Assignment 10
STRUCTURE OF MOLECULES AND MULTI-ATOM IONS ─ Part I
This Assignment emphasizes the structures of covalent molecules and multi-atom complex ions. Being able to
characterize the three-dimensional shape, or geometry, of molecules is often necessary to understand their
reactivity. The importance of molecular shape is especially evident in biochemical reactions such as drug and
enzyme interactions.
Covalent molecules are formed when electrons are shared between two atoms. Each pair of shared electrons is
a covalent bond. Sharing may be expected in chemical combinations where the tendency to form simple ions
by the loss or gain of electrons is not strong. The shared pair of electrons may involve one electron from each
atom, or both electrons may come from one atom and none from the other atom. The tendency is for each atom
to complete its valence shell of electrons—the shell containing the outermost electrons surrounding the nucleus
of an atom. Although a complete valence shell consists of 2 electrons for hydrogen and 8 electrons for all other
atoms, there are many examples in which more or less than eight electrons are involved. In counting electrons
in the outermost shell of an atom, both electrons in a shared pair are counted for each of the two bonded atoms.
When a covalent molecule is formed, the electrons of the participating atoms must undergo adjustments in their
spatial arrangements, in order for the compound as a whole to achieve the most stable condition (minimum
energy state for the molecule). According to VSEPR (Valence Shell Electron Pair Repulsion) Theory, the
molecular geometry can be predicted by arranging electron pairs such that the repulsions between electron pairs
are minimized.
For example, to arrange four electron pairs around an atom, a tetrahedral configuration is
created. The resulting geometric figure uses all three dimensions of space (x, y, z) as does a
tetrahedron.
In arranging three electron pairs around an atom, a trigonal coplanar
configuration, which uses only two dimensions (say x and y) is created.
The z dimension is not needed for this flat triangular-shaped figure.
Tetrahedral and trigonal coplanar are two examples of the five common electron pair shapes, which are
summarized in Table 10-1 (see page 17).
By predicting how the electrons are positioned about each central atom (the electron pair shape), we can deduce
the geometric arrangement of the atoms comprising the structure (the molecular shape).
03-111-10-1
STRUCTURAL INFORMATION FROM LEWIS DOT DIAGRAMS
For covalent structures a verbal distinction is made between a central atom (any atom which is bonded
covalently to two or more other atoms) and the outer atoms (called ligands)which are covalently bonded to that
central atom. In organic chemistry the outer atoms are usually called substituents.
Lewis dot diagrams (also called electron dot diagrams) are one method of representing connectivity within a
molecule or complex ion. In these diagrams, atoms are represented by their atomic symbols surrounded by dots
for their valence electrons. Guidelines for interpreting Lewis dot diagrams are given below.
Example: Interpret the electron dot diagram for cyanic acid,
(
H
O
C
N
)
In this molecule the O is a central atom; it is bonded to two ligands (the H atom, and the C within the
CN cluster of atoms).
Likewise, the C is a central atom; it is bonded to two ligands also (the N atom and the O within the OH
cluster of atoms).
Note that every central atom has its own set of ligands.
Electron dot descriptions show all of the valence electrons in a molecule or complex ion. Only valence
electrons are depicted. Each depicted electron pair then can be classified into one of three categories:
• a lone pair ── a pair of electrons that belongs to one atom only and is not used for bonding; or
• a sigma-bond (σ-bond) ── the "first" pair of electrons shared between a pair of atoms; or
• a pi-bond (π-bond) ── the "second" or "third" pair of electrons shared between those same
specified atoms.
In HOCN depicted above, the electron pairs are classified as follows:
The O atom has two lone pairs; the N atom has one lone pair.
There are a total of three σ-bonds in the molecule ── there is one σ-bond between H and O,
one between O and C, and one between C and N.
The other two pairs of electrons between C and N are two π-bonds. Note that the given dot
picture depicts these π-bonds explicitly; that is to say, the three pairs of electrons between C
and N explicitly show a triple bond, interpretable as one σ-bond and two π-bonds.
For covalent substances, the σ-bonds and lone pairs around a central atom are always explicitly and completely
depicted in a dot picture. π-bonds, in contrast, may be awkwardly or incompletely represented because for
many substances π-bonds are delocalized ── i.e., the π-bond spreads over more than two atoms. The difficulty
of explicitly depicting these π-bonds is an inherent limitation of the electron dot description. Various
techniques have been devised to accommodate this limitation, including the writing of "resonance" structures (a
topic which is addressed later in this Assignment). Since the presence or absence of π-bonding does NOT
influence (in any crucial way) the predictions of geometrical shape, π-bonds can generally be ignored in the
deductive procedure about shape.
Such a detailed analysis of the electron dot diagram for a covalent substance enables one to deduce the
molecular shape by the following procedure.
03-111-10-2
STEP-BY-STEP PROCEDURE FOR PREDICTING THE GEOMETRY OF COVALENT SUBSTANCES.
(In Assignment 10 the electron dot description for the molecule or complex will be provided as given data. The
structure analysis proceeds as follows, beginning with this dot picture.)
a) For any central atom, identify the total number of electron pairs which surround the central atom ── count
only the σ-bonding pairs and lone pairs; ignore any π-bonding pairs.
b) Draw a diagram to show how these electron pairs should arrange geometrically in space around the central
atom. The arrangement in space of the total number of electron pairs is called the electron pair shape.
(See Table 10-1 and the accompanying page of suggestions about drawing these sketches. NOTE: The
geometrical shapes and names in Table 10-1 need to be committed to memory.)
Give the name for the diagrammed electron pair shape.
c) Construct a second diagram to illustrate the three-dimensional arrangement of the atoms in the molecule.
This molecular shape is determined as follows:
Start with an exact duplicate of the electron pair shape for the central atom; then attach all the ligands at
appropriate positions. Place the ligands to show that they are sharing the electron pairs with the central
atom ── according to the following criteria:
If all electron pairs around the central atom are shared pairs, the molecular shape and the
electron pair shape are the same.
If the central atom contains any lone pairs, designate the lone pairs to be at positions that
minimize as much as possible any 90° interactions between lone pairs, because lone pairs repel
other lone pairs in a rather space-dominating way.
The possible molecular shapes and their corresponding names are shown in Table 10-2.
(NOTE: These shapes and names must be committed to memory also.)
If the entity you are depicting is a complex ion, show the ionic charge in the completed diagram.
Give the name for the molecular shape, in accord with Table 10-2. The molecular shape refers to the
arrangement of the atom centers (as bonded in the completed diagram) and does NOT describe any lone
pair locations.
d) Judge how many π-bonds, if any, are occurring (or should occur) in the substance.
For most substances, any π-bonds will be revealed explicitly by the depictions of double or triple bonds in
the given dot-picture. (See the HOCN example, above.)
Certain substances with delocalized π-bond(s), however, may or may not have their π-bonds depicted
explicitly in the given dot-picture. For this situation, the necessity of the π-bond's presence must be
deduced by logic. (See the nitrite ion example below.) The general guideline is: If the central atom in the
dot diagram is shown with less than an octet of electrons, then the ligands (collectively) will provide—out
of their own octets—enough π-bonds to the central atom to complete the octet.
The effectiveness of π-bonding is greatest with 2nd row elements such as C, N, and O.
03-111-10-3
It is important to recognize, in the EXAMPLES below, that each substance is held together entirely by covalent
bonds, with neighboring atoms sharing the same pairs of electrons. This sharing is quite unlike the ionic
bonding which will be discussed in Assignment 20. A covalent bond succeeds because the negatively charged
electron pair is attracted electrostatically by the positively charged nuclei of the two neighboring atoms, thereby
drawing the two atoms close together. The shared electrons are located in-between the two nuclei, thereby
holding the two nuclei in continuing proximity, the electrons acting as the link between the nuclei. Electron dot
descriptions show the shared pair as a pair of dots positioned between the elements' symbols, or alternatively, as
a solid line connecting the symbols.
Thus,
H
O
C
N
may also be written as
H
O
C
N
EXAMPLES
(Note that on the quiz for this problem assignment you will be provided with the dot
picture and be expected to deduce and provide the following information from the dot picture.)
1) Given information:
a) O is the central atom.
Electron pairs about O?
H
O
H
(water molecule, H2O)
Four pairs total (2 σ-pairs, 2 lone pairs)
b) Electron pair shape?
Tetrahedral shape
c) Molecular shape?
Angular shape for the molecule (109o predicted for ∠H-O-H)
(Lone pair/lone pair repulsion causes actual angle to be 104o.)
(On two of the electron pairs, H atoms are attached; the other
two pairs remain as lone pairs. Choice of position for the lone
pairs here is arbitrary, since all 4 locations are equivalent
geometrically. When describing molecular shape, ignore the
lone pairs and describe the rest of the molecule.)
d) π-bonding?
None (the oxygen center already has an octet).
H
2) Given information:
a) N is central atom.
Electron pairs around N?
b) Electron pair shape?
H
N
H
(ammonia molecule, NH3)
Four pairs total (3 σ-pairs, 1 lone pair)
Tetrahedral shape
c) Molecular shape?
(In naming the shape, ignore the lone pair.)
Pyramidal shape
(109o H-N-H angles predicted; but actually 107o)
d) π-bonding?
None
03-111-10-4
H
3) Given information:
H
N
(ammonium ion, NH4+)
H
H
a) N is the central atom.
Electron pairs about N?
Four pairs (all four shared)
b) Electron pair shape?
c) Molecular shape?
Tetrahedral shape
Tetrahedral (Note that the electron pair shape and the molecular
shape coincide here.)
(Note that the complex ion's overall charge, 1+, needs to appear on the
complete, three-dimensional diagram.)
d) π-bonding?
4)
None
(sulfite ion, SO32-)
Given information:
a) S is the central atom.
Electron pairs about S?
Four pairs (three σ-pairs, one lone pair)
b) Electron pair shape?
Tetrahedral shape
c) Molecular shape?
d) π-bonding?
Pyramidal
(Choice of location for lone pair is arbitrary.)
None
03-111-10-5
5) Given information:
a) C is the central atom.
Electron pairs about C?
O
C
O
(molecule of carbon dioxide, CO2)
A total of four pairs are being shared to only two outer atoms ── in
only two directions out from the central atom. By Step-by-Step
Procedure the electron pairs around C are classified as: no lone pairs,
two σ-bonds, and two π-bonds. Recall that multiple shared pairs
(π-bonds) between two atoms are ignored in predictions about
shape.
Since no lone pairs are present, the two σ-pairs go opposite each other, say along the x axis. (The πbonds accompany the σ-bonds without setting new bonding directions.)
b) Electron pair shape?
Linear.
c) Molecular shape?
Linear (see above)
or, as a more precise diagram,
since the dot picture shows a double bond between each O and C.
(Each double bond is one σ-bond and one π-bond.)
d) π-bonding?
Two π-bonds must occur in the actual molecule
(Regardless which diagram is used, a chemist understands that two
π-bonds will be present to satisfy carbon's octet.)
(monohydrogenperoxide ion, HO2−)
6) Given information:
a) Central atom? O
Total central pairs?
Four pairs
b) Electron pair shape?
Tetrahedral shape
Angular shape (109o predicted for ∠HOO)
c) Molecular shape?
(Actual ∠HOO for the bond is
approximately 100o)
d) π-bonding?
None
03-111-10-6
7) Given information:
(propanal,
CH3CH2COH)
a) There are three C central atoms, which will be distinguished by using the labels: C1 ── C2 ── C3 .
b) Electron pairs about C1 ?
Electron pair shape?
b) Electron pairs about C2 ?
Electron pair shape?
b) Electron pairs about C3 ?
For C1
Four pairs (four σ-bonds)
Tetrahedral shape
For C2
Four pairs (four σ-bonds)
Tetrahedral shape
For C3
Four total, but only three bonding directions
(i.e., there are three σ-bonds and one π-bond).
Electron pair shape?
Trigonal coplanar shape (the π-bond is ignored);
one π-bond present (explicitly depicted),
completing the bonding between C3 and the O
substituent.
c) Molecular shapes?
Tetrahedral around C1
Tetrahedral around C2
Trigonal coplanar around C3
A REMINDER: As you work problems and draw structures, be VERY CAREFUL to mark all electron dots
darkly and precisely and to indicate clearly the electrical charge of each component and structure.
03-111-10-7
RESONANCE STRUCTURES
In the previous examples, we have worked with structures in which the π-bonding, if present, was more or
less localized. That is, the electron pair in the π-bond spends most of its time associated with only two
nuclei. There are many situations in which bonding can be delocalized; i.e., that bonding electron pairs can
be distributed over more than two nuclei. There are two fundamentally different ways which chemists use to
represent delocalized bonding: resonance structures derived from valence bond theory; and molecular
orbitals derived from molecular orbital theory. Both of these theories are based in quantum mechanics, and
both of these theories will be treated somewhat more extensively later in this course. In this Assignment, we
will show you how to use resonance structures to represent delocalized π-bonds, and we will show you how
to recognize when such treatment is appropriate.
For example, let's work the procedure for the nitrite ion, NO2−:
O NO
N is at the center and is surrounded by three electron pairs (one lone pair and two σ-bonds). The electron
pair shape is trigonal coplanar, and the molecular shape is angular.
N
trigonal coplanar
electron pair shape
N
O
O
angular molecular
shape
We are, however, not done with nitrite yet. There are two issues to consider when deciding on the need for
resonance structures in this course: 1) Do any atoms (besides hydrogen) in the structure have fewer electrons
than an octet? 2) Is there a minimum of "charge separation" in the structure? (That is, Coulomb’s law will
disfavor large separation of positive and negative charge. This charge separation can be assessed by
calculating something called formal charge.)
To answer question #1, we look at a given atom in a structure. Count the electrons in the lone pairs on that
atom and all of the electrons in bonds connected to that atom, and see if the total is less than eight.
To answer question #2, we have to calculate the formal charge of each atom. The procedure in your
textbook (Oxtoby, Gillis, and Nachtrieb) for calculating formal charge (FC) is:
FC = group number − (# of lone pair electrons) − 1/2(# of bonding pair electrons)
NOTE! The sum of the formal charges of all of the atoms in a molecule or ion must equal the total
charge of the molecule or ion. This allows you to conveniently check if you have correctly assigned the
FC’s of the entire structure.
So let's work these procedures for nitrite: Issue #1 (octet rule): Although the oxygen atoms each have an
octet, the nitrogen atom does not. Issue #2 (formal charge): Consider the formal charges of the atoms in the
nitrite ion as we have written it so far:
03-111-10-8
this vertical line represents a lone pair
+1
I
N
O
-1
O
-1
The formal charge on N is +1 because the group number of nitrogen from the periodic table is 5, and:
FC = 5 − 2 lone pair electrons − 1/2 (4 electrons total in two σ-bonds to N) = +1
The formal charge on each O is −1 because the group number of oxygen from the periodic table is 6, and:
FC = 6 − 6 lone pair electrons − 1/2 (2 electrons in the σ-bond drawn to O) = −1
Notice that the total of the formal charges equals the total charge of the ion
(+1 −1 −1 = −1). If we share one of the oxygen's lone pairs with the nitrogen, the following structure arises:
0
II
N
O
-1
O
0
Notice that we now have a π-bond between the nitrogen and the oxygen on the right. This π-bonding pair of
electrons was a lone pair on the right hand oxygen in the previous structures.
The formal charge on N is now 0 because the group number of nitrogen from the periodic table is 5, and:
FC = 5 − 2 lone pair electrons − 1/2 (6 electrons total in two σ-bonds and one π-bond to N) = 0
The formal charge on the left O is still −1 because the group number of oxygen from the periodic table is 6,
and:
FC = 6 − 6 lone pair electrons − 1/2 (2 electrons in the σ-bond drawn to O) = −1
The formal charge on the right O is now 0 because the group number of oxygen from the periodic table is 6,
and:
FC = 6 − 4 lone pair electrons − 1/2 (4 electrons from the one σ-bond and one π-bond drawn to O)= 0
Observe that the total of the formal charges still equals the overall charge of the ion (0 −1 +0 = −1).
Structure II is preferable to structure I, because there is less separation of formal positive and negative
charge. However, structure II is not unique---there is no reason why we could not make the double bond
between the nitrogen and the oxygen on the left, and this gives us what we call resonance structures:
0
0
N
O
0
O
-1
N
O
-1
O
0
Neither of these two resonance structures is real, and neither is intended to show the bond-structure of the
nitrite ion. The nitrite ion is an average of these two structures; i.e., a structure which has one σ-bond and
0.5 π-bonds between each oxygen and nitrogen. The π-bonding here is delocalized, because the one π-bond
is shared by three atoms, rather than two.
03-111-10-9
Now, let's work the procedure for the nitrate ion, NO3−:
O
O
NO
N is at the center and is surrounded by three electron pairs (three σ-bonds). The electron pair shape is
trigonal coplanar, and the molecular shape is also trigonal coplanar.
O
N
N
O
trigonal coplanar
electron pair shape
O
trigonal coplanar
molecular shape
We are, however, not done with nitrate yet. Although the oxygen atoms each have an octet, the nitrogen
atom does not. Furthermore, consider the formal charges of the atoms in the nitrate ion as we have written it
so far:
O
-1
+2
N
O
-1
O
-1
The formal charge on N is +2 because the group number of nitrogen from the periodic table is 5, and:
FC = 5 − 1/2(6 electrons from each of the three σ-bonds drawn to N) = +2
The formal charge on each O is −1 because the group number of oxygen from the periodic table is 6, and:
FC = 6 − 6 lone pair electrons − 1/2 (2 electrons in the σ-bond drawn to O) = −1
Notice that the total of the formal charges equals the total charge of the ion (+2 −1−1−1 = −1). If we share
one of the oxygen's lone pairs with the nitrogen, the following structure arises:
O
-1
+1
N
O
-1
O
0
We now have a π-bond between the nitrogen and the oxygen on the right. This π-bonding pair of electrons
was a lone pair on the right hand oxygen in the previous structures. Note that the formal charge of the
nitrogen is now +1, the formal charge of the right hand oxygen is now zero, and the formal charge of the
other oxygens is still −1. This structure is preferable to the structure above, because there is less separation
of formal positive and negative charge. However, this structure is not unique---there is no reason why we
could not make the double bond between the nitrogen and either of the other oxygens:
03-111-10-10
O
-1
O
+1
O
N
N
O
0
O
-1
-1
+1
+1
N
O
-1
0
O
-1
O
0
O
-1
None of these three resonance structures is real, and none is intended to show the bond-structure of the
nitrate ion. The nitrate ion is an average of these three structures; i.e., a structure which has one σ-bond and
0.33 π-bonds between each oxygen and nitrogen. The π-bonding here is delocalized, because the one π-bond
is shared by four atoms, rather than two.
Now, let's see how these procedures can give us insight into chemical reactivity as well as chemical
structures by working through boron trifluoride---a neutral molecule.
F
F BF
B is at the center and is surrounded by three electron pairs (three σ-bonds). The electron pair shape is
trigonal coplanar, and the molecular shape is also trigonal coplanar.
F
B
B
F
trigonal coplanar
electron pair shape
F
trigonal coplanar
molecular shape
We are, however, not done with BF3 yet. Although the fluorine atoms each have an octet, the boron atom
does not. Also, our formal charge situation is different than it was for nitrite or nitrate:
0
F
0
III
B
F
0
F
0
The formal charge on B is 0 because the group number of boron from the periodic table is 3, and:
FC = 3 − 1/2(6 electrons from each of the three σ-bonds drawn to B) = 0
The formal charge on each F is 0 because the group number of fluorine from the periodic table is 7, and:
FC = 7 − 6 lone pair electrons − 1/2 (2 electrons in the σ-bond drawn to F) = 0
Notice that the total of the formal charges equals the total charge of the molecule. If we share one of the
fluorine's lone pairs with the boron, the following three resonance structures arise:
03-111-10-11
0
F
-1
F
+1
-1
B
B
F
0
0
F
-1
B
F
0
+1
F
F
0
F
+1
F
0
We now have a π-bond shared between the boron and the three fluorines. Note that the formal charge of the
boron is now -1, the formal charge of the singly bonded fluorines is still zero, and the formal charge of the
doubly bonded fluorine is +1.
The formal charge on B is −1 because the group number of boron from the periodic table is 3, and:
FC = 3 − 1/2 (8 electrons from the four bonds drawn to B) = −1
The formal charge on each singly bonded F is 0 because the group number of fluorine from the periodic table
is 7, and:
FC = 7 − 6 lone pair electrons − 1/2 (2 electrons in the σ-bond drawn to F) = 0
The formal charge on each doubly bonded F is +1 because the group number of fluorine from the periodic
table is 7, and:
FC = 7 − 4 lone pair electrons − 1/2 (4 electrons in the two bonds drawn to F) = +1
These resonance structures and structure III all suffer discomfort!!!! Structure III has a boron atom that
does not have an octet, which is bad, and the resonance structures with the shared π-bond have created a
separation of charge, which is bad!!! Also, as you are learning in class, boron is less electronegative than
fluorine, and the resonance structures are putting the positive charge on the more electronegative fluorine and
the negative charge on the less electronegative boron; this is also bad. Boron trifluoride is going to be a
reactive molecule under many conditions, because you cannot write a "happy" structure for it!
HYPERVALENCE or VALENCE SHELL EXPANSION
Elements in columns III through VIII of the periodic table that are in the third row or below may be
surrounded by more than an octet of electrons. Elements that have more than an octet of electrons are said to
be hypervalent or to have undergone valence shell expansion. To accurately predict the geometry of these
species you must use the VSEPR rules (valence shell electron pair repulsion).
You have already been using the VSEPR rules in this problem set, since the whole premise of VSEPR rules
is that electrons, all having negative charge, will Coulombically repel one another and separate themselves as
much as possible. Geometrically, the greatest separation one can achieve between two electron pairs is 180o,
resulting in a linear electron pair shape. The greatest separation one can achieve between three electron pairs
is 120o, resulting in a trigonal coplanar (usually called just trigonal planar) electron pair shape. The greatest
separation one can achieve between four electron pairs is 109.4o, resulting in a tetrahedral electron pair
shape.
For hypervalent compounds, the pertinent electron pair shapes from which all molecular shapes are derived
are the trigonal bipyramid and the octahedron (see Table 10-2). Where there is a possibility for more than
one molecular shape based on the appropriate electron pair shape, the proper geometry can be assessed as
follows:
03-111-10-12
1) Minimize the number of lone pair-lone pair (lp-lp) interactions that have an angle less than or equal to
90o. If one structure has the fewest lp-lp interactions, it is the correct structure, and you are done with
your analysis.
2) Minimize the number of lone pair-bonding pair (lp-bp) interactions that have an angle less than or equal
to 90o. If two or more structures have the same minimal number of lp-lp interactions, the correct
structure is the one with the fewest number of lb-bp interactions. You are done with your analysis.
3) If two or more structures have the same minimum number of lp-lp interactions and the same minimum
number of lp-bp interactions, then the correct structure will be the one that has the fewest number of
bonding pair-bonding pair (bp-bp) interactions.
The following examples will illustrate the meaning of this procedure.
You are given the dot structure for a molecule of phosphorus pentafluoride, PF5:
F
F
F
P
F
F
P is at the center and there are five σ-bonds. The electron pair shape will be trigonal bipyramidal and the
molecular shape will be trigonal bipyramidal. There is no need for π-bonding, because all atoms have at
least an octet.
F
P
F
F
P
F
F
trigonal bipyramidal
electron pair shape
trigonal bipyramidal
molecular shape
Let’s see if the trigonal bipyramidal molecular shape satisfies our VSEPR rules by comparing it to a square
pyramidal shape, another conceivable geometry for a central atom with five σ-bonds.
F
F
P
F
F
F
F
F
F
trigonal bipyramidal
molecular shape
P
F
F
square pyramidal
molecular shape
1) Since neither phosphorus atom has lone pairs, there cannot be any lp-lp interactions.
2) Since neither phosphorus atom has lone pairs, there cannot be any lp-bp interactions.
3) Since the trigonal bipyramid has six bp-bp interactions and the square pyramid has eight bp-bp
interactions, the trigonal bipyramid has the lowest energy and is preferred.
03-111-10-13
To see the bp-bp interactions that are at 90o or less, look below:
F
F
F
F
F
P
F
P
F
F
F
F
Here, the double headed arrows are indicating 90o bp-bp interactions.
Consider now a molecule of iodine trichloride, ICl3. You are given the following dot structure:
Cl
I
Cl
Cl
I is at the center and it has three σ-bonds and two lone pairs. The electron pair shape will be trigonal
bipyramidal and the molecular shape will be T-shaped. There is no need for π-bonding, because all atoms
have at least an octet.
Cl
I
Cl
I
Cl
trigonal bipyramidal
electron pair shape
T-shaped
molecular shape
Let’s see if the T-shaped molecular shape satisfies our VSEPR rules by comparing it to two other shapes.
Cl
Cl
I
Cl
Cl
I
Cl
Cl
Cl
I
Cl
Cl
1) The I in the left structure has no 90o lp-lp interactions (the lone pairs in this structure are 120o apart). The
I in the middle structure has no 90o lp-lp interactions (the lone pairs in this structure are 180o apart). The I in
the structure on the right DOES have one 90o lp-lp interaction---this structure is immediately disqualified,
because it has more lp-lp interactions than the other two.
03-111-10-14
2) The I in the left structure has four 90o lp-bp interactions. The I in the middle structure has six 90o lp-bp
interactions---now this structure is disqualified because it has more lp-bp interactions than the structure on
the left. Although it is not relevant to solving the problem, the structure on the right has three 90o lp-bp
interactions, but remember that it was already disqualified in step one above. Thus, the left structure of ICl3
is the correct one.
3) Although the correct structure (the one on the left) has already been determined, we will complete the
analysis of the bp-bp interactions for the practice. The left structure has two bp-bp interactions. The middle
structure has zero bp-bp interactions. The structure on the right has two bp-bp interactions.
Finally, let’s examine xenon tetrafluoride, XeF4, for which you are given the following dot structure:
F
F
F
Xe
F
Xe is at the center and it has four σ-bonds and two lone pairs. The electron pair shape will be octahedral and
the molecular shape will be square coplanar (or simply square planar). There is no need for π-bonding,
because all atoms have at least an octet.
F
Xe
F
Xe
F
F
square planar
molecular shape
octahedral electron
pair shape
Let’s see if the square planar molecular shape satisfies our VSEPR rules by comparing it to the other possible
shape.
F
F
F
Xe
F
F
F
F
Xe
F
1) The Xe in the square planar structure has no 90o lp-lp interactions (the lone pairs in this structure are 180o
apart). The Xe in the structure on the right DOES have one 90o lp-lp interaction---this structure is
immediately disqualified, because it has more lp-lp interactions than the other.
2). Although the correct structure (the square planar one on the left) has already been determined, we will
complete the analysis of the lp-bp interactions for the practice. The Xe in the left structure has eight 90o lpbp interactions. The Xe in the structure on the right has six 90o lp-bp interactions.
3) Although the correct structure (the square planar one on the left) was determined in step one, we will also
complete the analysis of the bp-bp interactions for the practice. The left structure left has four bp-bp
interactions. The structure on the right has five bp-bp interactions.
03-111-10-15
In mastering the diagrams in Tables 10-1 and 10-2 on the next two pages, it is important to memorize the name
associated with the shape and to understand the significance of the corresponding diagram. Each diagram is an attempt
to depict the three-dimensional arrangement of the electron pairs (specifically, the lone pairs and σ-pairs). Visualizing
three-dimensional structures is often quite difficult. Perhaps the following may help. The plane of the paper is usually
considered the xz plane (except in the case of 6 pairs). The central atom is represented here by A.
03-111-10-16
TABLE 10 - 1
( ELECTRON PAIR ARRANGEMENTS ABOUT A CENTRAL ATOM )
NOTE: In the diagrams, each electron pair is represented by a line:
a solid line lies in the plane of the paper;
a dashed line extends in back of the plane of the paper;
a wedge line extends in front of the plane of the paper.
The letter A represents the central atom of the molecule.
Number of
electron pairs**
(σ-bonds and lone)
Electron pair
shape
1
(Not appropriate)
2
Linear
3
Trigonal coplanar
4
Tetrahedral
5
Trigonal bipyramidal
6
Octahedral
**NOTE:
Diagram
of the
electron pair shape
In those cases where multiple bonds exist, ALL the electrons lying between any two
atoms are considered as ONE electron pair to determine the electron pair shape ──
since all of the electrons in the multiple bond are used to join only two atoms, and
since these two atoms occupy a SINGLE bonding position or bonding direction (e.g.,
see the carbon dioxide example, in which C in CO2 is considered as having 2 electron
pairs for purposes of designing the electron pair shape).
03-111-10-17
TABLE 10 ─ 2
Total of
lone pairs
and σ-pairs
( LIST OF MOLECULAR SHAPES ) Note: In the following diagrams, A represents
the central atom; Z represents each ligand,
bonded to the center by an electron pair.
Electron Number of
pr. shape
σ-bonds
Molecular
Shape
1
---
1
Linear
2
Linear
1 or 2
Linear
1
Linear
2
Angular
3
Trigonal
coplanar
1
Linear
2
Angular
3
Pyramidal
4
Tetrahedral
3
4
Trigonal
coplanar
(120o angles)
Tetrahedral
(109o angles)
5
Trigonal
bipyramidal
(angles are 90o,120o,180o)
6
Diagram of molecular shape
**In cases with five electron pairs, the lone pairs are placed
preferentially in equatorial positions (the 3 in-plane positions).
1
**Linear
2
**Linear
3
**T-shaped
4
**Teeter (see-saw)
(rotate to show teeter)
5
Trigonal bipyramidal
Octahedral
(90o,180o angles)
***In cases with six electron pairs, the lone pairs are placed
preferentially opposite from one another insofar as possible.
1
Linear
2
***Linear
3
***T-shaped
4
***Square coplanar
5
Square pyramidal
6
Octahedral
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