LEWIS STRUCTURES AND VSEPR
Group
Atom
IV
V
VI
VII
Carbon
Nitrogen
Oxygen
Fluorine
Hydrogen
Number of Bonds
Or Valence
4
3
2
1
1
Electron Configuration
Or Valence Electrons
2s2 2p2
2s2 2p3
2s2 2p4
2s2 2p5
1s1
Number of electrons needed for He/Ne
electron configuration
4
3
2
1
1
Rules for Drawing Lewis Structures (in priority order).
1.
Calculate the total number of valence electrons from each atom in the molecule.
a) positive ions: deduct one electron for each positive charge
b) negative ions: add one electron for each negative charge.
The number of valence electrons for atoms in groups I through VIII is equal to the Group Number
The number of valence electrons for transition metal atoms is equal to the oxidation state of the atom
2.
Arrange the atoms appropriately
a) the central atom is usually the least electronegative.
b) the molecule is usually symmetrical
c) the molecular formula is usually drawn indicating the order of atoms, i.e. CH3CNO and CH3NCO have a
different order of the carbon and nitrogen only.
d) if you cannot obtain a reasonable Lewis structure (considering the number of bonds formed for each type of
atom as in the table above), then merely try another arrangement of atoms.
3.
Add lone pairs of electrons to satisfy the octet rule for each atom. Atoms from Period 2 (C, N, O, F) and higher take
eight electrons. Be careful of:
a) Hydrogen (H), an atom from Period 1 takes two electrons only instead of the eight.
b) The electron deficient atoms, Be only takes a total of 4 electrons and B, 6 electrons instead of the usual
eight.
4.
Total the electrons in your structure, counting bonds (sticks joining atoms) as two electrons and lone pairs as two
electrons.
a) if the total electrons in your structure equals the total valence electrons, then you have a good structure.
b) if the total electrons exceeds the number of valence electrons, then add a double bond ( bond) for each
pair of electrons in excess of the valence. Note: Resonance occurs when there is more than one possible
position to place the double bond ( bond).
c) if the total electrons is less than the number of valence electrons, add electron pairs to the central atom until
you have achieved the total number of valence electrons. The central atom is the least electronegative and so is
the most likely candidate for the extra electrons.
Atoms from Period 1 can never take more than two electrons (H).
Atoms from Period II can never take more than eight electrons (C, N, O, F). Thus the central atom will be from
Period 3 or higher when you add electron pairs to it.
Hydrogen will only ever form one bond and for this course fluorine will only ever form one bond.
Formal Charge
When you count the electrons around each atom in a Lewis structure by assigning one electron from each bond (stick) to that
atom and assigning both electrons from a lone pair, the total electrons should equal the valence electrons of that atom. When
this total differs from the number of valence electrons, then the atom has a formal charge. Because we are localizing
electrons in a Lewis structure, when in reality all the electrons are smeared over the whole molecule (not uniformly smeared,
some regions will have more ‘electron’ that another), this artificial localization of charge results. However it does give us a
reasonable picture of the electron distribution in a molecule; a starting point for more sophisticated models.
Thus the formal charge on an atom in a molecule is given by:
Formal charge = (number of valence electrons)  (number of electrons in lone pairs)  (number of bonds)
Notes: a) an atom will only have a formal charge if the number of bonds it forms is more or less than its desired number
b) the sum of the formal charge in a molecule must equal the charge on that molecule (including zero charge).
c) the preference for carrying negative charge is O > N > C
d) the best structure has the lowest or least separation of formal charge
The Lewis structure with the least amount of formal charge is the more reasonable picture of the electron distribution in a
molecule or the stable structure. The formal charge in a Lewis structure may be minimized in a situation where an atom with
a negative formal charge is ‘bonded’ to an atom with a positive formal charge. A pair of electrons on the atom with the
negative formal charge is moved to form a double bond between the two atoms of opposite formal charge. The effect of this
relocation is to move an electron from the negatively charged to the positively charged atom, thus eliminating the two
charges. The total number of electrons in the molecule remains unchanged, the only difference is that two electrons have
been relocated from a lone pair to a bond pair. See the example below with ClO 2.
O
Cl
O
O
Cl
O
Resonance
Resonance occurs when there is a choice of placement of a double bond ( bond) in developing a Lewis structure.
Each placement will give rise to a different Lewis structure and each of these Lewis structures will contribute to our picture
of the real molecule. The sum of these Lewis structures is called a resonance hybrid. Each Lewis structure may contribute to
the resonance hybrid to a different extent.
Resonance structures for a molecule differ only in the positions of their electron pairs.
The resonance hybrid is more stable (has a lower energy) than any of the contributing structures. Less stable structures can
sometimes be ignored altogether, depending on their contribution.
Comparitive stabilities can often be evaluated on the basis of formal charge, using the following rules:
a) the most stable structure has the least formal charge.
b) the preference for a negative charge is oxygen > nitrogen > carbon.
c) structures in which adjacent atoms have the formal charges of the same sign are especially unstable.
Electronegativity
Electronegativity is the ability of an atom to attract electrons to itself. A Table of electronegativity values follows this
section on Lewis structures.
Polar bonds generally have higher bond energies (the energy necessary to separate two atoms) than nonpolar bonds
composed of the parent atoms. HF has a higher bond energy than either H 2 or F2. This is attributed to an ionic component in
the HF bond. Fluorine attracts the electrons to itself obtaining a partial negative charge and hyrogen being deficient in
electrons obtains a partial positive charge.
Pauling devised a scale for electronegativity based on this difference in bond energies. The electronegativity of fluorine is
defined as being 4.00.
The difference in electronegativity (EN) between two atoms is given as:

EN  X A  X B  0.102 D AB  D AA  D BB

1
2
where DAB is the bond dissociation energy between atoms A and B
and XA is the electronegativity of A and XB is the electronegativity of B.
Calculate the electronegativity of hydrogen given the following data:
Molecule
Bond Dissociation Energy (kJ.mol-1)
1
HF
568

H2
436

F2
158
kJ  436 kJ  158 kJ  2  0102
kJ 2  178
EN  X F  X H  0102
.  568 mol
.
30553
. mol
.
mol
mol 

but the electronegativity of fluorine is defined as 4.00 and so the electronegativity of hydrogen is:
XH  XF  EN  4.00  178
.  2.22
1
Bond Polarities and Electronegativity Differences
The greater the electronegativity difference between two atoms, the closer the electron density is located to the more
electronegative atom and the greater the degree of ionic character in the bond between them (in other words the less
electronegative atom develops a partial positive charge and the more electronegative atom, a partial negative charge). A
covalent bond with ionic character is called a polar bond and the polarity of a molecule is assessed by experimentally
measuring the dipole moment.
The dipole moment  is defined as the magnitude q multiplied by the distance d between their centres. The SI unit is the C.m
but a much smaller unit , the debye is commonly used for molecules. One debye (D) = 3.33 x 10 -30 C.m.
The dipole moment correlates directly with the electronegativity difference as shown in the table:
Diatomic molecule
H2
HI
HBr
HCl
HF
Dipole moment (D)
0
0.44
0.82
1.08
1.82
Electronegativity difference
0
0.4
0.8
1.0
1.8
Bond Energy (kJ.mol-1)
436
298
366
432
568
Thus we can use electronegativity differences to predict the polarity of molecules.
Bonds are classified into nonpolar, polar and ionic. Electronegativity differences can be used to make this classification, but
they are not infallible.
Bond Type
Nonpolar
Polar
Ionic
Electronegativity Difference
0
0 to 1.7
> 1.7
Valence Shell Electron Pair Repulsion Theory (VSEPR)
The shape of a molecule can be predicted by assuming the regions of electron density around an atom will be as far apart as
possible. This assumption, when applied to the valence electrons of a molecule is called VSEPR.
For now we will consider the central atom but later we will apply the theory to any atom in a molecule with more than one
bond.
The regions of electron density when applied to the valence electrons are the lone pairs and bond pairs (multiple bonds
count as one bond pair).
Rules:
a) draw the Lewis structure
b) count the number of lone pairs and bond pairs on the central atom.
c) obtain the general shape from the table below
d) alter the shape for repulsion between lone pairs
Number of VSEPR electron pairs
2
3
4
5
6
Angle between VSEPR Pairs
180º
120º
109.5º
120º & 90º
90º
Shape
linear
trigonal planar
tetrahedral
trigonal bipyramid
octahedral
Examples of all the general VSEPR shapes with bonding and nonbonding electron pairs.
Predicting the Polarities of Molecules
The polarity of a molecule with more than one bond will depend on the individual dipole moments of each bond and the
geometry of the molecule. Molecular polarity is then the vector sum of the bond dipole moments. It was shown above (Bond
Polarities and Electronegativity Differences) that bond dipoles correlate closely with bond electronegativity differences and
thus we are able to predict the polarity of a molecule from the vector sum of the electronegativity differences. The
electronegativity difference for a bond is treated as a vector with the magnitude of the vector representing the
electronegativity difference and direction of the vector representing the orientation of the bond in the molecule. The head
(arrow) of the vector points towards the most electronegative atom in the bond and thus the resultant vector points toward
the electron dense side of the molecule. The examples below show BF3 where all the vectors cancel to give a nonpolar
molecule and CHF3 where the resultant vector indicates a very polar molecule (0.35 + 1.43 = 1.78). The dipole moment for
BF3 is 0 D and for CHF3 1.65 D.
EN(F-B) = 3.98  2.04 = 1.94
EN(F-C) = 3.98  2.55 = 1.43
EN(C-H) = 2.55  2.20 = 0.35
H
F
_
C
B
F
F
F
F
F
The next topic deals with the magnitude and direction of the vector we assign to a lone pair of electrons on a molecule. The
direction of the vector is straightforward, as the lone pair is considered a region of electron density in VSEPR. We can
estimate the magnitude if we compare the molecules NH3, NCl3 and NF3. In each of these molecules we have three
substituents bonded to a nitrogen and the nitrogen has a lone pair. The object of this exercise is to estimate the magnitude of
the electronegativity vector for the lone pair on nitrogen.
The shapes of each of these molecules is tetrahedral with the three hydrogen electronegativity vectors (or chlorine or
fluorine) forming an inverted umbrella with the lone pair above (see (a) in the diagrams). The three substituent vectors will
form a resultant that is equal in magnitude to one of the substituent vectors but aligned directly underneath the lone pair
vector (see (b) and (c) in the diagrams). In the case of the hydrogen vectors this resultant vector will increase the magnitude
of the lone pair vector. For a quantitative example of how to find the resultant of three pyramidal vectors in a tetrahedral
molecule see 'Calculation of the resultant of the electronegativity vectors for the molecule ClFO3’ following.
EN(N-H) = 3.04  2.20 = 0.84
EN(Cl-N) = 3.16  3.04 = 0.12
EN(F-N) = 3.98  3.04 = 0.94
(a)
(b)
(c)
resultant
_
N
H
H
H
N
_
Cl
Cl
Cl
_
N
F
F
F
If we assign the variable x to the magnitude of the lone pair vector, and we assume that the dipole moment correlates directly
with the resultant electronegativity vector for the molecule, we can write the equation:
magnitude of lone pair vector (x) + magnitude of resultant of the 3 substituent vectors = dipole moment in debyes
Diagram (c) illustrates the two electronegativity vectors that are being summed to give the dipole moment.
Molecule
NH3
NCl3
NF3
EN
0.84
0.12
0.94
 (D)
1.47
0.39
0.235
Vector Equation
0.84 + x = 1.471
0.12 + x = 0.39
0.94  x = 0.235
Magnitude of Electronegativity Vector for Lone Pair on Nitrogen
0.63
0.51
0.71
So the lone pair electronegativity vector on nitrogen can be taken as the average of these three values which is 0.62.
But EN = Xlp  XN = 0.62 and since XN = 3.04 thus Xlp = 3.66 on nitrogen.
This is a reasonable value to apply to lone pairs on most atoms when using electronegativity vectors to assess polarities of
molecules.
Calculation of the resultant of the electronegativity vectors for the molecule ClFO 3
The resultant electronegativity vector for the molecule will pass along the fluorine/chlorine bond and through the centre of
the pyramid formed by the three oxygens and the chlorine. Thus the y axis will be in the direction of this resultant vector and
the x axis will be through the chlorine and at 90º to the y axis, in the same plane as the left side O. Since the FClO angle is
109.5, the angle between the OCl bond and the x axis is 19.5º (109.5 - 90) and the angle between the OCl bond and the y
axis is 70.5º (90 - 19.5). See the diagrams below.
EN(O-Cl) = 3.44  3.16 = 0.28
EN(F-Cl) = 3.98  3.16 = 0.82
Each O-Cl electronegativity vector will contribute an amount h to the resultant vector.
h
Cos(70.5) 
EN (O  Cl )
h = 0.28Cos(70.5) = 0.09346
The contribution for the three O-Cl vectors to the resultant is 3h = 0.280
This is the same as one O-Cl vector, which must be the case if you think about it (for the molecule ClO 4, the resultant of
three O-Cl vectors must be equal and opposite to the fourth as the molecule is nonpolar).
The resultant vector is then: EN(F-Cl)  3h = 0.82  0.28 = 0.54 in the direction of fluorine.
F
F
F
y axis
y axis
109.5 o
x axis
19.5 o
O
Cl
O
x axis
19.5 o
O
x axis
Cl
O
70.5 o
O
O
y axis
O
Cl
h
O
O
Electronegativity Values
1
H
2
He
2.20
3
Li
4
Be
5
B
6
C
7
N
8
O
9
F
10
Ne
0.98
1.57
2.04
2.55
3.04
3.44
3.98
11
Na
12
Mg
13
Al
14
Si
15
P
16
S
17
Cl
0.93
1.31
1.61
1.90
2.19
2.58
3.16
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
0.82
1.00
1.36
1.54
1.63
1.66
1.55
1.83
1.88
1.91
1.90
1.65
1.81
2.01
2.18
2.55
2.96
2.9
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
0.82
0.95
1.22
1.33
1.6
2.16
1.9
2.2
2.28
2.20
1.93
1.69
1.78
1.96
2.05
2.1
2.66
2.6
55
Cs
56
Ba
57
La
72
Hf
73
Ta
74
W
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
0.79
0.89
1.10
1.3
1.5
2.36
1.9
2.2
2.20
2.28
2.54
2.00
2.04
2.33
2.02
2.0
2.2
87
Fr
88
Ra
89
Ac
0.7
0.9
1.1
18
Ar