Lewis Structures
and Bonding
(If we did it after molecular shape- AKA
VSEPR- it would be a prequel to “What
shape are your molecules in?”)
World of Chemistry, Zumdahl
Chpt 12 pp 358-381 (Lewis)
1
You’ll find out about…
o  Lewis structures
How many bonds do each element make?
n  What can expand? What can be deficient?
n
o  Bonding (covalent)
n
Polarity
p  Electronegativity
and determining bond type
o  Isomers
o  Formal charge
o  Resonance
o  Resonance v. isomers
2
Lewis Structures
o  Remember that all atoms want a full
outer shell, so do Lewis structures
o  Remember that for a given Lewis
structure, the number of electrons
around the atoms must equal the total
number of electrons individually
assigned.
Ex: C has 4, H has 1, so CH4 must have 8
total
n  We also did NH4+ on the board
n
3
Drawing Lewis Structures
• Add the valence electrons.
• Identify the central atom (usually the one with the
highest molecular mass and closest to the center of the
periodic table).
• Place the central atom in the center of the molecule
and add all other atoms around it.
• Place one bond (two electrons) between each pair of
atoms.
• Complete the octet for the central atom.
• Complete the octets for all other atoms. Use double
bonds if necessary.
4
Determining formal charge
Formal charge can be determined by:
Normal number of electrons in outer shell
[(1/2 the number of bonded electrons) + lone electrons]
_____________________________________
= formal charge
5
Formal Charge, continued
o  Example: N in NH4+
FC =5- [(1/2 of 8)+ 0]= +1
o
H in NH4 +
o
FC =1- [(1/2 of 2)+ 0]= 0
o  Overall, the formal charge on NH4+ is + ,
so we write NH4+ or as [NH4]+
o
n
This bracketed version is typically used,
and is more precise for reasons we have
yet to get into, but we will.
6
Formal charge and stability
o  The most “happy” molecules tend to
have no formal charges
o  However, molecules may be “happy” if
they have no NET charge on them
n
if there is 1+ and 1-, so a net of +1 + (-1)=0
o  Structures that are the best have a
minimal formal charge and a full octet
(valence shell) around each atom
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Electronegativity and Bond
Type
o  Determine the absolute value between the
electronegativities of the atoms involved
If it is , ≥ 2 then the bond is ionic
n  If it is 0.5 ≥1.9 then the bond is polar covalent
n
p  Electrons
n
n
are shared unequally between the atoms
resulting in partial charges (dipoles)
If it is ≤ 0.4, then the bond is nonpolar covalent
p  Electrons
are shared equally between the atoms of
the bond
n
no partial charges
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Examples:
o  A bond from H-O
n  H has an electronegativity of 2.1
n
O has an electronegativity of 3.5
p  The difference is 1.4, so the bond is polar
n
The more electronegative element has a slightly negative charge
(here, O)
n
The less electronegative element has a slightly positive charge
(here, H)
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Ionic Bonding
Na(s) + ½Cl2(g) → NaCl(s)
ΔH°f = - 410.9 kJ
NaCl is more stable than its constituent elements.
Why?
Na has lost an electron to become Na+ and chlorine has
gained the electron to become Cl-.
Na+ (10 electrons) => 1s2 2s2 sp6 same as Ne
Cl- (18 electrons) => [Ne]3s2 3p6 same as Ar
That is, both Na+ and Cl- have noble gas configurations
(an octet of electrons surrounding the central ion). 10
Electronegativity and Bond Polarity
K-F
Ionic
ED = 3.2
Cl-F
Covalent
Polar
ED = 1.0
F-F
Covalent
NonPolar
ED = 0.0
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Covalent Bonding
Multiple Bonds
It is possible for more than one pair of electrons to be
shared between two atoms (multiple bonds):
One shared pair of electrons = single bond (e.g. H2);
Two shared pairs of electrons = double bond (e.g. O2);
Three shared pairs of electrons = triple bond (e.g. N2).
H H
O O
N N
Generally, bond distances decrease as we move from
single through double to triple bonds.
12
Bonds and Bond Order
o  A single bond has a bond order of 1
o  Multiple bonds
n
double: two shared pairs of electrons
p  Bond
n
triple: three shared pairs of electrons
p  Bond
n
order of a double bond is 2
order of a triple bond is 3
quadruple bonds do not occur (too hard to
share 8 electrons; steric hinderance)
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Bond Energy
o  Bond energy: the amount of energy
needed to break a bond
n
Usually measured in KJ/ mol
o  Bond energy for each type of bond (Ex:
C-C single, C=C double, C-O, C=O, H-H,
C-N, C=N, etc.) is different
n
Why? Different atomic radii for each
element, different electronegativity, etc)
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Bond Energy and Bond Order
o  Single bonds have lowest bond energy
o  Double bonds have higher bond energy
o  Triple bonds have highest bond energy
15
Bond energies of common
bonds (kJ/ mol)
o  C-C
o  C=C
o  C C
o  N N
376
720
862
945
16
Atomic Radii and
Bond Energy
o  Bond
o  H-F
o  H-Cl
o  H-Br
o  Cl-Cl
o  Br-Br
o  I-I
Bond Energy
570
432
366
243
193
151
17
Bond Order and Bond Length
Bond
Bond Order
Bond Enthalpy
Bond Length
C-C
1
348 kJ/mol
1.54 Å
C=C
2
614 kJ/mol
1.34 Å
C≡C
3
839 kJ/mol
1.20 Å
N-N
1
163 kJ/mol
1.47 Å
N=N
2
418 kJ/mol
1.24 Å
N≡N
3
941 kJ/mol
1.10 Å
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Isomers
o  Same formula, different arrangement of
atoms
o  Physically break bonds and MOVE atoms
o  Other example on the board; looked at
structural v molecular formulas and the
hybrid version with CH3OCH3 and
CH2OHCH3
Resonance Structures
o  Have the same
alignment of
atoms, but
different bonding
(electrons ONLY
are moved, both
in bonds and
lone pairs)
20
Resonance structures of BF3
Remember that none of
these is a real picture of
BF3, but the real picture
is a hybrid of all of these.
(BO=1.3 for all B-F
bonds)
Some of these are better
pictures of what really
happens than others: the
better ones are those
with the least formal
charge.
AGAIN: Formal charge and
stability
o  The most “happy” molecules tend to have no
formal charges
o  However, molecules may be “happy” if they
have no NET charge on them
n
if there is 1+ and 1-, so a net of +1 + (-1)=0
o  Resonance structures that are better structures
have a minimal formal charge and a full octet
(valence shell) around each atom
22
Resonance, Formal Charge
and Good Lewis Structures
o  Those that are not as good, but fulfill the octet
rule are considered to be minor contributors
n
They can exist, but not as much of the overall
picture looks like them
o  If it doesn’t fulfill the octet rule and it isn’t a
KNOWN exception*, and/ or it has crazy formal
charges, toss it. It doesn’t work.
(* exceptions: next slide)
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Words About Exceptions:
o They happen. A fair amount of the time with
CERTAIN ELEMENTS- not with everything.
o Hence, exceptions to the general rules, but
rules for those elements are now different.
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Words About Exceptions:
o  Q: Why do chemists do this to me?
n  A: Because they hate you.
o  Q: Really?
n  A: No, they don’t. The properties of the atoms set the rules,
based upon their electron configurations and other properties
discussed in class (IE, EA, AR, shielding, Zeff….) Chemists do
not make up rules to annoy you; they make up general rules to
fit MOST situations; elements sometimes don’t fit the
constraints.
o  Q: Will the exceptions mess me up?
n  A: Not if you can follow all of the other rules.
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Deficient* atoms:
*(less than a full shell)
o Boron: USUALLY keeps 6 electrons in its
outer shell. It just does, because of being small
and being a nonmetal.
o  Basically, every element below is a metal, and
gives up electrons to bond ionically.
o  But B is so small it wants more. So it bonds
covalently to get them. But it can’t go to a full
8- it’s too many.
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Expansion of the octet
o  P, S, halogens and noble gases (yeah, I
know- we’ll explain why later on)
heavier than Br (Br, I, At, Kr, Xe, Rn)
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Expand to fit…
o  Typically 5 or 6 PAIRS of electrons, and
these are added as lone pairs, not as
bonded pairs.
o  Add lone pairs on the central atom only
until the number of electrons needed is
reached.
o  Ex: PCl5, I3, SF4, XeF3
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An excerpt from a web site on expansion:
o  The concept of the Expanded Octet occurs in any system that
has an atom with more than four electron pairs attached to it.
o  Most commonly, atoms will expand their octets to contain a total
of five or six electron pairs, in total. In theory, it is possible to
expand beyond those number.
n
The large amounts of negative charge concentrated in small
volumes of space prevent those larger expanded octets from
forming.
o  When an atom expands its octet, it does so by making use of
empty d orbitals that are available in the valence level of the atom
doing the expanding.
o  Atoms that do not have empty valence level d orbitals will not be
able to expand their octets. The atom that expands its octet in a
structure will usually be located in the center of the structure and
the system will not use any multiple bonds in attaching atoms to
the central atom.
o  The process of expanding octets is strictly a last resort on the
part of atoms.
http://www.bcpl.net/~kdrews/molegeo/molegeo.html#Expanded
More theft: But this is well
stated at
http://www.towson.edu/~yau/ExpandedOctet.htm
o  “Expanded octet” refers to the Lewis structures where the central
atom ends up with more than an octet, such as in PCl5 or XeF4. In
drawing the Lewis structure for PCl5 , there is a total of 40 valence
electrons to put in (5 + 5x7 = 40). One can easily see that if the
central atom, P, is to be joined to five Cl atoms, P would have 10
electrons instead of the octet. (Remember that one does NOT string
out the elements to look thus: P-Cl-Cl-Cl-Cl-Cl. ß No !!)
o  Clearly there is a violation of the “Octet Rule”.
o  How do we know to allow this violation? It’s simply this: PCl5 exists.
Our rules have to be revised to accommodate observations. If PCl5
exists, then this violation must be permitted, since there is no other
way to explain it. Again, chemists don’t hate you. Nature just does
things that we didn’t plan on when we use the easiest explanations.30
Exceptions continued…
o  A word of caution: Does this mean that we can
violate the Octet Rule any time we wish? No!
We can violate the Octet Rule only when there
is no other way to explain it.
o  Limitations to the “expanded octet”:
o  This cannot occur with elements that do not
contain d-orbitals! Which elements would that
be? Elements smaller than neon (atomic
number 10) have electrons only in the first two
main energy levels (n = 1 and n = 2), and those
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energy levels do not contain d-orbitals.
More on exceptions…
o  Second of all, expanded octets generally occur
when there are too many electrons to fit in.
Since double bonds and triple bonds occur only
when there are insufficient number of electrons,
you would not normally apply the “expanded
octet” to central atoms with multiple bonds! In
other words, the extra electrons must be added
as lone pairs and NOT as double or triple
bonds!
32
Summary of limitations of
applying the “expanded octet”:
o  The central atom with an expanded octet
o  MUST have an atomic number larger than 10 (beyond
Ne)
n
(this means it has available d orbitals to extend into)
o  Extra electrons should be first placed on the outside
atoms to fulfill their octets.
n
n
n
After that, there are still extra electrons, start with
placing them as lone pairs on the central atom.
If the central atom does not have a positive formal
charge, do not go any further. You have the correct
Lewis structure.
Only if the central atom has a positive charge should
you move a lone pair from the outside atoms to
share (to neutralize the formal positive charge)
p
Do NOT indiscriminately double or triple bond!
33
Let’s make XeF4
o  Step One: Count the total number of valence electrons.
8 from Xe+ 4 F x (7e- per F) = 36e-, or 18 pairs.
o  Step Two: The first element is usually the central atom,
and then you cluster the other atoms around it.
o  Step Three: The 4 covalent bonds shown above account
for 4 pairs. As you put lone pairs onto the surrounding F,
you would account for 12 more pairs, giving you a total of
16 pairs. Where are you going to put two more pairs?
The only place would be on the central atom (on Xe) as
lone pairs. Xe would therefore have 4 atoms and a lone
pair (AX4E2).
34
o  Xenon now has twelve electrons instead of the
octet! This is called an “expanded octet”,
expanding beyond the octet.
o  Notice we do NOT put a double bond between the
Xe and one of the F! A mistake that students often
make. In the structure shown above, Xe has a
formal charge of zero, so there is no reason to do
any more to the structure.
35
o  This structure is incorrect because Xe now has a
formal charge of 2+ (6 electrons instead of 8) and
the two F with double bonds each has a formal
charge of −1 (8 electrons instead of 7). Compared
to the structure above which has no formal charges,
this is certainly not preferred.
o  In addition, in terms of the Octet Rule, F has
exceed the Octet and it is not an element than can
exceed the Octet Rule (atomic number les than 10).
36