Chapter 10
Shapes of Molecules
The Three Dimensional Reality of Molecular Shapes
The Three Dimensional Reality of Molecular Shapes
Each atom, bonding e- pair, and lone e- pair has its own position in 3D space.
Position determined by the attractive and repulsive forces that govern all matter.
Molecular Shape is Crucial to Life Processes:
Molecular Shape is Important in Nanotechnology:
Nanotechnology: “Building tiny machines at the molecular level.”
Buckyball (fullerenes) – a C60 structure discovered
in soot in 1985
Many, many uses: One use may be to “sneak” medicines
into cells or through the blood brain barrier, where some
substances (medicines) may not normally enter.
Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures
2D before 3D
Lewis dot structures consist of two parts:
(1) Element symbol – nucleus + inner electrons
Ex: The element lithium has an element symbol Li
(2) Surrounding dots – valence electrons (outer most shell)
Different elements can have the same number of dots
Be
Mg
Same Group
(Column)
Li
Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures
Electron types: Bonding Pairs & Lone Pairs
Bond types:
Single, Double, Triple
Steps for writing Lewis Structures for molecules that have only single bonds:
(1) Place the atoms relative to each other, with the atom with the lower group (column)
number in the center. Often, center atom is also atom with lower E.N.
F
Example: NF3
N
F
F
If atoms have same groups number, place atom with higher period (row) number in
the center.
O
Example: SO3
S
O
O
Continued
Steps for writing Lewis Structures for molecules that have only single bonds:
(2) Determine the total number of valence e- available
N  5 e-
Example: NF3
F  7 e- (x 3) = 21 e-
Total = 26 e-
(3) Draw a single bond from each surrounding atom to the central atom, and subtract
2 valence e-’s for each bond to find the # of e- remaining
F
Example: NF3
3 single bonds x 2 e- = 6 e-
N
F
e- remaining = total – bonded = 26 – 6 = 20 eF
(4) Distribute the remaining e- in pairs so that each atom has 8 e- (except H, has 2 e-)
1st: Place lone e- pairs on surrounding (more EN) atoms to give each full valence.
2nd: If any e- remain, place them on the central atom.
Example: NF3
(5) Check that each has a full valence shell. DONE.
Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures
Lewis Dot Structures are not 3D, so several different depictions are correct.
F
F
F
N
F
F
F
N
N
F
Etc…….
F
F
This method works for singly bonded compounds where C, N, and O, and
elements in higher periods (rows), are the central atom.
In nearly all compounds:
• H atoms form 1 bond
• C forms 4 bonds
• N forms 3 bonds
• O forms 2 bonds
• Halogens (F, Cl, Br, I,…) form 1 bond
Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures
In cases of multiple bonds (double, triple), there is an additional step:
(6) If a central atom still does not have an octet (after Step 4), make a
multiple bond by changing a lone pair from one of the surrounding
atoms into a bonding pair to the central atom.
Examples:
CH4
N2
In cases where there is a polyatomic ion, Lewis Dot Structure is shown
in square brackets:
Section 10.1: Resonance Structures
Have the same relative placement of atoms, but different locations of
bonding and lone e- pairs.
Example: Nitrite (NO2-)
Double bonds change location.
Lone pair on O atom changes location.
Neither structure depicts nitrite accurately: The two O,N bonds in this compound
have bond lengths and bond energies that lie somewhere between the O–N
and the O=N bond.
Section 10.1: Resonance Hybrids
Resonance structures are not real bonding depictions:
Structure I
Structure II
In reality, Structure I and Structure II
do not switch back and forth from one
instant to the next. The actual nitrite
molecule is an average of the two.
The e-’s are delocalized over the entire molecule.
(Just as e-’s in a metallic bond are delocalized
around the entire sea of electrons.)
Section 10.1: Resonance Hybrids
Sometimes implied (no indication).
Sometimes indicated by dotted lines.
Example: Ozone (O3)
Example: Benzene (C6H6)
Draw a Lewis Dot Structure for:
CCl4
CCl2F2
CH4O
NH3O
C2H6O (no O–H bonds)
10.6
10.8
Resonance Structures:
10.10
10.12
Section 10.1: Which is the more important resonance structure?
All resonance forms contribute equally when central atom has surrounding atoms that
are all the same.
When atoms surrounding the central atom are not the same: One resonance form may
“weight” the average (In other words, it “counts more” than the other forms.)
Determining the most important resonance form: Determine each atom’s
formal charge – the charge it would have if the bonding electrons were equally shared,
Section 10.1: Which is the more important resonance structure?
Formal charge – Determined for each atom in a compound
# valence e- - (# unshared valance e- + ½ # shared valence e-)
Example: Ozone (O3)
Formal charge for OA = 6 – (4 – ½*4) = 0
Formal charge for OB = 6 – (2 – ½*6) = +1
Formal charge for OC = 6 – (6 – ½*2) = -1
Formal charges must sum to the total charge on the chemical species.
In the case of ozone, both resonance structures (I and II) have the same formal
charges (but on different O atoms) so they contribute equally to the resonance
hybrid.
Section 10.1: Which is the more important resonance structure?
Formal charge – Determined for each atom in a compound
Recall – In nearly all compounds:
• H atoms form 1 bond
• C forms 4 bonds
• N forms 3 bonds
• O forms 2 bonds
• Halogens (F, Cl, Br, I,…) form 1 bond
When formal charge is 0, an atom has its usual # of bonds.
Ozone Example:
Formal charge for OA = 6 – (4 – ½*4) = 0
Formal charge for OB = 6 – (2 – ½*6) = +1
Formal charge for OC = 6 – (6 – ½*2) = -1
Section 10.1: Which is the more important resonance structure?
What about a case where the resonance structures do not contribute equally?
Occurs when there are different atoms around the central atom.
Example: Cyanate ion, NCO-
Criteria for choosing the more important resonance structure:
(1) Smaller formal charges (+ or -) are preferable to larger ones.
Resonance form I is out.
(2) The same nonzero formal charges on adjacent atoms are not preferred.
Not applicable to this example.
(3) A more negative formal charge should reside on a more electronegative atom.
O is more EN than N, so Resonance form III is the winner.
Section 10.1: Formal charge is not the same as the ON
What is the difference?
Formal charge - Bonding e-’s are assigned equally to the atoms, so that each
atom has ½
Formal charge = # valence e- - (# lone pair e- + ½ # bonding e-)
Oxidation number - Bonding e-’s are assigned completely to the more EN atom
O.N. = # valence e- - (# lone pair e- + # bonding e-)
Section 10.1: Exceptions (Limitations) to the Octet and Formal Charge Rules
(1) e- deficient molecules – gaseous molecules containing Be or B as central atom
Halogens much more EN than Be or B =
Formal charge rules make sharing of
extra lone pairs by halogens unlikely
Be: 4 e-
B: 6 e-
(2) Odd e- molecules– central atom has odd # of valence e-
free radicals – very reactive (b/c very unstable): often react with each other to pair
up their lone e- (make you age)
(3) Expanded valence shells – more than 8 valence e-; occurs only where d orbitals
are available  Row 3 or higher
(Review of orbital types: p289 – 295; s – 2 e-, p – 6 e-, d – 10 e-, f – 14 e-)
Resonance Structures:
10.14a
10.16b
Other suggested problems:
10.15
10.17
10.19
10.24
Formal charge = # valence e- - (# unshared valance e- + ½ # shared valence e-)
Criteria for choosing the more important resonance structure:
(1) Smaller formal charges (+ or -) are preferable to larger ones.
Resonance form I is out.
(2) The same nonzero formal charges on adjacent atoms are not preferred.
Not applicable to this example.
(3) A more negative formal charge should reside on a more electronegative atom.
O is more EN than N, so Resonance form III is the winner.
Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory
Molecular shape is important in many, many scientific disciplines.
Medicine: Receptors
Nanotechnology:
Membrane Transport
Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory
Molecular shape is important in many, many scientific disciplines.
Ecology: Talking trees
Jack Schultz, Chemical Ecologist
Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory
Lewis Dot Structures, 2D (Blueprint)
VSEPR, 3D (House)
Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory
Each group of valence electrons around a central atom is located as far
away as possible from the others in order to minimize repulsions.
e- group can be: a single bond, double bond, triple bond, lone pair, lone e-
Section 10.3: Molecular Shape and Molecular Polarity
In molecules with more than 2 atoms:
Shape and bond polarity determine molecular polarity.
Molecules with only 2 atoms.
Relative electronegativities of the two
atoms determine polarity.
Section 10.3: Molecular Shape and Molecular Polarity
In molecules with more than 2 atoms: Dipole moments
Dipole moments – a measure of molecular polarity
magnitude of partial charges on ends of a molecule (in coulombs)
x
distance between them
Behavior of Molecules With and Without Dipole Moments
Electric field: Polar molecules (which have a dipole moment)
orient with partial charges towards oppositely charged electric plates.
Molecules with out a dipole moment will not orient themselves in any
particular direction. *Molecules with no dipole moment can be polar.
Section 10.3: Molecular Shape and Molecular Polarity
Dipole moments: When molecular shape influences polarity
Large ∆EN between C (EN = 2.5) and O (EN = 3.5)  C = O bonds are polar
CO2 molecule is linear  Two identical bond polarities are counterbalanced (in other
words, they cancel each other out)
As a result, CO2 has no net dipole moment.
Section 10.3: Molecular Shape and Molecular Polarity
Dipole moments: When molecular shape influences polarity
H2O (like CO2) also has two identical molecules bonded to the central atom.
However, H2O (unlike CO2) has a dipole moment.
Bond polarities are not canceled out because of the shape of the water molecule:
V-shaped rather than linear.
The O end of the molecule is more negative than the H ends
Section 10.3: Molecular Shape and Molecular Polarity
When different molecules have the same shape, the nature of the atoms
Surrounding the central atom can have a major effect on polarity.
CCl4 – does not have a dipole
CHCl3 – has a dipole
Effect of Molecular Polarity on Behavior
Example: Boiling point of NH3 versus PH3
Why is NH3 boiling point higher?
Also determines reaction behavior: NH3 + H+  NH4+ (p392)
A closer look at molecular shapes: double bonds and lone pairs
Bond Angles:
Idealized
vs.
Actual
A closer look at molecular shapes: double bonds
Effect of double bonds on bond angle when surrounding atoms are different.
Rule: The double bond, with its
greater e- density, repels the
two single bonds more strongly
than they repel each other.
A closer look at molecular shapes: lone pairs
Effect of lone pairs on bond angle.
Rule: Lone pairs repel bonding pairs
more strongly than bonding pairs
repel each other.
A closer look at molecular shapes: a few more details
Bond angles for:
Equatorial groups = 90º
Axial groups = 120º
General Rule: The greater the
bond angle, the weaker the
repulsion.
In this case: Equatorial-equatorial
repulsions are weaker than axialrquatorial repulsions.
Implication: Lone pairs, which
exert stronger repulsions, will tend
to occupy equatorial positions.
A closer look at molecular shapes: a few more details
General Rule: The greater the
bond angle, the weaker the
repulsion.
Implication: If two lone pairs are
present, they will always occupy
opposite vertices (furthest apart)
More VSEPR Practice
(1) Lewis dot structure (dominant resonance form – calculate formal charge if needed)
(2) 3-D geometry (Table 10.9)
(3) Molecular polarity
Bond dipoles cancel?
Lone pairs present?
Different surrounding atoms?
GeH2
PCl5
SF4
ClF3
XeF2
SF6
BrF5
XeF4
NH4+
SO4Additional Optional Homework Problems:
10.30, 10.35, 10.38, 10.39, 10.51, 10.53, 10.54, 10.97
Chemical Reaction vs. Physical Interactions (Chapter 12)
Boiling point of NH3 versus PH3: Why is NH3 boiling point higher?
The N – H bonds between NH3 molecules matter,
not the N – H bonds within the NH3 molecule.
N has greater EN than P: Why?
EN is inversely proportion to atomic radius.
Why is EN is inversely proportion to
atomic radius?
e- shielding
Summary So Far: Overall Polarity of a Molecule
If a molecule is polar, it will:
(1) have a net dipole moment
(2) orient itself in an electric field
If a molecule is nonpolar, it will:
(1) not have a net dipole moment
(2) will move about randomly in an electric field
Steps used to determine molecular polarity:
(1) Draw the 2-D Lewis dot structure to determine the
number and types (single, double, triple) of bonds
present, and any lone pairs present.
*When dealing with several options (resonance structures),
determine the dominant resonant structure by:
1. calculating formal charge for the atoms of each
molecular possibility that you are evaluating
2. using the three criteria for selecting the dominant
resonance structure based on formal charge
(2) Determine the 3-D shape of the molecule
(3) Determine the overall polarity of the molecule
In addition to the wavelength of energy interaction with the molecule
The symmetry of the molecule (Lewis dot structure + VSEPR!!!)
will also determine whether a photon can be absorbed.
*Symmetry is NOT a net dipole moment:
• Symmetrical = mirror image
• Asymmetrical = not mirror image
Is HCl symmetrical?
Is CO2 symmetrical?