VALENCE BOND THEORY AND MOLECULAR ORBITAL THEORY
PART
I:
VALENCE
BOND
THEORY
A. INTRODUCTION
Valence Bond (VB) Theory proposes that “overlapping” of atomic VALENCE orbitals results in forming a
BOND between two atoms. Specifically:
Covalent bonds are formed by the overlap of valence atomic orbitals.
To simplify matters, valence bond theory assumes that lone pair orbitals and inner shell orbitals (that is, filled
atomic orbitals) are unchanged when bonds form, and that the shape of filled atomic orbitals is not distorted
by the presence of neighbouring atoms.
The bond orbitals formed by this overlap of atomic orbitals are said to be LINEAR COMBINATIONS of
ATOMIC ORBITALS (LCAO’s). This orbital overlap to form new bond orbitals is analogous to the situation
occurring when two water droplets on a glass surface touch each other and blend into a new, larger drop.
VALENCE BOND THEORY PREDICTS THE SHAPES OF BOND ORBITALS, HOW MOLECULES ARE
BUILT FROM SUCH ORBITALS AND THE SHAPES OF MOLECULES.
Atomic orbitals are expressed as a “wave functions” that are mathematical combinations of exponential, sine
and cosine functions. The overlap of atomic orbitals can be visualized by illustrating what happens when two
sine functions, “A” and “B”, are added. (The “+” and “–” in the diagrams simply indicate that the sine wave is
above or below the x–axis.)
+
+
+
Р
Р
A
A+ B
B
Р
In the example above, the sine waves are “in phase” and the amplitude of the resulting wave is increased: the
waves are said to reinforce each other.
Р
+
+
A
Р
B
A+B
In the example above, the sine waves are “out of phase” and the amplitude of the resulting wave is
decreased: the waves are said to cancel each other.
The wave functions of the “s” and “p” atomic orbitals may be represented as shown below.
s=
+
p=
+
Р
Here, the “+” and “–” have no physical meaning. They show that a particular electron “wave function” has
different phases at different points in space, roughly analogous to the way a “+” and “–” is assigned to the sine
functions, above.
The overlap of atomic orbitals operates similar to the way sine waves are added.
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EXAMPLE:
In the above examples, the “+” parts of the wave functions overlap, so their electron wave functions ADD and
a bond is formed. This is similar to the sine wave reinforcement on page 1.
On the other hand, when the s–orbital of one atom approaches a second atom's p–orbital “from the side”, the
following situation occurs.
+
+
–
In this case, equal amounts of s(+) + p(+) overlap AND s(+) + p(–) overlap give a NET CANCELLING of
electron wave functions, so that NO BOND forms. This is similar to the sine wave cancellation shown on
page 1.
ORBITAL DIAGRAMS
———————————————————————————————————————————————
A convenient method of showing how electrons fill atomic orbitals is the use of ORBITAL DIAGRAMS.
EXAMPLE: The orbital diagram showing the filling of selenium's atomic orbitals when it forms H 2Se is:
+
4s
4p
Se
where
1s
1s
HA HB
H2Se
and a re el ectrons wi th o ppos ite spins
, a nd HA and HB a re di ffere nt h yd rogen ato ms
.
HYBRIDIZATION
———————————————————————————————————————————————
Mixing atomic orbital wave functions creates new orbitals that are combinations or “hybrids” of atomic orbitals.
The hybridization of atomic orbitals is the basis of the orbital shapes and geometric orientations used in the
valence–shell electron–pair repulsion theory in the upcoming Section C.
The more two orbitals on different atoms overlap, the stronger the bond between the atoms.
The advantage of hybrid orbitals is that they overlap better with other orbitals and therefore form
stronger bonds.
(a) sp Hybrid Orbitals
————————————————————————————————————————————————
AP : VB AND MO THEORY
3
——————————————————————————————————————————————————
An “sp” hybrid orbital is formed by mixing an “s” with a “p” atomic orbital from the same atom. Because
2 orbitals are mixed together, 2 new orbitals are produced, only now they are “mixed” (or hybrid) orbitals.
(One cup of applesauce plus one cup of cranberries, put into a blender, yields 2 cups of “cranapple”
sauce.)
+
+
–
+
s
–
+
p
–
+
sp
+
+
sp
–
–
+
LINEAR
geo metry
ind ividua l hybri d orb ital s
both hybri d orb ital s
on the same ato m
The two hybrid orbitals take on a linear arrangement to minimize the repulsion between electrons in the
two orbitals.
Analogy: Two tear–drop shaped balloons tied end-to-end push each other until they are 180o apart.
EXAMPLE: Beryllium uses sp hybrid orbitals when bonding with hydrogen.
BeH2 =
+
H
+
–
–
+
Be
+
H(1s) + Be (sp) overlap
H
Background theory
The formation of sp hybrid orbitals requires unpairing the 2s electrons and promoting one into a stable
first excited state. The excited state is stabilized as a result of the net lowering of energy occurring
when electrons are allowed to occupy hybrid orbitals. (The energy of the hybrid orbitals is the
average of the energies of the atomic orbitals used to form the hybrids.)
form
hybrid
Be (groun d state)
two sp
hybrid
orbi tals
Be (1st excited s tate )
uno ccupie d
p o rbita ls
(b) sp2 Hybrid Orbitals
————————————————————————————————————————————————
An “sp2” hybrid orbital is formed by mixing an “s” with TWO “p” atomic orbitals from the same atom.
+
+
2s
+
–
+
2p
+
+
–
–
2p
(p orbitals are p erpen dicular
to each o ther)
+
–
+
+
–
+
sp2
in dividu al h yb rid o rbita ls
+
–
+
– –
+
TRIANGULAR
PLANAR
ge ometry
The three hybrid orbitals have a triangular planar arrangement to minimize the repulsion between
electrons in the three orbitals.
Analogy: Three tear–drop shaped balloons tied together push each other until they are 120 o apart.
EXAMPLE: Boron uses sp2 hybrid orbitals when bonding with hydrogen.
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H
H
B
H(1s) + B(sp2) overl ap
1s
H
Background theory
The formation of sp2 hybrid orbitals requires unpairing the 2s electrons and promoting one into a
stable first excited state. Again, the excited state is stabilized as a result of the lowering of energy
occurring when electrons occupy hybrid orbitals.
form
hybrid
B (g round sta te)
thre e sp 2
hybrid
orbi tals
B (1 st e xcite d state)
uno ccupied
p o rbita l
(c) sp3 Hybrid Orbitals
————————————————————————————————————————————————
An “sp3” hybrid orbital is formed by mixing an “s” with THREE “p” atomic orbitals from the same atom.
Again, since 4 atomic orbitals are mixed and 4 hybrid orbitals are produced.
+
+
+
2s
–
+
+
2p
+
+
+
–
–
2p
2p
+
+
TETRAHEDRAL
(4 sp3 o rbita ls)
Analogy: Four tear–drop shaped balloons tied together push each other until they are 109o apart.
EXAMPLE:
Carbon uses sp3 hybrid orbitals when bonding to hydrogen.
H
C
H
H
H
Background theory
The formation of sp3 hybrid orbitals requires unpairing the 2s electrons and promoting one into a
stable first excited state.
AP : VB AND MO THEORY
5
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form
hybrid
C (gro und state)
fou r sp 3 hybri d
C (1s t excited sta te)
(d) sp3d Hybrid Orbitals
————————————————————————————————————————————————
An “sp3d” hybrid orbital is formed by mixing an “s” orbital with THREE “p” orbitals and ONE “d” orbital from
the same atom.
+
3s + 3 p+ 3 p + 3p + 3d
+
+
TRIGONAL BIPYRAMID
ge ometry
+
Note that this hybrid makes use of an unoccupied 3d orbital with relatively low energy. In the case of As,
for example, the 3d orbitals are already filled, so that the hydrid orbitals are made as follows.
5 “sp3d” orbitals
4s + 4p + 4p + 4p + 4d
The use of sp3d orbitals (and sp3d2, below) is illustrated in Section C.
(e) sp3d2 Hybrid Orbitals
————————————————————————————————————————————————
An sp3d2 hybrid orbital is formed by mixing an “s” orbital with THREE “p” orbitals and TWO “d” orbitals
from the same atom.
+
+
3s + 3p + 3p + 3 p + 3d + 3d
+
+
+
OCTAHEDRAL
geo metry
+
EXERCISE:
1. Predict the angles between each of the following.
(a) two sp orbitals.
(b) two sp2 orbitals.
(c) two adjacent sp3d orbitals. Note that there are 2 different possible angles involved.
(d) two adjacent sp3d2 orbitals.
(e) NASTY (for math freaks only) Predict the angle between two sp3 orbitals (Hint: In the following
diagram A, B, C, D are the corners of a tetrahedron. The tetrahedron rests inside a cube and O
is at the exact center of the cube.)
A
B
O
C
D
B. LEWIS STRUCTURES
(A Review From Chemistry 11)
I. The Lewis Structures of Covalent Compounds that Obey the Octet Rule
————————————————————————————————————————————————
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Lewis Structures show how the VALENCE electrons are distributed in a molecule. The octet rule states
that most atoms, other than hydrogen, tend to attain an octet of electrons as a result of forming covalent
bonds.
EXAMPLE: The Lewis structure of H2O is shown below.
H O
H
For water, above, H and O contribute one electron each to the covalent bonds between them and share
the two electrons in the bond. Each H can then “lay claim” to a closed shell of 2 electrons. The O atom
has 4 electrons which it does not share with the H's in addition to the 4 electrons shared with the H's, for a
total of 8 electrons: a closed shell and a “full octet”.
Drawing the Lewis Structures of molecules follows a simple set of rules.
THE “RULES OF THE GAME”
A. Count up the total number of valence electrons in the molecule. Adjust this number by
SUBTRACTING one electron for every POSITIVE charge and ADDING one electron for every
NEGATIVE charge on the molecule.
B. Determine which atoms are bonded together and put 2 electrons into each bond.
Note: You will always be shown which atoms are connected to which other atoms.
C. Use the remaining valence electrons to complete the octets of the atoms surrounding the
central atom(s). Then place any remaining electrons, in pairs, on the central atom(s).
D. If a central atom has less than an octet of electrons, have a neighbour share electrons with
the “deficient” atom by putting an extra pair (or pairs) of electrons into the shared bond.
E. Tidy up: replace each pair of electrons engaged in a bond with a dash, “—”.
EXAMPLES:
NH 4
This molecule has 8 valence electrons: N has 5, each of the 4 H’s have 1, and one electron is
removed to make a +1 charge.
The atoms connected together are shown below (

indicates a bond).
H
H
N
H
H
First, assign 2 electrons to each bond.
H
H
N
H
H
A check shows that each H has its required 2 electrons (2 electrons is a full shell for H) and
the N has an octet of 8 electrons. The structure is finished, apart from tidying up by replacing
each pair of electrons in a bond with a “dash” and adding the charge.
H
H
N
+
H
H
CHO 2

This molecule has 18 valence electrons: C has 4, H has 1, each of the 2 O’s have 6, and one
AP : VB AND MO THEORY
7
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extra electron is added to make a –1 charge.
The atoms that are connected are shown below.
H
C
O
O
First, assign 2 electrons to each bond.
H
C
O
O
Second, fill the octets on each O attached to the central C.
H
C
O
O
All of the 18 available electrons have been distributed, but the central carbon atom does not
yet have a full octet (it only has 6 electrons). Arbitrarily, let one of the oxygens donate an extra
2 electrons to a bond shared with carbon.
H
C
O
or, a ltern atel y
H
C
O
O
O
Replacing each pair of electrons in a bond with a “dash” and adding the charge gives the
finished structure.
H
C
O
or, a ltern atel y
O
HOPO
H
C
O
O
This molecule has 18 valence electrons: H has 1, P has 5 and each O has 6.
The atoms are connected as shown below.
H
O
P
O
First, assign 2 electrons to each bond.
H
O
P
O
Second, fill the octets on the outer O's attached to the “central” P.
H
O
P
O
At this point, 16 out of the 18 electrons available are now used. The remaining 2 electrons are
put on the central P atom.
H
O
P
O
The central P atom only has 6 electrons and needs another 2 electrons to be donated by one
or the other of the attached oxygens.
• If the left oxygen donates 2 electrons the structure below results.
H
O
P
O
• If the right oxygen donates 2 electrons the structure below results.
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H
O
P
O
Note: Either answer is considered to be correct; they are alternate ways to represent the
Lewis structure and the actual situation within the molecule will act as if it were an
average of the two structures.
Tidy up by substituting a “dash” for each pair of bond electrons.
O
H
P
O
or
H
O
P
O
II. The Lewis Structures of Covalent Compounds that Violate the Octet Rule
————————————————————————————————————————————————
Electron–Deficient Molecules
In addition to H, the atoms Be, B and Al are exceptions to the tendency for covalently–bonded atoms
to complete their octets. These atoms have such low electronegativities that the best they can do is
to GAIN ONE EXTRA ELECTRON IN A COVALENT BOND FOR EVERY ELECTRON THEY
CONTRIBUTE TO THE BOND. (They do not have sufficient electronegativity to pull extra electrons
on an adjacent atom into covalent bonds.) Therefore:
Be has 2 valence electrons and can share a maximum of 4 electrons, and
B and Al have 3 valence electrons and can share a maximum of 6 electrons.
Definition: A molecule in which one or more atoms (other than hydrogen) does not possess a full
octet of electrons is called an ELECTRON–DEFICIENT molecule.
BF3 , below, is an example of an electron-deficient molecule because the central B atom only has 6
valence electrons after bonding.
EXAMPLE:
Draw the Lewis Structure of BF3 .
This molecule has 3 (B) + 3 x 7 (F) = 24 valence electrons.
The atoms are connected as shown below.
F
B
F
F
First, assign 2 electrons to each bond.
F
B
F
F
Second, complete the octets of the outer fluorines attached to the central B.
F
B
F
F
All 24 of the available valence electrons have now been assigned. No attempt is
made to try to fill the octet of the central boron: recall that atoms in column 3
have 3 valence electrons and only get up to a maximum of 6 electrons after
AP : VB AND MO THEORY
9
——————————————————————————————————————————————————
bonding occurs. The low electronegativity of B prevents it from attracting extra
electrons from F.
The structure is completed by substituting a dash for every pair of bond electrons.
F
F
B
F
Atoms Having an Expanded Octet of Valence Electrons
Elements in the third and fourth periods of the periodic table frequently attain more than an octet of
valence electrons when they form covalent compounds. (The extra electrons are placed in low–lying
d–orbitals.) Other than the fact that the central atom ends up with more than eight valence electrons,
the same rules are used for forming the Lewis Structures.
EXAMPLE:
Draw the Lewis Structure for PCl5 .
The chlorine atoms are connected to the central phosphorus atom.
Cl
Cl
P
Cl
Cl
Cl
The molecule has a total of 40 valence electrons. Two electrons are assigned to each
bond and the octet of each Cl is filled.
Cl
Cl
Cl
P
Cl
Cl
The phosphorus atom has an expanded octet of 10 valence electrons.
EXERCISE:
2. Assign Lewis structures to the following molecules.
(a)
(d)
H
H
Cl
(b)
H
H
C
C
H
H
H
(e)
I
I
H
H
C
C
H
H
(c) I
Cl
(f)
C
H
C
H
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(g) F
Be
F
(h)
O
O
(i)
N
N
(k)
H
S
Cl
H
O
(j)
Cl
C
H
(l)
C
H
C
C
H
(p) [ O
N
O ]–
(q) [ N
(s) [ H
N
H ]
(t)
H
H
(v) N
N
O ]
S
O
F
(w)
N
C
(u) Cl
S
S
Cl
N
O ]
F
N
(z)
O
O
(bb) H
O
F
H
O
H
(x) [ C
S
F
(y)
O
H
F
F
H
(r)
C
H
O
C
C
H
C
C
H
H
C
C
H
(aa) O
N
N
O
O
Br
O
H
(cc)
H
Br
Se
Br
Br
C. VALENCE–SHELL ELECTRON–PAIR REPULSION (VSEPR) THEORY
VALENCE BOND THEORY assumes that bonds form by overlapping orbitals on adjacent atoms, so that
electrons in overlapping atomic orbitals are shared by both atoms. The valence bond theory also assumes
that the more orbitals overlap, the more electrons are shared and the stronger the bond. This increased
sharing of electrons causes atoms to position themselves to allow the maximum possible orbital overlap.
A simple version of valence bond theory is called the VALENCE–SHELL ELECTRON–PAIR REPULSION
THEORY (VSEPR Theory).
The following additional definitions are used.
BONDING ORBITALS, BONDING PAIR.
These terms refer to orbitals that are directly involved in
AP : VB AND MO THEORY
11
——————————————————————————————————————————————————
forming bonds, and to the pairs of electrons that occupy such orbitals. The electrons in a bond may come
from either of the atoms involved in the bond.
LONE PAIR ORBITAL, LONE PAIR ELECTRONS. These terms refer to a valence orbital which contains
electrons NOT directly involved in a bond.
H

EXAMPLE:



O
H
bon d pa ir el ectrons in b ondi ng o rbita l

O

lon e pa ir orb ital
lon e pa ir el ectrons



OCCUPIED valence
orbitals ina molecule are somewhat different from orbitals in an atom and consist of
“lobes” extending out from theatom in ONE direction. The position of these lobes, relative to each other, can

be shifted around to minimize electrical repulsions existing between electrons in orbital lobes.
Because electrons repel each other, an atom's occupied valence orbitals stay as far from each other as
possible. Lone pair electrons are confined to a small region of space around a single atom, whereas bond
electrons move farther away from the central atom by shifting closer to the attached atom. Hence, bonding
electrons suffer less repulsion from other central–atom electrons because the bonding electrons retreat
farther from the central atom. As a result of differing amounts of repulsion for bond pair and lone pair
electrons, the following rules apply.
 repulsion 


 between 2  


LONE PAIRS 
 repulsion between 


a LONE PAIR and a 


 BOND PAIR

 repulsion 


 between 2 


BOND PAIRS 

TABLE OF MOLECULAR SHAPES
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Total n umbe r
of orbitals
A = attach ed a tom
C = central a tom
A
2
A
3
= b ondi ng o rbita l
= l one pair orbital
C
A
LINEAR
A
A
A
C
C
A
TRIANGULAR
PLANAR
V-SHAPED
A
C
C
A
A
4
A
A
TETRAHEDRON
C
A
A
A
TRIGONAL
PYRAMID
A
A
A
V-SHAPED
A
A
A
C
5
A
A
C
A
A
IRREGULAR
TETRAHEDRON
A
A
C
A
T-SHAPED
A
LINEAR
A
A
A
C
A
A
A
A
TRIGONAL
BIPYRAMID
6
C
A
A
OCTAHEDRON
A
A
C
A
A
C
A
SQUARE
PYRAMID
A
A
SQUARE
PLANAR
The VSEPR theory has a fairly general application, but is best suited to species where the central atom is a
REPRESENTATIVE element; that is, elements that are NOT transition metals.
The fact that some central atoms bond to 5 or 6 (or more) other atoms implies 10 or more electrons can exist
in the valence shell. The ability of a central atom to accommodate 10 or 12 valence electrons is a result of
involving one or more low–lying d–orbitals; this is where sp3d and sp3d2 orbitals come in.
THE RULES OF THE GAME
1. Write the Lewis structure for the molecule, as usual.
• Count ALL the valence electrons, adding or subtracting electrons to allow for the overall charge.
• Supply electrons to the surrounding atoms to bring them to a full–shell configuration.
• Make sure there is AT LEAST ONE BOND between each attached atom and the central atom.
2. Determine the number of left–over electrons on the central atom, that is “lone–pair electrons” (lpe),
as follows.
AP : VB AND MO THEORY
13
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# of lpe = (total valence electrons) – (# of e– donated to other atoms)
3. Determine the number of lone–pair orbitals (lpo) on the central atom required to hold the lpe as
follows.
# of lpo = (# of lpe) / 2
If the number of lpe is an ODD number, round the number of lpo UP (because the odd electron
occupies a full orbital.
4. Determine the total # of orbitals required as follows.
total # orbitals = (# of attached atoms) + (# of lpo)
5. If the central atom is not in columns I, II or III, and if the central atom has less than an octet of
electrons, form multiple bonds to the central atom.
6. Use the Table of Molecular Shapes (you will eventually have to memorize it) to assign a geometry
based on
(A) the total # of orbitals, and (B) the # of lpo.
The presence of multiple bonds does not affect the geometry of the molecule. Double and triple
bonds are considered to be “fat single bonds”.
EXAMPLE:
What is the shape of SF4?
•
•
•
•
•
•
•
SF4 has 34 e– (6 from S, 7 from each F)
F 
Assigning 8 e– to each F uses up 32 valence e–
 F S F 
There are 34 – 32 = 2 lpe


2


Therefore, there is /2 = 1 lpo

 
   F 
Overall, there are 5 orbitals: 1 lpo + 4 attached atoms

 


S has 10 valence e– (5 orbitals x 2 e–), so no double bonds are required


From the Table of Molecular Shapes we assign the following shape
  

S
EXAMPLE:

F
 
IRREGULAR
TETRAHEDRON
  
(based on a trigonal bipyramidal

 F
 
  F 
arrangem ent of orbitals)


 



What is the
shape of XeF2?

  
 

• XeF2 has 22 e–
(8 from Xe, 7 from each F)
• Assigning 8 e– to each F uses up 16 valence e–
F Xe
• There are 22 – 16 = 6 lpe

  
• Therefore, there are 6/2 = 3 lpo
• Overall, there are 5 orbitals: 3 lpo + 2 attached atoms




required

• Xe has 10 valence e– (5 orbitals x 2 e–), so no double bonds are


• From the Table of Molecular Shapes we assign the following shape
 



 
 
 F 






F



 F 
Xe
EXAMPLE:

F 

LINEAR
(bas ed o n a trigo nal bipyram idal
arran geme nt o f orb ital s)

What is the
shape of SO2?

– 

(6 from S, 6 from each O)
• SO2 has 18 e
• Assigning 8 e– to each O uses up 16 valence e–
 O




S





O


14
Hebden : Chemistry AP
——————————————————————————————————————————————————
•
•
•
•
•
There are 18 – 16 = 2 lpe
Therefore, there are 2/2 = 1 lpo
Overall, there are 3 orbitals: 1 lpo + 2 attached atoms
S only has 6 valence e–, so a double bond is formed
From the Table of Molecular Shapes we assign the following shape

O 




 O

S
V-SHAPED
(bas ed on a triangular planar
arrangement of orbital s)
 

POLAR AND NON–POLAR MOLECULES (A Partial Review of Chemistry 11)

————————————————————————————————————————————————
A dipolar bond exists when the atoms in the bond have different ELECTRONEGATIVITIES.
Example: In the molecule H–Cl, H has a smaller electronegativity than Cl. As a result, the electrons in
the bond are shared unequally: Cl has a greater tendency to attract electrons than does H, so
the shared electrons are displaced more toward the Cl.
H

Cl


where + means “a slight excess of positive charge”
– means “a slight excess of negative charge”
 
Whenever a bond exists between two different atoms, the bond is polar.
In addition to having polar bonds, a molecule must also possess one other property in order to be a
POLAR MOLECULE: THE MOLECULE MUST BE ASYMMETRICAL.
A molecule is ASYMMETRICAL if one end of the molecule is different from the other end.
The following molecules all have at least one polar bond, are ASYMMETRICAL, and therefore are
POLAR MOLECULES.
H
H
Cl
H
I
H
C
Cl
O
B
I
H
H
Br
The following molecules all have at least one dipolar bond BUT are SYMMETRICAL, so that they are
NON–POLAR MOLECULES.
H
Cl
Cl
F
Be
F
B
H
H
(borane)
Cl
C
Cl
C
Cl
Cl
Notice that the borane molecule is actually symmetric although it might appear that one “end” of the
molecule has 2 H's and the other “end” only has one. The 3 H's are attached at 120o angles around the
B. A better way to think of symmetry is as follows.
Pretend the central atom is a rock and each bond is a rope attached to the rock, pulling
outwards from the rock. If each attached atom has the same electronegativity, each
rope is pulled equally. If the attached atoms are symmetrically arranged around the
central atom, each attached atom's “pull” on the central “rock” CANCELS the pull from
the other “ropes” and the central rock/atom doesn’t move. This total cancellation of
AP : VB AND MO THEORY
15
——————————————————————————————————————————————————
“pull” on the electrons in the molecule creates a NONPOLAR MOLECULE.
Using the above analogy, you should be able so see why the presence of three hydrogen atoms
symmetrically arranged around a central B in BH3 produces a NONPOLAR MOLECULE.
By studying the Table of Molecular Shapes you should see that only the following shapes give rise to
POLAR MOLECULES (if the attached atoms are identical).
V–shaped
trigonal pyramid
irregular tetrahedron
T–shaped
square pyramid
EXERCISES:
3. Write the Lewis structures and predict the shapes of the following. Also, state the hybridization of
orbitals that exists on the central atom.
(a) SeF4
(e) IF5
(i) ClO 3
(m) SCl2
(q) ArF3
(u) FOO
(b) O3
(f) ClF3
(j) SbCl5
(n) SF6
(c) ICl5
(g) BF4
(k) I 3
(o) BeCl2
(l) ClO2
(p) NCl3
(d)
SiF62 
(h) AsF5


(r) XeO2F2
(v) IO 4
(s) KrF4
(w) ICl4
(t) PCl3
(x) BF3

4. Which of the above are free radicals? Recall that a FREE RADICAL is a species containing an ODD


number of electrons.

5. Which of the above are polar molecules?
6. Write Lewis structures for the following and state the shape of the molecule around each bold atom.
(a)
Cl
H
(c)
C
Cl
C
C
O
H
(e)
C
N
C
Cl
(d)
H2C
NH3
C
N
C H2
CH2
(f)
+
S
H
O
O
–
O
(b)
H
C
C
N
N
CH3
16
Hebden : Chemistry AP
——————————————————————————————————————————————————
D. SIGMA AND PI BONDS
Examine the diagram below, showing the orbitals involved in forming bonds in CH 3 – CH3 . Notice the
overlaps between the H(1s) and C(sp3) orbitals, and between the C(sp3) orbitals on each carbon atom.
H
H
H
sp3
C
H
C
H
sig ma () bo nd
H
Definition: A SIGMA (  ) BOND is a covalent SINGLE bond in which shared electrons are centered on a
line connecting the nuclei of the two atoms involved.
A question now arises: what happens to unhybridized p orbitals in atoms using sp and sp 2 hybridization? In
the molecule CH2=CH2 , below, all atoms lie in one plane. In addition, all the bonds lying in the plane are
sigma bonds. If you refer back to “sp2 Hybrid Orbitals – Background Theory” on page 4, you will notice that
the process of hybridization produces 3 sp2 hybrid orbitals and leaves one atomic (unhybridized) p orbital.
The unhybridized p orbitals on each carbon (shown shaded) are perpendicular to the plane.
H
H
sp2
C

1s
C
sp2
H
H

2p
2p
pi () b ond forme d by the o ve rlap of
un hybridi ze d orb ital s on ca rbon
The p orbitals are actually “fat” enough that they overlap and produce a pi bond that lies entirely above and
below the plane.
+
+
+
th is i s a sing le b ond wi th
2 “lobe s” of opp osite
ph ase (the 2 el ectrons in
th e pi bon d are spre ad
over the 2 lo bes)
hybri dize
–
–
–
Definition: A PI (  ) BOND is a covalent bond in which adjacent parallel p orbitals share an electron pair
occupying the space above and below the axis joining the atoms.
Whenever a molecule has a double or triple bond, the following statements are true:
AP : VB AND MO THEORY
17
——————————————————————————————————————————————————
• all single bonds are SIGMA bonds.
• all bonds over and above the first bond in a multiple bond are PI bonds.
H
H

C
 

C
H
C
C


H
H

H
double bond = 1  + 1 
triple bond = 1  + 2 
The presence of pi bonds leads to one of two possible arrangements of atoms around the pi–bonded atoms.
Case I: Double bonds create a PLANAR arrangement of atoms.
H
VSEPR theory p redicts that CH2=CH2
ha s a trian gula r pla nar a rrange ment of
atoms aroun d ea ch carb on.
H
C
C
H
H
The resulting molecule is PLANAR and has the following orbital arrangement:

H


 C
C
H
H

H
or

H
C
H
C
H
H
Case II: Triple bonds create a LINEAR arrangement of atoms.
VSEPR theory predicts H — C
C — H has a linear arrangement of atoms around each carbon.

H

C

 C
 H
or
H
C
C
H
The presence of p–bonds “locks” the atoms rigidly to each other, so that the 2 atoms involved
in the bond cannot “rotate” with respect to each other.
Note: “Sigma” is the Greek letter “s” and sigma bonds usually involve orbitals that are at least partially s–
orbitals in nature.
“Pi” is the Greek letter “p” and pi bonds always involve atomic p–orbitals.
“Delta” is the Greek letter “d” and delta bonds involve d–orbitals. (Delta bonds are rare).
EXERCISES:
7. Predict the type of orbitals (  ,  ) and arrangement of atoms involved in the following molecules.
18
Hebden : Chemistry AP
——————————————————————————————————————————————————
(a)
CH2=C=CH2
(c)
F–N=N–F
O=N–N=O
(e)
O
O
(b)
H–O–CH=CH–C N
(d)
H
CH3–C–CH3
(f)
C=O
H
8. Predict the hybridization of the bold atoms in exercise 6.
9. (a) In the simplest cases, sigma bonds are formed by overlap of s orbitals and pi bonds are formed
by overlap of p orbitals. Note that “sigma” is the Greek letter “s” and “pi” is the Greek letter “p”.
Which elements would you expect to be involved in delta bonding?
(b) Why is it that elements in the 1st and 2nd period can never form bonds involving sp 3d and sp3d2
orbitals?
(c) Which molecular shapes would you expect can never form with elements in the 2nd period?
10. Draw orbital diagrams for each of the following molecules and describe the types of hybrid orbitals
involved in the bonds.
(a) BeF2
(b) AlF3
(c) SnF2
(d) SnF4
(e) SnF62 
(f) SbF3
(g) SbF5
11. If the oxygen atom in water formed bonds involving unhybridized p orbitals, instead of sp3 hybrids,
what would the H–O–H angle be? Explain your answer.

12. Examine the structures of butane and cyclobutane, below.
H
H
H
H
H
C
H
C
90o
H
C
C
C
H
C
H
H
C
C
H
H
H
H
H
H
H
H
Which molecule has stronger –bonds between carbon atoms: butane or cyclobutane? Explain.
E. RESONANCE
The Lewis structure of SO2 can be written as

O
 
S

or as

O

O

S

O
  
These structures imply that SO2 should have one O–S single
 bond and one O=S double bond. An O–S
 double bond.
weaker and 

single bond is
longer
than
an
O=S
 Also, the stronger O=S double bond has a




 bond. In
higher vibrational
frequencythan the weaker O–S
spite of expectations, experiments reveal only one
kind of bond exists in SO2 . To resolve this problem, a new type of “average” structure called a RESONANCE
HYBRID is POSTULATED.
AP : VB AND MO THEORY
19
——————————————————————————————————————————————————
The actual bond behaves as if it were the average of a single and a double bond.

1
2

S
O
 

+
O

O


S

 O
S
O


 O
“1 1/ 2 bo nd”



 of “bond order”
 between two

A plot of 
“bond
length”asa function
(the # of bonds
atoms) is shown below.













b ond
le ngth
1
1.5
2
bo nd o rder
RESONANCE is a way of representing the actual electron distribution of a molecule or polyatomic ion as a
composite or average of two or more Lewis structures, called the resonance or contributing structures (none
of which has a real independent existence).
The two RESONANCE STRUCTURES of SO2 are shown to be related to each other by using a doubleheaded “resonance arrow”, as shown below.


S
S


O
O
O 
O
 





The structures appear to show electrons moving from bond to bond; that is,“resonating” between two possible
structures.
is no such electron
the electrons involved are simply “smeared” over the
In fact, there
 movement:






atoms involved.


Definition: A RESONANCE HYBRID is the ACTUAL bond structure of a molecule or polyatomic ion. It is
given by the composite or average of all the contributing or resonance structures.

Fo r examp le, the re sona nce
hybri d of SO2 i s:
S
O


 O
th e do tted lin e indicates
a parti al b ond
The bond order and bond length of a 
particular bond
in a resonance hybrid is estimated by averaging the

bond order and bond length of the individual
resonance
 structures.

EXAMPLE: The nitrate ion has the following resonance structures.
 O 
  O

 
 
N
 O 

 

 
 O
O

N
 O 

O
  


 
O
N

 
O 
Averaging the shaded bond orders: average bond order = (1+1+2)/3 = 4/3






O bond length)



1x
(N


2
x
(N
O
bond length)









=

and average
bond length
3
All bonds in the molecule actually have the same bond order and bond length.
EXERCISE:

20
Hebden : Chemistry AP
——————————————————————————————————————————————————
13. The bond length of C=O is 120 pm (pm = “picometers” = 10–12 m); the bond length of C–O is 143 pm.
(a) The formate ion, HCO 2 , has 2 resonance structures, shown below.

O 


C

O

O

C
O




 H

 



 H


Estimate the average bond length and bond order of the C–O bond in the resonance hybrid.
(b) The carbonate ion, CO 23  , has three resonance structures. Estimate the average C–O bond
length and bond order in the carbonate ion resonance hybrid.

F. DRAWING THE RESONANCE
STRUCTURES OF A NEUTRAL MOLECULE
Resonance structures are possible whenever a central atom is surrounded by at least one double bond and at
least one single bond. (The single bond must not be to a hydrogen atom.) The important features of
resonance structures are shown by writing the Lewis structures of SO3 .
 O 
S


 


O 
 O

Because any of the THREE S–O bonds can be chosen to receive the extra electrons used to form the double

bond, there are THREE possible resonance
structures
involving one double bond.


 


RULE: The number of choices for resonance structures involving double bonds equals the number of
equivalent choices that can be made as to where to place the double bonds.
The three resonance structures produced this way are shown below.
 O 


 

O 

S
 O 


 

O


O

S
 O 

 O


 

 O

EXERCISES:







2



the following
 in decreasing bond
 order: CO3
14. List
, CO, CO2 .

 

S

O
 



15. Draw the resonance structures for each of the following molecules. No lone pair electrons are shown.
(a)
NO2
(b)
H
H
C
C
H
C
(c)
H
H
H
2Р
CO3
G. FORMAL CHARGE
(f)
O=C=O
H
H
O
N
O
C
C
O
H
C
C
H
(e)
C
C
C
(d)
N
O
AP : VB AND MO THEORY
21
——————————————————————————————————————————————————
The “formal charge” on an ATOM in a MOLECULE is defined as follows.
FORMAL CHARGE = (# of valence e– ) – (# of bonds) – (# of lone pair e– )
where: a single bond counts as “1 bond”, a double bond as “2 bonds”, and a triple bond as “3
bonds” (makes sense, doesn't it?)
Examples:
# of valence
electrons
O
SO2 = 
S
# of bonds
# of lone pair
electrons
6
–
1
–
6
=
–1
6
–
3
–
2
=
+1
6
–
2
–
4
=
0


 

 O 



 

Note that the overall charge is: –1 + 1 + 0 = 0
 O 
6
–
2
–
4
=
0
N
5
–
4
–
0
=
+1
 O 
6
–
2
–
4
=
0
+
NO2 =

 

Formal
charge
Note that the overall charge is: 0 + 1 + 0 = +1

 

EXERCISE:
16. Calculate the formal charges of the non–hydrogen atoms in the following molecules.
(a) H

N


N


 H

(d) 
 Br




N
C
C


(b) 
 Br
H




 
 O

C
N
C

C
C 


H
H





O

C
N

(c) CH3
N




(e) CH3


O

CH

CH2
C

O
 
22
Hebden : Chemistry AP
——————————————————————————————————————————————————
H. RESONANCE STRUCTURES AND FORMAL CHARGES REVISITED
Although formal charges usually are not shown when drawing resonance structures, nevertheless it is often
useful to include the formal charge.
Including formal charges with the resonance structures of SO 3 produces the structure shown below.
Р


Р

O

2+
A = 
O
S
 O 





 bonds produces two more structures.
Moving electrons from oxygens
the single
into

Р
+

  



B =
O
S O
and
C = O
S O





 O 
 O 




Resonance structures having
lower formal charges make a greater
contribution to the actual bond structure in






a molecule. Hence,
structure C is actually the preferred
way to write the Lewis structure for SO 3 , although


 

 

ALL possible resonance
structures play a part in determining
the actual structure.
Structure C violates the octet rule by using low–lying d–orbitals to assign 12 valence electrons to S. Note that
second row elements such as C and N do NOT have d–orbitals available and never exceed an octet.
NO 2 has the following resonance structures, listed in more–or–less decreasing contribution to the overall
structure.




O

–
+
N

O


O


+
+
+
–
+
N
+


 



O
O
N
O
–
+

     
      
  2+  

O
N
O
O
N
O
O
N

O 
 

    




–
  
 


 
O

–
 3+ 
N
O
  
  


 

 

–
O

+
N
+
+
O

 O
–

N
–

O

+

O

–
+
–
–
–
  
N
O

3+
  3+   

 N O   O N O 
O
           




2+
O  N O
    

–
+




 
 
 
 
 
EXERCISE:


 

17. Draw some of the resonance structures of the following and include the formal charges for all non–
hydrogen atoms. Which of your resonance structures make(s) the largest contribution to the overall
structure?



(a) O = C = O


(b)



+
(d) CH2 = CH Р CH2



O

N
 O 
  




(e)
H
C C

H





O

H
(c) 
O


O

  



(f)



N


O

CH3 Р O Р CH = CH2

AP : VB AND MO THEORY
23
——————————————————————————————————————————————————
PART
II:
MOLECULAR
ORBITAL
THEORY
I . BACKGROUND THEORY AND TERMINOLOGY
In atomic systems, the probability of finding an electron around the nucleus is a function of the ATOMIC
ORBITALS (AO's) involved.
The VALENCE BOND THEORY is simplified by assuming that atomic orbitals are essentially unchanged
except when they overlap to make bond orbitals. This simplified theory works well for predicting molecular
shapes. Unfortunately, a simplified theory does not work well when examining the energies of electrons in a
molecule, or when examining systems with double or triple bonds.
The MOLECULAR ORBITAL (MO) THEORY assumes that one or more orbitals exist in a molecule such that
valence electrons are spread over an ENTIRE SET OF NUCLEI.
Disclaimer: The full MO theory is manageable only for diatomic molecules and quickly becomes
unmanageable for larger systems. (Triatomic and larger systems require massive
amounts of computer time and advanced mathematical analysis to create a meaningful
set of molecular orbitals.) Obviously, we will ONLY look at simple diatomic systems.
Further, we will ONLY deal with elements in the first 2 rows of the periodic table.
When two atomic orbitals are mixed to create a molecular orbital, the mixing takes place in two ways.
1. AO's formed by ADDING atomic orbitals
————————————————————————————————————————————————
This is a situation where waves reinforce each other.
(a) SIGMA BONDING MO's: 
In this case, the greatest “electron density” lies directly between the nuclei.
i. Adding s–orbitals
+
1s
1s bonding MO
1s
ii. Adding 2px–orbitals
Note: Assume the following convention for 3–dimensional axes.
y
x
The nuclei of the two atoms in a bond
are assumed to l ie along the x-axis .
z
+
2px
2px
2p bonding MO
Notice that in each case the MO encompasses both nuclei because an atomic orbital is an orbital
confined to an atom, whereas a molecular orbital is an orbital spread out over an entire molecule.
24
Hebden : Chemistry AP
——————————————————————————————————————————————————
(b) PI BONDING MO's: 
In this case, the greatest electron density lies above and below an imaginary line connecting the
nuclei.
+
2py
or 2pz
2p bonding MO
2py
or 2pz
2. AO's formed by SUBTRACTING the atomic orbitals
————————————————————————————————————————————————
This is a situation where waves cancel each other.
(a) SIGMA ANTIBONDING MO's:
*
These orbitals resemble atomic orbitals that are repelling each other.
i. Subtracting s–orbitals
+
1s
*
1s
1s
antibonding MO
ii. Subtracting 2px–orbitals

+
2px
2px
*
2p
antibonding MO
(b) PI ANTIBONDING MO's: *
These orbitals resemble two parallel p–orbitals that arerepelling each other.
+
2py
2py
or 2pz
or 2pz
* antibonding MO
 2p

AP : VB AND MO THEORY
25
——————————————————————————————————————————————————
II. THE MO'S AND ENERGY LEVELS OF MOLECULES MADE FROM H AND He
The energy of
1s is lower than the energy of 1s* because the repulsion between antibonding electrons
increases the energy in a
1s*
orbital whereas the pairing of electron spins and attraction of bonding
electrons to opposing nuclei lowers the energy of electrons in a

1s*

Ene rgy
1s orbital.
1s
When two atoms with s–orbitals form a bond,
 the MO's formed are described by the “MO diagram” below.
 *
1s
1s
ind ividua l atom

A

1s
1s
ind ividua l atom
B
RESUL TING
MOL ECULE
Notice that overlap of s–orbitals to form –orbitals leads to a net lowering of energy, relative to the energies
of the separate atomic 1s orbitals. This energy decrease stabilizes the –orbital and favours the formation of
a –bond. On the other hand, electrons in *–orbitals have a higher energy than the separate atomic
orbitals. This implies that electrons in a *–orbital do not form stable bonds and tend to favour returning to
isolated atomic orbitals.
Definition:
BOND ORDER =
1
2
[ (# of bonding e–) – (# of antibonding e–) ]
THE MO DIAGRAM FOR THE H2 MOLECULE

————————————————————————————————————————————————
As was done with atomic energy levels, electrons are added to MO diagrams according to the following
principles.
(a) Add electrons to the lowest energy levels first.
(b) Each energy level holds up to TWO electrons.
(c) When the lowest available energy level(s) are full, subsequent electrons are put in the next– higher
level.
(d) Electrons placed in degenerate orbitals remain unpaired as long as possible.
Using the above principles, the MO diagram for H2 is constructed.
1s*
H(1s)
1s
H(1s)


H2
Definition: A DIAMAGNETIC species possesses NO UNPAIRED electrons (such as H2 above).
A PARAMAGNETIC species possesses AT LEAST ONE unpaired electron.
A paramagnetic species is attracted to a magnetic field; a diamagnetic species is repelled by a magnetic
field.
EXERCISES:
26
Hebden : Chemistry AP
——————————————————————————————————————————————————
18. Draw MO diagrams for the following molecules, predict their bond order and predict whether the
molecule is diamagnetic or paramagnetic.
(a) H2
(b) H2
(c) H22 
(d) He 2
(e) He 2
(f) He2
2

19. Based only on the bond orders, rank the relative BOND STRENGTHS of He 2
2 , He 2 and He2
from highest to lowest. One or more of these molecules does NOT exist. Which one(s), and why
 is this so?






III. THE MO's AND ENERGY LEVELS OF MOLECULES INVOLVING SECOND
ROW ELEMENTS
The general energy level scheme for the MO’s of two second row elements, A and B, is shown below.
Molec ule A–B
Atom A
Atom B
*
2p
 2p*
2p*

2p
2p
2p




 2p

2s
2s
1s*

1s
1s


2s*
2s

ATOMIC
ORBITAL S
 2p
MOLECUL AR
ORBITAL S
1s
ATOMIC
ORBITAL S
IV. WRITING MO CONFIGURATIONS
The procedure for writing MO configurations is similar to that for writing electron configurations using atomic
orbitals on an isolated atom.
* )2(2p)2 ]
B2 has 10 electrons: B2 [ (1s)2 ( 1s* )2(2s)2( 2s
N2 has 15 electrons:
* )2(2p)42p)2(  * )1 ]
N2 [ (1s)2 ( 1s* )2(2s)2( 2s
2p
EXERCISE:


20. Write the MO configuration for each of the following.





AP : VB AND MO THEORY
27
——————————————————————————————————————————————————
(a) Li 2
(b) Be2
(c) Be2
(d) C2
(e) C22 
————————————————————————————————————————————————
Since
of filled 1s 
and 1s* MO's
normal atomic orbitals, the core electrons are
 a “core” 
 closely resembles

omitted and MO configurations are written in terms of VALENCE ELECTRONS as shown below.
Li2 [ (1s)2 ( 1s* )2(2s)2 ]

becomes:
Li2 [ (2s)2 ]

Similarly, C2 has 12 electrons
and its full notation
* )2(2p)4 ]
C2 [ (1s)2 ( 1s* )2(2s)2( 2s
becomes:
Definition:
EXAMPLE:
* )2(2p)4 ]
C2 [ (2s)2( 2s


A HOMONUCLEAR
diatomic
molecule has 2 identical atoms.
A HETERONUCLEAR diatomic molecule has 2 different atoms.

We can write the MO's for CO, a 10 valence–electron system, as:
* )2(2p)4 (2p)2 ] .
CO [ (2s)2( 2s
EXERCISES:
21. If a specific number
 of atomic orbitals are combined, how many molecular orbitals are formed?
22. What type of molecular orbitals are formed by adding atomic orbitals?
What type of molecular orbitals are formed by subtracting atomic orbitals?
23. How does the energy of a bonding MO compare to the energies of the original atomic orbitals from
which the MO is made? How does the energy of an antibonding MO compare to the energies of the
original atomic orbitals from which the MO is made?
24. Draw and label a molecular orbital energy diagram to show how two 1s atomic orbitals combine to
form molecular orbitals.
25. What does the symbol “1s” represent? What does the symbol “ 1s* ” represent?
26. Draw and label a molecular orbital energy diagram to show how two complete sets of 2p atomic
orbitals combine to form molecular orbitals.
27. What is the bond order of a covalent single bond?

28. Write the MO configurations of the following species, determine the bond order for each species, put
the species in decreasing order of bond strength, suggest which species has the shortest bond
length, and state which species are paramagnetic.
(a) O2
(b) O 2
(c) O 2
(d) O2
(e) O2
2
2
29. Give the MO configuration for NO, and state whether the molecule is paramagnetic. Is the molecule
NO+ more or less stable than NO? Why?




30. Give the MO configuration of each of the following molecules, determine the bond order of each
molecule, and list the molecules in order of decreasing bond strength.
(a) CN
(b) CN+
(c) CN–
31. Which of  or  does not allow free rotation about a bond? Why does this situation occur?
32. Suggest a reason why cyclohexyne does not exist but cyclohexene does.
28
Hebden : Chemistry AP
——————————————————————————————————————————————————
CH2
C
C
CH2
CH2
HC
CH2
HC
CH2
CH2
CH2
cyclohexyne
33. (a)
(b)
(c)
(d)
(e)
(f)
CH2
cyclohexene
Write the Lewis structure for carbon monoxide, CO.
Draw the orbital diagram for carbon monoxide.
Draw a diagram to show how the atomic orbitals on CO overlap.
Draw the orbital diagram for the ion HCO–.
Draw a diagram to show how the atomic orbitals on HCO– overlap.
What is the most likely HCO bond angle?
34. Ethanal, or formaldehyde, has the Lewis structure shown below.
O
 
C

 
 H
H
Give a complete VB description of the bonding in ethanal, including the molecular shape,
hybridization on carbon, types of orbital overlaps and types of orbitals formed by orbital overlap.
V.
A BRIEF COMMENT ON “DELOCALIZED MOLECULAR ORBITALS”
When a molecular orbital is spread over more than two atoms, the resulting MO is called a DELOCALIZED
MOLECULAR ORBITAL. The term “delocalized” comes from the fact that the electrons in the orbital are “free
to wander” somewhat, and are not forced to be localized between two atoms. The best example of
delocalization is the BENZENE molecule, C6H6 . The resonance structures of benzene are shown below.
H
H
C
C
H
H
C
C
C
H
H
C
C
C
C
H
H
C
C
C
H
H
H
H
The hybrid structure shown below is based on unhybridized p–orbitals.
H
H
C
H
C
C
C
C
H
H
C
H
The delocalized  MO corresponding to the above diagram is shown below.
AP : VB AND MO THEORY
29
——————————————————————————————————————————————————
H
H
C
H
C
C
C
C
H
C
H
H
The resulting  MO is a ring spread out over all the carbons. Such a ring is said to be an AROMATIC RING,
and molecules possessing such a ring are said to be AROMATIC MOLECULES. The electrons in the ring are
exceptionally unreactive because of the stability of the MO. The benzene ring is sometimes shown as
follows, to emphasize the delocalized nature of the  MO.
H
H
C
C
H
H
C
C
C
C
H
H
VI. A COMPARISON OF VALENCE BOND AND MOLECULAR ORBITAL THEORIES
The Essential Difference: In VB theory, the electrons occupy orbitals in ATOMS whereas in MO theory the
electrons occupy orbitals in MOLECULES.
In more detail:
VB theory
• is the simpler of the two models and often has trouble making precise predictions of
molecular properties.
• is basically concerned with the bonding existing between two atoms.
• the bonds between 2 atoms are formed by modifying the atomic orbitals on the atoms
involved.
• the electrons are considered to be LOCALIZED on an atom (eg. lone pairs) or in the
space between the atoms, and are individually placed in orbitals.
MO theory
• requires sophisticated mathematical treatment for anything other than simple molecules,
but does produce extremely accurate results.
• defines the possible orbitals in terms of delocalization over several atoms.
• treats all the electrons in the molecule as a whole; the electrons are added in sets to
delocalized orbitals.