Chapter 9 Molecular Geometry and Bonding Theories

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Transcript Chapter 9 Molecular Geometry and Bonding Theories

Chemistry 100
Chapter 9
Molecular Geometry and
Bonding Theories
Molecular Geometry
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The three-dimensional arrangement
of atoms in a molecule  molecular
geometry
Lewis structures can’t be used to
predict geometry
Repulsion between electron pairs
(both bonding and non-bonding)
helps account for the molecular
structure!
The VSEPR Model
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Electrons are negatively charged, they want to
occupy positions such that electron

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Electron interactions are minimised as much as possible
Valence Shell Electron-Pair Repulsion Model
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treat double and triple bonds as single domains
resonance structure - apply VSEPR to any of them
formal charges are usually omitted
Molecules With More Than One
Central Atom
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We simply apply VSEPR to each ‘central atom’
in the molecule.
• Carbon #1 –
tetrahedral
• Carbon #2 – trigonal
planar
Dipole Moments
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The HF molecule has a bond dipole – a charge
separation due to the electronegativity
difference between F and H.
The shape of a molecule and the magnitude of
the bond dipole(s) can give the molecule an
overall degree of polarity  dipole moment.
+H-F
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Homonuclear diatomics  no dipole moment
(O2, F2, Cl2, etc)
Triatomic molecules (and greater). Must look
at the net effect of all the bond dipoles.
In molecules like CCl4 (tetrahedral) BF3
(trigonal planar) all the individual bond dipoles
cancel  no resultant dipole moment.
Bond Dipoles in Molecules
More Bond Dipoles
Valence Bond Theory and
Hybridisation
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Valence bond theory
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description of the covalent bonding and
structure in molecules.
Electrons in a molecule occupy the
atomic orbitals of individual atoms.
The covalent bond results from the
overlap of the atomic orbitals on the
individual atoms
The Bonding in H2
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Hydrogen molecule
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a single bond indicating the overlap of the
1s orbitals on the individual atoms
cylindrical symmetry with respect to the
line joining the atomic centres, i.e., a 
bond
H
H
Overlap Region
1s (H1) – 1s(H2)  bond
The Bonding in H2
H
H
The Cl2 Molecule
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In the chlorine molecule, we
observe a single bond indicating the
overlap of the 3p orbitals on the
individual atoms.
Cl
Bonding description
2)
Cl
3pz (Cl 1) – 3pz (Cl
Is This a  Bond?
Cl
Cl
Hybrid Atomic Orbitals
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Look at the bonding picture in
methane (CH4).
• Bonding and geometry in polyatomic molecules
may be explained in terms of the formation of
hybrid atomic orbitals
• Bonds  overlap of the hybrid atomic orbitals on
central atoms with appropriate half-filled atomic
orbital on the terminal atoms.
The CH4 Molecule
The Formation of the sp3 Hybrids
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Mix 3 “pure” p
orbitals and a
“pure” s orbital
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form an sp3
“hybrid” orbital.
Rationalize the
bonding around the
C central atom.
sp2 Hybridisation
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What if we try to rationalise the
bonding picture in the BH3 (a trigonal
planar molecule)?
We mix 2 “pure” p orbitals and a “pure” s
orbital to form “hybrid” or mixed sp2
orbitals.
These three sp2 hybrid orbitals lie in the
same plane with an angle of 120
between them.
A Trigonal Planar Molecule
H
H
Overlap
regions
B
Overlap region
H
sp Hybridisation
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What if we try to rationalize the
bonding picture in the BeH2 species
(a linear molecule)?
We mix a single “pure” p orbital and
a “pure” s orbital to form two
“hybrid” or mixed sp orbitals
These sp hybrid orbitals have an
angle of 180 between them.
A Linear Molecule
The BeH2 molecule
Overlap Regions
H
Be
H
Double Bonds
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Look at ethene C2H4.
Each central atom is an AB3 system,
the bonding picture must be
consistent with VSEPR theory.
The  Bond
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Additional feature

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an unhybridized p
orbital on adjacent
carbon atoms.
Overlap the two
parallel 2pz orbitals
(a -bond is
formed).
Bond overlaps in C2H4
There are three
different types of bonds
[sp2 (C ) – 1s (H) ] x 4 
type
[sp2 (C 1 ) – sp2 (C 2 ) ]
 type
[2pz (C 1 ) – 2pz (C 2 ) ]
 type

The C2H4 Molecule
The Bond Angles in C2H4
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Bond angles HCH =
HCC  120.
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 bond is perpendicular
to the plane containing
the molecule.
Double bonds –

Rationalize by assuming
sp2 hybridization exists
on the central atoms!
Any double bond  one 
bond and a  bond
The Triple Bond in C2H2
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Bond angles HCH = HCC =
180.

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 bonds are perpendicular to
the molecular plane.
Triple bond  one  bond and
two  bonds
Triple bond rationalized by
assuming sp hybridization exists on
the central atoms!
Bond Overlaps in C2H2
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There are again three different types of bonds
[sp (C ) – 1s (H) ] x 2  type
[sp (C 1 ) – sp (C 2 ) ]
 type
[2py (C 1 ) – 2py (C 2 ) ]  type
[2pz (C 1 ) – 2pz (C 2 ) ]  type
Bond Overlaps in H2CO
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There are again three different types of bonds
[sp (C) – 1s (H) ] x 2  type
[sp2 (C) – sp2 (O) ]
 type
[2p (C) – 2p (O) ]
 type
Key Connection – VSEPR and Valence
Bond Theory!!
sp3d Hybridisation
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How can we use the hybridisation concept to
explain the bonding picture PCl5.
There are five bonds between P and Cl (all 
type bonds).
5 sp3d orbitals  these orbitals overlap with the
3p orbitals in Cl to form the 5  bonds with the
required VSEPR geometry  trigonal bipyramid.
Bond overlaps
[sp3d (P ) – 3pz (Cl) ] x 5
 type
sp3d2 Hybridisation
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Look at the SF6 molecule.
6 sp3d2 orbitals  these orbitals
overlap with the 2pz orbitals in F to
form the 6  bonds with the
required VSEPR geometry 
octahedral.
Bond overlaps
[sp3d2 (S ) – 2pz (F) ] x 6
 type
Notes for Understanding
Hybridisation
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Applied to atoms in molecules only
Number hybrid orbitals = number of atomic
orbitals used to make them
Hybrid orbitals have different energies and
shapes from the atomic orbitals from which
they were made.
Hybridisation requires energy for the promotion
of the electron and the mixing of the orbitals 
energy is offset by bond formation.
Delocalised Bonding
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In almost all the cases where we described the
bonding n the molecule, the bonding electrons
have been totally associated with the two
atoms that form the bond  they are localised.
What about the bonding situation in benzene,
the nitrate ion, the carbonate ion?
In benzene, the C-C  bonds are formed from
the sp2 hybrid orbitals. The unhybridised 2pz
orbital on C overlaps with another 2pz orbital on
the adjacent C atom.
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Three  bonds are formed. These 
bonds extend over the whole
molecule (i.e. the  bonds are
delocalised).
The  electrons are free to move
around the benzene ring.
Any species where we had several
resonance structures, we would
have delocalisation of the electrons.
Delocalised Electrons in Molecules
Molecular Orbital (M.O.) Theory
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Valence bond and the concept of the
hybridisation of atomic orbitals does not
account for a number of fundamental
observations of chemistry.
To reconcile these and other differences, we
turn to molecular orbital theory (MO theory).
MO theory – covalent bonding is described in
terms of molecular orbitals
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the combination of atomic orbitals that results in
an orbital associated with the whole molecule.
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Recall the wave properties of
electrons.
constructive interference  the two e- waves interact
favourably; loosely analogous to a build-up of edensity between the two atomic centres.
destructive interference  unfavourable interaction of
e- waves; analogous to the decrease of e- density
between two atomic centres.
Constructive and Destructive
Interference
Constructive
+
Destructive
+
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ybonding = C1 ls (H 1) + C2 ls (H 2)
yanti = C1 ls (H 1) - C2 ls (H 2)
Bonding Orbital  a centro-symmetric orbital
(i.e. symmetric about the line of symmetry of
the bonding atoms).
Bonding M’s have lower energy and greater
stability than the AO’s from which it was
formed.
Electron density is concentrated in the region
immediately between the bonding nuclei.
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Anti-bonding orbital  a node (0 electron
density) between the two nuclei.
In an anti-bonding MO, we have higher energy
and less stability than the atomic orbitals from
which it was formed.
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As with valance bond theory (hybridisation)
2 AO’s  2 MO’s
Bonding and Anti-Bonding M.O.’s
from 1s atomic Orbitals
* 1s
1s
1s
Energy
1s
The MO’s in the H2 Atom
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The situation for two 2s orbitals is the same!
The situation for two 3s orbital is the same.
Let’s look at the following series of molecules
H2, He2+, He2
bond order = ½ {bonding - anti-bonding e-‘s}.
Higher bond order  greater bond stability.