Bonding
Ionic Bonds: Formed between ions. Transfer
of electrons occurring.
Covalent Bonds: Molecules form between
atoms that share electrons.
Metallic Bonds: Form multiple bonds using the
free flowing electrons of a metal.
Atoms tend to gain and lose electrons until they have 8 valence
electrons similar to noble gases. This is a stable electron
configuration as shown experimentally by the high ionization
energy and low electron affinity.
An
atom with a low ionization
energy reacts with an atom with
high electron affinity.
The electron moves.
Opposite charges hold the atoms
together.
𝑄1𝑄2
𝐸=𝑘
𝑑
 Q is the charge.
 d is the distance between the
centers.
 If charges are opposite, E is negative
 exothermic
 Same
charge, positive E, requires
energy to bring them together.
 Lattice
energy - the energy
associated with making a solid ionic
compound from its gaseous ions.
M
+(g) + X-(g)  MX(s)
 This
is the energy that “pays” for
making ionic compounds.
 Energy is a state function so we can
get from reactants to products in a
round about way.
Na(s)  Na(g)
DH = 109 kJ/mol
Na(g)  Na+(g) + e
DH = 495 kJ/mol
½F2(g)  F(g)
DH = 77 kJ/mol
F(g) + e F (g)
DH = -328 kJ/mol

First sublime Na

Ionize Na(g)

Break F-F Bond

Add electron to F

Formation of solid

Lattice energy
Na+ + F- NaF (s)
DH = -575 kJ/mol
Na(s) + ½F2(g)  NaF(s)
DH = -928 kJ/mol
 Electrons
are shared by atoms.
 Ionic and Covalent are extremes.
 In between are polar covalent bonds.
 The electrons are not shared evenly.
 One end is partially positive, the other
negative. Indicated using small delta d.
-
+
d+
d-
H-F
-
d+
d-
H-F
d+
d+
H-F
d-
H-F
d+
d+
d-
d-
H-F
d-
H-F
H-F
d+
+
d-
d+
d-
H-F
d+
d-
H-F
Zero
Intermediate
Large
Bond
Type
Covalent
Polar
Covalent
Ionic
Covalent Character
decreases
Ionic Character increases
Electronegativity
difference
 The
electrons in each atom are attracted to the
nucleus of the other.
 The electrons repel each other,
 The nuclei repel each other.
 They reach a distance with the lowest possible
energy.
 The distance between is the bond length.
Energy
0
Bond Length
Internuclear Distance
Energy
Bond Energy
0
Internuclear Distance
 The
ability of an electron to attract shared
electrons to itself.
 Tends to increase left to right.
 Decreases as you go down a group.
 Noble gases aren’t discussed.
 Difference in electronegativity between
atoms tells us how polar.
A
molecule with a center of negative charge
and a center of positive charge is dipolar
(two poles),
 We say that is has a dipole moment.
 Center of charge doesn’t have to be on an
atom.
 Will line up in the presence of an electric
field.
d+
d-
H-F
LiF
75%
% Ionic Character
LiBr
50%
25%
HCl
Electronegativity difference
 The
forces that cause a group of atoms
to behave as a unit.
 Why?
 Due to the tendency of atoms to achieve
the lowest energy state.
 It takes 1652 kJ to dissociate a mole of
CH4 into its ions
 Since each hydrogen is hooked to the
carbon, we get the average energy = 413
kJ/mol
 We
made some simplifications in describing the
bond energy of CH4
 Each C-H bond has a different energy.
CH4  CH3 + H
CH3  CH2 + H
CH2  CH + H
CH C + H
 Each
DH
DH
DH
DH
= 435 kJ/mol
= 453 kJ/mol
= 425 kJ/mol
= 339 kJ/mol
bond is sensitive to its environment.
 single
bond one pair of electrons is shared.
 double bond two pair of electrons are shared.
 triple bond three pair of electrons are shared.
 More bonds, shorter bond length.
can find DH for a reaction.
 It takes energy to break bonds, and end up with
atoms (+).
 We get energy when we use atoms to form
bonds (-).
 If we add up the energy it took to break the
bonds, and subtract the energy we get from
forming the bonds we get the DH.
 Energy and Enthalpy are state functions.
 We
2 CH2 = CHCH3
+
2NH3
+ O2

2 CH2 = CHC  N + 6 H2O
C-H 413 kJ/mol
O-H 467 kJ/mol
C=C 614kJ/mol
O=O 495 kJ/mol
N-H 391 kJ/mol
CN 891 kJ/mol
C-C 347 kJ/mol


Simple model, easily applied.
A molecule is composed of atoms that
are bound together by sharing pairs of
electrons using the atomic orbitals of
the bound atoms.
Three Parts
1) Valence electrons using Lewis
structures
2) Prediction of geometry using VSEPR
3) Description of the types of orbitals
Shows how the valence electrons are
arranged.
 One dot for each valence electron.
 A stable compound has all its atoms with a
noble gas configuration.
 Hydrogen follows the duet rule.
 The rest follow the octet rule.
 Bonding pair is the one between the symbols.

Sum the valence electrons.
2. Use a pair to form a bond between each pair of
atoms.
3. Arrange the rest to fulfill the octet rule (except
for H and the duet).
4. Place left over electrons on the central atom.
5. If there are not enough electrons to give the
central atom an octet, try multiple bonds.
 H2 O
 CO2
 CN1.
 The
difference between the number of
valence electrons on the free atom and
that assigned in the molecule.
 We count half the electrons in each bond
as “belonging” to the atom.
 Molecules try to achieve as low a formal
charge as possible.
 Negative formal charges should be on
electronegative elements.
 SO42-
 Sometimes
there is more than one
valid structure for an molecule or
ion.
 Use double arrows to indicate it is
the “average” of the structures.
 It doesn’t switch between them.
 Localized electron model is based on
pairs of electrons, doesn’t deal with
odd numbers.
 NO3 NO2-
 XeO
3
-3
 NO
4
 SO Cl
2 2
Odd numbers of electrons
1.

NO
Atoms with fewer than octet
2.

Be and B often do not achieve octet

BF3
Atoms with more than octet
3.



Third row and larger elements can exceed the octet
SF6
Use 3d orbitals?
 When
we must exceed the octet, extra
electrons go on central atom.
 ClF
3
 XeO
3
 ICl
4
 BeCl2
 Lewis
structures tell us how the atoms are
connected to each other.
 They don’t tell us anything about shape.
 The shape of a molecule can greatly affect
its properties.
 Valence Shell Electron Pair Repulsion Theory
allows us to predict geometry
 Molecules take a shape that puts electron
pairs as far away from each other as
possible.
 Have
to draw the Lewis structure to
determine electron domains.




bonding
nonbonding lone pair
Lone pair takes more space.
Multiple bonds count as one domain.
 The


bond angles
underlying structure
 The

number of domains determines
number of atoms determines
actual shape
Electron Bond
Domains Angles
2
180°
Underlying
Shape
Linear
3
120°
4
109.5°
Tetrahedral
5
90° &
120°
6
90°
Trigonal
Bipyramidal
Octagonal
Trigonal Planar
Electron Domain
 Just count the number of electron domains
and the geometry corresponds to that
number.
Molecular Shape
 Electron domain geometry is many times not
the shape of the molecule. Here we are
worried about just the atoms.
 Multiple
molecular geometries can result
from the same electron geometry.
 Non-bonding
pairs take up more space than
bonding pairs and therefore change the
angles within the atom.
 Multiple
bonds create a high electron density
on one side of the molecule which also
changes angles.
 Linear
 Trigonal
Planar
 Bent
 Tetrahedral
 Trigonal
pyramidal
 Trigonal bipyramidal
 Seesaw
 T-shaped
 Octahedral
 Square pyramidal
 Square planar
 Individual
bonds can have bond dipoles but
the overall molecules polarity may be
different.
 Individual
bond dipoles (vectors) can be
added to give the entire molecular dipole.
Beryllium: Has no single occupied orbitals
therefore should not bond.
If it absorbs energy to excite an electron to 2p
it has 2 electrons that can form a bond.
The s and p orbitals can combine to make a
hybrid orbital called the sp hybridized orbital.
-sp orbitals have a linear position and is
consistent with molecular geometries.
-it has 2 degenerate orbitals.
 Other
hybrid orbitals that can be made
include sp2 and sp3. This occurs when there
is more than one p orbital overlapping with
an s orbital.
 Sigma
(σ) bonds overlap head to head and
electron density is in line with the
internuclear axis.
 Pi
(π) bonds overlap side to side and electron
density is above and below internuclear axis.
 Single
bonds are all sigma bonds.
 Multiple bonds have one sigma bond and the
rest are pi.
 Pi
electrons are
not localized
over one N-O
bond but
delocalized over
all 3 bonds.
 This is related to
resonance.
 In
MO theory, we invoke the wave nature of
electrons.
 If waves interact constructively, the resulting
orbital is lower in energy: a bonding molecular
orbital.

In hydrogen the
electrons go into the
bonding orbital because
it is lower in energy.

The bond order is one half
the difference between
the number of bonding and
antibonding electrons.
For hydrogen, with two
electrons in the bonding
MO and none in the
antibonding MO, the bond
order is 2-0/2 = 1

 In
the case of He2,
the bond order
would be
1
(2 − 2) = 0
2
• Therefore, He2
does not exist.
 For
atoms with both s
and p orbitals, there
are two types of
interactions:
The s and the p orbitals
that face each other
overlap in  fashion.
 The other two sets of p
orbitals overlap in 
fashion.

 The
resulting MO
diagram looks like this
(Fig. 9.41).
 There are both  and 
bonding molecular
orbitals and * and *
antibonding molecular
orbitals.