Section Six pp. 1
Section Six: Chemical Bonding
Reading: Tro Chapters 9 and 10
Recommended Problems: Mastering Chemistry Online Homework
Chemical Bonding: An Introduction
1.
chemical bond: described as a link between atoms
2.
ionic bond: results from the electrostatic attraction of oppositely charged ions,
where complete transfer of one or more electrons is made from one atom to
another
Examples: NaCl, MgBr2, CaO; recall that the two bonded atoms should
have a ∆χ (electronegativity) = 2.0 in order to be classified as ionic.
Polarizability: all ionic bonds possess some covalent character. Ionic
bonds acquire more covalent character as the distortion of the electron
cloud on the anion increases. Furthermore, compounds composed of
highly polarizing cations and highly polarizable anions have significant
covalent character in their bonding.
3.
covalent bond: consists of a pair of electrons shared between two atoms
Examples: F2, CH4, SO2, P4, S8
Electronegativity: two bonded atoms should have a Pauling
electronegativity difference generally less than 1.5 to be classified as
covalent.
Polar Covalent: arises between two atoms with partial electric charges
arising from their differences in electronegativity
4.
octet rule: in covalent bond formation, atoms go as far as possible towards
completing their octets by sharing electron pairs.
EXCEPTIONS: Group I, II, and III elements
5.
Lewis dot structure: shows the atoms by their chemical symbols, the covalent
bonds by lines, and the lone pairs by pairs of dots
Section Six pp. 2
6.
single vs. double vs. triple bonds:
Bond Strength (Order): Triple > Double > Single bond (measured as
Average Dissociation Energy in units of kJ/mol; greater dissociation
energy = stronger bond)
I-I(g) → 2I(g)
D = 139 kJ/mol
C≡O(g) → C(g) + O(g)
D = 1062 kJ/mol
Bond Length: defined as the distance between the centers of two atoms
joined by a covalent bond; multiple bonds are shorter than single bonds
(e.g., “rubber band” analogy)
7.
Resonance: delocalization of electrons; more than one legitimate Lewis dot
structure can be drawn
Example: benzene
8.
Formal Charge example:
Rules for writing Lewis dot structures for molecules:
1.
Write the skeletal structure of the compound showing which atoms are bonded to
what other atoms. Consider the following useful tips:
A.
B.
2.
Determine the sum of the valence electrons for all atoms in the molecule. For
polyatomic ions,
A.
B.
3.
The least electronegative atom usually occupies the central position in a
molecule.
Molecules are often symmetric.
add an electron for every negative charge, and
subtract an electron for every positive charge.
A pair of bonding electrons between atoms is designated with a solid line, which
represents TWO electrons. Remember that atoms can be bonded in multiple
manners (i.e. double and triple bonds).
Section Six pp. 3
4.
Arrange the rest of the electrons (dots) around the atoms so that every atom has
eight electrons (an octet). Remember that if the central atom is from row 3 or
higher of the periodic table, it may constitute an exception to the octet rule (i.e. it
can possess more than 8 surrounding electrons). Also recall that elements in
Groups I, II, and III do not obey the octet rule either. The general rule for these
atoms is that the number of valence electrons = number of bonds.
5.
Calculate the formal charge for each atom in your molecule; recall that the best
Lewis dot structure is the one that minimizes formal charge amongst all the atoms
(note: this may not necessarily mean “0”, but perhaps as close to “0” as possible).
Problem #1: Draw the best possible Lewis dot structures for each of the following
compounds or ions shown below, and include resonance hybrids where appropriate:
A.
carbon tetrabromide
B.
AsH3
C.
formate ion, HCO2-
D.
ethanol
E.
CH3NH2
F.
CN-
G.
SF6
H.
XeF4
I.
ClF3
J.
AsF5
Section Six pp. 4
K.
AsO4-3
L.
IO4-
M.
Sulfuric Acid
N.
Phosphoric Acid
O.
benzene (an aromatic organic compound)
Problem #2: The CO32- ion has three possible Lewis dot structures.
A.
B.
Draw the corresponding electron-dot diagrams and assign formal charges to all
the atoms present.
Consider the following bond lengths:
C-O 1.43 Å
C=O 1.23 Å
C≡O 1.09 Å
Rationalize the experimental observation that all three C-O bonds have identical
bond lengths of 1.36 Å.
Structural isomers are compounds that possess the same chemical formula but
different connectivity among the various atoms. For example, consider the formula
C2H6O, which can be written in two different ways: (1) CH3CH2OH or (2) CH3OCH3:
Can you draw various structural isomers for the hexane molecule?:
Section Six pp. 5
Now consider a bond in which both electrons originate from one of the atoms,
known as the coordinate covalent bond. While this type of bond is common to all
transition metal complexes (as we will see during the spring semester), it can also be
formed among main group elements, particularly among Lewis Acid/Base complexes.
Recall that we define a Lewis base as an electron pair donor, and a Lewis acid as an
electron pair acceptor. The form of the reaction is therefore:
Acid + :base → complex
For example, boron trichloride and ammonia react in a Lewis Acid-Base reaction as
follows:
Problem #3: Show that the formation of Al2Cl6 from AlCl3 molecules is a Lewis
acid/base reaction.
Valence Shell Electron Pair Repulsion Theory (VSEPR) states that electron
pairs repel one another, whether they are in chemical bonds (bonding pairs) or unshared
(lone pairs). Electron pairs assume orientations about an atom to minimize repulsions.
The schematic below summarizes four out of the five main electronic geometries (NOTE:
linear is not shown) along with their respective molecular geometries. You are expected
to know ALL OF THESE!!! Please note that this model is idealized; molecules are not
always in these “perfect” conformations.
“Electronic” Geometry
“Molecular” Geometry
(# unshared pair electrons)
Trigonal Planar
Tetrahedral
Trigonal Planar
Angular (bent)
Tetrahedral
(0)
(1)
(0)
Trigonal
pyramidal
(1)
Angular/bent
(2)
Section Six pp. 6
Trigonal Bipyramidal
Trigonal Bipyramidal
(0)
Seesaw
T-shaped
(1)
Linear
(2)
(3)
Octahedral
Octahedral
(0)
Square Pyramidal
(1)
Square Planar
(2)
Furthermore, while VSEPR provides a simple means for predicting shapes of
molecules, it does not explain why bonds exist between atoms. Instead, we turn to
Valence Bond Theory, relying on hybridization to further describe the overlap of atomic
orbitals which form molecular orbitals:
Atomic Orbital Set
s, p
s, p, p
s, p, p, p
s, p, p, p, d
s, p, p, p, d, d
Hybrid Orbital Set
Two sp
Three sp2
Four sp3
Five dsp3
Six d2sp3
Electronic Geometry
Linear
Trigonal Planar
Tetrahedral
Trigonal Bipyramidal
Octahedral
Each single bond in a molecule represents a σ bond; each subsequent bond within each
single (σ) bond represents a π bond. Once the framework of a molecule is set up using
the appropriate hybrid orbitals for σ bonds, the remaining orbitals may mix together to
form π bonds. Consider the orbital overlap in C2H4:
Section Six pp. 7
Problem #4: Consider all the following compounds/ions from the previous problem
along with a few new structures. Use VSEPR to predict the shape for each molecule.
Predict the approximate bond angles where appropriate and determine whether the
molecule is polar or nonpolar. Finally, determine the hybridization of the central atom.
How many sigma (σ) and pi (π) bonds are contained within each compound?
A.
carbon tetrabromide
B.
AsH3
C.
formate ion, HCO2-
D.
ethanol
E.
CH3NH2
F.
CN-
G.
SF6
H.
XeF4
I.
ClF3
J.
AsF5
K.
AsO4-3
L.
IO4-
M.
Sulfuric Acid
N.
Phosphoric Acid
O.
CH2Br2
P.
CS2
Section Six pp. 8
Q.
NO2-
S.
C2H2Br2
R.
PCl3
The Born-Haber Cycle
A.
For a given solid, the difference in molar enthalpy between the solid and a
gas of widely separated ions is called the lattice enthalpy (energy) of the
solid. That is, lattice enthalpy (or energy) is the energy required to
separate gaseous ions from a solid.
MX(s) → M+(g) + X-(g)
B.
C.
D.
E.
The lattice enthalpy of a solid CANNOT be measured directly.
However, it can be calculated by utilizing the first law of thermodynamics
and, in particular, the fact that enthalpy is a state function.
This calculation invokes the use of a Born-Haber cycle, a closed path of
steps, one of which is the formation of a solid lattice from its gaseous ions.
The cycle traces the enthalpy changes that occur:
1.
beginning with the pure elements,
2.
atomize the pure elements to form gaseous atoms,
3.
ionize the atoms to form gaseous ions,
4.
allow the ions to form an ionic solid,
5.
AND convert the solid back into the pure elements.
Use your appendix for pertinent lattice enthalpy values when appropriate.
Problem #5: Devise and use a Born-Haber cycle to calculate the lattice enthalpy of
potassium chloride.
Section Six pp. 9
Problem #6:
Calculate the lattice enthalpy of calcium chloride.
Problem #7:
Calculate the lattice enthalpy of magnesium bromide.
Bond Energies (Enthalpies) – the difference between the standard molar enthalpies of a
molecule X-Y and its fragments X and Y. NOTE: This is an APPROXIMATION!
ΔH° = ΣBE (bonds broken) – ΣBE (bonds made)
A.
B.
All bond enthalpies listed in your textbook are POSITIVE because heat
must be supplied to break a bond.
Therefore, bond breaking is always ENDOTHERMIC, and bond
formations is always EXOTHERMIC.
Section Six pp. 10
C.
D.
Assume that the average bond energy applies regardless of the specific
molecular environment; intermolecular interactions are expected to be
minimal and hence are NOT taken into account.
These calculations are limited to cases where ALL reactants/products are
in the gas phase!
Problem #8: Estimate the enthalpy change of the reaction between gaseous iodoethane
and water vapor:
CH3CH2I(g) + H2O(g) → CH3CH2OH(g) + HI(g)
Bond Energies (kJ/mol):
Problem #9:
C-O
C-I
360
238
H-O
H-I
463
299
Estimate the standard enthalpy of the following reaction:
CCl3CHCl2(g) + 2HF(g) → CCl3CHF2(g) + 2HCl(g)
Bond Energies (kJ/mol):
C-F
H-F
485
565
C-Cl 339
H-Cl 431
Molecular Orbital Theory
Just as atomic orbitals are solutions to the quantum mechanical treatment of
atoms, molecular orbitals (MOs) are solutions to the molecular problem. Hence, another
method often used to describe bonding is the molecular orbital model. In this model, the
electrons are assumed to be delocalized rather than always located between a given pair
of atoms (i.e. the orbitals extend over the entire molecule). Drawings for the various
atomic orbitals overlapping to form molecular orbitals will be given in class!
There is still one fundamental difficulty encountered with this model when
dealing with polyelectronic atoms – the electron correlation problem. Since one cannot
account for the details of the electron movements, one cannot deal with the electronelectron interactions in a specific way. We can only make approximations that allow the
solution of the problem but do not destroy the model’s physical integrity. The success of
these approximations can only be measured by comparing predictions from the theory
with experimental observations.
Section Six pp. 11
The following vocabulary terms are crucial in terms of your understanding of
Molecular Orbital (MO) Theory. Consider the following:
1.
bonding molecular orbitals: lower in energy than the atomic orbitals of which it is
composed. Electrons in this type of orbital favor the molecule; that is, they will
favor bonding.
2.
antibonding molecular orbitals: higher in energy than the atomic orbitals of which
it is composed. Electrons in this type of orbital will favor the separated atoms.
3.
bond order: the difference between the number of bonding electrons and the
number of antibonding electrons, divided by 2. Bond order is an indication of
strength. Larger bond orders indicate greater bond strength.
4.
sigma (σ) molecular orbitals: The electron probability of both bonding and
antibonding molecular orbitals is centered along the line passing through the two
nuclei, where the electron probability is the same along any line drawn
perpendicular to the bond axis at a given point on the axis. They are designated
σs for the bonding MO and σs* for the antibonding MO.
5.
pi (π) molecular orbitals: p orbitals that overlap in a parallel fashion also produce
bonding and antibonding orbitals, where the electron probability lies above and
below the line between the nuclei. They are designated πp for the bonding MO
and πp* for the antibonding MO.
The following are some useful ideas about molecular orbitals and how electrons
are assigned to them:
1.
2.
The number of MOs formed is equal to the number of atomic orbitals combined.
Of the two MOs formed when two atomic orbitals are combined, one is a bonding
MO at a lower energy than the original atomic orbitals. The other is an
antibonding MO at a higher energy.
3.
In ground-state configurations, electrons enter the lowest energy MOs available.
4.
The maximum number of electrons in a given MO is two (Pauli Exclusion
Principle).
5.
In ground-state configurations, electrons enter MOs of identical energies singly
before pairing begins (Hund’s Rule).
Consider the MO diagrams for the diatomic molecules and ions of the first-period
elements:
Section Six pp. 12
σ 1s*
1s
σ 1s
σ 1s*
1s
1s
H2+
H2
σ 1s*
1s
σ 1s
1s
σ 1s
σ 1s*
1s
He2+
1s
1s
σ 1s
He2
Now consider two possible molecular orbital energy-level diagrams for diatomic
molecules of the second-period elements:
Z>8
σ 2p*
π 2p*
2p
2p
π 2p
σ 2p
σ 2s*
2s
σ 2s
2s
Z<7
σ 2p*
π 2p*
2p
2p
σ 2p
π 2p
σ 2s*
2s
σ 2s
2s
Section Six pp. 13
What if the two diatomic elements (or ions) are different? Then you must take
electronegativity into account when constructing the molecular orbital energy diagram:
σ 2p*
π 2p*
2p
2p
π 2p
σ 2p
σ 2s*
2s
σ 2s
2s
Finally, consider a diatomic molecule where one of the bonded atoms is hydrogen:
σ 2p*
1s
π nb
2p
σ 2p
2s
σ nb
Problem #10: Consider the C22- ion for the following problem.
A.
B.
C.
Draw the Molecular Orbital diagram. Make sure to include the proper
atomic orbitals for each ion as well as properly label all bonding and
antibonding molecular orbitals.
Calculate the bond order for the ion based on the Molecular Orbital
diagram.
Determine whether the ion is diamagnetic or paramagnetic? Justify your
answer based on the Molecular Orbital diagram.
Section Six pp. 14
Problem #11: Draw the Molecular Orbital energy diagram for the O2+ ion.
Problem #12: Draw the Molecular Orbital energy diagram for the CO molecule.
Problem #13: Draw the Molecular Orbital energy diagram for the HBr molecule.
Section Six pp. 15
Valence Band Theory
Metallic Conductor: An electronic conductor in which the electrical conductivity
decreases as the temperature is raised. The resistance of the metal to conduct electricity
decreases as the temperature is raised because when heated, the atoms vibrate more
vigorously, passing electrons collide with the vibrating atoms, and hence do not pass
through the solid as readily.
An electronic conductor in which the electrical conductivity
Semiconductor:
increases as the temperature is raised. There are two types of semiconductors: n-type and
p-type (see schematic below).
n-type: Doping with an element of extra negative charge (electrons) into a system.
There is NO extra room for these electrons in the valence band; consequently,
they are promoted into the conduction band, where they have access to many
vacant orbitals within the energy band they occupy and serve as electrical carriers.
p-type: Doping with an element of less electrons in order to create electron
vacancies or positive holes in the system. Because the valence band is
incompletely filled, under the influence of an applied field, electrons can move
from occupied molecular orbitals to the few that are vacant, thereby allowing
current to flow.
Insulator:
Does NOT conduct electricity.
Superconductor:
A solid that has zero resistance to an electric current. Some metals
become superconductors at very low temperatures, and other compounds turn into
superconductors at relatively high temperatures.
Conduction
band
Valence
band
Insulator
* electrons are not mobile
Conduction
band
Energy
Band Gap (large
for insulators)
Energy
Energy
Conduction
band
Valence
band
n-type semiconductor
Valence
band
p-type semiconductor
* Example: Si doped with As * Example: Si doped with In