1 OpenStax-CNX module: m Molecular Structure and Polarity OpenStax College This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 By the end of this section, you will be able to: Abstract Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory Explain the concepts of polar covalent bonds and molecular polarity Assess the polarity of a molecule based on its bonding and structure Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure 1). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = m) or picometers (1 pm = m, 100 pm = 1 Å). Figure 1: Bond distances (lengths) and angles are shown for the formaldehyde molecule, H 2CO. 1 VSEPR Theory Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of Version 1.7: Jun 22, :27 pm
2 OpenStax-CNX module: m the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible. VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the nal arrangement that atoms adopt in a particular molecular structure. As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF 2 molecule. The Lewis structure of BeF 2 (Figure 2) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180 (Figure 2). Figure 2: The 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 3 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; ve regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.
3 OpenStax-CNX module: m Figure 3: The basic electron-pair geometries predicted by VSEPR theory maximize the space around any region of electron density (bonds or lone pairs). 1.1 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 3 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 dierentiate 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 dierent when there are lone pairs present on the central atom. For example, the methane molecule, 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). On the other hand, the ammonia molecule, 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 inuences the shape of the molecule (Figure 5).
4 OpenStax-CNX module: m Figure 4: The molecular structure of the methane molecule, 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. Figure 5: (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 As seen in Figure 5, small distortions from the ideal angles in Figure 3 can result from dierences 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 dierent kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair (1) This order of repulsions determines the amount of space occupied by dierent 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
5 OpenStax-CNX module: m smallest is: lone pair > triple bond > double bond > single bond (2) Consider formaldehyde, H 2 CO, which is used as a preservative for biological and anatomical specimens (Figure 1). 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 bond angles, but we see that the double bond causes slightly larger angles (121 ), and the angle between the single bonds is slightly smaller (118 ). In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a at, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure 5) 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 HNH bond angles in NH 3 are slightly smaller than the angle in a regular tetrahedron (Figure 3) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion (Figure 5). Figure 6 illustrates the ideal molecular structures, which are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs. Figure 6: The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (rst column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modications of the corresponding electron-pair geometry. According to VSEPR theory, the terminal atom locations (Xs in Figure 6) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the rst 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.
6 OpenStax-CNX module: m For trigonal bipyramidal electron-pair geometries, however, there are two distinct X positions, as shown in Figure 7: 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 6, the axial position is surrounded by bond angles of 90, whereas the equatorial position has more space available because of the 120 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 ClF 3 molecule (Figure 7). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure. Figure 7: (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. (bd) The two lone pairs (red lines) in 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 apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure 6). 1.2 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 6, rst column). 4. Use the number of lone pairs to determine the molecular structure (Figure 6). 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.
7 OpenStax-CNX module: m Example 1 Predicting Electron-pair Geometry and Molecular Structure: CO 2 and BCl 3 Predict the electron-pair geometry and molecular structure for each of the following: (a) carbon dioxide, CO 2, a molecule produced by the combustion of fossil fuels (b) boron trichloride, BCl 3, an important industrial chemical Solution (a) We write the Lewis structure of CO 2 as: This shows us two regions of high electron density around the carbon atomeach 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. The electron-pair geometry and molecular structure are identical, and CO 2 molecules are linear. (b) We write the Lewis structure of BCl 3 as: Thus we see that 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 BCl bonds lie in a plane with 120 angles between them. BCl 3 also has a trigonal planar molecular structure (Figure 8). Figure 8 The electron-pair geometry and molecular structure of BCl 3 are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles (120 ), unlike the Lewis structure shown above. Check Your Learning Carbonate, 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?
8 OpenStax-CNX module: m note: The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three CO bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density. Example 2 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 NH 4+ cation. Solution We write the Lewis structure of NH 4+ as: We can see that 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 6). Therefore, the electron pair geometry of NH 4+ is tetrahedral, and the molecular structure is also tetrahedral (Figure 9). Figure 9: The ammonium ion displays a tetrahedral electron-pair geometry as well as a tetrahedral molecular structure.
9 OpenStax-CNX module: m Check Your Learning Identify a molecule with trigonal bipyramidal molecular structure. note: Any molecule with ve electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. PF 5 is a common example. The next several examples illustrate the eect of lone pairs of electrons on molecular structure. Example 3 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. Solution The Lewis structure of H 2 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 10), as indicated in Figure 6. Thus, the electron-pair geometry is tetrahedral and the molecular structure is bent with an angle slightly less than In fact, the bond angle is Figure 10: (a) H 2O has four regions of electron density around the central atom, so it has a tetrahedral electron-pair geometry. (b) Two of the electron regions are lone pairs, so the molecular structure is bent. Check Your Learning The hydronium ion, H 3 O +, forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation. note: electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal
10 OpenStax-CNX module: m Example 4 Predicting Electron-pair Geometry and Molecular Structure: SF 4 Sulfur tetrauoride, SF 4, is extremely valuable for the preparation of uorine-containing compounds used as herbicides (i.e., SF 4 is used as a uorinating agent). Predict the electron-pair geometry and molecular structure of a SF 4 molecule. Solution The Lewis structure of SF 4 indicates ve regions of electron density around the sulfur atom: one lone pair and four bonding pairs: We expect these ve regions to adopt a trigonal bipyramidal electronpair geometry. To minimize lone pair repulsions, the lone pair occupies one of the equatorial positions. The molecular structure (Figure 11) is that of a seesaw (Figure 6). Figure 11: (a) SF4 has a trigonal bipyramidal arrangement of the ve regions of electron density. (b) One of the regions is a lone pair, which results in a seesaw-shaped molecular structure. Check Your Learning Predict the electron pair geometry and molecular structure for molecules of XeF 2. note: The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear.
11 OpenStax-CNX module: m Example 5 Predicting Electron-pair Geometry and Molecular Structure: XeF 4 Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and uorine. Predict the electron-pair geometry and molecular structure of the XeF 4 molecule. Solution The Lewis structure of 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 6), which is the electron-pair geometry. To minimize repulsions, the lone pairs should be on opposite sides of the central atom (Figure 12). The ve atoms are all in the same plane and have a square planar molecular structure. Figure 12: (a) 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. 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? note: electron pair geometry: trigonal bipyramidal; molecular structure: linear
12 OpenStax-CNX module: m 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 inuences 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. Example 6 Predicting Structure in Multicenter Molecules The Lewis structure for the simplest amino acid, glycine, H 2 NCH 2 CO 2 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: Solution Consider each central atom independently. The electron-pair geometries: ˆ nitrogen-four regions of electron density; tetrahedral ˆ carbon (CH 2 )-four regions of electron density; tetrahedral ˆ carbon (CO 2 )three regions of electron density; trigonal planar ˆ oxygen (OH)four regions of electron density; tetrahedral The local structures: ˆ nitrogen-three bonds, one lone pair; trigonal pyramidal ˆ carbon (CH 2 )four bonds, no lone pairs; tetrahedral
13 OpenStax-CNX module: m ˆ carbon (CO 2 )three bonds (double bond counts as one bond), no lone pairs; trigonal planar ˆ oxygen (OH)two bonds, two lone pairs; bent (109 ) 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: note: electron-pair geometries: nitrogen-tetrahedral; carbon (CH)tetrahedral; carbon (CH 3 ) tetrahedral; carbon (CO 2 )trigonal planar; oxygen (OH)tetrahedral; local structures: nitrogen trigonal pyramidal; carbon (CH)tetrahedral; carbon (CH 3 )tetrahedral; carbon (CO 2 )trigonal planar; oxygen (OH)bent (109 ) note: The molecular shape simulator 1 lets you build various molecules and practice naming their electron-pair geometries and molecular structures. Example 7 Molecular Simulation Using molecular shape simulator 2 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: HC N (3)
14 OpenStax-CNX module: m 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? Solution The molecular structure is linear. Check Your Learning Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to nd a chemical formula that would match the structure you have drawn. note: 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. XeOF 4 is a molecule that adopts this structure. 2 Molecular Polarity and Dipole Moment As discussed previously, polar covalent bonds connect two atoms with diering electronegativities, leaving one atom with a partial positive charge (δ+) and the other atom with a partial negative charge (δ), 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 (µ) and is given by the formula shown here, where Q is the magnitude of the partial charges (determined by the electronegativity dierence) and r is the distance between the charges: µ = Qr (4) This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure 13). 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 dierence between the two atoms. Figure 13: (a) There is a small dierence in electronegativity between C and H, represented as a short vector. (b) The electronegativity dierence 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
15 OpenStax-CNX module: m 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. For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as Br 2 and N 2 have no dierence 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 dierence 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 CO 2 (Figure 14). 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 CO 2 molecule is linear with polar C=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 14), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity dierence 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 14: 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 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 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:
16 OpenStax-CNX module: m Although the CO bond is polar, C and S have the same electronegativity values as shown in here 3, so there is no CS dipole. Thus, the two bonds do not have of the same bond dipole moment, and the bond moments do not cancel. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end. Chloromethane, CH 3 Cl, is another example of a polar molecule. Although the polar CCl and CH bonds are arranged in a tetrahedral geometry, the CCl bonds have a larger bond moment than the CH bond, and the bond moments do not completely cancel each other. All of the dipoles have a downward component in the orientation shown, since carbon is more electronegative than hydrogen and less electronegative than chlorine: When we examine the highly symmetrical molecules BF 3 (trigonal planar), CH 4 (tetrahedral), PF 5 (trigonal bipyramidal), and SF 6 (octahedral), in which all the polar bonds are identical, the molecules are nonpolar. The bonds in these molecules are arranged such that their dipoles cancel. However, just because a molecule contains identical bonds does not mean that the dipoles will always cancel. Many molecules that have identical bonds and lone pairs on the central atoms have bond dipoles that do not cancel. Examples include H 2 S and NH 3. A hydrogen atom is at the positive end and a nitrogen or sulfur atom is at the negative end of the polar bonds in these molecules: must: To summarize, to be polar, a molecule 3 "Covalent Bonding", Figure 3 <
17 OpenStax-CNX module: m 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. 2.1 Properties of Polar Molecules Polar molecules tend to align when placed in an electric eld with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure 15). 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 15: (a) Molecules are always randomly distributed in the liquid state in the absence of an electric eld. (b) When an electric eld is applied, polar molecules like HF will align to the dipoles with the eld direction. note: The molecule polarity simulation 4 provides many ways to explore dipole moments of bonds and molecules. 4
18 OpenStax-CNX module: m Example 8 Polarity Simulations Open the molecule polarity simulation 5 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 eld, similar to Figure 15. 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. Solution (a) Molecular dipole moment points immediately between A and C. (b) Molecular dipole moment points along the AB bond, toward A. Check Your Learning Determine the partial charges that will give the largest possible bond dipoles. note: 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 dierence 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. 3 Key Concepts and Summary 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 dierence 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 elds, whereas nonpolar molecules do not. 4 Chemistry End of Chapter Exercises Exercise 1 (Solution on p. 23.) Explain why the HOH molecule is bent, whereas the HBeH molecule is linear. Exercise 2 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? Exercise 3 (Solution on p. 23.) Explain the dierence between electron-pair geometry and molecular structure. Exercise 4 Why is the HNH angle in NH 3 smaller than the HCH bond angle in CH 4? Why is the HNH angle in NH 4+ identical to the HCH bond angle in CH 4? 5
19 OpenStax-CNX module: m Exercise 5 (Solution on p. 23.) Explain how a molecule that contains polar bonds can be nonpolar. Exercise 6 As a general rule, MX 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. NH 3 (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they? Exercise 7 (Solution on p. 23.) Predict the electron pair geometry and the molecular structure of each of the following molecules or ions: (a) SF 6 (b) PCl 5 (c) BeH 2 (d) CH 3 + Exercise 8 Identify the electron pair geometry and the molecular structure of each of the following molecules or ions: (a) IF 6 + (b) CF 4 (c) BF 3 (d) SiF 5 (e) BeCl 2 Exercise 9 (Solution on p. 23.) What are the electron-pair geometry and the molecular structure of each of the following molecules or ions? (a) ClF 5 (b) ClO 2 (c) TeCl 4 2 (d) PCl 3 (e) SeF 4 (f) PH 2 Exercise 10 Predict the electron pair geometry and the molecular structure of each of the following ions: (a) H 3 O + (b) PCl 4 (c) SnCl 3 (d) BrCl 4 (e) ICl 3 (f) XeF 4 (g) SF 2 Exercise 11 (Solution on p. 23.) Identify the electron pair geometry and the molecular structure of each of the following molecules: (a) ClNO (N is the central atom) (b) CS 2 (c) Cl 2 CO (C is the central atom) (d) Cl 2 SO (S is the central atom) (e) SO 2 F 2 (S is the central atom) (f) XeO 2 F 2 (Xe is the central atom) (g) ClOF 2+ (Cl is the central atom)
20 OpenStax-CNX module: m Exercise 12 Predict the electron pair geometry and the molecular structure of each of the following: (a) IOF 5 (I is the central atom) (b) POCl 3 (P is the central atom) (c) Cl 2 SeO (Se is the central atom) (d) ClSO + (S is the central atom) (e) F 2 SO (S is the central atom) (f) NO 2 (g) SiO 4 4 Exercise 13 (Solution on p. 23.) Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments? (a) ClF 5 (b) ClO 2 (c) TeCl 4 2 (d) PCl 3 (e) SeF 4 (f) PH 2 (g) XeF 2 Exercise 14 Which of the molecules and ions in Exercise contain polar bonds? Which of these molecules and ions have dipole moments? (a) H 3 O + (b) PCl 4 (c) SnCl 3 (d) BrCl 4 (e) ICl 3 (f) XeF 4 (g) SF 2 Exercise 15 (Solution on p. 23.) Which of the following molecules have dipole moments? (a) CS 2 (b) SeS 2 (c) CCl 2 F 2 (d) PCl 3 (P is the central atom) (e) ClNO (N is the central atom) Exercise 16 Identify the molecules with a dipole moment: (a) SF 4 (b) CF 4 (c) Cl 2 CCBr 2 (d) CH 3 Cl (e) H 2 CO Exercise 17 (Solution on p. 23.) The molecule XF 3 has a dipole moment. Is X boron or phosphorus? Exercise 18 The molecule XCl 2 has a dipole moment. Is X beryllium or sulfur? Exercise 19 (Solution on p. 23.) Is the Cl 2 BBCl 2 molecule polar or nonpolar?
21 OpenStax-CNX module: m Exercise 20 There are three possible structures for PCl 2 F 3 with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them. Exercise 21 (Solution on p. 23.) Describe the molecular structure around the indicated atom or atoms: (a) the sulfur atom in sulfuric acid, H 2 SO 4 [(HO) 2 SO 2 ] (b) the chlorine atom in chloric acid, HClO 3 [HOClO 2 ] (c) the oxygen atom in hydrogen peroxide, HOOH (d) the nitrogen atom in nitric acid, HNO 3 [HONO 2 ] (e) the oxygen atom in the OH group in nitric acid, HNO 3 [HONO 2 ] (f) the central oxygen atom in the ozone molecule, O 3 (g) each of the carbon atoms in propyne, CH 3 CCH (h) the carbon atom in Freon, CCl 2 F 2 (i) each of the carbon atoms in allene, H 2 CCCH 2 Exercise 22 Draw the Lewis structures and predict the shape of each compound or ion: (a) CO 2 (b) NO 2 (c) SO 3 (d) SO 3 2 Exercise 23 (Solution on p. 23.) A molecule with the formula AB 2, in which A and B represent dierent atoms, could have one of three dierent shapes. Sketch and name the three dierent shapes that this molecule might have. Give an example of a molecule or ion for each shape. Exercise 24 A molecule with the formula AB 3, in which A and B represent dierent atoms, could have one of three dierent shapes. Sketch and name the three dierent shapes that this molecule might have. Give an example of a molecule or ion that has each shape. Exercise 25 (Solution on p. 24.) Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate: (a) CS 3 2 (b) CS 2 (c) CS (d) predict the molecular shapes for CS 3 2 and CS 2 and explain how you arrived at your predictions Exercise 26 What is the molecular structure of the stable form of FNO 2? (N is the central atom.) Exercise 27 (Solution on p. 24.) A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure? Exercise 28 Use the simulation 6 to perform the following exercises for a two-atom 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 eld and describe what happens. 6
22 OpenStax-CNX module: m (c) With a small partial negative charge on A, turn on the electric eld and describe what happens. (d) Reset all, and then with a large partial negative charge on A, turn on the electric eld and describe what happens. Exercise 29 (Solution on p. 24.) Use the simulation 7 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 O 3. Explain your observations. (b) Look at the bond dipoles for NH 3. Use these dipoles to predict whether N or H is more electronegative. (c) Predict whether there should be a molecular dipole for NH 3 and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis. Exercise 30 Use the Molecule Shape simulator 8 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. Exercise 31 (Solution on p. 24.) Use the Molecule Shape simulator 9 to explore real molecules. On the Real Molecules tab, select H 2 O. Switch between the real and model modes. Explain the dierence observed. Exercise 32 Use the Molecule Shape simulator 10 to explore real molecules. On the Real Molecules tab, select model mode and S 2 O. What is the model bond angle? Explain whether the real bond angle should be larger or smaller than the ideal model angle
23 OpenStax-CNX module: m Solutions to Exercises in this Module Solution to Exercise (p. 18) The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear. Solution to Exercise (p. 18) Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry. Solution to Exercise (p. 19) As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar. Solution to Exercise (p. 19) (a) Both the electron geometry and the molecular structure are octahedral. (b) Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. (d) Both the electron geometry and the molecular structure are trigonal planar. Solution to Exercise (p. 19) (a) electron-pair geometry: octahedral, molecular structure: square pyramidal; (b) electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electronpair geometry: trigonal bypyramidal, molecular structure: seesaw; (f) electron-pair geometry: tetrahedral, molecular structure: bent (109 ) Solution to Exercise (p. 19) (a) electron-pair geometry: trigonal planar, molecular structure: bent (120 ); (b) electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: tetrahedral, molecular structure: tetrahedral; (f) electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal Solution to Exercise (p. 20) All of these molecules and ions contain polar bonds. Only ClF 5, ClO 2, PCl 3, SeF 4, and PH 2 have dipole moments. Solution to Exercise (p. 20) SeS 2, CCl 2 F 2, PCl 3, and ClNO all have dipole moments. Solution to Exercise (p. 20) P Solution to Exercise (p. 20) nonpolar Solution to Exercise (p. 21) (a) tetrahedral; (b) trigonal pyramidal; (c) bent (109 ); (d) trigonal planar; (e) bent (109 ); (f) bent (109 ); (g) CH 3 CCH tetrahedral, CH 3 CCH linear; (h) tetrahedral; (i) H 2 CCCH 2 linear; H 2 CCCH 2 trigonal planar Solution to Exercise (p. 21)
24 OpenStax-CNX module: m Solution to Exercise (p. 21) (a) (b) ; (c) ; ; (d) CS 3 2 includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS 2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear Solution to Exercise (p. 21) The Lewis structure is made from three units, but the atoms must be rearranged: Solution to Exercise (p. 22) The molecular dipole points away from the hydrogen atoms. Solution to Exercise (p. 22) The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as In the real mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of
25 OpenStax-CNX module: m Glossary Denition 1: axial position location in a trigonal bipyramidal geometry in which there is another atom at a 180 angle and the equatorial positions are at a 90 angle Denition 2: bond angle angle between any two covalent bonds that share a common atom Denition 3: bond distance (also, bond length) distance between the nuclei of two bonded atoms Denition 4: bond dipole moment separation of charge in a bond that depends on the dierence in electronegativity and the bond distance represented by partial charges or a vector Denition 5: 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 Denition 6: electron-pair geometry arrangement around a central atom of all regions of electron density (bonds, lone pairs, or unpaired electrons) Denition 7: equatorial position one of the three positions in a trigonal bipyramidal geometry with 120 angles between them; the axial positions are located at a 90 angle Denition 8: linear shape in which two outside groups are placed on opposite sides of a central atom Denition 9: molecular structure structure that includes only the placement of the atoms in the molecule Denition 10: octahedral shape in which six outside groups are placed around a central atom such that a three-dimensional 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 Denition 11: polar molecule (also, dipole) molecule with an overall dipole moment Denition 12: tetrahedral shape in which four outside groups are placed around a central atom such that a three-dimensional shape is generated with four corners and angles between each pair and the central atom Denition 13: trigonal bipyramidal shape in which ve outside groups are placed around a central atom such that three form a at triangle with 120 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 Denition 14: trigonal planar shape in which three outside groups are placed in a at triangle around a central atom with 120 angles between each pair and the central atom Denition 15: 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 Denition 16: vector quantity having magnitude and direction
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ATER 10 TE AE MLEULE 10.1 To be the central atom in a compound, the atom must be able to simultaneously bond to at least two other atoms. e,, and cannot serve as central atoms in a Lewis structure. elium
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ATER 10 TE AE MLEULE EMIAL ETI BED READIG RBLEM B10.1 lan: Examine the Lewis structure, noting the number of regions of electron density around the carbon and nitrogen atoms in the two resonance structures.
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Lewis Structures & the VSEPR Model A Directed Learning Activity for Hartnell College Chemistry 1 Funded by the Title V STEM Grant #P031S090007 through Hartnell College For information contact email@example.com
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