Recap – Last Lecture
Cl2
Covalent Bond
HCl
Polar Covalent Bond
NaCl
Ionic Bond
Lewis Structures
•  Lewis Structures provide a way to
predict stable bonding arrangements in
covalent molecules.
•  Based on maximum number of
electrons allowed in the valence shell:
Cl
Cl
Na+
Cl-
–  H
2 electrons
–  C, N, O, F 8 electrons
–  Period 3 & below: up to 18 electrons but
usually 8, 10 or 12
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Rules for Lewis Structures
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Period 2 – CH4
1.  C is at the centre
2.  Total number of valence electrons =
4 (C) + 4×1 (H) = 8
1.  Arrange the atoms.
–  Place the least electronegative atom (not H) in the centre.
2.  Count the total number of valence electrons.
3.  Four C-H bonds require 4×2 electrons:
–  Remember to add or subtract e- for anions and cations.
3.  Allocate two electrons between each pair of atoms
which are assumed to be covalently bonded.
4.  Use remaining valence electrons to form lone pairs.
–  Start with the surrounding atoms (centre atom last).
5.  Check number of electrons on the central atom.
–  If too few, move lone pairs from the (least electronegative)
surrounding atoms into the bonding region (make double
bonds).
4.  Electrons remaining = 8 (valence) – 8 (bonding) = 0
5.  Carbon has 8 electrons: stable
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Period 3 – SF4
Polyatomic ions – NH4+
1.  S is at the centre
2.  Total number of valence electrons
•  6 (S) + 4×7 (F) = 34
3.  Four S-F bonds require 8 electrons:
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1.  N is at the centre
2.  Total number of valence electrons
•  5 (N) + 4×1 (H) -1 (positive charge) = 8
F
F
S
F
3.  Four N-H bonds require 4×2 electrons:
F
5.  Fluorine has 8 electrons: stable
Sulfur has 10 electrons: stable
H
S
F
N
H
H
F
F
+
H
4.  Electrons remaining = 34 (valence) – 8 (bonding) = 26
•  Place 3 lone pairs on each fluorine (24 electrons)
•  Place remaining electrons on sulfur (2 electrons)
4.  Electrons remaining = 8 (valence) – 8 (bonding) = 0
5.  Nitrogen has 8 electrons: stable
F
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Molecules with multiple bonds
O2
Ozone
O3
Total number of valence electrons: 2 x 6 = 12
Total number of valence electrons: 3 x 6 = 18
Small rings are highly ‘strained’ and do not
normally form.
Try a different arrangement of oxygen atoms
Each oxygen has 8 electrons: stable
N2
Total number of valence electrons: 2 x 5 = 10
Another valid Lewis structure would be:
Each nitrogen has 8 electrons: stable
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Resonance
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Resonance
•  Benzene: C6H6
•  When two or more Lewis structures are possible for a
molecule it is said to exhibit resonance.
•  This is indicated by double headed arrows: ↔
•  This does not mean the molecule flips back and forth
between the possible structures – the true structure is
an average of the possible resonance structures.
•  Bonds are neither single nor double – they are
intermediate between the two. They have intermediate
length and strength.
Benzene is best described as being a resonance hybrid
of these two bonding arrangements.
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Resonance
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Resonance – polyatomic ions
•  Benzene: C6H6
•  Nitrate, NO3-
Bond Length
Bond Strength
C–C
154 pm
356 kJ mol-1
C=C
133 pm
636 kJ mol-1
benzene
139 pm
518 kJ mol-1
Total number of valence electrons:
5 + (3 x 6) + 1 = 24
•  Benzene is also about 140 kJ mol-1 more stable than
predicted for 1,3,5-cyclohexatriene (the structure with
localised double bonds).
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12
2
NO3- Nitrate ion
Resonance
If resonance occurs…
•  The actual structure is an average of all possible Lewis
structures.
•  The actual structure is a single structure.
•  The actual structure is more stable than predicted from
the Lewis structure.
•  There is no easy way of drawing the actual structure!
•  All N-O bond lengths the same and intermediate
between N-O and N=O (bond order = 1 1/3)
•  Negative charge distributed evenly across the three
oxygen atoms (1/3 on each oxygen)
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Predicting the best structure
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Example: POCl3
Total number of valence electrons: 5 + 6 + (3 x 7) = 32
•  Draw the Lewis structure, including lone pairs and
multiple bonds.
•  Check for the presence of ‘equivalent’ resonance
structures.
•  Check number of bonds on C, N, O & F as a prompt
for ‘non-equivalent’ resonance structures.
•  Choose best structure and check again for resonance
structures.
Oxygen has a valency of 2 so, when uncharged, forms
two bonds.
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Example: POCl3
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Applications
•  Streptonigrin is an anti-biotic and anti-tumor
agent.
•  It contains flat benzene like rings.
•  P now has 10 electrons, but this is OK as valence
electrons are in the third shell.
•  O has the two bonds expected from its valency.
•  These are resonance structures (only position of the
electrons have moved) but not ‘equivalent’.
•  Right hand structure is a better representation of the17
real structure.
•  These rings can slot between
adjacent base pairs in DNA, rather
like putting a letter in a letterbox,
and disrupt the replication of DNA
in tumor cells.
http://www.sigmaaldrich.com/content/dam/sigmaaldrich/structure0/131/mfcd00063401.eps/_jcr_content/
renditions/mfcd00063401-medium.png
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Learning Outcomes:
Questions to complete for next lecture:
•  By the end of this lecture, you should:
−  be able to draw out plausible Lewis structures for
simple polyatomic molecules.
−  assign bond orders based on sharing of electron
pairs and resonance structures.
−  identify the structure that most closely resembles
the actual structure given non-identical
resonance structures.
1.  Draw the Lewis structures of NH2-, NH3 and NH4+ and
indicate how the number of bonds relates to the valency
of nitrogen.
2.  Draw the Lewis structure of PCl3 and PCl5 and indicate
the number of electrons associated with the phosphorus
atom.
3.  Draw the ‘best’ Lewis structure of SO2 and SO3 and give
a reason for your choice.
4.  Draw the ‘best’ Lewis structure of HPO32- and give a
reason for your choice (note H is bonded to the P).
−  be able to complete the worksheet (if you
haven’t already done so…).
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