Bonding and
Molecular Structure
Goals:
1. Understand the differences between ionic and
covalent bonds.
2. Draw Lewis electron dot structures for small
molecules and ions.
3. Use the valance shell electron-pair repulsion
theory (VSEPR) to predict the shapes of simple
molecules and ions and to understand the
structures of more complex molecules.
4. Use electronegativity to predict the charge
distribution in molecules and ions and to define
the polarity of bonds.
5. Predict the polarity of molecules.
Chemical Bonding
Problems and questions —
How is a molecule or polyatomic ion held together?
Why are atoms distributed at strange angles?
Why are molecules not flat?
Can we predict the structure?
How is structure related to chemical and physical
properties?
There are 2 extreme forms of connecting or
bonding atoms:
Ionic—complete transfer of ________________from one
atom to another.
Covalent—some valence electrons _________ between
atoms
Most bonds are somewhere in between!
Chemical Bonding
• Wherever two atoms or ions are
strongly attached to each other :
chemical bond between them.
– Ionic, covalent, metallic.
• Ionic bond: electrostatic force between
ions of opposite charge.
• Covalent bond: sharing of electrons
between two atoms.
Lewis Symbols
• Electrons involved in chemical
bonding are the _______________
(those in the outermost occupied
shell of an atom).
• Lewis electron-dot symbols: simple
way of showing the valance electrons.
• Lewis symbol:
– Chemical symbol for the element + a dot
for each valance electron.
– Dots are placed on the four sides of the
atomic symbol. Each side can
accommodate up to two electrons.
– The number of valance electrons in any
representative element is the same as
the group number of the element.
Sulfur:
[Ne]3s23p4
•
••
S•
••
The Octet Rule
• Atoms tend to gain, lose, or share electrons
until they are surrounded by eight valance
electrons.
• They have the same number of electrons as
the ________ closest to them in the period
table – stable electron arrangements (high
ionization energies, low affinity for additional
e-s, general lack of chemical reactivity.
• An octet of electrons consists of full s and p
subshells in an atom:
• ns2np6
• There are some exceptions…
Ionic Bonding
Na (s) + 1/2 Cl2(g) ---> NaCl (s)
Metal of low IE
Nonmetal
of high EA
DHfo = -410.9 kJ
Ionic Bonding
Na (s) + 1/2 Cl2(g) ---> NaCl (s)
DHfo = -410.9 kJ
• An electron has been lost by a sodium atom and
gained by a chlorine atom – electron transfer from
the Na atom to the Cl atom.
• NaCl is a typical ionic compound: it consists of a metal
of low ionization energy and a nonmetal of high
electron affinity.
• Lewis electron-dot representation:
Energetics of Ionic Bond
Formation
• The principal reason that ionic compounds are stable is
the attraction between ions of unlike charge.
• This attraction draws the ions together, releasing energy
and causing the ions to form a solid array – lattice.
• A measure of how much stabilization results from
arranging oppositely charged ions in an ionic solid:
• Lattice energy – is the energy required to completely
separate a mole of a solid ionic compound into its gaseous
ions.
• NaCl  Na+ (g) + Cl- (g) DHlattice = + 788 kJ/mol
• Highly endotermic process – the reverse process – the
coming together is highly exothermic.
• Eel = kQ1Q2 / d For a given arrangement of ions, the
lattice energy increases as the charges on the ions
increase and their radii decrease.
Arrange the following ionic compounds in order
of increasing lattice energy: NaF, CsI, and CaO.
Covalent Bonding
The bond arises from the mutual attraction of 2
nuclei for the same electrons.
Electron sharing results.
HA + HB
HA
HB
Bond is a balance of attractive and repulsive forces.
Bond Formation
• A bond can result from a “head-to-head”
overlap
atoms.
of atomic orbitals on neighboring
Overlap of H (1s) and Cl (2p)
Note that each atom has a single, unpaired electron.
Electron Distribution in
Molecules
• Electron distribution is depicted
with Lewis
electron dot
structures
•
Valence electrons are distributed
as shared or ______ PAIRS and
unshared or ______ PAIRS.
••
G. N. Lewis
1875 - 1946
H
Cl
••
shared or
bond pair
•
•
lone pair (LP)
Valance Electrons
Electrons are divided between core and valence
electrons:
Valance e-: e- in the outermost shell
B 1s2 2s2 2p1
Core = [He] , valence = 2s2 2p1
Br [Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 , valence = 4s2 4p5
The Octet Rule
• Each atom (except H) has a share in four pairs of
electrons, so each has achieved a noble gas configuration.
• Each atom is surrounded by an octet of e-s.
• Octet Rule: The tendency of molecules and polyatomic
ions to ______________________________________
_______________.
• No. of valence electrons of an atom = group number
• For Groups 1A-4A, no. of bond pairs = group number
• For Groups 5A -7A, BP’s = 8 – group number
• Except for H (and sometimes atoms of 3rd and
higher periods),
BP’s + LP’s = 4
Building an Electron Dot
Structure
Ammonia, NH3
1. Decide on the central atom: Central atom is atom of
lowest affinity for electrons; never H.
Therefore, N is central
2. Count total valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
The trick is to count the e-s!
Students should be familiar with writing
Lewis structures.
Ammonia, NH3
3. Form a single bond between the central
atom and each surrounding atom. Each
line represent 1 pair of shared e-.
4. Remaining electrons form LONE PAIRS
to complete octet as needed.
N: 3 BOND PAIRS and 1 LONE PAIR.
H N H
H
••
H N H
H
Note that N has a share in 4 pairs
(8 electrons), while H shares 1 pair.
Practice with the Lab Experiment # 11
Sulfite ion, SO32Step 1. Central atom
=S
Step 2. Count total valence electrons
Notice charge in ion: add or substract eaccording to charge!
Negative charge = 2 e- more
Sulfite ion, SO32Step 3. Form bonds
Step 4. Assign as lone pairs to outside
atoms (never to H), if some left, then
to the center atom.
Step 5. Check all atoms have an octet.
If the central atom as fewer than eight
e-, move one or more of the lone pairs
on the terminal atoms and form
multiple bonds. – See next example.
Carbon Dioxide, CO2
Step 1. Central atom
=C
Step 2. Count total valence electrons
Step 3. Form bonds
Step 4. Assign as lone pairs to outside
atoms (never to H), if some left, then
to the center atom.
Step 5. Check all atoms have an octet.
Carbon Dioxide, CO2
Step 5. Check all atoms have an octet. Form multiple
bonds if needed.
So that C has an octet, we shall form DOUBLE
BONDS between C and O.
The second bonding pair forms a
pi (π) bond.
Sulfur Dioxide, SO2
Step 1. Central atom
=S
Step 2. Count total valence electrons
Step 3. Form bonds
Step 4. Assign as lone pairs to outside
atoms (never to H), if some left, then
to the center atom.
Step 5. Check all atoms have an octet.
Sulfur Dioxide, SO2
Step 5. Check all atoms have an octet. Form double bond so
that S has an octet — but note that there are two ways
of doing this.
This leads to the following structures.
These equivalent structures are called
RESONANCE STRUCTURES. The true
electronic structure is a HYBRID of the two.
Formal Atom Charges
• Atoms in molecules often bear a charge (+ or -).
• The predominant resonance structure of a
molecule is the one with charges as close to 0
as possible.
• Formal charge
= Number of valance e- (group number)
– 1/2 (no. of bonding electrons)
- (no. of LP electrons)
Formal Atom Charges
Formal charge = Valance e- – 1/2 (BP e-) – LP eFor C,
= 4 – ½ (8) – 0 = 0
For O,
= 6 – ½ (4) – 4 = 0
Yellow = negative & red = positive
Relative size = relative charge
Formal Atom Charges
Formal charge = Group number – 1/2 (BP e-) – LP eThiocyanate Ion, SCNFor S,
6 - ½ (2) - 6 = -1
For C,
= 4 – ½ (8) – 0 = 0
For N,
5 - ½ (6) - 2 = 0
•
•
••
S
C
••
••
•
•
S
N
••
••
C
•
•
N
••
•
•
S
C
•
•
N
••
Which is the most important resonance form?
All atoms are negative, but most negative is S.
Violations of the Octet Rule
Usually occurs with B and elements of
higher periods.
BF3
SF4
Boron Trifluoride, BF3
The B atom has a share
in only 6 pairs of
electrons (or 3 pairs). B
atom in many molecules is
electron deficient.
Calc’d partial charges in BF3
F is negative and B is
positive
Sulfur Tetrafluoride, SF4
5 pairs around the S
atom. A common
occurrence outside the
2nd period.
Which of the following species will have a Lewis
structure most like that of a sulfate ion, SO42-? Assume
that the Lewis structure has no double bonds.
a)
b)
c)
d)
NH3
CBr4
SO3
H2CO
Which of the following species are isoelectronic
to SO4-2?
CO32- ; PO4-3 ; NH4+
Free Radicals
• Free radicals – atomic or molecular specie with ______
____________.
• Free radicals are generally _____________.
• H, Cl
• O2 - Oxygen itself is a free radical, but one with two unpaired
electrons; reduction by adding one electron to give superoxide
involves the pairing of two of the electron spins, leaving one unpaired
O2 + 1e-  O2·• NO – vascular system (11 valance electrons – odd number)
• Vitamin C, Vitamin E – Antioxidants
– They produce very stable radicals (stabilized by resonance
structures)
Vit E
MOLECULAR GEOMETRY
VSEPR
• Valence Shell Electron Pair Repulsion
theory.
• Most important factor in determining
geometry is ________________________
electron pairs.
• Molecule adopts the shape
that ___________ the
electron pair repulsions.
Exp 11: Molecular Geometry
• I suggest you practice with:
3. SO2
13. CS2
29. NO2+
7. NO322. CH336. PO339. NH4+
27. H3O+
43. H2O
10.CH2Cl2
28. O3
46. CH3+
48. HCN
1. Give the total number of valance electrons.
2. Draw the correct Lewis dot structure.
3. Give the geometry using VSEPR theory.
4. Give the polarity of the MOLECULE or ION.
Electron Pair Geometries
Structure Determination by
VSEPR
Ammonia, NH3
1. Draw electron dot structure H
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the
corners of a tetrahedron.
••
N H
H
lone pair of electrons
in tetrahedral position
N
H
H
H
The ELECTRON PAIR GEOMETRY is tetrahedral.
Structure Determination by
VSEPR
Ammonia, NH3
The electron pair geometry is
tetrahedral.
lone pair of electrons
in tetrahedral position
N
H
H
H
The MOLECULAR GEOMETRY — the positions
of the atoms — is PYRAMIDAL.
Structure Determination by
VSEPR
Water, H2O
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the
corners of a tetrahedron.
The electron pair
geometry is
TETRAHEDRAL.
Structure Determination by
VSEPR
Water, H2O
The molecular
geometry is
BENT.
The electron
pair geometry
is tetrahedral.
Geometries for
Four Electron Pairs
Students should be familiar with
predicting molecular geometries.
The Hydronium Ion
H3O+
Draw the Lewis dot structure and predict the
geometry.
Molecular Geometries for Five
Electron Pairs
All based on trigonal
bipyramid.
Sulfur Tetrafluoride, SF4
Lone pair is in the
equator because _____
__________________.
90
••
o
S
F
F
••
S
•• F ••
••
F
F
••
•F
•
••
•• ••
F
••
120
o
••
F
••
••
Molecular Geometries for Six
Electron Pairs
All are based on the 8sided octahedron
Sulfur Hexafluoride, SF6
90
o
F
F
S
F
Octahedron
F
F
F
6 electron pairs
90
o
Electronegativity
• Concept proposed by Linus Pauling (1901-1994).
The only person to receive two unshared Nobel prizes
(for Peace and Chemistry).
Chemistry areas: bonding, electronegativity, protein
structure.
Electronegativity
•
•
•
•
•
•

Is the ability of _________________to
____________________.
It is related to its ionization energy and
electron affinity (properties of isolated
atoms).
Electronegativity values are unitless.
____________, the most electronegative,
has a value of 4.0
__________, the least electronegative has a
value of 0.7.
The values for all other elements lie between
these two extremes.
Electronegativity
• Used for atoms in molecules.
• It is related to its ionization energy and electron affinity
(properties of isolated atoms).
Bond Polarity
HCl is ______ because
it has a positive end and
a negative end.
+d
-d
••
••
H Cl
••
Cl has a greater share in
bonding electrons than
does H.
Cl has slight negative charge (-d) and H has
slight positive charge (+ d)
Bond Polarity
+d
-d
••
••
H Cl
••
Due to the bond polarity, the H—Cl
bond energy is __________ than
expected for a “pure” covalent bond.
BOND
“pure” bond
real bond
ENERGY
339 kJ/mol calc’d
432 kJ/mol measured
Difference = 92 kJ. This difference is
proportional to the difference in
ELECTRONEGATIVITY, .
Bond Polarity
• Three molecules with polar,
covalent bonds.
• Each bond has one atom with a
slight negative charge (-d) and
and another with a slight
positive charge (+ d)
• Relative values of  determine
BOND POLARITY (and point of
attack on a molecule).
• Dipole moment – _________
______________________
______________________
______________________.
• They are added as VECTORS (in
math)
Which bond is more polar?
O—H
O—F

3.5 - 2.1
3.5 - 4.0
D
____
_____
______ is more polar than ____
-d
O
+d
H
+d
O
and polarity is “reversed.”
-d
F
Net Dipole Moment
Molecules—such as HI and H2O— can
be POLAR (or dipolar).
They have a DIPOLE MOMENT. The polar HCl
molecule will turn to align with an electric field.
Net Dipole Moment
Compare CO2 and H2O. Which one is polar?
The magnitude of the dipole is given in Debye units. Named for
Peter Debye (1884 - 1966). Rec’d 1936 Nobel prize for work
on x-ray diffraction and dipole moments.
Charge distribution
Molecular Polarity
Molecules will be polar if
a) bonds are polar
AND
b) the molecule is NOT “symmetric”
Students should be familiar with
identifying polar vs. nonpolar molecules.
Consider AB3 molecules: BF3, Cl2CO,
and NH3. Find the polarity.
Molecular Polarity
C-H bond is polar.
Methane is symmetrical
and is NOT polar.
C—F bond is very polar.
Molecule is not symmetrical
and so, it is polar.
Use VSEPR theory to predict the molecular
geometry of HCN, is it polar, or nonpolar?
Bond Properties
• What is the effect of bonding and structure on
molecular properties?
Free rotation
around C–C
single bond
No rotation
around C=C
double bond
• Bond Order – Number of bonding e- pairs between a pair
of atoms (1, 2, 3… fractional bond orders).
– Bond Length
– Bond Strength
Bond Order
______ bond
Acrylonitrile
Bond order =
______ bond
______ bond
Total # of
e - pairs used for a type of bond
Total # of X - Y links
Fractional bond orders occur in molecules with
_______________ structures (book example: O3).
Nitrite ion, NO2••
••
N
N
The N—O bond
•• • •••
••
••
O
O• • O
O
order = _____
••
••
••
••
Bond Order
Bond order is proportional to two important bond
properties:
(a) bond length
(b) bond strength
414 kJ
123 pm
110 pm
745 kJ
Bond Length
• Bond length is
the distance
between _____
____________
___________.
Bond Length
Bond length depends on
size of ___________.
Bond length depends on
_____________.
H—F
H—Cl
H—I
Bond distances measured
in Angstrom units where
1 A = 10-2 pm.
Bond Strength
• —measured by the energy req’d to break a
bond. See Tables 9.9 and 9.10
BOND
H-H
C-C
C=C
CC
NN
STRENGTH
(kJ/mol)
436
346
602
835
945
The GREATER the number of bonds (bond order) the
______ the bond strength and the ________ the bond.
Bond Strength
Bond
Order
Length
Strength
HO—OH
1
142 pm
O=O
2
121
498
1.5
128
?
••
O
••O
••
•• •
O•
••
210 kJ/mol
Using Bond Energies
Estimate the energy of the reaction
H—H(g) + Cl—Cl(g) ----> 2 H—Cl(g)
Net energy = ∆Hrxn =
= energy required to break bonds
- energy evolved when bonds are made
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Using Bond Energies
Estimate the energy of the reaction
H—H = 436 kJ/mol
H—H + Cl—Cl ----> 2 H—Cl
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Remember
• Go over all the contents of your
textbook.
• Practice with examples and with
problems at the end of the chapter.
• Practice with OWL tutor.
• Work on your assignment for Chapter 9.
• Practice with the quiz on CD of
Chemistry Now.