VBT by B R Thorat

1. Dr. B. R. Thorat
Department of Chemistry
Govt. of Maharashtra
Ismail Yusuf College, Jogeshwari (E), Maharashtra 400060

2. Atomic/molecular bonding
Covalent bonding Pi-Bonding Ionic/electrovalent-BondingCo-ordinate-Bonding
Orbital overlapping
The essential conditions for the overlap of electron waves of orbitals are:
The orbitals entering into combination or
overlapping must have only one electron: The
orbitals containing a pair of electrons are not
capable of overlapping to form covalent bond. In
fact, half-filled orbitals on one atom have a
tendency to combine with half-filled orbitals on
other atom, and the resulting orbital acquires a
pair of electrons of opposite spins
The atoms with valence or bonding orbital
(half-filled) should approach sufficiently close
to one another with the axis of their orbitals in
proper alignment: The strength of a covalent
bond depends upon the extent of overlapping.
Greater the overlapping between the atomic
orbitals stronger is the bond formed between the
two atoms
Electron transfer
Bond formation release energy
Bonding energy

3. Atomic
Orbitals
Interact each others
Energy released
Bonding / stabilizing Energy
Atomic nuclei or electron
repulsive force have been
exactly balances by the force
of attraction or overlapping
(nuclei – electron)
Bond length
Bond Energy
(kJ/mol)
Bond length
(A)
Bond Energy
(kJ/mol)
Bond length (A)
H-H 103.2 0.74 C-N 70.3 1.52
C-H 98.7 1.10 C=N 148 1.29
C-C 83 1.54 C≡N 213 1.15
C=C 146 1.34 C-O 84 1.43
C≡C 200 1.20 C=O 177 1.24

6. Potential energy diagram – Hydrogen molecule
When two atoms approach each other along the bonding axis,
the electrons and nucleus of one atom interact with the
electrons and nucleus of the other atom – energy decreases.
the energy of the system is lowered because of the interactions,
then a chemical bond is form. If the energy of the system is
raised by the interactions, then a chemical bond does not form
interaction energy is usually
calculated as a function of the
inter-nuclear distance between the
two bonding atoms
1
2
3
4
most stable point on the curve
occurs at the minimum of the
interaction energy
bond energy
For the H2 molecule, the
bond length is 0.74 Å and
bond energy is 103.2
kcal/mole.

7. Formation of H2
1. Actually, when two H atoms are far separated - do not have any force of interaction – attractive and repulsive.
2. Attraction Force - The nucleus of one H is attracted towards the electrons of the other H atoms & vice-versa - energy is
released (exothermic).
3. Repusive Force - The nuclei of the one atom as well as their electron repel the nucleus and electron of other atom. Energy
is needed to overcome the force (endothermic).
if the magnitude of the attractive
forces is more than that of the
repulsive forces, a stable molecule
will be formed
if the repulsive forces are
more than the attractive
forces, then the atoms will
not combine
potential energy decreases
potential energy increases
Molecule does not formed
stable hydrogen molecule is formed
s–s overlapping - formation of sigma
bond between two hydrogen atoms,
433 kJ/mol energy is released

8. cA = cB = 1
+. +. . .+
bondingg
Amplitudes of wave
functions added
g = N [A + B]
Constructive interference

9. Amplitudes of wave
functions
subtracted.
Destructive interference
Nodal plane perpendicular to the H-H bond
axis (en density = 0)
Energy of the en in this orbital is higher.
+. -. ..
node
antibonding
u = N [A - B]
cA = +1, cB = -1 u
+ -
A-B

10. When orbitals of two atoms come close to form bond, their
overlap may be positive, negative or zero depending upon
the sign (phase) and direction of orientation of amplitude
of orbital wave function in space.
Orbitals forming bond should have same sign (phase) and
orientation in space. This is called positive overlap.
The covalent bond may be classified into two types depending upon the types of overlapping: (i) Sigma(σ) bond, and (ii) pi(π)
bond.
Formation of homoatomic dimoleules

11. The formation of He2 involves the combination of two He atoms with ground state electronic configurations of 1s2.
Each He atom has two paired electrons and so VB theory predicts that the formation of a He-He bond is not possible due to
completely filled atomic orbitals.
Formation of He2
+
s-orbitals
He2 moleculeNo overlapping

12. Formation of Be2
Formation of Li2
Ground state electronic configurations of 1s22s12p0. Each Li atom has one unpaired electrons and so VB theory predicts the
formation of an Li-Li single bond.
+
s-orbitals
Li2 molecule
Ground state electronic configurations of 1s22s22p0. Each Be atom has two paired electrons and so VB theory predicts that the
formation of an Be-Be bond is not possible due to completely filled atomic orbitals.
+
s-orbitals
Be2 moleculeNo overlapping

13. Formation of B2
The ground state electronic configuration of B is [He]2s22p1, and the presence of the single unpaired electron in p-orbital
indicates the formation of an B-B single bond by p-p overlapping of atomic orbitals.
Formation of C2
Ground state electronic configurations of 1s22s22p2. Each C atom has two unpaired electrons and so VB theory predicts the
formation of a C=C double bond but C is tetravalent which not explain by orbital overlapping.
+
C2 molecule
p-orbitals
But C is tetravalent

14. Formation of N2
Ground state electronic configurations of 1s22s22p5. Each N atom has three unpaired electrons and so VB theory predicts the
formation of an N≡N triple bond
Formation of O2
Ground state electronic configurations of 1s22s22p4. Each O atom has two unpaired electrons and so VB theory predicts the
formation of an O=O double bond.

15. Formation of F2
The ground state electronic configuration of F is [He]2s22p5, and the presence of the single unpaired electron indicates the
formation of an F-F single bond by p-p overlapping of atomic orbitals.
+
2P 2P
F2 molecule
covalent
bond
Formation of Ne2
The ground state electronic configuration of neon is [He]2s22p6 in which all orbitals are completely filled therefore neon atom
does not undergo overlapping of atomic orbitals.

16. Single Lewis structure cannot describe a molecular structure accurately, a number of structures with similar energy,
positions of nuclei, bonding and non bonding pairs of electrons are considered to represent the structure.
Resonance or Canonical structures
Each such structure is called as canonical structure. A resonance hybrid consists of many canonical structures.
The molecule CO has only 10 valence shell electrons with which to achieve an octet around each atom.
The structure |C≡O| makes use of exactly 10 electrons, and that makes it possible to place an octet (three shared pairs and
one unshared pair) around each atom. This can be explaining with the concept called resonance.

17. Simple procedure for deciding how to place the electrons is as follows:
1. Determine the total number of valence shell electrons for all atoms (N) that are available to be distributed in the
structure.
2. Multiply the number of atoms present by eight to determine the number of electrons required to complete octet
around each atom (S).
3. Calculate the difference (S–N) gives the number of electrons that must be shared in the structure.
4. If possible, change the distribution of electrons to give favorable formal charges (discussed later in this chapter) on
the atoms.
For CO, the total number of valence shell electrons is 10, and to give octets around two atoms required 16 electrons.
Therefore, 16 - 10 = 6 electrons must be shared by the two atoms. Six electrons are equivalent to three pairs or three
covalent bonds.

18. For a molecule such as SO2, the number of valence shell electrons is 18, whereas three atoms would require 24 electrons to
make three octets.
Therefore, 24 - 18 = 6, the number of electrons that must be shared, which gives a total of three bonds between the sulfur
atom and the two oxygen atoms.
Hence, the sulfur atom must have an unshared pair of electrons that is localized on it and the three pairs that it is sharing.
But the resonance structure of SO2 is neutral. The true structure of SO2 is one that lies halfway between the two
structures shown above. The molecule is all of the time is resonance hybrid of the structures shown.
The hybrid orbital type is sp2, which accounts for the bond angle being 119.5°. There is one p orbital not used in the
hybridization that is perpendicular to the plane of the molecule, which allows for the π bonding to the two oxygen atoms
simultaneously. The π bond is described as being delocalized, and this can be shown as follows:
S
O O
(I)
S
O O
(II)
S
O
O
Unhybride orbital
Hybridized orbital

19. Rules to drawing resonance structures
Resonance relates to different ways of placing electrons in the structures, not ways of arranging the atoms themselves.
Some important guidelines for drawing resonance structures are listed below –
1. The atoms must be present in the same relative positions in all structures. e.g. SO2 molecule has a bent or angular
structure.
2. Structures showing the molecule with some other geometry (e.g., linear) are not permitted.
3. Structures in which maximum number of electrons used in bonding (consistent with the octet rule) contribute more to
the true structure.
4. All resonance structures of molecules or ions drawn must have the same number of unpaired electrons if there are
any.
5. Negative formal charges normally reside on the atoms having higher electronegativity.
6. Like charges on adjacent atom – less stable.

20. NO2 molecule
The NO2 molecule has a total of 17 valence shell electrons; there are eight pairs of electrons and one unpaired
electron. Structures drawn for NO2 must reflect this.
N
O O
(I)
N
O O
(II)
N-O bond length = 118 pm
bond angle = 1340
There are seven electrons present on nitrogen atom and oxygen atoms have complete octets because of higher
electronegativity. The NO2 get dimerize by pairing of electrons of two molecules as shown by the equation –
N
O
O
N
O
O
+ N
O
O
N
O
O
In NO2 molecule, the nonbonding orbital located on the nitrogen atom contains only one electron, so repulsion between that
non-bonding orbital and the shared electron pairs is small. Therefore, the bond angle is larger because the repulsion
between the bonding pairs is not balanced by the repulsion between nonbonding and bonding orbitals.

22. Formal Charge
In order to determine the formal charge on each atom in a structure, the total number of valence electrons among each atom
should be known. This is done as per following rules –
1. Any unshared pairs of electrons belong to the atom on which they are located.
2. Shared electron pairs are divided equally between the atoms sharing them.
3. The total number of electrons on an atom in a structure is the sum from steps 1 and 2.
4. Compare the total number of electrons that appear on each atom to the number of valence shell electrons and
normally both are same. If the number of electrons in the valence shell is greater than indicated in step 3, the
atom appears to have lost one or more electrons and has a positive formal charge. If the number indicated in step
3 is larger than the number in the valence shell, the atom appears to have gained one or more electrons and has a
negative formal charge.
5. Structures that have formal charges with the same sign on adjacent atoms will contribute less in to the true structure.
6. The sum of formal charges on the atoms must total the overall charge on the species.

23. Carbon monoxide molecule
If the shared pairs are divided equally, each atom appears to have three electrons from bonding pairs.
Each atom in the structure appears to have a total of five electrons – each have one lone pair of electrons.
Carbon normally has four electrons in its valence shell, so in the structure carbon has a formal charge of -1.
An oxygen atom normally has six valence shell electrons, so it appears that the oxygen atom has lost one electron, giving it
a formal charge of +1.

24. Formal charges can be used to predict the stable arrangement of atoms in many molecules.
Nitrous oxide, N2O, might have the structures shown as follows:
Nitrous oxide
Structure III will contribute approximately 0% to the
true structure because –
it same formal charges are present on adjacent atoms.
it has a positive formal charge on the oxygen atom
while a nitrogen atom has a -2 formal charge,
it has higher overall formal charges
The relative contributions from structures I and II in true
structure is somewhat more difficult.
the negative formal charge on the oxygen
atom, it also has a triple bond between the
two nitrogen atoms
two double bonds even though it
places a negative formal charge on a
nitrogen atom
Structure I places a negative formal charge on the terminal nitrogen atom, while structure II places a negative formal charge
on the oxygen atom on the opposite end of the molecule. If the two structures contribute equally, these effects should cancel,
which would result a molecule is not polar.
In fact, the dipole moment of N2O is only 0.17 D, so structures I and II must make approximately equal contributions.

25. Isocyanate ion
There are 16 valence shell electrons were distributed around three atoms. To provide total of three octets, 24 electrons would
be needed. Therefore, eight electrons must be shared, which means that there will be a total of four bonds, two from the
central atom to each of the terminal atoms.
the carbon atom has a
formal charge of 0
the oxygen atom has a
formal charge of +2
the nitrogen atom has a
formal charge of +1
Both structures II and III have an arrangement of atoms that places a positive formal charge on atoms that are higher in
electronegativity than carbon. Consequently, the most stable arrangement of atoms is as shown in structure I.
formal charges –
-1, 0, and 0
formal charges –
0, 0, and -1
formal charges –
-2, 0, and +1
contribute approximately
0%
It is concluded that structures I and II probably contribute about equally to the true structure. Other triatomic molecules having
16 electrons and linear geometry are – CO2, SCN-, OCN-, N2O and 𝑁𝑂2
−
.

26. 𝑪𝑶3
2−
There are several important chemical species that consist of four atoms and have a total of 24 valence shell electrons. Because
four atoms would require a total of 32 electrons for each to have an octet, then total eight electrons must be shared in four
bonds.
C
O
OO
Bond length (C-O) = 132 pm
C
O
OO
C
O
OO
SO3
S
O
OO
Bond length (S-O) = 143 pm
S
O
OO
-1
-1
+2 +1
-1
0
0 0
(I) (II)
HNO3
SO3 anion

27. Bonding in Polyatomic Species: The role of Hybridization
The overlapping of atomic orbitals cannot adequately explain the bonding in many other molecules such as compound formed
from carbon and hydrogen. Actually carbon is tetravalent, but VBT predicted divalent structures.
VBT
Carbon has only two half-filled orbitals and should
therefore form only two bonds with two hydrogen
atoms. We would therefore predict that carbon and
hydrogen should form a molecule with the formula
CH2 and with a bond angle of 900 (corresponding
to the angle between any two p orbitals).
Facts or Limitations of VBT
The stable compound formed from carbon
and hydrogen is CH4 (methane), which has
bond angles of 109.50.
Valence bond theory explains the bonding in CH4 and
many other polyatomic molecules by incorporating an
additional concept called orbital hybridization.
C
2p
2p
c
the overlapping of atomic orbitals which produces
chemical bonds are simply the s, p and d atomic
orbitals but valence bond theory treats these electrons
in a molecule before undergoes overlapping
Have
Different
Shape &
Size

28. Hybridization is a mathematical procedure in which the standard atomic orbitals having same or nearly same energy are
mixed and redistribted to form new set of atomic orbitals called hybrid orbitals.
Hybrid orbitals are still localized on individual atoms, but their shapes and energies differ from those of standard atomic
orbitals.
In hybrid orbitals, the electron probability density is more concentrated in a single directional lobe, allowing greater
overlap with the orbitals of other atoms.
the more bonds that an central of interior atom forms, the greater the tendency of its orbitals to hybridize.
The bigger lobe of hybrid orbital involves itself in the process of an overlap with orbitals of other
atoms to form bonds while the smaller lobes of hybrid orbitals are neglected while considering bond
formation.
The number of standard atomic orbitals added together always equals the number of Hybrid
orbitals formed.
Hybrid orbitals are generally used for the formation of sigma bonding (if half filled) or
remains as it is (completely filled, unshared pair of electrons).
Hybridization

29. Hybridization

30. sp Hybridization
One s and one p orbitals of the central bonding atom get “mix,” or hybridize, forming two equal set of energy hybrid orbitals,
each of them is called sp hybrid orbital and other orbitals (if available) remains unhybridized which either empty or half filled
used for the formation of pi-bonding.
e.g. BeCl2, C2H2, etc.
The two hybrid orbitals have a linear geometry with 1800 angles between them. The unhybridized p orbitals are oriented in
the plane perpendicular to the two hybridized sp-orbitals.

31. sp2 Hybridization
One s and two p orbitals of the central bonding atom get “mix,” or hybridize, forming three equal set of energy hybrid orbitals,
each of them is called sp2 hybrid orbital and other orbitals (if available) remains unhybridized which either empty or half filled
used for the formation of pi-bonding. E.g. BF3, C2H4, etc.
The three hybrid orbitals have a trigonal planar geometry with 1200 angles between them. The unhybridized p orbital is
perpendicular to the three hybridized orbitals.

32. sp3 Hybridization
One s and three p orbitals of the central bonding atom get “mix,” or hybridize, forming four equal set of energy hybrid orbitals,
each of them is called sp3 hybrid orbital. E.g. CH4, NH3, H2O, etc.
All hybrid orbitals are have same energy and are oriented tetrahydrally with bond angle 109.50. The hybrid orbital may be half
filled, empty or completely filled which are able to for bond formation. However, if the central atom of a molecule contains lone
pairs, hybrid orbitals can also accommodate them. For example, the nitrogen orbitals in ammonia are sp3 hybrids.

33. sp3 Hybridization as in CH4
The ground state electronic configuration of carbon is 1s22s22p2. One of the electrons of 2s orbital is get excited to empty 2p
orbital by absorbing energy. The VB theory proposes that the one s and all three p orbitals of the central atom mix and form
four sp3 hybrid orbitals, which point toward the vertices of a tetrahedron. The C atom in methane is sp3 hybridized.
Its four valence electrons half-fill the four sp3 hybrids, which overlap the half-filled 1s orbitals of four H atoms and form four
C-H bonds. Due to the tetrahedral disposition of sp3 hybrid orbitals, the orbitals are inclined at an angle of 109.5º. Thus all the
HCH angles are equal to 109.5º.

34. sp3 Hybridization as in NH3
One s and all three p orbitals of the central nitrogen atom mix and form four sp3 hybrid orbitals, which point toward the vertices
of a tetrahedron. The nitrogen forming three N-H bonds and one hybrid orbital contains lone pair of electrons therefore the
geometry of ammonia is trigonal pyramidal.
The lone pair is attracted more towards the N-atom than the bond pairs which
belongs to the H-atoms and N-atom jointly. This is because of the fact that the
lone pair belongs only to the N-atom and hence its electron cloud is more
concentrated near the N-atom. The lone pair is, therefore, capable of exerting a
greater repulsion on a bond pair than a bond pair can repel another bond pair.

35. sp3 Hybridization as in H2O
One s and all three p orbitals of the central oxygen atom mix and form four sp3 hybrid orbitals, which point toward the
vertices of a tetrahedron. The oxygen forming two O-H bonds and two hybrid orbital contains lone pair of electrons therefore
the geometry of water is bent shape.
In water molecule there are two lone pairs in the vicinity of the
central O-atom along with two bond pairs with hydrogen. Here
we would expect the two lone pairs to repel each other more
strongly than do a lone pair and a bond pair, and of course, even
more strongly than two bond pairs.

36. Bonding in hydrides of P, As, Sb
In NH3, the H-N-H bond angles are about 1070 whereas in PH3 the H-P-H angles are about 930. Although increasing the extent
of hybridization reduces repulsion (nonbonding-bonding electron pairs) by giving bond angles closer to tetrahedral angle, but it
decrease the effectiveness of the overlap of the small hydrogen 1s orbital with the sp3 orbitals on the central atom. Therefore,
orbital overlapping is effective than hybrid orbitals.
The bond angles in AsH3 and SbH3 are even slightly smaller than those in PH3. Similar trend is observed for the series H2O,
H2S, H2Se, and H2Te.
In general, for the hydrogen compounds of the heavier members of each group, there is little tendency for the central atom to
form sp3 hybrids. Thus, the H-X-H bond angles indicate that the central atom uses essentially pure p-orbitals in bonding to the
hydrogen atoms, and the bond angles in these compounds are approximately 900.

37. Hybridization of Elements involving d Orbitals
The elements occurring in the third row of the periodic table (or below) can exhibit expanded octets. The equivalent concept
in valence bond theory is hybridization involving the d orbitals.
For third period elements, the 3d orbitals are involved in hybridization because their energies are close to the energies of the
3s and 3p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence the
hybridization involving either 3s, 3p and 3d (Outer d-orital) or 3d, 4s and 4p (Inner d-orital) is possible.
However, since the difference in energies of 3p and 4s orbitals is significant, no hybridization involving 3p, 3d and 4s orbitals
is possible.
Hybridization with inner d-orbital – dsp3, d2sp3, d2sp2. Hybridization with outer d-orbital - sp3d, sp3d2.

38. sp3d Hybridization
The hybridization of one s orbital, three p orbitals, and one d orbital results the formation of five sp3d hybrid orbitals.
The five new sp3d hybrid orbitals will be rearranged such that three of them in a plane at an angle of 1200 to one another and
the other two in a direction perpendicular to the plane i.e. trigonal bipyramidal arrangement with two bond angles 1800 or
1200 and 900.
E.g. AsF5, PCl5, SF4, ICl3, etc.

39. sp3d2 Hybridization
The hybridization of one s orbital, three p orbitals, and two d orbital results the formation of six sp3d2 hybrid orbitals. The six
sp3d2 hybrid orbitals have a octahedral or square bipyramidal arrangement with two bond angle 1800 or 900 i.e. four sp3d2
hybrid orbitals were dispersed in a plane at an angle of 900 each and the rest two are directed up and below this plane in a
direction perpendicular to it.
E.g. SF6.

40. sp2d or dsp2 hybridization
The hybridization of orbitals belonging to the same energy level (say 3s, 3p, and 3d orbitals) of an atom. There is very little
energy difference between orbitals such as 3d, 4s and 4p orbitals may undergo dsp2 hybridization.
The d orbital 𝑑(𝑥2−𝑦2)involved in this type of hybridization has the same planar character as the two p orbitals (Px and Py)
have and the hybrid orbitals will also be planar (XY-plane), dispersed in such a way so as to be farthest apart i.e., subtending
an angle of 90º between them.
This gives a square planar arrangement for them and the hybridization is, therefore, called Square planar hybridization.

41. sp2d2 or d2sp2 hybridization
The hybridization of two d orbitals and high energy one s orbital and two p orbitals results the formation of six sp2d2 hybrid
orbitals. The five sp2d2 hybrid orbitals have a square pyramidal arrangement with two bond angle 1800 or 900.

42. Equivalent and Non-Equivalent hybrid orbitals
Hybridization the orbitals should have nearly same energy. But it is not true in the case of hybridization involving d
orbitals. d orbitals have much more higher energy than p orbitals and involved in hybridization.
A huge positive charge developed on central atom which shrinks the d orbital and reduce the energy gap.
Equivalent hybridization is a type of hybridization in which all hybrid orbitals are of same shape, size and energy. All bond
angles are same.
e.g.: sp hybridization, sp2 hybridization, sp3 hybridization etc. They have same shape, same size, same energy and same
bond angle.
In Non Equivalent Hybridization all hybrid orbitals do not have same shape, size and energy. Bond angles are also
different. Some hybrid orbitals may have some similarity but not all.
e.g.: sp3d hybridization. In this type of hybridization there are 5 hybrid orbitals, There are two possibilities of this - Trigonal
bi-pyramidal and Square pyramidal.
Trigonal bi-pyramidal: In this there are two
types of hybrid orbital. Bond angle between
these two types is 90o. Example: PCl5.
Square pyramidal: In this there are two types of hybrid orbital - Four equivalent orbitals developed
from (s, px, py and dx
2
-y
2) and one is the pure pz orbitals lying perpendicular to the xy-plane.
Three equivalent orbitals
developed from(s, px and py)
: Lie in the equivalent plane and Bond
angle is 1200.
Two equivalent orbitals
developed from (pz and d2
z)
: Lie in the axial plane and Bond angle is
1800.
pz

43. Hybridization of CN-

44. Hybridization of CO3
2-

45. Hybridization of ClO4
-

46. Hybridization of PF5

47. Hybridization of SF6