Transcript Document

Chapter 9
The Shapes of Molecules
10-1
The Shapes of Molecules
Depicting Molecules and Ions with Lewis Structures
Using Lewis Structures and Bond Energies to
Calculate Heats of Reaction
Valence-Shell Electron-Pair Repulsion (VSEPR)
Theory and Molecular Shape
Molecular Shape and Molecular Polarity
10-2
Rules for Lewis Structures
• 1. Attach atoms in reasonable fashion with single bonds
– Consider ABn formulas, atom with lower group number
goes in center with others attached to it. If same group
number, place atom with highest period in center.
– The atom with the lowest E.N. goes in middle
– Consider # of free valence e-, this is often the number
of bonds this atom will make. Atoms making more
bonds will be central in molecules.
• 2. Sum valence electrons (adding for neg charge and sub
for pos.)
• 3. Complete octets of peripheral atoms (not He like atoms)
• 4. Place leftover e- on central atom
• 5. If necessary use multiple bonds to fill center atom's octet
• Last Slide
10-3
Sum of
valence e-
:
: F:
: F:
:
Atom
placement
For NF3
:
Molecular
formula
N
Remaining
valence eLewis
structure
10-4
:
: F:
N
5e-
F 7e- X 3 = 21eTotal
26e-
SAMPLE PROBLEM 10.1
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
.
SOLUTION:
:
: Cl :
:Cl C
: F:
:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
10-5
F
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
:
PLAN: Follow the steps outlined in Figure 10.1
F:
:
PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
SAMPLE PROBLEM 10.2
PROBLEM:
SOLUTION:
Writing Lewis Structure for Molecules with
More than One Central Atom
Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
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
10-6
H
SAMPLE PROBLEM 10.3
Writing Lewis Structures for Molecules with
Multiple Bonds.
Write Lewis structures for the following:
(a) Ethylene (C2H4), the most important reactant in the
manufacture of polymers
(b) Nitrogen (N2), the most abundant atmospheric gas
PROBLEM:
PLAN:
For molecules with multiple bonds, there is a Step 5 which follows the
other steps in Lewis structure construction. If a central atom does not
have 8e-, an octet, then e- can be moved in to form a multiple bond.
(a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom.
SOLUTION:
H
H
:
H
C
C
H
H
H
H
C
C
H
(b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make
the octet around each N.
N
.
:
N
.
:
:
:.
.:
10-7
N
.
N
N
:
N
.
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
A
O
O
O
C
O
O
O
A
O
C
Resonance structures have the same relative atom placement but a
difference in the locations of bonding and nonbonding electron pairs.
10-8
SAMPLE PROBLEM 10.4
PROBLEM:
PLAN:
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
N
N
N
O
O
O
10-9
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
O
O
N does not have an
octet; a pair of ewill move in to form
a double bond.
O
O
O
O
N
N
N
O
O
O
O
O
Resonance (continued)
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
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.
10-10
Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
For OA
# valence
e-
O
# nonbonding
# bonding
e-
# valence e- = 6
O
=6
e-
For OC
=4
= 4 X 1/2 = 2
Formal charge = 0
A
O
# nonbonding e- = 6
C
# bonding e- = 2 X 1/2 = 1
For OB
# valence
Formal charge = -1
e-
=6
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
10-11
Resonance (continued)
EXAMPLE: NCO- has 3 possible resonance forms -
N C
O
N C
A
N C
O
B
O
C
formal charges
-2
0
N C
+1
O
-1
0
N C
0
O
0
0
N C
-1
O
Forms B and C have negative formal charges on N and O; this makes them
more important than form A.
Form C has a negative charge on O which is the more electronegative
element, therefore C contributes the most to the resonance hybrid.
10-12
Exceptions to the Octet Rule
• Electron deficient – have fewer than eight
– ex BeCl2, BF3
– may attain an octet by coordinate covalent bond
• Odd number of electrons – aka free radicals
– ex NO2
– May attain an octet by pairing with another free radical
• Expanded Octets – only on period 3 and higher
– Expanded octets form when an atom can decrease (or
maintain at 0) it’s formal charge
– ex SF6, PCl5, H2SO4, SO2, SO3, SO4
10-13
SAMPLE PROBLEM 10.5
PROBLEM:
PLAN:
Writing Lewis Structures for Exceptions to the
Octet Rule.
Write Lewis structures for (a) H3PO4 and (b) BFCl2. In (a)
decide on the most likely structure.
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
10-14
+1
0
O H 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
Bond Energies and Hrxn
• In chemical reactions, reactant bonds are broken
and new product bonds are formed. The overall
energy change of a reaction is the energy change
of this process, plus the energy associated with
changes of state.
• In gaseous reactions (no state changes)
• Hºrxn = Hºbonds broken + Hºbonds formed
10-15
Figure 10.2
Using bond energies to calculate H
0
rxn
Enthalpy, H
H0rxn = H0reactant bonds broken + H0product bonds formed
H01 = + sum of BE
H0rxn
10-16
H02 = - sum of BE
Figure 10.3
Using bond energies to calculate H
0
rxn
of methane
BOND BREAKAGE
4BE(C-H)= +1652kJ
2BE(O2)= + 996kJ
H0(bond breaking) = +2648kJ
Enthalpy,H
2[-BE(C O)]= -1598kJ
4[-BE(O-H)]= -1868kJ
H0(bond forming) = -3466kJ
H0rxn= -818kJ
10-17
BOND FORMATION
SAMPLE PROBLEM 10.6
PROBLEM:
Calculating Enthalpy Changes from Bond
Energies
Use Table 9.2 (button at right) to calculate H0rxn for the
following reaction:
CH4(g) + 3Cl2(g)
PLAN:
CHCl3(g) + 3HCl(g)
Write the Lewis structures for all reactants and products and
calculate the number of bonds broken and formed.
SOLUTION:
H
Cl
H C H
+
3
Cl
H
H C Cl
+
3 H
Cl
bonds broken
10-18
Cl
bonds formed
Cl
SAMPLE PROBLEM 10.6
Calculating Enthalpy Changes from Bond
Energies
continued
bonds broken
4 C-H
bonds formed
= 4 mol(413 kJ/mol) = 1652 kJ
3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ
3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ
1 C-H = 1 mol(-413 kJ/mol) = -413 kJ
H0bonds broken = 2381 kJ
3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ
H0bonds formed = -2711 kJ
H0reaction = H0bonds broken + H0bonds formed = 2381 kJ + (-2711 kJ) = - 330 kJ
10-19
Molecular Shapes
Lewis structures show which atoms are
connected where, and by how many bonds,
but they don't properly show 3-D shapes of
molecules.
To find the actual shape of a molecule, first
draw the Lewis structure, and then use
VSEPR Theory.
10-20
Valence Shell Electron-Pair Repulsion Theory
or VSEPR
• molecular shape is determined by the repulsions of
electron pairs
– Electron pairs around the central atom stay as far apart
as possible.
• electron pair geometry - based on number of
regions of electron density
– Consider non-bonding (lone pairs) as well as bonding
electrons.
– Electron pairs in single, double and triple bonds are
treated as single electron clouds.
• molecular geometry - based on the electron pair
geometry, this is the shape of the molecule
10-21
Figure 10.4
A balloon analogy for the mutual repulsion of electron groups.
10-22
Figure 10.5
Electron-group repulsions and the five basic molecular shapes.
10-23
Figure 10.6
The single molecular shape of the linear electron-group
arrangement.
Examples:
CS2, HCN, BeF2
10-24
Figure 10.7
The two molecular shapes of the trigonal planar electrongroup arrangement.
Class
Examples:
SO2, O3, PbCl2, SnBr2
Shape
Examples:
SO3, BF3, NO3-, CO32-
10-25
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons are
bonding electrons of the same order.
H
Effect of Double Bonds
1200
ideal
1200
O
1160
real
Sn
Cl
Cl
950
10-26
C
H
greater
electron
density
Effect of Nonbonding Pairs
lone pairs repel bonding pairs
more strongly than bonding
pairs repel each other
H
larger EN
C
H
1220
O
Figure 10.8
The three molecular shapes of the tetrahedral electrongroup arrangement.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H 2O
PF3
OF2
ClO3
SCl2
H 3 O+
10-27
Figure 10.9
10-28
Lewis structures and molecular shapes
Figure 10.10
The four molecular shapes of the trigonal bipyramidal
electron-group arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
ClF3
XeF2
BrF3
I3 -
IF2-
10-29
Figure 10.11
The three molecular shapes of the octahedral electrongroup arrangement.
SF6
IOF5
BrF5
TeF5
-
XeOF4
10-30
XeF4
ICl4-
Figure 10.12
Molecular
formula
The steps in determining a molecular shape.
Step 1
Lewis
structure
See Figure
10.1
Step 2
Electron-group
arrangement
Count all e- groups around central
atom (A)
Step 3
Bond
angles
Note lone pairs and double
bonds
Count bonding and
Step 4
nonbonding egroups separately.
Molecular
shape
(AXmEn)
10-31
SAMPLE PROBLEM 10.7
PROBLEM:
Predicting Molecular Shapes with Two, Three,
or Four Electron Groups
Draw the molecular shape and predict the bond angles (relative
to the ideal bond angles) of (a) PF3 and (b) COCl2.
SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair
The shape is based upon the tetrahedral arrangement.
F
P
F
F
P
F
F
F
<109.50
The type of shape is
AX3E
10-32
The F-P-F bond angles should be <109.50 due
to the repulsion of the nonbonding electron
pair.
The final shape is trigonal pyramidal.
SAMPLE PROBLEM 10.7
Predicting Molecular Shapes with Two, Three,
or Four Electron Groups
continued
(b) For COCl2, C has the lowest EN and will be the center atom.
There are 24 valence e-, 3 atoms attached to the center atom.
Cl
C
O
Cl
The shape for an atom with three atom
attachments and no nonbonding pairs on the
central atom is trigonal planar.
O
O
C
Cl
10-33
C does not have an octet; a pair of nonbonding
electrons will move in from the O to make a
double bond.
Cl
The Cl-C-Cl bond angle will
be less than 1200 due to
the electron density of the
C=O.
124.50
C
Cl
1110
Cl
Type AX3
SAMPLE PROBLEM 10.8
PROBLEM:
SOLUTION:
Predicting Molecular Shapes with Five or Six
Electron Groups
Determine the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
(a) SbF5 - 40 valence e-; all electrons around central
atom will be in bonding pairs; shape is AX5 - trigonal
bipyramidal.
F
F
F
F
Sb
F
F
F
Sb
F
F
F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central
atom. Shape is AX5E, square pyramidal.
F
F
F
10-34
Br
F
F
Figure 10.13
The tetrahedral
centers of ethane.
10-35
Figure 10.13
The tetrahedral
centers of ethanol.
10-36
SAMPLE PROBLEM 10.9
PROBLEM:
PLAN:
Predicting Molecular Shapes with More Than
One Central Atom
Determine the shape around each of the central atoms in
acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis
structure.
SOLUTION:
tetrahedral
H
H C
H
O
C
tetrahedral
H
C H
H
trigonal planar
O
H
C
H C
C
H
HH
>1200
H
<1200
10-37
Figure 10.14
The orientation of polar molecules
in an electric field.
Electric field OFF
10-38
Electric field ON
SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules
PROBLEM:
From electronegativity (EN) values (button) 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
(c) Carbonyl sulfide, COS (atom sequence SCO)
PLAN: Draw the shape, find the EN values and combine the concepts to
determine the polarity.
SOLUTION:
(a) NH3
The dipoles reinforce each
other, so the overall
molecule is definitely polar.
ENN = 3.0
H
ENH = 2.1
N
H
H
H
N
H
H
bond dipoles
10-39
H
N
H
H
molecular
dipole
SAMPLE PROBLEM 10.10 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.
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar(EN) so the molecule is polar overall.
S
10-40
C
O
10-41
10-42