484. $s p^{3}$ Hybridization -
In a methane molecule, the $1 s$ orbital of each of the four hydrogen atoms overlaps with one of the four $s p^{3}$ orbitals of the carbon atom to form a sigma ( $\sigma$ ) 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, $\mathrm{CH}_{4}$.
The structure of ethane, $\mathrm{C}_{2} \mathrm{H}_{6}$, 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 5.17). However, in ethane an $s p^{3}$ orbital of one carbon atom overlaps end to end with an $s p^{3}$ orbital of a second carbon atom to form a $\sigma$ bond between the two carbon atoms. Each of the remaining $s p^{3}$ hybrid orbitals overlaps with an $s$ orbital of a hydrogen atom to form carbon-hydrogen $\sigma$ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure 5.17. The orientation of the two $\mathrm{CH}_{3}$ groups is not fixed relative to each other. Experimental evidence shows that rotation around $\sigma$ bonds occurs easily.

FIGURE 5.17 (a) In the ethane molecule, $\mathrm{C}_{2} \mathrm{H}_{6}$, each carbon has four $s p^{3}$ orbitals. (b) These four orbitals overlap to form seven $\sigma$ bonds.

An $s p^{3}$ 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 $s p^{3}$ hybridized with one hybrid orbital occupied by the lone pair.


484. $s p^{3}$ Hybridization -
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 $s p^{3}$ 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 $109.5^{\circ}$, but the observed angles in ammonia ( $107.3^{\circ}$ ) and water ( $104.5^{\circ}$ ) are slightly smaller. Other examples of $s p^{3}$ hybridization include $\mathrm{CCl}_{4}, \mathrm{PCl}_{3}$, and $\mathrm{NCl}_{3}$.
$s p^{3} d$ and $s p^{3} d^{2}$ 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 $\boldsymbol{s p}^{\mathbf{3} \boldsymbol{d}}$ 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 $\boldsymbol{s p}^{3} \boldsymbol{d}^{2}$ 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).


484. $s p^{3}$ Hybridization -
In a molecule of phosphorus pentachloride, $\mathrm{PCl}_{5}$, there are five $\mathrm{P}-\mathrm{Cl}$ bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the $3 s$ orbital, the three $3 p$ orbitals, and one of the $3 d$ orbitals to form the set of five $s p^{3} d$ hybrid orbitals (Figure 5.19) that are involved in the $\mathrm{P}-\mathrm{Cl}$ bonds. Other atoms that exhibit $s p^{3} d$ hybridization include the sulfur atom in $\mathrm{SF}_{4}$ and the chlorine atoms in $\mathrm{ClF}_{3}$ and in $\mathrm{ClF}_{4}{ }^{+}$. (The electrons on fluorine atoms are omitted for clarity.)


484. $s p^{3}$ Hybridization -
FIGURE 5.18 The three compounds pictured exhibit $s p^{3} d$ hybridization in the central atom and a trigonal bipyramid form. $\mathrm{SF}_{4}$ and $\mathrm{ClF}_{4}{ }^{+}$have one lone pair of electrons on the central atom, and $\mathrm{ClF}_{3}$ has two lone pairs giving it the T -shape shown.

(a)

(b)

FIGURE 5.19 (a) The five regions of electron density around phosphorus in $\mathrm{PCl}_{5}$ require five hybrid $s p^{3} d$ orbitals. (b) These orbitals combine to form a trigonal bipyramidal structure with each large lobe of the hybrid orbital pointing at a vertex. As before, there are also small lobes pointing in the opposite direction for each orbital (not shown for clarity).

The sulfur atom in sulfur hexafluoride, $\mathrm{SF}_{6}$, exhibits $s p^{3} d^{2}$ 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 $3 s$ orbital, the three $3 p$ orbitals, and two of the $3 d$ orbitals form six equivalent $s p^{3} d^{2}$ hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit $s p^{3} d^{2}$ hybridization include the phosphorus atom in $\mathrm{PCl}_{6}{ }^{-}$, the iodine atom in the interhalogens $\mathrm{IF}_{6}{ }^{+}, \mathrm{IF}_{5}, \mathrm{ICl}_{4}{ }^{-}, \mathrm{IF}_{4}{ }^{-}$and the xenon atom in $\mathrm{XeF}_{4}$.


484. $s p^{3}$ Hybridization -
FIGURE 5.20 (a) Sulfur hexafluoride, $\mathrm{SF}_{6}$, has an octahedral structure that requires $s p^{3} d^{2}$ hybridization. (b) The six $s p^{3} d^{2}$ orbitals form an octahedral structure around sulfur. Again, the minor lobe of each orbital is not shown for clarity.


485. 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 5.21. 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 5.21 that corresponds to this geometry.

| Regions of Electron Density | Arrangement |  | Hybridization |  |
| :---: | :---: | :---: | :---: | :---: |
| 2 | ---------- | linear | $s p$ |  |
| 3 |  | trigonal planar | $s p^{2}$ |  |
| 4 |  | tetrahedral | $s p^{3}$ |  |
| 5 |  | trigonal bipyramidal | $s p^{3} d$ |  |
| 6 |  | octahedral | $s p^{3} d^{2}$ |  |

FIGURE 5.21 The shapes of hybridized orbital sets are consistent with the electron-pair geometries. For example, an atom surrounded by three regions of electron density is $s p^{2}$ hybridized, and the three $s p^{2}$ orbitals are arranged in a trigonal planar fashion.


485. Assignment of Hybrid Orbitals to Central Atoms -
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 $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angle in $\mathrm{H}_{2} \mathrm{O}, 104.5^{\circ}$, which is more consistent with $s p^{3}$ hybrid orbitals $\left(109.5^{\circ}\right)$ on the central atom than with $2 p$ orbitals $\left(90^{\circ}\right)$. Sulfur is in the same group as oxygen, and $\mathrm{H}_{2} \mathrm{~S}$ has a similar Lewis structure. However, it has a much smaller bond angle $\left(92.1^{\circ}\right)$, which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for $\mathrm{H}_{2} \mathrm{Te}$, the observed bond angle $\left(90^{\circ}\right)$ is consistent with overlap of the $5 p$ orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures.


487. Assigning Hybridization -
Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, $\mathrm{SO}_{4}{ }^{2-}$ ?


488. Solution -
The Lewis structure of sulfate shows there are four regions of electron density. The hybridization is $s p^{3}$.


489. Check Your Learning -
What is the hybridization of the selenium atom in $\mathrm{SeF}_{4}$ ?


490. Answer: -
The selenium atom is $s p^{3} d$ hybridized.


492. Assigning Hybridization -
Urea, $\mathrm{NH}_{2} \mathrm{C}(\mathrm{O}) \mathrm{NH}_{2}$, is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of the carbon atom in urea?


493. Solution -
The Lewis structure of urea is

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 $s p^{2}$ (Figure 5.21), which is the hybridization of the carbon atom in urea.


494. Check Your Learning -
Acetic acid, $\mathrm{H}_{3} \mathrm{CC}(\mathrm{O}) \mathrm{OH}$, is the molecule that gives vinegar its odor and sour taste. What is the hybridization of the two carbon atoms in acetic acid?


495. Answer: -
$\mathrm{H}_{3} \mathrm{C}, s p^{3} ; \underline{\mathrm{C}}(\mathrm{O}) \mathrm{OH}, s p^{2}$


496. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe multiple covalent bonding in terms of atomic orbital overlap
- Relate the concept of resonance to $\pi$-bonding and electron delocalization

The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of $\sigma$ and $\pi$ bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, $\mathrm{C}_{2} \mathrm{H}_{4}$, shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.

The three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the $\sigma$ bonds from each carbon atom are formed using a set of $s p^{2}$ hybrid orbitals that result from hybridization of two of the $2 p$ orbitals and the $2 s$ orbital (Figure 5.22). These orbitals form the $\mathrm{C}-\mathrm{H}$ single bonds and the $\sigma$ bond in the $\mathrm{C}=\mathrm{C}$ double bond (Figure 5.23). The $\pi$ bond in the $\mathbf{C}=\mathbf{C}$ double bond results from the overlap of the third (remaining) $2 p$ orbital on each carbon atom that is not involved in hybridization. This unhybridized $p$ orbital (lobes shown in red and blue in Figure 5.23) is perpendicular to the plane of the $s p^{2}$ hybrid orbitals. Thus the unhybridized $2 p$ orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure 5.23) and form a $\pi$ bond.

FIGURE 5.22 In ethene, each carbon atom is $s p^{2}$ hybridized, and the $s p^{2}$ orbitals and the $p$ orbital are singly occupied. The hybrid orbitals overlap to form $\sigma$ bonds, while the $p$ orbitals on each carbon atom overlap to form $a \pi$ bond.

(a)

(b)


496. LEARNING OBJECTIVES -
(a)

(b)

FIGURE 5.23 In the ethene molecule, $\mathrm{C}_{2} \mathrm{H}_{4}$, there are (a) five $\sigma$ bonds. One $\mathrm{C}-\mathrm{C} \sigma$ bond results from overlap of $s p^{2}$ hybrid orbitals on the carbon atom with one $s p^{2}$ hybrid orbital on the other carbon atom. Four $\mathrm{C}-\mathrm{H}$ bonds result from the overlap between the C atoms' $s p^{2}$ orbitals with $s$ orbitals on the hydrogen atoms. (b) The $\pi$ bond is formed by the side-by-side overlap of the two unhybridized $p$ orbitals in the two carbon atoms. The two lobes of the $\pi$ bond are above and below the plane of the $\sigma$ system.

In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of $s p^{2}$ hybrid orbitals tilted relative to each other, the $p$ orbitals would not be oriented to overlap efficiently to create the $\pi$ bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between $\sigma$ and $\pi$ bonds; rotation around single ( $\sigma$ ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the $\sigma$ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the $\pi$ bonding orbitals, essentially breaking the $\pi$ bond.


496. LEARNING OBJECTIVES -
In molecules with $s p$ hybrid orbitals, two unhybridized $p$ orbitals remain on the atom (Figure 5.24). We find this situation in acetylene, $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$, which is a linear molecule. The $s p$ hybrid orbitals of the two carbon atoms overlap end to end to form a $\sigma$ bond between the carbon atoms (Figure 5.25 ). The remaining $s p$ orbitals form $\sigma$ bonds with hydrogen atoms. The two unhybridized $p$ orbitals per carbon are positioned such that they overlap side by side and, hence, form two $\pi$ bonds. The two carbon atoms of acetylene are thus bound together by one $\sigma$ bond and two $\pi$ bonds, giving a triple bond.

FIGURE 5.24 Diagram of the two linear sp hybrid orbitals of a carbon atom, which lie in a straight line, and the two unhybridized $p$ orbitals at perpendicular angles.

(a)

(b)

FIGURE 5.25 (a) In the acetylene molecule, $\mathrm{C}_{2} \mathrm{H}_{2}$, there are two $\mathrm{C}-\mathrm{H} \sigma$ bonds and $\mathrm{a} \mathrm{C} \equiv \mathrm{C}$ triple bond involving one $\mathrm{C}-\mathrm{C} \sigma$ bond and two $\mathrm{C}-\mathrm{C} \pi$ bonds. The dashed lines, each connecting two lobes, indicate the side-by-side overlap of the four unhybridized $p$ orbitals. (b) This shows the overall outline of the bonds in $\mathrm{C}_{2} \mathrm{H}_{2}$. The two lobes of each of the $\pi$ bonds are positioned across from each other around the line of the $\mathrm{C}-\mathrm{C} \sigma$ bond.


496. LEARNING OBJECTIVES -
Hybridization involves only $\sigma$ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of $\pi$ bonds are possible. Since the arrangement of $\pi$ bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.

For example, molecule benzene has two resonance forms (Figure 5.26). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is $s p^{2}$. The electrons in the unhybridized $p$ orbitals form $\pi$ bonds. Neither resonance structure completely describes the electrons in the $\pi$ bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. (See the next module.)

FIGURE 5.26 Each carbon atom in benzene, $\mathrm{C}_{6} \mathrm{H}_{6}$, is $s p^{2}$ hybridized, independently of which resonance form is considered. The electrons in the $\pi$ bonds are not located in one set of $p$ orbitals or the other, but rather delocalized throughout the molecule.


498. Assignment of Hybridization Involving Resonance -
Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, $\mathrm{SO}_{2}$, is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in $\mathrm{SO}_{2}$ ?


499. Solution -
The resonance structures of $\mathrm{SO}_{2}$ are

The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is $s p^{2}$.


500. Check Your Learning -
Another acid in acid rain is nitric acid, $\mathrm{HNO}_{3}$, which is produced by the reaction of nitrogen dioxide, $\mathrm{NO}_{2}$, with atmospheric water vapor. What is the hybridization of the nitrogen atom in $\mathrm{NO}_{2}$ ? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)


501. Answer: -
$s p^{2}$


502. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Outline the basic quantum-mechanical approach to deriving molecular orbitals from atomic orbitals
- Describe traits of bonding and antibonding molecular orbitals
- Calculate bond orders based on molecular electron configurations
- Write molecular electron configurations for first- and second-row diatomic molecules
- Relate these electron configurations to the molecules' stabilities and magnetic properties

For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule $\mathrm{O}_{2}$, presents a problem with respect to its Lewis structure. We would write the following Lewis structure for $\mathrm{O}_{2}$ :

This electronic structure adheres to all the rules governing Lewis theory. There is an $\mathrm{O}=\mathrm{O}$ double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, $\mathrm{O}_{2}$ is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity, as in Figure 5.1. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of $\mathrm{O}_{2}$ indicates that all electrons are paired. How do we account for this discrepancy?

Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure 5.27), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.

FIGURE 5.27 A Gouy balance compares the mass of a sample in the presence of a magnetic field with the mass with the electromagnet turned off to determine the number of unpaired electrons in a sample.


502. LEARNING OBJECTIVES -
FIGURE 5.27 A Gouy balance compares the mass of a sample in the presence of a magnetic field with the mass with the electromagnet turned off to determine the number of unpaired electrons in a sample.

Experiments show that each $\mathrm{O}_{2}$ molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion.


503. LINK TO LEARNING -
Water, like most molecules, contains all paired electrons. Living things contain a large percentage of water, so they demonstrate diamagnetic behavior. If you place a frog near a sufficiently large magnet, it will levitate. You can see videos (http://openstax.org/l/16diamagnetic) of diamagnetic floating frogs, strawberries, and more.

Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table 5.1 summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.

| Valence Bond Theory | Molecular Orbital Theory |
| :--- | :--- |
| considers bonds as localized between one pair of
 atoms | considers electrons delocalized throughout the
 entire molecule |
| creates bonds from overlap of atomic orbitals ( $s, p, d . .)$.
 and hybrid orbitals $\left(s p, s p^{2}, s p^{3} \ldots\right)$ | combines atomic orbitals to form molecular
 orbitals $\left(\sigma, \sigma^{*}, \pi, \pi^{*}\right)$ |
| forms $\sigma$ or $\pi$ bonds | creates bonding and antibonding interactions
 based on which orbitals are filled |
| predicts molecular shape based on the number of
 regions of electron density | predicts the arrangement of electrons in
 molecules |
| needs multiple structures to describe resonance |  |

TABLE 5.1


503. LINK TO LEARNING -
TABLE 5.1

Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, $\Psi$, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital ( $\Psi^{2}$ ). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.

We will consider the molecular orbitals in molecules composed of two identical atoms ( $\mathrm{H}_{2}$ or $\mathrm{Cl}_{2}$, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.

The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure 5.28). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.

FIGURE 5.28 (a) When in-phase waves combine, constructive interference produces a wave with greater amplitude. (b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude.


503. LINK TO LEARNING -
FIGURE 5.28 (a) When in-phase waves combine, constructive interference produces a wave with greater amplitude. (b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude.

There are two types of molecular orbitals that can form from the overlap of two atomic $s$ orbitals on adjacent atoms. The two types are illustrated in Figure 5.29. The in-phase combination produces a lower energy $\boldsymbol{\sigma}_{\boldsymbol{s}}$
molecular orbital (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy $\sigma_{s}^{*}$ molecular orbital (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a $\sigma_{S}$ orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the $\sigma_{s}^{*}$ orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.

FIGURE 5.29 Sigma ( $\sigma$ ) and sigma-star ( $\sigma^{*}$ ) molecular orbitals are formed by the combination of two $s$ atomic orbitals. The dots $(\cdot)$ indicate the locations of nuclei.


504. LINK TO LEARNING -
You can watch animations (http://openstax.org/l/16molecorbital) visualizing the calculated atomic orbitals combining to form various molecular orbitals at the Orbitron website.

In $p$ orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a twodimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When $p$ orbitals overlap end to end, they create $\sigma$ and $\sigma^{*}$ orbitals (Figure 5.30). If two atoms are located along the $x$-axis in a Cartesian coordinate system, the two $p_{x}$ orbitals overlap end to end and form $\sigma_{p x}$ (bonding) and $\sigma_{p x}^{*}$ (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with $s$-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.


504. LINK TO LEARNING -
FIGURE 5.30 Combining wave functions of two $p$ atomic orbitals along the internuclear axis creates two molecular orbitals, $\sigma_{p}$ and $\sigma_{p}^{*}$.
The side-by-side overlap of two $p$ orbitals gives rise to a pi ( $\boldsymbol{\pi}$ ) bonding molecular orbital and a $\boldsymbol{\pi}^{*}$ *
antibonding molecular orbital, as shown in Figure 5.31. In valence bond theory, we describe $\pi$ bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the $p$ orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the $\pi$ orbital by this same
shape, and a $\pi$ bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.

FIGURE 5.31 Side-by-side overlap of each two $p$ orbitals results in the formation of two $\pi$ molecular orbitals. Combining the out-of-phase orbitals results in an antibonding molecular orbital with two nodes. One contains the internuclear axis, and one is perpendicular to the axis. Combining the in-phase orbitals results in a bonding orbital. There is a node (blue) containing the internuclear axis with the two lobes of the orbital located above and below this node.


504. LINK TO LEARNING -
In the molecular orbitals of diatomic molecules, each atom also has two sets of $p$ orbitals oriented side by side ( $p_{y}$ and $p_{z}$ ), so these four atomic orbitals combine pairwise to create two $\pi$ orbitals and two $\pi^{*}$ orbitals. The $\pi_{p y}$ and $\pi_{p y}^{*}$ orbitals are oriented at right angles to the $\pi_{p z}$ and $\pi_{p z}^{*}$ orbitals. Except for their orientation, the $\pi_{p y}$ and $\pi_{p z}$ orbitals are identical and have the same energy; they are degenerate orbitals. The $\pi_{p y}^{*}$ and $\pi_{p z}^{*}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic $p$ orbitals in two atoms: $\sigma_{p x}$ and $\sigma_{p x}^{*}, \pi_{p y}$ and $\pi_{p y}^{*}, \pi_{p z}$ and $\pi_{p z}^{*}$.


506. Molecular Orbitals -
Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.


507. Solution -
(a) is an in-phase combination, resulting in a $\sigma_{3 p}$ orbital
(b) will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.
(c) is an out-of-phase combination, resulting in a $\pi_{3 p}^{*}$ orbital.


508. Check Your Learning -
Label the molecular orbital shown as $\sigma$ or $\pi$, bonding or antibonding and indicate where the node occurs.


509. Answer: -
The orbital is located along the internuclear axis, so it is a $\sigma$ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.


511. Walter Kohn: Nobel Laureate -
Walter Kohn (Figure 5.32) is a theoretical physicist who studies the electronic structure of solids. His work combines the principles of quantum mechanics with advanced mathematical techniques. This technique, called density functional theory, makes it possible to compute properties of molecular orbitals, including their shape and energies. Kohn and mathematician John Pople were awarded the Nobel Prize in Chemistry in 1998 for their contributions to our understanding of electronic structure. Kohn also made significant contributions to the physics of semiconductors.

FIGURE 5.32 Walter Kohn developed methods to describe molecular orbitals. (credit: image courtesy of Walter Kohn)

Kohn's biography has been remarkable outside the realm of physical chemistry as well. He was born in Austria, and during World War II he was part of the Kindertransport program that rescued 10,000 children from the Nazi regime. His summer jobs included discovering gold deposits in Canada and helping Polaroid explain how its instant film worked. Dr. Kohn passed away in 2016 at the age of 93.


513. Computational Chemistry in Drug Design -
While the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (see Figure 5.33). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway.

FIGURE 5.33 The molecule shown, HIV-1 protease, is an important target for pharmaceutical research. By designing molecules that bind to this protein, scientists are able to drastically inhibit the progress of the disease.


514. Molecular Orbital Energy Diagrams -
The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure 5.34). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six $2 p$ atomic orbitals results in three bonding orbitals (one $\sigma$ and two $\pi$ ) and three antibonding orbitals (one $\sigma^{*}$ and two $\pi^{*}$ ).

We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure 5.34). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\mathrm{Be}_{2}{ }^{+}$) would have the molecular electron configuration $\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{1}$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons.


514. Molecular Orbital Energy Diagrams -
FIGURE 5.34 This is the molecular orbital diagram for the homonuclear diatomic $\mathrm{Be}_{2}{ }^{+}$, showing the molecular orbitals of the valence shell only. The molecular orbitals are filled in the same manner as atomic orbitals, using the Aufbau principle and Hund's rule.


515. Bond Order -
The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.

When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1 , a double bond has a bond order of 2 , and a triple bond has a bond order of 3 . We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.

In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:

$$
\text { bond order }=\frac{\text { (number of bonding electrons) }- \text { (number of antibonding electrons) }}{2}
$$

The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.


516. Bonding in Diatomic Molecules -
A dihydrogen molecule $\left(\mathrm{H}_{2}\right)$ forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the $\sigma_{1 s}$ bonding orbital. A dihydrogen molecule, $\mathrm{H}_{2}$, readily forms because the energy of a $\mathrm{H}_{2}$ molecule is lower than that of two H atoms. The $\sigma_{1 s}$ orbital that contains both electrons is lower in energy than either of the two 1 s atomic orbitals.

A molecular orbital can hold two electrons, so both electrons in the $\mathrm{H}_{2}$ molecule are in the $\sigma_{1 s}$ bonding orbital; the electron configuration is $\left(\sigma_{1 s}\right)^{2}$. We represent this configuration by a molecular orbital energy diagram (Figure 5.35) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.

FIGURE 5.35 The molecular orbital energy diagram predicts that $\mathrm{H}_{2}$ will be a stable molecule with lower energy than the separated atoms.

A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have

$$
\text { bond order in } \mathrm{H}_{2}=\frac{(2-0)}{2}=1
$$


516. Bonding in Diatomic Molecules -
A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have

$$
\text { bond order in } \mathrm{H}_{2}=\frac{(2-0)}{2}=1
$$

Because the bond order for the $\mathrm{H}-\mathrm{H}$ bond is equal to 1 , the bond is a single bond.
A helium atom has two electrons, both of which are in its $1 s$ orbital. Two helium atoms do not combine to form a dihelium molecule, $\mathrm{He}_{2}$, with four electrons, because the stabilizing effect of the two electrons in the lowerenergy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of $\mathrm{He}_{2}$ as $\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}$ as in Figure 5.36. The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.

$$
\text { bond order in } \mathrm{He}_{2}=\frac{(2-2)}{2}=0
$$

A bond order of zero indicates that no bond is formed between two atoms.

FIGURE 5.36 The molecular orbital energy diagram predicts that $\mathrm{He}_{2}$ will not be a stable molecule, since it has equal numbers of bonding and antibonding electrons.

The Diatomic Molecules of the Second Period
Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: $\mathrm{Li}_{2}, \mathrm{Be}_{2}, \mathrm{~B}_{2}, \mathrm{C}_{2}, \mathrm{~N}_{2}, \mathrm{O}_{2}, \mathrm{~F}_{2}$, and $\mathrm{Ne}_{2}$. However, we can predict that the $\mathrm{Be}_{2}$ molecule and the $\mathrm{Ne}_{2}$ molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table 5.2).


516. Bonding in Diatomic Molecules -
We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies.

Consistent with Hund's rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place.

As we saw in valence bond theory, $\sigma$ bonds are generally more stable than $\pi$ bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, $\sigma$ orbitals are usually more stable than $\pi$ orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure 5.37. Looking at $\mathrm{Ne}_{2}$ molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the $p$ orbitals (Li through N) we observe a different pattern, in which the $\sigma_{p}$ orbital is higher in energy than the $\pi_{p}$ set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.

FIGURE 5.37 This shows the MO diagrams for each homonuclear diatomic molecule in the second period. The orbital energies decrease across the period as the effective nuclear charge increases and atomic radius decreases. Between $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$, the order of the orbitals changes.

This switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The $\sigma_{\mathrm{s}}$ wavefunction mathematically combines with the $\sigma_{p}$ wavefunction, with the result that the $\sigma_{s}$ orbital becomes more stable, and the $\sigma_{p}$ orbital becomes less stable (Figure 5.38). Similarly, the antibonding orbitals also undergo s-p mixing, with the $\sigma_{\mathrm{s}^{*}}$ becoming more stable and the $\sigma_{\mathrm{p}^{*}}$ becoming less stable.


516. Bonding in Diatomic Molecules -
FIGURE 5.38 Without mixing, the MO pattern occurs as expected, with the $\sigma_{p}$ orbital lower in energy than the $\pi_{p}$ orbitals. When s-p mixing occurs, the orbitals shift as shown, with the $\sigma_{p}$ orbital higher in energy than the $\pi_{p}$ orbitals. $s$-p mixing occurs when the $s$ and $p$ orbitals have similar energies. The energy difference between $2 s$ and $2 p$ orbitals in $\mathrm{O}, \mathrm{F}$, and Ne is greater than that in Li, Be, B, C, and N. Because of this, $\mathrm{O}_{2}, \mathrm{~F}_{2}$, and $\mathrm{Ne}_{2}$ exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure 5.37. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the $\sigma_{p}$ orbital is raised above the $\pi_{p}$ set.

Using the MO diagrams shown in Figure 5.37, we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table 5.2, $\mathrm{Be}_{2}$ and $\mathrm{Ne}_{2}$ molecules would have a bond order of 0 , and these molecules do not exist.

Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements


516. Bonding in Diatomic Molecules -
Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements

| Molecule | Electron Configuration | Bond Order |
| :--- | :--- | :--- |
| $\mathrm{Li}_{2}$ | $\left(\sigma_{2 s}\right)^{2}$ | 1 |
| $\mathrm{Be}_{2}$ (unstable) | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}$ | 0 |
| $\mathrm{~B}_{2}$ | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{2}$ | 1 |
| $\mathrm{C}_{2}$ | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}$ | 2 |

TABLE 5.2


516. Bonding in Diatomic Molecules -
TABLE 5.2

| Molecule | Electron Configuration | Bond Order |
| :--- | :--- | :--- |
| $\mathrm{N}_{2}$ | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}\left(\sigma_{2 p x}\right)^{2}$ | 3 |
| $\mathrm{O}_{2}$ | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p x}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}\left(\pi_{2 p y}^{*}, \pi_{2 p z}^{*}\right)^{2}$ | 2 |
| $\mathrm{~F}_{2}$ | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p x}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}\left(\pi_{2 p y}^{*}, \pi_{2 p z}^{*}\right)^{4}$ | 1 |
| $\mathrm{Ne}_{2}$ (unstable) | $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p x}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}\left(\pi_{2 p y}^{*}, \pi_{2 p z}^{*}\right)^{4}\left(\sigma_{2 p x}^{*}\right)^{2}$ | 0 |

TABLE 5.2


516. Bonding in Diatomic Molecules -
TABLE 5.2

The combination of two lithium atoms to form a lithium molecule, $\mathrm{Li}_{2}$, is analogous to the formation of $\mathrm{H}_{2}$, but the atomic orbitals involved are the valence $2 s$ orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the $\sigma_{2 s}$ bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the $\mathrm{Li}_{2}$ molecule to be stable. The molecule is, in fact, present in appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table 5.2 with a bond order greater than zero are also known.

The $\mathrm{O}_{2}$ molecule has enough electrons to half fill the $\left(\pi_{2 p y}^{*}, \pi_{2 p z}^{*}\right)$ level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for $\mathrm{O}_{2}$ is in accord with the fact that the oxygen molecule has two unpaired electrons (Figure 5.40). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory.


518. Band Theory -
When two identical atomic orbitals on different atoms combine, two molecular orbitals result (see Figure 5.29). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are inphase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase.

In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically $>10^{23}$ atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When $N$ valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, $N / 2$ (filled) bonding orbitals and $N / 2$ (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure 5.39 shows the bands for three important classes of materials: insulators, semiconductors, and conductors.

FIGURE 5.39 Molecular orbitals in solids are so closely spaced that they are described as bands. The valence band is lower in energy and the conduction band is higher in energy. The type of solid is determined by the size of the "band gap" between the valence and conduction bands. Only a very small amount of energy is required to move electrons from the valance band to the conduction band in a conductor, and so they conduct electricity well. In an insulator, the band gap is large, so that very few electrons move, and they are poor conductors of electricity. Semiconductors are in between: they conduct electricity better than insulators, but not as well as conductors.


518. Band Theory -
In order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is "easy" to overcome, so they are good conductors of electricity. In an insulator, the band gap is so "large" that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when "moderate" amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics.

Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to $46 \%$ of the energy in sunlight could be converted into electricity using solar cells.


519. EXAMPLE 5.7 -
Molecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons
Draw the molecular orbital diagram for the oxygen molecule, $\mathrm{O}_{2}$. From this diagram, calculate the bond order for $\mathrm{O}_{2}$. How does this diagram account for the paramagnetism of $\mathrm{O}_{2}$ ?


520. Solution -
We draw a molecular orbital energy diagram similar to that shown in Figure 5.37. Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure 5.40.

FIGURE 5.40 The molecular orbital energy diagram for $\mathrm{O}_{2}$ predicts two unpaired electrons.
We calculate the bond order as

$$
\mathrm{O}_{2}=\frac{(8-4)}{2}=2
$$

Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the ( $\Pi_{2 p y}, \Pi_{2 p z}$ )* molecular orbitals.


521. Check Your Learning -
The main component of air is $N_{2}$. From the molecular orbital diagram of $N_{2}$, predict its bond order and whether it is diamagnetic or paramagnetic.


522. Answer: -
$\mathrm{N}_{2}$ has a bond order of 3 and is diamagnetic.


524. Ion Predictions with MO Diagrams -
Give the molecular orbital configuration for the valence electrons in $\mathrm{C}_{2}{ }^{2-}$. Will this ion be stable?


525. Solution -
Looking at the appropriate MO diagram, we see that the $\pi$ orbitals are lower in energy than the $\sigma_{p}$ orbital. The valence electron configuration for $\mathrm{C}_{2}$ is $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}$. Adding two more electrons to generate the $\mathrm{C}_{2}{ }^{2-}$ anion will give a valence electron configuration of $\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}\left(\sigma_{2 p x}\right)^{2}$. Since this has six more bonding electrons than antibonding, the bond order will be 3 , and the ion should be stable.


526. Check Your Learning -
How many unpaired electrons would be present on a $\mathrm{Be}_{2}{ }^{2-}$ ion? Would it be paramagnetic or diamagnetic?


527. Answer: -
two, paramagnetic


528. LINK TO LEARNING -
Creating molecular orbital diagrams for molecules with more than two atoms relies on the same basic ideas as the diatomic examples presented here. However, with more atoms, computers are required to calculate how the atomic orbitals combine. See three-dimensional drawings (http://openstax.org/l/16orbitaldiag) of the molecular orbitals for $\mathrm{C}_{6} \mathrm{H}_{6}$.


529. 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 region of space in which an electron has a high probability of being found in a molecule
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
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 ( $\boldsymbol{\pi}$ bond) covalent bond formed by side-byside overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis
s-p mixing change that causes $\sigma_{p}$ orbitals to be less stable than $\pi_{p}$ orbitals due to the mixing of $S$ and $p$-based molecular orbitals of similar energies.
sigma bond ( $\sigma$ bond) covalent bond formed by overlap of atomic orbitals along the internuclear axis
$\boldsymbol{s p}$ hybrid orbital one of a set of two orbitals with a linear arrangement that results from combining one $s$ and one $p$ orbital


529. Key Terms -
$\boldsymbol{s p}$ hybrid orbital one of a set of two orbitals with a linear arrangement that results from combining one $s$ and one $p$ orbital
$\boldsymbol{s} \boldsymbol{p}^{\mathbf{2}}$ hybrid orbital one of a set of three orbitals with a trigonal planar arrangement that results from combining one $s$ and two $p$ orbitals
$\boldsymbol{s p} \boldsymbol{p}^{\mathbf{3}}$ hybrid orbital one of a set of four orbitals with a tetrahedral arrangement that results from combining one $s$ and three $p$ orbitals
$\boldsymbol{s p}^{\mathbf{3}} \boldsymbol{d}$ 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
$\boldsymbol{s p}^{\mathbf{3}} \boldsymbol{d}^{\mathbf{2}}$ 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 $\sigma$ or $\pi$ bonds, within which pairs of electrons are shared
$\boldsymbol{\pi}$ 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
$\boldsymbol{\pi}$ * 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
$\boldsymbol{\sigma}$ bonding orbital molecular orbital in which the electron density is found along the axis of the bond
$\boldsymbol{\sigma}^{*}$ 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


530. Key Equations -
bond order $=\frac{(\text { number of bonding electron) }-(\text { number of antibonding electrons })}{2}$


531. Summary - 531.1. 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 $\sigma$ bond. When they overlap in a fashion that creates a node along this axis, they form a $\pi$ bond. Dipole moments can be used to determine partial separations of charges between atoms.


531. Summary - 531.2. 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 ( $\sigma$ ) 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 $s p$ hybridization; three, $s p^{2}$ hybridization; four, $s p^{3}$ hybridization; five, $s p^{3} d$ hybridization; and six, $s p^{3} d^{2}$ hybridization. $\operatorname{Pi}(\pi)$ bonds are formed from unhybridized atomic orbitals ( $p$ or $d$ orbitals).


531. Summary - 531.3. Multiple Bonds
Multiple bonds consist of a $\sigma$ bond located along the axis between two atoms and one or two $\pi$ bonds. The $\sigma$ bonds are usually formed by the overlap of hybridized atomic orbitals, while the $\pi$ 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 $\pi$ bonds
can vary.


531. Summary - 531.4. 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 $\sigma$ 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 $\pi$ 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.


532. Exercises - 532.1. Valence Bond Theory
1. Explain how $\sigma$ and $\pi$ bonds are similar and how they are different.
2. Use valence bond theory to explain the bonding in $\mathrm{F}_{2}$, HF , and ClBr . Sketch the overlap of the atomic orbitals involved in the bonds.
3. Use valence bond theory to explain the bonding in $\mathrm{O}_{2}$. Sketch the overlap of the atomic orbitals involved in the bonds in $\mathrm{O}_{2}$.
4. How many $\sigma$ and $\pi$ bonds are present in the molecule HCN?
5. A friend tells you $N_{2}$ has three $\pi$ bonds due to overlap of the three $p$-orbitals on each $N$ atom. Do you agree?
6. Draw the Lewis structures for $\mathrm{CO}_{2}$ and CO , and predict the number of $\sigma$ and $\pi$ bonds for each molecule.
(a) $\mathrm{CO}_{2}$
(b) CO


532. Exercises - 532.2. Hybrid Atomic Orbitals
7. Why is the concept of hybridization required in valence bond theory?
8. Give the shape that describes each hybrid orbital set:
(a) $s p^{2}$
(b) $s p^{3} d$
(c) $s p$
(d) $s p^{3} d^{2}$
9. Explain why a carbon atom cannot form five bonds using $s p^{3} d$ hybrid orbitals.
10. What is the hybridization of the central atom in each of the following?
(a) $\mathrm{BeH}_{2}$
(b) $\mathrm{SF}_{6}$
(c) $\mathrm{PO}_{4}{ }^{3-}$
(d) $\mathrm{PCl}_{5}$
11. A molecule with the formula $A B_{3}$ could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
12. Methionine, $\mathrm{CH}_{3} \mathrm{SCH}_{2} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{CO}_{2} \mathrm{H}$, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?


532. Exercises - 532.2. Hybrid Atomic Orbitals
13. Sulfuric acid is manufactured by a series of reactions represented by the following equations:
$\mathrm{S}_{8}(\mathrm{~s})+8 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{SO}_{2}(g)$
$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$
$\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(l)$
Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:
(a) circular $\mathrm{S}_{8}$ molecule
(b) $\mathrm{SO}_{2}$ molecule
(c) $\mathrm{SO}_{3}$ molecule
(d) $\mathrm{H}_{2} \mathrm{SO}_{4}$ molecule (the hydrogen atoms are bonded to oxygen atoms)
14. Two important industrial chemicals, ethene, $\mathrm{C}_{2} \mathrm{H}_{4}$, and propene, $\mathrm{C}_{3} \mathrm{H}_{6}$, are produced by the steam (or thermal) cracking process:

$$
2 \mathrm{C}_{3} \mathrm{H}_{8}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{CH}_{4}(g)+\mathrm{H}_{2}(g)
$$


532. Exercises - 532.2. Hybrid Atomic Orbitals
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.
15. Analysis of a compound indicates that it contains $77.55 \% \mathrm{Xe}$ and $22.45 \% \mathrm{~F}$ by mass.
(a) What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise).
(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?
16. Consider nitrous acid, $\mathrm{HNO}_{2}$ (HONO).
(a) Write a Lewis structure.
(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the $\mathrm{HNO}_{2}$ molecule?
(c) What is the hybridization on the internal oxygen and nitrogen atoms in $\mathrm{HNO}_{2}$ ?
17. Strike-anywhere matches contain a layer of $\mathrm{KClO}_{3}$ and a layer of $\mathrm{P}_{4} \mathrm{~S}_{3}$. 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. $\mathrm{KClO}_{3}$ contains the $\mathrm{ClO}_{3}{ }^{-}$ion. $\mathrm{P}_{4} \mathrm{~S}_{3}$ is an unusual molecule with the skeletal structure.


532. Exercises - 532.2. Hybrid Atomic Orbitals
(a) Write Lewis structures for $\mathrm{P}_{4} \mathrm{~S}_{3}$ and the $\mathrm{ClO}_{3}{ }^{-}$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 $\mathrm{P}_{4} \mathrm{~S}_{3}$ and the $\mathrm{ClO}_{3}{ }^{-}$ion.
18. 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.)

|  | H | H |  |  |  | H | H |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
|  | H | C | C | C | C | C | H |  |
|  | H | H |  |  |  |  |  |  |

19. Write Lewis structures for $\mathrm{NF}_{3}$ and $\mathrm{PF}_{5}$. On the basis of hybrid orbitals, explain the fact that $\mathrm{NF}_{3}, \mathrm{PF}_{3}$, and $\mathrm{PF}_{5}$ are stable molecules, but $\mathrm{NF}_{5}$ does not exist.
20. In addition to $\mathrm{NF}_{3}$, two other fluoro derivatives of nitrogen are known: $\mathrm{N}_{2} \mathrm{~F}_{4}$ and $\mathrm{N}_{2} \mathrm{~F}_{2}$. What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?


532. Exercises - 532.3. Multiple Bonds
21. The bond energy of a $\mathrm{C}-\mathrm{C}$ single bond averages $347 \mathrm{~kJ} \mathrm{~mol}^{-1}$; that of a $\mathrm{C} \equiv \mathrm{C}$ triple bond averages 839 kJ $\mathrm{mol}^{-1}$. Explain why the triple bond is not three times as strong as a single bond.
22. For the carbonate ion, $\mathrm{CO}_{3}{ }^{2-}$, draw all of the resonance structures. Identify which orbitals overlap to create each bond.
23. A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, $\mathrm{H}_{3} \mathrm{CCN}$. It is present in paint strippers.
(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.
(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form $\sigma$ bonds.
(c) Describe the atomic orbitals that form the $\pi$ bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
24. For the molecule allene, $\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}$, give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?
25. Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:
(a) ClNO ( N is the central atom)
(b) $\mathrm{CS}_{2}$
(c) $\mathrm{Cl}_{2} \mathrm{CO}$ ( C is the central atom)
(d) $\mathrm{Cl}_{2} \mathrm{SO}$ ( S is the central atom)
(e) $\mathrm{SO}_{2} \mathrm{~F}_{2}$ (S is the central atom)
(f) $\mathrm{XeO}_{2} \mathrm{~F}_{2}$ ( Xe is the central atom)
(g) $\mathrm{ClOF}_{2}{ }^{+}$( Cl is the central atom)
26. Describe the molecular geometry and hybridization of the $N, P$, or $S$ atoms in each of the following compounds.


532. Exercises - 532.3. Multiple Bonds
26. Describe the molecular geometry and hybridization of the $N, P$, or $S$ atoms in each of the following compounds.
(a) $\mathrm{H}_{3} \mathrm{PO}_{4}$, phosphoric acid, used in cola soft drinks
(b) $\mathrm{NH}_{4} \mathrm{NO}_{3}$, ammonium nitrate, a fertilizer and explosive
(c) $\mathrm{S}_{2} \mathrm{Cl}_{2}$, disulfur dichloride, used in vulcanizing rubber
(d) $\mathrm{K}_{4}\left[\mathrm{O}_{3} \mathrm{POPO}_{3}\right]$, potassium pyrophosphate, an ingredient in some toothpastes
27. For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:
(a) ozone $\left(\mathrm{O}_{3}\right)$ central O hybridization
(b) carbon dioxide $\left(\mathrm{CO}_{2}\right)$ central C hybridization
(c) nitrogen dioxide $\left(\mathrm{NO}_{2}\right)$ central N hybridization
(d) phosphate ion $\left(\mathrm{PO}_{4}{ }^{3-}\right)$ central P hybridization
28. For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:
(a) Hybridization of each carbon


532. Exercises - 532.3. Multiple Bonds
(b) Hybridization of sulfur

(c) All atoms

29. Draw the orbital diagram for carbon in $\mathrm{CO}_{2}$ showing how many carbon atom electrons are in each orbital.


532. Exercises - 532.4. Molecular Orbital Theory
30. Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two $s$ orbitals and from two $p$ orbitals.
31. How are the following similar, and how do they differ?
(a) $\sigma$ molecular orbitals and $\pi$ molecular orbitals
(b) $\psi$ for an atomic orbital and $\psi$ for a molecular orbital
(c) bonding orbitals and antibonding orbitals
32. If molecular orbitals are created by combining five atomic orbitals from atom $A$ and five atomic orbitals from atom B combine, how many molecular orbitals will result?
33. Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
34. Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.
35. Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?
36. Calculate the bond order for an ion with this configuration:

$$
\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p x}\right)^{2}\left(\pi_{2 p y}, \pi_{2 p z}\right)^{4}\left(\pi_{2 p y}^{*}, \pi_{2 p z}^{*}\right)^{3}
$$


532. Exercises - 532.4. Molecular Orbital Theory
37. Explain why an electron in the bonding molecular orbital in the $\mathrm{H}_{2}$ molecule has a lower energy than an electron in the $1 s$ atomic orbital of either of the separated hydrogen atoms.
38. Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
(a) $\mathrm{Na}_{2}{ }^{2+}$
(b) $\mathrm{Mg}_{2}{ }^{2+}$
(c) $\mathrm{Al}_{2}{ }^{2+}$
(d) $\mathrm{Si}_{2}{ }^{2+}$
(e) $\mathrm{P}_{2}{ }^{2+}$
(f) $\mathrm{S}_{2}{ }^{2+}$
(g) $\mathrm{F}_{2}{ }^{2+}$
(h) $\mathrm{Ar}_{2}{ }^{2+}$
39. 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) $\mathrm{H}_{2}, \mathrm{H}_{2}{ }^{+}, \mathrm{H}_{2}^{-}$
(b) $\mathrm{O}_{2}, \mathrm{O}_{2}{ }^{2+}, \mathrm{O}_{2}{ }^{2-}$
(c) $\mathrm{Li}_{2}, \mathrm{Be}_{2}{ }^{+}, \mathrm{Be}_{2}$
(d) $\mathrm{F}_{2}, \mathrm{~F}_{2}{ }^{+}, \mathrm{F}_{2}^{-}$
(e) $\mathrm{N}_{2}, \mathrm{~N}_{2}{ }^{+}, \mathrm{N}_{2}{ }^{-}$
40. For the first ionization energy for an $\mathrm{N}_{2}$ molecule, what molecular orbital is the electron removed from?
41. 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 $\mathrm{H}_{2}$
(b) N and $\mathrm{N}_{2}$
(c) O and $\mathrm{O}_{2}$
(d) C and $\mathrm{C}_{2}$


532. Exercises - 532.4. Molecular Orbital Theory
(a) H and $\mathrm{H}_{2}$
(b) N and $\mathrm{N}_{2}$
(c) O and $\mathrm{O}_{2}$
(d) C and $\mathrm{C}_{2}$
(e) B and $\mathrm{B}_{2}$
42. Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?
43. A friend tells you that the $2 s$ orbital for fluorine starts off at a much lower energy than the $2 s$ orbital for lithium, so the resulting $\sigma_{2 s}$ molecular orbital in $\mathrm{F}_{2}$ is more stable than in $\mathrm{Li}_{2}$. Do you agree?
44. True or false: Boron contains $2 s^{2} 2 p^{1}$ valence electrons, so only one $p$ orbital is needed to form molecular orbitals.
45. What charge would be needed on $F_{2}$ to generate an ion with a bond order of 2 ?
46. Predict whether the MO diagram for $S_{2}$ would show s-p mixing or not.
47. Explain why $\mathrm{N}_{2}{ }^{2+}$ is diamagnetic, while $\mathrm{O}_{2}{ }^{4+}$, which has the same number of valence electrons, is paramagnetic.
48. Using the MO diagrams, predict the bond order for the stronger bond in each pair:
(a) $\mathrm{B}_{2}$ or $\mathrm{B}_{2}{ }^{+}$
(b) $\mathrm{F}_{2}$ or $\mathrm{F}_{2}{ }^{+}$
(c) $\mathrm{O}_{2}$ or $\mathrm{O}_{2}{ }^{2+}$
(d) $\mathrm{C}_{2}{ }^{+}$or $\mathrm{C}_{2}{ }^{-}$


533. CHAPTER 6 <br> Composition of Substances and Solutions -
Figure 6.1 The water in a swimming pool is a complex mixture of substances whose relative amounts must be carefully maintained to ensure the health and comfort of people using the pool. (credit: modification of work by Vic Brincat)


534. CHAPTER OUTLINE - 534.1. Formula Mass
6.2 Determining Empirical and Molecular Formulas
6.3 Molarity
6.4 Other Units for Solution Concentrations

INTRODUCTION Swimming pools have long been a popular means of recreation, exercise, and physical therapy. Since it is impractical to refill large pools with fresh water on a frequent basis, pool water is regularly treated with chemicals to prevent the growth of harmful bacteria and algae. Proper pool maintenance requires regular additions of various chemical compounds in carefully measured amounts. For example, the relative amount of calcium ion, $\mathrm{Ca}^{2+}$, in the water should be maintained within certain limits to prevent eye irritation and avoid damage to the pool bed and plumbing. To maintain proper calcium levels, calcium cations are added to the water in the form of an ionic compound that also contains anions; thus, it is necessary to know both the relative amount of $\mathrm{Ca}^{2+}$ in the compound and the volume of water in the pool in order to achieve the proper calcium level. Quantitative aspects of the composition of substances (such as the calcium-containing compound) and mixtures (such as the pool water) are the subject of this chapter.


535. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Calculate formula masses for covalent and ionic compounds

Many argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton's atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.


536. Formula Mass -
An earlier chapter of this text described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance's formula.


537. Formula Mass for Covalent Substances -
For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform $\left(\mathrm{CHCl}_{3}\right)$, a covalent compound once used as a surgical anesthetic and now primarily used in the production of the "anti-stick" polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 6.2 outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu .

| Element | Quantity | Average atomic
 mass (amu) | Subtotal
 $(\mathbf{a m u})$ |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| C | 1 | $\times$ | 12.01 | $=$ | 12.01 |
| H | 1 | $\times$ | 1.008 | $=$ | 1.008 |
| Cl | 3 | $\times$ | 35.45 | $=$ | 106.35 |
| Molecular mass |  |  |  |  | 119.37 |

FIGURE 6.2 The average mass of a chloroform molecule, $\mathrm{CHCl}_{3}$, is 119.37 amu , which is the sum of the average atomic masses of each of its constituent atoms. The model shows the molecular structure of chloroform.

Likewise, the molecular mass of an aspirin molecule, $\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}$, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure 6.3).


537. Formula Mass for Covalent Substances -
| Element | Quantity | Average atomic
 mass (amu) | Subtotal
 $(\mathbf{a m u})$ |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| C | 9 | $\times$ | 12.01 | $=$ | 108.09 |
| H | 8 | $\times$ | 1.008 | $=$ | 8.064 |
| O | 4 | $\times$ | 16.00 | $=$ | 64.00 |
| Molecular mass |  |  |  |  | 180.15 |

FIGURE 6.3 The average mass of an aspirin molecule is 180.15 amu . The model shows the molecular structure of aspirin, $\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}$.


539. Computing Molecular Mass for a Covalent Compound -
Ibuprofen, $\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}$, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?


540. Solution -
Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:

| Element | Quantity | Average atomic
 mass (amu) | Subtotal
 (amu) |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| C | 13 | $\times$ | 12.01 | $=$ | 156.13 |
| H | 18 | $\times$ | 1.008 | $=$ | 18.144 |
| O | 2 | $\times$ | 16.00 | $=$ | 32.00 |
| Molecular mass |  |  |  |  | 206.27 |


541. Check Your Learning -
Acetaminophen, $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2}$, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?


542. Answer: -
151.16 amu


543. Formula Mass for Ionic Compounds -
Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound's formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the "molecular mass."

As an example, consider sodium chloride, NaCl , the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, $\mathrm{Na}^{+}$, and chloride anions, $\mathrm{Cl}^{-}$, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (see Figure 6.4).

| Element | Quantity |  | Average atomic <br> mass (amu) | Subtotal |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| Na | 1 | $\times$ | 22.99 | $=$ | 22.99 |
| Cl | 1 | $\times$ | 35.45 | $=$ | 35.45 |
|  |  |  |  |  |  |
| Formula mass |  |  |  |  | 58.44 |

FIGURE 6.4 Table salt, NaCl , contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu .


543. Formula Mass for Ionic Compounds -
FIGURE 6.4 Table salt, NaCl , contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu .

Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing
or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.


545. Computing Formula Mass for an Ionic Compound -
Aluminum sulfate, $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass ( amu ) of this compound?


546. Solution -
The formula for this compound indicates it contains $\mathrm{Al}^{3+}$ and $\mathrm{SO}_{4}{ }^{2-}$ ions combined in a $2: 3$ ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, $\mathrm{Al}_{2} \mathrm{~S}_{3} \mathrm{O}_{12}$. Following the approach outlined above, the formula mass for this compound is calculated as follows:

| Element | Quantity | Average atomic
 mass (amu) | Subtotal
 $(\mathbf{a m u})$ |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Al | 2 | $\times$ | 26.98 | $=$ | 53.96 |
| S | 3 | $\times$ | 32.06 | $=$ | 96.18 |
| O | 12 | $\times$ | 16.00 | $=$ | 192.00 |
| Molecular mass 342.14 |  |  |  |  |  |


547. Check Your Learning -
Calcium phosphate, $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?


548. Answer: -
310.18 amu


548. Answer: - 548.1. Determining Empirical and Molecular Formulas
A previous chapter of this text discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.


549. Percent Composition -
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound's formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound's percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

$$
\begin{aligned}
& \% \mathrm{H}=\frac{\text { mass } \mathrm{H}}{\text { mass compound }} \times 100 \% \\
& \% \mathrm{C}=\frac{\text { mass } \mathrm{C}}{\text { mass compound }} \times 100 \%
\end{aligned}
$$

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C , the percent composition would be calculated to be $25 \% \mathrm{H}$ and $75 \% \mathrm{C}$ :

$$
\begin{aligned}
& \% \mathrm{H}=\frac{2.5 \mathrm{~g} \mathrm{H}}{10.0 \mathrm{~g} \text { compound }} \times 100 \%=25 \% \\
& \% \mathrm{C}=\frac{7.5 \mathrm{~g} \mathrm{C}}{10.0 \mathrm{~g} \text { compound }} \times 100 \%=75 \%
\end{aligned}
$$


551. Calculation of Percent Composition -
Analysis of a $12.04-\mathrm{g}$ sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain $7.34 \mathrm{~g} \mathrm{C}, 1.85 \mathrm{~g} \mathrm{H}$, and 2.85 g N . What is the percent composition of this compound?


552. Solution -
To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

$$
\begin{aligned}
& \% \mathrm{C}=\frac{7.34 \mathrm{~g} \mathrm{C}}{12.04 \mathrm{~g} \text { compound }} \times 100 \%=61.0 \% \\
& \% \mathrm{H}=\frac{1.85 \mathrm{~g} \mathrm{H}}{12.04 \mathrm{~g} \text { compound }} \times 100 \%=15.4 \% \\
& \% \mathrm{~N}=\frac{2.85 \mathrm{~g} \mathrm{~N}}{12.04 \mathrm{~g} \text { compound }} \times 100 \%=23.7 \%
\end{aligned}
$$

The analysis results indicate that the compound is $61.0 \% \mathrm{C}, 15.4 \% \mathrm{H}$, and $23.7 \% \mathrm{~N}$ by mass.


553. Check Your Learning -
A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain $3.01 \mathrm{~g} \mathrm{C}, 4.00 \mathrm{~g} \mathrm{O}$, and 17.81 g Cl . What is this compound's percent composition?


554. Answer: -
12.1\% C, 16.1\% O, 71.79\% Cl


555. Determining Percent Composition from Molecular or Empirical Formulas -
Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia $\left(\mathrm{NH}_{3}\right)$, ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$, and urea $\left(\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}\right)$. The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of $\mathrm{NH}_{3}$ contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 $\times 1.008 \mathrm{amu})=3.024 \mathrm{amu}$. The formula mass of ammonia is therefore $(14.01 \mathrm{amu}+3.024 \mathrm{amu})=17.03 \mathrm{amu}$, and its percent composition is:

$$
\begin{aligned}
& \% \mathrm{~N}=\frac{14.01 \mathrm{amu} \mathrm{~N}}{17.03 \mathrm{amu} \mathrm{NH}} 3 \mathrm{am} \\
& \% \mathrm{H}=\frac{3.024 \mathrm{amu} \mathrm{H}}{17.03 \mathrm{amu} \mathrm{NH}_{3} \times 100 \%=82.27 \%}=17.76 \%
\end{aligned}
$$

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 6.4. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound's elements.


557. Determining Percent Composition from a Molecular Formula -
Aspirin is a compound with the molecular formula $\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}$. What is its percent composition?


558. Solution -
To calculate the percent composition, the masses of $\mathrm{C}, \mathrm{H}$, and O in a known mass of $\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}$ are needed. It is convenient to consider 1 mol of $\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}$ and use its molar mass ( $180.159 \mathrm{~g} / \mathrm{mole}$, determined from the chemical formula) to calculate the percentages of each of its elements:


558. Solution -
$$
\begin{aligned}
& \% \mathrm{C}=\frac{9 \mathrm{~mol} \mathrm{C} \times \text { molar mass } \mathrm{C}}{\mathrm{molar} \text { mass } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}} \times 100=\frac{9 \times 12.01 \mathrm{~g} / \mathrm{mol}}{180.159 \mathrm{~g} / \mathrm{mol}} \times 100=\frac{108.09 \mathrm{~g} / \mathrm{mol}}{180.159 \mathrm{~g} / \mathrm{mol}} \times 100 \\
& \% \mathrm{C}=60.00 \% \mathrm{C} \\
& \% \mathrm{H}=\frac{8 \mathrm{~mol} \mathrm{H} \times \text { molar mass } \mathrm{H}}{\mathrm{molar} \text { mass } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}} \times 100=\frac{8 \times 1.008 \mathrm{~g} / \mathrm{mol}}{180.159 \mathrm{~g} / \mathrm{mol}} \times 100=\frac{8.064 \mathrm{~g} / \mathrm{mol}}{180.159 \mathrm{~g} / \mathrm{mol}} \times 100 \\
& \% \mathrm{H}=4.476 \% \mathrm{H} \\
& \% \mathrm{O}=\frac{4 \text { mol O } \times \text { molar mass } \mathrm{O}}{\text { molar mass } \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}} \times 100=\frac{4 \times 16.00 \mathrm{~g} / \mathrm{mol}}{180.159 \mathrm{~g} / \mathrm{mol}} \times 100=\frac{64.00 \mathrm{~g} / \mathrm{mol}}{180.159 \mathrm{~g} / \mathrm{mol}} \times 100 \\


558. Solution -
& \% \mathrm{O}=35.52 \%
\end{aligned}
$$


558. Solution -
Note that these percentages sum to equal $100.00 \%$ when appropriately rounded.
Check Your Learning
To three significant digits, what is the mass percentage of iron in the compound $\mathrm{Fe}_{2} \mathrm{O}_{3}$ ?


559. Answer: -
69.9\% Fe


560. Determination of Empirical Formulas -
As previously mentioned, the most common approach to determining a compound's chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H . The corresponding numbers of atoms (in moles) are:

$$
\begin{aligned}
& 1.71 \mathrm{~g} \mathrm{C} \times \frac{1 \mathrm{~mol} \mathrm{C}}{12.01 \mathrm{~g} \mathrm{C}}=0.142 \mathrm{~mol} \mathrm{C} \\
& 0.287 \mathrm{~g} \mathrm{H} \times \frac{1 \mathrm{~mol} \mathrm{H}}{1.008 \mathrm{~g} \mathrm{H}}=0.284 \mathrm{~mol} \mathrm{H}
\end{aligned}
$$

Thus, this compound may be represented by the formula $\mathrm{C}_{0.142} \mathrm{H}_{0.284}$. Per convention, formulas contain wholenumber subscripts, which can be achieved by dividing each subscript by the smaller subscript:

$$
\mathrm{C}_{\frac{0.142}{0.142}} \mathrm{H}_{\frac{0.284}{0.142}} \text { or } \mathrm{CH}_{2}
$$

(Recall that subscripts of " 1 " are not written but rather assumed if no other number is present.)
The empirical formula for this compound is thus $\mathrm{CH}_{2}$. This may or not be the compound's molecular formula
as well; however, additional information is needed to make that determination (as discussed later in this section).


560. Determination of Empirical Formulas -
Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O . Following the same approach yields a tentative empirical formula of:

$$
\mathrm{Cl}_{0.150} \mathrm{O}_{0.525}=\mathrm{Cl}_{\frac{0.150}{0.150}} \mathrm{O}_{\frac{0.525}{0.150}}=\mathrm{ClO}_{3.5}
$$

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding $\mathrm{Cl}_{2} \mathrm{O}_{7}$ as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

1. Deriving the number of moles of each element from its mass
2. Dividing each element's molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Figure 6.5 outlines this procedure in flow chart fashion for a substance containing elements A and X.

FIGURE 6.5 The empirical formula of a compound can be derived from the masses of all elements in the sample.


562. Determining a Compound's Empirical Formula from the Masses of Its Elements -
A sample of the black mineral hematite (Figure 6.6), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

FIGURE 6.6 Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)


563. Solution -
This problem provides the mass in grams of each element. Begin by finding the moles of each:

$$
\begin{aligned}
& 34.97 \mathrm{~g} \mathrm{Fe}\left(\frac{\mathrm{~mol} \mathrm{Fe}}{55.85 \mathrm{~g}}\right)=0.6261 \mathrm{~mol} \mathrm{Fe} \\
& 15.03 \mathrm{~g} \mathrm{O}\left(\frac{\mathrm{~mol} \mathrm{O}}{16.00 \mathrm{~g}}\right)=0.9394 \mathrm{~mol} \mathrm{O}
\end{aligned}
$$

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

$$
\begin{aligned}
& \frac{0.6261}{0.6261}=1.000 \mathrm{~mol} \mathrm{Fe} \\
& \frac{0.9394}{0.6261}=1.500 \mathrm{~mol} \mathrm{O}
\end{aligned}
$$

The ratio is 1.000 mol of iron to 1.500 mol of oxygen $\left(\mathrm{Fe}_{1} \mathrm{O}_{1.5}\right)$. Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

$$
2\left(\mathrm{Fe}_{1} \mathrm{O}_{1.5}\right)=\mathrm{Fe}_{2} \mathrm{O}_{3}
$$

The empirical formula is $\mathrm{Fe}_{2} \mathrm{O}_{3}$.


564. Check Your Learning -
What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?


565. Answer: -
$\mathrm{N}_{2} \mathrm{O}_{5}$


566. LINK TO LEARNING -
For additional worked examples illustrating the derivation of empirical formulas, watch the brief video (http://openstax.org/l/16empforms) clip.


567. Deriving Empirical Formulas from Percent Composition -
Finally, with regard to deriving empirical formulas, consider instances in which a compound's percent composition is available rather than the absolute masses of the compound's constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.


569. Determining an Empirical Formula from Percent Composition -
The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29\% C and $72.71 \% \mathrm{O}$ (Figure 6.7). What is the empirical formula for this gas?

FIGURE 6.7 An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: "Dual Freq"/Wikimedia Commons)


570. Solution -
Since the scale for percentages is 100 , it is most convenient to calculate the mass of elements present in a sample weighing 100 g . The calculation is "most convenient" because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element's mass percentage. This numerical equivalence results from the definition of the "percentage" unit, whose name is derived from the Latin phrase per centum meaning "by the hundred." Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

$$
\begin{aligned}
27.29 \% \mathrm{C} & =\frac{27.29 \mathrm{~g} \mathrm{C}}{100 \mathrm{~g} \text { compound }} \\
72.71 \% \mathrm{O} & =\frac{72.71 \mathrm{~g} \mathrm{O}}{100 \mathrm{~g} \text { compound }}
\end{aligned}
$$

The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element's mass by its molar mass:

$$
\begin{aligned}
& 27.29 \mathrm{~g} \mathrm{C}\left(\frac{\mathrm{~mol} \mathrm{C}}{12.01 \mathrm{~g}}\right)=2.272 \mathrm{~mol} \mathrm{C} \\
& 72.71 \mathrm{~g} \mathrm{O}\left(\frac{\mathrm{~mol} \mathrm{O}}{16.00 \mathrm{~g}}\right)=4.544 \mathrm{~mol} \mathrm{O}
\end{aligned}
$$

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

$$
\begin{aligned}
& \frac{2.272 \mathrm{~mol} \mathrm{C}}{2.272}=1 \\
& \frac{4.544 \mathrm{~mol} \mathrm{O}}{2.272}=2
\end{aligned}
$$

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is $\mathrm{CO}_{2}$.


571. Check Your Learning -
What is the empirical formula of a compound containing $40.0 \% \mathrm{C}, 6.71 \% \mathrm{H}$, and $53.28 \% \mathrm{O}$ ?


572. Answer: -
$\mathrm{CH}_{2} \mathrm{O}$