Lecture 21: Ionic to Covalent
• Reading: Zumdahl 13.4-13.6
• Outline
– Binary Ionic Compounds
– Partial Ionic Compounds
– Covalent Compounds
Electronegativity
• Electronegativity: The ability of an atom in a
molecule to attract shared electrons to itself.
Electronegativity
• Electronegativity can be defined in many ways.
Pauling model is the most widely used.
• Idea: compare the bond energy of an “HX” molecule
to that of the average of an HH bond and an XX bond:
Expected energy = [(H-H energy)(X-X energy)]1/2
D = (H-X)experimental - (H-X)expected
D = 0: covalent
D > 0: ionic character
Electronegativity (cont.)
• Pauling used this approach to develop a scale, where
F = 4.0 (flourine has largest electronegativity).
F=4
O = 3.4
Cl = 3.2
C = 2.6
H = 2.2
Na = 0.9
Electronegativity (cont.)
• The key idea is this: the greater the electronegativity
difference between two atoms, the more ionic the
bond.
• Example: Which of the following compounds is
expected to demonstrate intermediate bonding
behavior (i.e., polar covalent).
Delect
Cl-Cl
O-H
Na-Cl
0
1.2
2.3
Dipole Moments
• The above discussion involved bonds in
which electrons were shared, but shared
unequally in polar, covalent bonds
• In the HF example, when placed in an
electric field the HF atoms will align.
• This observation demonstrates that the
centers of negative and positive charge do
not coincide.
Dipole Moments (cont.)
• When the centers of negative and positive charge are
separated, we say that the molecule has a dipole moment.
Dipole Moments (cont.)
• The dipole moment (m) is defined as:
m = QR
Charge magnitude
R
+ center
Separation distance
Dipole Moments (cont.)
• The units of dipole moment are generally the Debye (D):
1 D = 3.336 x 10-30 C.m
• Example, the dipole moment of HF is 1.83 D. What would
it be if HF formed an ionic bond (bond length = 92 pm)?
M = (1.6 x 10-19 C)(9.2 x 10-11 m)
= 1.5 x 10-29 C.m x (1D/3.336 x 10-30 C.m)
= 4.4 D
Dipole Moments (cont.)
• Molecular geometry is a critical factor in determining if a
molecule has a dipole moment:
Dipole Moments (cont.)
• Molecular geometry is a critical factor in determining if a
molecule has a dipole moment:
No net dipole moment. Dipoles add vectorially!
Properties of Ions
• When will a stable bond be formed?
• When one exams a series of stable compounds, it
becomes evident that in the majority of
compounds, bonding is achieved such that atoms
can achieve a nobel-gas configuration
• Example: NaCl versus Na+ClNa: [Ne]3s1
Cl: [Ne]3s23p5
Na+: [Ne]
Cl-: [Ne]3s23p6 = [Ar]
Properties of Ions (cont.)
• In this example involving NaCl,
we have a metal (Na) bonding to
a non-metal (Cl).
• Metal/non-metal binding
generally results in ionic bonding.
Properties of Ions (cont.)
• One can use this tendency to satisfy the “octet rule” to predict
the stoichiometry of ionic compounds.
• Example: Ca and O
Ca: [Ar]4s2
O: [He]2s22p4
2 e-
Ca2+: [Ar]
O2-: [He]2s22p6 = [Ne]
Formula: CaO
Properties of Ions (cont.)
• Ions on figure correspond
to nobel-gas electron
configurations.
• To form ionic binary
compounds, one simply
combines in proportions
such that total charge is
zero.
• This approach is not to be
applied to transition metals.
Properties of Ions (cont.)
• Note that size decreases
for isoelectronic species.
• Mainly a consequence of
increased charge of nucleus.
Partial Ionic Compounds
• From last lecture, if two atoms forming a bond have
differing electronegativities, they will form a bond
having ionic character.
• But where is the dividing line between “ionic” bonding
and “polar covalent” bonding?
• In the end, total ionic bonding is probably never
achieved, and all “ionic” bonds can be considered polar
covalent, with varying degrees of ionic character.
Dipole Moments
• The dipole moment (m) is defined as:
m = QR
Charge magnitude
R
+ center
Separation distance
Dipole Moments (cont.)
• Example, the dipole moment of HF is 1.83 D. What would
it be if HF formed an ionic bond (bond length = 92 pm)?
M = (1.6 x 10-19 C)(9.2 x 10-11 m)
= 1.5 x 10-29 C.m x (1D/3.336 x 10-30 C.m)
= 4.4 D
Partial Ionic Compounds (cont.)
• We can define the ionic character of bonds as follows:
% Ionic Character =
(dipole moment X-Y)experimental
(dipole moment X+Y-)calculated
x 100%
Partial Ionic Compounds (cont.)
Covalent
Ionic
Increased Ionic Character
Polar Covalent
Covalent Compounds
• In covalent bonding, electrons are “shared” between bonding
partners.
• In ionic bonding, Coulombic interactions resulted in the
bonding elements being more stable than the separated atoms.
• What about covalent bonds…what is the “driving force”?
Covalent Compounds (cont.)
• Back to H2.
Covalent Compounds (cont.)
• The same concept can be envisioned for other covalent
compounds:
Think of the covalent bond as the
electron density existing
between the C and H atoms.
Covalent Compounds (cont.)
• We can quantify the degree of stabilization by seeing how
much energy it takes to separate a covalent compound into
its atomic constituents.
C(g) + 4H(g)
CH4(g)
q
Covalent Compounds (cont.)
• Since we broke 4 C-H bonds with 1652 kJ in, the bond
energy for a C-H bond is:
1652 kJ mol
4
 413 kJ mol
• We can continue this process for a variety of compounds to

develop
a table of bond strengths.
Covalent Compounds (cont.)
• Example: It takes 1578 kJ/mol to decompose CH3Cl into
its atomic constituents. What is the energy of the C-Cl
bond?
CH3Cl: 3 C-H bonds and 1 C-C bond.
3 (C-H bond energy) + C-Cl bond energy = 1578 kJ/mol
413 kJ/mol
1239 kJ/mol + C-Cl bond energy = 1578 kJ/mol
C-Cl bond energy = 339 kJ/mol
Covalent Compounds (cont.)
• We can use these bond energies to determine DHrxn:
DH = sum of energy required to break bonds (positive….heat
into system) plus the sum of energy released when the
new bonds are formed (negative….heat out from system).
DHrxn   Dbonds broken   Dbonds formed
Covalent Compounds (cont.)
• Example: Calculate DH for the following reaction using
the bond enthalpy method.
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
Go to Table 13.6:
4 x C-H 413
2 x O=O 495
4 x O-H
2 x C=O
467
745
Covalent Compounds (cont.)
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
DHrxn   Dbonds broken   Dbonds formed
= 4D(C-H) + 2D(O=O) - 4D(O-H) - 2D(C=O)

= 4(413) + 2(495) - 4(467) - 2(745)
= -716 kJ/mol
• Exothermic, as expected.
Covalent Compounds (cont.)
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
• As a check:
o
DHrxn
 DH of (prod.)  DH of (react.)

0
= DH°f(CO2(g)) + 2DH°f(H2O(g))
- DH°f(CH4(g)) - 2 DH°f(O2(g))
= -393.5 kJ/mol + 2(-242 kJ/mol) - (-75 kJ/mol)
= -802.5 kJ/mol