AP Chemistry
Unit #7
Chemical Bonding &
Molecular Shape
Zumdahl Chapters 8 & 9
TYPES OF BONDING
BONDING
INTRA
(Within (inside) compounds)
STRONG
Ionic
(Metal + non-metal)
Giant ionic lattice formed
Covalent
(Non-metals)
Discrete molecules formed
Dative or Co-ordinate
(Electron deficient species)
Discrete molecules formed
INTER
(Interactions between the molecules of a compound)
WEAK
Hydrogen Bonding
(H attached to N, O or F)
Stronger permanaent dipoles
van der Waals Forces
(Attractions between dipoles)
Permanent or induced
Dipole-Dipole
(Polar molecules)
Permanent dipoles
London Dispersion Forces
(Non-polar molecules)
Induced dipoles
I. Ionic Bonding
A. Lewis Structures & The Octet Rule
1. Lewis Symbols – models used to show the valence electron configuration
for elements, molecules and compounds
Ex: Cl
O2
2. Octet Rule – all atoms tend to gain, lose or share electrons in order to
have eight electrons in their valence shell
B. Energy of Ionic Bond Formation (link)
1. Lattice Energy – the energy released when separated gaseous ions are
packed together to form an ionic solid
2. Magnitude of Lattice Energy – the energy released when ions form a solid
is directly proportional to the charge on the ions and inversely related to
the size of the ions; smaller ions with greater charge have greater lattice
energy than larger ions with lesser charge.
Where k is the Coulomb’s law constant with a value of 8.99 x 109 Jm/C2
Q1 and Q2 are the magnitudes of the charge on the ions in coulombs, and d
is the distance between ion centers in meters
Sample Ex 8.1 – Arrange the following compounds in order of increasing
lattice energy: LiF, KBr, MgO
C. Electron Configurations of Ions (link)
1. Cations – remove one or more electrons from the last number in the
electron configuration of the neutral element
2. Anions – add one or more electrons to the last number in the electron
configuration of the neutral element
D. Polyatomic Ion – a charged particle that contains two or more
covalently bonded atoms
E. Size of Ions
1. Isoelectronic Series – a group of ions (and often a noble gas) that
contain the same number of electrons
Example: O-2, F-, Na+, Mg+2, Al+3, Ne
 10 electrons each
 As nuclear charge increases, size __________________
Sample Problem – Arrange the ions below in order of increasing size:
Sr+2, As-3, Se-2, Rb+, Br−
II. Covalent Bonding – ___________________________________________
___________________________________________________________________
A. Lewis Structures of Covalent Bonds – ______________________
_____________________________________________________
_____________________________________________________
Sample Problem – Draw the Lewis structures for water, carbon dioxide, and
ammonia.
B. Multiple Bonds
1. Single Bond – one pair, or 2 valence electrons shared between atoms
2. Double Bond – two pair, or 4 valence electrons shared between atoms
3. Triple Bond – three pair, or 6 valence electrons shared between atoms
III. Bond Polarity and Electronegativity
A. Bond Polarity
1. Nonpolar Covalent Bond – a covalent bond in which the electrons are
shared equally between atoms
2. Polar Covalent Bond – a covalent bond in which the electrons are NOT
shared equally between atoms because of a difference in
electronegativity
3. Coordinate Covalent Bond – a bond formed when one atom provides
both electrons in a shared pair
Ex: NH4+ (draw the Lewis dot structure for this molecule)
B. Electronegativity
IV. Drawing Lewis Structures
A. Procedure for drawing Lewis Structures
1. Count up the number of valence electrons
2. Write the atom symbols and connect the atoms with single bonds
3. Distribute electrons (in pairs) to complete octets of atoms
4. If there are not enough electrons, make multiple bonds to complete
octets
5. If there are extra electrons, put them on the central atom
B. Sample Exercise 8.6 – Draw the Lewis Structure for phosphorus
trichloride.
C. Sample Exercise 8.7 – Draw the Lewis Structure for HCN
V. Resonance Structures
A. Resonance Structures/Resonance Forms – resonance occurs
when more than one valid Lewis structure can be written for a particular
molecule; the actual structure of the molecule is an “average” of its
resonance structures
B. Formal Charge – the difference between the number of valence
electrons on the free atom and the number of valence electrons assigned to
the atom in the molecule.
C. Sample Exersise 8.9 – Draw 2 resonance structures for the
nitrite ion.
D. Sample Problem – Which has the shorter sulfur–oxygen
bonds, SO3 or SO32- and why?
VI. Exceptions to the Octet Rule
A. Odd Number of Electrons –
B. Less than an Octet –
C. More than an Octet –
D. Sample exercise 8.10 – Draw the Lewis Structure for ICl4-
VII. BOND ENERGIES (Look @ Table 8.4 and 8.5 on p. 372 1st)
ΔH = sum of the ENERGIES REQUIRED to break old bonds plus the sum of
the ENERGIES RELEASED in the forming of new bonds
ΔH = ΣD (bonds broken) – ΣD (bonds formed)
energy required
energy released
Where D = bond energy per mole of bonds (always has a + sign)
Ex:
H2(g) + F2(g) → 2HF(g)
ΔH = DH-H + DF-F → 2DH-F
= 1 mol (432 kJ/mol) + 1 mol (154 kJ/mol) – 2 mol (565 kJ/mol)
= -544 kJ
Sample Exercise 8.5: ΔH From Bond Energies
Calculate ΔH for the rxn. of methane w/ chlorine and fluorine to give Freon12 (CF2Cl2).
Solution:
Reactant bonds broken:
CH4 = 4 mol C-H = 4 mol (413 kJ/mol) = 1652 kJ
2Cl2 = 2 mol Cl-Cl = 2 mol (239 kJ/mol) = 478 kJ
2F2 = 2 mol F-F = 2 mol (154 kJ/mol) = 308 kJ
Total energy required = 2438 kJ
Product bonds formed:
CF2Cl2 = 2 mol C-F = 2 mol (485 kJ/mol) = 970 kJ
2 mol C-Cl – 2 mol (339 kJ/mol) = 678 kJ
HF = 2 mol H-F = 2 mol (565 kJ/mol) = 1130 kJ
HCl = 2 mol H-Cl = 2 mol (427 kJ/mol) = 854 kJ
Total energy released = 3632 kJ
ΔH = ΣD (bonds broken) – ΣD (bonds formed)
= 2438 kJ – 3632 kJ
= -1194 kJ
Therefore, 1194 kJ of energy is released per mole of CF2Cl2 formed.
PROBLEMS: p.404 #47,49
A. Energy of Ionic Bond Formation
1. Lattice Energy –
2. Magnitude of Lattice Energy –
E=k(Q1Q2/d)
Sample Ex: Arrange the following compounds in order of increasing lattice energy:
LiF, KBr, MgO.
IIX. POLAR vs. NONPOLAR MOLECULES:
Molecule = A ________________ bonded substance; always 2 or more
________________ bonded together
*______________ compound = _____________ compound
A.) POLAR Molecules:
 _______________ molecules
 _____________ sharing of _______________
 Doesn’t pass the “mirror test”
 Can’t be folded to reflect itself
 2 atoms  different elements/electronegativities
 More than 2 atoms  unbonded e- or ________ pairs around the
___________ atom
Ex: HCl
Ex: H2O
B.) NONPOLAR Molecules:
 _________________ molecules
 __________ sharing of _________________ or
 DOES pass the “mirror test”
 CAN be folded to reflect itself
 2 Atoms  same element/electronegativities
 More than 2 atoms  ___ unbonded e- or ________ pairs
around the ____________ atom
Ex: Cl2
Ex: CO2
Ex: CCl4
BEWARE! There are often POLAR BONDS inside NONPOLAR MOLECULES
(look back at the previous 2 examples)
C. Intermolecular forces, A.K.A. IMF’s
 ONLY IN COVALENT MOLECULES, NEVER IONIC COMPOUNDS!
 _______ forces that act BETWEEN _______________ that hold
molecules to EACH OTHER
 Only exist in ____________ & _________ states
 Called WEAK forces because they are much weaker than CHEMICAL
BONDS
*REMEMBER: IMF’s occur BETWEEN molecules, whereas BONDING
occurs WITHIN molecules
http://www.northland.cc.mn.us/biology/Biology1111/animations/hydrogenbonds.html
Just Remember…
IMF’s ARE
NOT
BONDS!!!
Type of IMF
London dispersion
forces
(LDF’s)
Description/Example(s)
 Weakest of all the IMF’s
 Only important for NONPOLAR molecules
 _________________________________________
 Electron-electron repulsion creates BRIEF DIPOLES in
atoms/molecules
http://antoine.frostburg.edu/chem/senese/101/liquids/faq/h-bonding-vs-london-forces.shtml
Dipole
(dipole-dipole)
 Molecules such as HCl have both POSITIVE and a
NEGATIVE ends, or POLES
 Two poles = __________
 Results from an UNEQUAL/ASYMMETRICAL sharing of
electrons
 DIPOLE-DIPOLE = two molecules with permanent dipoles
are attracted to one another _______________
 DIPOLE MOMENT = measure of the ___________ of the
dipole within a molecule (POLARITY)
 The GREATER the difference in ELECTRONEGATIVITY
between atoms, the GREATER the POLARITY/DIPOLE
MOMENT
 The HIGHER the dipole moment, the STRONGER the
intermolecular forces (IMF’s)
 The stronger the IMF’s, the higher the m.p. and b.p.
http://chemmovies.unl.edu/ChemAnime/DIPOLED/DIPOLED.html
Hydrogen Bonds
 Specific type of ________ interaction
 In a POLAR BOND, hydrogen is basically reduced to a BARE
PROTON w/ almost no ATOMIC RADIUS
 __________________ of all IMF’s by far
 Only occur in molecules containing Hydrogen AND
____________, _____________, or _____________
http://programs.northlandcollege.edu/biology/Biology1111/animations/hydrogenbonds.html
IX. Lewis Structures & the Octet Rule
Lewis Structures: Comments About the Octet Rule

The second-row elements C, N, O, and F should always be assumed to obey the octet rule

The second-row elements B and Be often have fewer than eight electrons around them in their
compounds. These electron-deficient compounds are very reactive.

The second-row elements never exceed the octet rule, since their valence orbitals (2s and 2p)
can accommodate only eight electrons.

Third-row and heavier elements often satisfy the octet rule but can exceed the octet rule by
using their empty valence d orbitals.

When writing the Lewis structure for a molecule, satisfy the octet rule for the atoms first. If
electrons remain after the octet rule has been satisfied, then place them on the elements
having available d orbitals (elements in Period 3 or beyond).
Rules Governing Formal Charge

To calculate the formal charge on an atom:
1.
Take the sum of the lone pair electrons and ½ the shared electrons. This is the number of
valence electrons assigned to the atom in the molecule.
2. Subtract the number of assigned electrons from the number of valence electrons on the
free neutral atom to obtain the formal charge.

The sum of the formal charges of all atoms in a given molecule or ion must equal the overall
charge on that species.

If nonequivalent Lewis structures exist for a species, those with formal charges closest to zero
and with any negative formal charges on the most electronegative atoms are considered to best
describe the bonding in the molecule or ion.
*Delocalized bonding: bond pairs that appear to move between two or more different pairs of atoms; occurs in
molecules with resonance structures.
X. The Localized Electron Model
VSEPR Model (Valence Shell Electron Pair Repulsion)
 Used to predict shapes of molecules
Steps to Apply the VSEPR Model
1. Draw the Lewis structure for the model.
2. Count the electron pairs and arrange them in the way that minimizes
repulsion (that is, put the pairs as far apart as possible).
3. Determine the positions of the atoms from the way the electron
pairs are shared.
4. Determine the name of the molecular structure from the positions
of the atoms.
Summary of the VSEPR Model
The rules for using the VSEPR model to predict molecular structure:
1. Determine the Lewis structure(s) for the molecule.
2. For molecules with resonance structures, use any of the structures
to predict the molecular structure.
3. Sum the electron pairs around the central atom.
4. In counting pairs, count each multiple bond as a single effective pair.
5. The arrangement of the pairs is determined by minimizing electronpair repulsions. These arrangements are shown in Table 8.6 (page
394) in your textbook.
6. Lone pairs require more space than bonding pairs do. Choose an
arrangement that gives the lone pairs as much room as possible.
Recognize that the lone pairs may produce a slight distortion of the
structure at angles less than 120 degrees.
Chapter 9.1: Hybridization & the Localized Electron Model
I.
sp3 Hybridization: Consider the molecule methane (CH4)
(Assume valence e- only in bonding)
H = 1s
C = 2s & 2p orbitals
Problems:
1. CH4 is supposed to have 4 identical C-H bonds
2. Three 2p orbitals “want” to be perpendicular, but CH4 is tetrahedral
Solution: The bonds in the 2s and 2p orbitals “mix” to form 4 equivalent
orbitals  HYBRIDIZATION
Combine one s and three p orbitals  four sp3 hybrid orbitals
*All 4 orbitals are identical in shape (1 lg. lobe, 1 sm. lobe)
Principle: Whenever an atom is surrounded by 4 effective pairs, a
tetrahedral set of sp3 hybrid orbitals is required (the atom
becomes sp3 hybridized).
*The atom achieves minimum energy by hybridizing:
Sample Exercise 9.1: The Localized Electron Model I
Describe the bonding in the ammonia (NH3) molecule using the localized
electron model.
Solution:
1. Draw the Lewis structure
2. Determine VSEPR shape
3. Determine hybridization
 4 e- pairs = tetrahedral arrangement = sp3 hybridization
 Three sp3 orbitals are bonding
 One sp3 orbital is lone pair
PROBLEMS: p.441 #11
II. sp2 Hybridization: Consider the molecule ethylene (C2H4)
1. Draw the Lewis structure
2. Determine VSEPR shape
3. Determine hybridization
*Double bond acts like 1 effective pair  C2H4 has 3 effective pairs
around each C atom  trigonal planar arrangements (120º)
Principle: Whenever an atom is surrounded by 3 effective pairs,
a trigonal planar set of sp2 hybrid orbitals is required.
1. combine one s and two p orbitals  sp2 hybridization
2. remaining p orbital on each carbon exists in plane perpendicular to
sp2 orbitals
3. sp2 orbitals form a sigma (σ) bond
4. one remaining 2p orbital (not hybridized) forms a pi (π) bond by
sharing e- in space above and below the σ bond
Solution:
 3 e- pairs = trigonal planar arrangement = sp2 hybridization
 Three sp2 orbitals form sigma (σ) bonds
 One remaining p orbital forms pi a (π) bond
Click to view animation: http://www.youtube.com/watch?v=ZsfCPPl13So&feature=related&safe=active
Click to view animation: http://www.youtube.com/watch?NR=1&v=C2W-yDPcpl4&feature=endscreen&safe=active
III. sp Hybridization: Consider the molecule carbon dioxide (CO2)
Principle: Whenever an atom is surrounded by 2 effective pairs, a linear set
of sp hybrid orbitals is required.
1. Combine one s and one p orbital  two sp hybrid orbitals
2. One s orbital & one p orbital hybridize, forming a set of two sp
orbitals
3. Remaining two p orbitals will form π bonds
Sample Exercise 9.2: The Localized Electron Model II
Describe the bonding in the N2 molecule.
Solution:
 Each N is surrounded by 2 effective e- pairs (linear)
 2 effective e- pairs = sp hybridization
 Each N has two sp orbitals, one bonding (σ), one lone
(σ bond is “buried” in the diagram below)
 2 remaining p orbitals form each form π bonds
PROBLEMS: p.441 #13
IV. dsp3 Hybridization: Consider PCl5, which exceeds the octet rule
Principle: Whenever an atom is surrounded by 5 effective pairs, a set of
dsp3 hybrid orbitals is required in a trigonal bipyramind
arrangement.
4. One d orbital, one s orbital & three p orbitals hybridize,
forming a set of five dsp3 orbitals
In this case, each Cl atom is surrounded by 4 electron pairs
 each Cl requires a tetrahedral arrangement
 tetrahedral arrangement requires four sp3 orbitals
 each sp3 orbital on Cl shares electrons with a dsp3 orbital on P to
form σ bonds
 the remaining sp3 orbitals on Cl hold lone pairs
Sample Exercise 9.3: The Localized Electron Model III
Describe the bonding in
the triiodide ion (I3-).
Solution:
 The central iodine atom has five pairs of electrons
 A set of five electron pairs requires a trigonal bipyramidal
arrangement, which requires a set of dsp3 orbitals
 The outer atoms have four pairs of electrons, which class for a
tetrahedral arrangement and sp3 hybridization
 The central iodine is dsp3 hybridized. Three of these hybrid orbitals
hold lone pairs, and two of them overlap with sp3 orbitals of the
other two iodine atoms to form σ bonds
PROBLEMS: p.442 #21
V.
d2sp3 Hybridization: Consider sulfur hexafluoride (SF6), which has six
pairs of electrons around its central atom
Principle: Whenever an atom is surrounded by
6 effective pairs, they require a set of d2sp3 hybrid orbitals
arranged octahedrally.
 Each of the d2sp3 orbitals on the sulfur atom is used to bond to a
fluorine atom
 Each of the fluorine atoms has four electron pairs and they are
assumed to be sp3 hybridized
Sample Exercise 9.4: The Localized Electron Model IV
How is the xenon atom in XeF4 hybridized?
Solution:
 The Xe atom has six pairs of electrons around it, arranged
octahedrally to minimize repulsions
 An octahedral set of six electron pairs requires that Xe become
d2sp3 hybridized
 Four of the d2sp3 orbitals form σ bonds with fluorine atoms
 The remaining two d2sp3 orbitals hold lone pairs
PROBLEMS: p.442 #20
XI. Molecular Orbitals
A. Molecular Orbital Theory –
B. Bonding Molecular Orbital – The Hydrogen Molecule
C. Bond Order –
Sample EX:What is the bond order of the He2+ ion? Would you expect
this atom to be stable relative to the separated He atom and the He+
ion?
D. Second Period Diatomic Molecules
1. Paramagnetic –
2. Diamagnetic –