Transcript Molecular geometry

Chapter Ten
Chemical Bonding II :
Molecular Shapes,
Valence Bond Theory,
and Molecular Orbital Theory
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Chapter 10
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Contents
1.
2.
3.
4.
5.
6.
7.
8.
Artificial Sweeteners: Fooled by Molecular Shape
VSEPR Theory: The Five Basic Shapes
VSEPR Theory: The Effect of Lone Pairs
VSEPR Theory: Predicting Molecular Geometries
Molecular Shape and Polarity
Valence Bond Theory: Orbital Overlap as a
Chemical Bond
Valence Bond Theory: Hybridization of Atomic
Orbitals
Molecular Orbital Theory: Electron Delocalization
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1. Artificial Sweeteners: Fooled by Molecular Shape
 The sugar molecule (the key) enters the active site
(the lock ─ sugar receptor protein) of taste cell,
resulting in ion channels opened, nerve signal
transmission, and reaching the brain a sweet taste.
 Artificial sweeteners such as aspartame and
saccharin bind to the active more strongly than
sugar.
 Similarities in the shape of sucrose and artificial
sweeteners give those sweeteners the ability to
stimulate a sweet taste sensation.
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
*****
Valence Shell Electron Pair Repulsion (VSEPR)
theory: A theory that allows prediction of the
shapes of molecules or polyatomic ion based on
the idea that electrons ˗ either as lone pairs or as
bonding pairs ˗ repel one another.
• Electron geometry: The geometrical
arrangement of electron groups in a molecule.
• Molecular geometry: The geometrical
arrangement of atoms in a molecule.
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
i)
ii)
*****
VSEPR theory proceeding
Write a best Lewis structure
Determine VSEPR notation:
AXmEn:
A: Central atoms
X: Terminal atoms
E: Lone pairs electrons
H2O for example:
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AX2E2
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iii) Determine the electron geometry
*****
 An electron group can be:
- either single bond or a multiple bond
- a (resonance) hybrid bond
- a lone pairs of electron
- a unpaired single-electron
 Repulsion force in general:
LP vs. LP > LP vs. BP > BP vs. BP
* Lone Pairs (LP), Bonding Pairs (BP)
Angle for repulsion forces: 90° > 120° > 180°
 For central (interior) atom belong to third-period or
higher element with VSEPR notation such as AX5,
AX4E, AX3E2, AX6, AX5E, AX4E2 require an expanded
octet such as 3d orbital.
 Multiple bond occupy more space than single bond
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*****
iv) Determine the molecular geometry
 Structures for the central atom without lone-pair
electrons (AXn type), electron geometry and
molecular geometry are identical.
 Structures for the central atom with lone-pair
electrons (AXnEm type) type), electron
geometry and molecular geometry are
different.
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* Count only electron groups around the central atom. Each of the
following is considered one electron group: a lone pair, a single
bond, a double bond, a triple bond, or a single electron.
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2.
VSEPR Theory: The Five Basic Shapes
(All electrons around the central atom are bonding
group)

Two Electron Groups (AX2): Linear
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
Three Electron Groups (AX3): Trigonal Planar
* double bond contains
more electron density
than the single bond
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
Four Electron Groups (AX4): Tetrahedral

Five Electron Groups (AX5): Trigonal Bipyramidal
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
Six Electron Groups (AX6): Octahedral
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Example 10.1 VSEPR Theory and the Basic Shapes
Determine the molecular geometry of NO3−.
Solution
NO3− has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three
resonance structures:
Use any one of the resonance structures to determine the number of electron groups
around the central atom. The nitrogen atom has three electron groups.
The electron geometry is trigonal planar:
The molecular geometry is also trigonal planar.
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3.
VSEPR Theory: The Effect of Lone Pairs
(Some electrons around the central atom are lone
pairs)

Four Electron Groups with Lone Pairs
AX3E
AX2E2
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*
Effect of Lone Pairs on Molecular Geometry
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
Five Electron Groups with Lone Pairs
AX4E
AX3E2
AX2E3
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
Six Electron Groups with Lone Pairs
AX5E
AX4E2
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4. VSEPR Theory: Predicting Molecular Geometries
Example 10.2 Predicting Molecular Geometries
Predict the geometry and bond angles of PCl3.
Solution
Step 1 PCl3 has 26 valence electrons.
Lewis structure for the molecule:
Step 2 The central atom (P) has four electron groups.
Step 3 Three of the four electron groups
around P are bonding groups and
one is a lone pair.
Step 4 The electron geometry is tetrahedral
The molecular geometry is trigonal pyramidal
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Example 10.3 Predicting Molecular Geometries
Predict the geometry and bond angles of ICl4−.
Solution
Step 1 ICl4− has 36 valence electrons.
Lewis structure for the molecule:
Step 2 The central atom (I) has six electron groups.
Step 3 Four of the six electron groups
around I are bonding groups and
two are lone pairs.
Step 4 The electron geometry is octahedral
The molecular geometry is square planar.
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
Representing Molecular Geometries on Paper
Examples:
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
Predicting the Shapes of Larger Molecules
Example:
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Example 10.4 Predicting the Shape of Larger Molecules
Predict the geometry about each interior atom in methanol (CH3OH) and make a
sketch of the molecule.
Solution
The Lewis structure of CH3OH.
Three-dimensional sketch of the molecule:
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

*****
Memo for VSEPR
Without lone-pair electrons
VSEPR
Notation
AX2
AX3
AX4
Electron
Geometry
Linear
Trigonal planar
Tetrahedral
Molecular
Geometry
Linear
Trigonal planar
Tetrahedral
AX5
AX6
Trigonal bipyramidal
Octahedral
Trigonal bipyramidal
Octahedral
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
*****
With lone-pair electrons
VSEPR
Notation
AX2E
AX3E
AX2E2
Electron
Geometry
Trigonal planar
Tetrahedral
Tetrahedral
Molecular
Geometry
Bent
Trigonal pyramidal
Bent
AX4E
Trigonal bipyramidal
Seesaw
AX3E2
AX2E3
Trigonal bipyramidal
Trigonal bipyramidal
T-shaped
Linear
AX5E
AX4E2
Octahedral
Octahedral
Square pyramidal
Square planar
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5. Molecular Shape and Polarity
 Bond dipole versus Molecular dipole
 Bond dipole: A separation of positive and negative
charge in an individual bond.
 Molecular dipole:
• For diatomic molecule: molecular dipole is
identical to bond dipole.
• For a molecule consisted by three or more
atoms, molecular dipole is estimated by the
vector sum of individual bond dipole
moment (net dipole moment).
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 Polar molecule versus Nonpolar molecule
 Polar molecule: A molecule in which the
molecular dipole is nonzero.
 Nonpolar molecule: A molecule in which the
molecular dipole is zero.
 Molecular polarity prediction
• Draw the Lewis structure for the molecule and
determine its molecular geometry.
• Determine if the molecule contains polar bonds
by electronegativity values.
• Determine if the polar bonds add together to form
a net dipole moment.
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
*****
Examples
CO2
Molecular geometry: linear
(net) dipole moment: m = 0 D
Nonpolar molecule
H2O
Molecular geometry: bent
(net) dipole moment: m = 1.84 D
Polar molecule
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Example 10.5 Determining if a Molecule Is Polar
Determine if NH3 is polar.
Solution
Lewis structure:
Determine if the molecule contains polar bonds.
The electronegativities of nitrogen and hydrogen
are 3.0 and 2.1, respectively.
Determine if the polar bonds add together to
form a net dipole moment. The three dipole
moments sum to a net dipole moment.
Ans: The molecule is polar.
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
Polarity effects of the intermolecular forces
Example 1: For H2O
Example 2: Like dissolves like
Polar molecules interact
strongly with other polar
molecules excluding the
nonpolar molecules and
separating into distinct regions.
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 Quantum-Mechanical Approximation Technique
 Perturbation theory (used in valence bond
theory): A complex system (such as a molecule)
is viewed as a simpler system (such as two atoms)
that is slightly altered or perturbed by some
additional force or interaction (such as the
interaction between the two atoms).
 Variational method (used in molecular orbital
theory): The energy of a trial function (educated
function) within the Schrodinger equation is
minimized.
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 Schrodinger equation revisited
H = E
•H (Hamiltonian operator), a set of mathematical
operations that represent the total energy (kinetic
and potential) of the electron within the atom.
•E is the actual energy of the electron.
• is the wave function , a mathematical function
that describes the wavelike nature of the electron.
Perturbation theory: Approach by small changes
to a known system in which Hamiltonian operator is
modified.
Variational method: Approach by combining
systems of comparable weighting in which wave
function is modified.
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
Valence bond theory versus molecular orbital
theory
 Valence bond theory (VB): An advanced model
of chemical bonding in which electrons reside in
quantum-mechanical orbitals localized on
individual atoms that are a hybridized blend of
standard atomic orbitals; chemical bonds result
from an overlap of these orbitals.
 Molecular orbital theory (MO): An advanced
model of chemical bonding in which electrons
reside in molecular orbitals delocalized over the
entire molecule. In the simplest version, the
molecular orbitals are simply linear combinations
of atomic orbitals.
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6. Valence Bond Theory: Orbital Overlap as a
Chemical Bond
 Valence bond theory describes that covalent bonds are
formed when atomic orbitals on different atoms overlap.
 Simple Atomic Orbitals (AO’s) Overlap
Bonding in H2 for example
• A covalent bond is formed by the
pairing of two electrons with
opposing spins in the region of
overlap of atomic orbitals between two
atoms.
• This overlap region has a high
electron charge density.
• The overall energy of the system is
lowered.
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 Acceptable simple Atomic Orbitals (AO’s) Overlap
Bonding in H2S for example
• Predicted H˗S˗H angle is 90o, actual H˗S˗H angle is 92o,
therefore, the simple AO overlap is acceptable for H2S
molecule.
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 Unacceptable simple Atomic Orbitals (AO’s)
Overlap
C
Example 1: CH4
Example 2: NH3 and H2O
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Chapter 10
• Ground-state
electron
configuration of C
for example, it
should form only
2 bonds
• Actually, the
central atom of
H2S, H2O, NH3,
and CH4, are sp3
hybridization
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*****
7. Valence Bond Theory: Hybridization of Atomic
Orbitals
 Hybridization: A mathematical procedure in which
standard atomic orbitals are combined to form new,
hybrid orbitals.
•
•
•
Hybridizing is mixing different types of orbitals in the
valence shell to make a new set of degenerate orbitals
such as sp, sp2, sp3, sp3d, sp3d2.
Hybrid orbitals minimize the energy of the molecule
by maximizing the orbital overlap in a bond.
Those central atoms are available hybridized,
however, those terminal atoms are supposed to be
unhybridized.
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
General statements regarding hybridization *****
• Hybridization is employed for central atom
only, thus, the hybrid orbital describes the
electron geometry for central atom.
• Number of hybrid orbitals = Number of
standard atomic orbitals combined = Number of
σ bond + Number of lone pairs.
• Number of hybridization obitals of a central
atom = 2 → sp; = 3 → sp2; = 4 → sp3; = 5 →
sp3d; = 6 → sp3d2.
• Hybrid orbitals may overlap with standard
atomic orbitals or with other hybrid orbitals to
form σ bond.
• Molecular geometry is described by the relative
atomic position around central atom.
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
sp3 hybridization (C for example)
one s orbital with three p orbitals combine to
form four sp3 hybrid orbitals (degenerate).
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*****
Examples of sp3 hybridization (for central atom)
Central Moleatom
cule
Standard
orbitals
Hybrid
Orbital
σ
C
H
H C H
H
2s
O
σ
σ
sp3
2p
lone
N
σ
Geometry
σ
σ
σ
..
H N H
H
H
2s
2p
sp3
lone lone σ
.. .
O.
H
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2s
σ
sp3
2p
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
sp2 hybridization (B for example)
one s orbital with two p orbitals combine to form
three sp2 hybrid orbitals
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Examples of sp2 hybridization (for central atom)
Central
atom
Molecule
Standard
orbitals
B
C
F B F
F
H
H
2s
2p
H
H
2s
σ
σ
2p
sp2
σ
C C
Unhybridized
Orbital
Hybrid
Orbital
σ
2p
σ
σ
σ
π
2p
sp2
lone
*****
σ
π
.. ..
N
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H N N H
2s
2p
Chapter 10
sp2
2p
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
sp hybridization (Be for example)
one s orbital with one p orbitals combine to form
two sp hybrid orbitals
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*****
Examples of sp hybridization (for central atom)
Central
atom
Molecule
Standard
orbitals
Hybrid
Orbital
σ
Be
Cl Be Cl
2s
2p
C
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H C C H
2s
2p
Chapter 10
σ
sp
σ
Unhybridized
Orbital
σ
sp
2p
π
π
2p
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
*****
About Multiple Covalent Bond
 σ (sigma) bond: The first covalent bond formed
by end-to-end overlap of standard or
hybridized orbitals between the bonded atoms:
s + s, s + p, p + p (end-to-end), s + hybrid orbital
p + hybrid orbital, hybrid orbital + hybrid orbital
 π (Pi) bond: The second (and third, if present)
bond in a multiple bond, results from side-by-side
overlap of unhybridized p orbitals:
p + p (side-by-side)
 Summary:
- Single bonds: one σ bond
- Double bond: one σ bond and one π bond
- Triple bond: one σ bond and two π bonds
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
Sigma Bonding and Pi Bonding
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 VB theory of bonding in ethylene (H2C=CH2)
example of sp2 hybridization and a double bond
• Lewis structure
• A π-bond has two
lobes (above and
below plane), but
is one bond,
side-by-side
overlap of 2p–2p
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Continued
• All six atoms in
C2H4 lie in the
same plane
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 VB theory of bonding in Acetylene (HCCH)
example of sp hybridization and a triple bond
• Lewis structure
• Two π-bonds
from 2p–2p
overlap
forming a
cylinder of πelectron
density
around the
two carbon
atoms
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Continued
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 VB theory of bonding in Formaldehyde (H2C=O)
example of sp2 hybridization and a double bond
• Lewis structure
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Continued
σ σ
σ
π
• Valence bond
model
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 Bond Rotation
* Rotation around s bond does not require breaking the bond,
however, p bond interact above and below the internuclear
axis, so rotation around the axis requires the breaking the
bond.
* In general, bond energy of p bonds are weaker than that of
s bonds because the side-to-side orbital overlap tends to be
less efficient than the end-to-end orbital overlap.
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 cis versus trans Isomers
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 Expanded Octet hybridization
 sp3d hybridization, AsF5 for example
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Continued
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 sp3d2 hybridization, SF6 for example
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Continued
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*****

Example for hybridization/electron geometry types
versus molecular geometry
Number of
σ + lone
Hybridization
VSEPR
notation
Electron
geometry
Molecular
geometry
Example
2
sp
AX2
Linear
linear
Cl-Be-Cl
3
sp2
AX3
AX2E
Trigonal planar
Trigonal planar
Angular
BCl3
SO2
4
sp3
AX4
AX3E
AX2E2
Tetrahedral
Tetrahedral
Trigonal pyramidal
Angular
CH4
NH3
H2O
5
sp3d
AX5
AX4E
AX3E2
AX2E3
Trigonal bipyramidal
Trigonal bipyramidal
Seesaw
T-shaped
Linear
PBr5
SF4
ClF3
XeF2
6
sp3d2
AX6
AX5E
AX4E2
Octahedral
Octahedral
Square pyramidal
Square planar
SF6
BrF5
XeF4
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
Procedure for Hybridization and Bonding
Scheme
*****
1. Write the Lewis structure for the molecule.
2. Use VSEPR theory to predict the electron
geometry about the central atom.
3. Select the correct hybridization for the central
atom based on the electron geometry.
4. Sketch the molecule, beginning with the central
atom and its orbitals. Show overlap with the
appropriate orbitals on the terminal atoms.
5. Label all bonds using the σ or π notation
followed by the type of overlapping orbitals.
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Example 10.7 Hybridization and Bonding Scheme
Write a hybridization and bonding scheme for acetaldehyde,
Solution
Step 1 Lewis structure
Step 2 The leftmost carbon atom, tetrahedral electron geometry.
The rightmost carbon atom, trigonal planar geometry.
Step 3 The leftmost carbon atom, sp3 hybridized
The rightmost carbon atom, sp2 hybridized.
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Continued
Step 4 The appropriate orbitals on the terminal atoms.
Step 5 Type of overlapping orbitals.
*****
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8. Molecular Orbital Theory: Electron Delocalization
Chemical Bond
 Molecular Orbital (MO): A model of chemical
bonding in which electrons reside in molecular
orbitals delocalized over the entire molecule.
• The molecular orbitals are linear combinations of
atomic orbitals (LCAO).
• Because the orbitals are wave functions, the waves
can combine either constructively or
destructively.
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 MOs formed by combining two 1s AOs
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 MOs formed by combining two set 2p AOs
 σ2p and σ2p*:
end-to-end
overlap of
AOs
 π2p and π2p*:
side-by-side
overlap of
AOs
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*****
 Summarizing LCAO–MO Theory:
• The total number of MOs formed from a particular
set of AOs always equals the number of AOs in the
set.
• When two AOs combine to form two MOs, one MO
is lower in energy (the bonding MO) and the other is
higher in energy (the antibonding MO).
• When assigning the electrons of a molecule to MOs,
fill the lowest energy MOs first with a maximum of
two spin-paired electrons per orbital.
• When assigning electrons to two MOs of the same
energy, follow Hund’s rule—fill the orbitals singly
first, with parallel spins, before pairing.
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 Applications of MOs
• Estimate the bond order:
Bond Order (BO) =
(Σ bonding e– － Σ antibonding e–)/2
• Predict the existence of molecule
• Estimating bond length and bond energy
• Predicting magnetic properties
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*****
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
*****
1st Period Homonuclear Diatomic MOs
H2 and He2 for example:
σ1s*
AOs of H
(two 1s AOs)
σ1s
MOs of H2
BO = (2−0)/2 = 1
H2 molecule does exist
Diamagnetic
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σ1s*
AOs of He
(two 1s AOs)
σ1s
MOs of He2
BO = (2−2)/2 = 0
He2 molecule does not exist
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
2nd Period Homonuclear Diatomic MOs
 Effects of 2s–2p
Mixing: Increasing
energy difference,
decreasing the degree
of mixing.
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*****
Continued
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 Predicting magnetic properties by MOs
Lewis structure  Experiment
For O2:
showed O2 is
..
..
paramagnetic
.. O O ..
 MO prove O2
have unpaired
electrons
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

2nd Period Heteronuclear Diatomic MOs
NO for example
• Oxygen is more electronegative
than nitrogen, so its atomic
orbitals are lower in energy than
nitrogen’s atomic orbitals.
• The lower energy atomic orbital
makes a greater contribution to
the bonding molecular orbital
and the higher energy atomic
orbital makes a greater
contribution to the antibonding
molecular orbital.
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
HF for example
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Example 10.10 Molecular Orbital Theory
Draw an MO energy diagram and determine the bond order for the N2− ion.
Do you expect the bond to be stronger or weaker than in the N2 molecule? Is
N2− diamagnetic or paramagnetic?
Solution
The N2− ion has 11 valence electrons
(5 for each nitrogen atom plus 1 for
the negative charge). Assign the electrons
to the molecular orbitals beginning with
the lowest energy orbitals and following
Hund’s rule.
The bond order is 2.5, bond order for N2 molecule is 3, the bond is weaker.
N2− ion has one unpaired electron and is therefore paramagnetic.
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Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic
Molecules and Ions
Use molecular orbital theory to determine the bond order of the CN− ion. Is the
ion paramagnetic or diamagnetic?
Solution
Number of valence electrons
= 4 (from C) + 5 (from N) +
1 (from negative charge) = 10
Write an energy level diagram
Since the MO diagram has no unpaired electrons, the ion is diamagnetic.
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
Polyatomic Molecules
Example: O3
Lewis model
VB model
MO model
Example: C6H6
Lewis model
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MO model
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End of Chapter 10
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