This document discusses chemical bonding, including the different types of bonds and how they form. It describes ionic bonds as the electrostatic attraction between oppositely charged ions that forms when electrons are transferred. Covalent bonds form through the overlap of atomic orbitals between atoms, with sigma bonds forming from head-on overlap and pi bonds from lateral overlap. The document also discusses concepts like hybridization, which involves the mixing of atomic orbitals to allow bond angles that match experimental observations. Valence bond theory and molecular orbital theory are introduced as ways to explain bonding at the atomic level.

1. 29/11/2010
TOPIC TWO
CHEMICAL BONDING
29/11/2010 mov 1
CHEMICAL BOND
• This is the force of attraction that binds two or more atoms
together.
• Types of chemical bonds
1) Ionic
2) Covalent
3) Metallic
4) Coordinate
5) Van-derwaal’s
6) Hydrogen
29/11/2010 mov 2
1

2. 29/11/2010
Bond formation
• Why do atoms form chemical bonds ?
so that the system can achieve the lowest possible
potential energy
• Example covalent bonding in H2
H••H
29/11/2010 mov 3
Ionic bond
• This is the electrostatic attraction between oppositely
charged ions. It involves complete transfer of one or more
electrons from a highly electropositive to a highly
electronegative element.
• The atom that gives out electron becomes positively charged,
whereas the one that gains electrons becomes negatively
charged
Cation anion
• Formed between elements with the biggest difference in
electronegativity
Example:
29/11/2010 mov 4
2

3. 29/11/2010
Coulomb’s Law
• Especially prevalent in compounds formed between group
1A and 2A elements with group 6A and 7A elements.
• Ionic bond formation is influenced by the of electrostatic
force of attraction (F), which is directly proportional to
product of charge and inversely proportional to distance
between the ions
F = k Q1xQ2/r2
k = (2.31 x 10-19 J nm)
therefore, strong lattices are favoured when the ions have a
high charge to size ratio
29/11/2010 mov 5
Conditions for formation of ionic bond
1) Low ionisation potential of the metal –the lower the
ionisation potential the greater the ease to form cation
2) High electron affinity of the non-metal-the higher the
electron affinity, the greater the ease to firm anion
3) High lattice energy – the greater the lattice energy the
stronger the bond formed
29/11/2010 mov 6
3

4. 29/11/2010
Characteristics of ionic compounds
1) They have high melting and boiling points due to strong
electrostatic force of attraction between the ions in the solid
2) They are non-volatile due to high melting and boiling points
3) Electrical conductivity:
1) They are poor conductors of electricity in solid state due to strong
electrostatic force of attraction-hence ions are immobile
2) Good conductors of electricity in aqueous state-ions are mobile
3) Good conductors of electricity in molten or fused state
4) Soluble in polar solvents but in soluble in organic solvents
29/11/2010 mov 7
Formation of Ionic Compounds
• Electron affinity: is the energy change that occurs when
an electron is accepted by an atom in the gaseous state
(kJ/mol)
X(g) + e- X-(g)
• The higher the electron affinity, the greater the tendency of
the atom to accept an electron
• Chlorine has the greatest electron affinity of any element:
Cl(g) + e- Cl-(g) ∆H = -349 kJ
29/11/2010 mov 8
4

5. 29/11/2010
Formation of Ionic Compounds
• Lattice energy: this is the energy required to completely
separate one mole of a solid ionic compound into gaseous
ions.
• or the energy released when an ionic solid forms from its
ions
• It has a negative sign (-)
• Calculation of lattice Energy is based on two methods
1) Born-lande Equation (Theoretical method)
2) The Born-Haber Cycle (Experimental method)
29/11/2010 mov 9
Born-lande Equation (Theoretical
method)
• It is based on the coulombic interaction within the ionic
crystal:
1) Electrostatic force of attraction between oppositely charged ions
(cation and anion)
2) Repulsive interaction due to interpenetration of electron charge
clouds
• Based on these two interactions, Born-Lande theoretically
derived the equation for calculating lattice energy.
• The equation is called Born-Lande Equation and is given
by:
• U = -NoAZ+Z-e2 (1-1/n)
ro
29/11/2010 mov 10
5

6. 29/11/2010
Born-lande Equation Cont.
• Where,
No = Avogadro’s number = 6.022x1023 atom-1
A = Madlung constant
Z+ = Charge on cation
Z- = charge on anion
e- = Electronic charge
ro = Sum of radii of cation and anion
n = Born exponent >1
• Significance of Born-Lande Equation
1) From U α Z+Z-, the higher the charge on cation and anion,
the greater the magnitude of lattice energy
e.g. U (LiF) < U(CaF2) < U(MgS)
29/11/2010 mov 11
Born-lande Equation Cont.
2) U α 1/ro
The smaller the size of ions, the higher the lattice
energy e.g.
U (NaF) > U(NaCl) > U(NaBr) > U(NaI)
Exercise:
1) Zinc oxide, ZnO, is a very effective sun screen. How would the lattice energy of ZnO compare
with that of NaCl
2) The precious gem ruby is aluminium oxide, Al2O3, containing traces of Cr3+. The compound
Al2Se3 is used in the fabrication of some semiconductor devices. Which of the two has a
larger lattice energy?
29/11/2010 mov 12
6

7. 29/11/2010
Born-Haber Cycle
• Direct experimental determination of lattice energy is difficult,
hence determined indirectly by a cyclic process called Born-
Haber Cycle.
• Born-Haber cycle relates lattice energies of ionic compounds
with other thermodynamic data such as sublimation energy
(SE), ionization energies (IE), electron affinities (Ea),
dissociation energy (DE) and other atomic and molecular
properties
• The cycle is based on Hess’s law that total amount of heat
evolved or absorbed in a chemical reaction is constant
whether the reaction is carried out in a single step or
multiple steps
29/11/2010 mov 13
Born-Haber Cycle
e.g. ∆Hf = ∆Hs + IP + ½ ∆HD + Ea + U
Ea
IP
½ ∆HD
U
∆Hs
∆Hf
29/11/2010 mov 14
7

8. 29/11/2010
Ionic Bond Formation
• requires Coulombic attractive energy (lattice energy) to be
sufficiently large to overcome ionization energy of the
element that forms the cation.
• balance between energy input (ionization energies) and
stability gained from formation of the solid.
• The main impetus for the formation of an ionic compound
rather than a covalent compound results from the strong
mutual attraction among the ions
29/11/2010 mov 15
Ionic Bond Formation cont.
• Bonding in MgCl2 is ionic; Ionization energies ( I.E.)
Mg(g) Mg2+(g) + 2ebonding
∆H = I.E.1 +I.E.2
= 753 +1435 = +2180 kJ/mol
• bonds in AlCl3 are polar covalent
Al(g) Al3+(g) + 3e-
∆H = I.E.1 + I.E. 2 + I.E. 3
= 580 +1815 + 2740 = +4125 kJ/mol
29/11/2010 mov 16
8

9. 29/11/2010
Ionic Bond Formation cont.
• Bonds in AlCl3 are polar covalent
• energy input (ionization energies) out weighs stability
gained from formation of an ionic solid
Al (g) Al3+(g) + 3e-
∆H = I.E.1 + I.E. 2 + I.E.3
= 580 +1815 + 2740 = +4125 kJ/mol
29/11/2010 mov 17
Application of Born-Haber Cycle
1) To determine lattice energy of unknown crystals
eg. Calculate the lattice energy for formation of NaCl crystal based on the
data below
1) ∆Hf (NaCl) = -99 Kcal/mol
2) ∆Hs (Na) = 26 Kcal/g atom
3) IP (Na) = 117 Kcal/atom
4) ∆HD (Cl2) = 54 Kcal/mol
5) Ea (Cl) = -84 Kcal/mol
(Ans = -185 Kcal/mol)
29/11/2010 mov 18
9

10. 29/11/2010
Application of Born-Haber Cycle
cont.
2) To determine electron affinity of elements which are
difficult to determine by other methods
eg. Determine electron affinity of iodine given the following thermodynamic data:
1) ∆Hf (Nal) = -68.8 Kcal/mol
2) ∆Hs (Na) = 25.9 Kcal/atom
3) IP (Na) = 118.4 Kcal/atom
4) ∆HD (l2) = 25.5 Kcal/mol
5) U (Nal) = -165.4 Kcal/mol
(Ans = -73.2 Kcal/atom)
29/11/2010 mov 19
Application of Born-Haber Cycle
cont.
• Explain why group IIA oxides and chlorides are stable in higher oxidation state
despite the fact that the formation of Mg2+, Ca2+ etc require more energy than
Mg+ and Ca+
Ans. the higher lattice energy involved compensates for
the energy required for formation of Mg2+, Ca2+ making
MgCl2, CaCl2, Al2O3 more stable.
29/11/2010 mov 20
10

11. 29/11/2010
Valence Bond Theory
• It is based on linear combination of atomic orbitals.
• Main features:
1) A covalent bond is formed by the overlap of half filled atomic
orbitals of different atoms
2) Overlapping atomic orbitals must have electrons with opposite
spins
3) The bonded electron pair is localized between the two linked
atoms
4) The stability of covalent bond is due to exchange of valence
electrons between participating atoms, which lowers the
potential energy of the bonded atoms
5) Each atom of the covalent compound tends to acquire a noble
gas configuration by sharing electrons
• s
29/11/2010 mov 21
Bonding Types
• Two types of bonds result
from orbital overlap:
• sigma σ bonds
• from head-on overlap
• lie along the bond axis
• account for the first bond
• Can freely rotate around
bond
29/11/2010 22
11

12. 29/11/2010
Bonding Types
• pi π bonds
• from lateral overlap by
adjacent p or d orbitals
• pi bonds are
perpendicular to bond
axis
• account for the second
and third bonds in a
multiple bond
• Cannot undergo
rotation around bond
29/11/2010 mov 23
Valence Bond Theory cont.
• One pair of electrons can occupy this overlapping area
• Electron density is maximizes in overlapped region
• Example: H2 bonds form because atomic valence orbitals
overlap
• Each hydrogen contributed 1s orbital
1s 1s
29/11/2010 mov 24
12

13. 29/11/2010
Valence Bond Theory
• HF involves overlaps between the s orbital of H and the
2p orbital of F
1s 2s 2p
29/11/2010 mov
25
VB Theory And H2S
• Assume that the
unpaired e- in S and H
are free to form a
paired bond
• We may assume that
the H-S bond forms
between an s and a p
orbital
29/11/2010 mov
26
13

14. 29/11/2010
VB Theory and NH3 cont.
According to Valence Bond Theory:
Which orbitals overlap in the formation of NH3?
• Ground state of nitrogen
2s ↑↓ 2p _↑ ↑ _↑__
mov 27
Difficulties With VB Theory
• Most experimental bond angles do not support those
predicted by mere atomic orbital overlap
• For example: C 1s22s22p2 and H 1s1
• Experimental bond angles in methane are 109.5° and all
are the same
• p orbitals are 90° apart, and not all valence e- in C are in
the p orbitals
• How can multiple bonds form?
29/11/2010 mov
28
14

15. 29/11/2010
Hybridization
• The mixing of atomic orbitals to allow formation of bonds
that have realistic bond angles
• The new shapes that result are called “hybrid orbitals”
• The number of hybrid orbitals required = the number of
bonding domains + the number of non-bonding domains on
the atom
29/11/2010 mov
29
Hybrid between s and p Orbitals
Two sp Orbitals in Linear Arrangement Formed by
Hybridization of a single s and a single p Orbital
29/11/2010 mov 30
15

16. 29/11/2010
Hybrid orbitals
• Naming of the hybrid orbital is based on the combination
of the orbitals used to form the new hybrid
• The name shows what type of atomic orbitals, and how
many of each were used in formation of hybrid orbitals
e.g. sp, sp2, sp3 etc.
• One atomic orbital is used for every hybrid formed
(orbitals are conserved)
29/11/2010 mov
31
Hybrids From s & p Atomic Orbitals
explain VSEPR Geometry
Hybrid Atomic Orbitals Electron Geometry
Used
sp3 s + p x + py + pz Tetrahedral, bond
angles 109.5˚
mov 32
16

17. 29/11/2010
Hybrids From s & p Atomic Orbitals
explain VSEPR Geometry
Hybrid Atomic Orbitals Electron Geometry
Used
sp2 s + p x + py Trigonal planar, bond
angles 120 ˚
29/11/2010 mov 33
Hybrids From s & p Atomic Orbitals
explain VSEPR Geometry
Hybrid Atomic Orbitals Electron Geometry
Used
sp s + px Linear,
bond angles 180 ˚
29/11/2010 mov 34
17

18. 29/11/2010
Hybrid Orbitals in BeH2
• Ground state of Be
2s↑↓ 2p _ ___ [He]2s2
No half filled orbitals available for bonding
• For hybrid sp hybrid orbitals
sp ↑ ↑ 2p ___
• Now bond can form between 1s orbital of hydrogen and
sp hybrid orbital or berylllium
sp ↑↓ ↑↓ 2p ___
29/11/2010 mov 35
29/11/2010 mov 36
18

19. 29/11/2010
Consider the CH4 molecule
• Ground state for carbon
2s ↑ ↓ 2p ↑ _ ↑ ___ [He]2s2 2p2
• Form sp3 hybrid orbitals
↑ ↑_ ↑ ↑ .
• Each sp3 hybrid can now overlap with 1s orbital of
hydrogen
↑↓ ↑↓_ ↓↑ ↑↓ .
29/11/2010 mov
37
Bonding in CH4
• The 4 hybrid orbitals are
evenly distributed
around the C
• The H s-orbitals overlap
the sp3 hybrid orbitals to
form the bonds.
29/11/2010 mov
38
19

20. 29/11/2010
Bonding in NH3
• The 4 hybrid orbitals are
evenly distributed
around the N
• The H s-orbitals overlap
the sp3 hybrid orbitals to
form three bonds bonds
• The remaining lone pair
occupies the last hybrid
orbital
29/11/2010 mov
39
Hybridization for form sp2 orbitals
29/11/2010 mov 40
20

21. 29/11/2010
Ethene and Double Bonds
Sigma Bonds σ : Direct overlap of
orbitals between the two nuclei
Direct overlap of atomic orbitals is not
affected by rotation around that bond
C--C
29/11/2010 mov 41
C=C double bond consists of one sigma
σ bond and one pi π bond
29/11/2010 mov 42
21

22. 29/11/2010
Orbitals used for bonding in Ethene
Sigma bond: sp2 overlaps sp2
Pi bond: unhybridized p orbital overlaps
unhybridized p orbital
29/11/2010 mov 43
Formation of sp Hybrid Orbitals
29/11/2010 mov 44
22

23. 29/11/2010
Hybrid Orbitals is CO2 molecule
First bond is sigma bond
Second bond is pi bond
29/11/2010 mov 45
CO2 Molecule
• Unhybridized p orbitals are used to form pi bonds between
carbon and oxygen
• sp hybrid orbitals of carbon overlap sp2 hybrid orbitals from
oxygen to form sigma bonds
29/11/2010 mov 46
23

24. 29/11/2010
• Triple bond in N2 molecule consists of one sigma bond
and 2 pi bonds
• Sigma bond stems from sp hybrid overlap
• Pi bonds come from unhybridized p orbital overlap
29/11/2010 mov 47
Expanded Octet Hybridization
• Can be predicted from the geometry as well
• In these situations, d orbitals are be needed to provide
room for the extra electrons
• One d orbital is added for each pair of electrons in
excess of the standard octet
29/11/2010 mov
48
24

25. 29/11/2010
Expanded Octet hybridization
• dsp3 hybridization gives rise to trigonal bipyramid
geometry of PCl5
29/11/2010 mov 49
d2sp3 hybridization gives rise to octahedral geometry
29/11/2010 mov 50
25

26. 29/11/2010
Consider SF6
• Ground state for sulfur
3s ↑ ↓ 3p ↑ ↓ ↑ ↑_ 3d _ _ _ _ _
• Six hybrid orbitals needed
sp3d2 ↑ ↑ ↑ ↑ ↑ ↑. 3d _ _ _
• Each sp3d2 hybrid can now overlap with 2p orbital of fluorine
↑↓ ↑↓_ ↓↑ ↑↓ . ↓↑ ↓↑
29/11/2010 mov 51
Bonding is XeF4
• Placing lone pairs at axial positions lets them be as far
as possible from one another
• Square planar geometery
29/11/2010 mov 52
26

27. 29/11/2010
• Hybrid orbital can also hold nonbonding
electrons
• Usually results in polar molecules
29/11/2010 mov 53
Consider the SF4 molelcule
F
Lewis Structure :
F S F
F
• Four bonding + 1 nonbonding pairs around sulphur
• Five hybrid orbitals needed
sp3d ↑↓ ↑ ↑ ↑ ↑ . 3d _ _ _ _
• Four half filled orbitals available to overlap with 2p
orbital of fluorine
sp3d ↑↓ ↓↑ ↓ ↑ ↓↑ ↓ ↑ . 3d _ _ _ _
29/11/2010 mov 54
27

28. 29/11/2010
Geometry of SF4
F • sp3d requires trigonal bipyramid
geometry
• Nonbonding pair goes on equatorial
: position
S F
F
• Distorted tetrahedron geometry
F
29/11/2010 mov 55
Bonding in Ethene C2H4
• Carbon forms sp2 hybrid orbitals, and one
unhybridized p orbital
29/11/2010 mov 56
28

29. 29/11/2010
29/11/2010 mov 57
O
Sigma and Pi Bonding
C
H H
29/11/2010 mov
58
29

30. 29/11/2010
H−C≡C −H
• Each C has
a triple bond
and a single
bond
• Requires 2
hybrid
orbitals, sp
• unhybridize
d p orbitals
used to form
the pi bond
29/11/2010 mov
59
Types of bonds in Acetylene
29/11/2010 mov 60
30

31. 29/11/2010
Summary of Multiple Bonds
• Molecular skeleton held together by σ bonds. First bond
between two atoms always σ.
• Hybrid orbitals are used to form σ bonds, and to hold
nonbonding electrons
• Number of hybrid orbitals needed = # atoms bonded +
# of nonbonding pairs
• π bonds are formed using non-hybridized p or d orbitals
• Double bond is one σ and one π bond
• Triple bond consists of one σ and two π bonds
29/11/2010 mov 61
Molecular Orbital Theory
• Modification of VB theory, considers that the orbitals
may exhibit interference.
• Waves may interfere constructively or destructively
• Bonding orbitals stabilize, antibonding destabilize.
29/11/2010 mov
62
31

32. 29/11/2010
Molecular Orbital Theory concept
1) Atomic orbitals having same energy and symmetry combine
to form molecular orbitals by linear combination of atomic
orbitals (LCAO)
• If ΨA and ΨB are the wave functions of atoms A and B,
then by LCAO, Ψ = ΨA ± ΨB
2) Molecular orbital formed by addition of two atomic orbital
wave functions is called Bonding Molecular Orbital i.e.
Ψb = ΨA + ΨB
Probability of finding electron in BMO is given by
Ψb2 = ΨA2 + ΨB2 + 2 ΨAΨB> ΨA2 + ΨB2
The energy of bonding molecular orbital is less than the
energy of individual atomic orbitals
29/11/2010 mov 63
Molecular Orbital Theory concept
3) Molecular orbital formed by subtraction of two atomic
orbital wave functions is called Anti-Bonding Molecular
Orbital i.e.
Ψa = ΨA + ΨB
Probability of finding electron in ABMO is given by
Ψa2 = ΨA2 + ΨB2 - 2 ΨAΨB< ΨA2 + ΨB2
The energy of anti-bonding molecular orbital is higher
than the energy of individual atomic orbitals
29/11/2010 mov 64
32

33. 29/11/2010
Molecular Orbital Theory concept
4) Molecular Orbitals that do not participate in bonding are
called Non Bonding molecular orbitals. They occur at same
energy as individual atomic orbitals
5) Atomic orbitals are monocentric, whereas molecular orbitals
are polycentric
6) The Stability of the bond is expressed in terms of bond
order- defined as one half the difference between the
number of electrons in bonding molecular orbitals (Nb) and
atibonding molecular orbitals (Na) i.e.
Bond Order = (Nb – Na)/2
29/11/2010 mov 65
Molecular Orbital Theory concept
• Bond order my be zero, fraction or positive but not
negative
• The greater the bond order, the stronger the bond
formed, hence the greater the stability of the molecule
BO α Bond Strength α Stability α -1/bond length α
1/reactivity
29/11/2010 mov 66
33

34. 29/11/2010
Molecular orbital Energy Diagram
• This is a potential energy diagram showing the atomic
orbitals combining and molecular orbitals formed.
• Electrons fill molecular orbitals in order of increasing energy
levels
• Sequence of molecular orbital energies is based on light (H
to N) and heavy (≥ O )molecules
• Light molecules: σ1s< σ*1s< σ2s< σ*2s< π2px = π2py<
σ2pz< π*2px = π*2py<σ*2pz
• Heavy Molecules: σ1s< σ*1s< σ2s< σ*2s< σ2pz< π2px =
π2py< π*2px = π*2py<σ*2pz
29/11/2010 mov 67
Differences between bonding and anti-
bonding molecular orbitals
BMO ABMO
1. BMO is formed by the addition of 1. ABMO is formed by the subtraction of
overlapping atomic orbitals i.e. overlapping atomic orbitals
Ψmo= ΨA+ΨB Ψ*mo= ΨA-ΨB
2. It has greater electron density in the 2. It has lower electron density in the
region between the two nuclei of bonded region between the two nuclei of the
atoms atoms
3. Electrons in BMO contribute to 3. Electrons in ABMO contribute to
attraction between the two atoms repulsion between the two atoms
4. It possesses lower energy than 4. It possesses higher energy than
associated atomic orbitals associated atomic orbitals
29/11/2010 mov 68
34

35. 29/11/2010
MO diagram for H2
• Show atomic energy level
diagram for each atom
• Show molecular orbitals
(bonding and antibonding*)
• 1 MO for each Atomic orbital.
• Show electron occupancy of
the orbitals.
29/11/2010 mov 69
Filling MO diagrams
1. Electrons fill the lowest-energy orbitals that are
available.
2. No more than two electrons, with spins paired, can
occupy any orbital.
3. Electrons spread out as much as possible, with spins
unpaired, over orbitals that have the same energy.
(# bonding e - ) − (# antibonding e - )
Bond Order =
2 electrons/bond
29/11/2010 mov
70
35

36. 29/11/2010
H2 vs He2
2−0
H 2 Bond Order = =1
2
2−2
He 2 Bond Order = =0
2
29/11/2010 mov 71
Molecular Orbitals Using p Orbitals
Two boron atoms have one set of p orbitals that
can directly overlap to for sigma bond.
Two parallel p orbitals can form pi bonds
29/11/2010 mov 72
36

37. 29/11/2010
• Two px orbitals overlap for form sigma bonding
and antibonding molecular orbitals
29/11/2010 mov 73
• Two p orbitals overlap to
form pi bonding and anti-
bonding orbitals
• Can happen both to py
pair and to pz pair,
resulting in two bonding
and two anti-bonding
orbitals
29/11/2010 mov 74
37

38. 29/11/2010
Molecular Orbital Diagram for B2
Total Bonding - Total antibondin g
Bond Order =
2
4-2
= =1
2
B2 should be a stable
molecule
29/11/2010 mov 75
Diatomic MO diagrams differ by group
Second period used s and
A) I - V p orbitals B) VI-VIIIA
29/11/2010 mov
76
38

39. 29/11/2010
Molecular Orbitals Explains Paramagnetic O2
• Paramagnetic; weakly attracted to magnetic field
• Usually a result of unpaired electron
• Simple Lewis structure has no unpaired electrons
• However, MO treatment shows two unpaired
electrons in π* orbitals
29/11/2010 mov 77
Molecular Orbital Diagrams for B2 to F2
29/11/2010 mov 78
39

40. 29/11/2010
MO Diagram for Group I-V
Draw the MO
diagram for N2
π∗2p
σ∗2p
2p σ2p 2p
π2p
σ∗2s
2s
2s
σ2s
29/11/2010 mov 79
MO Diagram for Group VI-VIII
Draw the MO
diagram for O2 σ∗2p
π∗2p
2p 2p
π2p
σ2p
σ∗2s
2s
2s
σ2s
29/11/2010 mov 80
40

41. 29/11/2010
Draw the MO
MO also works for
diagram for NO heteronuclear
diatomics
2p 2p
2s
2s
nitrogen
29/11/2010 mov
oxygen 81
MO’s and Free Radicals NO
σ∗2p
Free radicals are
molecules with
π∗2p an unpaired
electron
2p
σ2p
2p
π2p
σ∗2s
2s
2s
σ2s oxygen
nitrogen
29/11/2010 mov 82
41

42. 29/11/2010
Resonance Structures for O3 and
NO3
29/11/2010 mov 83
Figure (a) benzene molecule (b) two
resonance structures for benzene molecule
29/11/2010 mov 84
42

43. 29/11/2010
Delocalized Electrons
• Lewis structures use resonance to explain that the
actual molecule appears to have several equivalent
bonds, rather than different possible structures
• MO theory shows the electrons being delocalized in
the structure
29/11/2010 mov
85
Calculation of Hybridization in a molecule
• Steps:
1) Calculate the total number of valence electrons
2) Calculate the number of duplet or octate
Duplet = Total valence electrons/2
Octet = Total valence electrons/8
3) Calculate the number of lone pairs of electrons
Lone pairs = (Total valence electrons – 2xNumber of
duplets)/2
Lone pairs = (Total valence electrons – 8xNumber of
duplets)/8
4) Calculate the number of orbitals used = No. Of duplets +
No. Of lone pairs or electrons
5) If there are no lone pairs of electrons, the structure and
shape are ideal, else distortion in shape occurs
29/11/2010 mov 86
43

44. 29/11/2010
Calculation of Hybridization in a
molecule
1: Calculate the type of hybridisation, geometry and
shapes of the following molecules: H2O, SO42- ions
H2O
i) Total No. Electrons = 1x2+6 = 8
ii) Required number of orbitals = 2
iii) Required electrons for duplets = 2x2 = 4
iv) No. Of lone pair electrons = (i)-(iii)/2 = 8-4/2 = 2
v) Number of orbitals = (ii) + (iv) = 2+2 = 4
• Thus, oxygen exhibits SP3 hybridisation and its structure is tetrahedral
with distortion bond angle due to presence of lone pairs
29/11/2010 mov 87
Calculation of Hybridization in a
molecule
SO42-
i) Total No. Electrons = 6+6x4+2 = 32
ii) Required number of orbitals = 4
iii) Required electrons for octet = 4x8 = 32
iv) No. Of lone pair electrons = ((i)-(iii))/2 = (32-32)/2 = 0
v) Number of orbitals = (ii) + (iv) = 4+0 = 4
• Thus, SO42- involves SP3 hybridisation and its structure is a
regular tetrahedron
29/11/2010 mov 88
44

45. 29/11/2010
Calculating the percentage s- and p-
character
• It is based on the equation:
Cosθ = s/s-1 = p-1/s, where θ = bond angle
Example: Calculate the percentage s-orbital and p-orbital character of a hybrid
orbital if the bond angle between the hybrid orbital is 105o
Cosθ = s/s-1 = p-1/s, where θ is the bond angle
s/s-1 = cos 105o = cos (90 + 15) = -sin 15o = -0.2588
s = 0.2588/1.2588 = 0.2056
similarly, p-1/p = cos 105o = -0.2588
p = 0.7944
29/11/2010 mov 89
Calculating the number σ and π
bonds
• Steps:
1) The number of π electrons in a noncyclic molecule which
does not contain hydrogen atom is given by 6n+2-v, where
n = total number of atoms of molecules and v is the sum of
the valence electrons in the molecule
2) The number of π electrons in a noncyclic molecule
containing hydrogen atoms is given by 6n-6q+2-v, where q
= the number of hydrogen atoms in the molecule
3) If the molecule is of Ax n type, and none of the them is a
hydrogen atom, then,
Number of σ bonds = Total No. Valence electrons/8
4) If the molecule contains a hydrogen atom, then,
Number of σ bonds = total no. Valence electrons/2
29/11/2010 mov 90
45

46. 29/11/2010
Calculating the number σ and π bonds
• Example: Calculate the number of π and σ bonds in the following molecules:
CO2, CO32-, H2O, NH3, C2H2, C2H4
CO2
No. Of π electrons = 6x3+2-(4+2x6) = 4
No. π bonds = 4/2 = 2
No. σ bonds = (4+2x6)/8 = 2
C2H4
No. Of π electrons = 6x6-6x4+2-(2x4+4) = 2
No. π bonds = 2/2 = 1
No. σ bonds = (2x4+4x1)/2 = 6
29/11/2010 mov 91
Calculating the number σ and π bonds
• Calculate the number of π and σ bonds in the following molecules: CO2, CO32-,
H2O, NH3, C2H2, C2H4
H2O
No. Of π electrons = 6x3+6x2+2-8 = 0
No. π bonds = 0
No. σ bonds = (2+6)/2 = 4
CO32-
No. Of π electrons = 6x4+2-(4+3x6+2) = 2
No. π bonds = 2/2 = 1
No. σ bonds = (4+3x6+2)/8 = 3
29/11/2010 mov 92
46