77. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
_
VI. BONDING and MOLECULAR STRUCTURE
A. Review of Atomic Orbital Theory
1. Orbitals and Probability
a) Orbital: the region in space where there is a high probability of finding the electron.
D
E
N
S
I
T
Y
DISTANCE
b) Types, or shapes, of orbitals
c) Orbital orientations
78. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
Type
# Orientations
Designations
d) Any single orbital has a maximum capacity of two e- ‘s
2. Orbitals and Energy
a) Orbitals: are sub-energy levels of the principal levels predicted by Bohr.
b) Number of orbitals per principal level = n2
e.g. the 3rd energy level hold 18 e- ‘s (2  32). To hold 18 e- ‘s you need 9 orbitals since each
orbital can only hold two e- ‘s.
_
79. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
c) Energy level diagram
E
n
e
r
g
y
_
80. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
3. Electron Configuration Using the Energy Diagram
Example(1): Give the electron configuration of 1H.
Example(2): Give the electron configuration of 2He.
a) Pauli exclusion principle: no two electrons in an atom can be exactly alike.
There are 4 distinguishing factors for an electron:
i) principal energy level
ii) orbital type
iii) orbital orientation
iv) spin
b) Hund's rule: Given a chance electrons will remain unpaired.
Example(3): Give the electron configuration of 6C.
Example(4): Give the electron configuration of 8O.
Example(5): Give the electron configuration of 19K.
_
81. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
_
4. Electron Configuration and Periodic Table
a) Relationship of orbital theory to the structure of the periodic table
On the blank periodic table, write in the last occupied orbital for each element and the number
of electrons in the orbital level. (e.g. the complete configuration of 6C is 1s22s22p2. Under 6C
write ... 2p2 . (The three dots means all lower level orbitals are completely filled.) You will soon
see the relation between the orbitals and the structure of the table.
82. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
_
5. Electron Configuration Using the Periodic Chart
s
p
1
2
d
3
4
5
6
7
Example(6): Write the complete electron configuration of 11Na, using the periodic table to determine
the order of orbital filling.
Example(7): Write the complete electron configuration of 33As, using the periodic table to determine the
order of orbital filling.
6. Relative Energy of Orbitals Using the Periodic Table
Example(8): Which orbital has the higher energy, 3s or 3p?
Example(9): Which orbital has the highest energy, 3s, 3p, 3d, or 4s?
7. Valence Configuration
Example(10): Determine the valence configuration of 12Mg.
83. VI. BONDING and MOLECULAR STRUCTURE / A. Review of Atomic Orbital Theory
Example(11): Determine the valence configuration of 32Ge.
Example(12): Determine the valence configuration of 83Bi.
Be able to do valence configurations for all atoms in the main groups, using the periodic table
as your guide.
8. Number of Unpaired
Example(13): Determine the number of unpaired electrons in 27Co.
Example(14): Determine the number of unpaired electrons in 44Ru.
Example(15): Determine the number of unpaired electrons in 50Sn.
_
84. VI. BONDING and MOLECULAR STRUCTURE / B Valence Bond Theory
_
B. Valence Bond Theory
1. Valence Bond Theory: the bond between 2 atoms is the result of the overlap of their
atomic orbitals.
Example(1): a) How is the covalent bond formed between 2 hydrogen atoms.
b) How does this hold the atoms together.
c) Draw an energy diagram showing how the energy of the atoms varies with the distance
between them.
d) On the diagram indicate the bond length.
Example(2): Explain why a bond between two helium atoms cannot be formed in the same way.
85. VI. BONDING and MOLECULAR STRUCTURE / B Valence Bond Theory
2. Sigma Bonds: The increase in the electron density lies directly between the two
nuclei.
Example(3): Show how is the covalent bond is formed between two fluorine atoms.
Example(4): Show how is the covalent bond is formed between a hydrogen and a fluorine atom.
Example(5): Show how is the covalent bonds are formed between two hydrogen atoms and one
selenium atom.
_
86. VI. BONDING and MOLECULAR STRUCTURE / B Valence Bond Theory
3. Pi Bonds: The increase in the electron density lies above and below the axis
connecting the two nuclei.
Example(6): Show how the covalent bonds are formed between two oxygen atoms.
Example(7): Show how the covalent bonds are formed between two nitrogen atoms.
_
87
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
_
C Hybridization of Orbitals
1. Hybridized Orbitals: are the result of “mixing” different orbital types together. (The
wave functions are added together to give new wave functions.)
Example(1): a) In its ground state, how many bonds can a carbon atom form?
b) In it first excited state, how many bonds can a carbon atom form?
c) In CH4 all of the bond angles are equal, what can be concluded about the carbon’s
orbitals
2p ___ ___ ___
2p ___ ___ ___
2s ___
2s ___
___ ___ ___ ___
ground state
excited state
hybridized state
Single sp3 orbital
Example(2): a) In its ground state, how many bonds can a boron atom form?
b) In it first excited state, many bonds can a boron atom form?
c) In BCl3 all of the bond angles are equal, what can be concluded about the boron’s orbitals
2p ___ ___ ___
2p ___ ___ ___
2s ___
2s ___
___
___ ___ ___
ground state
excited state
hybridized state
88
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
_
Example(3): a) In its ground state, how many covalent bonds can a beryllium atom form?
b) In it first excited state, how many bonds can a beryllium atom form?
c) In BeH2 the bond angle between the H’s is 180º, what can be concluded about the
beryllium’s orbitals
2p ___ ___ ___
2p ___ ___ ___
___ ___
___ ___
2s ___
ground state
2s ___
excited state
hybridized state
Example(4): In its ground state sulfur has only 2 unpaired electrons. But S can form SCl6. How does S
achieve 6 unpaired electrons?
3d ___ ___ ___ ___ ___
3d ___ ___ ___ ___ ___
3p ___ ___ ___
2p ___ ___ ___
3s ___
2s ___
___ ___ ___
___ ___ ___ ___ ___
89
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
_
2. Predicting hybridization from dot formulas
Example(5): a) Do the dot formula of NH3.
b) The bond angle between the H’s is a little less than 109º. What can be concluded about
the orbitals that the N atom uses?
c) Draw the orbital levels for the ground state and the hybridized state for nitrogen in NH3.
90
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
_
Example(6): a) Do the dot formula of H2O.
b) The bond angle between the H’s is much greater than 90º, but a little less than 109º.
What can be concluded about the orbitals that the O atom uses?
c) Draw the orbital levels for the ground state and the hybridized state for oxygen in H2O.
3. Pi bonds in hybridized atoms.
Example(7): a) Do the dot formula of C2H4.
b) All of the bond angles around each carbon atom are about 120º. What can be concluded
about the hybridization of the carbon’s orbitals forming the sigma bonds?
c) Draw the orbital levels for the ground state and the hybridized state for the carbons.
d) How is the pi bond formed?
91
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
Example(8): a) Do the dot formula of C2H2.
b) The bond angle around each carbon atom are about 180º. What can be concluded
about the hybridization of the carbon’s orbitals forming the sigma bonds?
c) Draw the orbital levels for the ground state and the hybridized state for the carbons.
d) How are the two pi bond formed?
_
92
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
_
4. Summary of Central Atom Hybridization and Molecular Shapes
Areas of
Electrons
Around the
Central Atom
Hybridization
Attached
Atoms,
Lone Pairs
Example
Molecular
Shape
Example
FOUR
sp3
4 atoms, 0 pair
Tetrahedral
CCl4
FOUR
sp3
3 atoms, 1 pair
Trigonal
Pyramidal
PH3
FOUR
sp3
2 atoms, 2 pair
Bent
or
“V” Shape
SCl2
FOUR
sp3
1 atom, 3 pair
Linear
HF
THREE
sp2
3 atoms, 0 pair
BBr3
SO3
THREE
sp2
2 atoms, 1 pair
Trigonal
Planer
Bent
or
“V” Shape
BCl2‾
SO2
TWO
sp
2 atoms, 0 pair
Linear
BeCl2
FIVE
sp3d
5 atoms, 0 pair
Trigonal
Bipyramidal
PCl5
SIX
sp3d2
6 atoms, 0 pair
Octahedral
SF6
SIX
sp3d2
4 atoms, 2 pair
Square
Planer
XeF4
93
VI. BONDING and MOLECULAR STRUCTURE / C. Hybridization
_
5. Molecular Polarity
a) Polar Bond: the unequal sharing of electrons between two atoms. The greater the
EN difference between the two atoms, the more polar the bond.
b) Polar Molecule: has a partially positive end opposite a partially negative end
(a dipole).
Example(9): For each of the example molecules in the summary table above, determine if the molecule
is polar or nonpolar.
94
VI. BONDING and MOLECULAR STRUCTURE / D. Molecular Orbital Theory
_
D. Molecular Orbital Theory
1. Molecular orbital theory assumes that around a set of nuclei there are a series of
orbitals. These orbitals have different shapes and different energies. (This is analogous to
atomic orbital theory: around a nucleus there are a series of atomic orbitals, with different
shapes and energies.)
2. To approximate the molecular orbitals (MO’s), the atomic orbitals (AO’s) are combined
together mathematically. There are two ways to combine the AO wave functions, they can be
added or subtracted.
a) “s” AO’s
Two Separate Wave functions
Wave functions Added
Bonding Orbital
Wave functions Subtracted
Antibonding Orbital
95
VI. BONDING and MOLECULAR STRUCTURE / D. Molecular Orbital Theory
b) “p” AO’s can be combined together in the same way.
i) Sigma MO’s. If the AO’s are along the axis connecting the 2 nuclei, you get a
sigma bonding MO (σ 2p) and a sigma antibonding MO (σ* 2p).
Bonding Orbital
Antibonding Orbital
ii) Pi MO’s. If the AO’s are perpendicular to the axis connecting the 2 nuclei, you
get a pi bonding MO (π 2p) and a pi antibonding MO (π* 2p).
Bonding Orbital
Antibonding Orbital
3. Conservation of Number of Orbitals. If you start with 2 AO’s you get 2 MO’s. If you
start with 3 AO’s you get 3 MO’s, etc.
4. Energy Level Diagrams
a) H2
σ* 1s
1s
1s
σ 1s
_
96
VI. BONDING and MOLECULAR STRUCTURE / D. Molecular Orbital Theory
b) O2 and F2
σ* 2px
π* 2py
px
py
π* 2pz
pz
pz
π 2py
py
px
π 2pZ
σ 2px
σ* 2s
2s
2s
σ 2s
c) Li2 to N2
σ* 2px
π* 2py
π 2pz
σ 2px
π 2py
π 2pz
σ* 2s
σ 2s
_
97
VI. BONDING and MOLECULAR STRUCTURE / D. Molecular Orbital Theory
_
5. Molecular Electron Configuration and Bond Order (BO).
a) Determining molecular electron configuration from the above energy diagrams.
As with atomic electron configuration, one starts at the lowest energy level and fills it
completely before moving to the next highest available level. When 2 electrons occupy the
same energy level they must have opposite spins (Pauli’s Exclusion Principle).
When there is more than one MO with the same energy, the electrons stay unpaired if possible
(Hund’s Rule).
b) Determining bond order (BO)
BO = # bonding electrons – # antibonding electrons
2
Example(1): Determine the molecular electron configuration and the BO for H2
Example(2): Prove that the BO for He2 is zero.
Example(3): Determine the molecular electron configuration and the BO for N2
Example(4): a) Determine the molecular electron configuration and the BO for O2.
b) How many unpaired electrons are in the O2 molecule?