Shapes of Molecules and Bonding
Molecular geometry is governed by energy. Molecules receive such geometry as to
minimize their potential energy. A striking example is DNA.
Lewis Dot Structures
1. VAL (total number of valence electrons) = sum of the valence electrons of each element
(sum of the group numbers, GN) – charge (of an ion)
2. STAB (stable noble gas configuration) = sum of the electrons in the noble gas
configuration for each element (2 for H, 8 for C and beyond). Since an atom achieves its
greatest stability and lowest energy when it has a filled shell, the noble gas configuration of
ns2np6, we assume that an atom in a molecule or ion will also be most stable when it has its
noble gas complement of electrons around it – the ubiquitous “octet” rule.
3. BOND (total number of bonding electrons) = STAB – VAL and BP (number of bond
pairs) = BOND/2. Since electrons are shared between atoms that are bonded together, the
“octet” around each atom over counts the actual numbers of electrons involved. The amount
of over counting is the number of bonding electrons.
4. LONE (total number of lone pair electrons) = VAL – BOND and LP (number of lone
pairs) = LONE/2. Valence electrons are of two varieties: they are bonding electrons (BOND)
or nonbonding electrons, lone electrons (LONE).
5. Draw the Lewis structure using your calculated BP and LP. Remember that nature likes
symmetry, put the odd atom in the center. Draw all resonance structures.
6. (FC) formal charge on atom) = GN – LONE – BOND/2. The sum of the formal charges on
all atoms is equal to the charge on the molecule or ion.
7. Valence shell expands to reduce FC and maximize resonance if atom is in the third period
or beyond.
Example
1
NF3
VAL: 5 (N) + (3x7) (F) = 26 e −
STAB: 8 (N) + 3 (8) = 32 e −
BOND: STAB – VAL: 32 – 26 = 6 e − [/2 = 3 BP]
LONE: VAL – BOND: 26 – 6 = 20 e −
..
..
: F: . . : F:
N
: .F.:
Example
Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of
stratospheric ozone
VAL: 4(C) + 2 (7) (Cl) + 2 (7) (F) = 32 e −
STAB: 5(8) = 40 e −
BOND: STAB – VAL: 40 – 32 = 8 e − [/2 = 4 BP]
LONE: VAL – BOND: 32 – 8 = 16 e −
Carbon has the lowest EN and is the central atom. The other atoms are placed around it.
Make bonds and fill in remaining valence electrons placing 8 e − around each atom
Make 4 bonds and fill in remaining valence electrons placing 8e- around each atom.
Example
Write the Lewis structure for methanol, (molecular formula CH4O), an important industrial
alcohol that is being used as a gasoline alternative in car engines.
2
VAL: 4(1)(H) + 4(C) + 6 (O) = 14 valence e −
STAB: 4(2)+2(8) = 24 e −
BOND: STAB – VAL: 24 – 14 = 10 e − [/2 = 5 BP]
LONE: VAL – BOND: 14 – 10 = 4 e −
Hydrogen can have only one bond so C and O must be next to each other with H filling in the
bonds.
H
H
C
..
O
..
H
H
In reality this is what do in order to convert a molecular formula to a Lewis dot structure
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:
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 two e- (either single or nonbonded pair)
can be moved in to form a multiple bond.
PLAN:
(a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom.
SOLUTION:
H
H
C
C
H
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. :
.
:N
N:
.
.
:N
N:
Resonance: Delocalized Electron-Pair Bonding
O3 can be drawn in 2 ways
-
B
B
O
O
O
A
O
O
C
A
O
C
Neither structure depicts O3 accurately 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
O
O
C
O
A
O
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
4
Writing Resonance Structures
PROBLEM:
Write resonance structures for the nitrate ion, NO3-.
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
O
N
N
N
O
O
O
O
–
O
O
–
O
O
O
N does not have an
octet; a pair of ewill move in to form
a double bond.
–
O
N
N
O
–
N
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 =
# of valence e- ─ (# of unshared electrons + 1 # of shared electrons)
2
B
For OC
O
For OA
# valence e- = 6
O
O
# valence e = 6
# nonbonding e- = 4
1
=2
# bonding e- = 4 x
2
Formal charge = 0
A
For OB
C
# valence e- = 6
# nonbonding e- = 6
1
=1
# bonding e- = 2 x
2
Formal charge = -1
# nonbonding e- = 2
1
=3
# bonding e- = 6 x
2
Formal charge = +1
Selecting the Best Resonance Structure
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 a more electronegative atom (an atom
with a larger EN value).
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Formal Charge: Selecting the Best Resonance Structure
EXAMPLE: NCO- has 3 possible resonance forms:
–
N C
–
N C
O
A
O
–
N
B
C
O
C
formal charges
-2
N
0
+1
C
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 preferred 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.
Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2.
a)
VAL: 3(1)(H) + 5(P) + (4)(6)(O) = 32 valence e −
STAB: 3(2)+5(8) = 46 e −
BOND: STAB – VAL: 46 – 32 = 14 e − [/2 = 7 BP]
LONE: VAL – BOND: 32 – 14 = 18 e −
P is a Period-3 element and can have an expanded valence shell.
Lower FC
0
-1
0
0 H O
0
+1
O
0
P O H0
O
H
0
0
0
0
O 0
H O
P
0
O0
H
0
O 0H
0
7
b) BFCl2 will have only 1 Lewis structure.
F
Cl
B
Cl
Valence Shell Electron Pair Repulsion Geometries (VSEPR Model)
This model is based on the idea that bond and lone paira in the valence shell of an element
repel each other and seek to be as far apart as possible,
• It is remarkably successful to predict structures of molecules and ions of main group
elements
• It is less effective and it is used seldom to predict structures of compounds containing
transition metal elements.
1. Repulsions BP/BP < BP/LP < LP/LP
2. Number of electron pairs (steric number) ⇒ electronic geometry
3. Molecular geometry (shape) ⇒ bond angles, distortions
4. Dipole moment, polarity
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Number of electron pairs
2
Geometry (bond angle)
Linear (180o)
Number of lone pairs
0
3
Trigoanal planar (120o)
Bent
0
1
4
Tetrahedral (109.47o)
Trigonal pyramidal
Bent
0
1
2
5
Triginal bipyramidal (90o,
120o)
seesaw
T-shaped
Linear
0
Octahedral (90o)
Square Pyramidal
Square Planar
0
1
2
6
1
2
3
A - Central atom
X - Surrounding atom
E - Nonbonding valence electron-group
9
The single molecular shape of the
linear electron-group arrangement
The two molecular shapes of the trigonal planar
electron-group arrangement.
10
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.
Effect of Double Bonds
H
o
120
ideal
120o
C
H
larger EN
O
H
116o
122o
C
O
H
greater
electron
density
real
Effect of Nonbonding (Lone) Pairs
Lone pairs repel bonding pairs
more strongly than bonding pairs
repel each other.
Sn
Cl
Cl
95o
The three molecular shapes of the
tetrahedral electron-group arrangement
Lewis structures and molecular shapes.
11
The four molecular shapes of the
trigonal bipyramidal electron-group
arrangement
The three molecular shapes of the octahedral electrongroup arrangement
12
The steps in determining a molecular shape
Molecular formula → Lewis Dot Structure → electron group arrangement → note positions of
lone pairs and double bonds → count bonding and non-bonding electrons → molecular shape
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
Draw the molecular shape and predict the bond angles (relative
to the ideal angles) of (a) PF3 and (b) COCl2.
PROBLEM:
SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair
F
P
The shape is based upon the tetrahedral arrangement.
F
F
F
P
F
F
<109.5o
The F-P-F bond angles should be <109.5o due
to the repulsion of the nonbonding electron
pair.
The final shape is trigonal pyramidal.
The type of shape is
AX3E
(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
C does not have an octet; a pair of nonbonding
electrons will move in from the O to make a
double bond.
Type AX3
O
C
Cl
Cl
The shape for a molecule with a central atom
having three attachments and no nonbonding
pairs is trigonal planar.
O
o
124.5
The Cl-C-Cl bond angle will
C
be less than 120o due to
Cl
Cl
the electron density of the
C=O.
111o
13
Predicting Molecular Shapes with Five or Six Electron Groups
PROBLEM:
SOLUTION:
F
F
Sb
F
F
F
Determine the molecular shape and predict the bond angles
(relative to the ideal 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 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
Br
F
F
The orientation of polar molecules in an electric field
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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,
(b) Boron trifluoride,
NH3
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:
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
(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
120o
F
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
C
O
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