Transcript silbchp11

Chapter 11
Theories of Covalent Bonding
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VALENCE BOND THEORY:



DEVELOPED BY LINUS PAULING, who
received the Nobel Prize in 1954 for his
work
A view of chemical bonding in which bonds
arise from the overlap of atomic orbitals on
two atoms to give a bonding orbital of
electrons localized between the bonded
atoms
RULE: Realize that Valence Bond Theory
and all the others don't explain everything
The Central Themes of VB Theory
Basic Principle
A covalent bond forms when the orbitals of two atoms overlap
and the overlap region, which is between the nuclei, is
occupied by a pair of electrons.
(The two wave functions are in phase so the amplitude increases
between the nuclei.)
The Central Themes of VB Theory
Themes
A set of overlapping orbitals has a maximum of two electrons
that must have opposite spins.
The greater the orbital overlap, the stronger (more stable) the
bond.
The valence atomic orbitals in a molecule are different from
those in isolated atoms.
There is a hybridization of atomic orbitals to form molecular
orbitals.
Figure 11.1
Orbital overlap and spin pairing in three
diatomic molecules.
Hydrogen, H2
Hydrogen fluoride, HF
Regular atomic orbital
overlap can explain these
bonds.
Fluorine, F2
VALENCE BOND THEORY:
Ha to Hb: 1sa to 1sb
overlap radius = 74 pm
As overlap increases, strength of bond increases both electrons are mutually attracted to both
atomic nuclei.
At optimum distance between nuclei with
maximum overlap, a sigma s bond (strong
primary bond) forms. Max electron density is
along the axis of the bond
Ha to Fb: 1sa to 2pb
direct overlap or s bond
VALENCE BOND THEORY:
F to F: the picture looks like a 2p orbital
on one F is overlapping with a 2p
orbital on the other F atom, but actually
each F is sp3 hybridized & electrons are
localized between two atomic nuclei
VALENCE BOND THEORY:
We cannot use this direct overlap picture
for CH4’s bonding. The 2s and the three
2p orbitals on each C do not fit into the
CH4 molecule's 109o bond angles, since
the 2p orbitals are at 90° to each other
Valence Bond Theory states that HYBRID
orbitals of the outermost orbitals on an
atom are formed from the atoms’ atomic
orbitals
Hybrid Orbitals
Key Points
The number of hybrid orbitals obtained equals the number of
atomic orbitals mixed.
The type of hybrid orbitals obtained varies with the types of
atomic orbitals mixed.
Types of Hybrid Orbitals
sp
sp2
sp3
sp3d
sp3d2
Figure 11.2
The sp hybrid orbitals in gaseous BeCl2.
atomic
orbitals
on Be
hybrid
orbitals
You have to
know how to
draw this
energy hybrid
formation.
orbital box diagrams
Figure 11.2
The sp hybrid orbitals in gaseous BeCl2(continued).
orbital box diagrams with orbital contours
Figure 11.3
The sp2 hybrid orbitals in BF3.
You have to know
how to draw this
energy hybrid
formation.
Note the three sigma
bonds formed between B
and each F.
Figure 11.4
The sp3 hybrid orbitals in CH4.
You have to know how to draw
this energy hybrid formation.
Figure 11.5
The sp3 hybrid orbitals in NH3.
You have to know how to draw
this energy hybrid formation.
Figure 11.5 continued
You have to know how to
draw this energy hybrid
formation.
The sp3 hybrid orbitals in H2O.
VALENCE BOND THEORY
Expanded Valence Shells have hybrid
orbitals using s, p & d atomic orbitals.
Example: PCl5 P: [Ne]3s23p3
dsp3 hybridization results in 5 s bonds
and trigonal bipyramidal geometry
(You can write these as dsp3 or sp3d)
Figure 11.6
The sp3d hybrid orbitals in PCl5.
You have to know
how to draw this
energy hybrid
formation.
Figure 11.7
The sp3d2 hybrid orbitals in SF6.
You have to know how to draw this
energy hybrid formation.
Figure 11.8
The conceptual steps from molecular formula to the
hybrid orbitals used in bonding.
Step 1
Step 2
Step 3
Molecular
shape
and
e
Molecular
Lewis
Hybrid
group
formula
structure
orbitals
arrangement
Figure 10.1
Figure 10.12
Table 11.1
SAMPLE PROBLEM 11.1
PROBLEM:
Postulating Hybrid Orbitals in a Molecule
Use partial orbital diagrams to describe mixing of the atomic
orbitals of the central atom leads to hybrid orbitals in each of
the following:
(a) Methanol, CH3OH
SOLUTION:
(b) Sulfur tetrafluoride, SF4
The groups around C are
arranged as a tetrahedron.
(a) CH3OH
H
H
C O
H H
O also has a tetrahedral
arrangement with 2 nonbonding
e- pairs.
SAMPLE PROBLEM 11.1
Postulating Hybrid Orbitals in a Molecule
continued
2p
2s
2p
sp3
single C atom
hybridized
C atom
2s
sp3
hybridized
O atom
single O atom
(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.
F
F S
F
F
3d
3d
3p
sp3d
3s
S atom
hybridized
S atom
VALENCE BOND THEORY
There can be more than one central
atom, and each has its own
hybridization and geometry
C2H6 and H2O2 and CH3COOH
C2H6: both C's are sp3 hybridized and
can rotate around axis of bond.
H2O2: both O's are sp3 , etc.
Figure 11.9
both C are sp3 hybridized
The s bonds in ethane(C2H6).
s-sp3 overlaps to s bonds
sp3-sp3 overlap to form a s bond
relatively even
distribution of electron
density over all s
bonds
VALENCE BOND THEORY:
Multiple Bonds
H2CO: the Lewis structures shows a double bond
between C and O, but we know it does not have
twice the bond dissociation energy of a single
C-O bond
Pauling proposed that there was only one sigma s
bond between any two atoms, and the other
multiples were weaker pi p bonds
If there are only 3 s bonds around this carbon, it
can't be sp3 hybridized - instead we have sp2
hybrid orbitals
sp2 hybridization results in only 3 s bonds, and trig
planar geometry, with 120° angles
p bond is a sideways or parallel overlap of the p
atomic orbitals rather than the direct overlap of
s bonds
Figure 11.10
The s and p bonds in ethylene (C2H4).
overlap in one position - s
p overlap - p
electron density
Proper name is ethene.
VALENCE BOND THEORY
Look at acetylene: its geometry is
linear. C is forming a triple bond to
another C and a single bond to H, so
that's only two s bonds
Therefore sp hybridization results in
only 2 s bonds, and linear geometry
There are 2 p bonds from the parallel
overlap of the 2p orbitals remaining on
both C's
Figure 11.11
The s and p bonds in acetylene (C2H2).
overlap in one position - s
p overlap - p
SAMPLE PROBLEM 11.2
PROBLEM:
PLAN:
Describing the Bond in Molecules
Describe the types of bonds and orbitals in acetone, (CH3)2CO.
Use the Lewis structures to ascertain the arrangement of groups and
shape at each central atom. Postulate the hybrid orbitals taking note of
the multiple bonds and their orbital overlaps.
sp2
SOLUTION:
sp2
sp3 hybridized
O
sp3
hybridized H
C
O
2
O
sp
H
C
C
H H H H
sp2 hybridized
H
2
sp3
H sp
2
sp2 C sp
C
3
sp H
sp3
H sp3
sp3
sp3
H
3
sp 3 H
sp
sbonds
C
C
H3 C
CH3
pbond
H2CO hybrid orbitals and sigma and pi bond formation
Remember the C=C
double bond has
sigman and pi bonds.
The C has sigma bonds from its hybrid orbitals to the two H’s and the O.
The leftover p orbitals will form the pi bond.
Figure 11.13 from 4th ed.
Restricted rotation of p-bonded molecules in C2H2Cl2.
CIS
TRANS
This cis/trans arrangement will be important in chem 2, organic chem and
biology!
VALENCE BOND THEORY:
RESONANCE
Resonance Structures and p Bonding:
p resonance structures involve an
electron pair used alternately as a p
bond or a LP
Ozone: O3 O==O--O or O--O==O
All are sp2, trig planar, each has 3 sp2
orbitals and a p orbital remaining.
VALENCE BOND THEORY
Benzene: C6H6 has carbons with sp2 hybrids
and 120o angles, each C has 2 s bonds to
other C's, 1 s bond to H, and 1 p bond
electron available
Get "ring" of delocalized p e-s
SUMMARY: draw the Lewis structure;
determine arrangement of electron pairs
using VSEPR, specify the hybrid orbitals to
accommodate the e- pairs
Benzene sigma bond formation between C’s and C-Hs
The leftover p orbitals will form
alternating pi bonds as shown in
sketch.
MOLECULAR ORBITAL
THEORY:
- explains why H2 forms easily and He2 does not
- is an alternate way of viewing e- orbitals in molecules
where pure s and pure p orbitals combine to produce
orbitals that are delocalized over the molecule
- they can have different energies and are assigned
electrons just like we do in an atom - Pauli exclusion
principle and Hund's rule included
Pauling's Valence Bond Theory does not explain
everything
MO Theory doesn't either, but it does correctly predict the
electronic structure of certain molecules that do not
follow Lewis's approach, including the paramagnetism
of certain molecules, like O2
The Central Themes of MO Theory
A molecule is viewed on a quantum mechanical level as a
collection of nuclei surrounded by delocalized molecular orbitals.
Atomic wave functions are summed to obtain molecular wave
functions. The number of molecular orbitals produced is always = #
of atomic orbitals brought by the combining atoms (only orbitals on
different atoms are combined).
If wave functions reinforce each other, a bonding MO is formed
(region of high electron density exists between the nuclei).
If wave functions cancel each other, an antibonding MO is formed
(a node of zero electron density occurs between the nuclei).
The electrons of the molecule are placed in bonding or antibonding
orbitals of successively higher energy (just like Hund's rule).
Atomic orbitals combine most effectively with orbitals of the same type
and similar energy (s w/s, n=2 w/ n=2)
Figure 11.13
An analogy between light waves and atomic wave functions.
Amplitudes of wave
functions added
Amplitudes of
wave functions
subtracted.
Figure 11.14
Contours and energies of the bonding and antibonding
molecular orbitals (MOs) in H2.
The bonding MO is lower in energy and the antibonding MO is higher in
energy than the AOs that combined to form them.
MOLECULAR ORBITAL
THEORY
BOND ORDER: the number of bonding epairs shared by 2 atoms in a molecule
Fractional bond orders are possible in MO
Theory!
Silberberg method:
B.O. = ½(# of e- in bonding orbitals - # of ein antibonding orbitals)
Figure 11.15
The MO diagram for H2.
Filling molecular orbitals with electrons follows the
same concept as filling atomic orbitals.
Energy
s*1s
1s
1s
H2 bond order
= 1/2(2-0) = 1
s1s
AO
of H
MO
of H2
AO
of H
Figure 11.16
MO diagram for He2+ and He2.
s*1s
1s
1s
Energy
Energy
s*1s
1s
1s
s1s
AO of
He
MO of
He+
s1s
AO of
He+
He2+ bond order = 1/2
AO of
He
MO of
He2
AO of
He
He2 bond order = 0
SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams
PROBLEM:
Use MO diagrams to predict whether H2+ and H2- exist.
Determine their bond orders and electron configurations.
SOLUTION:
s
1s
bond order
= 1/2(1-0)
= 1/2
AO of H
s
MO of H2+
H2- does exist
H2 does exist
1s
AO of H
s
+
bond order
= 1/2(2-1)
= 1/2
1s
1s
AO of H-
AO of H
s
MO of H2-
Energy
Figure 11.17
Bonding in s-block homonuclear diatomic molecules.
s*2s
s*2s
2s
2s
s2s
2s
2s
s2s
Be2
Li2
Li2 bond order = 1
Be2 bond order = 0
Figure 11.18
Contours and energies of s and p MOs through
combinations of 2p atomic orbitals.
Or the pz orbitals
Figure 11.19
Relative MO energy levels for Period 2 homonuclear
diatomic molecules.
with 2s-2p
mixing
without 2s-2p
mixing
Memorize
this!
MO energy levels
for O2, F2, and Ne2
MO energy levels
for B2, C2, and N2
Figure 11.20
MO occupancy
and molecular
properties for B2
through Ne2
Figure 11.21
The paramagnetic
properties of O2
SAMPLE PROBLEM 11.4
PROBLEM:
Using MO Theory to Explain Bond Properties
As the following data show, removing an electron from N2 forms
an ion with a weaker, longer bond than in the parent molecules,
whereas the ion formed from O2 has a stronger, shorter bond:
N2
N2+
O2
O 2+
Bond energy (kJ/mol)
945
841
498
623
Bond length (pm)
110
112
121
112
Explain these facts with diagrams that show the sequence and occupancy of MOs.
SOLUTION:
N2 has 10 valence electrons, so N2+ has 9.
O2 has 12 valence electrons, so O2+ has 11.
SAMPLE PROBLEM 11.4
Using MO Theory to Explain Bond Properties
continued
N2+
N2
bonding e- lost
1/2(8-2)=3
O2 +
O2
s2p
s2p
p2p
p2p
s2p
s2p
p2p
p2p
s2s
s2s
s2s
s2s
1/2(7-2)=2.5
bond
orders
1/2(8-4)=2
antibonding
e- lost
1/2(8-3)=2.5
MO Theory Practice
1. Draw the bonding and antibonding
molecular orbitals for H2.
2. Do Valence Bond Theory (hybridization)
and MO Theory for both O2 and O22-.
Which theory works better to explain the
molecule and ion?
3. For N2, N2+ and N2- compare
a. Magnetic character
b. Net number of p bonds
c. Bond Order
d. Bond length
e. Bond strength
Answers
1. See picture in text.
2. VB Theory shows O2 as sp2 hybridized with
one s bond and one p bond. There are two
lone pairs on each O. O22- has one s bond,
and each O has three lone pairs. MO
Theory shows a bond order of 2 for O2 and
that it is paramagnetic. MO Theory shows a
bond order of 1 for O22- and diamagnetic.
But MO Theory fits the real data that O2 is
paramagnetic.
Answers con’t
N2
N2+
N2-
a.
Diamag
Paramag
Paramag
b.
2
1.5
1.5
c.
3
2.5
2.5
d.
Short
Longer
Longer
e.
Strong
Weaker
Weaker