Chapter 10
The Shapes of Molecules
The Shapes of Molecules
10.1 Depicting Molecules and Ions with Lewis Structures
10.2 Valence-Shell Electron-Pair Repulsion (VSEPR)
Theory and Molecular Shape
10.3 Molecular Shape and Molecular Polarity
1
Lewis Electron - Dot Symbols
For main group elements The A group number gives the number of valence electrons.
Place one dot per valence electron on each of the four sides
of the element symbol.
Pair the dots (electrons) until all of the valence electrons are used.
Example:
Nitrogen, N, is in Group 5A and therefore has 5 valence
electrons.
. N. .
.
. N.
:
.
. N:
.
:
.
:N .
.
The steps in converting a molecular formula into a Lewis structure.
Molecular
formula
Step 1
Atom
placement
Lewis structure – The nonmetal
atoms can acquire stable noblegas structure by sharing electrons
to form an electron-pair bond.
Place atom
with lowest
EN in center
Step 2
Add A-group
numbers
Sum of
valence e-
Step 3
Remaining
valence e-
Octet rule – the nonmetals achieve a noble-gas
structure by sharing eight electrons. This
generalization is referred to as Octet rule.
Draw single bonds. Subtract 2efor each bond.
Step 4
Give each atom
8e(2e for H)
Lewis
structure
Duet rule – Hydrogen atoms are surrounded by a duet
of electrons in molecules.
2
Molecular
formula
For NF3
:
: F:
: F:
:
Sum of
valence e-
:
Atom
placement
N
: F:
5e-
F 7e-
X 3 = 21e-
Total
26 e-
:
Remaining
valence e-
N
Lewis
structure
Single bond – a single electron pair is shared between two bonded atoms.
Double bond – two electron pairs are shared between two bonded atoms.
Triple bond – three electron pairs are shared between two bonded atoms.
SAMPLE PROBLEM
Writing Lewis Structures for Molecules with One Central Atom
PROBLEM:
Write a Lewis structure for CCl2F2, one of the compounds responsible
for the depletion of stratospheric ozone.
PLAN: Follow the steps outlined.
SOLUTION:
F
:
: Cl :
F:
:
:
: Cl C
: F:
:
Make bonds and fill in remaining valence electrons
placing 8e- around each atom.
F
:
Steps 2-4:
C has 4 valence e-, Cl and F each have 7. The sum is
4 + 4(7) = 32 valence e-.
Cl C
:
Step 1: Carbon has the lowest EN and is the central atom.
The other atoms are placed around it.
Cl
3
SAMPLE PROBLEM
Writing Lewis Structure for Molecules with More than One Central Atom
PROBLEM: Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
SOLUTION:
Hydrogen can have only one bond so C and O must be next
to each other with H filling in the bonds.
There are 4(1) + 4 + 6 = 14 valence e-.
C has 4 bonds and O has 2. O has 2 pair of nonbonding e-.
:
H
C
O
H
:
H
H
Resonance: Delocalized Electron-Pair Bonding
O3 can be drawn in 2 ways -
O
O
O
O
O
O
Neither structure is actually correct but can be drawn to represent a structure
which is a hybrid of the two - a resonance structure.
B
B
O
O
O
O
A
C
O
O
O
O
O
A
C
Resonance structures have the same relative atom placement but a difference
in the locations of bonding and nonbonding electron pairs.
is used to indicate that resonance occurs.
4
Resonance (continued)
Three criteria for choosing the more important resonance structure:
Smaller formal charges (either positive or negative) are preferable to
larger charges.
Avoid like charges (+ + or - - ) on adjacent atoms.
A more negative formal charge should exist on an atom with a larger EN
value.
SAMPLE PROBLEM
PROBLEM:
PLAN:
Write resonance structures for the nitrate ion, NO3-.
After Steps 1-4, go to 5 and then see if other structures can be drawn
in which the electrons can be delocalized over more than two atoms.
SOLUTION:
O
Writing Resonance Structures
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
N
N
N
O
O
O
O
O
N does not have an
octet; a pair of e- will
move in to form a
double bond.
O
O
O
O
N
N
N
O
O
O
O
O
5
Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge is the charge an atom would have if the bonding electrons were shared equally.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
For OC
O
For OA
# valence e- = 6
O
# nonbonding e- = 4
A
# bonding e- = 4 × 1/2 = 2
# valence e- = 6
O
# nonbonding e- = 6
C
For OB
# bonding e- = 2 X 1/2 = 1
# valence e- = 6
Formal charge = -1
Formal charge = 0
# nonbonding
e-
=2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
SAMPLE PROBLEM
Writing Lewis Structures for Octet Rule Exceptions
PROBLEM:
PLAN:
Write Lewis structures for (a) H3PO4 (pick the most likely
structure); (b) BFCl2.
Draw the Lewis structures for the molecule and determine if there is an
element which can be an exception to the octet rule. Note that (a) contains
P which is a Period-3 element and can have an expanded valence shell.
SOLUTION:
(a) H3PO4 has two resonance forms and formal charges indicate the
more important form.
-1
0
O
0 H O
P
O
0
H
0
+1
O H 0
0
0
0 H O
0
O
0
P
O H 0
0
0 O
H more stable
0
lower formal charges
(b) BFCl2 will have only
1 Lewis structure.
F
B
Cl
Cl
6
The orientation of polar molecules in an electric field.
Negative end
of HF
Polarity –is called bond polarity
or chemical polarity.
Polarity describes how equally
bonding electrons are shared
between atoms.
Polar – is a result of an
unsymmetrical electron
distribution. (HF, H2O)
Dipole – is the bond or molecule
which contains positive or
negative pole.
Positive pole
of the
electric field
Electric field
ON
Electric field
OFF
Nonpolar – is a symmetrical
distribution of electrons which
leads to the a bond or molecule
without positive or negative
pole.(CO2, BF3)
If the electric field is off, the polar molecule, HF, orientated
randomly.
If the electric field is on, the polar molecule, HF, aligns their
positive ends toward negative poles of the electric field.
SAMPLE PROBLEM
Predicting the Polarity of Molecules
PROBLEM:
From electronegativity (EN) values and their periodic trends, predict
whether each of the following molecules is polar and show the
direction of bond dipoles and the overall molecular dipole when
applicable:
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
PLAN: Draw the shape, find the EN values and combine the concepts to determine
the polarity.
SOLUTION:
The dipoles reinforce each
other, so the overall
molecule is definitely
polar.
(a) NH3
ENN = 3.0
H
ENH = 2.1
N
H
H
H
N
H
H
bond dipoles
H
N
H
H
molecular
dipole
7
SAMPLE PROBLEM
Predicting the Polarity of Molecules
continued
(b) BF3 has 24 valence e- and all electrons around the B will be involved in
bonds. The shape is AX3, trigonal planar.
F
B
F
F
1200
F (EN 4.0) is more
electronegative than B
(EN 2.0) and all of the
dipoles will be
directed from B to F.
Because all are at the
same angle and of the
same magnitude, the
molecule is nonpolar.
The Central Themes of VB Theory
Basic Principle
A covalent bond forms when the orbitals of two atoms overlap and the overlap
region, which is between the nuclei, is occupied by a pair of electrons.
The two wave functions are in phase so the amplitude increases between the
nuclei.
Themes
A set of overlapping orbitals has a maximum of two electrons that must have
opposite spins.
The greater the orbital overlap, the stronger (more stable) the bond.
The valence atomic orbitals in a molecule are different from those in isolated atoms.
There is a hybridization of atomic orbitals to form molecular orbitals.
8
Hybrid Orbitals
Key Points
The number of hybrid orbitals obtained equals the number of
atomic orbitals mixed.
The type of hybrid orbitals obtained varies with the types of
atomic orbitals mixed.
Types of Hybrid Orbitals
sp
sp2
sp3
sp3d
sp3d2
VSEPR
VSEPR--Valence
ValenceShell
ShellElectron
ElectronPair
PairRepulsion
RepulsionTheory
Theory
Each group of valence electrons around a central atom is located as far
away as possible from the others in order to maximize repulsions.
These repulsions maximize the space that each object attached to the
central atom occupies.
The result is five electron-group arrangements of minimum energy seen
in a large majority of molecules and polyatomic ions.
The electron-groups are defining the object arrangement,but the
molecular shape is defined by the relative positions of the atomic nuclei.
Because valence electrons can be bonding or nonbonding, the same
electron-group arrangement can give rise to different molecular shapes.
A - central atom
X -surrounding atom
E -nonbonding valence electron-group
AXmEn
integers
9
Figure 10.2
Electron-group repulsions and the five basic molecular shapes.
linear
tetrahedral
trigonal planar
trigonal bipyramidal
octahedral
The sp hybrid orbitals in gaseous BeCl2.
atomic
orbitals
hybrid
orbitals
orbital box diagrams
10
The sp2 hybrid orbitals in BF3.
The sp3 hybrid orbitals in CH4.
11
The sp3 hybrid orbitals in NH3.
The sp3 hybrid orbitals in H2O.
12
The sp3d hybrid orbitals in PCl5.
The sp3d2 hybrid orbitals in SF6.
13
SAMPLE PROBLEM
PROBLEM:
Postulating Hybrid Orbitals in a Molecule
Use partial orbital diagrams to describe mixing of the atomic orbitals
of the central atom leads to hybrid orbitals in each of the following:
(a) Methanol, CH3OH
PLAN:
(b) Sulfur tetrafluoride, SF4
Use the Lewis structures to ascertain the arrangement of groups and
shape of each molecule. Postulate the hybrid orbitals. Use partial
orbital box diagrams to indicate the hybrid for the central atoms.
SOLUTION:
H
(a) CH3OH
The groups around C are
arranged as a tetrahedron.
C O
H
H H
SAMPLE PROBLEM
O also has a tetrahedral
arrangement with 2 nonbonding epairs.
Postulating Hybrid Orbitals in a Molecule
continued
2p
2s
2p
sp3
single C atom
hybridized
C atom
2s
sp3
hybridized
O atom
single O atom
(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.
F
F S
F
F
3d
3d
3p
sp 3d
3s
S atom
hybridized
S atom
14
Lewis electron - dot symbols for elements in Periods 2 and 3.
15