Molecular shape is determined by the number of bonds that form around individual atoms.

Chapter 9 CH 180 Major Concepts: Molecular shape is determined by the number of bonds that form around individual atoms. Sublevels (s, p, d, & f) of separate atoms may overlap and result in hybrid orbitals Some aspects of bonding are better explained by an alternative model called molecular orbital theory 9.1 Molecular Shapes Lewis structures do not indicate the shape of molecules; they simply show the number and types of bonds between atoms. The bond angles of a molecule, together with the bond lengths, accurately define the shape and size of a molecule. In order to predict molecular shape, we assume that the valence electrons repel each other. Therefore, the molecule adopts the three dimensional geometry that minimizes this repulsion. We call this model the Valence Shell Electron Pair Repulsion (VSEPR) model. 1

9.2 The VSEPR Model A bonding pair of electrons defines a region in which electrons will most likely be found. Such a region is called an electron domain. A nonbonding pair (or lone pair) of electrons defines an electron domain that is located principally on one atom. For example, NH 3 has a total of four electron domains around the central nitrogen atom (three bonding pairs and one nonbonding pair): Each multiple bond also constitutes a single electron domain. Thus the resonance structure for O 3 has three electron domains around the central oxygen atom (a single bond, a double bond, and a nonbonding pair of electrons): VSEPR predicts that the best arrangement of electron domains is the one that minimizes the repulsions among them. Electron domain geometry is the arrangement of electron domains around the central atom. In contrast, the molecular geometry is the arrangement of only the atoms in a molecule or ion and does not include nonbonding electron pairs. 2

Electron domain geometry provides us with information regarding the shape and bond angles of the molecule. For example: O 3 and H 2 O are both bent molecules, but have different bond angles because of their electron domain geometry. There are five different electron domain geometries: linear trigonal planar tetrahedral trigonal bipyramidal octahedral (two electron domains) (three domains) (four domains) (five domains) (six domains) The molecular geometry is the arrangement of the atoms in space. To determine the shape of a molecule we must distinguish between lone pairs and bonding pairs. We use the electron domain geometry to help us predict the molecular geometry. (1) Draw the Lewis structure. (2) Count the total number of electron domains around the central atom. (3) Arrange the electron domains in one of the above geometries to minimize electron electron repulsion. Next, determine the three dimensional structure of the molecule. We ignore lone pairs in the molecular geometry. Describe the molecular geometry in terms of the angular arrangement of the bonded atoms. Multiple bonds are counted as one electron domain. 3

Total Electron Domains Electron Domain Geometry Bonding Domains Nonbonding Domains Molecular Geometry (Shapes) Predicted Bond angles Example 2 2 0 180 Linear Linear 3 3 0 120 Trigonal planar Trigonal planar 2 1 120 Bent 4

4 4 0 109.5 Tetrahedral 3 1 107 Tetrahedral Trigonal pyramidial 2 2 104.5 Bent 5

Practice Questions: (1) Use the VSEPR model to predict the molecular geometry of: a. O 3 b. SnCl 3 (2) Predict the electron domain geometry and the molecular geometry of: a. SeCl 2 b. CO 3 2 6

The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles We can refine the VSEPR model to predict and explain slight distortions of molecules from the ideal geometries. Consider three molecules that all have tetrahedron electron domain geometries, but their bond angles differ slightly: CH 4 NH 3 H 2 O Notice that the bond angles decrease as the number of nonbonding electron pairs increases. A bonding pair of electrons is attracted by both nuclei of the bonded atoms. By contrast, a nonbonding pair experiences less nuclear attraction, and its electron domain is spread out more than a bonding pair. As a result, electron domains for nonbonding electron pairs exert greater repulsive forces on adjacent domains and thus tend to compress the bond angles. Also, electron domains for multiple bonds exert a greater repulsive force on adjacent electron domains than do electron domains for single bonds. For example: You d expect the molecule Cl 2 CO to be triginol planar and have 120 bond angles, but.. the double bond forces the Cl atoms farther away than expected, thus changing the bond angles. 7

Molecules with Expanded Valence Shells As you may recall, a central atom from period 3 or higher may break the octet rule to take on more than 8 electrons. In these cases, there will be more than 4 electron domains. Molecules with 5 electron domains involve trigonal bipyramidial geometries. The most stable geometry has two domains along the axial positions (up / down), and three domains in equatorial positions (side / side). o The axial domain is at a 90 angle to the equatorial domains o The equatorial domains are 120 from each other. Nonbonding pairs will always occupy equatorial domains because this is the area a least electron repulsion Molecules with 6 electron domains involve octahedral geometries. The most stable geometry has two domains along the axial positions (up / down), and four domains in equatorial positions (side / side). o All domains are 90 from each other Nonbonding pairs can occupy any domain, but more than one nonbonding pair will take places opposite of each other 8

Total e Domains Electron Domain Geometry Bonding Domains Nonbonding Domains Molecular Geometry Predicted Bond angles Example 5 5 0 90 120 PCl 5 Trigonal bipyramidial Trigonal bipyramidial 4 1 SF 4 Seesaw 3 2 ClF 3 T shaped 2 3 XeF 2 Linear 9

6 6 0 90 SF 6 Octahedral 5 1 Octahedral BrF 5 Square pyramidial 4 2 XeF 4 Square planar 10

Practice Questions: (1) Use the VSEPR model to predict the molecular geometry of SF 4 and IF 5 (2) Predict the electron domain geometry and molecular geometry of ClF 3 and ICl 4 11

Shapes of Larger Molecules Although the molecules and ions whose structures we have thus far examined only contain a single central atom, the VSEPR model can be extended to more complex molecules. For example: 12

Practice Questions: (1) Eyedrops usually contain a water soluble polymer called poly(vinyl alcohol) which is based on the unstable organic molecule called vinyl alcohol: H H.. H O C = C H.. Predict the approximate values for the H O C and O C C bond angles in vinyl alcohol. (2) Predict the H C H and C C C bond angles in the following molecule, called propyne. H H C C C H H 13

(3) Draw the Lewis structure for each of the following molecules or ions, and predict their electron domain and molecular geometries: (a) PF 3, (b) CH 3 +, (c) BrF 3, (d) ClO 4, (e) XeF 2, (f) BrO 2 14

Lesson #1 Homework (1) (a) Methane (CH 4 ) and the perchlorate ion (ClO 4 ) are both described as tetrahedral. What does this indicate about their bond angles? (b) The NH 3 molecule is trigonal pyramidial. How many parameters need to be specified to define its geometry completely? (2) Describe the characteristic electron domain geometry of each of the following numbers of electron domains about the central atom: (a) 3, (b) 4, (c) 5, (d) 6. (3) Indicate the number of electron domains about a central atom, given the following angles between them: (a) 120, (b) 180, (c) 109.5, (d) 90. 15

(4) Draw the Lewis structure for each of the following molecules or ions, and predict their electron domain and molecular geometries: (a) PF 3, (b) CH 3 +, (c) BrF 3, (d) ClO 4, (e) XeF 2, (f) BrO 2. (5) Give the electron domain and molecular geometries for the following molecules and ions: (a) HCN, (b) SO 3 2, (d) SF 4, (d) PF 6, (e) NH 3 Cl +, (f) N 3 16

9.3 Molecular Shape and Molecular Polarity A measure of the separation and magnitude of the positive and negative charges in polar molecules is called a dipole moment. The overall dipole moment of a molecule is the sum of its bond dipoles. It is possible for a molecule with polar bonds to be either polar or nonpolar. In carbon dioxide: The bond dipoles are equal in magnitude, but exactly opposite each other. The overall dipole moment is zero, therefore making the molecule nonpolar. In water: The bonds are equal in magnitude, but do not exactly oppose each other. The molecule has a nonzero dipole moment overall, making the molecule polar. Polar molecules interact with electric fields. Nonpolar molecules do not. 17

Practice Questions: (1) Predict whether the following molecules are polar or nonpolar: a. SO 2 b. SF 6 (2) Determine whether the following molecules are polar or nonpolar: a. NF 3 b. BCl 3 18

9.4 Covalent Bonding and Orbital Overlap Lewis structures and VSEPR theory give us the shape and location of electrons in a molecule. They do not explain why a chemical bond forms. The combination of Lewis s notion of electron pair bonds and the idea of atomic orbitals leads to a model of chemical bonding called valence bond theory. A covalent bond forms when the orbitals on two atoms overlap. The shared region of space between the orbitals is called the orbital overlap. There are two electrons (usually one from each atom) of opposite spin in the orbital overlap. There is always an optimum distance between the two bonded nuclei in any covalent bond. 19

The diagram below shows how the potential energy of the system changes as two H atoms come together to form an H 2 molecule: During the overlap: As the distance between the atoms decrease (as they get closer), the overlap between the 1s orbital increases Because of the resultant increase in electron density between the nuclei, the potential energy of the system decreases (the strength of the bond increases). The observed bond length is at the distance where the attractive forces between the unlike charges (electrons and protons) are balanced by the repulsive forces between like charges (electron / electron and proton / proton of the nuclei). In this case 0.74 Ǻ 20

9.5 Hybrid Orbitals We can apply the idea of orbital overlap and valence bond theory to polyatomic molecules. The shape of hybrid orbitals are different from the shapes of the original atomic orbitals. sp Hybrid Orbitals To illustrate the process of hybridization, consider the BeF 2 molecule. The VSEPR model correctly predicts that it is linear with two identical Be F bonds. The electron configuration of F (1s 2 2s 2 2p 5 ) indicates there is an unpaired electron in a 2p orbital. This 2p electron pairs with the an unpaired electron from Be to form a polar covalent bond. Which orbitals on the Be atom overlap with those on the F atoms to form the Be F bonds? The orbital diagram for a ground state Be atom is: Because it has no unpaired electrons, the Be atom in its ground state is incapable of forming bonds with the fluorine atoms. It could form two bonds, however, by promoting one of the 2s electrons to a 2p orbital: The Be atom now has two unpaired electrons and therefore can form two polar covalent bonds with the F atoms. The two bonds will not be identical, however, because a Be 2s orbital would be used for one bond and a 2p will be used for the other. To identify the structure of BeF 2, we need to hybridize the 2s orbital and one of the 2p orbitals to form two new orbitals. 21

Like the p orbitals, the hybridized orbitals (sp) each have two lobes. Unlike the p orbitals, one lobe is much larger than the other. The two new orbitals are identical in shape, but their lobes point in the opposite directions. According to the valence bond model, a linear arrangement of electron domains implies sp hybridization For the Be atom of BeF 2, we write the orbital diagram for the formation of two sp hybrid orbitals as follows 22

sp 2 and sp 3 Hybrid Orbitals Whenever we mix a certain number of atomic orbitals, we get the same number of hybrid orbitals. Each of these hybrid orbitals is equivalent to the others but points in a different direction. In BF 3, a 2s electron on the B atom can be promoted to a vacant 2p orbital. Mixing the 2s and the two 2p orbitals yields three equivalent sp 2 hybrid orbitals: The three sp 2 hybrid orbitals lie in the same plane, 120 apart from one another. They are used to make three equivalent bonds with three fluorine atoms, leading to the trigonal planar geometry. 23

An s orbital can also mix with all three p orbitals in the same subshell. For example, the carbon atom in CH 4 forms four equivalent bonds with the four hydrogen atoms. The result creates four equivalent sp 3 hybrid orbitals. 24

Hybridization Involving d Orbitals Since there are only three p orbitals, trigonal bipyramidal and octahedral electron pair geometries must involve d orbitals. Trigonal bipyramidal electron pair geometries require sp 3 d hybridization. Octahedral electron pair geometries require sp 3 d 2 hybridization. Note that the electron pair VSEPR geometry corresponds well with the hybridization. Use of d orbitals in making hybrid orbitals corresponds well with the idea of an expanded octet. 25

Summary We need to know the electron domain geometry before we can assign hybridization. To assign hybridization: Draw a Lewis structure. Assign the electron domain geometry using VSEPR theory. Specify the hybridization required to accommodate the electron pairs based on their geometric arrangement. Name the geometry by the positions of the atoms. Steric Number Hybridization Basic shape (e domains) 1 s 2 sp Linear 3 sp 2 Δ planar 4 sp 3 Tetrahedral 5 sp 3 d Δ bipyramidial 6 sp 3 d 2 Octahedral 26

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Practice Questions: (1) Indicate the hybridization of orbitals employed by the central atom in: a. NH 2 b. SF 4 (2) Predict the electron domain geometry and the hybridization of the central atom in: a. SO 3 2 b. SF 6 28

Lesson #2 Homework (1) Predict whether each of the following molecules is polar or nonpolar: (a) CCl 4, (b) NH 3, (c) SF 4, (d) XeF 4, (e) CH 3 Br, (f) GaH 3 (2) Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) CS 2, (c) SO 3, (d) PCl 3, (e) SF 6, (f) IF 5. (3) (a) Draw a Lewis structure for silane (SiH 4 ), and predict its molecular geometry. (b) Is it necessary to promote an electron before forming hybrid orbitals for the Si atom? (c) What type of hybridization exists in SiH 4? 29

(4) Indicate the hybridization and bond angles associated with each of the following electron domain geometries: (a) linear, (b) tetrahedral, (c) trigonal planar; (d) octahedral, (e) trigonal bipyramidial (5) (a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized p orbitals remain in the valence shell? How many pi bonds can the atom form? (b) how many sigma and pi bonds are generally part of a triple bond? (c) How do multiple bonds introduce rigidity into molecules? (6) Popylene, C 3 H 6, is a gas that is used to form the important polymer called polypropylene. (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make sigma bonds in the molecule? (c) How many valence electrons are used to make pi bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule? 30

9.6 Multiple Bonds In the covalent bonds we have seen so far the electron density has been concentrated symmetrically about the internuclear axis. All single bonds are sigma bonds (σ). These include s or p orbitals or sp, sp 2, or sp 3 hybrid orbitals. To describe multiple bonding, we must consider a second kind of bond, one that results form the overlap between two p orbitals oriented perpendicularly to the internuclear axis. This sideways overlap produces a pi (π) bond. A double bond consists of one σ bond and one π bond. A triple bond has one σ bond and two π bonds. Often, the p orbitals involved in π bonding come from unhybridized orbitals. For example: ethylene, C 2 H 4, has: One σ and one π bond. Both C atoms are sp 2 hybridized (because of the trigonal planar geometry). This orbital is not hybridized because of the trigonal planar geometry 31

The unhybridized 2p orbital is directly perpendicular to the plane that contains the three sp 2 hybrid orbitals. Each sp 2 hybrid orbital on a carbon atom contains one electron. Thus 10 of the 12 valence electrons in the C 2 H 4 molecule are used to form five σ bonds. The remaining two valence electrons reside in the unhybridized 2p orbitals (one for each carbon). These two 2p orbitals can overlap with each other in a sideways fashion, producing a π bond (one lobe above the σ bond, and one lobe below.) Because π bonds require that portions of a molecule be planar, they can introduce rigidity into molecules. 32

Triple bonds can also be explained by using hybrid orbitals. For example: acetylene, C 2 H 2 : The electron domain geometry of each C is linear. Therefore, the C atoms are sp hybridized. The sp hybrid orbitals form the C C and C H σ bonds. There are two unhybridized p orbitals on each C atom. Both unhybridized p orbitals form the two π bonds; One π bond is above and below the plane of the nuclei; One π bond is in front and behind the plane of the nuclei. When triple bonds form (e.g., N 2 ), one π bond is always above and below and the other is in front and behind the plane of the nuclei. 33

Practice Questions: (1) Formaldehyde has the Lewis structure Describe how the bonds in formaldehyde are formed in terms of overlaps of appropriate hybridized and unhybridized orbitals. (2) Consider the acetonitrile molecule: H H C C N : H (a) Predict the bond angles around each carbon atom (b) Describe the hybridization at each of the carbon atoms (c) Determine the total number of σ and π bonds in the molecule. 34

Delocalized π Bonding So far all the bonds we have encountered are localized between two nuclei. In the case of benzene: There are six C C σ bonds and six C H π bonds. Each C atom is sp 2 hybridized (because of trigonal planar geometry). There is one unhybridized p orbital on each carbon atom, resulting in six unhybridized carbon p orbitals in a ring. In benzene there are two options for the three π bonds because it has two resonance structures: we can place them localized between carbon atoms or delocalized over the entire ring (i.e., the π electrons are shared by all six carbon atoms). Delocalization is best used for resonance structures. Experimentally, all C C bonds are the same length in benzene. Therefore, all C C bonds are of the same type (recall single bonds are longer than double bonds). OR 35

Another example is the localization and delocalization of π bonds in NO 3 36

Practice Questions: (1) Describe the bonding in the nitrate ion, NO 3. Does this ion have delocalized π bonds. (2) Which of the following molecules or ions will exhibit delocalized bonding; SO 3, SO 3 2, H 2 CO, O 3, NH 4 +? General Conclusions. Every pair of bonded atoms shares one or more pairs of electrons Two electrons shared between atoms on the same axis as the nuclei are σ bonds. σ bonds are always localized in the region between two bonded atoms. If two atoms share more than one pair of electrons, the additional pairs form πbonds. When resonance structures are possible, delocalization is also possible. 37

9.7 Molecular Orbitals Some aspects of bonding are not explained by Lewis structures, VSEPR theory, and hybridization. For example: Why does O 2 interact with a magnetic field? Why are some molecules colored? For these molecules, we use molecular orbital (MO) theory. Just as electrons in atoms are found in atomic orbitals, electrons in molecules are found in molecular orbitals. Molecular orbitals: Some characteristics are similar to those of atomic orbitals. Each contains a maximum of two electrons with opposite spins. Each has a definite energy. Electron density distribution can be visualized with contour diagrams. However, unlike atomic orbitals, molecular orbitals are associated with an entire molecule. 38

The Hydrogen Molecule For H 2, we will use the two 1s atomic orbitals (one for each H atom) to build molecular orbitals for the H 2 molecule. When two Atomic Orbitals overlap, two Molecular Orbitals form. Thus, the overlap of the 1s orbitals of two hydrogen atoms to form H 2 produces two MOs. The bonding molecular orbital (σ 1s ) has lower energy and concentrates electron density between the two hydrogen nuclei. This MO results from summing the two atomic orbitals so that orbital wave functions enhance each other, forming a stable covalent bond. The antibonding molecular orbital (σ* 1s ) has higher energy and very little electron density. Instead of enhancing each other in the region between the nuclei, the atomic orbitals cancel each other out in this region, and the greatest electron density is on the opposite sides of the nuclei. Sigma (σ) MOs have electron density in both molecular orbitals centered about the internuclear axis. 39

The energy level diagram, or MO diagram, shows the energies of the orbitals in a molecule. These diagrams show the interacting atomic orbitals in the left and the right columns and the MOs in the middle column. Note that since the antibonding orbitals have higher energy, they are placed above bonding orbitals. Like atomic orbitals, each MO can accommodate two electrons with their spins paired. In contrast, the hypothetical He 2 molecule would require four electrons to fill its molecular orbitals. Because only two electrons can be put in the σ 1s the other two must be put into σ* 1s The energy decrease from the two electrons in the bonding MO is offset by the energy increase from the two electrons in the antibonding MO*, thus He 2 is an unstable molecule. 40

Bond Order In molecular orbital theory the stability of a covalent bond is related to its bond order, defined as have the difference between the number of bonding electrons and the number of antibonding electrons: Bond order = ½ (no. of bonding electrons no. of antibonding electrons) Bond order = 1 for a single bond. Bond order = 2 for a double bond. Bond order = 3 for a triple bond. Fractional bond orders are possible. For example, consider the molecule H 2. H 2 has two bonding electrons. Bond order for H 2 is: ½ (bonding electrons antibonding electrons) = ½ (2 0) = 1 Therefore, H 2 has a single bond.. For example, consider the species He 2. He 2 has two bonding electrons and two antibonding electrons. Bond order for He 2 is: ½ (bonding electrons antibonding electrons) = ½ (2 2) = 0. Therefore, He 2 is not a stable molecule MO theory correctly predicts that hydrogen forms a diatomic molecule but that helium does not! As bond order increases, bond length decreases. bond energy increases. 41

Practice Questions: (1) What is the bond order of He 2 + ion? Would you expect this ion to be stable relative to the separated He atom and He + ion? (2) Determine the bond order of the H 2 ion. 42

9.8 Second Row Diatomic Molecules We look at homonuclear diatomic molecules (those composed of two identical atoms) [e.g., Li 2, Be 2, B 2 etc.]. Molecular Orbitals for Li 2 and Be 2 Each 1s orbital combines with another 1s orbital to give one σ 1s and one σ * 1s orbital, both of which are occupied (since Li and Be have 1s 2 electron configurations). Each 2s orbital combines with another 2s orbital give one σ 2s and one σ s orbital. The energies of the 1s and 2s orbitals are sufficiently different so that there is no cross mixing of orbitals (i.e., we do not get 1s + 2s). Consider the bonding in Li 2. There are a total of six electrons in Li 2. two electrons in σ 1s. two electrons in σ 1s. two electrons in σ 2s. zero electrons in σ * 2s. Therefore the bond order is ½ (4 2) = 1. Since the 1s AOs are completely filled, the σ 1s and σ * 1s are filled. We generally ignore core electrons in MO diagrams. Core electrons usually do not contribute significantly to bonding in molecule formation 43

Consider bonding in Be 2. There are a total of eight electrons in Be 2. two electrons in σ 1s. two electrons in σ * 1s. two electrons in σ 2s. two electrons in σ * 2s. Therefore the bond order is ½ (4 4) = 0. Be 2 does not exist. 44

Molecular Orbitals for 2p Atomic Orbitals There are two ways in which two p orbitals can overlap: End on so that the resulting MO has electron density on the axis between nuclei (i.e., σ type orbital). Sideways, so that the resulting MO has electron density above and below the axis between nuclei. These are called pi (π) molecular orbitals. 45

Electron configurations and Molecular Properties Two types of magnetic behavior are: paramagnetism (unpaired electrons in molecule) strong attraction between magnetic field and molecule diamagnetism (no unpaired electrons in molecule) weak repulsion between magnetic field and molecule Magnetic behavior is detected by determining the mass of a sample in the presence and absence of a magnetic field. A large increase in mass indicates paramagnetism. A small decrease in mass indicates diamagnetism. Experimentally, O 2 is paramagnetic. The Lewis structure for O 2 shows no unpaired electrons. The MO diagram for O 2 shows two unpaired electrons in the π * 2p orbital. Experimentally, O 2 has a short bond length (1.21 Å) and high bond dissociation energy (495 kj/mol). This suggests a double bond. The MO diagram for O 2 predicts both paramagnetism and the double bond (bond order = 2). 46

Practice Questions: (1) Predict the following properties of O 2 + : a. Number of unpaired electrons b. Bond order c. Bond enthalpy and bond length (2) Predict the magnetic properties and bond orders of: a. The peroxide ion, O 2 2 b. The acetylide ion C 2 2 47

Lesson #3 Homework (1) Consider the H 2 + ion. (a) Sketch the molecular orbitals of the ion, and draw its energy level diagram. (b) How many electrons are there in the H 2 + ion? (c) Write the electron configuration of the ion in terms of it s MOs. (d) What is the bond order in H 2 +? (e) Suppose that the ion is excited by light so that an electron moves from a lower energy to a higher energy MO. Would you expect the excited state H 2 + ion to be stable or fall apart? Explain. (2) (a) What is meant by the term diamagnetism? (b) How does a diatomic substance respond to a magnetic field? (c) Which of the following ions would you expect to be diamagnetic: N 2 2, O 2 2, Be 2 2+, C 2? (3) Determine the electron configurations for CN +, CN, and CN. Calculate the bond order for each, and indicate which ones are paramagnetic. 48