VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY
The basic assumptions of this theory are summarized below.
1) The electron pairs in the valence shell around the central atom of a molecule
repel each other and tend to orient in space so as to minimize the repulsions and
maximize the distance between them.
2) There are two types of valence shell electron pairs viz., i) Bond pairs and ii)
Lone pairs
Bond pairs are shared by two atoms and are attracted by two nuclei. Hence they
occupy less space and cause less repulsion.
Lone pairs are not involved in bond formation and are in attraction with only one
nucleus. Hence they occupy more space. As a result, the lone pairs cause more
repulsion.
The order of repulsion between different types of electron pairs is as follows:
Lone pair - Lone pair > Lone Pair - Bond pair > Bond pair - Bond pair
Note: The bond pairs are usually represented by a solid line, whereas the lone
pairs are represented by a lobe with two electrons.
3) In VSEPR theory, the multiple bonds are treated as if they were single bonds.
The electron pairs in multiple bonds are treated collectively as a single super pair.
The repulsion caused by bonds increases with increase in the number of bonded
pairs between two atoms i.e., a triple bond causes more repulsion than a double bond
which in turn causes more repulsion than a single bond.
4) The shape of a molecule can be predicted from the number and type of valence
shell electron pairs around the central atom.
When the valence shell of central atom contains only bond pairs, the molecule
assumes symmetrical geometry due to even repulsions between them.
However the symmetry is distorted when there are also lone pairs along with bond
pairs due to uneven repulsion forces.
5) Primary & Secondary effects on bond angle and shape:
i) The bond angle decreases due to the presence of lone pairs, which cause more
repulsion on the bond pairs and as a result the bond pairs tend to come closer.
ii) The repulsion between electron pairs increases with increase in
electronegativity of central atom and hence the bond angle increases. The bond pairs
are closer and thus by shortening the distance between them, which in turn increases
the repulsion. Hence the bonds tend to move away from each other.
The bond pairs tend to move
away from each other since
the distance between them is
shortened as they are more
localized
on
more
electronegative central atom.
However the bond angle decreases when the electronegativities of ligand atoms
are more than that of central atom. There is increase in the distance between bond
pairs since they are now closer to ligand atoms. Due to this, they tend to move closer
resulting in the decrease in bond angle.
The bond pairs tend to come
closer since the distance
between them is increased as
they are more localized on
more
electronegative ligand
atoms.
iii) The bond angle decreases with increase in the size of central atom.
On smaller central
atoms
the
bond
pairs are closer and
hence tend to move
away from each
other so as to
minimize repulsion.
Hence bond angle
will be more.
On bigger central
atoms, the bond
pairs
are
more
distant from each
other and hence
there
is
less
repulsion.
Hence
they tend to move
closer and thus by
decreasing the bond
angle.
However the bond angle increases with increase in the size of ligand atoms, which
surround the central atom.
There is less
repulsion
between smaller
ligand atoms and
they can move
closer to each
other and thus
decrease
the
bond angle.
These is more repulsion
between bigger ligand
atoms and hence they
tend to move away from
each other. Thus bond
angle increases.
iv) The bond angles are also changed when multiple bonds are present. It is due to
uneven repulsions.
6) When there are two or more resonance structures, the VSEPR theory is
applicable to any of such contributing structure.
RELATION BETWEEN NUMBER & TYPE OF VALENCE ELECTRON PAIRS WITH THE
SHAPE OF MOLECULE
* The shape of molecule and also the approximate bond angles can be predicted
from the number and type of electron pairs in the valence shell of central atom as
tabulated below.
In the following table the molecule is represented by "AXE" notation, where
A = Central atom
X = Ligand atom bonded to the central atom either by a single bond or by
multiple bond; indicating a bond pair.
E = Lone pair
Note:
* The sum of number of ligand atoms (X) and number of lone pairs (E) is also
known as steric number.
* The bond pairs are shown as green colored thick lines, whereas the lone pairs
are shown as point charges using green colored lobes.
Numbe
r
of
numb
Bond
er
pairs
Numbe
r
Shape
Formu
of
of
la
Lone
molecule
pairs
Approxima
te
Exampl
Bond
es
angles
1
0
-
Steric
1
AX
Linear
ClF, BrF,
2
2
0
AX2
Linear
180o
3
0
AX3
Trigonal
planar
120o
2
1
AX2E
Angular
120o
4
0
AX4
Tetrahedr
al
109o28'
3
4
3
2
1
2
AX3E
AX2E2
Trigonal
pyramidal
Angular
around
109o28'
around
109o28'
BrCl,
HF, O2
BeCl2,
HgCl2,
CO2
BF3,
CO32-,
NO3-,
SO3
SO2,
SnCl2,
O3,
NSF,
NO2CH4,
SiCl4,
NH4+,
PO43-,
SO42-,
ClO4NH3,
PCl3,
XeO3
H2O,
SCl2,
Cl2O,
OF2
5
0
AX5
Trigonal
bipyramid
al
120o & 90o
PCl5,
SOF4
AX4E
See saw
or
distorted
tetrahedr
al
-
SF4,
TeCl4
5
4
1
6
7
ClF3,
BrF3,
BrCl3
3
2
AX3E2
T-Shape
90o
2
3
AX2E3
Linear
180o
XeF2, I3-
6
0
AX6
Octahedr
al
90o
SF6
5
1
AX5E
Square
pyramidal
90o
ClF5,
BrF5, ICl5
4
2
AX4E2
Square
planar
90o
XeF4
AX7
Pentagon
al
bipyramid
al
72o & 90o
IF7
7
0
6
1
AX6E
Pentagon
al
pyramidal
around
72o & 90o
XeOF5-,
IOF52-
STEPS INVOLVED IN PREDICTING THE SHAPES OF MOLECULES USING VSEPR THEORY
* The first step in determination of shape of a molecule is to write the Lewis dot
structure of the molecule.
* Then find out the number of bond pairs and lone pairs in the valence shell of
central atom.
While counting the number of bond pairs, treat multiple bonds as if they were
single bonds. Thus electron pairs in multiple bonds are to be treated collectively as a
single super pair.
* Use the above table to predict the shape of molecule based on steric number and
the number of bond pairs and lone pairs.
APPLICATIONS & ILLUSTRATIONS OF VSEPR THEORY
1) Methane (CH4):
* The Lewis structure of methane molecule is:
* There are 4 bond pairs around the central carbon atom in its valence shell.
Hence it has tetrahedral shape with 109o28' of bond angles.
2) Ammonia (NH3):
* The Lewis structure of ammonia indicates there are three bond pairs and one
lone pair around the central nitrogen atom.
* Since the steric number is 4, its structure is based on tetrahedral geometry.
However, its shape is pyramidal with a lone pair on nitrogen atom.
* The bond angle is decreased from 109o28' to 107o48' due to repulsion caused by
lone pair on the bond pairs.
3) Water (H2O):
* It is evident from the Lewis structure of water molecule, there are two bond
pairs and two lone pairs in the valence shell of oxygen. Hence its structure is based on
tetrahedral geometry. However its shape is angular with two lone pairs on oxygen.
* The bond angle is decreased to 104o28' due to repulsions caused by lone pairs on
bond pairs. It can be noted that the bond angle decreases with increase in the number
of lone pairs on the central atom.
4) Sulfur tetrachloride (SCl4):
Since there are four bond pairs and one lone pair around sulfur in its valence shell,
the structure of SCl4 is based on trigonal bipyramidal geometry. It has seesaw shape
with a lone pair occupying the equatorial position.
The angles between P-Claxial and P-Clequatorial are less than 90o due to repulsion
exerted by the lone pair. The angle between P-Clequatorial bonds also decreases from its
usual value, 120o.
The lone pair occupies the equatorial position to minimize the repulsions.
Note: Usually the lone pairs, bulky groups and less electronegative atoms tend to
occupy equatorial position to minimize repulsions. This is because they experience
repulsion only from two groups at 90o, when they occupy the equatorial positions.
However the repulsion will be more when they occupy axial positions, since they
encounter three groups at 90o.
5) PF3Cl2:
There are only 5 bond pairs on phosphorus atom. Hence it has trigonal bipyramidal
shape. The chlorine atoms occupy the equatorial positions to minimize the repulsions
since they are not only bulkier and also less electronegative than fluorine atoms.
The bond pair of P-Cl is slightly more closer towards the P atom when compared to
the bond pair of P-F, since the chlorine atoms are comparatively less electronegative
than fluorine atoms. Hence there is comparatively more negative charge
accumulation towards P atom, which makes the P-Cl bonds to experience more
repulsion than P-F bonds. Hence they orient in equatorial positions at 120 o to
minimize repulsions.
Note that, here we are comparing the polarity of P-Cl bond with P-F bond. But one
should keep in mind that the bond pair of P-Cl bond is still closer to Cl, since it is
more electronegative than P atom.
6) Formaldehyde (HCHO):
There are three bond pairs around the central carbon atom. The double bond
between C and O is considered as a single super pair. Hence the shape of the
molecule is trigonal planar and the bond angles are expected to be equal to 120 o.
However, the C=O exerts more repulsion on the C-H bond pairs. Hence the ∠H-C-H
bond angle will be less than 120o and the ∠H-C-O is greater than 120o.
VALENCE BOND THEORY (VBT) & HYBRIDIZATION
The valence bond theory was proposed by Heitler and London to explain the
formation of covalent bond quantitatively using quantum mechanics. Later on, Linus
Pauling improved this theory by introducing the concept of hybridization.
The main postulates of this theory are as follows:
* A covalent bond is formed by the overlapping of two half filled valence atomic
orbitals of two different atoms.
* The electrons in the overlapping orbitals get paired and confined between the
nuclei of two atoms.
* The electron density between two bonded atoms increases due to overlapping.
This confers stability to the molecule.
* Greater the extent of overlapping, stronger is the bond formed.
* The direction of the covalent bond is along the region of overlapping of the
atomic orbitals i.e., covalent bond is directional.
* There are two types of covalent bonds based on the pattern of overlapping as
follows:
(i) σ-bond: The covalent bond formed due to overlapping of atomic orbital along
the inter nucleus axis is called σ-bond. It is a stronger bond and cylindrically
symmetrical.
Depending on the types of orbitals overlapping, the σ-bond is divided into
following types:
σs-s bond:
σp-p bond:
σs-p bond:
(ii) π-bond: The covalent bond formed by sidewise overlapping of atomic orbitals
is called π- bond. In this bond, the electron density is present above and below the
inter nuclear axis. It is relatively a weaker bond since the electrons are not strongly
attracted by the nuclei of bonding atoms.
Note: The 's' orbitals can only form σ-bonds, whereas the p, d & f orbitals can form
both σ and π-bonds.
ILLUSTRATIONS
1) H2 molecule:
* The electronic configuration of hydrogen atom in the ground state is 1s1.
* In the formation of hydrogen molecule, two half filled 1s orbitals of hydrogen
atoms overlap along the inter-nuclear axis and thus by forming a σs-sbond.
2) Cl2 molecule:
* The electronic
[Ne]3s2 3px2 3py2 3pz1.
configuration
of
Cl
atom
in
the
ground
state
is
* The two half filled 3pz atomic orbitals of two chlorine atoms overlap along the
inter-nuclear axis and thus by forming a σp-p bond.
3) HCl molecule:
* In the ground state, the electronic configuration of hydrogen atom is 1s1.
* And the ground state electronic configuration of Cl atom is [Ne]3s2 3px2 3py2 3pz1.
* The half filled 1s orbital of hydrogen overlap with the half filled 3p z atomic
orbital of chlorine atom along the inter-nuclear axis to form a σs-p bond.
4) O2 molecule:
* The electronic configuration of O in the ground state is [He] 2s2 2px2 2py1 2pz1.
* The half filled 2py orbitals of two oxygen atoms overlap along the inter-nuclear
axis and form σp-p bond.
* The remaining half filled 2pz orbitals overlap laterally to form a πp-p bond.
* Thus a double bond (one σp-p and one πp-p) is formed between two oxygen atoms.
5) N2 molecule:
* The ground state electronic configuration of N is [He] 2s2 2px1 2py1 2pz1.
* A σp-p bond is formed between two nitrogen atoms due to overlapping of half
filled 2px atomic orbitals along the inter-nuclear axis.
* The remaining half filled 2py and 2pz orbitals form two πp-p bonds due to lateral
overlapping. Thus a triple bond (one and two) is formed between two nitrogen
atoms.
NEED FOR MODIFICATION OF VBT
However the old version of valence bond theory is limited to diatomic molecules
only. It could not explain the structures and bond angles of molecules with more than
three atoms.
E.g. It could not explain the structures and bond angles of H2O, NH3 etc.,
However, in order to explain the structures and bond angles of molecules, Linus
Pauling modified the valence bond theory using hybridization concept.
HYBRIDIZATION: A hypothetical process of intermixing of two or more pure atomic
orbitals of an atom with almost same energy to give same number of identical and
degenerate new type of orbitals is known as hybridization.
The new orbitals formed are also known as hybrid orbitals.
What is intermixing?
The intermixing or hybridization of atomic orbitals is a mathematical concept
based on quantum mechanics. During this process, the wavefunctions, Ψ of atomic
orbitals of same atom are combined to give new wavefunctions corresponding to
hybrid orbitals.
What are the requirements for atomic orbitals to undergo hybridization?
* The atomic orbitals of same atom with almost same energy can only participate
in the hybridization.
* The full filled or half filled or even empty orbitals can undergo hybridization
provided they have almost equal energy.
Do the orbitals of different atoms undergo hybridization?
No! The hybridization is the mixing of orbitals of same atom only. The combination
of orbitals belonging to different atoms is called bonding.
What are hybrid orbitals? And what are its characteristics?
* The new orbitals that are formed due to intermixing of atomic orbitals are also
known as hybrid orbitals, which have mixed characteristics of atomic orbitals.
* The shapes of hybrid orbitals are identical. Usually they have one big lobe
associated with a small lobe on the other side.
* The hybrid orbitals are degenerate i.e., they are associated with same energy.
How many hybrid orbitals are formed?
* The number of hybrid orbitals formed is equal to the number of pure atomic
orbitals undergoing hybridization.
E.g. If three atomic orbitals intermix with each other, the number of hybrid
orbitals formed will be equal to 3.
How do the electrons are going to be filled in the hybrid orbitals?
* The hybrid orbitals are filled with those electrons which were present in the pure
atomic orbitals forming them.
* The filling up of electrons in them follows Pauli's exclusion principle and Hund's
rule.
What is the use of hybrid orbitals?
* The hybrid orbitals participate in the σ bond formation with other atoms.
Why atomic orbitals in a given atom undergo hybridization?
* The hybrid orbitals are oriented in space so as to minimize repulsions between
them. This explains why the atomic orbitals undergo hybridization before bond
formation.
The reason for hybridization is to minimize the repulsions between the bonds that
are going to be formed by the atoms by using hybrid orbitals.
Remember that the hybridization is the process that occurs before bond
formation.
And finally:
* The bond angles in the molecule are equal to or almost equal to the angles
between the hybrid orbitals forming the σ bonds. The shape of the molecule is
determined by the type of hybridization, number of bonds formed by them and the
number of lone pairs.
TYPES OF HYBRIDIZATION
During hybridization, the atomic orbitals with different characteristics are mixed
with each other. Hence there is no meaning of hybridization between same type of
orbitals i.e., mixing of two 's' orbitals or two 'p' orbitals is not called hybridization.
However orbital of 's' type can can mix with the orbitals of 'p' type or of 'd' type.
Based on the type and number of orbitals, the hybridization can be subdivided into
following types.
sp HYBRIDIZATION
* Intermixing of one 's' and one 'p' orbitals of almost equal energy to give two
identical and degenerate hybrid orbitals is called 'sp' hybridization.
* These sp-hybrid orbitals are arranged linearly at by making 180o of angle.
* They possess 50% 's' and 50% 'p' character.
JUMP TO ILLUSTRATIONS OF SP HYBRIDIZATION
sp2 HYBRIDIZATION
* Intermixing of one 's' and two 'p' orbitals of almost equal energy to give three
identical and degenerate hybrid orbitals is known as sp 2 hybridization.
* The three sp2 hybrid orbitals are oriented in trigonal planar symmetry at angles
of 120o to each other.
* The sp2 hybrid orbitals have 33.3% 's' character and 66.6% 'p' character.
JUMP TO ILLUSTRATIONS OF SP2 HYBRIDIZATION
sp3 HYBRIDIZATION
* In sp3 hybridization, one 's' and three 'p' orbitals of almost equal energy intermix
to give four identical and degenerate hybrid orbitals.
* These four sp3 hybrid orbitals are oriented in tetrahedral symmetry with 109 o28'
angle with each other.
* The sp3 hybrid orbitals have 25% ‘s’ character and 75% 'p' character.
JUMP TO ILLUSTRATIONS OF SP3 HYBRIDIZATION
sp3d HYBRIDIZATION
* In sp3d hybridization, one 's', three 'p' and one 'd' orbitals of almost equal energy
intermix to give five identical and degenerate hybrid orbitals, which are arranged in
trigonal bipyramidal symmetry.
Among them, three are arranged in trigonal plane and the remaining two orbitals
are present above and below the trigonal plane at right angles.
* The sp3d hybrid orbitals have 20% 's', 60% 'p' and 20% 'd' characters.
JUMP TO ILLUSTRATIONS OF SP3D HYBRIDIZATION
sp3d2 HYBRIDIZATION
* Intermixing of one 's', three 'p' and two 'd' orbitals of almost same energy by
giving six identical and degenerate hybrid orbitals is called sp 3d2hybridization.
* These six sp3d2 orbitals are arranged in octahedral symmetry by making
90o angles to each other. This arrangement can be visualized as four orbitals arranged
in a square plane and the remaining two are oriented above and below this plane
perpendicularly.
JUMP TO ILLUSTRATIONS OF SP3D2 HYBRIDIZATION
sp3d3 HYBRIDIZATION
* In sp3d3 hybridization, one 's', three 'p' and three 'd' orbitals of almost same
energy intermix to give seven sp3d3 hybrid orbitals, which are oriented in pentagonal
bipyramidal symmetry.
* Five among the sp3d3 orbitals are arranged in a pentagonal plane by making
72o of angles. The remaining are arranged perpendicularly above and below this
pentagonal plane.
sp HYBRIDIZATION
1) Beryllium Chloride (BeCl2):
* The electronic configuration of 'Be' in ground state is 1s 2 2s2. Since there are no
unpaired electrons, it undergoes excitation by promoting one of its 2s electron into
empty 2p orbital.
Thus in the excited state, the electronic configuration of Be is 1s2 2s1 2p1.
If the beryllium atom forms bonds using these pure orbitals, the molecule might be
angular. However the observed shape of BeCl2 is linear. To account for this, following
sp hybridization was proposed.
* In the excited state, the beryllium atom undergoes 'sp' hybridization by mixing a
2s and one 2p orbitals. Thus two half filled 'sp' hybrid orbitals are formed, which are
arranged linearly.
* These half filled sp-orbitals form two σ bonds with two 'Cl' atoms.
* Thus BeCl2 is linear in shape with the bond angle of 180o.
2) Acetylene (C2H2):
* The ground state electronic configuration of 'C' is 1s2 2s2 2px12py1. There are only
two unpaired electrons in the ground state. However, the valency of carbon is four
i.e., it forms 4 bonds. In order to form four bonds, there must be four unpaired
electrons. Hence carbon promotes one of its 2s electron into the empty 2p z orbital in
the excited state.
1s2
Thus in the excited
2s1 2px12py12pz1.
state,
the
electronic
configuration
of
carbon
is
* Each carbon atom undergoes 'sp' hybridization by using a 2s and one 2p orbitals in
the excited state to give two half filled 'sp' orbitals, which are arranged linearly.
* The two carbon atoms form a σsp-sp bond with each other by using sp-orbitals.
However there are also two unhybridized p orbitals i.e., 2p y and 2pz on each
carbon atom which are perpendicular to the sp hybrid orbitals. These orbitals form
two πp-p bonds between the two carbon atoms.
Thus a triple bond (including one σsp-sp bond & two πp-p bonds ) is formed between
carbon atoms.
* Each carbon also forms a σsp-s bond with the hydrogen atom.
* Thus acetylene molecule is linear with 180o of bond angle.
sp2 HYBRIDIZATION
1) Boron trichloride (BCl3):
* The electronic configuration of 'B' in ground state is 1s 2 2s2 2p1 with only one
unpaired electron. Since the formation of three bonds with chlorine atoms require
three unpaired electrons, there is promotion of one of 2s electron into the 2p sublevel
by absorbing energy.
Thus Boron atom gets electronic configuration: 1s2 2s2 2px12py1.
However to account for the trigonal planar shape of this BCl 3 molecule,
sp2 hybridization before bond formation was put forwarded.
* In the excited state, Boron undergoes sp 2 hybridization by using a 2s and two 2p
orbitals to give three half filled sp2 hybrid orbitals which are oriented in trigonal
planar symmetry.
* Boron forms three σsp-p bonds with three chlorine atoms by using its half filled
hybrid orbitals. Each chlorine atom uses it's half filled p-orbital for the σ-bond
formation.
sp2
* Thus the shape of BCl3 is trigonal planar with bond angles equal to 120o.
2) Ethylene (C2H4):
* During the formation of ethylene molecule, each carbon atom undergoes
sp2 hybridization in its excited state by mixing 2s and two 2p orbitals to give three
half filled sp2 hybrid orbitals oriented in trigonal planar symmetry.
There is also one half filled unhybridized 2p z orbital on each carbon perpedicular
to the plane of sp2 hybrid orbitals.
* The carbon atoms form a σsp2-sp2 bond with each other by using sp 2 hybrid
orbitals.
A πp-p bond is also formed between them due to lateral overlapping of
unhybridized 2pz orbitals.
Thus there is a double bond (σsp2-sp2 & πp-p) between two carbon atoms.
* Each carbon atom also forms two σsp2-s bonds with two hydrogen atoms.
* Thus ethylene molecule is planar with ∠HCH & ∠HCC bond angles equal to 120o.
* All the atoms are present in one plane.
sp3 HYBRIDIZATION
1) Methane (CH4):
* During the formation of methane molecule, the carbon atom undergoes
sp3 hybridization in the excited state by mixing one ‘2s’ and three 2p orbitals to
furnish four half filled sp3 hybrid orbitals, which are oriented in tetrahedral symmetry
in space around the carbon atom.
* Each of these sp3 hybrid orbitals forms a σsp3-s bond with one hydrogen atom.
Thus carbon forms four σsp3-s bonds with four hydrogen atoms.
* Methane molecule is tetrahedral in shape with 109o28' bond angle.
2) Ethane (C2H6):
* Just like in methane molecule, each carbon atom undergoes sp3 hybridization in
the excited state to give four sp3 hybrid orbitals in tetrahedral geometry.
sp3
* The two carbon atoms form a σsp3-sp3 bond with each other due to overlapping of
hybrid orbitals along the inter-nuclear axis.
Each carbon atom also forms three σsp3-s bonds with hydrogen atoms.
* Thus there is tetrahedral symmetry around each carbon with ∠HCH & ∠HCC bond
angles equal to 109o28'.
3) Ammonia (NH3) :
* The ground state electronic configuration of nitrogen atom is:
2s2 2px12py12pz1. Since there are three unpaired electrons in the 2p sublevel, the
nitrogen atom can form three bonds with three hydrogen atoms. This will give
ammonia molecule with 90o of bond angles. However, the bond angles are reported to
be 107o48'.
1s2
* Therefore, it was proposed that, the Nitrogen atom undergoes sp 3 hybridization
of a 2s and three 2p orbitals to give four sp 3 orbitals, which are arranged in
tetrahedral symmetry. It is clear that this arrangement will give more stability to the
molecule due to minimization of repulsions.
Among them three are half filled and one is full filled.
* Nitrogen atom forms 3 σsp3-s bonds with three hydrogen atoms by using three half
filled sp3 hybrid orbitals. There is also a lone pair on nitrogen atom belonging to the
full filled sp3 hybrid orbital. It occupied more space than the bond pairs.
* However, the ∠HNH bond angle is not equal to normal tetrahedral angle: 109 o28'.
The reported bond angle is 107o48'. The observed decrease in the bond angle is due to
the repulsion caused by lone pair over the bond pairs.
That is why, ammonia molecule is trigonal pyramidal in shape with a lone pair on
nitrogen atom.
4) Water molecule (H2O):
* The electronic configuration of oxygen is 1s2 2s2 2px22py12pz1. There are two
unpaired electrons in oxygen atom, which may form bonds with hydrogen atoms.
However the the bond angles in the resulting molecule should be equal to 90 o.
sp3
The experimental bond angles reported were equal to 104 o28'. To account this,
hybridization before the bond formation was proposed.
* During the formation of water molecule, the oxygen atom undergoes
hybridization by mixing a 2s and three 2p orbitals to furnish four sp 3 hybrid
orbitals oriented in tetrahedral geometry.
sp3
Among them, two are half filled and the remaining two are completely filled.
* Now the oxygen atom forms two σsp3-s bonds with hydrogen atoms by using half
filled hybrid orbitals.
* The reported bond angle is 104o28' instead of regular tetrahedral angle: 109o28'.
It is again due to repulsions caused by two lone pairs on the bond pairs.
Thus water molecule gets angular shape (V shape).
sp3d HYBRIDIZATION
1) Phosphorus pentachloride(PCl5):
1s2
* The ground state
2s22p6 3s23px13py13pz1.
electronic
configuration
of
phosphorus
atom
is:
* The formation of PCl5 molecule requires 5 unpaired electrons. Hence the
phosphorus atom undergoes excitation to promote one electron from 3s orbital to one
of empty 3d orbital.
* Thus the electronic
1s2 2s22p6 3s23px13py13pz1 3d1.
configuration
of
'P'
in
the
excited
state
is
* In the excited state, intermixing of a 3s, three 3p and one 3d orbitals to give five
half filled sp3d hybrid orbitals, which are arranged in trigonal bipyramidal symmetry.
i.e., Three orbitals are arranged in trigonal planar symmetry, whereas the
remaining two are arranged perpendicularly above and below this plane.
* By using these half filled sp3d orbitals, phosphorous forms five σsp3d-p bonds with
chlorine atoms. Each chlorine atom makes use of half filled 3p zorbital for the bond
formation.
* The shape of PCl5 molecule is trigonal bipyramidal with 120o and 90o of ∠Cl - P Cl bond angles.
sp3d2 HYBRIDIZATION
1) Sulfur hexa flouride (SF6):
* The electronic configuration of 'S' in ground state is 1s2 2s22p6 3s23px23py13pz1.
* In SF6 molecule, there are six bonds formed by sulfur atom. Hence there must be
6 unpaired electrons. However there are only 2 unpaired electrons in the ground state
of sulfur. Hence it promotes two electrons into two of the 3d orbitals (one from 3s
and one from 3px).
1s2
* Thus the electronic
2s22p6 3s13px13py13pz13d2.
configuration
of
'S'
in
its
2 nd excited
state
is
* In the second excited state, sulfur under goes sp 3d2 hybridization by mixing a 3s,
three 3p and two 3d orbitals. Thus formed six half filled sp 3d2hybrid orbitals are
arranged in octahedral symmetry.
Sulfur atom forms six σsp3d2-p bonds with 6 fluorine atoms by using these
orbitals. Each fluorine atom uses is half-filled 2pz orbitals for the bond
formation. SF6 is octahedral in shape with bond angles equal to 90o.
sp3d2
sp3d3 HYBRIDIZATION
1) Iodine heptafluoride (IF7):
* The electronic configuration of Iodine atom in the ground state is: [Kr]4d 105s25p5.
Since the formation of IF7 requires 7 unpaired electrons, the iodine atom promotes
three of its electrons (one from 5s orbital and two from 5p sublevel) into empty 5d
orbitals. This state is referred to as third excited state.
* The electronic configuration of Iodine in the third excited state can be written
as: [Kr]4d105s15p35d3.
In the third excited state, iodine atom undergoes sp 3d3 hybridization to give 7 half
filled sp3d3 hybrid orbitals in pentagonal bipyramidal symmetry. These will form 7
σsp3d3-p bonds with fluorine atoms.
Thus the shape of IF7 is pentagonal bipyramidal. The ∠F-I-F bond angles in the
pentagonal plane are equal to 72o, whereas two fluorine are present perpendicularly
to the pentagonal plane above and below.