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C H A P T E R

10Chemical Bonding II: MolecularGeometry and Hybridization of Atomic OrbitalsI N T R O D U C T I O N

IN CHAPTER 9 WE DISCUSSED BONDING IN TERMS OF THE LEWIS THE-

ORY. HERE WE WILL STUDY THE SHAPE, OR GEOMETRY, OF MOLECULES.

GEOMETRY HAS AN IMPORTANT INFLUENCE ON THE PHYSICAL AND CHEM-

ICAL PROPERTIES OF MOLECULES, SUCH AS MELTING POINT, BOILING POINT,

AND REACTIVITY. WE WILL SEE THAT WE CAN PREDICT THE SHAPES OF

MOLECULES WITH CONSIDERABLE ACCURACY USING A SIMPLE METHOD

BASED ON LEWIS STRUCTURES.

THE LEWIS THEORY OF CHEMICAL BONDING, ALTHOUGH USEFUL

AND EASY TO APPLY, DOES NOT TELL US HOW AND WHY BONDS FORM.

A PROPER UNDERSTANDING OF BONDING COMES FROM QUANTUM ME-

CHANICS. THEREFORE, IN THE SECOND PART OF THIS CHAPTER WE WILL

APPLY QUANTUM MECHANICS TO THE STUDY OF THE GEOMETRY AND

STABILITY OF MOLECULES.

10.1 MOLECULAR GEOMETRY

10.2 DIPOLE MOMENTS

10.3 VALENCE BOND THEORY

10.4 HYBRIDIZATION OF ATOMIC ORBITALS

10.5 HYBRIDIZATION IN MOLECULESCONTAINING DOUBLE AND TRIPLEBONDS

10.6 MOLECULAR ORBITAL THEORY

10.7 MOLECULAR ORBITAL CONFIGURATIONS

10.8 DELOCALIZED MOLECULAR ORBITALS

367



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Molecular geometry is the three-dimensional arrangement of atoms in a molecule. Amolecule’s geometry affects its physical and chemical properties, such as melting point,boiling point, density, and the types of reactions it undergoes. In general, bond lengthsand bond angles must be determined by experiment. However, there is a simple pro-cedure that allows us to predict with considerable success the overall geometry of amolecule or ion if we know the number of electrons surrounding a central atom in itsLewis structure. The basis of this approach is the assumption that electron pairs in the valence shell of an atom repel one another. The valence shell is the outermost electron-occupied shell of an atom; it holds the electrons that are usually involved inbonding. In a covalent bond, a pair of electrons, often called the bonding pair, is re-sponsible for holding two atoms together. However, in a polyatomic molecule, wherethere are two or more bonds between the central atom and the surrounding atoms, therepulsion between electrons in different bonding pairs causes them to remain as farapart as possible. The geometry that the molecule ultimately assumes (as defined bythe positions of all the atoms) minimizes the repulsion. This approach to the study ofmolecular geometry is called the valence-shell electron-pair repulsion (VSEPR)model, because it accounts for the geometric arrangements of electron pairs around acentral atom in terms of the electrostatic repulsion between electron pairs.

Two general rules govern the use of the VSEPR model:

• As far as electron-pair repulsion is concerned, double bonds and triple bonds canbe treated like single bonds. This approximation is good for qualitative purposes.However, you should realize that in reality multiple bonds are “larger” than singlebonds; that is, because there are two or three bonds between two atoms, the elec-tron density occupies more space.

• If a molecule has two or more resonance structures, we can apply the VSEPR modelto any one of them. Formal charges are usually not shown.

With this model in mind, we can predict the geometry of molecules (and ions) in asystematic way. For this purpose, it is convenient to divide molecules into two cate-gories, according to whether or not the central atom has lone pairs.

MOLECULES IN WHICH THE CENTRAL ATOM HAS NO LONE PAIRS

For simplicity we will consider molecules that contain atoms of only two elements, Aand B, of which A is the central atom. These molecules have the general formula ABx,where x is an integer 2, 3, . . . . (If x � 1, we have the diatomic molecule AB, whichis linear by definition.) In the vast majority of cases, x is between 2 and 6.

Table 10.1 shows five possible arrangements of electron pairs around the centralatom A. As a result of mutual repulsion, the electron pairs stay as far from one anotheras possible. Note that the table shows arrangements of the electron pairs but not thepositions of the atoms that surround the central atom. Molecules in which the centralatom has no lone pairs have one of these five arrangements of bonding pairs. UsingTable 10.1 as a reference, let us take a close look at the geometry of molecules withthe formulas AB2, AB3, AB4, AB5, and AB6.

368 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS

10.1 MOLECULAR GEOMETRY

The term “central atom” means anatom that is not a terminal atom

in a polyatomic molecule.

VSEPR is pronounced “vesper.”



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AB2: Beryllium Chloride (BeCl2)

The Lewis structure of beryllium chloride in the gaseous state is

Because the bonding pairs repel each other, they must be at opposite ends of a straight

ClOS QOBe ClSOQO

10.1 MOLECULAR GEOMETRY 369

TABLE 10.1 Arrangement of Electron Pairs about a Central Atom (A) in a Molecule and Geometry of Some Simple Molecules and Ions in which the Central Atom Has No Lone Pairs

NUMBER OF ARRANGEMENT MOLECULAR

ELECTRON PAIRS OF ELECTRON PAIRS* GEOMETRY* EXAMPLES

2 BOAOB BeCl2, HgCl2Linear Linear

3 BF3

Trigonal planar Trigonal planar

4 CH4, NH4�

Tetrahedral Tetrahedral

5 PCl5

Trigonal bipyramidal Trigonal bipyramidal

6 SF6

Octahedral Octahedral

*The colored lines are used only to show the overall shapes; they do not represent bonds.

A

B

B

B

B

B

B

90�

�

ð ð

�

�

ðð

90

A

B B

B

B

A

B

�

�120

ð

ð

�

�

ðA

90

B

B

A B

B

109.5�

Að

ð

�

�

B

A

B

B

�

�

�½

120A

180�

ð A ð



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line in order for them to be as far apart as possible. Thus, the ClBeCl angle is pre-dicted to be 180°, and the molecule is linear (see Table 10.1). The “ball-and-stick”model of BeCl2 is


AB3: Boron Trifluoride (BF3)

Boron trifluoride contains three covalent bonds, or bonding pairs. In the most stablearrangement, the three BF bonds point to the corners of an equilateral triangle with Bin the center of the triangle:

According to Table 10.1, the geometry of BF3 is trigonal planar because the three endatoms are at the corners of an equilateral triangle which is planar:

SF

B

F

F

O

SQ

S

QO

S

O

A

GD

Thus, each of the three FBF angles is 120°, and all four atoms lie in the same plane.

AB4: Methane (CH4)

The Lewis structure of methane is

Since there are four bonding pairs, the geometry of CH4 is tetrahedral (see Table 10.1).A tetrahedron has four sides (the prefix tetra means “four”), or faces, all of which areequilateral triangles. In a tetrahedral molecule, the central atom (C in this case) is lo-cated at the center of the tetrahedron and the other four atoms are at the corners. Thebond angles are all 109.5°.

H CA

AO

H

H

H

O



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AB5: Phosphorus Pentachloride (PCl5)

The Lewis structure of phosphorus pentachloride (in the gas phase) is

The only way to minimize the repulsive forces among the five bonding pairs is toarrange the PCl bonds in the form of a trigonal bipyramid (see Table 10.1). A trigonalbipyramid can be generated by joining two tetrahedrons along a common triangularbase:

ClSO

QO

OSS

AEHA

P

SO

SQ

Q

S

S O

Cl

ClCl

Cl

Q


The central atom (P in this case) is at the center of the common triangle with the sur-rounding atoms positioned at the five corners of the trigonal bipyramid. The atoms thatare above and below the triangular plane are said to occupy axial positions, and thosethat are in the triangular plane are said to occupy equatorial positions. The angle be-tween any two equatorial bonds is 120°; that between an axial bond and an equatorialbond is 90°, and that between the two axial bonds is 180°.

AB6: Sulfur Hexafluoride (SF6)

The Lewis structure of sulfur hexafluoride is

The most stable arrangement of the six SF bonding pairs is in the shape of an octahe-dron, shown in Table 10.1. An octahedron has eight sides (the prefix octa means “eight”).It can be generated by joining two square pyramids on a common base. The centralatom (S in this case) is at the center of the square base and the surrounding atoms areat the six corners. All bond angles are 90° except the one made by the bonds betweenthe central atom and the pairs of atoms that are diametrically opposite each other. Thatangle is 180°. Since the six bonds are equivalent in an octahedral molecule, we cannotuse the terms “axial” and “equatorial” as in a trigonal bipyramidal molecule.

OSS

AEHA

HE

SF

FF

F

FF

O

SSQ

OS

Q

QS

OS S

Q

Q

O



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MOLECULES IN WHICH THE CENTRAL ATOM HAS ONE OR MORE LONE PAIRS

Determining the geometry of a molecule is more complicated if the central atom hasboth lone pairs and bonding pairs. In such molecules there are three types of repulsiveforces— those between bonding pairs, those between lone pairs, and those between abonding pair and a lone pair. In general, according to the VSEPR model, the repulsiveforces decrease in the following order:

lone-pair vs. lone-pair � lone-pair vs. bonding- � bonding-pair vs. bonding-repulsion pair repulsion pair repulsion

Electrons in a bond are held by the attractive forces exerted by the nuclei of thetwo bonded atoms. These electrons have less “spatial distribution” than lone pairs; thatis, they take up less space than lone-pair electrons, which are associated with only oneparticular atom. Because lone-pair electrons in a molecule occupy more space, theyexperience greater repulsion from neighboring lone pairs and bonding pairs. To keeptrack of the total number of bonding pairs and lone pairs, we designate molecules withlone pairs as ABxEy, where A is the central atom, B is a surrounding atom, and E is alone pair on A. Both x and y are integers; x � 2, 3, . . . , and y � 1, 2, . . . . Thus thevalues of x and y indicate the number of surrounding atoms and number of lone pairson the central atom, respectively. The simplest such molecule would be a triatomicmolecule with one lone pair on the central atom and the formula is AB2E.

As the following examples show, in most cases the presence of lone pairs on thecentral atom makes it difficult to predict the bond angles accurately.

AB2E: Sulfur Dioxide (SO2)

The Lewis structure of sulfur dioxide is

Because VSEPR treats double bonds as though they were single, the SO2 molecule canbe viewed as consisting of three electron pairs on the central S atom. Of these, two arebonding pairs and one is a lone pair. In Table 10.1 we see that the overall arrangementof three electron pairs is trigonal planar. But because one of the electron pairs is a lonepair, the SO2 molecule has a “bent” shape.

Since the lone-pair versus bonding-pair repulsion is greater than the bonding-pair ver-sus bonding-pair repulsion, the two sulfur-to-oxygen bonds are pushed together slightlyand the OSO angle is less than 120°.

AB3E: Ammonia (NH3)

The ammonia molecule contains three bonding pairs and one lone pair:

As Table 10.1 shows, the overall arrangement of four electron pairs is tetrahedral. Butin NH3 one of the electron pairs is a lone pair, so the geometry of NH3 is trigonal pyra-

HO HOAN

H

O

OSJ G

OOS

S

SS S

OQ OQOO OSPOS




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FIGURE 10.1 (a) The relativesizes of bonding pairs and lonepairs in CH4, NH3, and H2O.(b) The bond angles in CH4,NH3, and H2O. Note thatdashed lines represent bond axesbehind the plane of the paper,wedged lines represent bondaxes in front of the plane of thepaper, and thin solid lines repre-sent bonds in the plane of thepaper.

midal (so called because it looks like a pyramid, with the N atom at the apex). Becausethe lone pair repels the bonding pairs more strongly, the three NH bonding pairs arepushed closer together:

Thus the HNH angle in ammonia is smaller than the ideal tetrahedral angle of 109.5°(Figure 10.1).

AB2E2: Water (H2O)

A water molecule contains two bonding pairs and two lone pairs:

The overall arrangement of the four electron pairs in water is tetrahedral, the same asin ammonia. However, unlike ammonia, water has two lone pairs on the central O atom.These lone pairs tend to be as far from each other as possible. Consequently, the twoOH bonding pairs are pushed toward each other, and we predict an even greater devi-ation from the tetrahedral angle than in NH3. As Figure 10.1 shows, the HOH angle is104.5°. The geometry of H2O is bent:

AB4E: Sulfur Tetrafluoride (SF4)

The Lewis structure of SF4 is

H H

OD G

SS

HO HOOQO

HGDA

N

H H

O


C

H

H

H

H

N

H

H

H

O

H

H

(a)

C

H

H

H

H

N

HH

O

H

(b)

H H

109.5° 107.3° 104.5°



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The central sulfur atom has five electron pairs whose arrangement, according to Table10.1, is trigonal bipyramidal. In the SF4 molecule, however, one of the electron pairsis a lone pair, so the molecule must have one of the following geometries:

In (a) the lone pair occupies an equatorial position, and in (b) it occupies an axial po-sition. The axial position has three neighboring pairs at 90° and one at 180°, while theequatorial position has two neighboring pairs at 90° and two more at 120°. The repul-sion is smaller for (a), and indeed (a) is the structure observed experimentally. Thisshape is sometimes described as a distorted tetrahedron (or a folded square, or seesawshape). The angle between the axial F atoms and S is 186°, and that between the equa-torial F atoms and S is 116°.

Table 10.2 shows the geometries of simple molecules in which the central atomhas one or more lone pairs, including some that we have not discussed.

GEOMETRY OF MOLECULES WITH MORE THAN ONE CENTRAL ATOM

So far we have discussed the geometry of molecules having only one central atom.The overall geometry of molecules with more than one central atom is difficult to de-fine in most cases. Often we can only describe the shape around each of the centralatoms. For example, consider methanol, CH3OH, whose Lewis structure is

The two central (nonterminal) atoms in methanol are C and O. We can say that the threeCH and the CO bonding pairs are tetrahedrally arranged about the C atom. The HCHand OCH bond angles are approximately 109°. The O atom here is like the one in wa-ter in that it has two lone pairs and two bonding pairs. Therefore, the HOC portion ofthe molecule is bent, and the angle HOC is approximately equal to 105° (Figure 10.2).

GUIDELINES FOR APPLYING THE VSEPR MODEL

Having studied the geometries of molecules in two categories (central atoms with andwithout lone pairs), let us consider some rules for applying the VSEPR model to alltypes of molecules:

• Write the Lewis structure of the molecule, considering only the electron pairs aroundthe central atom (that is, the atom that is bonded to more than one other atom).

HOAC

H

HOOQOOAH

F

Q

S

F

(b)

F

F

SS

F

F(a)

F

F

O

S

F

FF

F

SQO

Q

S

G DS SQ

OGD

O

Q

O


SF4

FIGURE 10.2 The geometry ofCH3OH.



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TABLE 10.2 Geometry of Simple Molecules and Ions in which the Central Atom Has One or More Lone Pairs

CLASS OF TOTAL NUMBER NUMBER OF NUMBER OF ARRANGEMENT OF

MOLECULE OF ELECTRON PAIRS BONDING PAIRS LONE PAIRS ELECTRON PAIRS* GEOMETRY EXAMPLES

AB2E 3 2 1 Bent SO2, O3

Trigonal planar

AB3E 4 3 1 Trigonal NH3

pyramidal

Tetrahedral

AB2E2 4 2 2 Bent H2O

Tetrahedral

AB4E 5 4 1 Distorted IF4�, SF4, XeO2F2

tetrahedron(or seesaw)

Trigonal bipyramidal

AB3E2 5 3 2 T-shaped ClF3


AB2E3 5 2 3 Linear XeF2, I3�


AB5E 6 5 1 Square BrF5, XeOF4

pyramidal

Octahedral

AB4E2 6 4 2 Square planar XeF4, ICl4�

Octahedral

*The colored lines are used to show the overall shape, not bonds.

�

š

B

A

B B

B

�

B

A

B

B B

B

ð

ð

ð

B

B

A

ð

ð

B

B

AB

A

B

ð

B

B B

š B

š

A

B

A

B

š

B B

A

B

š

B



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• Count the number of electron pairs around the central atom (bonding pairs and lonepairs). Treat double and triple bonds as though they were single bonds. Refer toTable 10.1 to predict the overall arrangement of the electron pairs.

• Use Tables 10.1 and 10.2 to predict the geometry of the molecule.• In predicting bond angles, note that a lone pair repels another lone pair or a bond-

ing pair more strongly than a bonding pair repels another bonding pair. Rememberthat there is no easy way to predict bond angles accurately when the central atompossesses one or more lone pairs.

The VSEPR model generates reliable predictions of the geometries of a variety of mo-lecular structures. Chemists use the VSEPR approach because of its simplicity.Although there are some theoretical concerns about whether “electron-pair repulsion”actually determines molecular shapes, the assumption that it does leads to useful (andgenerally reliable) predictions. We need not ask more of any model at this stage in thestudy of chemistry.

We will use the VSEPR model to predict the geometry of some simple moleculesand polyatomic ions in the following example.

EXAMPLE 10.1

Use the VSEPR model to predict the geometry of the following molecules and ions:(a) AsH3, (b) OF2, (c) AlCl4

�, (d) I3�, (e) C2H4.

Answer (a) The Lewis structure of AsH3 is

This molecule has three bonding pairs and one lone pair, a combination similar tothat in ammonia. Therefore, the geometry of AsH3 is trigonal pyramidal, like NH3.We cannot predict the HAsH angle accurately, but we know that it must be less than109.5° because the repulsion of the bonding electron pairs by the lone pair on Asis greater than the repulsion between bonding pairs.(b) The Lewis structure of OF2 is

Since there are two lone pairs on the O atom, the OF2 molecule must have a bentshape, like that of H2O. Again, all we can say about angle size is that the FOF an-gle should be less than 109.5° due to the stronger repulsive force between the lonepairs and the bonding pairs.(c) The Lewis structure of AlCl4

� is

Since the central Al atom has no lone pairs and all four AlOCl bonds are equiva-lent, the AlCl4

� ion should be tetrahedral, and the ClAlCl angles should all be 109.5°.(d) The Lewis structure of I3

� is

QS SOOOQOQ

�

I IIS

S

AOO

S S

QOAAl

Cl

Cl

Cl Cl

SS

S SO

O�

Q

Q

QSO OFOQO SOFOQ

OH As H

H

OAO




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Similar problems: 10.7, 10.8,10.9.

The central I atom has two bonding pairs and three lone pairs. From Table 10.2 wesee that the three lone pairs all lie in the triangular plane, and the I3

� ion should belinear.(e) The Lewis structure is C2H4 is

The CPC bond is treated as though it were a single bond. Because there are nolone pairs present, the arrangement around each C atom has a trigonal planar shapelike BF3, discussed earlier. Thus the predicted bond angles in C2H4 are all 120°.

Comment (a) The I3� ion is one of the few structures for which the bond angle

(180°) can be predicted accurately even though the central atom contains lone pairs.(b) In C2H4, all six atoms lie in the same plane. The overall planar geometry is notpredicted by the VSEPR model, but we will see why the molecule prefers to be pla-nar later. In reality, the angles are close, but not equal, to 120° because the bondsare not all equivalent.

PRACTICE EXERCISEUse the VSEPR model to predict the geometry of (a) SiBr4, (b) CS2, and (c) NO3

�.

In Section 9.3 we learned that hydrogen fluoride is a covalent molecule with a polarbond. There is a shift of electron density from H to F because the F atom is more elec-tronegative than the H atom. The shift of electron density is symbolized by placing acrossed arrow ( ) above the Lewis structure to indicate the direction of the shift.For example,

The consequent charge separation can be represented as

where � (delta) denotes a partial charge. This separation of charges can be confirmedin an electric field (Figure 10.3). When the field is turned on, HF molecules orient theirnegative ends toward the positive plate and their positive ends toward the negativeplate. This alignment of molecules can be detected experimentally.

A quantitative measure of the polarity of a bond is its dipole moment (�) whichis the product of the charge Q and the distance r between the charges:

� � Q r (10.1)

To maintain electrical neutrality, the charges on both ends of an electrically neutral di-

SOFOQH��

SOFOQH

GDP

G DH H

C C

H H120�

120�

120�

GDP

G DH H

C C

H H

10.2 DIPOLE MOMENTS 377

10.2 DIPOLE MOMENTS



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FIGURE 10.3 Behavior of po-lar molecules (a) in the absenceof an external electric field and(b) when the electric field isturned on. Nonpolar moleculesare not affected by an electricfield.

atomic molecule must be equal in magnitude and opposite in sign. However, in Equation(10.1), Q refers only to the magnitude of the charge and not to its sign, so � is alwayspositive. Dipole moments are usually expressed in debye units (D), named for PeterDebye.† The conversion factor is

1 D � 3.33 10�30 C m

where C is coulomb and m is meter.Diatomic molecules containing atoms of different elements (for example, HCl,

CO, and NO) have dipole moments and are called polar molecules. Diatomic mole-cules containing atoms of the same element (for example, H2, O2, and F2) are exam-ples of nonpolar molecules because they do not have dipole moments. For a moleculemade up of three or more atoms both the polarity of the bonds and the molecular geom-etry determine whether there is a dipole moment. Even if polar bonds are present, themolecule will not necessarily have a dipole moment. Carbon dioxide (CO2), for ex-ample, is a triatomic molecule, so its geometry is either linear or bent:

The arrows show the shift of electron density from the less electronegative carbon atomto the more electronegative oxygen atom. In each case, the dipole moment of the en-tire molecule is made up of two bond moments, that is, individual dipole moments inthe polar CPO bonds. The bond moment is a vector quantity, which means that it hasboth magnitude and direction. The measured dipole moment is equal to the vector sumof the bond moments. The two bond moments in CO2 are equal in magnitude. Since

P

NK

O C O

CO O

bent molecule(has a dipole moment)

Presultantdipole moment

linear molecule(no dipole moment)


†Peter Joseph William Debye (1884–1966). American chemist and physicist of Dutch origin. Debye made many significantcontributions in the study of molecular structure, polymer chemistry, X-ray analysis, and electrolyte solution. He was awardedthe Nobel Prize in Chemistry in 1936.

+– +

–

+–

+

– +– +

–

+–

+–

+–

+–+

–+–

+–

–+

+

–

+

+

–

+

–

+–

–

+

–

–

+

–

++

– +++

+

+

+

–

–

–

–

–

(b)(a)

+

In a diatomic molecule like HF,the charge Q is equal to �� and

��.



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FIGURE 10.4 Bond momentsand resultant dipole moments inNH3 and NF3.

they point in opposite directions in a linear CO2 molecule, the sum or resultant dipolemoment would be zero. On the other hand, if the CO2 molecule were bent, the twobond moments would partially reinforce each other, so that the molecule would havea dipole moment. Experimentally it is found that carbon dioxide has no dipole mo-ment. Therefore, we conclude that the carbon dioxide molecule is linear. The linear na-ture of carbon dioxide has been confirmed through other experimental measurements.

Next let us consider the NH3 and NF3 molecules shown in Figure 10.4. In bothcases the central N atom has a lone pair, whose charge density is away from the Natom. From Figure 9.5 we know that N is more electronegative than H, and F is moreelectronegative than N. For this reason, the shift of electron density in NH3 is towardN and so contributes a larger dipole moment, whereas the NF bond moments are di-rected away from the N atom and so together they offset the contribution of the lonepair to the dipole moment. Thus the resultant dipole moment in NH3 is larger than thatin NF3.

Dipole moments can be used to distinguish between molecules that have the sameformula but different structures. For example, the following molecules both exist; theyhave the same molecular formula (C2H2Cl2), the same number and type of bonds, butdifferent molecular structures:

Since cis-dichloroethylene is a polar molecule but trans-dichloroethylene is not, theycan readily be distinguished by a dipole moment measurement. Additionally, as we willsee in the next chapter, the strength of intermolecular forces is partially determined bywhether molecules possess a dipole moment. Table 10.3 lists the dipole moments ofseveral polar molecules.

Example 10.2 shows how we can predict whether a molecule possesses a dipolemoment if we know its molecular geometry.

P

trans-dichloroethylene� 0

G D

GC

Cl

H

C

Cl

H

P

cis-dichloroethylene� 1.89 D

G D

GDC

Cl

H

C

Cl

HD

resultantdipole moment

� �

10.2 DIPOLE MOMENTS 379

N

H

H

H

Resultant dipolemoment = 1.46 D

N

F

F

F

Resultant dipolemoment = 0.24 D

The VSEPR model predicts thatCO2 is a linear molecule.



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An analogy is an object that ispulled in the directions shown by

the three bond moments. If theforces are equal, the object

will not move.

EXAMPLE 10.2

Predict whether each of the following molecules has a dipole moment: (a) IBr,(b) BF3 (trigonal planar), (c) CH2Cl2 (tetrahedral).

Answer (a) Since IBr (iodine bromide) is diatomic, it has a linear geometry.Bromine is more electronegative than iodine (see Figure 9.5), so IBr is polar withbromine at the negative end.

Thus the molecule does have a dipole moment.(b) Since fluorine is more electronegative than boron, each BOF bond in BF3 (borontrifluoride) is polar and the three bond moments are equal. However, the symmetryof a trigonal planar shape means that the three bond moments exactly cancel oneanother:

Consequently BF3 has no dipole moment; it is a nonpolar molecule.(c) The Lewis structure of CH2Cl2 (methylene chloride) is

This molecule is similar to CH4 in that it has an overall tetrahedral shape. However,because not all the bonds are identical, there are three different bond angles: HCH,HCCl, and ClCCl. These bond angles are close to, but not equal to, 109.5°. Since chlo-rine is more electronegative than carbon, which is more electronegative than hydro-gen, the bond moments do not cancel and the molecule possesses a dipole moment:

OAO

AH

Cl

H

Cl

C

D

F

B

F FG

A

I BrO


TABLE 10.3 Dipole Moments of Some PolarMolecules

MOLECULE GEOMETRY DIPOLE MOMENT (D)

HF Linear 1.92HCl Linear 1.08HBr Linear 0.78HI Linear 0.38H2O Bent 1.87H2S Bent 1.10NH3 Pyramidal 1.46SO2 Bent 1.60



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Similar problems: 10.19, 10.21,10.22. Thus CH2Cl2 is a polar molecule.

PRACTICE EXERCISEDoes the AlCl3 molecule have a dipole moment?

The VSEPR model, based largely on Lewis structures, provides a relatively simple andstraightforward method for predicting the geometry of molecules. But as we noted ear-lier, the Lewis theory of chemical bonding does not clearly explain why chemical bondsexist. Relating the formation of a covalent bond to the pairing of electrons was a stepin the right direction, but it did not go far enough. For example, the Lewis theory de-scribes the single bond between the H atoms in H2 and that between the F atoms in F2

in essentially the same way—as the pairing of two electrons. Yet these two moleculeshave quite different bond dissociation energies and bond lengths (436.4 kJ/mol and74 pm for H2 and 150.6 kJ/mol and 142 pm for F2). These and many other facts can-not be explained by the Lewis theory. For a more complete explanation of chemicalbond formation we look to quantum mechanics. In fact, the quantum mechanical studyof chemical bonding also provides a means for understanding molecular geometry.

At present, two quantum mechanical theories are used to describe covalent bondformation and the electronic structure of molecules. Valence bond (VB) theory assumesthat the electrons in a molecule occupy atomic orbitals of the individual atoms. It per-mits us to retain a picture of individual atoms taking part in the bond formation. Thesecond theory, called molecular orbital (MO) theory, assumes the formation of mo-lecular orbitals from the atomic orbitals. Neither theory perfectly explains all aspectsof bonding, but each has contributed something to our understanding of many observedmolecular properties.

Let us start our discussion of valence bond theory by considering the formationof a H2 molecule from two H atoms. The Lewis theory describes the HOH bond interms of the pairing of the two electrons on the H atoms. In the framework of valencebond theory, the covalent HOH bond is formed by the overlap of the two 1s orbitalsin the H atoms. By overlap, we mean that the two orbitals share a common region inspace.

What happens to two H atoms as they move toward each other and form a bond?Initially, when the two atoms are far apart, there is no interaction. We say that the po-tential energy of this system (that is, the two H atoms) is zero. As the atoms approacheach other, each electron is attracted by the nucleus of the other atom; at the same time,the electrons repel each other, as do the nuclei. While the atoms are still separated, at-traction is stronger than repulsion, so that the potential energy of the system decreases(that is, it becomes negative) as the atoms approach each other (Figure 10.5). This trendcontinues until the potential energy reaches a minimum value. At this point, when thesystem has the lowest potential energy, it is most stable. This condition corresponds to

resultantdipole moment

GC

Cl

Cl

LLHH

A

&

10.3 VALENCE BOND THEORY 381

10.3 VALENCE BOND THEORY

Recall that an object has potentialenergy by virtue of its position.



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Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chem

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In the last twenty years the microwave oven has be-come a ubiquitous appliance. Microwave technologyenables us to thaw and cook food much more rapidlythan conventional appliances do. How do microwavesheat food so quickly?

In Chapter 7 we saw that microwaves are a form of electromagnetic radiation (see Figure 7.3).Microwaves are generated by a magnetron, whichwas invented during World War II when radar tech-nology was being developed. The magnetron is a hol-low cylinder encased in a horseshoe-shaped magnet.In the center of the cylinder is a cathode rod. The wallsof the cylinder act as an anode. When heated, thecathode emits electrons that travel toward the anode.The magnetic field forces the electrons to move in acircular path. This motion of charged particles gener-ates microwaves, which are adjusted to a frequencyof 2.45 GHz (2.45 109 Hz) for cooking. A “wave-guide” directs the microwaves into the cooking com-partment. Rotating fan blades reflect the microwavesto all parts of the oven.

The cooking action in a microwave oven resultsfrom the interaction between the electric field compo-nent of the radiation with the polar molecules—mostlywater—in food. All molecules rotate at room temper-ature. If the frequency of the radiation and that of the

molecular rotation are equal, energy can be trans-ferred from microwave to the polar molecule. As a re-sult, the molecule will rotate faster. The frequency2.45 GHz is particularly suitable for increasing the ro-tational energy of water molecules. The friction fromrapidly rotating water molecules eventually heats upthe surrounding food molecules.

The reason that a microwave oven can cook foodso fast is that the radiation is not absorbed by non-polar molecules and can therefore reach different partsof food at the same time. (Depending on the amountof water present, microwaves can penetrate food toa depth of several inches.) In a conventional oven,heat can affect the center of foods only by conduction(that is, by transfer of heat from hot air molecules tocooler molecules in food in a layer-by-layer fashion),which is a very slow process.

The following points are relevant to the operationof a microwave oven. Plastics and Pyrex glasswaresdo not contain polar molecules and are therefore notaffected by microwave radiation. (Styrofoam and cer-tain plastics cannot be used in microwaves becausethey melt from the heat of the food.) Metals, however,reflect microwaves, thereby shielding the food andpossibly returning enough energy to the microwaveemitter to overload it. Because microwaves can induce

Microwave Ovens—Dipole Moments at Work

A microwave oven. The microwaves generated by the magnetron are reflected to all parts of theoven by the rotating fan blades.

Anode

Cathode

Waveguide

Magnet

MagnetronRotating blades



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substantial overlap of the 1s orbitals and the formation of a stable H2 molecule. If thedistance between nuclei were to decrease further, the potential energy would rise steeplyand finally become positive as a result of the increased electron-electron and nuclear-nuclear repulsions.

In accord with the law of conservation of energy, the decrease in potential energyas a result of H2 formation must be accompanied by a release of energy. Experiments

10.3 VALENCE BOND THEORY 383

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a current in the metal, this action can lead to sparksjumping between the container and the bottom or wallsof the oven. Finally, although water molecules in iceare locked in position and therefore cannot rotate, weroutinely thaw food in a microwave oven. The reason

is that at room temperature a thin film of liquid waterquickly forms on the surface of frozen food and themobile molecules in that film can absorb the radiationto start the thawing process.

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Interaction between the electric field component of the microwave and a polar molecule. (a) Thenegative end of the dipole follows the propagation of the wave (the positive region) and rotatesin a clockwise direction. (b) If, after the molecule has rotated to the new position the radiationhas also moved along to its next cycle, the positive end of the dipole will move into the negativeregion of the wave while the negative end will be pushed up. Thus the molecule will rotate faster.No such interaction can occur with nonpolar molecules.



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FIGURE 10.5 Change in po-tential energy of two H atomswith changes in their distance ofseparation. At the point of mini-mum potential energy, the H2molecule is in its most stablestate, and the bond length is74 pm.

show that as a H2 molecule is formed from two H atoms, heat is given off. The con-verse is also true. To break a HOH bond, energy must be supplied to the molecule.

Thus valence bond theory gives a clearer picture of chemical bond formation thanthe Lewis theory does. Valence bond theory states that a stable molecule forms fromreacting atoms when the potential energy of the system has decreased to a minimum;the Lewis theory ignores energy changes in chemical bond formation.

The concept of overlapping atomic orbitals applies equally well to diatomic mol-ecules other than H2. Thus a stable F2 molecule forms when the 2p orbitals (contain-ing the unpaired electrons) in the two F atoms overlap to form a covalent bond.Similarly, the formation of the HF molecule can be explained by the overlap of the 1sorbital in H with the 2p orbital in F. In each case, VB theory accounts for the changesin potential energy as the distance between the reacting atoms changes. Because theorbitals involved are not the same kind in all cases, we can see why the bond energiesand bond lengths in H2, F2, and HF might be different. As we stated earlier, Lewis the-ory treats all covalent bonds the same way and offers no explanation for the differ-ences among covalent bonds.

The concept of atomic orbital overlap should apply also to polyatomic molecules.However, a satisfactory bonding scheme must account for molecular geometry. We willdiscuss three examples of VB treatment of bonding in polyatomic molecules.

sp3 HYBRIDIZATION

Consider the CH4 molecule. Focusing only on the valence electrons, we can representthe orbital diagram of C as


10.4 HYBRIDIZATION OF ATOMIC ORBITALS

0

Pote

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Distance of separation

The orbital diagram of the F atomis shown on page 272.



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FIGURE 10.6 Formation of sp3

hybrid orbitals.

2s 2p

Because the carbon atom has two unpaired electrons (one in each of the two 2p or-bitals), it can form only two bonds with hydrogen in its ground state. Although thespecies CH2 is known, it is very unstable. To account for the four COH bonds inmethane, we can try to promote (that is, energetically excite) an electron from the 2sorbital to the 2p orbital:

2s 2p

Now there are four unpaired electrons on C that could form four COH bonds. However,the geometry is wrong, because three of the HCH bond angles would have to be 90°(remember that the three 2p orbitals on carbon are mutually perpendicular), and yet allHCH angles are 109.5°.

To explain bonding in methane, VB theory uses hypothetical hybrid orbitals,which are atomic orbitals obtained when two or more nonequivalent orbitals of thesame atom combine in preparation for covalent bond formation. Hybridization is theterm applied to the mixing of atomic orbitals in an atom (usually a central atom) togenerate a set of hybrid orbitals. We can generate four equivalent hybrid orbitals forcarbon by mixing the 2s orbital and the three 2p orbitals:

sp3 orbitals

Because the new orbitals are formed from one s and three p orbitals, they are calledsp3 hybrid orbitals. Figure 10.6 shows the shape and orientations of the sp3 orbitals.These four hybrid orbitals are directed toward the four corners of a regular tetrahe-

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10.4 HYBRIDIZATION OF ATOMIC ORBITALS 385

sp3 is pronounced “s-p three.”

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FIGURE 10.7 Formation offour bonds between the carbonsp3 hybrid orbitals and the hydro-gen 1s orbitals in CH4.

dron. Figure 10.7 shows the formation of four covalent bonds between the carbon sp3

hybrid orbitals and the hydrogen 1s orbitals in CH4. Thus CH4 has a tetrahedral shape,and all the HCH angles are 109.5°. Note that although energy is required to bring abouthybridization, this input is more than compensated for by the energy released upon theformation of COH bonds. (Recall that bond formation is an exothermic process.)

The following analogy is useful for understanding hybridization. Suppose that wehave a beaker of a red solution and three beakers of blue solutions and that the vol-ume of each is 50 mL. The red solution corresponds to one 2s orbital, the blue solu-tions represent three 2p orbitals, and the four equal volumes symbolize four separateorbitals. By mixing the solutions we obtain 200 mL of a purple solution, which can bedivided into four 50-mL portions (that is, the hybridization process generates four sp3

orbitals). Just as the purple color is made up of the red and blue components of theoriginal solutions, the sp3 hybrid orbitals possess both s and p orbital characteristics.

Another example of sp3 hybridization is ammonia (NH3). Table 10.1 shows thatthe arrangement of four electron pairs is tetrahedral, so that the bonding in NH3 canbe explained by assuming that N, like C in CH4, is sp3-hybridized. The ground-stateelectron configuration of N is 1s22s22p3, so that the orbital diagram for the sp3-hy-bridized N atom is

sp3 orbitals

Three of the four hybrid orbitals form covalent NOH bonds, and the fourth hybrid or-bital accommodates the lone pair on nitrogen (Figure 10.8). Repulsion between thelone-pair electrons and electrons in the bonding orbitals decreases the HNH bond an-gles from 109.5° to 107.3°.

It is important to understand the relationship between hybridization and the VSEPRmodel. We use hybridization to describe the bonding scheme only when the arrange-ment of electron pairs has been predicted using VSEPR. If the VSEPR model predictsa tetrahedral arrangement of electron pairs, then we assume that one s and three p or-bitals are hybridized to form four sp3 hybrid orbitals. The following are examples ofother types of hybridization.

sp HYBRIDIZATION

The beryllium chloride (BeCl2) molecule is predicted to be linear by VSEPR. The or-bital diagram for the valence electrons in Be is

2s 2p

We know that in its ground state Be does not form covalent bonds with Cl because itselectrons are paired in the 2s orbital. So we turn to hybridization for an explanation ofBe’s bonding behavior. First we promote a 2s electron to a 2p orbital, resulting in

2s 2p

Now there are two Be orbitals available for bonding, the 2s and 2p. However, if two

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CH H

H

H

NH H

H

FIGURE 10.8 The sp3-hybridized N atom in NH3.Three sp3 hybrid orbitals formbonds with the H atoms. Thefourth is occupied by nitrogen’slone pair.



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FIGURE 10.9 Formation of sphybrid orbitals.

Cl atoms were to combine with Be in this excited state, one Cl atom would share a 2selectron and the other Cl would share a 2p electron, making two nonequivalent BeClbonds. This scheme contradicts experimental evidence. In the actual BeCl2 molecule,the two BeCl bonds are identical in every respect. Thus the 2s and 2p orbitals must bemixed, or hybridized, to form two equivalent sp hybrid orbitals:

sp orbitals empty 2p orbitals

Figure 10.9 shows the shape and orientation of the sp orbitals. These two hybrid or-bitals lie on the same line, the x-axis, so that the angle between them is 180°. Each ofthe BeCl bonds is then formed by the overlap of a Be sp hybrid orbital and a Cl 3porbital, and the resulting BeCl2 molecule has a linear geometry (Figure 10.10).

sp2 HYBRIDIZATION

Next we will look at the BF3 (boron trifluoride) molecule, known to have planar geom-etry based on VSEPR. Considering only the valence electrons, the orbital diagram ofB is

2s 2p

First, we promote a 2s electron to an empty 2p orbital:

2s 2p

Mixing the 2s orbital with the two 2p orbitals generates three sp2 hybrid orbitals:

sp2 orbitals empty 2p orbital

These three sp2 orbitals lie in the same plane, and the angle between any two of themis 120° (Figure 10.11). Each of the BF bonds is formed by the overlap of a boron sp2

hybrid orbital and a fluorine 2p orbital (Figure 10.12). The BF3 molecule is planar withall the FBF angles equal to 120°. This result conforms to experimental findings andalso to VSEPR predictions.

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y y y y

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FIGURE 10.10 The lineargeometry of BeCl2 can be ex-plained by assuming that Be issp-hybridized. The two sp hybridorbitals overlap with the two chlo-rine 3p orbitals to form two cova-lent bonds.

sp2 is pronounced “s-p two.”



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FIGURE 10.11 Formation ofsp2 hybrid orbitals.

You may have noticed an interesting connection between hybridization and theoctet rule. Regardless of the type of hybridization, an atom starting with one s andthree p orbitals would still possess four orbitals, enough to accommodate a total ofeight electrons in a compound. For elements in the second period of the periodic table,eight is the maximum number of electrons that an atom of any of these elements canaccommodate in the valence shell. It is for this reason that the octet rule is usuallyobeyed by the second-period elements.

The situation is different for an atom of a third-period element. If we use only the3s and 3p orbitals of the atom to form hybrid orbitals in a molecule, then the octet ruleapplies. However, in some molecules the same atom may use one or more 3d orbitals,in addition to the 3s and 3p orbitals, to form hybrid orbitals. In these cases the octetrule does not hold. We will see specific examples of the participation of the 3d orbitalin hybridization shortly.

To summarize our discussion of hybridization, we note that

• The concept of hybridization is not applied to isolated atoms. It is a theoreticalmodel used only to explain covalent bonding.

• Hybridization is the mixing of at least two nonequivalent atomic orbitals, for ex-ample, s and p orbitals. Therefore, a hybrid orbital is not a pure atomic orbital.Hybrid orbitals and pure atomic orbitals have very different shapes.

• The number of hybrid orbitals generated is equal to the number of pure atomic or-bitals that participate in the hybridization process.

• Hybridization requires an input of energy; however, the system more than recoversthis energy during bond formation.

• Covalent bonds in polyatomic molecules and ions are formed by the overlap of hy-brid orbitals, or of hybrid orbitals with unhybridized ones. Therefore, the hy-bridization bonding scheme is still within the framework of valence bond theory;electrons in a molecule are assumed to occupy hybrid orbitals of the individualatoms.

Table 10.4 summarizes sp, sp2, and sp3 hybridization (as well as other types that wewill discuss shortly).


FIGURE 10.12 The sp2 hybridorbitals of boron overlap with the2p orbitals of fluorine. The BF3molecule is planar, and all theFBF angles are 120°.

F F

F

B



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TABLE 10.4 Important Hybrid Orbitals and Their Shapes

PURE ATOMIC HYBRIDIZATION

ORBITALS OF OF THE NUMBER OF SHAPE OF

THE CENTRAL CENTRAL HYBRID HYBRID

ATOM ATOM ORBITALS ORBITALS EXAMPLES

s, p sp 2 BeCl2

Linear

s, p, p sp2 3 BF3

Planar

s, p, p, p sp3 4 CH4, NH4�

Tetrahedral

s, p, p, p, d sp3d 5 PCl5


s, p, p, p, d, d sp3d2 6 SF6

Octahedral

180�

120�

109.5�

90�

120�

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Table 7.3 gives the electron configurations of all the elements.

PROCEDURE FOR HYBRIDIZING ATOMIC ORBITALS

Before going on to discuss the hybridization of d orbitals, let us specify what we needto know in order to apply hybridization to bonding in polyatomic molecules in gen-eral. In essence, hybridization simply extends Lewis theory and the VSEPR model. Toassign a suitable state of hybridization to the central atom in a molecule, we must havesome idea about the geometry of the molecule. The steps are as follows:

1. Draw the Lewis structure of the molecule.2. Predict the overall arrangement of the electron pairs (both bonding pairs and lone

pairs) using the VSEPR model (see Table 10.1).3. Deduce the hybridization of the central atom by matching the arrangement of the

electron pairs with those of the hybrid orbitals shown in Table 10.4.

The following example illustrates this procedure.

EXAMPLE 10.3

Determine the hybridization state of the central (underlined) atom in each of the fol-lowing molecules: (a) HgCl2, (b) AlI3, and (c) PF3. Describe the hybridizationprocess and determine the molecular geometry in each case.

Answer (a) The ground-state electron configuration of Hg is [Xe]6s24f 145d10;therefore, the Hg atom has two valence electrons (the 6s electrons). The Lewis struc-ture of HgCl2 is

There are no lone pairs on the Hg atom, so the arrangement of the two electron pairsis linear (see Table 10.1). From Table 10.4 we conclude that Hg is sp-hybridizedbecause it has the geometry of the two sp hybrid orbitals. The hybridization processcan be imagined to take place as follows. First we draw the orbital diagram for theground state of Hg:

6s 6p

By promoting a 6s electron to the 6p orbital, we get the excited state:

6s 6p

The 6s and 6p orbitals then mix to form two sp hybrid orbitals:

sp orbitals empty 6porbitals

The two HgOCl bonds are formed by the overlap of the Hg sp hybrid orbitals withthe 3p orbitals of the Cl atoms. Thus HgCl2 is a linear molecule.(b) The ground-state electron configuration of Al is [Ne]3s23p1. Therefore, Al hasthree valence electrons. The Lewis structure of AlI3 is

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Similar problems: 10.31, 10.32.

There are three bonding pairs and no lone pair on the Al atom. From Table 10.1 wesee that the shape of three electron pairs is trigonal planar, and from Table 10.4 weconclude that Al must be sp2-hybridized in AlI3. The orbital diagram of the ground-state Al atom is

3s 3p

By promoting a 3s electron into the 3p orbital we obtain the following excited state:

3s 3p

The 3s and two 3p orbitals then mix to form three sp2 hybrid orbitals:

sp2 orbitals empty3p orbital

The sp2 hybrid orbitals overlap with the 5p orbitals of I to form three covalent AlOIbonds. We predict that the AlI3 molecule is trigonal planar and all the IAlI anglesare 120°.(c) The ground-state electron configuration of P is [Ne]3s23p3. Therefore, the P atomhas five valence electrons. The Lewis structure of PF3 is

There are three bonding pairs and one lone pair on the P atom. In Table 10.1 wesee that the overall arrangement of four electron pairs is tetrahedral, and from Table10.4 we conclude that P must be sp3-hybridized. The orbital diagram of the ground-state P atom is

3s 3p

By mixing the 3s and 3p orbitals, we obtain four sp3 hybrid orbitals.

sp3 orbitals

As in the case of NH3, one of the sp3 hybrid orbitals is used to accommodate thelone pair on P. The other three sp3 hybrid orbitals form covalent POF bonds withthe 2p orbitals of F. We predict the geometry of the molecule to be trigonal pyra-midal; the FPF angle should be somewhat less than 109.5°.

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sp3d2 is pronounced “s-p three dtwo.”

PRACTICE EXERCISEDetermine the hybridization state of the underlined atoms in the following com-pounds: (a) SiBr4 and (b) BCl3.

HYBRIDIZATION OF s, p, AND d ORBITALS

We have seen that hybridization neatly explains bonding that involves s and p orbitals.For elements in the third period and beyond, however, we cannot always account formolecular geometry by assuming that only s and p orbitals hybridize. To understandthe formation of molecules with trigonal bipyramidal and octahedral geometries, forinstance, we must include d orbitals in the hybridization concept.

Consider the SF6 molecule as an example. In Section 10.1 we saw that this mol-ecule has octahedral geometry, which is also the arrangement of the six electron pairs.Table 10.4 shows that the S atom is sp3d2-hybridized in SF6. The ground-state elec-tron configuration of S is [Ne]3s23p4:

3s 3p 3d

Since the 3d level is quite close in energy to the 3s and 3p levels, we can promote 3sand 3p electrons to two of the 3d orbitals:

3s 3p 3d

Mixing the 3s, three 3p, and two 3d orbitals generates six sp3d2 hybrid orbitals:

sp3d2 orbitals empty 3d orbitals

The six SOF bonds are formed by the overlap of the hybrid orbitals of the S atom andthe 2p orbitals of the F atoms. Since there are 12 electrons around the S atom, the octetrule is violated. The use of d orbitals in addition to s and p orbitals to form an expanded octet (see Section 9.9) is an example of valence-shell expansion. Second-period elements, unlike third-period elements, do not have 2d energy levels, so theycan never expand their valence shells. Hence atoms of second-period elements cannever be surrounded by more than eight electrons in any of their compounds.

Example 10.4 deals with valence-shell expansion in a third-period element.

EXAMPLE 10.4

Describe the hybridization state of phosphorus in phosphorus pentabromide (PBr5).

Answer The Lewis structure of PBr5 is

BrSO

QO

OSS

AEHA

P

SO

SQ

Q

S

S O

Br

BrBr

Br

Q

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Recall that when n � 2, � � 0 and1. Thus we can only have 2s and

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FIGURE 10.13 The sp2 hy-bridization of a carbon atom.The 2s orbital is mixed with onlytwo 2p orbitals to form threeequivalent sp2 hybrid orbitals.This process leaves an electron inthe unhybridized orbital, the 2pzorbital.

Table 10.1 indicates that the arrangement of five electron pairs is trigonal bipyra-midal. Referring to Table 10.4, we find that this is the shape of five sp3d hybrid or-bitals. Thus P must be sp3d-hybridized in PBr5. The ground-state electron configu-ration of P is [Ne]3s23p3. To describe the hybridization process, we start with theorbital diagram for the ground state of P:

3s 3p 3d

Promoting a 3s electron into a 3d orbital results in the following excited state:

3s 3p 3d

Mixing the one 3s, three 3p, and one 3d orbitals generates five sp3d hybrid orbitals:

sp3d orbitals empty 3d orbitals

These hybrid orbitals overlap the 4p orbitals of Br to form five covalent POBrbonds. Since there are no lone pairs on the P atom, the geometry of PBr5 is trigo-nal bipyramidal.

PRACTICE EXERCISEDescribe the hybridization state of Se in SeF6.

The concept of hybridization is useful also for molecules with double and triple bonds.Consider the ethylene molecule, C2H4, as an example. In Example 10.1 we saw thatC2H4 contains a carbon-carbon double bond and has planar geometry. Both the geom-etry and the bonding can be understood if we assume that each carbon atom is sp2-hybridized. Figure 10.13 shows orbital diagrams of this hybridization process. We assumethat only the 2px and 2py orbitals combine with the 2s orbital, and that the 2pz orbital remains unchanged. Figure 10.14 shows that the 2pz orbital is perpendicular to theplane of the hybrid orbitals. Now how do we account for the bonding of the C atoms?As Figure 10.15(a) shows, each carbon atom uses the three sp2 hybrid orbitals to formtwo bonds with the two hydrogen 1s orbitals and one bond with the sp2 hybrid orbitalof the adjacent C atom. In addition, the two unhybridized 2pz orbitals of the C atomsform another bond by overlapping sideways [Figure 10.15(b)].

A distinction is made between the two types of covalent bonds in C2H4. The threebonds formed by each C atom in Figure 10.15(a) are all sigma bonds (� bonds), co-valent bonds formed by orbitals overlapping end-to-end, with the electron density con-centrated between the nuclei of the bonding atoms. The second type is called a pi bond(� bond), which is defined as a covalent bond formed by sideways overlapping or-bitals with electron density concentrated above and below the plane of the nuclei ofthe bonding atoms. The two C atoms form a pi bond as shown in Figure 10.15(b). It

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10.5 HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS 393

10.5 HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS

Similar problem: 10.40.

Groundstate

Promotionof electron

sp2-Hybridizedstate

2s

2s

sp2 orbitals

2p

2p

2pz

FIGURE 10.14 Each carbonatom in the C2H4 molecule hasthree sp2 hybrid orbitals (green)and one unhybridized 2pz orbital(gray), which is perpendicular tothe plane of the hybrid orbitals.

120°

90°



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FIGURE 10.15 Bonding in eth-ylene, C2H4. (a) Top view of thesigma bonds between carbonatoms and between carbon andhydrogen atoms. All the atoms liein the same plane, making C2H4a planar molecule. (b) Side viewshowing how the two 2pz orbitalson the two carbon atoms over-lap, leading to the formation of api bond. (c) The interactions in(a) and (b) lead to the formationof the sigma bonds and the pibond in ethylene. Note that thepi bond lies above and belowthe plane of the molecule.

is this pi bond formation that gives ethylene its planar geometry. Figure 10.15(c) showsthe orientation of the sigma and pi bonds. Figure 10.16 is yet another way of lookingat the planar C2H4 molecule and the formation of the pi bond. Although we normallyrepresent the carbon-carbon double bond as CPC (as in a Lewis structure), it is im-portant to keep in mind that the two bonds are different types: One is a sigma bondand the other is a pi bond.

The acetylene molecule (C2H2) contains a carbon-carbon triple bond. Since themolecule is linear, we can explain its geometry and bonding by assuming that each Catom is sp-hybridized by mixing the 2s with the 2px orbital (Figure 10.17). As Figure10.18 shows, the two sp hybrid orbitals of each C atom form one sigma bond with ahydrogen 1s orbital and another sigma bond with the other C atom. In addition, twopi bonds are formed by the sideways overlap of the unhybridized 2py and 2pz orbitals.Thus the CqC bond is made up of one sigma bond and two pi bonds.

The following rule helps us predict hybridization in molecules containing multi-ple bonds: If the central atom forms a double bond, it is sp2-hybridized; if it forms twodouble bonds or a triple bond, it is sp-hybridized. Note that this rule applies only to


(a)

(b) (c)

π

π

σσ σ

σσH H

HH

2pz 2pz

C C

H 1s H 1s

H 1s H 1s

CC

C C

2pz 2pz

C CH

H

H

HC C

H

H

H

H

π

π

FIGURE 10.16 Another viewof pi bond formation in the C2H4molecule. Note that all six atomsare in the same plane. It is theoverlap of the 2pz orbitals thatcauses the molecule to assume aplanar structure.

Groundstate

Promotionof electron

sp-Hybridizedstate

2s

2s

sp orbitals

2p

2p

2pz

2py

FIGURE 10.17 The sp hy-bridization of a carbon atom.The 2s orbital is mixed with onlyone 2p orbital to form two sp hy-brid orbitals. This process leavesan electron in each of the twounhybridized 2p orbitals, namely,the 2py and 2pz orbitals.



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FIGURE 10.18 Bonding inacetylene, C2H2. (a) Top viewshowing the overlap of the sp or-bitals between the C atoms andthe overlap of the sp orbital withthe 1s orbital between the C andH atoms. All the atoms lie alonga straight line; therefore acety-lene is a linear molecule. (b) Sideview showing the overlap of thetwo 2py orbitals and of the two2pz orbitals of the two carbonatoms, which leads to the forma-tion of two pi bonds. (c) Forma-tion of the sigma and pi bondsas a result of the interactions in(a) and (b).

atoms of the second-period elements. Atoms of third-period elements and beyond thatform multiple bonds present a more complicated picture and will not be dealt withhere.

EXAMPLE 10.5

Describe the bonding in the formaldehyde molecule whose Lewis structure is

Assume that the O atom is sp2-hybridized.

Answer We note that the C atom (a second-period element) has a double bond;therefore, it is sp2-hybridized. The orbital diagram of the O atom is

sp2 2pz

Two of the sp2 orbitals of the C atom form two sigma bonds with the H atoms, andthe third sp2 orbital forms a sigma bond with a sp2 orbital of the O atom. The 2pz

orbital of the C atom overlaps with the 2pz orbital of the O atom to form a pi bond.The two lone pairs on the O atom are placed in its two remaining sp2 orbitals (Figure10.19).

PRACTICE EXERCISEDescribe the bonding in the hydrogen cyanide molecule, HCN. Assume that N issp-hybridized.

hhghgh

QCG

H

DH

POO

10.5 HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS 395

(b)

2pz

C

2pz

C

2py

2py

(a)

H 1s H 1s

C C

(c)

π

σ

H

C

H

π

π

π

σ

C


FIGURE 10.19 Bonding in theformaldehyde molecule. A sigmabond is formed by the overlap ofthe sp2 hybrid orbital of carbonand the sp2 hybrid orbital of oxy-gen; a pi bond is formed by theoverlap of the 2pz orbitals of thecarbon and oxygen atoms. Thetwo lone pairs on oxygen areplaced in the other two sp2 or-bitals of oxygen.

H

Hπ

σC O

πσ

σ

••••



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FIGURE 10.20 Liquid oxygencaught between the poles of amagnet, because the O2 mole-cules are paramagnetic.

Valence bond theory is one of the two quantum mechanical approaches that explainbonding in molecules. It accounts, at least qualitatively, for the stability of the cova-lent bond in terms of overlapping atomic orbitals. Using the concept of hybridization,valence bond theory can explain molecular geometries predicted by the VSEPR model.However, the assumption that electrons in a molecule occupy atomic orbitals of the in-dividual atoms can only be an approximation, since each bonding electron in a mole-cule must be in an orbital that is characteristic of the molecule as a whole.

In some cases, valence bond theory cannot satisfactorily account for observedproperties of molecules. Consider the oxygen molecule, whose Lewis structure is

According to this description, all the electrons in O2 are paired and the molecule shouldtherefore be diamagnetic. But experiments have shown that the oxygen molecule isparamagnetic, with two unpaired electrons (Figure 10.20). This finding suggests a fun-damental deficiency in valence bond theory, one that justifies searching for an alter-native bonding approach that accounts for the properties of O2 and other moleculesthat do not match the predictions of valence bond theory.

Magnetic and other properties of molecules are sometimes better explained by an-other quantum mechanical approach called molecular orbital (MO) theory. Molecularorbital theory describes covalent bonds in terms of molecular orbitals, which resultfrom interaction of the atomic orbitals of the bonding atoms and are associated withthe entire molecule. The difference between a molecular orbital and an atomic orbitalis that an atomic orbital is associated with only one atom.

BONDING AND ANTIBONDING MOLECULAR ORBITALS

According to MO theory, the overlap of the 1s orbitals of two hydrogen atoms leadsto the formation of two molecular orbitals: one bonding molecular orbital and one an-tibonding molecular orbital. A bonding molecular orbital has lower energy and greaterstability than the atomic orbitals from which it was formed. An antibonding molecu-lar orbital has higher energy and lower stability than the atomic orbitals from whichit was formed. As the names “bonding” and “antibonding” suggest, placing electrons

OQ QOOPO


10.6 MOLECULAR ORBITAL THEORY



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FIGURE 10.21 Constructive in-terference (a) and destructive in-terference (b) of two waves of thesame wavelength and amplitude.

in a bonding molecular orbital yields a stable covalent bond, whereas placing electronsin an antibonding molecular orbital results in an unstable bond.

In the bonding molecular orbital the electron density is greatest between the nu-clei of the bonding atoms. In the antibonding molecular orbital, on the other hand, theelectron density decreases to zero between the nuclei. We can understand this distinc-tion if we recall that electrons in orbitals have wave characteristics. A property uniqueto waves allows waves of the same type to interact in such a way that the resultantwave has either an enhanced amplitude or a diminished amplitude. In the former case,we call the interaction constructive interference; in the latter case, it is destructive in-terference (Figure 10.21).

The formation of bonding molecular orbitals corresponds to constructive inter-ference (the increase in amplitude is analogous to the buildup of electron density be-tween the two nuclei). The formation of antibonding molecular orbitals corresponds todestructive interference (the decrease in amplitude is analogous to the decrease in elec-tron density between the two nuclei). The constructive and destructive interactions be-tween the two 1s orbitals in the H2 molecule, then, lead to the formation of a sigmabonding molecular orbital (1s) and a sigma antibonding molecular orbital (�

1s):

where the star denotes an antibonding molecular orbital.In a sigma molecular orbital (bonding or antibonding) the electron density is con-

centrated symmetrically around a line between the two nuclei of the bonding atoms.Two electrons in a sigma molecular orbital form a sigma bond (see Section 10.5).Remember that a single covalent bond (such as HOH or FOF) is almost always asigma bond.

Figure 10.22 shows the molecular orbital energy level diagram— that is, the rel-ative energy levels of the orbitals produced in the formation of the H2 molecule—and

a sigma antibondingmolecular orbital

formed from1s orbitals

1s

a sigma bondingmolecular orbital

formed from1s orbitals

1s�

10.6 MOLECULAR ORBITAL THEORY 397

Wave 1 Wave 1

Wave 2 Wave 2

Sum of 1 and 2 Sum of 1 and 2

(a) (b)

The two electrons in the sigma molecular orbital are paired. ThePauli exclusion principle applies

to molecules as well as to atoms.

FIGURE 10.22 (a) Energy levels of bonding and antibonding molecular orbitals in the H2 mol-ecule. Note that the two electrons in the 1s orbital must have opposite spins in accord with thePauli exclusion principle. Keep in mind that the higher the energy of the molecular orbital, theless stable the electrons in that molecular orbital. (b) Constructive and destructive interactions be-tween the two hydrogen 1s orbitals lead to the formation of a bonding and an antibonding mole-cular orbital. In the bonding molecular orbital, there is a buildup between the nuclei of electrondensity, which acts as a negatively charged “glue” to hold the positively charged nuclei together.

Atom

1s

Molecule

1s�σ

Atom

1s

1sσ

Destructiveinteraction

Constructiveinteraction

Antibonding sigmamolecular orbital

Bonding sigmamolecular orbital

Ene

rgy

(a) (b)



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the constructive and destructive interactions between the two 1s orbitals. Notice thatin the antibonding molecular orbital there is a node between the nuclei that signifieszero electron density. The nuclei are repelled by each other’s positive charges, ratherthan held together. Electrons in the antibonding molecular orbital have higher energy(and less stability) than they would have in the isolated atoms. On the other hand, elec-trons in the bonding molecular orbital have less energy (and hence greater stability)than they would have in the isolated atoms.

Although we have used the hydrogen molecule to illustrate molecular orbital for-mation, the concept is equally applicable to other molecules. In the H2 molecule weconsider only the interaction between 1s orbitals; with more complex molecules weneed to consider additional atomic orbitals as well. Nevertheless, for all s orbitals, theprocess is the same as for 1s orbitals. Thus, the interaction between two 2s or 3s or-bitals can be understood in terms of the molecular orbital energy level diagram and theformation of bonding and antibonding molecular orbitals shown in Figure 10.22.

For p orbitals the process is more complex because they can interact with eachother in two different ways. For example, two 2p orbitals can approach each other end-to-end to produce a sigma bonding and a sigma antibonding molecular orbital, as shownin Figure 10.23(a). Alternatively, the two p orbitals can overlap sideways to generatea bonding and an antibonding pi molecular orbital [Figure 10.23(b)].

In a pi molecular orbital (bonding or antibonding), the electron density is concentratedabove and below an imaginery line joining the two nuclei of the bonding atoms. Twoelectrons in a pi molecular orbital form a pi bond (see Section 10.5). A double bondis almost always composed of a sigma bond and a pi bond; a triple bond is always asigma bond plus two pi bonds.

To understand properties of molecules, we must know how electrons are distributedamong molecular orbitals. The procedure for determining the electron configuration ofa molecule is analogous to the one we use to determine the electron configurations ofatoms (see Section 7.8).

RULES GOVERNING MOLECULAR ELECTRON CONFIGURATION AND STABILITY

In order to write the electron configuration of a molecule, we must first arrange themolecular orbitals in order of increasing energy. Then we can use the following guide-lines to fill the molecular orbitals with electrons. The rules also help us understand thestabilities of the molecular orbitals.

• The number of molecular orbitals formed is always equal to the number of atomicorbitals combined.

a pi antibondingmolecular orbital

formed from2p orbitals

p

a pi bonding

formed from2p orbitals

�2 �2p

molecular orbital

�


10.7 MOLECULAR ORBITAL CONFIGURATIONS



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• The more stable the bonding molecular orbital, the less stable the correspondingantibonding molecular orbital.

• The filling of molecular orbitals proceeds from low to high energies. In a stablemolecule, the number of electrons in bonding molecular orbitals is always greaterthan that in antibonding molecular orbitals because we place electrons first in thelower-energy bonding molecular orbitals.

• Like an atomic orbital, each molecular orbital can accommodate up to two elec-trons with opposite spins in accordance with the Pauli exclusion principle.

• When electrons are added to molecular orbitals of the same energy, the most sta-ble arrangement is predicted by Hund’s rule; that is, electrons enter these molecu-lar orbitals with parallel spins.

• The number of electrons in the molecular orbitals is equal to the sum of all the elec-trons on the bonding atoms.

10.7 MOLECULAR ORBITAL CONFIGURATIONS 399

Atom

2p

Molecule

2p�σ

Atom

2p

2pσ

Ene

rgy

(a)

Atom

2p

Molecule

2p�π

Atom

2p

2pπ

Ene

rgy

(b)

Antibonding sigmamolecular orbital

Bonding sigmamolecular orbital

Antibonding pimolecular orbital

Bonding pimolecular orbital

+

Destructive interaction

Constructive interaction

+

+

Destructive interaction

Constructive interaction

+

FIGURE 10.23 Two possible interactions between two equivalent p orbitals and the corre-sponding molecular orbitals. (a) When the p orbitals overlap end-to-end, a sigma bonding and asigma antibonding molecular orbital form. (b) When the p orbitals overlap side-to-side, a pi bond-ing and a pi antibonding molecular orbital form. Normally, a sigma bonding molecular orbital ismore stable than a pi bonding molecular orbital, since side-to-side interaction leads to a smalleroverlap of the p orbitals than does end-to-end interaction. We assume that the 2px orbitals takepart in the sigma molecular orbital formation. The 2py and 2pz orbitals can interact to form only� molecular orbitals. The behavior shown in (b) represents the interaction between the 2py or-bitals or the 2pz orbitals.



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FIGURE 10.24 Energy levelsof the bonding and antibondingmolecular orbitals in H2

�, H2,He2

�, and He2. In all thesespecies, the molecular orbitalsare formed by the interaction oftwo 1s orbitals.

HYDROGEN AND HELIUM MOLECULES

Later in this section we will study molecules formed by atoms of the second-periodelements. Before we do, it will be instructive to predict the relative stabilities of thesimple species H2

�, H2, He2�, and He2, using the energy-level diagrams shown in Figure

10.24. The 1s and �1s orbitals can accommodate a maximum of four electrons. The

total number of electrons increases from one for H2� to four for He2. The Pauli exclu-

sion principle stipulates that each molecular orbital can accommodate a maximum oftwo electrons with opposite spins. We are concerned only with the ground-state elec-tron configurations in these cases.

To evaluate the stabilities of these species we determine their bond order, definedas

bond order � �12

� (number of electrons�

number of electronsin bonding MOs in antibonding MOs) (10.2)

The bond order indicates the strength of a bond. For example, if there are two elec-trons in the bonding molecular orbital and none in the antibonding molecular orbital,the bond order is one, which means that there is one covalent bond and that the mol-ecule is stable. Note that the bond order can be a fraction, but a bond order of zero (ora negative value) means the bond has no stability and the molecule cannot exist. Bondorder can be used only qualitatively for purposes of comparison. For example, a bond-ing sigma molecular orbital with two electrons and a bonding pi molecular orbital withtwo electrons would each have a bond order of one. Yet, these two bonds must differin bond strength (and bond length) because of the differences in the extent of atomicorbital overlap.

We are ready now to make predictions about the stability of H2�, H2, He2

�, andHe2 (see Figure 10.24). The H2

� molecular ion has only one electron, in the 1s orbital.Since a covalent bond consists of two electrons in a bonding molecular orbital, H2

� hasonly half of one bond, or a bond order of �

12

�. Thus, we predict that the H2� molecule

may be a stable species. The electron configuration of H2� is written as (1s)

1.The H2 molecule has two electrons, both of which are in the 1s orbital. According

to our scheme, two electrons equal one full bond; therefore, the H2 molecule has abond order of one, or one full covalent bond. The electron configuration of H2 is (1s)

2.As for the He2

� molecular ion, we place the first two electrons in the 1s orbitaland the third electron in the �

1s orbital. Because the antibonding molecular orbital isdestabilizing, we expect He2

� to be less stable than H2. Roughly speaking, the insta-bility resulting from the electron in the �

1s orbital is balanced by one of the 1s elec-


1s�σ

1sσE

nerg

y

1s�σ

1sσ

1s�σ

1sσ

1s�σ

1sσ

H+2 H2 He+

2 He2

The quantitative measure of thestrength of a bond is bond disso-

ciation energy, or bond energy (Section 9.10).

The superscript in (1s)1 indicates

that there is one electron in thesigma bonding molecular orbital.



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FIGURE 10.25 Molecular or-bital energy level diagram for theLi2 molecule. The six electrons inLi2 (Li’s electron configuration is1s22s1) are in the 1s, �

1s, and2s orbitals. Since there are twoelectrons each in 1s and �

1s(just as in He2), there is no netbonding or antibonding effect.Therefore, the single covalentbond in Li2 is formed by the twoelectrons in the bonding molecu-lar orbital 2s.

trons. The bond order is �12

�(2 � 1) � �12

� and the overall stability of He2� is similar to that

of the H2� molecule. The electron configuration of He2

� is (1s)2(�

1s)1.

In He2 there would be two electrons in the 1s orbital and two electrons in the�

1s orbital, so the molecule would have a bond order of zero and no net stability. Theelectron configuration of He2 would be (1s)

2(�1s)

2.To summarize, we can arrange our examples in order of decreasing stability:

H2 � H2�, He2

� � He2

We know that the hydrogen molecule is a stable species. Our simple molecular orbitalmethod predicts that H2

� and He2� also possess some stability, since both have bond

orders of �12

�. Indeed, their existence has been confirmed by experiment. It turns out thatH2

� is somewhat more stable than He2�, since there is only one electron in the hydro-

gen molecular ion and therefore it has no electron-electron repulsion. Furthermore, H2�

also has less nuclear repulsion than He2�. Our prediction about He2 is that it would

have no stability, but in 1993 He2 gas was found to exist. The “molecule” is extremelyunstable and has only a transient existence under specially created conditions.

HOMONUCLEAR DIATOMIC MOLECULES OF SECOND-PERIOD ELEMENTS

We are now ready to study the ground-state electron configuration of molecules con-taining second-period elements. We will consider only the simplest case, that ofhomonuclear diatomic molecules, or diatomic molecules containing atoms of the sameelements.

Figure 10.25 shows the molecular orbital energy level diagram for the first mem-ber of the second period, Li2. These molecular orbitals are formed by the overlap of


Atom

Molecule

2s�σ

Atom

2s 2s

2sσ

Ene

rgy

1s�σ

1s 1s

1sσ



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1s and 2s orbitals. We will use this diagram to build up all the diatomic molecules, aswe will see shortly.

The situation is more complex when the bonding also involves p orbitals. Two porbitals can form either a sigma bond or a pi bond. Since there are three p orbitals foreach atom of a second-period element, we know that one sigma and two pi molecularorbitals will result from the constructive interaction. The sigma molecular orbital isformed by the overlap of the 2px orbitals along the internuclear axis, that is, the x-axis.The 2py and 2pz orbitals are perpendicular to the x-axis, and they will overlap side-ways to give two pi molecular orbitals. The molecular orbitals are called 2px

, �2py,

and �2pzorbitals, where the subscripts indicate which atomic orbitals take part in form-

ing the molecular orbitals. As shown in Figure 10.23, overlap of the two p orbitals isnormally greater in a molecular orbital than in a � molecular orbital, so we wouldexpect the former to be lower in energy. However, the energies of molecular orbitalsactually increase as follows:

1s �1s 2s �

2s �2py� �2pz

2px ��

2py� ��

2pz �

2px

The inversion of the 2pxorbital and the �2py

and �2pzcan be explained qualitatively.

The electrons in the 1s, �1s, 2s and �

2s orbitals tend to concentrate between the twonuclei. Since the 2px

orbital has a geometry similar to that of the s orbitals, its elec-trons will concentrate in the same region. Consequently, electrons in the 2px

orbitalwill experience greater repulsion from electrons in the filled s and s

� orbitals thanthose in the �2py and �2pz

orbitals.Note that the repulsion between the electrons in the 2px

and inner s orbitals de-creases from left to right in the second period. As a result, the 2px

orbital is higher inenergy than the �2py

and �2pzorbitals for B2, C2, and N2, but lower in energy than the

�2py and �2pzorbitals for O2 and F2.

With these concepts and Figure 10.26, which shows the order of increasing ener-gies for 2p molecular orbitals, we can write the electron configurations and predict themagnetic properties and bond orders of second-period homonuclear diatomic mole-cules. We will consider a few examples.

THE LITHIUM MOLECULE (Li2)

The electron configuration of Li is 1s22s1, so Li2 has a total of six electrons. Accordingto Figure 10.25, these electrons are placed (two each) in the 1s, �

1s, and 2s molec-ular orbitals. The electrons of 1s and �

1s make no net contribution to the bonding inLi2. Thus the electron configuration of the molecular orbitals in Li2 is (1s)

2(�1s)

2(2s)2.

Since there are two more electrons in the bonding molecular orbitals than in antibondingorbitals, the bond order is 1 [see Equation (10.2)]. We conclude that the Li2 moleculeis stable, and since it has no unpaired electron spins, it should be diamagnetic. Indeed,diamagnetic Li2 molecules are known to exist in the vapor phase.

THE CARBON MOLECULE (C2)

The carbon atom has the electron configuration 1s22s22p2; thus there are 12 electronsin the C2 molecule. From the bonding scheme for Li2, we place four additional carbonelectrons in the �2py

and �2pzorbitals. Therefore, C2 has the electron configuration




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(1s)2(�

1s)2(2s)

2(�2s)

2(�2py)2(�2pz)

2

Its bond order is 2, and the molecular should be diamagnetic. Again, diamagnetic C2

molecules have been detected in the vapor state. Note that the double bonds in C2 areboth pi bonds because of the four electrons in the two pi molecular orbitals. In mostother molecules, a double bond is made up of a sigma bond and a pi bond.

THE OXYGEN MOLECULE (O2)

As we stated earlier, valence bond theory does not account for the magnetic propertiesof the oxygen molecule. To show the two unpaired electrons on O2, we need to drawan alternative to the resonance structure present on p. 396:

This structure is unsatisfactory on at least two counts. First, it implies the presence ofa single covalent bond, but experimental evidence strongly suggests that there is a dou-ble bond in this molecule. Second, it places seven valence electrons around each oxy-gen atom, a violation of the octet rule.

The ground-state electron configuration of O is 1s22s22p4; thus there are 16 elec-trons in O2. Using the order of increasing energies of the molecular orbitals discussedabove, we write the ground-state electron configuration of O2 as

(1s)2(�

1s)2(2s)

2(�2s)

2(2px)2(�2py)

2(�2pz)2(��

2py)1(��

2pz)1

T TOQOQOO O


FIGURE 10.26 General molecular orbital energy-level diagram for the second-period homonu-clear diatomic molecules Li2, Be2, B2, C2, and N2. For simplicity, the 1s and 2s orbitals havebeen omitted. Note that in these molecules the 2px orbital is higher in energy than either the�2py or the �2pz orbitals. This means that electrons in the 2px orbitals are less stable than thosein �2py and �2pz. This aberration stems from the different interactions between the electrons inthe 2px orbital, on one hand, and �2py and �2pz orbitals, on the other hand, with the electronsin the lower-energy s orbitals. For O2 and F2, the 2px orbital is lower in energy than �2py and�2pz.

2py

�π

Molecule

2px

�σ

2pxσ

2pyπ

Ene

rgy

Atom

2px 2py 2pz

Atom

2px 2py 2pz

2pzπ

2pz

�π



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According to Hund’s rule, the last two electrons enter the ��2py and ��

2pz orbitals withparallel spins. Ignoring the 1s and 2s orbitals (since their net effects on bonding arezero), we calculate the bond order of O2 using Equation (10.2):


�(6 � 2) � 2

Therefore, the O2 molecule has a bond order of 2 and is paramagnetic, a predictionthat corresponds to experimental observations.

Table 10.5 summarizes the general properties of the stable diatomic molecules ofthe second period.

Example 10.6 shows how MO theory can help predict molecular properties ofions.

EXAMPLE 10.6

The N2� ion can be prepared by bombarding the N2 molecule with fast-moving elec-

trons. Predict the following properties of N2�: (a) electron configuration, (b) bond

order, (c) magnetic character, and (d) bond length relative to the bond length of N2

(is it longer or shorter?).

Answer (a) Since N2� has one fewer electron than N2, its electron configuration is

(see Table 10.5)

(1s)2(�

1s)2(2s)

2(�2s)

2(�2py)2(�2pz)

2(2px)1

(b) The bond order of N2� is found by using Equation (10.2):


�(9 � 4) � 2.5

(c) N2� has one unpaired electron, so it is paramagnetic.


TABLE 10.5 Properties of Homonuclear Diatomic Molecules of the Second-period Elements*

Li2 B2 C2 N2 O2 F2

�2px �

2px

��2py, �

�2pz ��

2py, ��2pz

2px�2py

, �2pz

�2py, �2pz 2px

�2s �

2s

2s 2s

Bond order 1 1 2 3 2 1Bond length (pm) 267 159 131 110 121 142Bond dissociation 104.6 288.7 627.6 941.4 498.7 156.9

energy (kJ/mol)Magnetic properties Diamagnetic Paramagnetic Diamagnetic Diamagnetic Paramagnetic Diamagnetic

*For simplicity the 1s and �1s orbitals are omitted. These two orbitals hold a total of four electrons. Remember that for O2 and F2, 2px is lower in energy than �2py

and �2pz.

hghghghghghg

hghghghghg

hghghghghghghh

hghghghghg

hghghh



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FIGURE 10.27 The sigmabond framework in the benzenemolecule. Each carbon atom issp2-hybridized and forms sigmabonds with two adjacent carbonatoms and another sigma bondwith a hydrogen atom.

(d) Since the electrons in the bonding molecular orbitals are responsible for hold-ing the atoms together, N2

� should have a weaker and, therefore, longer bond thanN2. (In fact, the bond length of N2

� is 112 pm, compared with 110 pm for N2.)

PRACTICE EXERCISEWhich of the following species has a longer bond length: F2 or F2

�?

So far we have discussed chemical bonding only in terms of electron pairs. However,the properties of a molecule cannot always be explained accurately by a single struc-ture. A case in point is the O3 molecule, discussed in Section 9.8. There we overcamethe dilemma by introducing the concept of resonance. In this section we will tackle theproblem in another way—by applying the molecular orbital approach. As in Section9.8, we will use the benzene molecule and the carbonate ion as examples. Note that indiscussing the bonding of polyatomic molecules or ions, it is convenient to determinefirst the hybridization state of the atoms present (a valence bond approach), followedby the formation of appropriate molecular orbitals.

THE BENZENE MOLECULE

Benzene (C6H6) is a planar hexagonal molecule with carbon atoms situated at the sixcorners. All carbon-carbon bonds are equal in length and strength, as are all carbon-hydrogen bonds, and the CCC and HCC angles are all 120°. Therefore, each carbonatom is sp2-hybridized; it forms three sigma bonds with two adjacent carbon atoms anda hydrogen atom (Figure 10.27). This arrangement leaves an unhybridized 2pz orbitalon each carbon atom, perpendicular to the plane of the benzene molecule, or benzenering, as it is often called. So far the description resembles the configuration of ethyl-ene (C2H4), discussed in Section 10.5, except that in this case there are six unhybridized2pz orbitals in a cyclic arrangement.

Because of their similar shape and orientation, each 2pz orbital overlaps two oth-ers, one on each adjacent carbon atom. According to the rules listed on p. 398, the in-teraction of six 2pz orbitals leads to the formation of six pi molecular orbitals, of whichthree are bonding and three antibonding. A benzene molecule in the ground state there-fore has six electrons in the three pi bonding molecular orbitals, two electrons withpaired spins in each orbital (Figure 10.28).

Unlike the pi bonding molecular orbitals in ethylene, those in benzene form de-localized molecular orbitals, which are not confined between two adjacent bondingatoms, but actually extend over three or more atoms. Therefore, electrons residing inany of these orbitals are free to move around the benzene ring. For this reason, thestructure of benzene is sometimes represented as

in which the circle indicates that the pi bonds between carbon atoms are not confinedto individual pairs of atoms; rather, the pi electron densities are evenly distributed

10.8 DELOCALIZED MOLECULAR ORBITALS 405

10.8 DELOCALIZED MOLECULAR ORBITALS

H H

H H

H

H

C C

C

C C

C




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throughout the benzene molecule. The carbon and hydrogen atoms are not shown inthe simplified diagram.

We can now state that each carbon-to-carbon linkage in benzene contains a sigmabond and a “partial” pi bond. The bond order between any two adjacent carbon atomsis therefore between 1 and 2. Thus molecular orbital theory offers an alternative to theresonance approach, which is based on valence bond theory. (The resonance structuresof benzene are shown on p. 349.)

THE CARBONATE ION

Cyclic compounds like benzene are not the only ones with delocalized molecular or-bitals. Let’s look at bonding in the carbonate ion (CO3

2�). VSEPR predicts a trigonalplanar geometry for the carbonate ion, like that for BF3. The planar structure of thecarbonate ion can be explained by assuming that the carbon atom is sp2-hybridized.The C atom forms sigma bonds with three O atoms. Thus the unhybridized 2pz orbitalof the C atom can simultaneously overlap the 2pz orbitals of all three O atoms (Figure10.29). The result is a delocalized molecular orbital that extends over all four nucleiin such a way that the electron densities (and hence the bond orders) in the carbon-to-


FIGURE 10.29 Bonding in the carbonate ion. The carbon atom forms three sigma bonds withthe three oxygen atoms. In addition, the 2pz orbitals of the carbon and oxygen atoms overlap toform delocalized molecular orbitals, so that there is also a partial pi bond between the carbonatom and each of the three oxygen atoms.

Top view Side view

(b)(a)

FIGURE 10.28 (a) The six 2pz orbitals on the carbon atoms in benzene. (b) The delocalizedmolecular orbital formed by the overlap of the 2pz orbitals. The delocalized molecular orbital pos-sesses pi symmetry and lies above and below the plane of the benzene ring. Actually, these 2pzorbitals can combine in six different ways to yield three bonding molecular orbitals and three an-tibonding molecular orbitals. The one shown here is the most stable.

O C

O

O

O C

O

O



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oxygen bonds are all the same. Molecular orbital theory therefore provides an accept-able alternative explanation of the properties of the carbonate ion as compared withthe resonance structures of the ion shown on p. 349.

We should note that molecules with delocalized molecular orbitals are generallymore stable than those containing molecular orbitals extending over only two atoms.For example, the benzene molecule, which contains delocalized molecular orbitals, ischemically less reactive (and hence more stable) than molecules containing “localized”CPC bonds, such as ethylene.

• � � Q r (10.1) Expressing dipole moment in terms of charge (Q) and distance of separation (r) between charges.

• bond order � �12

��number of electrons�

number of electronsin bonding MOs in antibonding MOs� (10.2)

1. The VSEPR model for predicting molecular geometry is based on the assumption that valence-shell electron pairs repel one another and tend to stay as far apart as possible.

2. According to the VSEPR model, molecular geometry can be predicted from the numberof bonding electron pairs and lone pairs. Lone pairs repel other pairs more forcefully thanbonding pairs do and thus distort bond angles from the ideal geometry.

3. Dipole moment is a measure of the charge separation in molecules containing atoms ofdifferent electronegativities. The dipole moment of a molecule is the resultant of whateverbond moments are present. Information about molecular geometry can be obtained fromdipole moment measurements.

4. There are two quantum mechanical explanations for covalent bond formation: valencebond theory and molecular orbital theory. In valence bond theory, hybridized atomic or-bitals are formed by the combination and rearrangement of orbitals from the same atom.The hybridized orbitals are all of equal energy and electron density, and the number ofhybridized orbitals is equal to the number of pure atomic orbitals that combine.

5. Valence-shell expansion can be explained by assuming hybridization of s, p, and d or-bitals.

6. In sp hybridization, the two hybrid orbitals lie in a straight line; in sp2 hybridization, thethree hybrid orbitals are directed toward the corners of a triangle; in sp3 hybridization, thefour hybrid orbitals are directed toward the corners of a tetrahedron; in sp3d hybridiza-tion, the five hybrid orbitals are directed toward the corners of a trigonal bipyramid; in sp3d2

hybridization, the six hybrid orbitals are directed toward the corners of an octahedron.7. In an sp2-hybridized atom (for example, carbon), the one unhybridized p orbital can form

a pi bond with another p orbital. A carbon-carbon double bond consists of a sigma bondand a pi bond. In an sp-hybridized carbon atom, the two unhybridized p orbitals can formtwo pi bonds with two p orbitals on another atom (or atoms). A carbon-carbon triple bondconsists of one sigma bond and two pi bonds.

8. Molecular orbital theory describes bonding in terms of the combination and rearrange-ment of atomic orbitals to form orbitals that are associated with the molecule as a whole.

9. Bonding molecular orbitals increase electron density between the nuclei and are lower inenergy than individual atomic orbitals. Antibonding molecular orbitals have a region ofzero electron density between the nuclei, and an energy level higher than that of the indi-vidual atomic orbitals.

10. We write electron configurations for molecular orbitals as we do for atomic orbitals, fill-ing in electrons in the order of increasing energy levels. The number of molecular orbitalsalways equals the number of atomic orbitals that were combined. The Pauli exclusionprinciple and Hund’s rule govern the filling of molecular orbitals.

SUMMARY OF FACTS AND CONCEPTS 407

SUMMARY OF KEY EQUATIONS

SUMMARY OF FACTS AND CONCEPTS



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11. Molecules are stable if the number of electrons in bonding molecular orbitals is greaterthan that in antibonding molecular orbitals.

12. Delocalized molecular orbitals, in which electrons are free to move around a whole mole-cule or group of atoms, are formed by electrons in p orbitals of adjacent atoms. Delocalizedmolecular orbitals are an alternative to resonance structures in explaining observed molecu-lar properties.


KEY WORDSAntibonding molecular

orbital, p. 397Bond order, p. 400Bonding molecular orbital,

p. 397Delocalized molecular orbital,

p. 405

Dipole moment (�), p. 377Homonuclear diatomic

molecule, p. 401Hybrid orbital, p. 385Hybridization, p. 385Molecular orbital, p. 396Nonpolar molecule, p. 378

Pi bond (� bond), p. 393Pi molecular orbital, p. 398Polar molecule, p. 378Sigma bond ( bond), p. 393Sigma molecular orbital,

p. 397Valence shell, p. 368

Valence-shell electron-pairrepulsion (VSEPR) model,p. 368

QUESTIONS AND PROBLEMSMOLECULAR GEOMETRY

Review Questions

10.1 How is the geometry of a molecule defined and whyis the study of molecular geometry important?

10.2 Sketch the shape of a linear triatomic molecule, a trig-onal planar molecule containing four atoms, a tetra-hedral molecule, a trigonal bipyramidal molecule,and an octahedral molecule. Give the bond angles ineach case.

10.3 How many atoms are directly bonded to the centralatom in a tetrahedral molecule, a trigonal bipyrami-dal molecule, and an octahedral molecule?

10.4 Discuss the basic features of the VSEPR model.Explain why the magnitude of repulsion decreases inthe following order: lone pair-lone pair � lone pair-bonding pair � bonding pair-bonding pair.

10.5 In the trigonal bipyramidal arrangement, why does alone pair occupy an equatorial position rather than anaxial position?

10.6 The geometry of CH4 could be square planar, withthe four H atoms at the corners of a square and theC atom at the center of the square. Sketch this geom-etry and compare its stability with that of a tetrahe-dral CH4 molecule.

Problems

10.7 Predict the geometries of the following species usingthe VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4,(d) TeCl4.

10.8 Predict the geometries of the following species:(a) AlCl3, (b) ZnCl2, (c) ZnCl4

2�.

10.9 Predict the geometry of the following molecules us-ing the VSEPR method: (a) HgBr2, (b) N2O (arrange-ment of atoms is NNO), (c) SCN� (arrangement ofatoms is SCN).

10.10 Predict the geometries of the following ions:(a) NH4

�, (b) NH2�, (c) CO3

2� (d) ICl2�, (e) ICl4

�,(f ) AlH4

�, (g) SnCl5�, (h) H3O�, (i) BeF4

2�.10.11 Describe the geometry around each of the three cen-

tral atoms in the CH3COOH molecule.10.12 Which of the following species are tetrahedral? SiCl4,

SeF4, XeF4, CI4, CdCl42�

DIPOLE MOMENTS

Review Questions

10.13 Define dipole moment. What are the units and sym-bol for dipole moment?

10.14 What is the relationship between the dipole momentand bond moment? How is it possible for a moleculeto have bond moments and yet be nonpolar?

10.15 Explain why an atom cannot have a permanent di-pole moment.

10.16 The bonds in beryllium hydride (BeH2) molecules arepolar, and yet the dipole moment of the molecule iszero. Explain.

Problems

10.17 Referring to Table 10.3, arrange the following mole-



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cules in order of increasing dipole moment: H2O,H2S, H2Te, H2Se.

10.18 The dipole moments of the hydrogen halides decreasefrom HF to HI (see Table 10.3). Explain this trend.

10.19 List the following molecules in order of increasingdipole moment: H2O, CBr4, H2S, HF, NH3, CO2.

10.20 Does the molecule OCS have a higher or lower di-pole moment than CS2?

10.21 Which of the following molecules has a higher di-pole moment?

10.22 Arrange the following compounds in order of in-creasing dipole moment:

PG

D G

DC

Br

H

C

H

Br

PG

D G

DC

Br H

C

H Br(a) (b)

10.29 How would you distinguish between a sigma bondand a pi bond?

10.30 Which of the following pairs of atomic orbitals of ad-jacent nuclei can overlap to form a sigma bond?Which overlap to form a pi bond? Which cannot over-lap (no bond)? Consider the x-axis to be the internu-clear axis, that is, the line joining the nuclei of thetwo atoms. (a) 1s and 1s, (b) 1s and 2px, (c) 2px and2py, (d) 3py and 3py, (e) 2px and 2px, (f ) 1s and 2s

Problems

10.31 Describe the bonding scheme of the AsH3 moleculein terms of hybridization.

10.32 What is the hybridization of Si in SiH4 and inH3SiOSiH3?

10.33 Describe the change in hybridization (if any) of theAl atom in the following reaction:

AlCl3 � Cl� 88n AlCl4�

10.34 Consider the reaction

BF3 � NH3 88n F3BONH3

Describe the changes in hybridization (if any) of theB and N atoms as a result of this reaction.

10.35 What hybrid orbitals are used by nitrogen atoms inthe following species? (a) NH3, (b) H2NONH2,(c) NO3

�

10.36 What are the hybrid orbitals of the carbon atoms inthe following molecules?(a) H3COCH3(b) H3COCHPCH2(c) CH3OCqCOCH2OH(d) CH3CHPO(e) CH3COOH

10.37 Specify which hybrid orbitals are used by carbonatoms in the following species: (a) CO, (b) CO2,(c) CN�.

10.38 What is the hybridization state of the central N atomin the azide ion, N3

�? (Arrangement of atoms: NNN.)10.39 The allene molecule H2CPCPCH2 is linear (the

three C atoms lie on a straight line). What are the hy-bridization states of the carbon atoms? Draw dia-grams to show the formation of sigma bonds and pibonds in allene.

10.40 Describe the hybridization of phosphorus in PF5.10.41 How many sigma bonds and pi bonds are there in

each of the following molecules?

C

(c)

P qOA

HCH3C OCCOA

H

H

PG

D G

DH

H

Cl

(b)

C C

H

H

CCl

(a)

AOO

AH

Cl

QUESTIONS AND PROBLEMS 409

(d)(a)

Cl

ClCl

Cl ClCl

ClCl

Cl

ClCl(b) (c)

VALENCE BOND THEORY

Review Questions

10.23 What is valence bond theory? How does it differ fromthe Lewis concept of chemical bonding?

10.24 Use valence bond theory to explain the bonding inCl2 and HCl. Show how the atomic orbitals overlapwhen a bond is formed.

10.25 Draw a potential energy curve for the bond forma-tion in F2.

HYBRIDIZATION

Review Questions

10.26 What is the hybridization of atomic orbitals? Why isit impossible for an isolated atom to exist in the hy-bridized state?

10.27 How does a hybrid orbital differ from a pure atomicorbital? Can two 2p orbitals of an atom hybridize togive two hybridized orbitals?

10.28 What is the angle between the following two hybridorbitals on the same atom? (a) sp and sp hybrid or-bitals, (b) sp2 and sp2 hybrid orbitals, (c) sp3 and sp3

hybrid orbitals



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10.42 How many pi bonds and sigma bonds are there in thetetracyanoethylene molecule?

MOLECULAR ORBITAL THEORY

Review Questions

10.43 What is molecular orbital theory? How does it differfrom valence bond theory?

10.44 Define the following terms: bonding molecular or-bital, antibonding molecular orbital, pi molecular or-bital, sigma molecular orbital.

10.45 Sketch the shapes of the following molecular orbitals:1s, �

ls, �2p, and ��2p. How do their energies com-

pare?10.46 Explain the significance of bond order. Can bond or-

der be used for quantitative comparisons of thestrengths of chemical bonds?

Problems

10.47 Explain in molecular orbital terms the changes inHOH internuclear distance that occur as the molec-ular H2 is ionized first to H2

� and then to H22�.

10.48 The formation of H2 from two H atoms is an ener-getically favorable process. Yet statistically there isless than a 100 percent chance that any two H atomswill undergo the reaction. Apart from energy consid-erations, how would you account for this observationbased on the electron spins in the two H atoms?

10.49 Draw a molecular orbital energy level diagram foreach of the following species: He2, HHe, He2

�.Compare their relative stabilities in terms of bond or-ders. (Treat HHe as a diatomic molecule with threeelectrons.)

10.50 Arrange the following species in order of increasingstability: Li2, Li2

�, Li2�. Justify your choice with a

molecular orbital energy level diagram.10.51 Use molecular orbital theory to explain why the Be2

molecule does not exist.10.52 Which of these species has a longer bond, B2 or B2

�?Explain in terms of molecular orbital theory.

10.53 Acetylene (C2H2) has a tendency to lose two protons(H�) and form the carbide ion (C2

2�), which is pres-ent in a number of ionic compounds, such as CaC2and MgC2. Describe the bonding scheme in the C2

2�

ion in terms of molecular orbital theory. Compare thebond order in C2

2� with that in C2.

PG

D G

DC

C

q

C

C

C

qq

q C

N

N

N

N

10.54 Compare the Lewis and molecular orbital treatmentsof the oxygen molecule.

10.55 Explain why the bond order of N2 is greater than thatof N2

�, but the bond order of O2 is less than that ofO2

�.10.56 Compare the relative stability of the following species

and indicate their magnetic properties (that is, dia-magnetic or paramagnetic): O2, O2

�, O2� (superoxide

ion), O22� (peroxide ion).

10.57 Use molecular orbital theory to compare the relativestabilities of F2 and F2

�.10.58 A single bond is almost always a sigma bond, and a

double bond is almost always made up of a sigmabond and a pi bond. There are very few exceptionsto this rule. Show that the B2 and C2 molecules areexamples of the exceptions.

DELOCALIZED MOLECULAR ORBITALS

Review Questions

10.59 How does a delocalized molecular orbital differ froma molecular orbital such as that found in H2 or C2H4?What do you think are the minimum conditions (forexample, number of atoms and types of orbitals) forforming a delocalized molecular orbital?

10.60 In Chapter 9 we saw that the resonance concept isuseful for dealing with species such as the benzenemolecule and the carbonate ion. How does molecu-lar orbital theory deal with these species?

Problems

10.61 Both ethylene (C2H4) and benzene (C6H6) contain theCPC bond. The reactivity of ethylene is greater thanthat of benzene. For example, ethylene readily reactswith molecular bromine, whereas benzene is nor-mally quite inert toward molecular bromine and manyother compounds. Explain this difference in reactiv-ity.

10.62 Explain why the symbol on the left is a better repre-sentation of benzene molecules than that on the right.

10.63 Determine which of these molecules has a more de-localized orbital and justify your choice.




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(Hint: Both molecules contain two benzene rings. Innaphthalene, the two rings are fused together. Inbiphenyl, the two rings are joined by a single bond,around which the two rings can rotate.)

10.64 Nitryl fluoride (FNO2) is very reactive chemically.The fluorine and oxygen atoms are bonded to the ni-trogen atom. (a) Write a Lewis structure for FNO2.(b) Indicate the hybridization of the nitrogen atom.(c) Describe the bonding in terms of molecular or-bital theory. Where would you expect delocalizedmolecular orbitals to form?

10.65 Describe the bonding in the nitrate ion NO3� in terms

of delocalized molecular orbitals.10.66 What is the state of hybridization of the central O

atom in O3? Describe the bonding in O3 in terms ofdelocalized molecular orbitals.

ADDITIONAL PROBLEMS

10.67 Which of the following species is not likely to havea tetrahedral shape? (a) SiBr4, (b) NF4

�, (c) SF4,(d) BeCl4

2�, (e) BF4�, (f ) AlCl4

�

10.68 Draw the Lewis structure of mercury(II) bromide. Isthis molecule linear or bent? How would you estab-lish its geometry?

10.69 Sketch the bond moments and resultant dipole mo-ments for the following molecules: H2O, PCl3, XeF4,PCl5, SF6.

10.70 Although both carbon and silicon are in Group 4A,very few SiPSi bonds are known. Account for theinstability of silicon-to-silicon double bonds in gen-eral. (Hint: Compare the atomic radii of C and Si inFigure 8.5. What effect would the larger size have onpi bond formation?)

10.71 Predict the geometry of sulfur dichloride (SCl2) andthe hybridization of the sulfur atom.

10.72 Antimony pentafluoride, SbF5, reacts with XeF4 andXeF6 to form ionic compounds, XeF3

�SbF6� and

XeF5�SbF6

�. Describe the geometries of the cationsand anion in these two compounds.

10.73 Draw Lewis structures and give the other informa-tion requested for the following molecules: (a) BF3.Shape: planar or nonplanar? (b) ClO3

�. Shape: planaror nonplanar? (c) H2O. Show the direction of the re-sultant dipole moment. (d) OF2. Polar or nonpolarmolecule? (e) NO2. Estimate the ONO bond angle.

10.74 Predict the bond angles for the following molecules:(a) BeCl2, (b) BCl3, (c) CCl4, (d) CH3Cl, (e) Hg2Cl2(arrangement of atoms: ClHgHgCl), (f) SnCl2,(g) H2O2, (i) SnH4.

10.75 Briefly compare the VSEPR and hybridization ap-proaches to the study of molecular geometry.

10.76 Describe the hybridization state of arsenic in arsenicpentafluoride (AsF5).

10.77 Draw Lewis structures and give the other informa-tion requested for the following: (a) SO3. Polar ornonpolar molecule? (b) PF3. Polar or nonpolar mol-ecule? (c) F3SiH. Show the direction of the resultantdipole moment. (d) SiH3

�. Planar or pyramidal shape?(e) Br2CH2. Polar or nonpolar molecule?

10.78 Which of the following molecules are linear? ICl2�,

IF2�, OF2, SnI2, CdBr2

10.79 Draw the Lewis structure for the BeCl42� ion. Predict

its geometry and describe the hybridization state ofthe Be atom.

10.80 The N2F2 molecule can exist in either of the follow-ing two forms:

(a) What is the hybridization of N in the molecule?(b) Which structure has a dipole moment?

10.81 Cyclopropane (C3H6) has the shape of a triangle inwhich a C atom is bonded to two H atoms and twoother C atoms at each corner. Cubane (C8H8) has theshape of a cube in which a C atom is bonded to oneH atom and three other C atoms at each corner.(a) Draw Lewis structures of these molecules.(b) Compare the CCC angles in these molecules withthose predicted for an sp3-hybridized C atom.(c) Would you expect these molecules to be easy tomake?

10.82 The compound 1,2-dichloroethane (C2H4Cl2) is non-polar, while cis-dichloroethylene (C2H2Cl2) has a di-pole moment:

The reason for the difference is that groups connectedby a single bond can rotate with respect to each other,but no rotation occurs when a double bond connectsthe groups. On the basis of bonding considerations,explain why rotation occurs in 1,2-dichloroethane butnot in cis-dichloroethylene.

PG

D G

D

H

Cl

C C

cis-dichloroethylene

Cl

H

OA

HCAHH

OCHA

ClCl

1,2-dichloroethane

OA

PG D

F

NN

F

PD

DNN

F

F




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10.83 Does the following molecule have a dipole moment?

(Hint: See the answer to Problem 10.39.)10.84 So-called greenhouse gases, which contribute to

global warming, have a dipole moment or can be bentor distorted into shapes that have a dipole moment.Which of the following gases are greenhouse gases?N2, O2, O3, CO, CO2, NO2, N2O, CH4, CFCl3

10.85 The bond angle of SO2 is very close to 120°, eventhough there is a lone pair on S. Explain.

10.86 3′-azido-3′-deoxythymidine, commonly known asAZT, is one of the drugs used to treat acquired immunedeficiency syndrome (AIDS). What are the hybridiza-tion states of the C and N atoms in this molecule?

10.87 The following molecules (AX4Y2) all have octahe-dral geometry. Group the molecules that are equiva-lent to each other.

A

CH3

CH2

OC

H

N

HOO O

C

CC

C

N

N

H

H

O

A A A A

AAH

H

CH

H

C

C

B

B

N

N

H

PG

D G

DC C

Cl

PC

Cl

H

10.88 The compounds carbon tetrachloride (CCl4) and sil-icon tetrachloride (SiCl4) are similar in geometry andhybridization. However, CCl4 does not react with wa-ter but SiCl4 does. Explain the difference in theirchemical reactivities. (Hint: The first step of the re-action is believed to be the addition of a water mol-ecule to the Si atom in SiCl4.)

10.89 Write the ground-state electron configuration for B2.Is the molecule diamagnetic or paramagnetic?

10.90 What are the hybridization states of the C and Natoms in this molecule?

10.91 Use molecular orbital theory to explain the differencebetween the bond energies of F2 and F2

� (see Problem9.108).

10.92 Referring to the Chemistry in Action on p. 382, an-swer the following questions: (a) If you wanted tocook a roast (beef or lamb), would you use a mi-crowave oven or a conventional oven? (b) Radar is ameans of locating an object by measuring the timefor the echo of a microwave from the object to re-turn to the source and the direction from which it re-turns. Would radar work if oxygen, nitrogen, and car-bon dioxide were polar molecules? (c) In early testsof radar at the English Channel during World War II,the results were inconclusive even though there wasno equipment malfunction. Why? (Hint: The weatheris often foggy in that region.)

10.93 The geometries discussed in this chapter all lend

NC H

O

NH2

H

HNC C

C




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themselves to fairly straightforward elucidation ofbond angles. The exception is the tetrahedron, be-cause its bond angles are hard to visualize. Considerthe CCl4 molecule, which has a tetrahedral geometryand is nonpolar. By equating the bond moment of aparticular COCl bond to the resultant bond momentsof the other three COCl bonds in opposite directions,show that the bond angles are all equal to 109.5°.

10.94 Referring to Table 9.4, explain why the bond disso-ciation energy for Cl2 is greater than that for F2. (Hint:The bond lengths of F2 and Cl2 are 142 pm and199 pm, respectively.)

10.95 Use molecular orbital theory to explain the bondingin the azide ion (N3

�).

Answers to Practice Exercises: 10.1 (a) Tetrahedral,(b) linear, (c) trigonal planar. 10.2 No. 10.3 (a) sp3, (b) sp2. 10.4sp3d2. 10.5 The C atom is sp-hybridized. It forms a sigma bondwith the H atom and another sigma bond with the N atom. Thetwo unhybridized p orbitals on the C atom are used to form twopi bonds with the N atom. The lone pair on the N atom is placedin the sp orbital. 10.6 F2

�.




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C H E M I S T R Y I N T H R E E D I M E N S I O N S

Buckyball and Other Large Allotropes of Carbon

414

In 1985 chemists at Rice University in Texas used a high-powered laser to vaporizegraphite in an effort to create unusual molecules believed to exist in interstellar space.Mass spectrometry revealed that one of the products was an unknown species with theformula C60. Because of its size and the fact that it is pure carbon, this molecule hasan exotic shape, which the researchers worked out using paper, scissors, and tape.Subsequent spectroscopic and X-ray measurements confirmed that C60 is shaped likea hollow sphere with a carbon atom at each of the 60 vertices. Geometrically, bucky-ball (short for “buckminsterfullerene”) is the most symmetrical molecule known. Inspite of its unique features, however, its bonding scheme is straightforward. Each car-bon is sp2-hybridized, and there are extensive delocalized molecular orbitals over theentire structure.

The discovery of buckyball generated tremendous interest within the scientificcommunity. Here was a new allotrope of carbon with an intriguing geometry and un-known properties to investigate. Since 1985 chemists have created a whole class offullerenes, with 70, 76, and even larger numbers of carbon atoms. Moreover, bucky-ball has been found to be a natural component of soot, and the fullerenes C60 and C70

turned up in a rock sample from northwestern Russia.Buckyball and its heavier members represent a whole new concept in molecular

architecture with far-reaching implications. Studies have shown that the fullerenes andtheir compounds can act as high-temperature superconductors and lubricants, as wellas catalysts. One fascinating discovery, made in 1991 by Japanese scientists, was theidentification of structural relatives of buckyball. These molecules are hundreds of

Geodesic domes were a favoritedesign element of the architectand philosopher R. BuckminsterFuller, for whom buckminster-fullerene was named because ofits structural resemblance toFuller’s domes.

The geometry of a buckyball C60 (top) resembles a soccer ball(bottom). Scientists arrived at this structure by fitting together pa-per cutouts of enough hexagons and pentagons to accommodate60 carbon atoms at the points where they intersect.

335 pm

Graphite is made up of layers of six-memberedrings of carbon.



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415

nanometers long with a tubular shape and an internal cavity about 15 nanometers indiameter. Dubbed “buckytubes” or “nanotubes” (because of their size), these moleculeshave two distinctly different structures. One is a single sheet of graphite that is cappedat both ends with a kind of truncated buckyball. The other is a scroll-like tube havinganywhere from 2 to 30 graphite-like layers. Buckytubes are several times stronger thansteel wires of similar dimensions and one day they might be used to make ultralightbicycles, rocket motor casings, and tennis racquets. They could also serve as castingmolds for very thin metal wires to be used in microscopic integrated circuits or as “bot-tles” for storing molecules.

In the first biological application of buckyball, chemists at the University ofCalifornia at San Francisco and Santa Barbara made a discovery in 1993 that couldhelp in designing drugs to treat AIDS. The human immunodeficiency virus (HIV) thatcauses AIDS reproduces by synthesizing a long protein chain, which is cut into smallersegments by an enzyme called HIV-protease. One way to stop AIDS, then, might beto inactivate the enzyme. When the chemists reacted a water-soluble derivative ofbucky-ball with HIV-protease, they found that it binds to the portion of the enzymethat would ordinarily cleave the reproductive protein, thereby preventing the HIV virusfrom reproducing. Consequently, the virus could no longer infect the human cells theyhad grown in the laboratory. The buckyball compound itself is not a suitable drug foruse against AIDS because of potential side effects and delivery difficulties, but it doesprovide a model for the development of such drugs.

An electron micrograph of a mul-tiwalled buckytube. The thicknessof 14 adjacent graphite-like lay-ers is about 5 nanometers.

Computer-generated model of thebinding of a buckyball derivativeto the site of HIV-protease thatnormally attaches to a proteinneeded for the reproduction ofHIV. The buckyball structure fitstightly into the active site, thuspreventing the enzyme from car-rying out its function.

The structure of a buckytube thatconsists of a single layer of car-bon atoms. Note that the trun-cated buckyball “cap,” whichhas been separated from the restof the buckytube in this view, hasa different structure than thegraphite-like cylindrical portion ofthe tube. Chemists have devisedways to open the cap in order toplace other molecules inside thetube.



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