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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).

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NH O C

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, ( ) 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.

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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.

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NH O C

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, may also be written as 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: (water molecule, H2O) a) O is the central atom. Electron pairs about O? 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). 2) Given information: (ammonia molecule, NH3) a) N is central atom. Electron pairs around N? Four pairs total (3 σ-pairs, 1 lone pair) b) Electron pair shape? 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

NH O C

H O H

H N H

H

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3) Given information: (ammonium ion, NH4+)

a) N is the central atom. Electron pairs about N? Four pairs (all four shared) b) Electron pair shape? Tetrahedral shape c) Molecular 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? None 4) Given information: (sulfite ion, SO3

2-) 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? Pyramidal (Choice of location for lone pair is arbitrary.) d) π-bonding? None

H N H

H

H

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5) Given information: (molecule of carbon dioxide, CO2) a) C is the central atom. Electron pairs about C? 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 π-

bonds 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.) 6) Given information: (monohydrogenperoxide ion, HO2

−) a) Central atom? O Total central pairs? Four pairs b) Electron pair shape? Tetrahedral shape c) Molecular shape? Angular shape (109o predicted for ∠HOO) (Actual ∠HOO for the bond is approximately 100o) d) π-bonding? None

O C O

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7) Given information: (propanal, CH3CH2COH) a) There are three C central atoms, which will be distinguished by using the labels: C1 ── C2 ── C3 . For C1 b) Electron pairs about C1 ? Four pairs (four σ-bonds) Electron pair shape? Tetrahedral shape For C2 b) Electron pairs about C2 ? Four pairs (four σ-bonds) Electron pair shape? Tetrahedral shape For C3 b) Electron pairs about 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.

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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

−:

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.

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:

O ON

N NO O

trigonal coplanarelectron pair shape angular molecular

shape

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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:

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:

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.

NO O-1 -1

+1this vertical line represents a lone pair

I

NO O-1 0

0II

NO O

-10

0N

O O-1 0

0

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Now, let's work the procedure for the nitrate ion, NO3−:

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.

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:

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:

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:

O ONO

N

O

NO O

trigonal coplanarelectron pair shape trigonal coplanar

molecular shape

O

NO O-1 -1

+2

-1

O

NO O-1 0

+1

-1

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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.

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.

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:

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:

O

NO O-1 0

+1

-1O

NO O-1 -1

+1

0O

NO O0 -1

+1

-1

F FBF

B

F

BF F

trigonal coplanarelectron pair shape trigonal coplanar

molecular shape

F

BF F

III

0 0

0

0

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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:

F

BF F0 +1

-1

0F

BF F0 0

-1

+1F

BF F+1 0

-1

0

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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:

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.

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.

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.

PF

FF

F F

P F PF

F

F

Ftrigonal bipyramidalelectron pair shape trigonal bipyramidal

molecular shape

F PF

F

F

Ftrigonal bipyramidalmolecular shape

FPF F

F

F

square pyramidalmolecular shape

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To see the bp-bp interactions that are at 90o or less, look below:

Consider now a molecule of iodine trichloride, ICl3. You are given the following dot structure:

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.

Let’s see if the T-shaped molecular shape satisfies our VSEPR rules by comparing it to two other shapes.

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.

F P

F

F

F

FF

PF F

F

F

Here, the double headed arrows are indicating 90o bp-bp interactions.

ICl

Cl Cl

I Cl I

Cl

Cltrigonal bipyramidalelectron pair shape T-shaped

molecular shape

Cl I

Cl

Cl

Cl ICl

Cl Cl ICl

Cl

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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:

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.

Let’s see if the square planar molecular shape satisfies our VSEPR rules by comparing it to the other possible shape.

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 lp-bp 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.

XeF

F

FF

XeF

XeF F

F

octahedral electronpair shape

square planarmolecular shape

FXe

F F

F FXe

FF

F

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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.

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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) 1

2

3

4

5

6

Electron pair shape (Not appropriate)

Linear

Trigonal coplanar

Tetrahedral

Trigonal bipyramidal

Octahedral

Diagram of the electron pair shape

**NOTE: 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).

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TABLE 10 ─ 2 ( LIST OF MOLECULAR SHAPES ) Total of lone pairs Electron Number of Molecular and σ-pairs pr. shape σ-bonds Shape Diagram of molecular shape 1 --- 1 Linear 2 Linear 1 or 2 Linear 3 Trigonal 1 Linear coplanar (120o angles) 2 Angular 3 Trigonal coplanar 4 Tetrahedral 1 Linear (109o angles) 2 Angular 3 Pyramidal 4 Tetrahedral 5 Trigonal **In cases with five electron pairs, the lone pairs are placed bipyramidal preferentially in equatorial positions (the 3 in-plane positions). (angles are 90o,120o,180o) 1 **Linear 2 **Linear 3 **T-shaped 4 **Teeter (see-saw) (rotate to show teeter) 5 Trigonal bipyramidal 6 Octahedral ***In cases with six electron pairs, the lone pairs are placed (90o,180o angles) preferentially opposite from one another insofar as possible. 1 Linear 2 ***Linear 3 ***T-shaped 4 ***Square coplanar 5 Square pyramidal 6 Octahedral �

Note: In the following diagrams, A represents the central atom; Z represents each ligand, bonded to the center by an electron pair.