412. VSEPR Theory -
FIGURE 4.15 The $\mathrm{BeF}_{2}$ molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom.

Figure 4.16 illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.

| Number of
 regions | Two regions of
 high electron
 density (bonds
 and/or
 unshared
 pairs) | Three regions
 of high
 electron
 density (bonds
 and/or
 unshared
 pairs) | Four regions
 of high
 electron
 density (bonds
 and/or
 unshared
 pairs) | Five regions of
 high electron
 density (bonds
 and/or
 unshared
 pairs) | Six regions of
 high electron
 density (bonds
 and/or
 unshared
 pairs) |
| :--- | :--- | :--- | :--- | :--- | :--- |
| Spatial
 arrangement |  |  |  |  |  |

FIGURE 4.16 The basic electron-pair geometries predicted by VSEPR theory maximize the space around any region of electron density (bonds or lone pairs).


413. Electron-pair Geometry versus Molecular Structure -
It is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure 4.16 describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms, not the electrons.

We differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure. The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom.

For example, the methane molecule, $\mathrm{CH}_{4}$, which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure 4.17). On the other hand, the ammonia molecule, $\mathrm{NH}_{3}$, also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure 4.18).

FIGURE 4.17 The molecular structure of the methane molecule, $\mathrm{CH}_{4}$, is shown with a tetrahedral arrangement of the hydrogen atoms. VSEPR structures like this one are often drawn using the wedge and dash notation, in which solid lines represent bonds in the plane of the page, solid wedges represent bonds coming up out of the plane, and dashed lines represent bonds going down into the plane.

(a)

(b)

(c)

FIGURE 4.18 (a) The electron-pair geometry for the ammonia molecule is tetrahedral with one lone pair and three single bonds. (b) The trigonal pyramidal molecular structure is determined from the electron-pair geometry. (c) The actual bond angles deviate slightly from the idealized angles because the lone pair takes up a larger region of space than do the single bonds, causing the HNH angle to be slightly smaller than $109.5^{\circ}$.


413. Electron-pair Geometry versus Molecular Structure -
As seen in Figure 4.18, small distortions from the ideal angles in Figure 4.16 can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is:


414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair -
This order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is:

$$
\text { lone pair }>\text { triple bond }>\text { double bond }>\text { single bond }
$$

Consider formaldehyde, $\mathrm{H}_{2} \mathrm{CO}$, which is used as a preservative for biological and anatomical specimens (Figure 4.14). This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with $120^{\circ}$ bond angles, but we see that the double bond causes slightly larger angles $\left(121^{\circ}\right)$, and the angle between the single bonds is slightly smaller ( $118^{\circ}$ ).

In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure 4.18) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angles in $\mathrm{NH}_{3}$ are slightly smaller than the $109.5^{\circ}$ angle in a regular tetrahedron (Figure 4.16) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion (Figure 4.18). Figure 4.19 illustrates the ideal molecular structures, which are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs.


414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair -
| Number of electron regions | Electron region geometries: 0 lone pair | 1 lone pair | 2 lone pairs | 3 lone pairs | 4 lone pairs |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | $x \overbrace{\mathrm{E}}^{180^{\circ}}$
 Linear |  |  |  |  |
| 3 |
 Trigonal planar |
 Bent or angular |  |  |  |
| 4 | <smiles>[Y]C([Y])([Y])C([Y])([Y])[Y10]</smiles>
 Tetrahedral | Trigonal pyramid | Bent or angular |  |  |
| 5 |
 Trigonal bipyramid |
 Sawhorse or seesaw |
 T-shape |
 Linear |  |
| 6 |
 Octahedral |  |
 Square planar |
 T-shape | Linear |

FIGURE 4.19 The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (first column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modifications of the corresponding electron-pair geometry.


414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair -
According to VSEPR theory, the terminal atom locations (Xs in Figure 4.19) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct $X$ positions, as shown in Figure 4.20: an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). As shown in Figure 4.19, the axial position is surrounded by bond angles of $90^{\circ}$, whereas the equatorial position has more space available because of the $120^{\circ}$ bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs.

Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the $\mathrm{ClF}_{3}$ molecule (Figure 4.20). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure.

FIGURE 4.20 (a) In a trigonal bipyramid, the two axial positions are located directly across from one another, whereas the three equatorial positions are located in a triangular arrangement. (b-d) The two lone pairs (red lines) in $\mathrm{ClF}_{3}$ have several possible arrangements, but the T -shaped molecular structure (b) is the one actually observed, consistent with the larger lone pairs both occupying equatorial positions.

When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electronpair geometry. The two lone pairs are on opposite sides of the octahedron ( $180^{\circ}$ apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure 4.19).


414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair -
Predicting Electron Pair Geometry and Molecular Structure
The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures:

1. Write the Lewis structure of the molecule or polyatomic ion.
2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.
3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure 4.19, first column).
4. Use the number of lone pairs to determine the molecular structure (Figure 4.19). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.

The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry.


416. Predicting Electron-pair Geometry and Molecular Structure: $\mathbf{C O}_{\mathbf{2}}$ and $\mathbf{B C l}_{\mathbf{3}}$ -
Predict the electron-pair geometry and molecular structure for each of the following:
(a) carbon dioxide, $\mathrm{CO}_{2}$, a molecule produced by the combustion of fossil fuels
(b) boron trichloride, $\mathrm{BCl}_{3}$, an important industrial chemical


417. Solution -
(a) We write the Lewis structure of $\mathrm{CO}_{2}$ as:

This shows us two regions of high electron density around the carbon atom-each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of $180^{\circ}$. The
electron-pair geometry and molecular structure are identical, and $\mathrm{CO}_{2}$ molecules are linear.
(b) We write the Lewis structure of $\mathrm{BCl}_{3}$ as:

Thus we see that $\mathrm{BCl}_{3}$ contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The $\mathrm{B}-\mathrm{Cl}$ bonds lie in a plane with $120^{\circ}$ angles between them. $\mathrm{BCl}_{3}$ also has a trigonal planar molecular structure (Figure 4.21).

FIGURE 4.21
The electron-pair geometry and molecular structure of $\mathrm{BCl}_{3}$ are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles ( $120^{\circ}$ ), unlike the Lewis structure shown above.


418. Check Your Learning -
Carbonate, $\mathrm{CO}_{3}{ }^{2-}$, is a common polyatomic ion found in various materials from eggshells to antacids. What are the electron-pair geometry and molecular structure of this polyatomic ion?


419. Answer: -
The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three $\mathrm{C}-\mathrm{O}$ bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density.


421. Predicting Electron-pair Geometry and Molecular Structure: Ammonium -
Two of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the $\mathrm{NH}_{4}{ }^{+}$cation.


422. Solution -
We write the Lewis structure of $\mathrm{NH}_{4}{ }^{+}$as:

We can see that $\mathrm{NH}_{4}{ }^{+}$contains four bonds from the nitrogen atom to hydrogen atoms and no lone pairs. We expect the four regions of high electron density to arrange themselves so that they point to the corners of a tetrahedron with the central nitrogen atom in the middle (Figure 4.19). Therefore, the electron pair geometry of $\mathrm{NH}_{4}{ }^{+}$is tetrahedral, and the molecular structure is also tetrahedral (Figure 4.22).

FIGURE 4.22 The ammonium ion displays a tetrahedral electron-pair geometry as well as a tetrahedral molecular structure.

Check Your Learning
Identify a molecule with trigonal bipyramidal molecular structure.


423. Answer: -
Any molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. $\mathrm{PF}_{5}$ is a common example.

The next several examples illustrate the effect of lone pairs of electrons on molecular structure.


425. Predicting Electron-pair Geometry and Molecular Structure: Lone Pairs on the Central Atom -
Predict the electron-pair geometry and molecular structure of a water molecule.


426. Solution -
The Lewis structure of $\mathrm{H}_{2} \mathrm{O}$ indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds:

We predict that these four regions are arranged in a tetrahedral fashion (Figure 4.23), as indicated in Figure 4.19. Thus, the electron-pair geometry is tetrahedral and the molecular structure is bent with an angle slightly less than $109.5^{\circ}$. In fact, the bond angle is $104.5^{\circ}$.

(a)

(b)

FIGURE 4.23 (a) $\mathrm{H}_{2} \mathrm{O}$ has four regions of electron density around the central atom, so it has a tetrahedral electronpair geometry. (b) Two of the electron regions are lone pairs, so the molecular structure is bent.


427. Check Your Learning -
The hydronium ion, $\mathrm{H}_{3} \mathrm{O}^{+}$, forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation.


428. Answer: -
electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal


430. Predicting Electron-pair Geometry and Molecular Structure: $\mathbf{S F}_{\mathbf{4}}$ -
Sulfur tetrafluoride, $\mathrm{SF}_{4}$, is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., $\mathrm{SF}_{4}$ is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a $\mathrm{SF}_{4}$ molecule.


431. Solution -
The Lewis structure of $\mathrm{SF}_{4}$ indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs:

We expect these five regions to adopt a trigonal bipyramidal electron-pair geometry. To minimize lone pair repulsions, the lone pair occupies one of the equatorial positions. The molecular structure (Figure 4.24) is that of a seesaw (Figure 4.19).

FIGURE 4.24 (a) SF4 has a trigonal bipyramidal arrangement of the five regions of electron density. (b) One of the regions is a lone pair, which results in a seesaw-shaped molecular structure.


432. Check Your Learning -
Predict the electron pair geometry and molecular structure for molecules of $\mathrm{XeF}_{2}$.


433. Answer: -
The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear.


435. Predicting Electron-pair Geometry and Molecular Structure: $\mathrm{XeF}_{4}$ -
Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the $\mathrm{XeF}_{4}$ molecule.


436. Solution -
The Lewis structure of $\mathrm{XeF}_{4}$ indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds:

These six regions adopt an octahedral arrangement (Figure 4.19), which is the electron-pair geometry. To minimize repulsions, the lone pairs should be on opposite sides of the central atom (Figure 4.25). The five atoms are all in the same plane and have a square planar molecular structure.

FIGURE 4.25 (a) $\mathrm{XeF}_{4}$ adopts an octahedral arrangement with two lone pairs (red lines) and four bonds in the electron-pair geometry. (b) The molecular structure is square planar with the lone pairs directly across from one another.


437. Check Your Learning -
In a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be?


438. Answer: -
electron pair geometry: trigonal bipyramidal; molecular structure: linear


439. Molecular Structure for Multicenter Molecules -
When a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a "local" geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion. For our purposes, we will only focus on determining the local structures.


441. Predicting Structure in Multicenter Molecules -
The Lewis structure for the simplest amino acid, glycine, $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}$, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached:


442. Solution -
Consider each central atom independently. The electron-pair geometries:

- nitrogen--four regions of electron density; tetrahedral
- carbon $\left(\mathrm{CH}_{2}\right)$--four regions of electron density; tetrahedral
- carbon $\left(\mathrm{CO}_{2}\right)$-three regions of electron density; trigonal planar
- oxygen ( OH )-four regions of electron density; tetrahedral

The local structures:

- nitrogen--three bonds, one lone pair; trigonal pyramidal
- carbon $\left(\mathrm{CH}_{2}\right)$-four bonds, no lone pairs; tetrahedral
- carbon $\left(\mathrm{CO}_{2}\right)$-three bonds (double bond counts as one bond), no lone pairs; trigonal planar
- oxygen ( OH )-two bonds, two lone pairs; bent (109 $)$


443. Check Your Learning -
Another amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached:


444. Answer: -
electron-pair geometries: nitrogen--tetrahedral; carbon ( $\underline{\mathrm{C}} \mathbf{H}$ )-tetrahedral; carbon $\left(\mathrm{CH}_{3}\right)$-tetrahedral; carbon $\left(\mathrm{CO}_{2}\right)$-trigonal planar; oxygen ( OH )-tetrahedral; local structures: nitrogen-trigonal pyramidal; carbon $(\mathrm{CH})$-tetrahedral; carbon $\left(\mathrm{CH}_{3}\right)$-tetrahedral; carbon $\left(\mathrm{CO}_{2}\right)$-trigonal planar; oxygen $(\mathrm{OH})$-bent $\left(109^{\circ}\right)$


445. LINK TO LEARNING -
The molecular shape simulator (http://openstax.org/l/16MolecShape) lets you build various molecules and practice naming their electron-pair geometries and molecular structures.


447. Molecular Simulation -
Using molecular shape simulator (http://openstax.org/l/16MolecShape) allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under "Options" on the right. We can also use the "Name" checkboxes at bottom-left to display or hide the electron pair geometry (called "electron geometry" in the simulator) and/or molecular structure (called "molecular shape" in the simulator).

Build the molecule HCN in the simulator based on the following Lewis structure:

$$
\mathrm{H}-\mathrm{C} \equiv \mathrm{~N}
$$

Click on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this?


448. Solution -
The molecular structure is linear.


449. Check Your Learning -
Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn.


450. Answer: -
Answers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. $\mathrm{XeOF}_{4}$ is a molecule that adopts this structure.


451. Molecular Polarity and Dipole Moment -
As discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge ( $\delta^{+}$) and the other atom with a partial negative charge ( $\delta^{-}$), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment. The magnitude of a bond dipole moment is represented by the Greek letter mu( $\mu$ ) and is given by the formula shown here, where Q is the magnitude of the partial charges (determined by the electronegativity difference) and $r$ is the distance between the charges:

$$
\mu=\mathrm{Qr}
$$

This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure 4.26). Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms.

(a)

(b)

FIGURE 4.26 (a) There is a small difference in electronegativity between C and H , represented as a short vector. (b) The electronegativity difference between $B$ and $F$ is much larger, so the vector representing the bond moment is much longer.

A whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a polar molecule (or dipole); otherwise the molecule is said to be nonpolar. The dipole moment measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure.


451. Molecular Polarity and Dipole Moment -
For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as $\mathrm{Br}_{2}$ and $\mathrm{N}_{2}$ have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For HF, there is a larger dipole moment because there is a larger difference in electronegativity.

When a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in $\mathrm{CO}_{2}$ (Figure 4.27). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the $\mathrm{CO}_{2}$ molecule is linear with polar $\mathrm{C}=\mathrm{O}$ bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the water molecule (Figure 4.27), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O , and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole).

FIGURE 4.27 The overall dipole moment of a molecule depends on the individual bond dipole moments and how they are arranged. (a) Each CO bond has a bond dipole moment, but they point in opposite directions so that the net $\mathrm{CO}_{2}$ molecule is nonpolar. (b) In contrast, water is polar because the OH bond moments do not cancel out.

The OCS molecule has a structure similar to $\mathrm{CO}_{2}$, but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule:


451. Molecular Polarity and Dipole Moment -
The C-O bond is considerably polar. Although C and $S$ have very similar electronegativity values, $S$ is slightly more electronegative than C , and so the $\mathrm{C}-\mathrm{S}$ bond is just slightly polar. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end.

Chloromethane, $\mathrm{CH}_{3} \mathrm{Cl}$, is a tetrahedral molecule with three slightly polar $\mathrm{C}-\mathrm{H}$ bonds and a more polar $\mathrm{C}-\mathrm{Cl}$ bond. The relative electronegativities of the bonded atoms is $\mathrm{H}<\mathrm{C}<\mathrm{Cl}$, and so the bond moments all point toward the Cl end of the molecule and sum to yield a considerable dipole moment (the molecules are relatively polar).

For molecules of high symmetry such as $\mathrm{BF}_{3}$ (trigonal planar), $\mathrm{CH}_{4}$ (tetrahedral), $\mathrm{PF}_{5}$ (trigonal bipyramidal), and $\mathrm{SF}_{6}$ (octahedral), all the bonds are of identical polarity (same bond moment) and they are oriented in geometries that yield nonpolar molecules (dipole moment is zero). Molecules of less geometric symmetry, however, may be polar even when all bond moments are identical. For these molecules, the directions of the equal bond moments are such that they sum to give a nonzero dipole moment and a polar molecule. Examples of such molecules include hydrogen sulfide, $\mathrm{H}_{2} \mathrm{~S}$ (nonlinear), and ammonia, $\mathrm{NH}_{3}$ (trigonal pyramidal).

To summarize, to be polar, a molecule must:

1. Contain at least one polar covalent bond.
2. Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel.


452. Properties of Polar Molecules -
Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented
toward the negative plate and the negative end toward the positive plate (Figure 4.28). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances.

FIGURE 4.28 (a) Molecules are always randomly distributed in the liquid state in the absence of an electric field. (b) When an electric field is applied, polar molecules like HF will align to the dipoles with the field direction.


453. LINK TO LEARNING -
The molecule polarity simulation (http://openstax.org/l/16MolecPolarity) provides many ways to explore dipole moments of bonds and molecules.


455. Polarity Simulations -
Open the molecule polarity simulation (http://openstax.org/l/16MolecPolarity) and select the "Three Atoms" tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure 4.28.

Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if:
(a) A and C are very electronegative and B is in the middle of the range.
(b) A is very electronegative, and $B$ and $C$ are not.


456. Solution -
(a) Molecular dipole moment points immediately between A and C.
(b) Molecular dipole moment points along the A-B bond, toward A.


457. Check Your Learning -
Determine the partial charges that will give the largest possible bond dipoles.


458. Answer: -
The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will.


459. Key Terms -
axial position location in a trigonal bipyramidal geometry in which there is another atom at a $180^{\circ}$ angle and the equatorial positions are at a $90^{\circ}$ angle
binary acid compound that contains hydrogen and one other element, bonded in a way that imparts acidic properties to the compound (ability to release $\mathrm{H}^{+}$ions when dissolved in water)
binary compound compound containing two different elements.
bond angle angle between any two covalent bonds that share a common atom
bond dipole moment separation of charge in a bond that depends on the difference in electronegativity and the bond distance represented by partial charges or a vector
bond distance (also, bond length) distance between the nuclei of two bonded atoms
bond length distance between the nuclei of two bonded atoms at which the lowest potential energy is achieved
covalent bond bond formed when electrons are shared between atoms
dipole moment property of a molecule that describes the separation of charge determined by the sum of the individual bond moments based on the molecular structure
double bond covalent bond in which two pairs of electrons are shared between two atoms
electron-pair geometry arrangement around a central atom of all regions of electron density (bonds, lone pairs, or unpaired electrons)
electronegativity tendency of an atom to attract electrons in a bond to itself
equatorial position one of the three positions in a trigonal bipyramidal geometry with $120^{\circ}$ angles between them; the axial positions are located at a $90^{\circ}$ angle
formal charge charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)
free radical molecule that contains an odd number of electrons
hypervalent molecule molecule containing at least one main group element that has more than eight electrons in its valence shell
inert pair effect tendency of heavy atoms to form ions in which their valence $s$ electrons are not lost
ionic bond strong electrostatic force of attraction between cations and anions in an ionic
compound
Lewis structure diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion
Lewis symbol symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion


459. Key Terms -
compound
Lewis structure diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion
Lewis symbol symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion
linear shape in which two outside groups are placed on opposite sides of a central atom
lone pair two (a pair of) valence electrons that are not used to form a covalent bond
molecular structure arrangement of atoms in a molecule or ion
molecular structure structure that includes only the placement of the atoms in the molecule
nomenclature system of rules for naming objects of interest
octahedral shape in which six outside groups are placed around a central atom such that a threedimensional shape is generated with four groups forming a square and the other two forming the apex of two pyramids, one above and one below the square plane
octet rule guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond
oxyacid compound that contains hydrogen, oxygen, and one other element, bonded in a way that imparts acidic properties to the compound (ability to release $\mathrm{H}^{+}$ions when dissolved in water)
polar covalent bond covalent bond between atoms of different electronegativities; a covalent bond with a positive end and a negative end
polar molecule (also, dipole) molecule with an overall dipole moment
pure covalent bond (also, nonpolar covalent bond) covalent bond between atoms of identical electronegativities
resonance situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed
resonance forms two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons
resonance hybrid average of the resonance forms shown by the individual Lewis structures
single bond bond in which a single pair of electrons is shared between two atoms
tetrahedral shape in which four outside groups are
placed around a central atom such that a threedimensional shape is generated with four corners and $109.5^{\circ}$ angles between each pair and the central atom


459. Key Terms -
tetrahedral shape in which four outside groups are
placed around a central atom such that a threedimensional shape is generated with four corners and $109.5^{\circ}$ angles between each pair and the central atom
trigonal bipyramidal shape in which five outside groups are placed around a central atom such that three form a flat triangle with $120^{\circ}$ angles between each pair and the central atom, and the other two form the apex of two pyramids, one above and one below the triangular plane
trigonal planar shape in which three outside
groups are placed in a flat triangle around a central atom with $120^{\circ}$ angles between each pair and the central atom
triple bond bond in which three pairs of electrons are shared between two atoms
valence shell electron-pair repulsion theory (VSEPR) theory used to predict the bond angles in a molecule based on positioning regions of high electron density as far apart as possible to minimize electrostatic repulsion
vector quantity having magnitude and direction


460. Key Equations -
formal charge $=\#$ valence shell electrons (free atom) $-\#$ lone pair electrons $-\frac{1}{2} \#$ bonding electrons


461. Summary - 461.1. Ionic Bonding
Atoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells.


461. Summary - 461.2. Covalent Bonding
Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.


461. Summary - 461.3. Chemical Nomenclature
Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a
nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to -ide. For example, $\mathrm{K}_{2} \mathrm{O}$ is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge. Thus, $\mathrm{FeCl}_{2}$ is iron(II) chloride and $\mathrm{FeCl}_{3}$ is iron(III) chloride. Some compounds contain polyatomic ions; the names of common polyatomic ions should be memorized. Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include $\mathrm{SF}_{6}$, sulfur hexafluoride, and $\mathrm{N}_{2} \mathrm{O}_{4}$, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the -ide suffix to $-i c$, and adding "acid;" HCl is hydrochloric acid. Oxyacids are named by changing the ending of the anion (-ate to -ic and -ite to -ous), and adding "acid;" $\mathrm{H}_{2} \mathrm{CO}_{3}$ is carbonic acid.


461. Summary - 461.4. Lewis Symbols and Structures
Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures-especially those containing second row elements-obey the octet rule, in which every atom (except H ) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient
molecules, and hypervalent molecules.


461. Summary - 461.5. Formal Charges and Resonance
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).


461. Summary - 461.6. Molecular Structure and Polarity
VSEPR theory predicts the three-dimensional
arrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom. A dipole moment measures a separation of charge. For one bond, the bond dipole moment is determined by the difference in electronegativity between the two atoms. For a molecule, the overall dipole moment is determined by both the individual bond moments and how these dipoles are arranged in the molecular structure. Polar molecules (those with an appreciable dipole moment) interact with electric fields, whereas nonpolar molecules do not.


462. Exercises - 462.1. Ionic Bonding
1. Does a cation gain protons to form a positive charge or does it lose electrons?
2. Iron(III) sulfate $\left[\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\right]$ is composed of $\mathrm{Fe}^{3+}$ and $\mathrm{SO}_{4}{ }^{2-}$ ions. Explain why a sample of iron(III) sulfate is uncharged.
3. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: $\mathrm{P}, \mathrm{I}, \mathrm{Mg}, \mathrm{Cl}, \mathrm{In}, \mathrm{Cs}, \mathrm{O}, \mathrm{Pb}, \mathrm{Co}$ ?
4. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: $\mathrm{Br}, \mathrm{Ca}, \mathrm{Na}, \mathrm{N}, \mathrm{F}, \mathrm{Al}, \mathrm{Sn}, \mathrm{S}, \mathrm{Cd}$ ?
5. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:
(a) P
(b) Mg
(c) Al
(d) O
(e) Cl
(f) Cs
6. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:
(a) I
(b) Sr
(c) K
(d) N
(e) S
(f) In
7. Write the electron configuration for each of the following ions:
(a) $\mathrm{As}^{3-}$
(b) $I^{-}$
(c) $\mathrm{Be}^{2+}$
(d) $\mathrm{Cd}^{2+}$
(e) $\mathrm{O}^{2-}$
(f) $\mathrm{Ga}^{3+}$
(g) $\mathrm{Li}^{+}$
(h) $\mathrm{N}^{3-}$
(i) $\mathrm{Sn}^{2+}$
(j) $\mathrm{Co}^{2+}$
(k) $\mathrm{Fe}^{2+}$
(l) $\mathrm{As}^{3+}$


462. Exercises - 462.1. Ionic Bonding
(i) $\mathrm{Sn}^{2+}$
(j) $\mathrm{Co}^{2+}$
(k) $\mathrm{Fe}^{2+}$
(l) $\mathrm{As}^{3+}$
8. Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater):
(a) Cl
(b) Na
(c) Mg
(d) Ca
(e) K
(f) Br
(g) Sr
(h) F
9. Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element:
(a) Al
(b) Br
(c) Sr
(d) Li
(e) As
(f) S
10. From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.)


462. Exercises - 462.2. Covalent Bonding
11. Why is it incorrect to speak of a molecule of solid NaCl ?
12. What information can you use to predict whether a bond between two atoms is covalent or ionic?
13. Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table:
(a) $\mathrm{Cl}_{2} \mathrm{CO}$
(b) MnO
(c) $\mathrm{NCl}_{3}$
(d) $\mathrm{CoBr}_{2}$
(e) $\mathrm{K}_{2} \mathrm{~S}$
(f) CO
(g) $\mathrm{CaF}_{2}$
(h) HI
(i) CaO
(j) IBr
(k) $\mathrm{CO}_{2}$
14. Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond.
15. From its position in the periodic table, determine which atom in each pair is more electronegative:
(a) Br or Cl
(b) N or O
(c) S or O
(d) P or S
(e) Si or N
(f) Ba or P
(g) N or K
16. From its position in the periodic table, determine which atom in each pair is more electronegative:
(a) N or P
(b) N or Ge
(c) S or F
(d) Cl or S
(e) H or C
(f) Se or P
(g) C or Si
17. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:
(a) C, F, H, N, O
(b) Br, Cl, F, H, I
(c) F, H, O, P, S
(d) Al, H, Na, O, P
(e) Ba, H, N, O, As
18. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:
(a) As, H, N, P, Sb
(b) Cl, H, P, S, Si


462. Exercises - 462.2. Covalent Bonding
(a) As, H, N, P, Sb
(b) Cl, H, P, S, Si
(c) $\mathrm{Br}, \mathrm{Cl}, \mathrm{Ge}, \mathrm{H}, \mathrm{Sr}$
(d) Ca, H, K, N, Si
(e) $\mathrm{Cl}, \mathrm{Cs}, \mathrm{Ge}, \mathrm{H}, \mathrm{Sr}$
19. Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom?
20. Which is the most polar bond?
(a) $\mathrm{C}-\mathrm{C}$
(b) $\mathrm{C}-\mathrm{H}$
(c) $\mathrm{N}-\mathrm{H}$
(d) $\mathrm{O}-\mathrm{H}$
(e) $\mathrm{Se}-\mathrm{H}$
21. Identify the more polar bond in each of the following pairs of bonds:
(a) HF or HCl
(b) NO or CO
(c) SH or OH
(d) PCl or SCl
(e) CH or NH
(f) SO or PO
(g) CN or NN
22. Which of the following molecules or ions contain polar bonds?
(a) $\mathrm{O}_{3}$
(b) $\mathrm{S}_{8}$
(c) $\mathrm{O}_{2}{ }^{2-}$
(d) $\mathrm{NO}_{3}{ }^{-}$
(e) $\mathrm{CO}_{2}$
(f) $\mathrm{H}_{2} \mathrm{~S}$
(g) $\mathrm{BH}_{4}{ }^{-}$


462. Exercises - 462.3. Chemical Nomenclature
23. Name the following compounds:
(a) CsCl
(b) BaO
(c) $\mathrm{K}_{2} \mathrm{~S}$
(d) $\mathrm{BeCl}_{2}$
(e) HBr
(f) $\mathrm{AlF}_{3}$
24. Name the following compounds:
(a) NaF
(b) $\mathrm{Rb}_{2} \mathrm{O}$
(c) $\mathrm{BCl}_{3}$
(d) $\mathrm{H}_{2} \mathrm{Se}$
(e) $\mathrm{P}_{4} \mathrm{O}_{6}$
(f) $\mathrm{ICl}_{3}$
25. Write the formulas of the following compounds:
(a) rubidium bromide
(b) magnesium selenide
(c) sodium oxide
(d) calcium chloride
(e) hydrogen fluoride
(f) gallium phosphide
(g) aluminum bromide
(h) ammonium sulfate
26. Write the formulas of the following compounds:
(a) lithium carbonate
(b) sodium perchlorate
(c) barium hydroxide
(d) ammonium carbonate
(e) sulfuric acid
(f) calcium acetate
(g) magnesium phosphate
(h) sodium sulfite
27. Write the formulas of the following compounds:
(a) chlorine dioxide
(b) dinitrogen tetraoxide
(c) potassium phosphide
(d) silver(I) sulfide
(e) aluminum fluoride trihydrate
(f) silicon dioxide
28. Write the formulas of the following compounds:
(a) barium chloride
(b) magnesium nitride
(c) sulfur dioxide
(d) nitrogen trichloride
(e) dinitrogen trioxide
(f) tin(IV) chloride
29. Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:
(a) $\mathrm{Cr}_{2} \mathrm{O}_{3}$
(b) $\mathrm{FeCl}_{2}$
(c) $\mathrm{CrO}_{3}$
(d) $\mathrm{TiCl}_{4}$
(e) $\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$
(f) $\mathrm{MoS}_{2}$


462. Exercises - 462.3. Chemical Nomenclature
(d) $\mathrm{TiCl}_{4}$
(e) $\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$
(f) $\mathrm{MoS}_{2}$
30. Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:
(a) $\mathrm{NiCO}_{3}$
(b) $\mathrm{MoO}_{3}$
(c) $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}$
(d) $\mathrm{V}_{2} \mathrm{O}_{5}$
(e) $\mathrm{MnO}_{2}$
(f) $\mathrm{Fe}_{2} \mathrm{O}_{3}$
31. The following ionic compounds are found in common household products. Write the formulas for each compound:
(a) potassium phosphate
(b) copper(II) sulfate
(c) calcium chloride
(d) titanium(IV) oxide
(e) ammonium nitrate
(f) sodium bisulfate (the common name for sodium hydrogen sulfate)
32. The following ionic compounds are found in common household products. Name each of the compounds:
(a) $\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}$
(b) $\mathrm{FeSO}_{4}$
(c) $\mathrm{CaCO}_{3}$
(d) MgO
(e) $\mathrm{NaNO}_{2}$
(f) KI
33. What are the IUPAC names of the following compounds?
(a) manganese dioxide
(b) mercurous chloride $\left(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\right)$
(c) ferric nitrate $\left[\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\right]$
(d) titanium tetrachloride
(e) cupric bromide $\left(\mathrm{CuBr}_{2}\right)$


462. Exercises - 462.4. Lewis Symbols and Structures
34. Write the Lewis symbols for each of the following ions:
(a) $\mathrm{As}^{3-}$
(b) $\mathrm{I}^{-}$
(c) $\mathrm{Be}^{2+}$
(d) $\mathrm{O}^{2-}$
(e) $\mathrm{Ga}^{3+}$
(f) $\mathrm{Li}^{+}$
(g) $\mathrm{N}^{3-}$
35. Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements:
(a) Cl
(b) Na
(c) Mg
(d) Ca
(e) K
(f) Br
(g) Sr
(h) F
36. Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed:
(a) MgS
(b) $\mathrm{Al}_{2} \mathrm{O}_{3}$
(c) $\mathrm{GaCl}_{3}$
(d) $\mathrm{K}_{2} \mathrm{O}$
(e) $\mathrm{Li}_{3} \mathrm{~N}$
(f) KF
37. In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:
(a)

$$
\leftM^{2+}\right^{2-}
$$

(b)

$$
\leftM^{3+}\right_{3}^{-}
$$

(c)

$$
\left[M^{+}\right]_{2}[: \ddot{x}:]^{2-}
$$

(d)

$$
\left[M^{3+}\right]_{2}[: \ddot{x}:]_{3}^{2-}
$$


462. Exercises - 462.4. Lewis Symbols and Structures
38. Write the Lewis structure for the diatomic molecule $P_{2}$, an unstable form of phosphorus found in hightemperature phosphorus vapor.
39. Write Lewis structures for the following:
(a) $\mathrm{H}_{2}$
(b) HBr
(c) $\mathrm{PCl}_{3}$
(d) $\mathrm{SF}_{2}$
(e) $\mathrm{H}_{2} \mathrm{CCH}_{2}$
(f) HNNH
(g) $\mathrm{H}_{2} \mathrm{CNH}$
(h) $\mathrm{NO}^{-}$
(i) $\mathrm{N}_{2}$
(j) CO
(k) $\mathrm{CN}^{-}$
40. Write Lewis structures for the following:
(a) $\mathrm{O}_{2}$
(b) $\mathrm{H}_{2} \mathrm{CO}$
(c) $\mathrm{AsF}_{3}$
(d) ClNO
(e) $\mathrm{SiCl}_{4}$
(f) $\mathrm{H}_{3} \mathrm{O}^{+}$
(g) $\mathrm{NH}_{4}{ }^{+}$
(h) $\mathrm{BF}_{4}{ }^{-}$
(i) HCCH
(j) ClCN
(k) $\mathrm{C}_{2}{ }^{2+}$
41. Write Lewis structures for the following:
(a) $\mathrm{ClF}_{3}$
(b) $\mathrm{PCl}_{5}$
(c) $\mathrm{BF}_{3}$
(d) $\mathrm{PF}_{6}{ }^{-}$
42. Write Lewis structures for the following:
(a) $\mathrm{SeF}_{6}$
(b) $\mathrm{XeF}_{4}$
(c) $\mathrm{SeCl}_{3}{ }^{+}$
(d) $\mathrm{Cl}_{2} \mathrm{BBCl}_{2}$ (contains a B-B bond)
43. Write Lewis structures for:
(a) $\mathrm{PO}_{4}{ }^{3-}$
(b) $\mathrm{ICl}_{4}{ }^{-}$
(c) $\mathrm{SO}_{3}{ }^{2-}$
(d) HONO


462. Exercises - 462.4. Lewis Symbols and Structures
(a) $\mathrm{PO}_{4}{ }^{3-}$
(b) $\mathrm{ICl}_{4}{ }^{-}$
(c) $\mathrm{SO}_{3}{ }^{2-}$
(d) HONO
44. Correct the following statement: "The bonds in solid $\mathrm{PbCl}_{2}$ are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in $\mathrm{PbCl}_{2}$ are located on the $\mathrm{Cl}^{-}$ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms."
45. Write Lewis structures for the following molecules or ions:
(a) $\mathrm{SbH}_{3}$
(b) $\mathrm{XeF}_{2}$
(c) $\mathrm{Se}_{8}$ (a cyclic molecule with a ring of eight Se atoms)
46. Methanol, $\mathrm{H}_{3} \mathrm{COH}$, is used as the fuel in some race cars. Ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.
47. Many planets in our solar system contain organic chemicals including methane $\left(\mathrm{CH}_{4}\right)$ and traces of ethylene $\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)$, ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, propyne $\left(\mathrm{H}_{3} \mathrm{CCCH}\right)$, and diacetylene $(\mathrm{HCCCCH})$. Write the Lewis structures for each of these molecules.
48. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, $\mathrm{Cl}_{2} \mathrm{CO}$. Write the Lewis structures for carbon tetrachloride and phosgene.


462. Exercises - 462.4. Lewis Symbols and Structures
49. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom:
(a) $1 s^{2} 2 s^{2} 2 p^{5}$
(b) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$
(c) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$
(d) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{4}$
(e) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{1}$
50. The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.
(a) the amino acid serine:


462. Exercises - 462.4. Lewis Symbols and Structures
(b) urea:

(c) pyruvic acid:

(d) uracil:

(e) carbonic acid:

51. A compound with a molar mass of about $28 \mathrm{~g} / \mathrm{mol}$ contains $85.7 \%$ carbon and $14.3 \%$ hydrogen by mass. Write the Lewis structure for a molecule of the compound.
52. A compound with a molar mass of about $42 \mathrm{~g} / \mathrm{mol}$ contains $85.7 \%$ carbon and $14.3 \%$ hydrogen by mass. Write the Lewis structure for a molecule of the compound.
53. Two arrangements of atoms are possible for a compound with a molar mass of about $45 \mathrm{~g} / \mathrm{mol}$ that contains $52.2 \% \mathrm{C}, 13.1 \% \mathrm{H}$, and $34.7 \%$ O by mass. Write the Lewis structures for the two molecules.
54. How are single, double, and triple bonds similar? How do they differ?


462. Exercises - 462.5. Formal Charges and Resonance
55. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) selenium dioxide, OSeO
(b) nitrate ion, $\mathrm{NO}_{3}{ }^{-}$
(c) nitric acid, $\mathrm{HNO}_{3}$ ( N is bonded to an OH group and two O atoms)
(d) benzene, $\mathrm{C}_{6} \mathrm{H}_{6}$ :

(e) the formate ion:

56. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) sulfur dioxide, $\mathrm{SO}_{2}$
(b) carbonate ion, $\mathrm{CO}_{3}{ }^{2-}$
(c) hydrogen carbonate ion, $\mathrm{HCO}_{3}{ }^{-}$( C is bonded to an OH group and two O atoms)
(d) pyridine:

(e) the allyl ion:

57. Write the resonance forms of ozone, $\mathrm{O}_{3}$, the component of the upper atmosphere that protects the Earth from ultraviolet radiation.
58. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, $\mathrm{NO}_{2}{ }^{-}$.
59. In terms of the bonds present, explain why acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, contains two distinct types of carbonoxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown:


462. Exercises - 462.5. Formal Charges and Resonance
60. Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
(a) $\mathrm{CO}_{2}$
(b) CO
61. Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
62. Determine the formal charge of each element in the following:
(a) HCl
(b) $\mathrm{CF}_{4}$
(c) $\mathrm{PCl}_{3}$
(d) $\mathrm{PF}_{5}$
63. Determine the formal charge of each element in the following:
(a) $\mathrm{H}_{3} \mathrm{O}^{+}$
(b) $\mathrm{SO}_{4}{ }^{2-}$
(c) $\mathrm{NH}_{3}$
(d) $\mathrm{O}_{2}{ }^{2-}$
(e) $\mathrm{H}_{2} \mathrm{O}_{2}$
64. Calculate the formal charge of chlorine in the molecules $\mathrm{Cl}_{2}, \mathrm{BeCl}_{2}$, and $\mathrm{ClF}_{5}$.
65. Calculate the formal charge of each element in the following compounds and ions:
(a) $\mathrm{F}_{2} \mathrm{CO}$
(b) $\mathrm{NO}^{-}$
(c) $\mathrm{BF}_{4}^{-}$
(d) $\mathrm{SnCl}_{3}-$
(e) $\mathrm{H}_{2} \mathrm{CCH}_{2}$
(f) $\mathrm{ClF}_{3}$
(g) $\mathrm{SeF}_{6}$
(h) $\mathrm{PO}_{4}{ }^{3-}$
66. Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:
(a) $\mathrm{O}_{3}$
(b) $\mathrm{SO}_{2}$
(c) $\mathrm{NO}_{2}{ }^{-}$
(d) $\mathrm{NO}_{3}{ }^{-}$
67. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?


462. Exercises - 462.5. Formal Charges and Resonance
(d) $\mathrm{NO}_{3}{ }^{-}$
67. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
68. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH ?
69. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
70. Draw the structure of hydroxylamine, $\mathrm{H}_{3} \mathrm{NO}$, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
71. Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:
(a) IF
(b) $\mathrm{IF}_{3}$
(c) $\mathrm{IF}_{5}$
(d) $\mathrm{IF}_{7}$
72. Write the Lewis structure and chemical formula of the compound with a molar mass of about $70 \mathrm{~g} / \mathrm{mol}$ that contains $19.7 \%$ nitrogen and $80.3 \%$ fluorine by mass, and determine the formal charge of the atoms in this compound.
73. Which of the following structures would we expect for nitrous acid? Determine the formal charges:


462. Exercises - 462.5. Formal Charges and Resonance
74. Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4}$, which has two oxygen atoms and two OH groups bonded to the sulfur.


462. Exercises - 462.6. Molecular Structure and Polarity
75. Explain why the HOH molecule is bent, whereas the HBeH molecule is linear.
76. What feature of a Lewis structure can be used to tell if a molecule's (or ion's) electron-pair geometry and molecular structure will be identical?
77. Explain the difference between electron-pair geometry and molecular structure.
78. Why is the $\mathrm{H}-\mathrm{N}-\mathrm{H}$ angle in $\mathrm{NH}_{3}$ smaller than the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle in $\mathrm{CH}_{4}$ ? Why is the $\mathrm{H}-\mathrm{N}-\mathrm{H}$ angle in $\mathrm{NH}_{4}{ }^{+}$identical to the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle in $\mathrm{CH}_{4}$ ?
79. Explain how a molecule that contains polar bonds can be nonpolar.
80. As a general rule, $M X_{n}$ molecules (where $M$ represents a central atom and $X$ represents terminal atoms; $n$ $=2-5)$ are polar if there is one or more lone pairs of electrons on $M . \mathrm{NH}_{3}(\mathrm{M}=\mathrm{N}, \mathrm{X}=\mathrm{H}, \mathrm{n}=3)$ is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they?
81. Predict the electron pair geometry and the molecular structure of each of the following molecules or ions:
(a) $\mathrm{SF}_{6}$
(b) $\mathrm{PCl}_{5}$
(c) $\mathrm{BeH}_{2}$
(d) $\mathrm{CH}_{3}{ }^{+}$
82. Identify the electron pair geometry and the molecular structure of each of the following molecules or ions:
(a) $\mathrm{IF}_{6}{ }^{+}$
(b) $\mathrm{CF}_{4}$
(c) $\mathrm{BF}_{3}$
(d) $\mathrm{SiF}_{5}{ }^{-}$
(e) $\mathrm{BeCl}_{2}$
83. What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?


462. Exercises - 462.6. Molecular Structure and Polarity
(d) $\mathrm{SiF}_{5}{ }^{-}$
(e) $\mathrm{BeCl}_{2}$
83. What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?
(a) $\mathrm{ClF}_{5}$
(b) $\mathrm{ClO}_{2}-$
(c) $\mathrm{TeCl}_{4}{ }^{2-}$
(d) $\mathrm{PCl}_{3}$
(e) $\mathrm{SeF}_{4}$
(f) $\mathrm{PH}_{2}{ }^{-}$
84. Predict the electron pair geometry and the molecular structure of each of the following ions:
(a) $\mathrm{H}_{3} \mathrm{O}^{+}$
(b) $\mathrm{PCl}_{4}^{-}$
(c) $\mathrm{SnCl}_{3}+$
(d) $\mathrm{BrCl}_{4}^{-}$
(e) $\mathrm{ICl}_{3}$
(f) $\mathrm{XeF}_{4}$
(g) $\mathrm{SF}_{2}$
85. Identify the electron pair geometry and the molecular structure of each of the following molecules:
(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) $\mathrm{ClOF}_{2}+(\mathrm{Cl}$ is the central atom $)$
86. Predict the electron pair geometry and the molecular structure of each of the following:
(a) $\mathrm{IOF}_{5}$ ( I is the central atom)
(b) $\mathrm{POCl}_{3}$ ( P is the central atom)
(c) $\mathrm{Cl}_{2} \mathrm{SeO}$ (Se is the central atom)
(d) $\mathrm{ClSO}^{+}$( S is the central atom)


462. Exercises - 462.6. Molecular Structure and Polarity
(c) $\mathrm{Cl}_{2} \mathrm{SeO}$ (Se is the central atom)
(d) $\mathrm{ClSO}^{+}$( S is the central atom)
(e) $\mathrm{F}_{2} \mathrm{SO}$ ( S is the central atom)
(f) $\mathrm{NO}_{2}^{-}$
(g) $\mathrm{SiO}_{4}{ }^{4-}$
87. Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?
(a) $\mathrm{ClF}_{5}$
(b) $\mathrm{ClO}_{2}^{-}$
(c) $\mathrm{TeCl}_{4}{ }^{2-}$
(d) $\mathrm{PCl}_{3}$
(e) $\mathrm{SeF}_{4}$
(f) $\mathrm{PH}_{2}{ }^{-}$
(g) $\mathrm{XeF}_{2}$
88. Which of these molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?
(a) $\mathrm{H}_{3} \mathrm{O}^{+}$
(b) $\mathrm{PCl}_{4}^{-}$
(c) $\mathrm{SnCl}_{3}-$
(d) $\mathrm{BrCl}_{4}{ }^{-}$
(e) $\mathrm{ICl}_{3}$
(f) $\mathrm{XeF}_{4}$
(g) $\mathrm{SF}_{2}$
89. Which of the following molecules have dipole moments?
(a) $\mathrm{CS}_{2}$
(b) $\mathrm{SeS}_{2}$
(c) $\mathrm{CCl}_{2} \mathrm{~F}_{2}$
(d) $\mathrm{PCl}_{3}$ ( P is the central atom)
(e) ClNO ( N is the central atom)
90. Identify the molecules with a dipole moment:
(a) $\mathrm{SF}_{4}$
(b) $\mathrm{CF}_{4}$
(c) $\mathrm{Cl}_{2} \mathrm{CCBr}_{2}$
(d) $\mathrm{CH}_{3} \mathrm{Cl}$
(e) $\mathrm{H}_{2} \mathrm{CO}$


462. Exercises - 462.6. Molecular Structure and Polarity
(c) $\mathrm{Cl}_{2} \mathrm{CCBr}_{2}$
(d) $\mathrm{CH}_{3} \mathrm{Cl}$
(e) $\mathrm{H}_{2} \mathrm{CO}$
91. The molecule $\mathrm{XF}_{3}$ has a dipole moment. Is X boron or phosphorus?
92. The molecule $\mathrm{XCl}_{2}$ has a dipole moment. Is X beryllium or sulfur?
93. Is the $\mathrm{Cl}_{2} \mathrm{BBCl}_{2}$ molecule polar or nonpolar?
94. There are three possible structures for $\mathrm{PCl}_{2} \mathrm{~F}_{3}$ with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them.
95. Describe the molecular structure around the indicated atom or atoms:
(a) the sulfur atom in sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4}\left[(\mathrm{HO})_{2} \mathrm{SO}_{2}\right]$
(b) the chlorine atom in chloric acid, $\mathrm{HClO}_{3}\left[\mathrm{HOClO}_{2}\right]$
(c) the oxygen atom in hydrogen peroxide, HOOH
(d) the nitrogen atom in nitric acid, $\mathrm{HNO}_{3}\left[\mathrm{HONO}_{2}\right]$
(e) the oxygen atom in the OH group in nitric acid, $\mathrm{HNO}_{3}\left[\mathrm{HONO}_{2}\right]$
(f) the central oxygen atom in the ozone molecule, $\mathrm{O}_{3}$
$(\mathrm{g})$ each of the carbon atoms in propyne, $\mathrm{CH}_{3} \mathrm{CCH}$
(h) the carbon atom in Freon, $\mathrm{CCl}_{2} \mathrm{~F}_{2}$
(i) each of the carbon atoms in allene, $\mathrm{H}_{2} \mathrm{CCCH}_{2}$
96. Draw the Lewis structures and predict the shape of each compound or ion:
(a) $\mathrm{CO}_{2}$
(b) $\mathrm{NO}_{2}{ }^{-}$
(c) $\mathrm{SO}_{3}$


462. Exercises - 462.6. Molecular Structure and Polarity
96. Draw the Lewis structures and predict the shape of each compound or ion:
(a) $\mathrm{CO}_{2}$
(b) $\mathrm{NO}_{2}{ }^{-}$
(c) $\mathrm{SO}_{3}$
(d) $\mathrm{SO}_{3}{ }^{2-}$
97. A molecule with the formula $A B_{2}$, in which $A$ and $B$ represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape.
98. A molecule with the formula $\mathrm{AB}_{3}$, in which $A$ and $B$ represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape.
99. Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate:
(a) $\mathrm{CS}_{3}{ }^{2-}$
(b) $\mathrm{CS}_{2}$
(c) CS
(d) predict the molecular shapes for $\mathrm{CS}_{3}{ }^{2-}$ and $\mathrm{CS}_{2}$ and explain how you arrived at your predictions
100. What is the molecular structure of the stable form of $\mathrm{FNO}_{2}$ ? ( N is the central atom.)
101. A compound with a molar mass of about $42 \mathrm{~g} / \mathrm{mol}$ contains $85.7 \%$ carbon and $14.3 \%$ hydrogen. What is its molecular structure?
102. Use the simulation (http://openstax.org/l/16MolecPolarity) to perform the following exercises for a twoatom molecule:
(a) Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A.
(b) With a partial positive charge on A, turn on the electric field and describe what happens.
(c) With a small partial negative charge on A, turn on the electric field and describe what happens.
(d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.


462. Exercises - 462.6. Molecular Structure and Polarity
(c) With a small partial negative charge on A, turn on the electric field and describe what happens.
(d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.
103. Use the simulation (http://openstax.org/l/16MolecPolarity) to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles.
(a) Sketch the bond dipoles and molecular dipole (if any) for $\mathrm{O}_{3}$. Explain your observations.
(b) Look at the bond dipoles for $\mathrm{NH}_{3}$. Use these dipoles to predict whether N or H is more electronegative.
(c) Predict whether there should be a molecular dipole for $\mathrm{NH}_{3}$ and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis.
104. Use the Molecule Shape simulator (http://openstax.org/l/16MolecShape) to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers.
105. Use the Molecule Shape simulator (http://openstax.org/l/16MolecShape) to explore real molecules. On the Real Molecules tab, select $\mathrm{H}_{2} \mathrm{O}$. Switch between the "real" and "model" modes. Explain the difference observed.
106. Use the Molecule Shape simulator (http://openstax.org/l/16MolecShape) to explore real molecules. On the Real Molecules tab, select "model" mode and $\mathrm{S}_{2} \mathrm{O}$. What is the model bond angle? Explain whether the "real" bond angle should be larger or smaller than the ideal model angle.


462. Exercises - 462.6. Molecular Structure and Polarity
236 4•Exercises


463. CHAPTER 5 <br> Advanced Theories of Bonding -
Figure 5.1 Oxygen molecules orient randomly most of the time, as shown in the top magnified view. However, when we pour liquid oxygen through a magnet, the molecules line up with the magnetic field, and the attraction allows them to stay suspended between the poles of the magnet where the magnetic field is strongest. Other diatomic molecules (like $\mathrm{N}_{2}$ ) flow past the magnet. The detailed explanation of bonding described in this chapter allows us to understand this phenomenon. (credit: modification of work by Jefferson Lab)


464. CHAPTER OUTLINE -
5.1 Valence Bond Theory
5.2 Hybrid Atomic Orbitals
5.3 Multiple Bonds
5.4 Molecular Orbital Theory

INTRODUCTION We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of $s, p$, and $d$ atomic orbitals with molecular shapes that show angles like $120^{\circ}$ and $109.5^{\circ}$ ? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ have fairly similar Lewis structures that contain lone pairs of electrons.

```
:N三N: :
```

Yet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen (shown in Figure 5.1) is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations.


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

- Describe the formation of covalent bonds in terms of atomic orbital overlap
- Define and give examples of $\sigma$ and $\pi$ bonds

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


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

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


466. Atomic Orbital Overlap -
The energy of the system depends on how much the orbitals overlap. Figure 5.2 illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure 5.2.

FIGURE 5.2 (a) The interaction of two hydrogen atoms changes as a function of distance. (b) The energy of the system changes as the atoms interact. The lowest (most stable) energy occurs at a distance of 74 pm , which is the bond length observed for the $\mathrm{H}_{2}$ molecule.

In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two $s$ orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure 5.3 illustrates this for two $p$ orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.

(a)

(b)


466. Atomic Orbital Overlap -
(a)

(b)

FIGURE 5.3 (a) The overlap of two $p$ orbitals is greatest when the orbitals are directed end to end. (b) Any other arrangement results in less overlap. The dots indicate the locations of the nuclei.

The overlap of two $s$ orbitals (as in $\mathrm{H}_{2}$ ), the overlap of an $s$ orbital and a $p$ orbital (as in HCl ), and the end-to-end overlap of two $p$ orbitals (as in $\mathrm{Cl}_{2}$ ) all produce sigma bonds ( $\boldsymbol{\sigma}$ bonds), as illustrated in Figure 5.4. A $\sigma$ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as $\sigma$ bonds in valence bond theory.

FIGURE 5.4 Sigma ( $\sigma$ ) bonds form from the overlap of the following: (a) two $s$ orbitals, (b) an $s$ orbital and a $p$ orbital, and (c) two $p$ orbitals. The dots indicate the locations of the nuclei.

A pi bond ( $\boldsymbol{\pi}$ bond) is a type of covalent bond that results from the side-by-side overlap of two $p$ orbitals, as illustrated in Figure 5.5. In a $\pi$ bond, the regions of orbital overlap lie on opposite sides of the internuclear axis.

Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.

FIGURE 5.5 $\mathrm{Pi}(\pi)$ bonds form from the side-by-side overlap of two $p$ orbitals. The dots indicate the location of the nuclei.


466. Atomic Orbital Overlap -
FIGURE 5.5 $\mathrm{Pi}(\pi)$ bonds form from the side-by-side overlap of two $p$ orbitals. The dots indicate the location of the nuclei.

While all single bonds are $\sigma$ bonds, multiple bonds consist of both $\sigma$ and $\pi$ bonds. As the Lewis structures below suggest, $\mathrm{O}_{2}$ contains a double bond, and $\mathrm{N}_{2}$ contains a triple bond. The double bond consists of one $\sigma$ bond and one $\pi$ bond, and the triple bond consists of one $\sigma$ bond and two $\pi$ bonds. Between any two atoms, the first bond formed will always be a $\sigma$ bond, but there can only be one $\sigma$ bond in any one location. In any multiple bond, there will be one $\sigma$ bond, and the remaining one or two bonds will be $\pi$ bonds. These bonds are described in more detail later in this chapter.

One $\sigma$ bond
No $\pi$ bonds

$$
: \ddot{o}=\ddot{\mathrm{O}}:
$$

One $\sigma$ bond One $\pi$ bond
$: N \equiv N:$


467. One $\sigma$ bond -
Two $\pi$ bonds


469. Counting $\sigma$ and $\pi$ Bonds -
Butadiene, $\mathrm{C}_{4} \mathrm{H}_{6}$, is used to make synthetic rubber. Identify the number of $\sigma$ and $\pi$ bonds contained in this molecule.


470. Solution -
There are six $\sigma \mathrm{C}-\mathrm{H}$ bonds and one $\sigma \mathrm{C}-\mathrm{C}$ bond, for a total of seven from the single bonds. There are two double bonds that each have a $\pi$ bond in addition to the $\sigma$ bond. This gives a total nine $\sigma$ and two $\pi$ bonds overall.


471. Check Your Learning -
Identify each illustration as depicting a $\sigma$ or $\pi$ bond:
(a) side-by-side overlap of a $4 p$ and a $2 p$ orbital
(b) end-to-end overlap of a $4 p$ and $4 p$ orbital
(c) end-to-end overlap of a $4 p$ and a $2 p$ orbital

(a)

(b)

(c)


472. Answer: -
(a) is a $\pi$ bond with a node along the axis connecting the nuclei while (b) and (c) are $\sigma$ bonds that overlap along the axis.


473. Dipole Moments and Ionic Character -
Now that we have seen the importance of understanding the connection between the location of electrons in atoms and the properties of elements, we can expand our understanding of the connection between atoms. This will be an introduction to more advanced aspects of the chemical bond, which is the very heart of chemistry itself. With the sole exception of the noble gases, atoms by themselves do not possess the most stable possible electron configuration. That is where the concept of chemical bonding comes into its own: atoms can attain a stable configuration by exchanging electrons with another atom, resulting in the formation of ions.
Ions, in turn, can associate by charge - simple Coulombic attraction - resulting in the formation of compounds we call ionic compounds. We will look at the ionic nature of bonds first, from a simple positive-negative attraction standpoint. Just as important is that some atoms bond by sharing rather than exchanging electrons; the sharing of electrons gives rise to the covalent bond. To add just one more dimension, some chemical species are neither completely ionic nor completely covalent; these species possess a permanent dipole, and are classified as polar.
In your introductory physics course, you will likely discuss the concept of Coulombic interactions in much more rigorous detail than we will do here. We are interested primarily in the differences in properties between species that arise from their relative covalent, ionic, or polar nature - not in a rigorous model of those properties. We are concerned with the connection between potential energy and force and the relative separation (or lack of separation) between charges. We begin by defining the electric or Coulomb force as the product of the charges divided by the square of the distance between those charges:

$$
F=\frac{Q_{1} Q_{2}}{d^{2}}
$$

Here, $Q$ is taken to be the fundamental constant of electron charge: one electron has a charge of $1.60218 \times 10^{-19} \mathrm{C}$. (We will work exclusively in the SI system, so distances will be measured in meters $(\mathrm{m})$ ).
And as you may recall, energy is force times distance, so

$$
E=\frac{Q_{1} Q_{2}}{d}
$$

To illustrate the trend in attractive force, we will consider first the attractive force between two ions of single charge separated by a distance of 2 d :


473. Dipole Moments and Ionic Character -
$$
E=\frac{Q_{1} Q_{2}}{d}
$$

To illustrate the trend in attractive force, we will consider first the attractive force between two ions of single charge separated by a distance of 2 d :

$$
F=\frac{(1)(-1)}{(2 d)^{2}}=-\frac{1}{4 d^{2}}
$$

And then the attractive force between two ions of double charge separated by a distance $d$ :

$$
F=\frac{(2)(-2)}{(d)^{2}}=-\frac{4}{d^{2}}
$$

The force of attraction increases with the charge and decreases with increased distance. If all matter were composed of ions that would be the end of the story, but it clearly is not. There are molecules for which the charge - either positive or negative - is permanently concentrated more on one atom than the other. Notice we say atom, because these compounds are not composed of ions but rather of atoms sharing electrons through covalent bonds.


474. Bond Dipole Moments -
The concept of a bond dipole moment allows us to examine the partial separation of charge between atoms. It is a simple model when applied to diatomic molecules, which will be more than sufficient for our purposes. The dipole moment of a bond is defined as the charge times the distance - charge once again being measured in multiples of the charge on an electron, or coulombs. The distance will always be in meters. Because we are considering very small charges and distances, and because it is the relative separation of charge rather than the actual value for it that we are interested in, we will introduce a new unit called the Debye, named after the physical chemist Peter Debye:

$$
1 \text { Debye }(\mathrm{D})=3.336 \times 10^{-30} \mathrm{C}-\mathrm{m} \quad \mu=Q \times d
$$

The usefulness of the Debye unit will be shown by example:
For HCl , the bond dipole moment is known to be 1.08 D
For HI, the bond dipole moment is known to be 0.44 D
Comparing the two, we can see that HI is less polar than HCl , which is what we would expect based on electronegativity values.

We have now made a transition between the concept of an ionic compound and a partially ionic one. Of course, the partially ionic compound must also by definition be partially covalent.


475. Partial Ionic Character -
The concept of the bond dipole moment helps bridge the concepts of ionic and covalent bonding. Because there is a separation of charge that is less complete than it is in an ionic bond, we can refer to polar bonds as being partially ionic in nature. In contrast to sodium chloride, hydrogen chloride shows partial charges (indicated with a delta notation) on the hydrogen and chlorine. As you would expect from the electronegativity values, hydrogen carries a partial positive charge, while chlorine carries a partial negative charge. Where do these charges come from?

It is easy to come up with the partial charges by comparing the actual dipole moments (which can be obtained experimentally, using spectroscopy) with the dipole expected in the limiting case (that is, if we were to consider the molecule ionic). The actual dipole moment is 1.03 D .


477. Finding the Partial Ionic Character -
What are the partial charges of an HCl molecule, whose bond length is 0.127 nm ?


478. Solution -
The bond dipole moment is $\left(1.60218 \times 10^{-19} \mathrm{C}\right)\left(0.127 \times 10^{-9} \mathrm{~m}\right)$ or $2.03 \times 10^{-29} \mathrm{C}-\mathrm{m}$. Converted to D , this is $\left(2.03 \times 10^{-29} \mathrm{C}-\mathrm{m}\right)\left(\frac{1 \text { Debye }}{3.336 \times 10^{-30} \mathrm{C}-\mathrm{m}}\right)$ or 6.09 D . Were HCl completely ionic, this would be its molecular dipole moment. To get the partial ionic character, we divide the experimentally measured bond moment by this limiting value: \%ionic character $\left.=\frac{\mu_{\exp }}{\mu_{\lim }} \times 100 \%=\frac{(1.03}{(6.09} \quad \mathrm{D}\right) ~ \times 100 \%=17 \%$. This means the bond is about $17 \%$ ionic - or, put another way, the positive charge in H is +0.17 and the partial negative charge on chlorine, -0.17.


479. Check Your Learning -
Repeat the calculation for HI , which has a dipole moment of 0.42 D and a bond length of 0.161 nm .
Answer: Calculated 7.73, percent 5.43

What does the result suggest about the relative polarity of the HI bond vs. that of the HCl bond? Does the calculated dipole and percent ionic character reconcile with the difference in electronegativity between Cl and I?

The electron configuration of an atom or ion is key to understanding the chemical behavior of an element. The atoms that make up the element combine in various ways, ranging from the mostly ionic ( NaCl ) to the partially ionic ( HCl ) to what we will call purely covalent. At the most fundamental level, all chemical bonds involve electrons, and a significant percentage of chemical and physical properties can be explained by considering the location and separation of charge in a species. By understanding the structure of matter at the atomic level, we can begin to build an understanding of the behavior of matter at both the microscopic and macroscopic levels.

An understanding of dipoles and partial ionic character is fundamental to understanding the interactions between particles, which we will examine in the chapter on liquids and solids. These intermolecular forces
become important in the liquid and solid states of matter.


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

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

Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration $1 s^{2} 2 s^{2} 2 p^{4}$, with two unpaired electrons (one in each of the two $2 p$ orbitals). Valence bond theory would predict that the two $\mathrm{O}-\mathrm{H}$ bonds form from the overlap of these two $2 p$ orbitals with the $1 s$ orbitals of the hydrogen atoms. If this were the case, the bond angle would be $90^{\circ}$, as shown in Figure 5.6 , because $p$ orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is $104.5^{\circ}$, not $90^{\circ}$. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.

FIGURE 5.6 The hypothetical overlap of two of the $2 p$ orbitals on an oxygen atom (red) with the 1 s orbitals of two hydrogen atoms (blue) would produce a bond angle of $90^{\circ}$. This is not consistent with experimental evidence. ${ }^{1}$


480. LEARNING OBJECTIVES -
Quantum-mechanical calculations suggest why the observed bond angles in $\mathrm{H}_{2} \mathrm{O}$ differ from those predicted by the overlap of the $1 s$ orbital of the hydrogen atoms with the $2 p$ orbitals of the oxygen atom. The mathematical expression known as the wave function, $\psi$, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a $2 s$ orbital and three $2 p$ orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 5.7). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle $\left(109.5^{\circ}\right)$. The observed angle of $104.5^{\circ}$ is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.

(a)

(b)

FIGURE 5.7 (a) A water molecule has four regions of electron density, so VSEPR theory predicts a tetrahedral arrangement of hybrid orbitals. (b) Two of the hybrid orbitals on oxygen contain lone pairs, and the other two overlap with the 1 s orbitals of hydrogen atoms to form the $\mathrm{O}-\mathrm{H}$ bonds in $\mathrm{H}_{2} \mathrm{O}$. This description is more consistent with the

[^3]experimental structure.
The following ideas are important in understanding hybridization:


480. LEARNING OBJECTIVES -
[^3]experimental structure.
The following ideas are important in understanding hybridization:

1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.
2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.
4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.
5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.
6. Hybrid orbitals overlap to form $\sigma$ bonds. Unhybridized orbitals overlap to form $\pi$ bonds.

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


481. sp Hybridization -
The beryllium atom in a gaseous $\mathrm{BeCl}_{2}$ molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the $\mathrm{BeCl}_{2}$ molecule that correspond to the two covalent $\mathrm{Be}-\mathrm{Cl}$ bonds. To accommodate these two electron domains, two of the Be atom's four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence $s$ orbital with one of the valence $p$ orbitals to yield two equivalent $\boldsymbol{s p}$ hybrid orbitals that are oriented in a linear geometry (Figure 5.8). In this figure, the set of $s p$ orbitals appears similar in shape to the original $p$ orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The $p$ orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point $180^{\circ}$ from each other. The two electrons that were originally in the $s$ orbital are now distributed to the two $s p$ orbitals, which are half filled. In gaseous $\mathrm{BeCl}_{2}$, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical $\sigma$ bonds.

Gives a linear arrangement

FIGURE 5.8 Hybridization of an $s$ orbital (blue) and a $p$ orbital (red) of the same atom produces two $s p$ hybrid orbitals (yellow). Each hybrid orbital is oriented primarily in just one direction. Note that each sp orbital contains one lobe that is significantly larger than the other. The set of two $s p$ orbitals are oriented at $180^{\circ}$, which is consistent with the geometry for two domains.


481. sp Hybridization -
We illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energylevel diagram in Figure 5.9. These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin.

FIGURE 5.9 This orbital energy-level diagram shows the $s p$ hybridized orbitals on Be in the linear $\mathrm{BeCl}_{2}$ molecule. Each of the two $s p$ hybrid orbitals holds one electron and is thus half filled and available for bonding via overlap with a Cl 3 p orbital.

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

Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit $s p$ hybridization. Other examples include the mercury atom in the linear $\mathrm{HgCl}_{2}$ molecule, the zinc atom in $\mathrm{Zn}\left(\mathrm{CH}_{3}\right)_{2}$, which contains a linear $\mathrm{C}-\mathrm{Zn}-\mathrm{C}$ arrangement, and the carbon atoms in HCCH and $\mathrm{CO}_{2}$.


482. (8) LINK TO LEARNING -
Check out the University of Wisconsin-Oshkosh website (http://openstax.org/l/16hybridorbital) to learn about visualizing hybrid orbitals in three dimensions.


483. $s p^{2}$ Hybridization -
The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three $\boldsymbol{s p}^{\mathbf{2}}$ hybrid orbitals and one unhybridized $p$ orbital. This arrangement results from $s p^{2}$ hybridization, the mixing of one $s$ orbital and two $p$ orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 5.10).

FIGURE 5.10 The hybridization of an $s$ orbital (blue) and two $p$ orbitals (red) produces three equivalent $s p^{2}$ hybridized orbitals (yellow) oriented at $120^{\circ}$ with respect to each other. The remaining unhybridized $p$ orbital is not shown here, but is located along the $z$ axis.

Although quantum mechanics yields the "plump" orbital lobes as depicted in Figure 5.10, sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure 5.11, to avoid obscuring other features of a given illustration. We will use these "thinner" representations whenever the true view is too crowded to easily visualize.

FIGURE 5.11 This alternate way of drawing the trigonal planar $s p^{2}$ hybrid orbitals is sometimes used in more crowded figures.

The observed structure of the borane molecule, $\mathrm{BH}_{3}$, suggests $s p^{2}$ hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure 5.12). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in $\mathrm{BH}_{3}$ as shown in the orbital energy level diagram in Figure 5.13. We redistribute the three valence electrons of the boron atom in the three $s p^{2}$ hybrid orbitals, and each boron electron pairs with a hydrogen electron when $\mathrm{B}-\mathrm{H}$ bonds form.

FIGURE $5.12 \quad \mathrm{BH}_{3}$ is an electron-deficient molecule with a trigonal planar structure.


483. $s p^{2}$ Hybridization -
FIGURE $5.12 \quad \mathrm{BH}_{3}$ is an electron-deficient molecule with a trigonal planar structure.

FIGURE 5.13 In an isolated $B$ atom, there are one $2 s$ and three $2 p$ valence orbitals. When boron is in a molecule with three regions of electron density, three of the orbitals hybridize and create a set of three $s p^{2}$ orbitals and one unhybridized $2 p$ orbital. The three half-filled hybrid orbitals each overlap with an orbital from a hydrogen atom to form three $\sigma$ bonds in $\mathrm{BH}_{3}$.

Any central atom surrounded by three regions of electron density will exhibit $s p^{2}$ hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure 5.14), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, $\mathrm{CH}_{2} \mathrm{O}$, and ethene, $\mathrm{H}_{2} \mathrm{CCH}_{2}$.

FIGURE 5.14 The central atom(s) in each of the structures shown contain three regions of electron density and are $s p^{2}$ hybridized. As we know from the discussion of VSEPR theory, a region of electron density contains all of the electrons that point in one direction. A lone pair, an unpaired electron, a single bond, or a multiple bond would each count as one region of electron density.


484. $s p^{3}$ Hybridization -
The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four $\boldsymbol{s p}^{\mathbf{3}}$ hybrid orbitals. The hybrids result from the mixing of one $s$ orbital and all three $p$ orbitals that produces four identical $s p^{3}$ hybrid orbitals (Figure 5.15). Each of these hybrid orbitals points toward a different corner of a tetrahedron.

FIGURE 5.15 The hybridization of an $s$ orbital (blue) and three $p$ orbitals (red) produces four equivalent $s p^{3}$ hybridized orbitals (yellow) oriented at $109.5^{\circ}$ with respect to each other.

A molecule of methane, $\mathrm{CH}_{4}$, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits $s p^{3}$ hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in $\mathrm{CH}_{4}$ in Figure 5.16. The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the $\mathrm{C}-\mathrm{H}$ bonds form.

FIGURE 5.16 The four valence atomic orbitals from an isolated carbon atom all hybridize when the carbon bonds
in a molecule like $\mathrm{CH}_{4}$ with four regions of electron density. This creates four equivalent $s p^{3}$ hybridized orbitals. Overlap of each of the hybrid orbitals with a hydrogen orbital creates a $\mathrm{C}-\mathrm{H} \sigma$ bond.