of triad

2. Newland’s law of octave

3. Lother Meyer’s curve

4. Mendeleev’s periodic table

5. Modern periodic table

Lecture – 2

1. Position of elements in periodic table

2. Screening effect

3. Atomic radius

4. Ionic Radius

Lecture – 3

1. Ionization energy

2. Electron affinity

Lecture – 4

1. Electron affinity

2. Electronegativity

LECTURE - 1Introduction

In the very beginning elements were classified into metals & non metals. After thediscovery of very large number of elements the need for their classification in smallgroups was felt. Attempts were made to classify the elements according to one propertyor other. After the establishment of Dalton’s atomic theory, the chemists took atomicmass as the fundamental property of the element & tried to seek relationship betweenthe properties of the elements & their atomic masses.

  1. DOBEREINER’S LAW OF TRIADS

(a) He classified the elements in a group of three elements called triads.

(b) According to Dobereiner  the atomic mass of the central element was nearly the

arithmetic mean of atomic masses of other two element.(c) He classified some common elements as following :

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1.1 Drawbacks of Dobereiner’s Law of Triads

(a) In some of the triads all the three elements possessed nearly the same atomicmasses.

eg. Fe Co Ni, Ru Rh Pd, Os Ir Pt

(b) The major drawback of his classification was that all the known elements

could not be arranged as triads.

  2. NEWLAND’S LAW OF OCTAVES

(a) John A.R. Newland proposed another classification of grouping the elements of similar properties.

(b) According to his law, when elements are arranged in their increasing atomic massesthe properties of every eighth element was resembled to the first one. On the basisof above observation Newland proposed his law which is known as Newland’s law of octaves due to similarity of the musical vowel Sa, Re, Ga, Ma, Pa, Dha, Ne, Sa.

(c) According to his law, the eighth element sodium had similar properties to Lithiumwhich was the first element and next so on.

2.1 Drawbacks of Newland’s Law of Octaves

(a) Newland’s octave type of system worked only for the lighter elements.

(b) He failed in the case of heavier elements as ‘Fe’, which has been placed alongwith ‘O’ & ‘S’.

(c) After the discovery of Inert gases, he could not decide the position of inertelements.

  3. LOTHER MEYER’S CURVE

(a) In 1870, Lother Meyer  studied the physical properties such as molar volume,melting point, boiling point of various elements to arrive at his periodic table of elements.

(b) Lother Meyer plotted a graph between atomic volume and atomic weight of variouselements.

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(c) According to this graph he reached some following conclusions :

(i) Alkali metals, having largest atomic volume and electro positive charactertake the peak position of the curve.

(ii) The alkaline earth metals which have less strong electropositive characteroccupy the descending position of the curve.

(i ii ) The halogens which have most electronegative character take the ascendingposition of the curve.

(iv) Transition metals and metalloids which have both type of character metallicand nonmetallic occupy the bottom position of the curve.

(v) On the basis of above conclusion he gave the rule that the physical propertiesof the elements are periodic function of their atomic weights.

3.1 Drawbacks of Lother meyer’s curve

(a) It was an ideal classification but it was not easy to keep in mind the various

positions of the curve.(b) He could not determine the position of all the known elements.

(c) He also could not specify the position of He and Li.

  4. MENDELEEV’S PERIODIC TABLE

(a) Dmitry Ivanovich Mendeleev gave a periodic law which states that the propertiesof elements (Physical & Chemical) are a periodic function of their atomic masses.

(b) On the basis of this law, he arranged all the known elements in the form of a tableknown as periodic table.

(c) If the elements arranged according to their atomic weight, exhibit an evident

periodicity of properties.

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(d) It consists nine groups & seven periods.

(e) He gave a systematic way to study of the elements according to this periodic table.

4.1 Merits

(a) In this periodic table some places were left vacant for new elements whichwere not discovered at that time.

(b) The magnitude of their atomic weight determine the character of the element.

(c) The properties and names of unknown elements were predicted on the basisof periodic law eg. For Al eka boron, for Ge eka silicon.

4.2 Defects

(a) The position of hydrogen in his periodic table was not well defined.

(b) Some pairs do not obey the rule of increasing atomic masses.

(c) In this periodic table, some elements placed together which had differentproperties and some elements placed differently, which had same properties.

(d) He could not explain the cause of periodicity.

(e) He could not separate the position of lanthanides and actinides.

  5. MODERN PERIODIC TABLE

5.1 Modern Periodic Law

(a) Moseley gave this law. According to it physical and chemical properties of theelement are periodic function of their atomic numbers.

(b) Long form of the periodic table was given by Bohr and depends on Bohr-Burryscheme of electronic configuration. Which was proposed by Rang, Werner,Bohr, Burry and Others.

(c) The long form of the periodic table consist of 7 horizontal rows called periodslike Mendeleev’s periodic table and 18 vertical columns called groups.

(d) The long form of the periodic table is just the graphical representation of 

Aufbau’s principle.

(e) Every period starts with filling of electron in new principal quantum numberand completes till then the outer most shell.

(f) It is studied in the following four portions :

(i) The left portion : This portion includes the IA and IIA group memberswhich have electropositive character and these are highly reactive metal.

(Exception – Hydrogen). All the groups in this portion are included ins-block.

(ii) The right portion :  The extreme right portion consists the elements of groups IIIA, IVA, VA, VIA, VIIA and zero groups & their isotopes. This portion

consists all the non-metals & metalloids and some metal elements. Allthe groups of this portion includes in the p-block.

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(iii) The middle portion : This portion includes the elements of IIIB, IVB, VB,VIB, VIIB, VIII, IB & IIB. It accommodates all metals which are the elementsof 3d, 4d, 5d & 6d series called transition series accept IIB elements i.e.(Zn, Cd, Hg).

(iv) The bottom portion :  This portion constitutes of two series such as

Lanthanide and Actinide. Each series consists 14 elements of 4f and 5f series are called inner transition series and elements are called as innertransition elements.

5.2 Advantages of long form of periodic table :(a) It is based on the concept of atomic number of elements, which are fundamental

property of elements.

(b) Due to similar outermost configuration every element shows similar propertiesof that particular group.

(c) It is easy to reproduce, remember them and grouping of elements in blocks.

(d) Position of isotopes were considered.

(e) The transition elements having separate property are set at their particular position.

(g) Periodicity in properties :When elements are arranged in the increasing order of atomic number,elements having similar properties recur at regular intervals in the periodictable. This type of property is called periodicity.

Cause of Periodicity :(a) The cause of periodicity in properties is due to the same outermost electronic

configuration coming at regular intervals.

(b) In the periodic table, elements with similar properties recur at intervals of 2,8, 8, 18, 18, 32 and 32. These numbers are called as magic numbers.

5.3 Defects of long form of periodic table(a) Position of hydrogen is not settled properly.

(b) Inner transition element are separately placed at the bottom of the periodic

table.

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  FDescription of periods & groups

PeriodsIn the periodic table seven horizontal rows are present, which is called periods. Aperiod consists of a series of elements having same valence shell. The sevenhorizontal period as follows :

(a) 1st period :

 This period is called shortest period because it consists only 2  elements(Hydrogen {1s1 } and Helium {1s2 }) in K shell or 1 orbit. i.e. It consists only twoelectron or in turn two elements.

(b) 2nd & 3rd period :

 These periods are called short periods and each period consists 8 elements i.e.In this period second shell or 2s & 2p are completed. The second period includes

3Li to

10Ne and the third period includes

11Na to

18Ar.

(c) 4th & 5th period :

 These periods are called long periods and consists 18 elements in each periodi.e. 18 electrons namely

19K to

36Kr &

37Rb to

59Xe. These periods consists d-

block element, which is known was transition element.

(d) 6th period :

 This period is called longest period and it consists of 32 elements. In the 32 elements14 elements is in Lanthanide series, which is called Inner transition elements.

(e) 7th period :

At present this period consists 26 element. It is called incomplete period. Seven of the

element (106 to 112) have been reported as heavy elements, which are very unstable.

Groups

(a) The vertical column of the long form of periodic table are called groups. There are 18vertical column. In the modern phenomena we say that 18 groups but according to oldnaming of groups, we say that there is 16 groups as IA to IIB, IB to VIIB, VIII and ‘0’ group.

(b) In the IA, IIA and IIIA to VIIA all their inner shells are complete but their outermostshell are incomplete. Elements of these groups are called Normal elements.Elements of ‘zero’ group are called inert gases, which consist ns2 np6 type of electronic

configuration. The electronic configuration of normal element is ns1 to ns

2np

5.

(c) Elements of group IB, IIB and VIII and IIIB to VIIB have incomplete filled penultimate

& outermost shells. The elements of these groups are called transition elements. Their general electronic configuration is (n – 1) d1 – 10 ns0 – 2.

(d) There are two more rows placed at the bottom of the periodic table for the sake of convenience.

(e) The outermost, penultimate & antipenultimate shells of both the series haveincompletely filled. They are called rare earth metals and also called Lanthanides andActinides. Their general outermost electronic configuration is (n – 2) f 1 – 14 (n – 1) d0 – 1 ns2.

(f) Zero group elements have general electronic configuration ns2 np6 and they have allcompletely filled shell. These are very less reactive. On the basis of less reactivitywe call them inert gases. According to above discussion the modern periodic table isdivided into four main blocks. Which are known as s, p, d and f blocks. The elements,which consist in these blocks are called s, p, d and f block elements.

FTO

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Question 1. In modern periodic table 6th period contain elements(a) 8 (b) 18 (c) 10 (d) 32

Ans. (d)

Question 2. Uranium is a member of (a) Transition series (b) III-period

(c) Actinide series (d) VI-periodAns. (c)

Question 3. The number of elements in fourth long period is(a) 18 (b) 8 (c) 32 (d) 10

Ans. (a)

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LECTURE - 2

  6. GENERAL CLASSIFICATION OF ELEMENTS

In addition to above four blocks elements can also be classified as following :6.1 Noble gases :

Elements of group 18 are called noble gases. These are also called as inert gasesbecause their outermost ns & np orbitals are completely filled (except He has 1s2)and these gases are non-reactive in nature under ordinary conditions.

6.2 Representative elements :

All the s and p block elements are known as representative elements except zero group.

6.3 Transition elements :

All the d-block elements (except IIB gr.) are called transition element. It comprisesinto 4th, 5th, 6th and 7th period. They lie between s and p-block elements.

6.4 Inner transition elements :All the f-block elements or 4f and 5f block elements are called Inner transitionelements. Total number of these elements is 28. They lie between IIIB & IVB andplaced at the bottom of periodic table.

6.5 Typical elements :

Elements of second & third period are known as typical elements.

6.6 Diagonal Relationship :

 The first three members of second period (Li, Be & B) not only show similaritieswith the members of their own groups but show similarities with the elementsdiagonally placed in higher groups. this resemblance is termed diagonalrelationship.

IALi

Na

IIABe

Mg

IIIAB

Al

IVAC

Si

VAN

P

VIAO

S

VIIAF

Cl3rd period

2nd period

  7. PREDICTION OF PERIOD, GROUP & BLOCK

1. Block :Last electron enters into which orbital.

2. Period :Maximum value of principal quantum no.

3. Groups :s-block elements - no. of valence electron.

p-block elements - 10 + no. of valence electron.

d-block elements - (n – 1) d + no. of s electron.

f-block elements - IIIB.

eg. 1. Atomic no. 17

Electronic configuration - 1s22s22p63s23p5

Block - p Period - 3rd

Group - 10 + 7 = 17th group

2. Fe = [Ar] 4s23d6

d-block, 4th period8th group

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  FTO

Some Important Points* Elements after Uranium (92) are known as Transuranic, post uranic elements.* Elements of 2nd and 3rd period are Typical Elements as they summarise the properties

of their respective groups.* Elements of 3rd period - Bridge elements.

HLiNa

AKRbCs

BCuAgAu

IB Coinage elementIA

* Largest group in P.T = IIIB.* Largest period in P.T = 6th period.* Most abundant inert gas = Argon.* Most abundant gas in atmosphere =N

2.

* Most abundant element in earth crust = O2.

* Most abundant metal in earth crust = Al.* Pt and Au – purely nobel metals.* Highest M.P. among non metals = Carbon.* Highest M.P. among metal = W (Tungsten)* Highest density – Ir, Os.

* Lowest density in metal – Li.* Lowest density in element – H

2.

* Liq. elements – Bi, Hg, Cs, Fr, Ga.* Gaseous elements – He, Ne, Ar, Kr, Xe, Rn

H2, Cl

2, F

2, O

2, N

2

* Volatile ‘d’ block – Zn, Cd, Hg.* Most poisonous metal – (Pu) plutonium 94* Element sublimes on Heating – I

2.

* Metal in Hb – Fe.* Metal in chlorophyll – Mg.

Best conductor of heat and electricity – AgI, Cu.

Best conductor of heat and electricity among non metals – GraphiteBad conductor of heat and electricity – Lead (Metal), Sulphur (non metal)

* Elements of second period show similarities with the elements of third period lying diagonaly – Diagonal relationship.

Li

Na

Be

Mg

B

Al

C

Si

N

P

O

S

 – almost similar ionic size – almost similar E.N. – almost similar polarising power of ions.

or charge ratio sizeDiagonal relationship vanishes with increase in atomic number.

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  8. NOMENCLATURE

Atomic No. > 100

0 1 2 3 4 5 6 7 8 9

nil un bi tri quad pent hex sept oct enn* But ending is done by ium.

101 un-nil-un-ium Unn

102 un-nil-bi-ium unb

Q. Find out the position of elements having atomic number.

(a) 17 (b) 37 (c) 27 (d) 33 (e) 43

Sol. (a)

[Ne] 3s23p

5, p-block, 3

rd period, valence e

 –  = 7 group = 17

Q. An element has 4 e –  in its valence shell with n = 4 what is its at no. and position.

Sol. 4s2 3d10 4p2 at. no. = 32

block – p Ge

Group – 14

Period – 4

Q. What will be the at. no. of and position of element placed just above in the same group

with an element of at. no. 49 and 52.

Sol. Ga – 31 Se – 34

Atomic No. 49 524s2 3d10 4p1 4s2 3d10 4p4

Block – p p

Group – 13 16

Period – 4 4

Q. A metal ion with +3 charge and has 5 valence e –  in 3d subshell. Identify element.

Fe3+ = 3d5

Q. If there were 10 periods in PT. Then how many element would this period comprise of :

1 2 × (1)2 2

2, 3 2 × (2)2 8 each

4, 5 2 × (3)2 18 each

6, 7 2 × (4)2 32 each

8, 9 2 × (5)2 50 each

10 2 × (6)2 72 Ans.

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  9. PERIODIC PROPERTIES

9.1 Screening or Shielding effect

(a) The effect produced by the electrons present in the inner shells is called the

screening or shielding effect. It is due to the inner electron shield or act as ascreen for the interaction between valence electrons and nucleus.

(b) On account of this, the real force of attraction between the nucleus and valenceelectrons is some what decreased by the screening or shielding effect of innershell’s electrons.

 TNC – Total nuclear charge effective for all the e – .

ENC – The net nuclear charge actually felt by valence e –  called as effective nuclear charge

or

Charge of nucleus which is effective for valence e –  only.

ENC = TNC – Screening constant

orENC = TNC –

Z* = Z –

(c) On the above discussion, the value of screening effect depends upon the

(i) The number of inner shell’s electrons, i.e. increase in shielding effect respectively.

(ii) Shape of orbital –– s > p > d > f.

(d) The screening effect of particular element is constant so, the screeningconstant is defined by the symbol of ‘ ’. It’s calculated by Slater’s rule.

(e) Calculation of screening constant ‘ ’ by Slater’s rule as follows :

  FTO

For ns and np orbital’s electrons.

(i) First write the electronic configuration of that element orderly whose screeningconstant will be determined.

(ii) The screening constant for last electron of nth  orbit is equal to zero due to thatelectrons do not produce screening effect.

(iii ) For all the ns and np electrons the screening constant is 0.35 each but for 1selectron the screening constant is 0.30 only.

(iv) The screening constant for all the (n – 1)th shell’s electrons are 0.85 each for s andp-orbital and 1 for d and f electrons.

(v) The screening constant for all the (n – 2)th shell’s electrons are 1.0 each.

9.2 Atomic radius :

 The distance between the nucleus of the atom and outermost shell of the electrons

is called radius of an atom.

9.2.1 Types of atomic radius :

(A) Covalent radius :

If bonding present in between atoms in a molecule is covalent, the radius iscalled covalent radius. It is the half of the distance between the nucleus of 

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two like atoms bonded together by a single covalent bond.

AA

rA rA

d  –A

(a) For homonuclear molecule like A2, B

2 etc.

covalent radius for A – A molecule ;

AA

dr

2where r

A = single bond covalent radius

dA – A

 = Distance between the nucleus of two like atom.

(b) For heterogenous molecule like A – B etc.

If covalent bond are formed by different type of atoms i.e. one atom is moreelectronegative than the other combined atom. Then, the covalent radiusis calculated by the relation which is called Stevenson formula as follows :

A B A Bd r r 0.09 x

where rA = radius of an atom ‘A’ 

rB = radius of an atom ‘B’ 

x = Electronegativity difference between two atoms. A BA B

dr .

2

(c) The bond length is decreased with increase in bond order respectively.

(d) The actual value of bond length is determined by single bond.

(B) Metallic radius :

It is defined as one half of the inter nuclear distance between the nuclei of twoadjacent atoms in the metallic lattice. Metallic radius of an element is greaterthan the covalent radius.

rArA

(C) Vander Waal’s radius :

(a) It is one half of the distance between the nucleus of two adjacent atomsbelonging to two neighbouring molecules of an element in the solid state.

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A2 A2

rcovalent rvander Waal's

(b) The vander waal’s radius constitute the atomic radii of noble gases.

(c) It is greater than that of all known radius.

(d) Vander waal’s radius is generally two times than covalent radius.

(e) Vander waal's force molecular weight

Trends :

(i) In a period :

While we move from left to right along a period, atomic radius generally

decreases.(ii) In a group :

In a given group as one moves from top to bottom atomic radius generallyincreases.

Order of radius and forces :

1. van der Waal’s radius > Metallic radius > Covalent radius

2. van der Waal’s forces < Metallic forces < Covalent forces

30. Which of the following have highest vander waal’s radius

(a) F2

(b) Cl2

(c) Br2

(d) none of these

Ans. (c)

31. Which of the following atom has smallest size

(a) He (b) F (c) H (d) none of these

(D) Ionic radius :

(a) Ionic radius is the distance between the nucleus and outermost shell of an ion.

(b) Ionic distance is also called as the effective distance between the nucleusof an ion and which it has its influence on its outermost electron cloud onwhich it has its influence.

(c) Types of ions – (i) cation (ii) Anion

(d) The radius of cation or positive ion is always smaller than their parent atombecause a cation is formed by the removal of one or more electrons from an atom.

eg. Atomic radius of Fe = 1.26 ÅIonic radius of Fe+2 = 0.76 Å

Ionic radius of Fe+3 = 0.64 Å

D-1Cation formation :

(a) A cation formed by the loss of electron where the outer shell is completelydisappeared since the inner shells do not extend` so far in space. So,the cation is much smaller than the metal atom.

eg. Na (2, 8, 1) Na+ ( 2, 8) Na+ < Na

(b) Whenever, a cation is formed the ratio of Z/e is increased with theresult of the effective nuclear charge is increased and the electrons

are pulled towards the nucleus. Consequently the cation becomessmaller.

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1Cationic radius

ve charge

D-2Anion formation :

(a) An anion is formed by the gain of electron, where one or more electronsare added, the electron cloud expands and the ionic size increases.

(b) In the formation of anion, the effective nuclear charge decreases, whichis the result of in expansion of electron cloud. Thus, the anion becomeslarger than the corresponding atom.

Anionic radius ve charge

Example – 1 : A given compound A2 whose total d

A – A is 1.4 Å. The atomic (covalent)

radius of an atom A is

(a) 0.7 Å (b) 0.5 Å (c) 0.2 Å (d) 0.1 Å

Solution : A AA

dr 1.4/2 0.7

2

Order of Radii for isoelectronic species

O2– 

F – 

Na+

Mg2+

Al3+

Z 8 9 11 12 13

e 10 10 10 10 10

Zeff 8

109

101110

1210

1310

O2–  > F –  > Ne > Na+ > Mg2+ > Al3+

9.2.2 Factors affecting the values of Atomic Radii :

(A) Nuclear attraction – (N.A.) means Zeff 

 :

1N.A.

Atomic Radius

If, the nuclear attraction on the outermost orbit increases, the atomic radiusdecreases respectively.

(B) Principal quantum number (n) :

n Atomic radius

On increasing principal quantum number, the atomic radius increasesrespectively.

(C) Screening effect :

Screening effect Atomic radius

On the increasing of screening effect the atomic radius will increasesimultaneously.

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(D) Multiplicity of bond :

Covalent radii of an atom depends upon the number of bonds formed by it, i.e.

1Number of bond

Atomic Radius

Eg. Radius of Carbon

H3C – CH

30.77 Å

H2C = CH

20.67 Å

HC CH 0.60 Å

9.2.3 Periodicity in Atomic radius :

(A) In a period :

(a) As we move from left to right along a period, the atomic radius decreasesregularly in the period of representative elements, because the principalquantum number remains the same in a period but the nuclear charge

increases. Due to increase of above property the force of attraction towardsnucleus increase, which brings contraction of size.

(b) In the case of noble gas elements, the atomic radius increasesexceptionally. It is due to the fact that the noble gases are considered vanderwaal’s radius, which always has the higher value than covalent radius.

(B) In a group :

As we move from top to bottom in a group the atomic radius increases as theatomic number increases. This is due to that the number of energy shellsincrease, which out weighs the effect of increased nuclear charge. Further,the atomic radius gradually increase down the group.

Important information regarding atomic radiuseg. The largest cation of the periodic table = Cs+

eg. The smallest cation of the periodic table = H+

eg. The largest anion of the periodic table = I – 

eg. The smallest anion of the periodic table = H – 

eg. The biggest element of periodic table = Fr

eg. The smallest element of periodic table = H

Exceptional Cases

1. Group IIIA Al Ga

3s

2

 3p

1

  4s

2

 3d

10

 4p

1

Poor SE of ‘d’ thus ENC inc size decrease.

Transition Element :

3d Series

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vertical coloumn at radius should be inc.

3d < 4d < 5d but actually the order is

3d < 4d 5d

 Ti < Zr Hf – due to Lanthanide Contraction.

 There are14 elements in lanthanides. Decrease in size from Ce – Lu.the e –   enters into f’ orbital and which shows poor SE. Thus, the effect of nuclear

charge brings the valence shell nearer to the nucleus & hence the size of atom or iongoes on decreases as we move in the series.

Q. Arrange the following in dec order of size.1. Na Mg K Al K > Na > Mg > Al

2. Ca Sr Ba Ra Ra > Ba > Sr > Ca

3. C, S, O, Se Se > S > C > O

4. P, S, N P > S > N

Q. Select the largest species from the following1. S2– , Ar, K+  S2– 

2. Ne, Mg, Ca, Na Ca

3. W or Au W

4. Si and Sn Sn

5. Ce and Lu Ce

Q. Largest size order1. N3–  < P3–  < As1– 

2. K+, Ca2+ Ar

3. Pb4+, Pb, Pb2+

4. Na+, O2– , Ne, Cl – 

LECTURE - 3

  10. IONISATION POTENTIAL OR IONISATION ENERGY (I.P.) OR (I.E.) :

 The minimum amount of required energy to remove the most loosely boundedelectron from the outermost orbit of an isolated atom in the gaseous state is calledas ionisation potential or ionisation energy.

M(g)

 + Energy (IE1) M+

(g) + e – 

10.1 Successive I.E. : The I.E. after the I.E.

1 is called successive I.E. as I.E.

2, I.E.

3 ...... etc. i.e. the

minimum amount of energy required to remove more electron from uni-di-tripositive ion is called as I.E.

2, I.E.

3, I.E.

4 and it is known as successive I.E. :

M+

(g) + I.E.2  M2+

(g) + e – 

M2+(g) + I.E.3  M3+(g) + e – 

M3+(g) + I.E.4  M4+(g) + e – 

Note : I.E.4 > I.E.

3 > I.E.

2 > I.E.

1 always.

It is measured in unit of electron volt (eV) per atom or kilojoule per mole orkilo calorie per mole of atoms.

1 eV/atom = 23.05 kcal mol –1 = 96.49 kJ mol –1

10.2 Factors affecting ionisation potential :(A) Atomic size or radius :

1I.P. Atomic size

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Larger the atomic size, smaller is the I.P. It is due to that the size of atomincreases the outermost electrons lie farther away from the nucleus andnucleus loses the attraction on that electrons and hence can be easily removed.

(B) Screening Effect :Higher is the screening effect on the outermost electrons contains less attractionfrom the nucleus and can be easily removed, which is leading to the lower value

of I.P.

1Screening Effect

I.P.

(C) Nuclear charge :

I.P. increases with the increase in nuclear charge between outermost electronsand nucleus.

eff I.P. Z where, Zeff 

 = Effective nuclear charge

(D) Half filled and fully filled electronic configuration or effect of stable

electronic configuration :According to Hund’s rule, an atom has extra stability which contains exactlyhalf or fully filled electronic configuration. Hence, the removal of an electronfrom that atom needs more energy than expected.

eg. (i) I.E. of N > I.E. of O

(1s22s22p3) (1s22s22p4)

(ii ) All inert gases having fully filled electronic configuration.

(E) Penetration effect of the electrons :

Ionization energy increases as the increase of penetration effect of electronin an atom. In a multielectron atom, the electrons of s-orbital in the same

orbit is much closer to nucleus than the p-orbital. Therefore, s electrons willexperience more attraction than that of p electrons. Hence, their removal isdifficult leading to higher I.E. Like wise I.E. of p-electrons will be more thanI.E. of d-electrons and so on.

In general, the ionisation energy follows the following order :

s p d f orbital of the same orbit.

eg. (i) I.E. of Be > I.E. of B

  (1s22s2) (1s22s22p1)

10.3 Periodicity in ionisation potential :

(A) In a period :

Generally, on moving from left to right along a period, the I.E. increases withincreasing of in atomic number. It breaks, where an electron of a slightlyhigher energy is removed from the same shell.

eg. Element Li Be B C N O F Ne

I.E. 500 900 800 1086 1403 1314 1681 2081

 Thus, the break occur at ‘B’ and ‘O’.

Reason : Higher stability of Half and fully filled orbitals.

(a) Transition elements :

Increase in nuclear charge helps in holding the valence electrons in nsorbitals more firmly. While, the shielding effect of (n – 1)d electrons on nselectron acts in opposite direction. The net result is that there is a slight

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change in I.P.

(B) In a group :

On moving top to bottom in a group the value of I.P. decreases. It is due to thefollowing factors :

(a) The nuclear charge decreases.

(b) Addition of principal energy shell.Exceptional Cases1. IE

1 Be > B but IE

2 of B > Be.

2. IE1 of N > O.

3. For IIIA

B > Al < Ga > In < T 

  poor SE of d poor SE of ‘f’ 

4. For IVA

C > Si > Ge > Sn < Pb

 

Poor screening effect of4f orbital, increases I.E.

5. For transition elements

Generally in period IE inc, but the inc is not so regular (Sc, Ti, V, Cr) differ only slightlyfrom each other and Fe, Co, Ni, Cu values are fairly close to each other from Cu – Znagain inc.

Q. In which of the following have higher difference in the value of IInd and IIIrd IP

(a) 1s2 2s2 2p6 3s2 3p1 (b) 1s2 2s2 2p6 3s2

(c) 1s2 2s2 2p6 3s1 (d) 1s2 2s2 2p6 3s2 3p2

Q. Arrange the following in inc order of IE1

1. Cl, S, Se 2. Si, P, S, Cl 3. Mg, Al, Na, Be

Q. Which has largest fifth IE Ge or AS ?Q. Which has largest IIIrd IE Ga or Ge ?

Q. C N O F – IE2 – O > F > N > C.

Example – 2 : I.E. of one H atom i s 2.18 × 10 –18 J. The I.E. of H atom in kJ mole –1 is

(a) 1505 kJ mole –1 (b) 1310 kJ mole –1

(c) 1608 kJ mole –1 (d) none of these

Solution :182.18 10 J 6.02

I.E.1 atom mole

 × 1023

= 1.31 × 106 J mole –1 = 1310 kJ mole –1

Q. What is the order of electron infinity of the following

N, Be, Ne

Ans. N > Be > Ne

10.4 Applications of Ionisation potential :

(A) Metallic and Nonmetallic character :

M.C. = Metallic character N.M.C. = Non metallic character

1M.C.

I.P.

N.M.C. I.P.

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Lower I.P. causes metallic character of an element.

Higher I.P. causes non metallic character of an element.

(B) Reactivity :

1Reactivity

I.P.Elements which have low I.P. values are more reactive. Thus, alkali metalsare more reactive due to lower I.P.

(C) Reducing capacity :

1Reducing capacity

I.P.

Reducing Elements or reductant : Element which donate its electron veryeasily to another element are called as reducing element or reductant andproperty is called reducing property or capacity of that element.

So, Lower the I.P. – Higher the reducing capacity.(D) Oxidising Capacity : Oxidising Capacity I.P.

Oxidising element or oxidant : Elements which have greater affinity toattract or accept the electrons from other element are called as oxidisingelement or oxidant and this type of property is called oxidising property orcapacity of that element.

So, higher the I.P. –– higher the oxidising capacity.

(E) Types of bond :

(a) Elements which possess low I.P. form ionic bond.

(b) Elements which possess higher I.P. form covalent compound.

Q : Arrange elements in order of increasing oxidising properties.

F, N, O

LECTURE - 4

  11. Electron affinity :

(a) The amount of energy released when an extra electron is added to a neutralgaseous atom to form a monovalent anion is called as electron affinity or first

electron affinity.eg. A(g) + e –   A –  + Energy (EA

1)

(b) It is measured as the capacity of addition of extra electron to an atom. It isexothermic process for, E.A.

I  H = –ve.

(c) E.A.II is endothermic process because both have the same charge with a lot of 

repulsion. Thus the addition of electron of A –   requires energy to remove theforce of repulsion.

H = +ve

A –  + e –  + Energy A –2

(d) Its values are expressed either in electron volt (eV) or KJ mole –1.

(e) Its values are also determined by Born - Haber cycle indirectly.

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11.1 Factors affecting E.A. :

(A) Atomic radius or size of the atom :

1E.A.

Atomic size

E.A. decreases with increase in the size of atom because of the increasein size of the atom, the distance between the last orbit and nucleus alsoincreases. As a result of this, the force of attraction between nucleus andincoming electron decreases.

(B) Nuclear charge :

Nuclear ch arg e E.A.

E.A. increases with increase in the nuclear charge because the force of attraction increase on incoming electron with increase in nuclear charge.

(C) Half filled and fully filled electronic configuration :

Atoms having above configuration is most stable and they do not showtendency to except the extra electron.

eg. E.A. of noble gases are zero.

11.2 Periodicity in electron affinity :

(A) In a period :

When we move from left to right along a period, the atomic size decreasesand nuclear charge increases. Thus the incoming electron attracts strongly.Hence, the value of E.A., increases in a period from left to right.

Exception : N, Ne and alkaline earth metals and alkaline metals.

In Be, Ne and alkaline earth metals = fully filled orbitalsN = Half filled orbitals

(B) In a group :

As we move from top to bottom in a group the value of E.A. decreases becausethe atomic size increases with the nuclear charge but the effect of increasing atomic size is greater than that of effect of nuclear charge. Thus, the force of attraction between the incoming electron and nucleusdecreases.

Exception : The E.A. of 2nd period is lower than E.A. of 3rd period elements.

Reason : In the 2nd period elements incoming electron enters into the 2p

orbital where in 3rd period elements incoming electrons enters into the 3porbital which experience more repulsion. Hence, E.A. (F) < E.A. (Cl).

Note : Halogens have the highest E.A.

Reason : (i) Small size and high nuclear charge.

(i i) To gain noble gases configuration to accept only one electron[ns2np5  ns2np6].

Exceptional Case

1. Cl > F > Br > I

2. S > Se > O

3. F > O > C > B > N > Be4. Cl > S > Si > P

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  FTO(i) In electron gain enthalpy, the minus sign indicates that energy is released. Only

numerical values are taken when periodicity or other properties are considered.

(ii ) In some books, the energy released when an electron is added to an isolated

gaseous atom is termed electron affinity (Ae). However, the electron gain enthalpy

and electron affinity differ slightly in numerical terms. Numerically, electron gainenthalpy is higher than that of electron affinity by 5/2 RT as electron affinity isdefined at absolute zero. The two terms are related to each other as

eg e eg

5H A RT ( H Electron gain enthalpy)

2

  12. ELECTRONEGATIVITY :

(a) The tendency of an atom in a compound or molecule to attract the sharedelectrons towards itself is called as electronegativity of the atom.

(b) Electronegativity and electron-affinity both are measured the electron attractingpower.

Difference between electron affinity and electronegativity.

  Electronegativity Electron affinity

1. It is the capacity of an atom in It is the capacity of an isolated atom

molecule or compound to attract in free state to attract an electron.

the bonded electrons.

2. It is the relative value of an atom. It is the absolute value of an atom.

3. It varies regularly in a period It does not change regularly in aor a group. period or a group.

4. It has no units but merely a Its units are electron volts per

number. atom or kilojoules per mole or

kilocalories per mole.

12.1 Factors affecting electronegativity :

(A) Nuclear attraction : Nuclear attraction Electronegativity

Electronegativity is proportional to the nuclear attraction. As the nuclearattraction on bonded electron increases the electronegativity also increasessimultaneously.

(B) Atomic radius :

1Atomic radius

electronegativity

If atomic radius increases in a molecule or compound, the nuclearattraction decreases on bonded electron. As a result of that theelectronegativity also decreases.

(C) Oxidation state :

(a) The value of electronegativity increases with the increasing value of 

(+)ve oxidation state.

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eg.2 4 7Mn Mn Mn

ve oxidation state

value of electronegativity

(+)ve oxidation state electronegativity

(b) The value of electronegativity decreases with the increasing value of (–)ve oxidation state.

eg.1 2 3A A A

ve oxidation state

value of electronegativity

1ve oxidation state

electronegativity

(D) Percentage s character :

When the percentage s character increases for an atom in a compoundthen the value of electronegativity also increases with them. So,

Percentage s character Electronegativity

percentage – s character for :

3 1sp 100 25 %

4

2 1sp 100 33.33 %

3

1sp 100 50 %

2So, the order of electronegativity as follows :

sp > sp2 > sp3

Q : Arrange the following in the order of their increasing acids strength

HClO, HClO2, HClO

3, HClO

4

12.2 Periodicity in electronegativity :

(A) In a period : As we move from left to right along a period the electronegativityincreases, this is due to decrease in size with increase in nuclear charge.

* The inert gases have zero electronegativity, due to fully filled outermostshell electronic configuration.

* The alkali metals have lowest value of electronegativity where thehalogens have highest value of electronegativity.

(B) In a group :

As we move from top to bottom within a group, the value of electronegativitydecreases due to increase in atomic size.

eg.Elements

F

Cl

BrI

Electronegativity

4.0

3.0

2.82.5

Reason : (a) Less nuclear attraction.

(b) Higher atomic radius.

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* The value of electronegativity of 5d series is greater than that of thevalue of 4d series. It is due to lanthanide contraction, because the nuclearcharges of 5d series elements increase with 32 unit against 4d.

* The highest electronegative element in the periodic table is ‘F’.

* The lowest electronegative element in the periodic table is ‘Cs’ and ‘Fr’.

 The group of elements which show same electronegative character is

following :

Elements

N and Cl

H and P

C, S and I

Cs and Fr

Electronegativity

3.0

2.1

2.5

0.7

(a)

(b)

(c)

(d)

Example – 3 : Give the correct order of electronegativity of central atom in followingcompounds(a) CH

3 – CH

3(b) CH

2 = CH

2(c) CH CH

The correct order is(a) a > b > c (b) c > a > b (c) c > b > a (d) b > c > a

Solution : %–S– character in CH3 – CH

3= sp3 = 25%

in CH2 = CH

2= sp2 = 33.33%

in CH CH = sp = 50%

So, the electronegativity order is :

CH CH > CH2 = CH

2 > CH

3 – CH

3

12.3 Applications of Electronegativity :(A) Metallic and non-metallic character :

Metallic character = M.C.

Non-metallic character = N.M.C.

1M.C.

E.N.N.M.C. E.N.

When E.N. of an element increases, the M.C. decreases and NMC increasesdue to decrease in atomic size.

(B) Bond energy :

Bond energy E.N.

Where, E.N. = electronegativity difference between two bonded atoms.

When the electronegativity difference increases between two bonded atomsthe bond make stronger and the bond energy of that bond increases.

eg.Compound

H – F

H – Cl

H – Br

H – I

 The increasing order of bond energy is :

H – I < H – Br < H – Cl < H – F

(C) Bond length (B.L.) :

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1Bond length

E.N.

When the E.N. increases between two bonded atoms, the bond lengthdecrease due to higher shifting of bonded electrons.

Example – 4 : Which of the following compound has highest value of bond length

(a) CsF (b) CsBr (c) CsI (d) CsCl

Solution :  The E.N. of CsI is minimum.

(D) Nature of bond :

Nature of the bond between two atoms can be decided from theelectronegativity difference of the two bonded atoms.

(a) The difference of ENA – EN

B = 0 i.e. EN

A = EN

B then, the bond is purely

covalent.

(b) The percentage ionic character of a bond more is its stability. Percentageionic character may be calculated by using Henry and Smyth equation :

% Ionic character = 16 (XA – X

B) + 3.5 (X

A – X

B)2

where, XA = Electronegativity of A XB = Electronegativity of B  FTO

(F) Nature of hydroxides :

(a) If XA – X

0 > X

0 – X

H in A – OH type of hydroxide then, A – OH A+ + OH – 

type of bond dissociation occured. Hence, the hydroxide is basic.

eg. Li – OH Li+ + OH –  [XLi = 2.5, X

OH = 1.4]

  (Basic hydroxide)

(b) If XA – X

0 < X

0 – X

H in A – OH type of hydroxide then, A – OH AO –  + H+

type of bond dissociation occured. Hence, the hydroxide is acidic.

eg. Cl – OH C1O – 

 + H+

[XCl = 0.5, XOH = 1.4](Acidic Hydroxide)

(G) Nature of Oxides :

(a) (X0 – X

A) predicts the nature of the oxides formed by the element A.

If X0 – X

A is high means (X

0 – X

A) > 2.3, the oxides shows basic nature.

eg. Na2O

If X0 – X

A is equal to 2.3, the oxide shows amphoteric nature. eg. Al

2O

3

If X0 – X

A is low means (X

0 – X

A) < 2.3, the oxides shows acidic nature.

eg. SO2

(b) In a period from left to right acidic nature of oxides is increased.

eg. Na2

O MgO Al2

O3

SiO2

P2

O5

SO2

Cl2

O

Basic Amphoteric Acidic

(c) In a group from top to bottom the acidic character decreases.

12.4 Measurements of Electronegativity :

(A) Mulliken Scale : Mulliken gave the electronegativity as the average valueof ionisation potential and electron affinity of an atom.

m

I.P. E.A.X

2

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(B) Pauling Scale : Pauling scale is based on value of bond energy.

pX 0.208 E   A B A A B Bor X .208 E E E

Where, E = Resonance energy = Actual bond energy – A BBE E

If KCal/mol is given

IP EA IP EAEN ev/atom (pauling scale)

5.6 23.06 130

If KJ/mol is given

IP EA IP EAEN ev/atom.

5.6 96.46 540

PM   8.2

  FTO

(C) Relation between Mulliken and Pauling :

P mX 0.395 X 0.615

(D) Allred Roshow electron negativity : He defined E.N. as the attractive forcebetween a nucleus and an electron at a distance equal to the covalentradius.

2eff.

AR 2

Z eX

rWhere Z

eff  = Effective nuclear charge

r = Covalent radius of the atom.

(E) Relation between Allred Roshow and Pauling :

eff p 2

ZX 0.359 0.744

r

  FTO

Exceptional Case

1. B Al Ga In Tl

2.0 1.5 1.6 1.7 1.8

 Dream on !!

š›œ•š›œ•


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