This document discusses chemical bonding, including ionic bonds, covalent bonds, and molecular orbital theory. It covers the factors that influence ionic bond formation, such as ionization energy and lattice energy. Born-Haber cycles are used to calculate the lattice energy of ionic compounds like NaCl. The document also discusses polar covalent bonds and how electronegativity differences between atoms determine bond polarity. Hybridization of atomic orbitals and VSEPR theory are introduced.

1. CHEMICAL BONDING - I
Dr V. LATHA
AP / CHEMISTRY
SRI SRNM COLLEGE, SATTUR
www.srnmcollege.ac.in

2. SYLLABUS
2
UNIT II CHEMICAL BONDING - I
 Ionic bond – factors influencing the formation of ionic bond –
Lattice energy – Born Haber cycle. Calculation of Lattice energy for
NaCl.
 Covalent character of ionic bond – Polarization of ions – Fajan‘s
rules – Effects of Polarization.
 Theories of covalent bond – Valence bond approach – Postulates –
types of overlap of atomic orbitals-MO theory of H2, O2, N2.
 Atomic orbitals-s-s overlap ,s-p overlap p-p overlap sigma bond and
π bond –examples of oxygen and nitrogen Molecules -comparison
between σ and π bonds. Hybridization – sp, sp2,sp3, sp3d and
sp3d2hybridization with examples.-BeCl2 , BCl3 , CH4, PCl5 and SF6
 VSEPR theory – shapes of simple molecules such as CH4, NH3and
H2O.
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

3. OVERVIEW
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Ionic bond – Lattice energy – Born Haber cycle.
 Covalent character of ionic bond – Fajan‘s rules
 Theories of covalent bond – VB and MO theory
 Overlapping of atomic orbitals.
 Hybridization – sp, sp2,sp3, sp3d and sp3d2
hybridization
 VSEPR theory –CH4, NH3and H2O.

4. CHEMICAL BOND
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Atoms or ions are held together in molecules or
compounds by chemical bonds.
 A bond is a force that holds groups of two or more
atoms together and makes them function as a unit.
Why and How atoms attach together - This will
help us understand how to:
1. Predict the shapes of molecules.
2. Predict properties of substances.
3. Design and build molecules with particular
sets of chemical and physical properties.

5. CHEMICAL BOND
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 All chemical reactions involve breaking of some bonds and
formation of new ones which yield new products with different
properties.

6. Three Major Types of Bonding
 Ionic Bonding
 forms ionic compounds
 transfer of valence e-
 Metallic Bonding
 Covalent Bonding
 forms molecules
 sharing of valence e-
 This is our focus this chapter
6 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

7. Chemical Bonds - Three types:
 Ionic
Electrostatic attraction
between ions
Covalent
Sharing of
electrons
Metallic
Metal atoms
bonded to
several other
atoms
7 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

8. Ionic Bonding Lattice energy – Born Haber cycle
 When a metal and a non-metal get together.
 Bond formed between two ions by the transfer of
electrons.
 Always formed between metal cations and non-metals
anions
 The oppositely charged ions stick like magnets
[METALS ]
+ [NON-METALS ]
-
Lost e-
Gained e-
8 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

9. Formation of Ions from Metals
 Ionic compounds result when metals react with
nonmetals
 Metals lose electrons to match the number of valence
electrons of their nearest noble gas
 Positive ions form when the number of electrons are
less than the number of protons
Group 1 metals  ion 1+
Group 2 metals  ion 2+
 Group 13 metals  ion 3+
9 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

10. Formation of Sodium Ion
Sodium atom Sodium ion
Na  – e  Na +
2-8-1 2-8 ( = Ne)
11 p+ 11 p+
11 e- 10 e-
0 1+
10 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

11. Formation of Magnesium Ion
Magnesium atom Magnesium ion

Mg  – 2e  Mg2+
2-8-2 2-8 (=Ne)
12 p+ 12 p+
12 e- 10 e-
0 2+
11 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

12. Some Typical Ions with Positive
Charges (Cations)
Group 1 Group 2 Group 13
H+ Mg2+ Al3+
Li+ Ca2+
Na+ Sr2+
K+ Ba2+
12 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

13. Learning Check
A. Number of valence electrons in aluminum
1) 1 e- 2) 2 e- 3) 3 e-
B. Change in electrons for octet
1) lose 3e- 2) gain 3 e- 3) gain 5 e-
C. Ionic charge of aluminum
1) 3- 2) 5- 3) 3+
13 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

14. Solution
A. Number of valence electrons in aluminum
3) 3 e-
B. Change in electrons for octet
1) lose 3e-
C. Ionic charge of aluminum
3) 3+
14 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

15. Learning Check
Give the ionic charge for each of the following:
A. 12 p+ and 10 e-
1) 0 2) 2+ 3) 2-
B. 50p+ and 46 e-
1) 2+ 2) 4+ 3) 4-
C. 15 p+ and 18e-
2) 3+ 2) 3- 3) 5-
15 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

16. Ions from Nonmetal Ions
In ionic compounds, nonmetals in 15, 16, and 17
gain electrons from metals
Nonmetal add electrons to achieve the octet
arrangement
Nonmetal ionic charge:
3-, 2-, or 1-
16 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

17. Fluoride Ion
unpaired electron octet
    1 -
: F  + e : F :
   
2-7 2-8 (= Ne)
9 p+ 9 p+
9 e- 10 e-
0 1 -
ionic charge
17 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

18. Ionic Bond - Recap
 Between atoms of metals and nonmetals
with very different electronegativity
 Bond formed by transfer of electrons
 Produce charged ions all states.
Conductors and have high melting point.
 Examples; NaCl, CaCl2, K2O
18 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

19. 19 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

20. • Ionic bond – electron from Na is transferred to Cl,
this causes a charge imbalance in each atom.
• The Na becomes (Na+) and the Cl becomes (Cl-),
charged particles or ions.
20 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

21. 21 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

22. Factors influencing Ionic Bond
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1. Ionization energy: (LOW)
 The lesser the ionization energy, the greater is the ease of the
formation of a cation.
2. Electron affinity: (HIGH)
 The higher the energy released during this process, the easier will be
the formation of an anion.
3. Lattice energy: (HIGH)
 The energy required to completely separate a mole of a solid ionic
compound into its gaseous ions.
(OR)
 It is defined as the amount of energy released when cations and
anions are brought from infinity to their respective equilibrium sites
in the crystal lattice to form one mole of the ionic compound.
 The higher the lattice energy, the greater is the tendency of the
formation of an ionic bond.
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

23. Lattice Energy
Eel = 
Q1Q2
d
23 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

24. Lattice Energy
 Lattice energy, then, increases with the charge on the
ions.
• It also increases with decreasing size of ions.
24 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

25.  NaCl – I gp
 MgCl2 - II gp >
 NaF >
 NaCl >
 NaBr>
 NaI>
Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur
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26. Energetics of Ionic Bonding
By accounting for all
three energies
(ionization energy,
electron affinity, and
lattice energy), we can
get a good idea of the
energetics involved in
such a process.
Na(s) + 1/2Cl2(g) -----> NaCl(s)
26 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

27. Born – Haber Cycle – Formation of
Sodiumchloride
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Step 1: Conversion from solid to gaseous (ΔsubH) –
Sublimation = 180.7KJ/mol
 Step 2: Gaseous sodium atom ionize to sodium ion -
(ΔionH) – Ionization = 493.8 KJ/mol
 Step 3: Dissociation of gaseous chlorine molecule into
chlorine atoms (ΔdissH) – Dissociation = 120.9 KJ/mol
 Step 4: Gaseous chlorine atom ionize to chloride ion -
(ΔionH) – Ionization = -379.5 KJ/mol
 Step 5: NaCl is formed from sodium ions and Chloride
ions. (ΔlatH) – Lattice Energy = - 754.8 KJ/mol

28. Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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29. Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 So the overall energy change can be
computed by sum of all energy changes:
ΔH = 180.7+ 493.8 + 120.9 –
379.5 – 754.8 = -410.9 KJ/mol

30. RECAP
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 The formation of ionic compound occur by:
Low ionization energy of the metal
High electron affinity of non metal
High Lattice energy

31. Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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32. Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Covalent character of ionic bond –
Polarization of ions – Fajan’s
rules – Effects of Polarization.

33. Equal Sharing of electrons
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Br – Br
 N – N
 O- O
 F- F
 C-C
 H -Cl

34. Types of Chemical Bonds:
Polar Covalent Bonding
 Polar covalent bonding occurs when there is an
unequal sharing of electrons.
 Example: HF, which has an unequal charge
distribution or a partial charge. Electrons are
more attracted to F.
 The partial charge is represented with a delta,
δ.
34 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

35. Bond Polarity
NaCl
HCl
Cl2
Ionic and covalent bonding are the extremes of types of bonding.
Ionic > Polar Covalent > Non-Polar Covalent
35 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

36. Electronegativity & Polarity
 The polarity of a bond depends on the difference between the
electronegativity values of the atoms forming the bond.
Difference of Zero Non-polar Covalent Bond
Intermediate Difference
(Between 1.0 and 2.0)
Polar Covalent Bond
Large Difference
(greater than 2.0)
Ionic Bond
36 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

37.  Covalent bond, sharing
electrons,
 But electron sharing not
always equal.
• Fluorine pulls harder on the shared electrons
than hydrogen does.
• Therefore, the fluorine end has more electron
density than the hydrogen end.
• But how do you know who pulls hardest?
37 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

38. Polar Covalent Bonds
 When two atoms share electrons
unequally, a bond dipole results.
 The dipole moment, , produced
by two equal but opposite
charges separated by a distance,
r, is calculated:
 = Qr
 It is measured in debyes (D).
38 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

39. Polar Covalent Bonds
The greater the
difference in
electronegativity,
the more polar is
the bond.
39 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

40. Covalent Bond Strength
 The strength of a bond is measured by determining
how much energy is required to break the bond.
 This is the bond enthalpy.
 The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is 242 kJ/mol.
H = 242 kJ/mol
40 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

41. Average Bond Enthalpies
 Average bond enthalpies are positive, because bond
breaking is an endothermic process.
41 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

42. Average Bond Enthalpies
NOTE:
These are average bond
enthalpies, not absolute
bond enthalpies; the
C—H bonds in methane,
CH4, will be a bit
different than the
C—H bond in
chloroform, CHCl3.
42 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

43. Enthalpies of Reaction
• Can use bond enthalpies to
estimate H for a reaction
Hrxn = (bond enthalpies of
bonds broken) 
(bond enthalpies of bonds
formed)
This is a fundamental idea in chemical reactions. The heat
of a reaction comes from breaking bonds and remaking
bonds.
43 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

44. Enthalpies of Reaction
CH4(g) + Cl2(g) 
CH3Cl(g) + HCl(g)
In this example, one
C—H bond and one
Cl—Cl bond are broken;
one C—Cl and one H—Cl
bond are formed.
44 Dr.V.Latha, AP/Chemistry, Sri. SRNM
College, Sattur

45. Enthalpies of Reaction
So,
Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)]
= (655 kJ)  (759 kJ)
= 104 kJ
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
45 Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur

46. FAJAN’S RULE
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Ionic compounds possess some covalent character, which
depends on following factors:
 Greater charge on cation, more is covalent
character.
 Smaller cation or bigger anion shows
greater covalent character.
 d-elements forms more of covalent
character compounds.

47. Covalent character increases in compounds with
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Greater charge on cation
 Smaller cation
 Larger anion
 D- block element

48. FAJAN’S RULE
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Polarising power
 It is the extent to which a cation can polarise an anion.
 It is proportional to charge density. = Charge/volume
Charge density is the ratio of charge to volume.
 More the charge density, greater is the polarising power
for that cation.
Polarisability
 It is the extent to which an ion can be polarised.
 Polarisation is the distortion of a spherically symmetric
electron cloud to an unsymmetric cloud.

49. Postulates of Fajans’ Rule
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 Size of the ion: Smaller the size of cation, the larger the size of
the anion, greater is the covalent character of the ionic bond.
 The charge of Cation: Greater the charge of cation, greater is
the covalent character of the ionic bond.
 Electronic configuration: For cations with same charge and
size, the one, with (n-1)dn nso which is found in transition
elements have greater covalent character than the cation with
ns2 np6 electronic configuration.

50. Explanation of Fazans’ Rule
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Rule 1:
 The first rule speaks about the polarising power of
the cation.
 If the cation is smaller, then we can say that the
volume of the ion is less.
 If the volume is less, we can conclude that
the charge density of the ion would be high.
 Since the charge density is high, the polarising
power of the ion would be high.
 This makes the compound to be more covalent.

51. Explanation of Fazans’ Rule
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Rule 2:
 The second rule speaks about the polarizability of the
anion.
 Larger the anion, less is the effective nuclear charge.
 Since the last electron is loosely bound in large
anions, it can easily be polarised by a cation, thereby
making the compound more covalent.

52. Explanation of Fazans’ Rule
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
52
Rule 3:
 For cations with same charge and size, the one, with
(n-1)dn nso which is found in transition elements have
greater covalent character than the cation with
ns2 np6 electronic configuration.
 Example:
 HgCl2 and CaCl2 - Hg2+ and Ca2+ are of almost equal size.
 The electronic configuration of Hg2+ is 6s0 5d10.
 This configuration is called pseudo-octet because d-orbital
is fully filled, but the element does not have 8 electrons or
an octet.

53. Explanation of Fazans’ Rule
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 We know that d orbitals are not good at shielding, so we can
say that the anion (Cl–) would be more polarised because the d
orbital is poor at shielding making HgCl2 more covalent than
CaCl2 because Ca2+ ion has a noble gas configuration.
 Since the anion is the same, we have to compare the cations.
According to Fajans’ rules, smaller the cation, more is the
covalency. Therefore, LiCl is the most covalent.

54. Example: Which compound contains more
covalent character ?
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 AlF3 , AlI3 –
 LiI, LiCl, LiBr, LiF
 LiF, NaF, KF

55. Aluminum Iodide (AlI3)
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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 This is an ionic bond which was formed by transfer of
electrons.
 The iodine being bigger has a lesser effective nuclear
charge. Thus, the bonding electrons are attracted lesser
towards the Iodine nucleus.
 On the contrary, the aluminium having three positive
charges attracts the shared pair of electrons towards itself.
 This leads to insufficient charge separation for it to be
ionic and so it results in the development of covalent
character in AlI3.

56. Aluminium Fluoride (AlF3)
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 This is an ionic bond which was also formed by transfer of
electron.
 But here the fluorine being smaller attracts the shared pair
of an electron more towards itself and so there is sufficient
charge separation to make it ionic.

57. QUIZ TIME
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Illustration 1:
Which compound should theoretically the most ionic and
the most covalent amongst the metal halides?
Solution:
 The smallest metal ion and the largest anion should
technically be the most covalent
 Therefore, LiI is the most covalent.
 The largest cation and the smallest anion should be the
most ionic. Therefore, CsF should be the most ionic.

58. Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Illustration 2: Arrange the following according to the
increasing order of covalency:
 NaF, NaCl, NaBr, NaI
 LiF, NaF,KF,RbF,CsF
Solution:
 1. Since the cation is the same, compare the anions.
Amongst the anions, larger the size more would be the
covalency. Therefore the order is: NaF < NaCl < NaBr <
NaI
 2. Here the anion is the same, so we compare with cations.
Smaller the cation more is the covalency. Therefore, the
order is: CsF < RbF < KF < NaF < LiF

59. FAJAN’S RULE- RECAP
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur
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Ionic Characteristic Covalent Characteristic
Large Cation Small Cation
Small Anion Large Anion
Small-charge Large Charge
Fajans’ Rule can be summarized as:

60. 60
THANK YOU
Dr.V.Latha, AP/Chemistry, Sri. SRNM College, Sattur