CHEMISTRY 115
VSEPR THEORY AND VALENCE BOND THEORY
I.
VSEPR THEORY:
The tables which follow show how VSEPR Theory and Valence Bond Theory are both
dependent upon the numbers of electron regions surrounding a central atom. In order to
apply either method, it is necessary to understand how we determine the number of
electron regions that exist on a particular atom.
Bonding Regions: An electron region is any set of electrons that is effectively localized in
a particular region of space. Any bond between atoms forms such a localized region,
regardless of how many electrons are involved in the bond. It makes no difference
whether the bond between two atoms is a single, double or triple bond, it will still be
directed along the line between the two atoms and thus localized in that direction. For
this reason, when we count the number of regions that contain electrons on an atom, each
bond to another atom will count as a single region – no matter how many electrons are
involved in that bond. For example, if an atom is bonded to 3 different other atoms, then
there will be 3 bonding regions on that atom.
Lone Pair Regions: Each lone pair on an atom counts as a single region of space that
contains electrons. Lone pairs belong solely to the atom on which they reside and each
lone pair will act as an independent group relative to any other lone pairs that may occur
on the same atom. If an atom has a single unpaired electron on it, then that electron will
occupy its own lone pair region. For example, if an atom has 3 lone pairs on it, then it
will have 3 lone pair regions and if another atom has 2 lone pairs and 1 unpaired electron,
(a “lone electron”, then that atom will also have 3 lone pair regions when we count.
The sum of the bonding regions and lone pair regions gives us the total number of
regions containing electrons on that atom. This is illustrated below with fictitious
molecules of the form AB2. Under each drawing, the number of electron regions on atom
A is listed below the drawing. The 1st value gives the number of bonding regions on
atom A, the 2nd value gives the number of lone pair regions on atom A and the last value
gives the total number of electron regions on atom A:
B
A
B
B
A
B
B
A
B
B
A
# bonding
2
2
2
2
#lone pair
0
0
1
2
Total #
2
2
3
4
B
Note that each bonding region counts as 1 region, no matter how many bonds connect the
two atoms – i.e., regardless of whether the bond is a single, double or triple bond. Thus
atom A has 2 regions due to bonding in each of the four molecules that are shown. In
–1–
Raynor: 3/6/2016
addition to the bonding regions, each lone pair or lone electron counts as a single
additional region. Thus the picture on the far right has 4 regions containing electrons – 2
regions which involve bonds to other atoms, 1 region containing a lone pair of electrons,
(shown as a bar), and 1 containing a single lone electron.
Both VSEPR Theory and Valence Bond Theory are dependent upon how many electron
regions occur on the central atom in a structure. Once you know this number, everything
else becomes dependent upon it. In VSEPR Theory, we use this number to determine
what the electron geometry will be – i.e., the geometry formed when the electron regions
are as far apart from each other as possible in 3D space. In Valence Bond Theory, this
number determines the hybridization that will occur by mixing atomic orbitals on the
central atom.
Electron Geometries: Each electron region will repel all the other regions since they all
contain negatively charged electrons. The strongest repulsions will be those between 2
lone pair sets of electrons with each other. This is because lone pair electrons are held
very close to their parent atom, due to the attraction of the positively charged nucleus for
the negatively charged electrons. In contrast, the electrons that are involved in bonds will
be spread out along the entire length of the bond. Their repulsions on other electron
regions will be less than the repulsions from lone pair electrons. However, all regions
will repel others, which ultimately forces the electron regions to position themselves as
far apart from each other as is possible. For example, if there are 2 regions containing
electrons on the central atom in a structure, (either in bonds to other atoms or in lone
pairs of electrons), then the farthest apart those electrons can get from one another is on
opposite sides of the atom. This creates a linear geometry of the electrons to occur, with
angles of 180  between them. [Note – an angle is defined using 3 positions in space. In
this case, the angles between electron geometries are defined by the position of the first
electron region, the atom it belongs to, and the position of the 2nd electron region.] If
there are 3 regions, then the farthest apart those electrons can get from one another is in a
trigonal planar arrangement, with angles of 120 between each set of electrons. If there
are 4 regions, then the farthest apart those electrons can get from one another is to form a
3-dimensional tetrahedral electron geometry about the central atom, with angles of 109.5
 between each set of electrons. When 5 regions occur, then the optimal geometry, (i.e.,
the one that gives the lowest repulsions between electrons), is called a trigonal
bipyramid geometry. This consists of 3 regions that form a trigonal planar geometry,
with angles of 120 between them, plus two more electron regions that are on opposite
sides of the atom, along the line that is perpendicular to the plane formed by the trigonal
planar groups. The two final regions will form 90 angles with each electron region in the
trigonal planar arrangement and angles of 180 with each other. [See drawings in
textbook or in the table at the end.] Finally, if 6 regions occur, then an octahedral
electron geometry will be formed by the electron regions. This geometry starts with a
square planar arrangement of 4 of the regions, (with 90 and 180 angles between each
pair), plus one region above and one below the square planar groups. All angles are 90
and 180 between pairs of electron groups. [Again, see pictures in text and at the end.]
–2–
Raynor: 3/6/2016
Molecular Geometries: The molecular geometry is the geometry formed by the atoms in
the molecule – i.e., it is described by the form taken by all the atoms – not by the positions
of the electrons. However, the electron geometry determines the starting point. It sets
for us the directions in which bonds can found. If all of the electron regions about a
central atom are due to bonding regions, then the molecular geometry will be identical to
the electron geometry. However, if one or more of the regions involve lone pairs of
electrons, then the molecular geometry will not be the same as the electron one.
However, the bond angles that occur will be identical to those expected for the angles
between electron regions, except they will be somewhat modified by the differences in
repulsion between lone pair electrons with bonding pairs. As discussed earlier,
repulsions of lone pair electrons with other lone pair electrons will be strongest, those of
lone pair electrons with bonding electrons will be next-strongest, and the repulsions
between two sets of bonding electrons will be weakest. This is summarized below using
“lp” to mean lone pairs of electrons and “bp” to mean bonding pairs of electrons. The
weakest interactions are at the left and strongest on the right in the relations shown
below:
Repulsion Strengths:
bp – bp
<
bp – lp
<
lp – lp
The effect of this is to cause lone pairs to repel the bonding pairs, causing the bond angles
to be slightly smaller than the angles occurring when all regions contain bonds.
Note: in the descriptions below, we will describe the molecular geometries in terms of the
numbers of one pair regions. The number of bonding regions is the difference between
this value and the total number of regions occurring for that electron geometry. For
example, in the description when 3 regions contain electrons, when 0 regions contain lone
pair regions, then 3 regions contain bonds and when 1 region contains a lone pair, then
the 2 remaining regions will involve bonds, etc.
-----------------------------------------------------------------------------------------------------------------------# Regions = 2: (AB2) Electron geometry = Linear, see drawing below. Angle = 180.
B
A
B
0 Lone Pair Regions (2 bond regions): Molecular Geometry = Electron Geometry =
Linear. Bond angle = 180. [Example is BeCl2]
-----------------------------------------------------------------------------------------------------------------------# Regions = 3: (AB3) Electron geometry = trigonal planar, see drawing below. Angles =
120.
B
A
B
B
0 Lone Pair Regions (3 bond regions): Molecular Geometry = Electron Geometry =
Trigonal Planar. All bond angles = 120. [Example is BH3]
–3–
Raynor: 3/6/2016
1 Lone Pair Region (2 bond regions): Molecular Geometry = Bent. Bond Angle =
< 120. [Example is SO2]
-----------------------------------------------------------------------------------------------------------------------# Regions = 4: (AB4) Electron geometry = tetrahedral. Angles = 109.5. This is a 3-D
structure, as shown below, where dotted lines indicate the line is going away from you,
behind the page, and a triangle indicates the line is coming up closer to you, out of the
page. Two different ways of drawing this structure are shown below. In the left one, the
3 B positions sit at the points of an equilateral triangle, with the one at the top sitting
directly above the midpoint of the triangle. In the drawing on the right, the B positions
sit at the 4 corners of a cube, with the corners on the top opposite to those at the bottom.
Both forms yield identical structures. All B-A-B angles are identical and all are equal to
109.5. The atom positions in the left-hand drawing are labeled as positions B1, B2, B3 and
B4.
B1
A
B4
B3
B2
0 Lone Pair Regions (4 bond regions): Molecular Geometry = Electron Geometry =
Tetrahedral. All bond angles = 109.5. [Example is CH4]
1 Lone Pair Region (3 bond regions): Molecular Geometry = Trigonal Pyramid.
Bond Angle = < 109.5. Structure is that obtained if you put the lone pair at position
B1, which means bonds run from atom A to the positions at B2, another from A to B3
and the third from A to B4. The repulsion from the lone pair at B1 onto the bonding
electrons causes the bond angle to become slightly smaller than 109.5 – perhaps
becoming 108 or 107. [Example is NH3]
2 Lone Pair Regions (2 bond regions): Molecular Geometry = Bent. Bond Angle =
< 109.5. Structure is most clearly seen if you put the lone pairs at positions B3 and
B4 and the bonds from A to B1 and A to B2. What remains is one bond angle from B1
to A to B2. The repulsions from the lone pairs at B3 and B4 will cause a slight
decrease in the bond angle from 109.5 to perhaps 105 or 106. [Example is H2O]
-----------------------------------------------------------------------------------------------------------------------# Regions = 5: (AB5) Electron geometry = trigonal bipyramid. Angles = 90, 120 and
180. This is also a 3-D structure, as shown below, where the trigonal planar arrangement
occurs among positions B1, B2 and B3. Positions B4 and B5 sit at opposite sides of the
equilateral triangle formed by positions B1 – B3. All the angles are as follows:
B1–A–B2 = B1–A–B3 = B2–A–B3 = 120;
B4–A–B1 = B4–A–B2 = B4–A–B3 = B5–A–B1 = B5–A–B2 = B5–A–B3 = 90;
B4–A–B5 = 180.
–4–
Raynor: 3/6/2016
What makes this structure unique is the fact that the positions on the equilateral triangle,
B1, B2 and B3, are different in environment than those at B4 and B5. For example, the
position at B1 is involved in 2- 120 interactions and 2 90 interactions, however the
position at B4 has no 120 interactions. It has 3- 90 and 1-180 interaction, instead.
B4
B3
B2
A
B1
B5
0 Lone Pair Regions (5 bond regions): Molecular Geometry = Electron Geometry =
Trigonal Bipyramid. Bond angles = 90, 120, 180. [Example is PCl5]
1 Lone Pair Region (4 bond regions): Molecular Geometry = “See-Saw”. Bond
angles = < 90, < 120 and < 180. Because the trigonal planar positions are in a
different environment than are the other two positions, the total energy of the
molecule with the lone pair placed at B1, (and bonds at all other positions), will be
different from the energy with the lone pair at B5 instead, (and bonds at all other
sites). Since 90 angle interactions between electrons will place the electrons much
closer together than will 120 angles, we want the structure that has the fewest 90
interactions between the lone pair and bonding pairs. It is easy to see that this
means that the lone pair should be placed at one of the trigonal planar positions, B1,
B2 or B3. It is easiest to see what the resulting molecular geometry looks like if we
place the lone pair at the B1 position. The rest of the positions will therefore involve
bonds between atom A and the atoms at positions B2 – B5, creating a geometry that
looks like a child’s see-saw., which is then the term used for the molecular geometry.
[Example is SF4]
2 Lone Pair Regions (3 bond regions): Molecular Geometry = T-Shaped. Bond
Angles = < 90 and < 180. The lowest energy structure places both lone pairs at two
of the trigonal planar positions. The structure is most clearly seen if you put the
lone pairs at positions B2 and B3 and the bonds at all other positions. The regions
that remain all are in the same plane as one another and appear to form a T, hence
the name of the structure. [Example is ClF3]
3 Lone Pair Regions (2 bond regions): Molecular Geometry = Linear. Bond Angle
= 180. The lowest energy structure places all three lone pairs in the trigonal planar
positions, B1, B2 , B3 and the bonds at the apical positions, yielding B4–A–B5 = 180 =
linear structure. [Example is XeF2]
-----------------------------------------------------------------------------------------------------------------------# Regions = 6: (AB6) Electron geometry = octahedral. Angles = 90 and 180. This is
also a 3-D structure, as shown below, where the square planar arrangement occurs
among positions B1, B2, B3 and B4. Positions B5 and B6 sit at opposite sides of the square
–5–
Raynor: 3/6/2016
formed by positions B1 – B4. All the angles are 90 or 180, depending on whether the
positions are next to one another, (90), or opposite to one another, (180).
B5
B3
B2
A
B4
B1
B6
0 Lone Pair Regions (6 bond regions): Molecular Geometry = Electron Geometry =
Octahedral. Bond angles = 90, 180. [Example is SF6]
1 Lone Pair Region (4 bond regions): Molecular Geometry = Square Pyramid.
Bond angles = < 90 and < 180. Since it does not matter which position is used for
the lone pair, the easiest to see is when position B6 is used as the lone pair positions,
with bonds occurring at the other 5 positions. If you draw lines from position B5 to
positions B1 through B4 and connect the lines along the square, i.e., from B1 to B2, B2
to B3, B3 to B4 and B4 to B1, you will discover that the structure you have created is a
4-sided pyramid – i.e., a pyramid with a square bottom – hence the name square
pyramid to describe the molecular geometry. [Example is BrF6]
2 Lone Pair Regions (3 bond regions): Molecular Geometry = Square Planar.
Bond angles = exactly 90 and 180. The lone pairs will provide the least repulsion
on each other if they are on opposite sides of the central atom. If we choose
positions B5 and B6 for the lone pairs, it is easiest to see the geometry that remains
for the 4 bonding regions – you can see that the atoms at positions B1 though B4 will
end up on the 4 corners of a square. This geometry is called the square planar
geometry for that reason. Since the lone pair electrons are on opposite sides of the
central atom, the repulsions from below on any bond will be exactly matched by the
repulsions from above. Thus the bond angles remain unchanged – they will be
exactly 90 and 180. [Example is XeF4]
-----------------------------------------------------------------------------------------------------------------------The electron geometries and molecular geometries that can arise from them are all
summarized in the table on the next page.
–6–
Raynor: 3/6/2016
Electron and Molecular Geometries from VSEPR Theory
# eregions
AB2
2
AB3
Electron
Geometry
# lone
pairs
Molecular
Geometry
Bond
Angles
linear
0
linear
180
trigonal planar
0
trigonal planar
B
A
B
trigonal planar
1
bent (angular)
< 120
B
B
A
120
3
AB2E
Drawing
B

A
B B
B
B A
B B

B A
B B

A
B B
B B
B A B
B
B B
B A B

AB4
tetrahedral
0
tetrahedral
109.5
AB3E
tetrahedral
1
trigonal pyramid
< 109.5
AB2E2
tetrahedral
2
bent (angular)
< 109.5
AB5
trigonal bipyramid
0
trigonal bipyramid
90, 120, 180
AB4E
trigonal bipyramid
1
seesaw
< 90, < 120,
<180
AB3E2
trigonal bipyramid
2
T-shaped
< 90, < 180
B A B
B
AB2E3
trigonal bipyramid
3
linear
180
AB6
octahedral
0
octahedral
90, 180
AB5E
octahedral
1
square pyramidal
< 90, < 180
octahedral
2
square planar
90, 180
B A B

BB
B A B
BB
BB
B A
BB
BB
A
BB
4
5
6
AB4E2
–7–
Raynor: 3/6/2016
II.
VALENCE BOND THEORY:
Valence Bond Theory uses the assumption that as bonds or lone pair regions are formed
on an atomic center, the valence atomic orbitals will mix together to form new orbitals
that are oriented along the optimal directions – those which will minimize the electronelectron repulsions occurring between them. Thus Valence Bond Theory is dependent
upon the number of electron regions on the atom in the same way that VSEPR theory
depends upon them.
One key factor to remember is the Conservation of Orbitals. In the same way that you
can’t create or destroy energy or mass, you cannot create or destroy the number of
orbitals in a system. Thus if you start with a total of q orbitals, you must end with a total
of q orbitals. This leads to a simple numbers game for discovering the types of
hybridization that will occur among atomic orbitals.
To demonstrate why this is important, consider what happens with the atomic orbitals on
carbon when methane, CH4, is formed. Carbon has the [He]2s22p2 ground state
configuration, as shown schematically below:







Note that the 2s orbital is lower in energy than the 2p one. Thus even if we promote one
of the 2s electrons up into the 2p orbital to allow for bonding with 4 H atoms, (as shown
in the right-hand drawing), the bond formed using the 2s-type electron should be lower
in energy than the ones involving the 2p orbitals. However, all four C-H bonds in CH4
are identical in length and energy to one another. Furthermore, they form a tetrahedral
molecular geometry, with all angles equal to 109.5. If bonds were formed between
hydrogen and the 2p orbitals directly, the bonds should end up at 90 angles, since the
three p-orbitals are oriented at 90 to one-another.
We can account for carbon forming 4 identical bonds with hydrogen when methane is
formed if we assume that the atomic orbitals on carbon first mix with one another to form
new orbitals, each or which is oriented along one of the 4 tetrahedral orientations around
the carbon atom. Mathematically, we can show that we can create 4 identical new
orbitals aligned along each of the 4 tetrahedral directions, if we mix together the 2s
orbital with all three 2p orbitals. [Note – since we are mixing together 4 atomic orbitals,
we must end up creating 4 new orbitals, due to the Conservation of Orbitals rule referred
to earlier.]
Similar outcomes occur when there are 2 regions containing electrons, or 3, or any of the
numbers of regions discussed earlier. In every case, the new hybrid orbitals that are
created end up being aligned along the directions prescribed for the electron geometry
that is expected to form. For example, when 5 orbitals mix to create 5 new ones, they will
become oriented along the 5 different directions in a trigonal bipyramid electron
geometry, etc.
–8–
Raynor: 3/6/2016
When we mix atomic orbitals to create new hybrid ones, we describe the new hybrid
orbitals in terms of the atomic orbitals used in the mixture. For example, with the carbon
atom in CH4, we mixed the 2s orbital with all three of the 2p orbitals to create the hybrid
orbitals. We call the new hybrid orbitals sp3 hybrids – which specifically indicates that
we mixed one s-type orbital with 3 p-type orbitals to create the 4 new hybrid orbitals.
Finally, when we create hybrid orbitals for bonding to other atoms, we want to generate
the lowest energy orbitals we can in order to stabilize the energy of the final molecule.
Thus we always mix the lowest energy orbital, (which is always an s-type orbital), with
the orbitals next up in energy, (which are always p-type orbitals, since we are discussing
non-metal to non-metal bonding). We will use d-type orbitals in our hybrids only if we
need to create more than 4 hybrid orbitals. For example, when the electron geometry is
trigonal bipyramid, we will have 5 regions around the central atom that contain electrons
and we will need to mix together 5 atomic orbitals to obtain 5 hybrid orbitals. This will
require that we use the lowest energy s-type orbital, all three p-type orbitals and one dtype orbital to create 5 new sp3d hybrid orbitals. For octahedral electron geometries it
follows that we must mix the s-type orbital, all three p-type orbitals and 2 d-type orbitals,
producing 6 new sp3d2 hybrid orbitals. [Remember: we need empty spaces available in
our new orbitals that can accept electrons. Thus we will need to use the lowest energy
empty d-orbitals when we create hybrid orbitals. If a d-orbital is filled with electrons, it
cannot accept additional ones and thus form a bond. For example, if we have the I3– ion,
which has the following Lewis Drawing,
I I I
1
the central I atom will have its electrons arranged in a trigonal bipyramid electron
geometry. We will need to mix its valence 5s and 5p-orbitals with one of the empty 5d
orbitals to create its new sp3d hybrid orbitals.
Hybrid orbitals are oriented directly along the direction of bonding, (or the direction of a
lone pair of electrons). They can then overlap with an orbital from another atom to form
sigma-type bonds – i.e., bonds that are directed along the bond-axis between 2 atoms –
and give the framework for the single bonds and lone pairs on an atom. If a double or
triple bond forms between atoms, the 2nd or 3rd bond between them arises from overlap
of un-hybridized p-type atomic orbitals. Consider, for example, the double bond
between carbon atoms in ethylene. Each C atom is involved in 2 single bonds to H atoms
and one double bond to the other C atom. There are thus 3 regions around each C atom
that contain electrons and we expect mixing of the 2s orbital with 2 or the 2p orbitals to
produce the 3 sp2 hybrid orbitals at 120 angles from one another. The third p-orbital is
the one perpendicular to the molecule’s plane. The lobes on the remaining p-orbital on
the left-hand carbon can overlap with the lobes on the remaining p-orbital on the righthand carbon to form a pi-type bond. The sigma-type bonds and their overlap with the 1stype orbitals on H and their overlap with each other are illustrated below:
–9–
Raynor: 3/6/2016
H
H
C
C
H
C
C
H
On the right-hand side, the overlap of the unhybridized 2p orbitals is illustrated. These
lobes actually occur with one lobe coming out of the page and the other going into the
page in the left-hand drawing – i.e., they are oriented perpendicular to the plane of the
molecule. The overlap of the lobes on these p-type orbitals creates one large lobe above
the bond axis and one large lobe below the bond axis, and is called a pi-type orbital and
thus bonds involving such orbitals are called pi-bonds. These pi-type orbitals can accept
up to 2 electrons, when forming a pi-bond between two atoms.
The figure below illustrates how hybrid orbitals are generated from the valence atomic
orbitals:
# Regions = 2:

s
p
sp
p
# Regions = 3:

s
p
sp2
p
# Regions = 4:

s
p
sp3
# Regions = 5:

s
p
d
sp3d
d
# Regions = 6:

s
p
d
– 10 –
sp3d2
d
Raynor: 3/6/2016
The table below summarizes the types of hybrid orbitals that form as a function of the
number of regions around an atom that contain electrons.
# Electron Regions
2
3
4
5
6
# Hybrid Orbitals Needed
2
3
4
5
6
# Atomic Orbitals Used
2
3
4
5
6
(1, 1, 0)
(1, 2, 0)
(1, 3, 0)
(1, 3, 1)
(1, 3, 2)
Type of Hybrid Orbitals
sp
sp2
sp3
sp3d
sp3d2
# and Type of Unused Orbitals
2p
1p
0 p, 5 d
4d
3d
# of (s, p, d) Used
Note how simple it is to remember how to generate this information. The number of
electron regions tells us how many atomic orbitals will mix together and how many new
hybrid orbitals will form. The type of hybrids formed can be easily determined by
remembering that the valence s-type orbital is always used in the mixture, (since it will be
lowest in energy). Next, we will mix in as many p-orbitals as needed to create the
number of hybrids necessary. However, if more than 4 atomic orbitals are needed, (i.e.,
more than the number of valence s-type and p-type orbitals available), then any
additional atomic orbitals will have to come from empty d-type orbitals that are close in
energy to the valence s-type and p-type orbitals. If multiple bonds are formed between
atoms, (i.e., double or triple bonds), then one of the bonds will involve overlap of hybrid
orbitals on the two atoms and any additional bonds will form from the overlap of
unhybridized atomic orbitals on the two centers. Consider, for example, the C atom in
CO2 and the S atom in SO42–. Their Lewis structures are shown below:
O
C
O
O
L
M
O S
M
M
N O
O
OP
P
P
Q
2
In CO2, the C atom has 2 regions about it that contain electrons and thus forms 2 new sphybrid orbitals, one directed towards the left-hand O atom and one toward the righthand O atom. The 2nd bond in the left-hand double bond arises from the overlap of one
of the unhybridized 2p orbitals on the C atom with a 2p orbital oriented in the same
direction on the left-hand O atom and the 2nd bond in the right-hand double bond
involves the other unhybridized 2p orbital on the C atom with a p orbital oriented in the
same direction on the right-hand O atom. [One pair of p orbitals will be oriented updown and the other will be oriented in-out, relative to the way the molecule is drawn. [In
SO42–, there are 4 regions containing electrons on the S atom. Thus the molecule will have
a tetrahedral molecular geometry as well as a tetrahedral electron geometry. Since all 4
3p-type atomic orbitals on sulfur will be used in creating the 4 sp3 hybrid orbitals on the S
atom, the double bonds must involve overlap of unhybridized 3d-type orbitals on the
– 11 –
Raynor: 3/6/2016
sulfur atom with 2p-type orbitals in the appropriate directions on the O atoms. Thus,
even though no d-type orbitals were needed in the formation of the hybridized orbitals,
they were necessary for the formation of the two multiple bonds between sulfur and
oxygen. O, S and Se are all in Group 16, but with different n quantum numbers for their
valence atomic orbitals. We can thus predict that the SeO42– anion could potentially be a
stable anion and its Lewis structure and geometry would be the same as that for SeO42–,
since the only difference would be that Se would use its 4s, 4p and 4d orbitals to generate
its hybrid orbitals and for the pi-type orbitals used in the multiple bonds. We can also
predict that making the substitution of O for S in the structure to create O 52– would be
highly unlikely to form a stable anion. The only allowed Lewis Diagram for O52– would
be the one where all bonds are single bonds, since oxygen cannot exceed its octet.
Although the O atom could use its 2s and 2p orbitals to create the sp3 hybrid orbitals that
are needed, it does not have any empty d-type orbitals that are close in energy to its
valence orbitals. Thus there are no unhybridized orbitals that could be used to create the
double bonds. The Lewis structure with all single bonds has a +2 formal charge on the
central atom. This means that if O52– were able to form at all, it would probably be very
unstable. In fact, I don’t believe the O52– ion exists!
– 12 –
Raynor: 3/6/2016