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# Advanced Theories of Covalent Bonding

## Chapter Outline

8.1 Valence Bond Theory

8.2 Hybrid Atomic Orbitals

8.3 Multiple Bonds

8.4 Molecular Orbital Theory

Introduction

We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of s, p, and d atomic orbitals with molecular shapes that show angles like 120° and 109.5°? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both N2 and O2 have fairly similar Lewis structures that contain lone pairs of electrons.

Two Lewis diagrams are shown. The diagram on the left shows two nitrogen atoms, represented by the letter N connected by three lines and with a lone pair of electrons on each end of the structure. The diagram on the right shows two oxygen atoms, depicted by the letter O, connected by two lines. Two pairs of electrons surround each oxygen to the top and ends of the structure.

Yet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen (shown in Figure 18.1) is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations.

## 8.1 Valence Bond Theory

### Summary

By the end of this section, you will be able to:

• Describe the formation of covalent bonds in terms of atomic orbital overlap
• Define and give examples of σ and π bonds

As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts three-dimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding.

There are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an s orbital, a dumbbell shape for a p orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization.

Valence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: (1) an orbital on one atom overlaps an orbital on a second atom and (2) the single electrons in each orbital combine to form an electron pair. The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap.

The energy of the system depends on how much the orbitals overlap. Figure 8.2 illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure.

The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the H2 molecule shown in Figure 8.2, at the bond distance of 74 pm the system is 7.24 × 10−19 J lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, we know from our earlier description of thermochemistry that bond energies are often discussed on a per-mole basis. For example, it requires 7.24 × 10−19 J to break one H–H bond, but it takes 4.36 × 105 J to break 1 mole of H–H bonds. A comparison of some bond lengths and energies is shown in Table 8.1. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first C–H bond in CH4 requires 439.3 kJ/mol, while breaking the first C–H bond in H–CH2C6H5 (a common paint thinner) requires 375.5 kJ/mol.

In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two s orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure 8.3 illustrates this for two p orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.

The overlap of two s orbitals (as in H2), the overlap of an s orbital and a p orbital (as in HCl), and the end-to-end overlap of two p orbitals (as in Cl2) all produce sigma bonds (σ bonds), as illustrated in Figure 8.4. A σ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as σ bonds in valence bond theory.

A pi bond (π bond) is a type of covalent bond that results from the side-by-side overlap of two p orbitals, as illustrated in Figure 8.5. In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.

While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures in suggest, O2 contains a double bond, and N2 contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter.

As seen in Table 8.1, an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond.

### Example 8.1

Counting σ and π Bonds

This figure shows a molecule composed of four carbon atoms. There is a double bond between carbons one and two and three and four, while a single bond holds carbon two and three together. Carbons one and four are also bonded to two hydrogens with a single bond while carbons two and three are each bonded to one hydrogen each by a single bond.

Butadiene, C6H6, is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule.

Solution

There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall.

Identify each illustration as depicting a σ or π bond:

(a) side-by-side overlap of a 4p and a 2p orbital

(b) end-to-end overlap of a 4p and 4p orbital

(c) end-to-end overlap of a 4p and a 2p orbital

Three diagrams are shown and labeled “a,” “b,” and “c.” Diagram a depicts two peanut-shaped orbitals lying vertically side-by-side and overlapping. One orbital is smaller than the other. Diagram b shows two peanut-shaped orbitals lying end-to-end and overlapping. Diagram c shows two unequally sized peanut-shaped orbitals lying end-to-end and overlapping.

(a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis.

## Exercises

### Question 8.1

8.1

Explain how σ and π bonds are similar and how they are different.

### Question 8.2

8.2

Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.

(a) Use the bond energy found in Table 8.1 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)

(b) Use the enthalpy of reaction and the bond energies for $H_2$ and $Cl_2$ to solve for the energy of one mole of HCl bonds.

$H_2(g) + Cl_2(g) ⇌ 2HCl(g)\ \ \ \ \ \ \ \ \ \ \ ΔH°_{rxn }= −184.7 kJ/mol$ table: Representative Bond Energies and Lengths

### Question 8.3

8.3

Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.

### Question 8.4

8.4

Use valence bond theory to explain the bonding in $F_2$, HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.

### Question 8.5

8.5

Use valence bond theory to explain the bonding in $O_2.$ Sketch the overlap of the atomic orbitals involved in the bonds in $O_2.$

### Question 8.6

8.6

How many σ and π bonds are present in the molecule HCN?

### Question 8.7

8.7

A friend tells you $N_2$ has three π bonds due to overlap of the three p-orbitals on each N atom. Do you agree?

### Question 8.8

8.8

Draw the Lewis structures for $CO_2$ and CO, and predict the number of σ and π bonds for each molecule.

(a) $CO_2$

(b) CO

## 8.2 Hybrid Atomic Orbitals

### Summary

By the end of this section, you will be able to:

• Explain the concept of atomic orbital hybridization
• Determine the hybrid orbitals associated with various molecular geometries

Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration 1s22s22p4, with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure, because p orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.

Quantum-mechanical calculations suggest why the observed bond angles in H2O differ from those predicted by the overlap of the 1s orbital of the hydrogen atoms with the 2p orbitals of the oxygen atom. The mathematical expression known as the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a 2s orbital and three 2p orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 8.7). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.

The following ideas are important in understanding hybridization:

1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.

2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.

3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.

4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.

5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.

6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.

In the following sections, we shall discuss the common types of hybrid orbitals.

### sp Hybridization

The beryllium atom in a gaseous BeCl2 molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure 8.8). In this figure, the set of sp orbitals appears similar in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds.

We illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energy-level diagram in Figure 8.9. These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin.

When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the sp orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds.

Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit sp hybridization. Other examples include the mercury atom in the linear HgCl2 molecule, the zinc atom in Zn(CH3)2, which contains a linear C–Zn–C arrangement, and the carbon atoms in HCCH and CO2.

Check out the University of Wisconsin-Oshkosh website to learn about visualizing hybrid orbitals in three dimensions.

### sp2 Hybridization

The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three sp2 hybrid orbitals and one unhybridized p orbital. This arrangement results from sp2 hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 8.10).

Although quantum mechanics yields the “plump” orbital lobes as depicted in Figure 8.10, sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure 8.11, to avoid obscuring other features of a given illustration. We will use these “thinner” representations whenever the true view is too crowded to easily visualize.

The observed structure of the borane molecule, BH3, suggests sp2 hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure 8.12). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH3 as shown in the orbital energy level diagram in Figure 8.13. We redistribute the three valence electrons of the boron atom in the three sp2 hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds form.

Any central atom surrounded by three regions of electron density will exhibit sp2 hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure 8.14), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, CH2O, and ethene, H2CCH2.

### sp3 Hybridization

The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four sp3 hybrid orbitals. The hybrids result from the mixing of one s orbital and all three p orbitals that produces four identical sp3 hybrid orbitals (Figure 8.15). Each of these hybrid orbitals points toward a different corner of a tetrahedron.

A molecule of methane, CH4, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits sp3 hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH4 in Figure 8.16. The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds form.

In a methane molecule, the 1s orbital of each of the four hydrogen atoms overlaps with one of the four sp3 orbitals of the carbon atom to form a sigma (σ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH4.

The structure of ethane, C2H6, is similar to that of methane in that each carbon in ethane has four neighboring atoms arranged at the corners of a tetrahedron—three hydrogen atoms and one carbon atom (Figure 8.17). However, in ethane an sp3 orbital of one carbon atom overlaps end to end with an sp3 orbital of a second carbon atom to form a σ bond between the two carbon atoms. Each of the remaining sp3 hybrid orbitals overlaps with an s orbital of a hydrogen atom to form carbon–hydrogen σ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure 8.17. The orientation of the two CH3 groups is not fixed relative to each other. Experimental evidence shows that rotation around σ bonds occurs easily.

An sp3 hybrid orbital can also hold a lone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is sp3 hybridized with one hybrid orbital occupied by the lone pair.

The molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is sp3 hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.5°) are slightly smaller. Other examples of sp3 hybridization include CCl4, PCl3, and NCl3.

### sp3d and sp3d2 Hybridization

To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the s orbital, the three p orbitals, and one of the d orbitals), which gives five sp3d hybrid orbitals. With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives six sp3d2 hybrid orbitals. These hybridizations are only possible for atoms that have d orbitals in their valence subshells (that is, not those in the first or second period).

In a molecule of phosphorus pentachloride, PCl5, there are five P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form the set of five sp3d hybrid orbitals (Figure 8.19) that are involved in the P–Cl bonds. Other atoms that exhibit sp3d hybridization include the sulfur atom in SF4 and the chlorine atoms in ClF3 and in ClF4+. (The electrons on fluorine atoms are omitted for clarity.)

The sulfur atom in sulfur hexafluoride, SF6, exhibits sp3d2 hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the 3s orbital, the three 3p orbitals, and two of the 3d orbitals form six equivalent sp3d2 hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit sp3d2 hybridization include the phosphorus atom in PCl6, the iodine atom in the interhalogens IF6+, IF5, ICl4, IF4 and the xenon atom in XeF4.

### Assignment of Hybrid Orbitals to Central Atoms

The hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure 8.21. These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To find the hybridization of a central atom, we can use the following guidelines:

1. Determine the Lewis structure of the molecule.

2. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region.

3. Assign the set of hybridized orbitals from Figure that corresponds to this geometry.

It is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries. The model works well for molecules containing small central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data. For example, we have discussed the H–O–H bond angle in H2O, 104.5°, which is more consistent with sp3 hybrid orbitals (109.5°) on the central atom than with 2p orbitals (90°). Sulfur is in the same group as oxygen, and H2S has a similar Lewis structure. However, it has a much smaller bond angle (92.1°), which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for H2Te, the observed bond angle (90°) is consistent with overlap of the 5p orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures.

### Example 8.2

Assigning Hybridization

Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, SO42−?

Solution

The Lewis structure of sulfate shows there are four regions of electron density. The hybridization is sp3.

What is the hybridization of the selenium atom in SeF4?

The selenium atom is sp3d hybridized.

### Example 8.3

Assigning Hybridization

Urea, NH2C(O)NH2, is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of each nitrogen and carbon atom in urea?

Solution

The Lewis structure of urea is

The nitrogen atoms are surrounded by four regions of electron density, which arrange themselves in a tetrahedral electron-pair geometry. The hybridization in a tetrahedral arrangement is sp3 (Figure 8.21). This is the hybridization of the nitrogen atoms in urea.

The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is sp2 (Figure 8.21), which is the hybridization of the carbon atom in urea.

Acetic acid, H3CC(O)OH, is the molecule that gives vinegar its odor and sour taste. What is the hybridization of the two carbon atoms in acetic acid?

H3C, sp3; C(O)OH, sp2

## Exercises

### Question 8.9

8.9

Why is the concept of hybridization required in valence bond theory?

### Question 8.10

8.10

Give the shape that describes each hybrid orbital set:

(a) $sp^2$

(b) $sp^3d$

(c) sp

(d) $sp^3d^2$

### Question 8.11

8.11

Explain why a carbon atom cannot form five bonds using $sp^3d$ hybrid orbitals.

### Question 8.12

8.12

What is the hybridization of the central atom in each of the following?

(a) $BeH_2$

(b) $SF_6$

(c) $PO_4^{3−}$

(d) $PCl_5$

### Question 8.13

8.13

A molecule with the formula $AB_3$ could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.

### Question 8.14

8.14

Methionine, $CH_3SCH_2CH_2CH(NH_2)CO_2H,$ is an amino acid found in proteins. Draw a Lewis structure of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?

### Question 8.15

8.15

Sulfuric acid is manufactured by a series of reactions represented by the following equations:

$S_8(s)+8O_2(g)⟶8SO_2(g)$

$2SO_2(g)+O_2(g)⟶2SO_3(g)$

$SO_3(g)+H2O(l)⟶H_2SO_4(l)$

Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:

(a) circular $S_8$ molecule

(b) $SO_2$ molecule

(c) $SO_3$ molecule

(d) $H_2SO_4$ molecule (the hydrogen atoms are bonded to oxygen atoms)

### Question 8.16

8.16

Two important industrial chemicals, ethene, $C_2H_4,$ and propene, $C_3H_6$, are produced by the steam (or thermal) cracking process:

$2C_3H_8(g)⟶C_2H_4(g)+C_3H_6(g)+CH_4(g)+H_2(g)$

For each of the four carbon compounds, do the following:

(a) Draw a Lewis structure.

(b) Predict the geometry about the carbon atom.

(c) Determine the hybridization of each type of carbon atom.

### Question 8.17

8.17

For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass.

(a) What is the formula of the compound?

(b) Write a Lewis structure for the compound.

(c) Predict the shape of the molecules of the compound.

(d) What hybridization is consistent with the shape you predicted?

### Question 8.18

8.18

Consider nitrous acid,$HNO_2 (HONO).$

(a) Write a Lewis structure.

(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the $HNO_2$ molecule?

(c) What is the hybridization on the internal oxygen and nitrogen atoms in $HNO_2$?

### Question 8.19

8.19

Strike-anywhere matches contain a layer of $KClO_3$ and a layer of $P4_S_3.$ The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. $KClO_3$ contains the $ClO_3^−$ ion. $P_4S_3$ is an unusual molecule with the skeletal structure.

(a) Write Lewis structures for $P_4S_3$ and the $Cl_O^{3–}$ ion.

(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.

(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.

(d) Determine the oxidation states and formal charge of the atoms in $P_4S_3$ and the $Cl_O^{3–}$ ion.

### Question 8.20

8.20

Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)

8.21

### Question 8.39

8.39

Explain why an electron in the bonding molecular orbital in the $H_2$ molecule has a lower energy than an electron in the 1s atomic orbital of either of the separated hydrogen atoms.

### Question 8.40

8.40

Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.

(a) $Na_2^{2+}$

(b) $Mg_2{2+}$

(c) $Al_2{2+}$

(d) $Si_2{2+}$

(e) $P_2^{2+}$

(f) $S_2^{2+}$

(g) $F_2^{2+}$

(h)$Ar_2^{2+}$

### Question 8.41

8.41

Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.

(a) $H_2, H_2^+, H_2^−$

(b) $O_2, O_2^{2+}, O_2^{2−}$

(c) $Li_2, Be_2^+, Be_2$

(d) $F_2, F_2+, F_2^−$

(e) $N_2, N_2+, N_2^−$

### Question 8.42

8.42

For the first ionization energy for an $N_2$ molecule, what molecular orbital is the electron removed from?

### Question 8.43

8.43

Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:

(a) H and $H_2$

(b) N and $N_2$

(c) O and $O_2$

(d) C and $C_2$

(e) B and $B_2$

### Question 8.44

8.44

Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?

### Question 8.45

8.45

A friend tells you that the 2s orbital for fluorine starts off at a much lower energy than the 2s orbital for lithium, so the resulting $σ_{2s}$ molecular orbital in $F_2$ is more stable than in $Li_2.$ Do you agree?

### Question 8.46

8.46

True or false: Boron contains $2s^22p^1$ valence electrons, so only one p orbital is needed to form molecular orbitals.

### Question 8.47

8.47

What charge would be needed on $F_2$ to generate an ion with a bond order of 2?

### Question 8.48

8.48

Predict whether the MO diagram for $S_2$ would show s-p mixing or not.

### Question 8.49

8.49

Explain why $N_2^{2+}$ is diamagnetic, while $O_2^{4+}$, which has the same number of valence electrons, is paramagnetic.

### Question 8.50

8.50

Using the MO diagrams, predict the bond order for the stronger bond in each pair:

(a) $B_2 or B_2^+$

(b) $F_2 or F_2^+$

(c) $O_2 or O_2^{2+}$

(d) $C_2^+ or C_2^−$

## Key Terms

antibonding orbital: molecular orbital located outside of the region between two nuclei; electrons in an antibonding orbital destabilize the molecule

bond order: number of pairs of electrons between two atoms; it can be found by the number of bonds in a Lewis structure or by the difference between the number of bonding and antibonding electrons divided by two

bonding orbital: molecular orbital located between two nuclei; electrons in a bonding orbital stabilize a molecule

degenerate orbitals: orbitals that have the same energy

diamagnetism: phenomenon in which a material is not magnetic itself but is repelled by a magnetic field; it occurs when there are only paired electrons present

homonuclear diatomic molecule: molecule consisting of two identical atoms

hybrid orbital: orbital created by combining atomic orbitals on a central atom

hybridization: model that describes the changes in the atomic orbitals of an atom when it forms a covalent compound

linear combination of atomic orbitals: technique for combining atomic orbitals to create molecular orbitals

molecular orbital diagram: visual representation of the relative energy levels of molecular orbitals

molecular orbital theory: model that describes the behavior of electrons delocalized throughout a molecule in terms of the combination of atomic wave functions

molecular orbital: region of space in which an electron has a high probability of being found in a molecule

node: plane separating different lobes of orbitals, where the probability of finding an electron is zero

overlap: coexistence of orbitals from two different atoms sharing the same region of space, leading to the formation of a covalent bond

paramagnetism: phenomenon in which a material is not magnetic itself but is attracted to a magnetic field; it occurs when there are unpaired electrons present

pi bond (π bond): covalent bond formed by side-by-side overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis

sigma bond (σ bond): covalent bond formed by overlap of atomic orbitals along the internuclear axis

s-p mixing: change that causes σp orbitals to be less stable than πp orbitals due to the mixing of s and p-based molecular orbitals of similar energies.

sp hybrid orbital: one of a set of two orbitals with a linear arrangement that results from combining one s and one p orbital

sp2 hybrid orbital: one of a set of three orbitals with a trigonal planar arrangement that results from combining one s and two p orbitals

sp3 hybrid orbital: one of a set of four orbitals with a tetrahedral arrangement that results from combining one s and three p orbitals

sp3d hybrid orbital: one of a set of five orbitals with a trigonal bipyramidal arrangement that results from combining one s, three p, and one d orbital

sp3d2 hybrid orbital: one of a set of six orbitals with an octahedral arrangement that results from combining one s, three p, and two d orbitals

valence bond theory: description of bonding that involves atomic orbitals overlapping to form σ or π bonds, within which pairs of electrons are shared

π bonding orbital: molecular orbital formed by side-by-side overlap of atomic orbitals, in which the electron density is found on opposite sides of the internuclear axis

π* bonding orbital: antibonding molecular orbital formed by out of phase side-by-side overlap of atomic orbitals, in which the electron density is found on both sides of the internuclear axis, and there is a node between the nuclei

σ bonding orbital: molecular orbital in which the electron density is found along the axis of the bond

σ* bonding orbital: antibonding molecular orbital formed by out-of-phase overlap of atomic orbital along the axis of the bond, generating a node between the nuclei

## Summary

8.1 Valence Bond Theory

Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.

8.2 Hybrid Atomic Orbitals

We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization; three, sp2 hybridization; four, sp3 hybridization; five, sp3d hybridization; and six, sp3d2 hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).

8.3 Multiple Bonds

Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary.

8.4 Molecular Orbital Theory

Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called σ MOs. They can be formed from s orbitals or from p orbitals oriented in an end-to-end fashion. Molecular orbitals formed from p orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called π orbitals.

We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund’s rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory.

Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).

The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.

Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side-by-side to form the π bond.

No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.

Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.

There are no d orbitals in the valence shell of carbon.

trigonal planar, sp2; trigonal pyramidal (one lone pair on A) sp3; T-shaped (two lone pairs on A sp3d, or (three lone pairs on A) sp3d2

(a) Each S has a bent (109°) geometry, sp3

(b) Bent (120°), sp2

(c) Trigonal planar, sp2

(d) Tetrahedral, sp3

17. (a) XeF2

(b)

(c) linear (d) sp3d

(a)

(b)

(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp3; (d) Oxidation states P +1, S −1(1/3) , Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1

Phosphorus and nitrogen can form sp3 hybrids to form three bonds and hold one lone pair in PF3 and NF3, respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp3d hybrid orbitals to bind five fluorine atoms in NF5. Phosphorus has d orbitals and can bind five fluorine atoms with sp3d hybrid orbitals in PF5.

A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.

(a)

(b) The terminal carbon atom uses sp3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2p orbital on carbon and a nitrogen 2p orbital.

(a) sp2; (b) sp; (c) sp2; (d) sp3; (e) sp3; (f) sp3d; (g) sp3

(a) sp2, delocalized; (b) sp, localized; (c) sp2, delocalized; (d) sp3, delocalized

Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.

(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.

An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.

Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.

The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.

(a) H2 bond order = 1, H2 + bond order = 0.5, H2 − bond order = 0.5, strongest bond is H2; (b) O2 bond order = 2, O2 2+ bond order = 3; O2 2− bond order = 1, strongest bond is O2 2+; (c) Li2 bond order = 1, Be2 + bond order = 0.5, Be2 bond order = 0, Li2 and Be2 + have the same strength bond; (d) F2 bond order = 1, F2 + bond order = 1.5, F2 bond order = 0.5, strongest bond is F2 +; (e) N2 bond order = 3, N2 + bond order = 2.5, N2 bond order = 2.5, strongest bond is N2

(a) H2; (b) N2; (c) O; (d) C2; (e) B2

Yes, fluorine is a smaller atom than Li, so atoms in the 2s orbital are closer to the nucleus and more stable.

2+