Curriculum Module Chemical Bonding

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AP® Chemistry Curriculum Module:

Chemical Bonding

2010

Curriculum Module

The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the College Board is composed of more than 5,700 schools, colleges, universities and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,800 colleges through major programs and services in college readiness, college admission, guidance, assessment, financial aid and enrollment. Among its widely recognized programs are the SAT®, the PSAT/NMSQT®, the Advanced Placement Program® (AP®), SpringBoard® and ACCUPLACER®. The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities and concerns. Visit the College Board on the Web: www.collegeboard.com. Pages 28–34: Images from Chemistry: The Molecular Nature of Matter and Change, 4th edition, by Martin S. Silberberg. © 2006 by The McGraw-Hill Companies, Inc. Used by permission. Page 45: Image from BROWN, THEODORE E.; LEMAY, H. EUGENE; BURSTEN, BRUCE E, CHEMISTRY: THE CENTRAL SCIENCE, 10th edition,©2006. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey. Page 49: Image from WHITTEN. *EMTX: GENERAL CHEMISTRY 6E, 6E. © 2001 Brooks/Cole, a part of Cengage Learning, Inc. Reproduced by permission. www.cengage.com/permissions

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Contents I. Introduction. .................................................................... 4 II. Lewis Structures.............................................................. 6 III. Valence Shell Electron Pair Repulsion Theory. ............17 IV. Valence Bond Theory and the Hybridization Model.... 29 V. Bond and Molecular Polarity..........................................37 VI. Intermolecular Attractions. .......................................... 44 VII. Further Bonding Issues — Beyond AP®........................ 58 About the Authors..................................................................... 62

Chemical Bonding

I. Introduction Arden P. Zipp This AP® Chemistry Curriculum Module will address several aspects of chemical bonding, which include: • Writing and using Lewis structures • Predicting molecular shapes with the valence shell electron pair repulsion (VSEPR) theory • Describing molecules and orbitals with the valence bond (VB) theory • Predicting the polarity of bonds and molecules • Describing the types and effects of intermolecular forces Each of these topics covers material that has been examined on past AP Chemistry Exams and, no doubt, will be on future ones. Individuals with a great deal of experience with AP Chemistry, including teachers, readers, members of the AP Chemistry Development Committee and contributors to previous resources for this course, wrote the selections. Before exploring the topics mentioned above, it should be remembered that chemistry is based on experiment, and even the most abstract aspects of chemical bonding have been developed to account for behavior observed in nature. This behavior includes the structures of individual molecules and the forces between them. Information about the structures of molecules can be gathered by a variety of spectroscopic methods, especially infrared (IR) spectroscopy; by diffraction methods; or by measurements of dipole moments. IR spectra can provide insight into the strengths of bonds between atoms, since the frequencies of IR vibrations increase with bond strength (but also depend on atomic masses). Diffraction methods, of which X-ray diffraction is the most common, can determine the positions of atoms within molecules. Atomic positions yield conclusions about bond lengths (and bond strengths, which are inversely related to bond lengths) as well as about bond angles. Dipole moment measurements can be used to draw conclusions about the polarity of molecules and the distribution of electrons and/or the arrangement of atoms within them. A relatively new technique called photoelectron spectroscopy (PES) can actually provide information about the energies of electrons within molecules. Conclusions about the forces between molecules are usually obtained from a consideration of the physical properties they exhibit in bulk, such as melting points, boiling points, vapor pressures and enthalpy changes associated with these phenomena. Melting points, boiling points and enthalpy changes increase with an increase in intermolecular forces, while the vapor pressure decreases.

4 © 2010 The College Board.

AP Chemistry Curriculum Module: Chemical Bonding

The various sections of this curriculum module present the means to describe and/or predict different features of these experimental results. In the first essay, Adele Mouakad describes how Lewis structures can be used to obtain an estimate of the strengths of the bonds in molecules, since the strength of a bond increases with the number of electron pairs between the constituent atoms. In addition, Lewis structures also provide the basis for VSEPR structures, which offer the means to determine the bond angles of both neutral and ionic species, as set forth in the second essay. The valence bond theory offers an alternate means to determine bond angles (through the geometries of atomic or hybrid orbitals) and bond lengths/strengths (via the formation of multiple bonds — sigma and pi); this is addressed in Marian DeWane’s essay. In the fourth essay, David Hostage discusses the use of electronegativities to obtain bond polarities; and, along with molecular geometries, he outlines the application of these to establish the polarity of a molecule. In the final essay, Valerie Ferguson describes dispersion forces, dipole-dipole interactions and hydrogen bonds, along with their origins and their relative strengths. The authors hope you enjoy reading these selections and find them useful.

5 © 2010 The College Board.

Chemical Bonding

II. Lewis Structures Adele Mouakad

Introduction In 1916, Gilbert N. Lewis, one of the most famous American chemists, formulated the idea that a covalent bond consisted of a shared pair of electrons. His ideas on chemical bonding were expanded upon by Irving Langmuir and became the inspiration for the studies on the nature of the chemical bond by Linus Pauling. Lewis used dots to represent the valence electrons and introduced the concept that when atoms bonded they did so by pairing electrons. He also suggested that one of the driving forces of chemical bonding was the acquisition of eight electrons by each atom in analogy with the very stable noble gases. This is referred to as the octet rule, which is the guiding principle when writing Lewis (electron dot) structures. Although the structures of molecules as written do not imply a sense of the shape of the molecule or ion considered, they can be used in conjunction with the VSEPR theory (described in the next essay in this volume) to determine shapes. The Lewis dot structures for molecules are developed from those of individual atoms and are given for atoms of the elements in the first four periods (excluding the transition elements) in the graphic below.

The structures are written to show the number of electrons present in the highest energy shell (valence) rather than the actual electron configurations in the free atoms. (For example, the two electrons in Be are actually paired, as are two of the three in B and two of the four in C.) All the elements in a given group or family have the same Lewis dot structure because they all have the same number of valence electrons. Lewis structures are used to represent molecules. The covalent bond in the hydrogen molecule, H2, can be represented as follows:

H:H or H – H 6 © 2010 The College Board.

AP Chemistry Curriculum Module: Chemical Bonding

This is called a single covalent bond; a pair of electrons is shared between the two hydrogen atoms. The dash is also used to represent a pair of bonding electrons. Hydrogen cannot follow the octet rule, since it can have only two electrons after sharing, which is the configuration of helium. In some textbooks, this is referred to as the duet rule. In a similar fashion, two fluorine atoms can pair electrons to form an F2 molecule.

F

+

F



F

F

Electron pair shared between the two Fluorine atoms

A line or pair of dots between two atoms represents a bonding pair of electrons, while a pair of dots on the periphery of a molecule is referred to as a nonbonding or lone pair.

Multiple Bonding Some molecules require more than single bonds to provide each atom with the required octet. Examples include the oxygen molecule O2 (12 valence electrons), which requires a double bond.

:Ö::Ö: or :Ö = Ö: Nitrogen (N2) requires a triple bond between the two nitrogen atoms to provide each nitrogen with an octet of electrons.

:N:::N: or :N  N: Multiple bonds are formed primarily by the carbon, nitrogen and oxygen atoms. The rules below can be used to develop Lewis structures for more complex species.

Rules for Writing Lewis Structures 1. Decide on the sequence of atoms. (i) For ABn formulas, the single atom usually is the central atom in the molecule (e.g., in CH4, NH3 and H2O, the central atoms are C, N and O, respectively). H and F are never central and other halogens seldom are. (ii) Compounds with symmetrical formulas often adopt corresponding structures (e.g., the atomic arrangement in H2O2 is H-O-O-H). 7 © 2010 The College Board.

Chemical Bonding

2. Determine the total number of valence electrons by: (i) Adding the valence electrons for all atoms. For example, the CH4 molecule contains eight valence electrons (four from the C and one from each of the 4 H atoms) while NF3 has 26 (five from the N and 3 x 7 for the three Fs). (ii) Adding an extra electron for every -1 charge. For example, the NO3- ion has a total of 24 valence electrons: five from the N and 3 x 6 for the three Os, plus one for the negative charge. (iii) Subtracting an electron for every +1 charge. For example, the NF4+ ion has a total of 32 electrons: five from the N, 4 x 7 for the four Fs, minus one for the positive charge. 3. Draw a single bond between the central atom and each peripheral atom, subtracting two electrons for each bond from the total; for example, the structure for CH4 is:

4. Distribute the remaining electrons in pairs so that each atom has eight, as in the Lewis structure for NF3. (A Lewis structure is not complete without the lone pairs.)

5. If, after step 4, a central atom does not have 8 e-, a lone pair should be changed to a bonding pair to make a multiple bond, as in ethane, C2H4.

6. Limit the number of electrons to eight around any second period element (Li - F). 7. Do not use lone pairs of electrons from halogens to form double bonds with other elements. Halogens cannot donate electrons to double bonds, but halogens can accept them under certain conditions (e.g., in oxyanions such as OClO3-). 8. If appropriate, more than eight electrons can be put around elements from the third or higher periods. For example, phosphorus can use its three unpaired 8 © 2010 The College Board.

AP Chemistry Curriculum Module: Chemical Bonding

electrons to form species such as PCl3, but it can also use all five electrons to form molecules like PCl5, in which there are 10 electrons around the P.

Many other molecules contain central atoms with more than an octet of electrons. The list includes compounds of sulfur, selenium, tellurium, the halogens (except for fluorine) and the noble gases: krypton, xenon and radon.

Resonance Some molecules can be represented by more than one Lewis dot structure. Two (or more) different Lewis dot structures for a given species are called resonance structures or resonance forms. It is important to understand that these resonance forms do not actually exist, even briefly, but are simply the best approximations to the actual molecular structure that can be made within the constraints of the Lewis theory. Resonance forms are useful, however, because the structure of a molecule described by resonance is intermediate between the structures of its resonance forms. For such molecules, the various resonance forms are typically written as connected by double-headed arrows to show their relationship. Ozone, a high energy allotrope of oxygen with the formula O3, provides a good example of resonance. In order to provide each of the three oxygen atoms in ozone (18 electrons) with an octet of electrons (as required by the Lewis theory), two of the oxygen atoms must be connected by a single bond and the other by a double bond. There are, however, two different ways this can be done, depending on where the double bond is placed, as shown in the diagram below.

Experiments indicate that the two bonds in the O3 molecule do not have different properties, as they would if the single and double bonds were fixed in position. Rather, the two bonds are equivalent with properties intermediate between those of a single and a double bond. 9 © 2010 The College Board.

Chemical Bonding

Carbon dioxide, CO2, is another molecule for which a number of resonance forms can be written. First, there is the most familiar form of CO2 in which the two oxygen atoms are bonded to the carbon by two double bonds. In addition, two other resonance forms can be written, each containing a single and a triple bond. Other familiar examples of species that are best described by resonance include the carbonate (CO32-) and the nitrate (NO3-) ions, each of which contains 24 valence electrons with the lone atom at the center of the ion. Three equivalent resonance forms with a double bond and two single bonds can represent each species. As discussed above, the actual structure of each of these ions is intermediate between the extremes represented by the resonance forms.

Formal Charge Formal charge represents the difference between the number of valence electrons an atom possesses in its free state and the number assigned to it in a given Lewis structure. The number of valence electrons assigned to an atom in the bonded state can be found by counting all of the electrons belonging exclusively to that atom (i.e., nonbonding electrons) and one-half of the electrons in the bonds to it. The formal charge is obtained by subtracting the number obtained in this way from the number in the free atom. This is summarized in the following equation:



Formal charge = Number of valence - nonbonding + 1/2 bonding electrons originally electrons electrons



The formal charge concept is a useful one because it may be helpful in determining the relative stabilities of several resonance forms or even in establishing the best arrangement of atoms in molecules for which a central atom is not immediately apparent. The cyanate ion, NCO-, with two reasonable resonance forms (labeled a = cyanate and and b = isocyanate) provides a useful example.

The formal charge for each of the atoms in structure (a) would be: FC: N = 5 e- - [2 nonbond e- + ½ (6 bond e-)] = N = 5 - 5 = -0 FC: C = 4 e- - [ ½ (8 bond e-)] FC: O = 6 e- - [6 nonbond e- + ½ (2 bond e-)]

= C = 4 - 4 = 0 = O = 6 -7 = -1

Similarly, the formal charge for each of the atoms in structure (b) would be: FC: N = 5 e- - [4 nonbond e- + ½ (4 bond e-)] = N = 5 - 6 = -1 10 © 2010 The College Board.

AP Chemistry Curriculum Module: Chemical Bonding

FC: C = 4 e- - [ ½ (8 bond e-)] FC: O = 6 e- - [4 nonbond e- + ½ (4 bond e-)]

= C = 4 - 4 = 0 = O = 6 - 6 = 0

Structure A (Cyanate Ion) would be more stable than Structure B (Isocyanate Ion) because the formal charges have been minimized at 0, 0 and -1. However, Structure A (Cyanate Ion) is also more stable than Structure B (Isocyanate Ion) because the -1 formal charge is located on O, which has a greater attraction for electrons than the N. Now examine the unstable fulminate ion:

The fulminate ion has a similar molecular formula to the cyanate ion even though its structural formula is different. When it is analyzed in terms of formal charge, a determination as to why it is unstable can be made. For Structure A (Cyanate Ion): FC: N = 5 e- - [ 0 nonbond e- + ½ (8 bond e-)]

= 5 – 4 = +1

FC: C = 4 –

= 4 – 6 = -2

[ 4 bond e- + ½ ( 4 bond e-)]

FC: O = 6 – [ 4 non e- + ½ ( 4 bond e-)]

= 6 – 6 = 0

The formal charges are N +1, C - 2 and O 0, with nitrogen having a +1 charge and carbon a –2 charge. In this case the charges are not a minimum and carbon, the least electronegative of the three atoms, has formal charge of -2. In the cyanate ion the formal charges are 0, 0 and -1, with oxygen, the most electronegative atom of the three, being the -1. In evaluating the stabilities of resonance forms, the following rules apply: (1) minimize nonzero formal charges; (2) avoid nonzero formal charges on adjacent atoms; and (3) place negative formal charges on atoms with greater electron attractions and positive formal charges on atoms with lower electron attractions. Determine the formal charges on the atoms in the hypothetical species with the atomic arrangement CON- and use these to explain why this ion is unknown. The reader should discover using the same electron arrangements as in structures (a), (b) and (c), the formal charges are (a) C -2, O +2, N -1; (b) C -3, O +1, N +1; (c) C 0, O +1, N -2. All three resonance forms have formal charges, with positive formal charges on the O atom.

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Chemical Bonding

Conclusion Based on the ideas described in the discussion above, the reader should be able to: (a) Write Lewis electron dot structures for a variety of molecules (b) Describe the bonds in those structures as single, double or triple bonds (c) Write and discuss resonance forms where appropriate (d) Determine formal charges for atoms within molecules and ions (e) Use formal charges to predict the relative stabilities for several resonance forms

Sample Questions from AP® Chemistry Examinations 1982 D (a) Draw the Lewis electron-dot structures for CO32-, CO2 and CO, including resonance structures where appropriate. (b) Which of the three species has the shortest C-O bond length? Explain the reason for your answer. (c) Account for the fact that the carbon-oxygen bond length in CO32– is greater than the carbon-oxygen bond length in CO2. Answers:

:O :

:

C

C

C

© 2010 The College Board.

:

O

:

:

:

: :

: 12

O

:

O

:

:

:O :

:

O

:

:

:

O

:

:

O

:O:

:

(a)

AP Chemistry Curriculum Module: Chemical Bonding

(b) CO has the shortest bond because there is a triple bond, or because the bond order is 3, meaning it has the greatest number of shared electrons between the carbon and the oxygen atom. (c) The carbonate ion exhibits resonance. Therefore, due to the delocalized electrons, the C-O bond is intermediate in size between a single bond and a double bond; whereas in carbon dioxide, the carbon-oxygen bond is a double bond.

1989 D

CF4   XeF4   ClF3 (a) Draw a Lewis electron-dot structure for each of the molecules above and identify the shape of each. Answer:

Figure 13:

(a)

1990 D Use simple structure and bonding models to account for each of the following: (a) The bond length between the two carbon atoms is shorter in C2H4 than in C2H6. (b) All the bond lengths in SO3 are identical and are shorter than a sulfur-oxygen single bond. Answers: (a) C2H6 has only single bonds. C2H4 has a double bond between the two carbon atoms. The more electrons that are involved in bonding, the shorter the bond length will be. (b)

: :

:

:

:

:

:

:

:

::

:

:

:

:

© 2010 The College Board.

::

:

:

:

13

:O : S O O

:

:

:

:

:

O

:

O

S

:O: S O O :

::

:

:O :

Chemical Bonding

SO3 exhibits resonance. Due to the delocalized electrons, the electrons in the resonance structures are distributed evenly between the sulfur and oxygen atoms. Due to this, there is only one form of the SO3 molecule, and the bond lengths between the oxygen and the sulfur are intermediate between characteristic singleand double- bond lengths.

1992 D

NO2   NO2-   NO2+ Nitrogen is the central atom in each of the species given above. (a) Draw the Lewis electron-dot structure for each of the three species. Answer:

+

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AP Chemistry Curriculum Module: Chemical Bonding

1999 D Answer the following question using principles of chemical bonding and molecular structure. (b) Consider the molecules CF4 and SF4. Draw the complete Lewis electron-dot structure for each molecule.

F F C F F F

F S

F

F

2005 D (c) Two Lewis structures can be drawn for the OPF3 molecule, as shown below. Structure 1   Structure 2

: O:

..

: O: .. .. .. .. :F : : F P F..: .. .. .. P F :F : .. F: ..: (i) How many sigma bonds and how many pi bonds are in Structure 1? (ii) Which one of the two structures best represents a molecule of OPF3? Justify your answer in terms of formal charge.

Answers: (i) One Ψ bond and four σ bonds (ii) Structure 1. In Structure 1, oxygen has a formal charge of 0 (six valence electrons – six assigned electrons), each fluorine is 0 (seven valence electrons – seven assigned electrons), and phosphorus is 0 (five valence electrons – five assigned electrons). 15 © 2010 The College Board.

Chemical Bonding

In Structure 2, oxygen has a formal charge of –1 (six valence electrons – seven assigned electrons), each fluorine is 0 (seven valence electrons – seven assigned electrons), and phosphorus is +1 (five valence electrons – four assigned electrons). Structures where all the atoms have formal charges of 0 are preferred.

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AP Chemistry Curriculum Module: Chemical Bonding

III. Valence Shell Electron Pair Repulsion Theory Arden P. Zipp The valence shell electron pair repulsion (VSEPR) theory was developed as a way to predict molecular geometries (based on electron dot structures) without invoking the geometry or hybridization of orbitals. The basic principle of the VSEPR theory is that electrons repel one another because of their like (negative) charges and that the shapes of covalently bonded species can be determined by the repulsion of electrons (either bonding or nonbonding). This theory offers the simplest means available to account for (or to predict) the structures of molecules and ions, which can be divided into two categories. These categories include species in which: a. Only bonding electrons surround the central atom. b. The central atom is surrounded by both bonding and nonbonding electrons. Species in the former category are designated as AXn species in many textbooks, where A represents the central atom, X represents the atoms or groups surrounding it, and n represents the number of such groups. The geometries of such species are often referred to as “ideal” because the positions of the surrounding substituents (atoms or groups) are not perturbed by the presence of nonbonding electrons. The effect of nonbonding electrons is discussed below. As a first approximation, the sizes of the atoms and the number of bonds (e.g., single, double or triple) have little effect on the geometry.

Ideal Geometries Two bonding pairs: These will be the farthest away from one another when they are on opposite sides of the central atom to give a linear species with a 180˚ angle between the bonds.

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Chemical Bonding

Three bonding pairs: Repulsion will be minimized with the three pairs at the corners of an equilateral triangle with angles of 120˚.

1200

Four bonding pairs: Although four pairs would form a square if constrained to one plane, the interaction among them is decreased if two pairs rotate out of the plane to form a three-dimensional shape called a tetrahedron. Every bond angle in this species is 109.5˚.

109.50

Five bonding pairs: In this species the five electron pairs are not equivalent, even though they are all bonding pairs. The lowest energy configuration has three pairs in an equilateral triangle (designated as the equatorial plane), with the other two at 90˚ above and below the plane of the triangle (axial). There are two different bond angles: 120˚ for the equatorial groups and 90˚ for the axial substituents (relative to the equatorial ones).

900

1200

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AP Chemistry Curriculum Module: Chemical Bonding

Six bonding pairs: Pairs will be found at the corners of an eight-sided geometric figure called an octahedron, with the angles between adjacent pairs of 90˚.

900

A summary of these results is presented in the table below. Formulas, Geometries, Bond Angles and Example Species for AX2 - AX6 Formula

Location of e- pairs

Bond Angles

Example

AX2

linear

180o

BeH2

AX3

triangular

120o

AX4

tetrahedral

109.5

AX5

trigonal

90o

bipyramidal

120

octahedral

90o

AX6

BH3 o

CH4 PCl5

o

SF6

Teaching Tips • Tying off inflated balloons and twisting the tails together can demonstrate the fact that these shapes occur as a result of repulsion. (Note: Six balloons are often too cumbersome to allow their attachment in this way.) • The first four shapes can be illustrated with a “personal” approach by using the hands to represent two bonding pairs on opposite sides of the body (which represent the central atom), by using the hands together and the feet spread for three pairs, and by using both hands extended and both feet extended for four pairs. • You can portray the advantage of the tetrahedron over the square by using a piece of clothesline. The length of the clothesline required with the ends in one’s hands and the middle under the feet will be less when the feet are under the shoulders (“square”) than with the hands twisted in front and behind the head (“tetrahedron”). 19 © 2010 The College Board.

Chemical Bonding

Because the spatial relationships of molecules are best appreciated by examining them in three dimensions, a hands-on activity using sets of molecular models is suggested.

Effect of Nonbonded Electrons When a molecule contains nonbonding as well as bonding electrons around the central atom, it is necessary to distinguish between the geometries of the electrons and those of the groups of atoms around the central atom (referred to as the molecular geometry). In these cases, the geometries of the bonded atoms will obviously differ from those found in species that contain an equal number of bonding electrons. Chemists are usually interested in the positions of the atoms, which can be determined experimentally. The positions of the nonbonding electrons are usually assigned by deduction from the location of the atoms surrounding the central atom. The molecular geometry of a species that contains nonbonding as well as bonding electrons can be determined by starting with the ideal geometry of the corresponding species with the same total number of electrons. The geometry will differ from the ideal one by the absence of one or more substituents. In addition, the bond angle(s) between the remaining bonded atoms will be less than the ideal angle(s) due to the greater repulsion of the nonbonding electrons. (See below.)

Structures based on AX3 Two bonding pairs, one nonbonding pair: If one of the electron pairs in such a species is nonbonding, the molecule is described as bent, with an angle between the two bonded pairs that is less than 120˚.

20 © 2010 The College Board.

AP Chemistry Curriculum Module: Chemical Bonding

Structures based on AX4 Three bonding pairs, one nonbonding pair: When one of the electron pairs is nonbonding, the central atom and the three bonded atoms describe a triangular (trigonal) pyramid.

> 109.50

< 109.50

Atomsonly only Atoms Trigonal TrigonalPyramid Pyramid

Atoms Atomsand and nonbonding non-bondingpair pair

Two bonding pairs, two nonbonding pairs: Because the parent AX4 species is nonplanar, this configuration can only lead to a bent geometry.

>109.5°

90°

< 90° 120°

Atoms and nonbonding pair

Atoms only, seesaw

Three bonding pairs, two nonbonding pairs: As above, the two nonbonding pairs occupy equatorial positions, leaving one equatorial and two axial positions for the bonding pairs, which gives the resulting species a T-shape.

Atoms and nonbonding pairs

Atoms only, T-shape

Two bonding pairs, three nonbonding pairs: With three nonbonding pairs in the equatorial plane and the bonding pairs in axial positions, a species with such a formula will have a linear molecular geometry.

Atoms and nonbonding pairs

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Atoms only, linear

AP Chemistry Curriculum Module: Chemical Bonding

Structures based on AX6 Five bonding pairs, one nonbonding pair: Since all six positions of an octahedron are equivalent, replacing any of the six bonding pairs with a nonbonding pair will yield the same structure, referred to as a square pyramid.

Atoms Atomsand and nonbonding non-bondingpair pair

Atoms Atoms only only Square Squarepyramid pyramid

Four bonding pairs, two nonbonding pairs: When a second nonbonding pair is introduced to the square pyramid above, the repulsion between the nonbonding pairs will be minimized when they are opposite one another. This molecular geometry is square planar.

Atoms Atomsand and nonbonding non-bondingpairs pairs

Atoms Atoms only only Linear Linear

As mentioned above, nonbonding electrons may decrease the angles between bonding electrons relative to those in ideal structures where all bonding electrons are equivalent (except for those in the trigonal bipyramid). This decrease can be understood on the grounds that bonding electron pairs are attracted by two positively charged nuclei, whereas nonbonding electrons are attracted to only one nucleus, allowing them to approach that nucleus more closely. As a consequence, the repulsion between nonbonding and bonding pairs of electrons is greater than that between bonding pairs. These results are summarized in the table below.

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Chemical Bonding

Characteristics of Species Containing Both Bonding and Nonbonding Electron Pairs Formula

Ideal Bonding Nonbonding Molecular Bond Example Geometry Pairs pairs Geometry Angle*

AX2E

trigonal planar

2

1

bent

120o

O3

AX3E

tetrahedral

3

1

trigonal pyramidal

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