Molecular geometry and chemical bonding

-Dr. Damodar Koirala
Amar Singh Model Secondary School
H
109.4o
O
1
C
H
H
H
109.4
S
H
H 92o
O
H
H 104.5o
N
H
H H
107.5o
..

 VSEPR theory
 Valance bond theory
 Hybridization theory
 Structure of different molecules
Dr. Damodar Koirala | koirala2059@gmail.com
2

 Valence Shell Electron Pair Repulsion theory
 Proposed by Sidgwick and Powell
 Developed by Gillespie and Nyholm in 1957
 Theory:
 Theory:
• “The electron pairs arrange themselves so that they
are as far apart as they can be. Depending on the
number of electrons pairs in molecule, it will have a
different shape.”
3

 This theory is based on the idea that the geometry of a
molecule is mostly determined by the repulsion among
the pairs of electrons around the central atom
 NOTE:
• 1bond = 1 pair electron
• 1bond = 1 pair electron
• 1 lone pair (non-bonding) = 1 pair electron
4
Molecules Central atom lp bp total
H2O O 2 2 4
BeH2 Be 0 2 2
BF3 B 0 3 3
NH3 N 1 3 4

1. The shape of molecule depends upon the number of
valence shell electron pairs (bonded or non-bonded)
around the central atom
2. Pairs of electrons is the valence shell repel one
another since their electron clouds are negatively
another since their electron clouds are negatively
charged
3. These pairs of electrons tends to occupy such space
that minimize repulsion and thus maximize distance
between them
5

4. Molecules or ions in which the central atom has
2,3,4,5 and 6 electron pairs have linear, trigonal
planar, tetrahedral, trigonal bipyramidial and
octahedral geometry respectively
5. The repulsive interaction of electron pairs decrease in
5. The repulsive interaction of electron pairs decrease in
the order lp-lp > lp-bp > bp-bp. Hence the presence
of lone pair of electrons distorts the regular geometry
of a molecule
6

6. The magnitude of repulsion between electron pair
also depends on electronegativity difference between
the central atom and the atom attached to it, which
further distort the regular molecular geometry as:
a. The increase in EN difference by increase of EN of
a. The increase in EN difference by increase of EN of
the central atom increases the bond angle
b. The increase in EN difference by increase of EN of
the other atom attached with it decreases the bond
angle
7. A multiple bond is treated as if it s a single bond or
single electron pair
7

 Determining the shape of regular molecule (molecules
with no lone pair of electrons)
Step 1: Draw the molecule using lewis notation
Step 2: Count the number of electron pairs around the
central atom
central atom
Step 3: Determine the basic geometry using point
number 4 of postulate, which is
“Molecules or ions in which the central atom has
2,3,4,5 and 6 electron pairs have linear, trigonal planar,
tetrahedral, trigonal bipyramidial and octahedral
geometry respectively”
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 Be is the central atom with two electron pairs around it
 According to VSEPR theory , to be a minimum
repulsion in between two electron pairs and to be in
maximum stable these bond pair of electron should be
directed in equal and opposite direction.Therefore,
BeCl2 has a linear structure with bond angle
9
Be Cl
Cl
Lewis structure
Linear structure
Be Cl
Cl
180o

 B is the central atom with three electron pairs around it
repulsion in between three electron pairs and to be in
directed in equal and opposite direction.Therefore, BF3
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directed in equal and opposite direction.Therefore, BF3
has a trigonal planar structure with bond angle 120o
B F
F
F
B
F
F
F
120o
Lewis structure Trigonal planar
structure

 C is the central atom with four electron pairs around it
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CH4 has a tetrahedral structure with bond angle 109.4o
C H
H
H
H
Lewis structure Tetrahedral structure
C
H
H
H
H
109.4o

 P is the central atom with two electron pairs around it
 According to VSEPR theory , to be a minimum repulsion
in between two electron pairs and to be in maximum
stable these bond pair of electron should be directed in
equal and opposite direction.Therefore, PCl5 has a
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equal and opposite direction.Therefore, PCl5 has a
triangular bipyramidal with bond angles 90o and 120o
P
Cl
Cl
Cl
Cl
Cl P
Cl
Cl
Cl
Cl
Cl
120o
90o
Lewis structure
Triangular bipyramidal

 S is the central atom with two electron pairs around it
directed in equal and opposite direction.Therefore, SF6
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directed in equal and opposite direction.Therefore, SF6
has a square bipyramidal (octahedral) structure with
bond angle 90o and 120o
S
F
F
F F
F
F
Lewis structure octahedral
S
F F
F
F
F F
120o
90o

 Determining the shape of regular molecule
Step 1: Draw the molecule using lewis notation
Step 2: Count the number of electron pairs (both lp and
bp) around the central atom
bp) around the central atom
Step 3: Determine the expected geometry considering
all electron pair as bond pair and then determine the
actual geometry using postulates point 4 and the
distorted geometry using point 5 and 6
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 N is the central atom with four electron pairs around it.
Among them 3 pairs are bp and 1 is lp.
 According to VSEPR theory, the expected geometry is
tetrahedral but due to the greater repulsion between lp
and bp the tetrahedral structure of ammonia is
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and bp the tetrahedral structure of ammonia is
distorted with bond angle 107.5o
N
H
H H
Lewis structure Distorted tetrahedral
..
N
H
H H
107.5o
..

 O is the central atom with four electron pairs around it.
Among them 2 pairs are bp and 2 pairs are lp.
tetrahedra but due to the greater repulsion between lp-
lp and also between lp-bp the tetrahedral structure of
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water is distorted with bond angle 104.5o
O H
H
O
H
H 104.5o

 S is the central atom with four electron pairs around it.
Among them 2 pairs are bp and 2 pairs are lp.
tetrahedra but due to the greater repulsion between lp-
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hydrogen sulphide is distorted with bond angle 92o
S H
H S
H
H 92o

 The central atoms in both molecule are connected to 3
hydrogen atoms.The difference in structure can be
attributed to the number on valance electrons in
central atom.
 The valance shell of the central atom of BH3 has 3 pair
 The valance shell of the central atom of BH3 has 3 pair
of electrons where as that of NH3 has 4 pair of
electrons.
 According to VSEPR theory,“Molecules or ions in
which the central atom has 3 and 4 electron pairs have
trigonal planar and tetrahedral geometry respectively”
 Hence, BH3 is trigonal planner and NH3 is tetrahedral.
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 Both molecules have same central atom (N) with same
number of valence electron and same number of
bonding and non-bonding electrons.The difference in
structure can be attributed to the electronegativity of
surrounding atom.
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 The surrounding atom in NH3 is hydrogen atom and
that in NCl3 is chlorine atom. Also, EN(Cl) > EN(H)
 According to VSEPR theory,“The increase in EN
difference by increase of EN of the other atom attached
with it decreases the bond angle”
 Hence, both have distorted tetrahedral geometry but
the bond angle H-N-H is greater that Cl-N-Cl.

 Both molecules have central atoms with same number
of valence electron and same number of bonding and
non-bonding electrons. In addition the surrounding is
also same.The difference in structure can be attributed
to the electronegativity of central atom.
 The central atom in H2O is oxygen atom and that in H2S
is sulphur atom. Also, EN(O) > EN(S)
 According to VSEPR theory,“The increase in EN
difference by increase of EN of the central atom
increases the bond angle”
the bond angle H-O-H (104.5o)is greater that in case of
H-S-H (92o)
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 Both molecules have central atoms with same number of
valence electron and same surrounding atoms.The
difference in structure can be attributed to the difference in
number of bonding and non-bonding electrons.
 The central atom in H2O has 2 pair of non-bonding electrons
and that in NH3 has 1 pair of non-bonding electrons.
 According to VSEPR theory,“ the extent of electron pair
repulsion is lp-lp > lp-bp > bp-bp”. i.e H2O has more
repulsion than in NH3
 Hence, both have distorted tetrahedral geometry but the
bond angle H-O-H (104.5o)is smaller that in case of H-N-H
(107.5o)
21

 VSEPR theory gives the geometry of simple molecules
 It does not explain them and also has limited
application
 To overcome this, two important theories based on
quantum principle are introduced
quantum principle are introduced
• Valance bond
• Molecular orbital (not studied)
22

 According to this theory a covalent bond is formed
between two atoms by the overlap of half filled valence
atomic orbital's of each atom containing one unpaired
electron.
 Two types of overlapping: Sigma bonds occur when the
orbitals of two shared electrons overlap head-to-head
orbitals of two shared electrons overlap head-to-head
and pi bonds occur when two orbitals overlap when
they are parallel.
23
Sigma bond pi bond

σ-bond п bond
It is formed by direct
overlapping along the
orbital axis
It is formed by sidewise
overlapping of pure
p-orbitals
orbital axis p-orbitals
It is strong because orbital
overlap to greater extent
It is weak beause orbital
overlap to less extent
There is free rotation of
atom about σ-bond
There is no free rotation of
atom about п-bond
There can be only σ-bond
without п-bond
There can not be only п-
bond without σ-bond
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Concept
 Atomic orbitals combines to form new set of equivalent
orbitals called hybrid orbital
 The process of intermixing of the orbitals of slightly
 The process of intermixing of the orbitals of slightly
different energies so as to redistribute their energies,
resulting in the formation of new set of orbitals of
equivalent energies and shape is called hybridization
 Eg: when one 2s and three 2p orbitals of carbon
hybridize, there is formation of four new sp3 hybrid
orbitals
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Salient features
 Number of mixing orbitals = no. of hybride orbitals
 Hybridized orbitals are always equivalent in energy
and shape
and shape
 Forms stable bond than pure atomic orbital
 Types of hybridization of cental atom indicates the
geometry of molecule
26

Important conditions for hybridization
 Orbitals present in valence shell of atom are
hybridized
 Orbitals undergoing hybridization should have almost
 Orbitals undergoing hybridization should have almost
equal energy
 Half-filled and full-filled orbitals can participate in
hybridization
 Empty orbital can not participate in hybridization
27

i) sp3 hybridization
-when 1 s-orbital and 3 p-orbitals mixed up together to form
four equivalent orbitals it id known as sp3 hybridization
-all together 4 sp3 orbitals are formed
-structure of sp3 hybridized central atom
-all saturated organic compound has carbon with sp3
hybridization. The bond angle is 109.28o
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hybridization
s p p p sp3 sp3 sp3 sp3

ii) sp2 hybridization
-when 1 s-orbital and 2 p-orbitals mixed up together to form
four equivalent orbitals it id known as sp2 hybridization
-all together 3 sp2 orbitals are formed
-structure of sp2 hybridized central atom
-the p-orbital that did not participate in hybridization remains
as pure p-orbital. It is either empty or may have electron
-unsaturated double bonded organic compound has carbon
with sp2 hybridization. The bond angle is 120o .
29
hybridization
s p p p sp2 sp2 sp2 p

iii) sp hybridization
-when 1 s-orbital and 1 p-orbitals mixed up together to form four
equivalent orbitals it id known as sp hybridization
-all together 2 sp orbitals are formed
-structure of sp hybridized central atom
- the p-orbital that did not participate in hybridization remains as
pure p-orbital. It is either empty or may have electron
-unsaturated triple bonded organic compound has carbon with sp
hybridization. The bond angle is 180o
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hybridization
s p p p sp sp p p

In BeCl2, beryllium (Be) is the central atom with sp
hybridization
Be atom
(ground state)
1s 2s 2px 2py 2pz
31
Be atom
(excited state)
sp hybridization
Two ‘sp’ hybrid orbitals

The central atom is sp hybridized and there is no lone
pair of electrons in central atom. Hence, the structure of
BeCl2 is linear with bond angle 180o.
32
Be Cl
Cl
180o
Linear
structure
sp Be sp
Cl P Cl
P

In BF3, boron (B) is the central atom with sp2 hybridization
B atom
(ground state)
1s 2s 2px 2py 2pz
33
B atom
(excited state)
sp2 hybridization
Three ‘sp2’ hybrid orbitals

The central atom is sp2 hybridized and there is no lone
pair of electrons in central atom. Hence, the structure of
BeF3 is trigonal planar with bond angle 120o.
F
F
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B
F
F
F
120o
Trigonal planar
structure
sp2
B
P
P
P
sp2
sp2
F
F

In CH4, carbon (C) is the central atom with sp3 hybridization
Carbon atom
(ground state)
1s 2s 2px 2py 2pz
35
Carbon atom
(excited state)
sp3 hybridization
Four sp3 hybrid orbitals

Since the central atom is sp3 hybridized, and there is not
lone pair of electrons in central atom, the structure of
CH4 is tetrahedral with bond angle 109.5o.
H
H
s
36
C
H
H
H
H
109.4o
tetrahedral
structure
sp
3
C
H
H
H s
s
s

In NH3, nitrogen (N) is the central atom with sp3 hybridization
Nitrogen atom
(ground state)
1s 2s 2px 2py 2pz
37
sp3 hybridization
Four ‘sp3’ hybrid orbitals

Since the central atom is sp3 hybridized, the structure of
NH3 is tetrahedral. However, there is 1 lone pair of
electrons in central atom. Since,
lp-bp > bp-bp repulsion
structure of NH3 is distorted tetrahedral with bond angle
38
N
H
H H
107.5o
..
Distorted tetrahedral
structure
(lp)sp3
N
H
H
H s
s
s
..

In H2O, oxygen (O) is the central atom with sp3 hybridization
Oxygen atom
(ground state)
1s 2s 2px 2py 2pz
39
sp3 hybridization
Four ‘sp3’ hybrid orbitals

Since the central atom is sp3 hybridized, the structure of
NH3 is tetrahedral. However, there is 1 lone pair of
electrons in central atom. Since,
lp-lp > lp-bp > bp-bp repulsion
structure of H2O is distorted tetrahedral with bond angle
40
O
H
H 104.5o
Trigonal planar
structure
(lp)sp3
O
H
H
s
s

In ethane, carbon has sp3 hybridization
Carbon atom
(ground state)
1s 2s 2px 2py 2pz
41
Carbon atom
(excited state)
sp3 hybridization
Four sp3 hybrid orbitals

42
 All are sigma bonds

In ethene, carbon has sp2 hybridization
Carbon atom
(ground state)
1s 2s 2px 2py 2pz
43
Carbon atom
(excited state)
sp2 hybridization
three sp2 hybrid orbitals
Unhybridized p-orbital
One pure p-orbital
(used in pi-bond )

Sigma bonds in ethene
44
pi bond in ethene

Orbitals structure in ethene
45

In ethyne, carbon has sp hybridization
Carbon atom
(ground state)
1s 2s 2px 2py 2pz
46
Carbon atom
(excited state)
sp hybridization
three sp2 hybrid orbitals
Unhybridized p-orbital
two pure p-orbital
(used in pi-bond )

ii) sp hybridization: structure of ethyne
47
Orbitals structure in ethyne: there are two pi bonds

 Both molecules have same central atom (N) with same
sp3 hybridization and same number of bonding and
non-bonding electrons.The difference in structure can
be attributed to the electronegativity of surrounding
atom.
48
 The surrounding atom in NH3 is hydrogen atom and
that in NCl3 is chlorine atom. Also, EN(Cl) > EN(H)
 We know,“The increase in EN difference by increase of
EN of the other atom attached with it decreases the
bond angle”
the bond angle H-N-H is greater that Cl-N-Cl.

 Both molecules have central atoms with same (sp3)
hybridization and same number of bonding and non-
bonding electrons. In addition the surrounding is also
same.The difference in structure can be attributed to
the electronegativity of central atom.
 The central atom in H2O is oxygen atom and that in H2S
is sulphur atom. Also, EN(O) > EN(S)
 We know,“The increase in EN difference by increase of
EN of the central atom increases the bond angle”
the bond angle H-O-H (104.5o)is greater that in case of
H-S-H (92o)
49

 Both molecules have central atoms with same (sp3)
hybridization and same surrounding atoms.The difference
in structure can be attributed to the difference in number of
bonding and non-bonding electrons.
and that in NH3 has 1 pair of non-bonding electrons.
 We know,“ the extent of electron pair repulsion is lp-lp > lp-
bp > bp-bp”. i.e H2O has more repulsion than in NH3
 Hence, both have distorted tetrahedral geometry but the
bond angle H-O-H (104.5o)is smaller that in case of H-N-H
(107.5o)
50

Molecular geometry and chemical bonding

Molecular geometry and chemical bonding

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