Transcript CHEMISTRY I PRE-AP CHAPTER 15 & 16 BONDING

AP CHEMISTRY
CHAPTER 9
BONDING
Hybridization
When drawing Lewis structures to
explain bonding, we have been using the
Localized Electron Model of bonding.
This assumes that the electrons stay with
(or close to) the atoms from which they
originated and that bonds are formed by
the overlap of atomic orbitals. This
model needs to be developed a little
further to explain experimental data.
In methane, CH4, carbon has one 2s and
three 2p orbitals available for bonding.
Hydrogen has one 1s orbital available. We
could imagine that one of the bonds in
methane would be formed from the overlap
of a hydrogen 1s orbital and a carbon 2s
orbital and the other three bonds would be
formed from the overlap of a hydrogen 1s
orbital and a carbon 2p orbital. This would
make 2 different kinds of bonds.
Experimental data shows that all four bonds
are identical.
Let’s look at the bonding in methane, CH4. Carbon has an electron
configuration of 1s22s22p2. This means that there is one 2s orbital and
three 2p orbitals available for bonding. Hydrogen has an electron
configuration of 1s1. It uses a 1s orbital for bonding. Remember that
s orbitals are spherical and p orbitals are “peanut” or “dumbbell”
shaped.
If these orbitals are used in their normal form for bonding, the result
would be as shown below. One bond is different from the other three.
However, experimental evidence shows that all 4 bonds in
CH4 are identical! This model does not work!
Bonding orbitals for C
Hybridization is an addition to the
localized electron model that explains
this. Carbon needs 4 identical orbitals to
form 4 identical bonds. Imagine
throwing the one 2s orbital and the three
2p orbitals from carbon into a blender.
Whrrrrr!! Out comes four new orbitals
that are a homogenous blend of the four
that went in!!
The new orbitals have properties of the
s and of the p orbitals. Since they were
made from one s and three p orbitals, we
can call them sp3 orbitals. We put four
atomic orbitals in and got four hybrid
orbitals out. The hybrid sp3 orbitals each
overlap with the 1s orbitals from each
hydrogen atom and make four identical
bonds having 109.5o bond angles.
Atoms with four effective electron pairs
have sp3 hybridization, even if they have
unshared pairs as in H2O and NH3. The
electron pairs have a tetrahedral
arrangement.
We can look at the hybridization
in the bonding of ethene (C2H4).
H
H
C=C
H
H
Each carbon atom has three effective
pairs and needs three equal orbitals. To
get three hybrid orbitals we throw one s
and two p orbitals into the blender.
Whrrrr!!! Out come three identical
orbitals made from one s and two p
orbitals. The three new orbitals are
called sp2. This type of hybridization
results in trigonal planar electron pair
arrangements and 120o bond angles.
The overlap of the 1s orbitals from each
hydrogen with the sp2 hybrid orbitals of
carbon results in sigma bonds. In a
sigma bond, the orbitals overlap head-on
and the electrons are in the region
between the two nuclei. All single bonds
are sigma bonds. Each double bond and
each triple bond contains one sigma
bond.
What happened to the p orbital that
wasn’t hybridized? The remaining p
orbitals from each carbon atom overlap
above and below the plane of the nuclei
resulting in a sideways overlap that is
called a pi bond.
A pi bond can form only if there is also a
sigma bond between the same two
atoms. Because the electrons in a pi
bond are not directly between the nuclei,
a pi bond is weaker than a sigma bond.
A double bond contains one sigma plus
one pi bond and is thus stronger and
shorter than a single bond.
Let’s look at the hybridization
and bonding in ethyne
(acetylene), C2H2.
H-CC-H
Each carbon in ethyne has two effective
electron pairs. We can throw one s and one
p orbital from carbon into the blender.
Whrrrr!!! Out come two sp orbitals. The s
orbital from each hydrogen atom overlaps
with one of the sp orbitals to form a sigma
bond. The other sp orbital overlaps with one
of the sp orbital on the other carbon to form
a sigma bond between the two C atoms.
The remaining two p orbitals from each
carbon atom overlap outside the plane of
the nuclei to form two pi bonds. A triple
bond is one sigma and two pi bonds and
is thus stronger and shorter than either a
single or a double bond. Linear
molecules and 180o bond angles result
from sp hybridization.
We learned that certain elements can
exceed the octet rule by utilizing unfilled d
orbitals. These d orbitals may also be
involved in hybridization. For example,
PF5 has five effective electron pairs. These
five pairs require five equal bonds. We
must throw one s, three p and one d orbital
into the blender to get five dsp3 orbitals.
This form of hybridization results in
trigonal bipyramidal shapes and 90o and
120o bond angles.
With SF6 we have six effective
electron pairs. We must
hybridize one s, three p, and two
2
3
d orbitals to get six d sp
orbitals. This results in the
octahedral shape with 90o bond
angles.
News Flash! Recent experimental data (2011) show that d
orbitals
are not involved in hybridization. Until textbooks catch
up, you may still see dsp3 and d2sp3 in test questions.
The predicted shapes have not changed.
You will not see dsp3 and d2sp3 on the AP test this year!
You will simply not be asked to explain the hybridization
of something with 5 or 6 electron pairs. You may still see
dsp3 and d2sp3 on other types of tests.
Sigma bonds (σ)- formed by the end-toend overlap of atomic orbitals between
two nuclei
Pi bonds(π)- formed by the
side-to-side overlap of unhybridized p
orbitals outside the plane of two nuclei
σ bonds are stronger than π bonds
because the electrons in a σ bond
are between two nuclei.
Single bonds= 1 sigma bond
Double bonds =
1 sigma bond + 1 pi bond
Double bonds are stronger and shorter than single bonds.
Triple bonds =
1 sigma bond + 2 pi bonds
Triple bonds are stronger and shorter than double bonds.
Practice:
For each of the following, determine:
a)hybridization of central atom(s)
b)bond angles
c)molecular shape
d)# of sigma bonds and # of pi bonds
e) molecular polarity
SbCl5
Cl Cl Cl
Sb
Cl Cl
−
NO3
O
O
N
O
CHCl3
H
Cl - C - Cl
Cl
:N  C
:N  C
CN:
C=C
CN:
MOLECULAR ORBITAL
BONDING THEORY (MO)
The MO model is more complex than the
localized electron model but works better
in many instances.
A molecular orbital (MO) is formed
from valence atomic orbitals. Two
atomic orbitals combine to form two
MOs.
One MO has a higher energy than the
atomic orbitals and is called an
antibonding MO. The other MO has a
lower energy (more stable) and is called
a bonding MO.
In a bonding orbital, electrons have the
greatest probability of being between the
nuclei. In an antibonding orbital,
electrons are mainly outside the space
between the nuclei.
We can have  and  MOs.
Orbitals are conserved. The number of
MOs will be the same as the number of
AOs used to construct them.
Bond order may be determined as a measure
of the stability of a molecule. Using the
localized electron model (Lewis structures),
the bond order is equal to the number of
bonds. For example, a triple bond has a bond
order of 3.
The larger the bond order, the stronger the
bond.
PARAMAGNETISM AND
DIAMAGNETISM
Substances that contain unpaired
electrons are weakly attracted into
magnetic fields and are said to be
paramagnetic. Substances in which all
of the electrons are paired are very
weakly repelled by a magnetic field and
are said to be diamagnetic.
Paramagnetism is much weaker than
ferromagnetism (Fe, Co, Ni)
Figure 9.37
Diagram of the
Kind of
Apparatus Used
to Measure the
Paramagnetism
of a Sample
The Lewis structure of a molecule does
not always correctly predict
paramagnetism. The oxygen molecule is
an example of this.
O=O
Experimental data shows that oxygen is
paramagnetic (has unpaired electrons).
The more complex molecular orbital
model will correctly predict oxygen’s
paramagnetism.
Paramagnetic or Diamagnetic?
NO2
17 electrons
O=N−O
ResonanceTo eliminate the need for resonance
we can use the localized electron model
to describe the  bonding and the MO
model to describe the  bonding.
In benzene, each C is sp2
hybridized. The 6 additional p orbitals
perpendicular to the ring are used to
form  MOs. The electrons in these 
MOs are delocalized above and below
the plane of the ring.
Figure 9.48
The Pi System for Benzene