Chapter 9 Molecular Geometries and Bonding Theories

Chapter 9 Molecular Geometries and Bonding Theories

Lecture Presentation Chapter 9 Molecular Geometry and Bonding Theories (With Chapter 8) MC 7 out of 60 FRQ every year 2015 Pearson Education, Inc. James F. Kirby Quinnipiac University Hamden, CT

9.1 Molecular Shapes Lewis Structures show bonding and lone pairs, but do not denote shape. However, we use Lewis Structures to help us determine shapes. Here we see some common shapes for molecules with two or three atoms connected to a central atom. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. What Determines the Shape of a Molecule?

The shape of a molecule is determined by its bond angles. Molecules and ions can be represented by the general notation ABn. - Central atom (A) is bonded to n outside atoms (B). Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

What Determines the Shape of a Molecule? Simply put, electron pairs, whether they be bonding or nonbonding, repel each other. By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of the molecule. This is the Valence-Shell Electron-Pair

Repulsion (VSEPR) model. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories In addition to tetrahedral, another common shape for AB4 molecules is square planar. All five atoms lie in the same plane, with the B atoms at the corners of a square and the A atom at the center of the square. Which shape in Figure 9.3 could lead to a square-planar shape upon removal of one or more atoms? a. Trigonal planar

b. Tetrahedral c. Trigonal bipyramidal d. Octahedral 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories 9.2 Valence-Shell Electron-Pair Repulsion (VSEPR) Model The best arrangement of a given number of electron domains is the one that minimizes the

repulsions among them. (The balloon analogy in the figure to the left demonstrates the maximum distances, which minimize Molecular Geometries repulsions.) and Bonding Theories 2015 Pearson Education, Inc. The VSPER Model 2015 Pearson Education, Inc.

We can refer to the directions to which electrons point as electron domains. This is true whether there is one or more electron pairs pointing in that direction. - Each nonbonding pair, single bond, and multiple bond produces an electron domain around the central atom. All one must do is count the # Molecular Geometries

of electron domains in theand Bonding Theories Lewis diagram Suppose a particular AB3 molecule has the resonance structure Does this structure follow the octet rule? a. No, there are 10 electrons around A. b. Yes, there are 8 electrons around A. c. Yes, there are three bonds and one electron pair around A. d. Yes, there are four bonds around A. 2015 Pearson Education, Inc.

Molecular Geometries and Bonding Theories Suppose a particular AB3 molecule has the resonance structure How many electron domains are there around the A atom? a. One electron domain b. Two electron domains c. Three electron domains d. Four electron domains 2015 Pearson Education, Inc.

Molecular Geometries and Bonding Theories Electron-Domain Geometries The Table shows the electron-domain geometries for two through six electron domains around a central atom. To determine the electron-domain geometry, count the total number of lone

pairs, single, double, and triple bonds Molecular on Geometries the central atom. and Bonding Theories 2015 Pearson Education, Inc. Molecular Geometries Once you have determined the electron-domain geometry, use the arrangement of the bonded atoms to determine the molecular geometry. Tables 9.2 and 9.3 show the potential molecular geometries.

We will look at each electron domain Molecular to see what molecular geometries are possible. Geometries and Bonding Theories 2015 Pearson Education, Inc. Molecular Geometries The electron-domain geometry is often not the shape of the molecule, however. The molecular geometry is that defined by the position of only the atoms in the molecules,Molecular not Geometries

the nonbonding pairs. and Bonding Theories 2015 Pearson Education, Inc. From the standpoint of the VSEPR model, what do nonbonding electron pairs, single bonds, and multiple bonds have in common? a. There are no common features. b. Each occurs about the central atom only. c. Each represents a single electron domain. d. All exist when a particular Lewis structure is drawn. Molecular Geometries

and Bonding Theories 2015 Pearson Education, Inc. With each electron domain, then, there might be Molecular more than one molecular geometry. Geometries and Bonding Theories 2015 Pearson Education, Inc. Linear Electron Domain In the linear domain, there is only one molecular geometry: linear.

NOTE: If there are only two atoms in the molecule, the molecule will be linear no matter what the electron domain is. (H-H, H-Cl) 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories Trigonal Planar Electron Domain There are two molecular geometries: trigonal planar, if all electron domains are bonding, and

bent, if one of the domains is a nonbonding pair. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories Tetrahedral Electron Domain There are three molecular geometries: tetrahedral, if all are bonding pairs, Molecular Geometries trigonal pyramidal, if one is a nonbonding pair, andand

Bonding Theories bent, 2015 Pearson Education,if Inc.there are two nonbonding pairs. Sample Exercise 9.1 Using the VSEPR Model Use the VSEPR model to predict the molecular geometry of (a) O3, (b) SnCl3. Solution Analyze We are given the molecular formulas of a molecule and a polyatomic ion, both conforming to the general formula ABn and both having a central atom from the p block of the periodic table. (Notice that for O3, the A and B atoms are all oxygen atoms.) Plan To predict the molecular geometries, we draw their Lewis structures and count electron domains around the central atom to get the electron-domain geometry. We then obtain the molecular geometry

from the arrangement of the domains that are due to bonds. Solve (a) We can draw two resonance structures for O 3: Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.1 Using the VSEPR Model Continued Because of resonance, the bonds between the central O atom and the outer O atoms are of equal length. In both resonance structures the central O atom is bonded to the two outer O atoms and has one nonbonding pair. Thus, there are three electron domains about the central O atoms. (Remember that a double bond counts as a single electron domain.) The arrangement of three electron domains is trigonal

planar (Table 9.1). Two of the domains are from bonds, and one is due to a nonbonding pair. So, the molecular geometry is bent with an ideal bond angle of 120 (Table 9.2). Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.1 Using the VSEPR Model Continued Comment As this example illustrates, when a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular geometry. (b) The Lewis structure for SnCl3 is Molecular

Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.1 Using the VSEPR Model Continued The central Sn atom is bonded to the three Cl atoms and has one nonbonding pair; thus, we have four electron domains, meaning a tetrahedral electron-domain geometry (Table 9.1) with one vertex occupied by a nonbonding pair of electrons. A tetrahedral electrondomain geometry with three bonding and one nonbonding domains leads to a trigonal-pyramidal molecular geometry (Table 9.2). Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. Sample Exercise 9.1 Using the VSEPR Model Continued Practice Exercise 1 Consider the following AB3 molecules and ions: PCl3, SO3, AlCl3, SO32, and CH3+. How many of these molecules and ions do you predict to have a trigonal-planar molecular geometry? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Practice Exercise 2 Predict the electron-domain and molecular geometries for (a) SeCl2, (b) CO32. Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. Nonbonding Pairs and Bond Angle Nonbonding pairs are physically larger than bonding pairs. Therefore, their repulsions are greater; this tends to compress bond angles. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

One resonance structure of the nitrate ion is The bond angles in this ion are 120. Is this observation consistent with the preceding discussion of the effect of multiple bonds on bond angles? a. No, the domain with the double bond should push the other two electron domains resulting in the bond angle between single bonds to less than 120. b. Yes, the existence of resonance with three resonance structures equalizes Molecular Geometries

repulsions between electron domains resulting in all bond angles equaling and Bonding 120. Theories 2015 Pearson Education, Inc. Multiple Bonds and Bond Angles Double and triple bonds have larger electron domains than single bonds. They exert a greater repulsive force than single bonds, making their bond angles

greater. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Expanding beyond the Octet Rule Remember that some elements can break the octet rule and make more than four bonds (or have more than four electron domains). The result is two more possible electron domains: five = trigonal bipyramidal; six = octahedral (as was seen in the slide on electron-domain geometries).

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. It might seem that a square-planar geometry of four electron domains around a central atom would be more favorable than a tetrahedron. Can you rationalize why the tetrahedron is preferred, based on angles between electron domains? a. Bond angles are not determined by a particular arrangement of electron domains.

b. A tetrahedral arrangement of electron domains results in greater electron repulsions and a less favorable geometry than electron domains in a square planar geometry. c. A tetrahedral arrangement of electron domains results in smaller electron repulsions and a more favorable geometry than electron domains in a square planar geometry. d. A tetrahedral arrangement of electron domains results in no greater Molecular

Geometries electron repulsions and no greater favorable geometry than electron and Bonding domains in a square planar geometry. Theories 2015 Pearson Education, Inc. Trigonal Bipyramidal Electron Domain There are two distinct positions in this geometry: Axial Equatorial Lone pairs occupy

equatorial positions. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Trigonal Bipyramidal Electron Domain Lower-energy conformations result from having nonbonding electron pairs in equatorial, rather than axial, positions in this geometry. Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. Trigonal Bipyramidal Electron Domain There are four distinct molecular geometries in this domain: Trigonal bipyramidal Seesaw T-shaped Linear 2015 Pearson Education, Inc.

Molecular Geometries and Bonding Theories Octahedral Electron Domain All positions are equivalent in the octahedral domain. There are three molecular geometries: Octahedral Square

pyramidal Molecular Square planar Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Use the VSEPR model to predict the molecular geometry of (a) SF4, (b) IF5. Solution Analyze The molecules are of the ABn type with a central p-block atom. Plan We first draw Lewis structures and then use the VSEPR model to determine the electron-domain geometry and molecular geometry.

Solve (a) The Lewis structure for SF4 is Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Continued The sulfur has five electron domains around it: four from the SF bonds and one from the nonbonding pair. Each domain points toward a vertex of a trigonal bipyramid. The domain from the nonbonding pair will point toward an equatorial position. The four bonds point toward the remaining four positions, resulting in a molecular geometry that is described as seesaw-shaped:

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Continued Comment The experimentally observed structure is shown on the right. We can infer that the nonbonding electron domain occupies an equatorial position, as predicted. The axial and equatorial SF bonds are slightly bent away from the nonbonding domain, suggesting that the bonding domains are pushed by the nonbonding domain, which exerts a greater repulsion (Figure 9.7). (b) The Lewis structure of IF5 is

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Continued The iodine has six electron domains around it, one of which is nonbonding. The electron-domain geometry is therefore octahedral, with one position occupied by the nonbonding pair, and the molecular geometry is square pyramidal (Table 9.3):

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Continued Comment Because the nonbonding domain is larger than the bonding domains, we predict that the four F atoms in the base of the pyramid will be tipped up slightly toward the top F atom. Experimentally, we find that the angle between the base atoms and the top F atom is 82, smaller than the ideal 90 angle of an octahedron. Practice Exercise 1 A certain AB4 molecule has a square-planar molecular geometry. Which of the following statements about

the molecule is or are true?: (i) The molecule has four electron domains about the central atom A. (ii) The BAB angles between neighboring B atoms is 90. (iii) The molecule has two nonbonding pairs of electrons on atom A. (a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2 Predict the electron-domain and molecular geometries of (a) BrF3, (b) SF5+. 2015 Pearson Education, Inc. Molecular

Geometries and Bonding Theories Shapes of Larger Molecules For larger molecules, look at the geometry about each atom rather than the molecule as a whole. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Shapes of Larger Molecules

This approach makes sense, especially because larger molecules tend to react at a particular site in the molecule. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.3 Predicting Bond Angles Eyedrops for dry eyes usually contain a water-soluble polymer called poly(vinyl alcohol), which is based on the unstable organic molecule vinyl alcohol: Predict the approximate values for the HOC and OCC bond angles in vinyl alcohol.

Solution Analyze We are given a Lewis structure and asked to determine two bond angles. Plan To predict a bond angle, we determine the number of electron domains surrounding the middle atom in the bond. The ideal angle corresponds to the electron-domain geometry around the atom. The angle will be compressed somewhat by nonbonding electrons or multiple bonds. Solve In HOC, the O atom has four electron domains (two bonding, two nonbonding). The electrondomain geometry around O is therefore tetrahedral, which gives an ideal angle of 109.5. The HOC angle is compressed somewhat by the nonbonding pairs, so we expect this angle to be slightly less than 109.5. To predict the OCC bond angle, we examine the middle atom in the angle. In the molecule, there are three atoms bonded to this C atom and no nonbonding pairs, and so it has three electron domains about it. The predicted electron-domain geometry is trigonal planar, resulting in an ideal bond angle of 120. Because of the larger size of the C C domain, the bond angle should be slightly greater than 120. Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. Sample Exercise 9.3 Predicting Bond Angles Continued Practice Exercise 1 The atoms of the compound methylhydrazine, CH 6N2, which is used as a rocket propellant, are connected as follows (note that lone pairs are not shown): What do you predict for the ideal values of the CNN and HNH angles, respectively? (a) 109.5 and 109.5 (b) 109.5 and 120 (c) 120 and 109.5 (d) 120 and 120 (e) None of the above Practice Exercise 2 Predict the HCH and CCC bond angles in propyne:

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. 9.3 Molecular Shape & Polarity Ask yourself: COVALENT or IONIC? If COVALENT: Are the BONDS polar? a. NO: The molecule is NONPOLAR! b. YES: ContinueDo the AVERAGE position of + and coincide? 1) YES: The molecule is NONPOLAR. 2) NO: The molecule is POLAR.

NOTE: Different atoms attached to the central Molecular atom have different polarity of bonds. Geometries and Bonding Theories 2015 Pearson Education, Inc. Comparison of the Polarity of Two Molecules In Chapter 8 we discussed bond dipoles. But just because a molecule possesses polar bonds does not mean the molecule as a whole will be

polar A NONPOLAR molecule Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Comparison of the Polarity of Two Molecules A POLAR molecule By looking at the bond polarity and the

symmetry of the molecule, we can determine if the molecule is polar. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Molecule Polarity Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. The molecule O=C=S is linear and has a Lewis structure analogous to that of CO2. Would you expect this molecule to be nonpolar? a. Yes, because COS has different elements than in CO2. b. No, because COS is linear. c. Yes, because O and S have different electronegativities, and CO and CS bond dipoles do not cancel each other. d. No, because O and S have similar electronegativities, and CO and CS bond dipoles cancel each other. Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. Polarity Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.4 Polarity of Molecules Predict whether these molecules are polar or nonpolar: (a) BrCl, (b) SO2, (c) SF6. Solution Analyze We are given three molecular formulas and asked to predict whether the molecules are polar.

Plan A molecule containing only two atoms is polar if the atoms differ in electronegativity. The polarity of a molecule containing three or more atoms depends on both the molecular geometry and the individual bond polarities. Thus, we must draw a Lewis structure for each molecule containing three or more atoms and determine its molecular geometry. We then use electronegativity values to determine the direction of the bond dipoles. Finally, we see whether the bond dipoles cancel to give a nonpolar molecule or reinforce each other to give a polar one. Solve (a) Chlorine is more electronegative than bromine. All diatomic molecules with polar bonds are polar molecules. Consequently, BrCl is polar, with chlorine carrying the partial negative charge: The measured dipole moment of BrCl is = 0.57 D. Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. Sample Exercise 9.4 Polarity of Molecules Continued (b) Because oxygen is more electronegative than sulfur, SO 2 has polar bonds. Three resonance forms can be written: For each of these, the VSEPR model predicts a bent molecular geometry. Because the molecule is bent, the bond dipoles do not cancel, and the molecule is polar: Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Sample Exercise 9.4 Polarity of Molecules Continued (c) Fluorine is more electronegative than sulfur, so the bond dipoles point toward fluorine. For clarity, only one SF dipole is shown. The six SF bonds are arranged octahedrally around the central sulfur: Because the octahedral molecular geometry is symmetrical, the bond dipoles cancel, and the molecule is nonpolar, meaning that = 0. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.4 Polarity of Molecules Continued

Practice Exercise 1 Consider an AB3 molecule in which A and B differ in electronegativity. You are told that the molecule has an overall dipole moment of zero. Which of the following could be the molecular geometry of the molecule? (a) Trigonal pyramidal (b) Trigonal planar (c) T-shaped (d) Tetrahedral (e) More than one of the above Practice Exercise 2 Determine whether the following molecules are polar or nonpolar: (a) SF4, (b) SiCl4. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Intermolecular Force Intermolecular Force forces of attraction between molecules. 1. Dipole-dipole: strong force of attraction between polar molecules. 2. Hydrogen bonding: very strong dipole-dipole caused between polar molecules containing hydrogen bonded to N, O, or F with an unshared pair of electrons. - Examples: HF, NH3, and H2O (H-NOF) 3. London dispersion forces: weak force of attraction between nonpolar molecules caused by momentary Molecular dipoles of constantly moving electrons. Geometries

and Bonding - Larger molecules have stronger momentary dipoles. Theories 2015 Pearson Education, Inc. 9.4 Covalent Bonding & Orbital Overlap Electrons in an atom are found in atomic orbitals. What happens to these orbital when atoms share electrons in a molecule? Lewis structures and VSPER theory give us the shape and location of electrons in a molecule, but they do not explain why a chemical bond forms. Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. Overlap and Bonding We use 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) o opposite spin in the orbital Molecular

Geometries overlap. and Bonding Theories 2015 Pearson Education, Inc. Overlap and Bonding 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. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Overlap and Bonding Increased overlap brings the electrons and nuclei closer together while simultaneously minimizing electronelectron repulsion. However, if atoms get too close, the internuclear repulsion greatly raises

the energy. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Single Covalent Bond A sharing of two valence electrons. Two specific atoms are joined. Atomic orbitals overlap and electrons are centralized between the two atoms in a new molecular orbital. This type of bond is called a sigma () bond.) bond. Molecular

Geometries and Bonding Theories 2015 Pearson Education, Inc. Sigma bonding orbitals Form s orbitals on separate atoms. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sigma bonding orbitals From px orbitals (complete overlap of lobes)

on separate atoms. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. There are two different ways to overlap! Sigma bonds () bond.) (used in single bonds) Pi bonds ()) (2nd and 3rd bond in multiple bonds) Molecular

Geometries and Bonding Theories 2015 Pearson Education, Inc. Pi bonding orbitals p orbitals (py and pz are a sideways overlap) on separate atoms. Pi bonding molecular orbital 2015 Pearson Education, Inc. Molecular Geometries and Bonding

Theories Sigma and pi bonds All single bonds are sigma bonds A double bond is one sigma and one pi bond. A triple bond is one sigma and two pi bonds. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Atomic Overlapping Orbitals Bonds between atoms are formed by electron pairs in overlapping atomic orbitals.

Example: H2 (H-H) - Use 1s orbitals for bonding Example: H2O - Use two 2p orbitals for bonding? This would imply bent, 104.50 angle between H atoms! 2015 Pearson Education, Inc. From VSPER: Molecular Geometries and Bonding Theories

9.5 VSEPR and Hybrid Orbitals If overlapping orbitals dont always seem to work, then how can we explain other geometries? Valence Bond Theory (Hybrids Orbitals) Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. 9.5 VSEPR and Hybrid Orbitals

VSEPR predicts shapes of molecules very well. How does that fit with orbitals? Lets use H2O as an example: If we draw the best Lewis structure to assign VSEPR, it becomes bent. If we look at oxygen, its electron configuration is 1s22s22p4. If it shares two electrons to fill its valence shell, they should be in 2p. Wouldnt that make the angle 90? Molecular Geometries

Why is it 104.5? and Bonding Theories 2015 Pearson Education, Inc. Hybridization the blending of orbitals Valence bond theory is based on two assumptions: 1. The strength of a covalent bond is proportional to the amount of overlap between atomic orbitals; the greater the overlap, the more stable the bond. 2. An atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded

Molecular Geometries atoms. (this is hybridization) and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbitals Hybrid orbitals form by mixing of atomic orbitals to create new orbitals of equal energy, called degenerate orbitals. CCl4 , NH3 (Bothe have 4 electron domains so they need 4 new hybrid orbitals! Where do they come from? Molecular

Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbitals You can see why they must be equal in energy if we look at methane. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Carbon ground state configuration

Can you see a problem with this? (Hint: How many unpaired electrons does this carbon atom have available for bonding) It appears that carbon has only TWO electrons available for bonding and that is not enough! Molecular How does carbon overcome this problem so Geometries and Bonding Theories that it may form four bonds? 2015 Pearson Education, Inc. Carbons Empty Orbital The first thought

that chemists had was that carbon promotes one of its to the empty 2p orbital. 2s electrons Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. However, they quickly recognized a problem with such an arrangement Three of the carbon-hydrogen bonds would

involve an electron pair in which the carbon Molecular electron was a 2p, matched with the lone 1s Geometries and Bonding electron from a hydrogen atom. Theories 2015 Pearson Education, Inc. However, they quickly recognized a problem with such an arrangement The fourth bond is between a 2s electron from the carbon and the lone1s hydrogen electron. Such a bond would have slightly less energy

than the other bonds in a methane molecule. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. This bond would be slightly different in character than the other three bonds in methane. This difference would be measurable to a chemist by determining the bond length and bond energy.

But is this what they observe? Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. The simple answer is No. Measurements show that all four bonds in methane are equal. Thus, we need a new explanation for the bonding in methane. Chemists have proposed an explanation called hybridization. Hybridization is the combining of two or more

orbitals of nearly equal energy within the same atom into orbitals of equal energy. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. In the case of methane, they call the hybridization sp3, meaning that an s orbital is combined with three p orbitals to create four equal hybrid orbitals. These new orbitals have slightly MORE energy than the 2s orbital and slightly LESS energy than the 2p orbitals.

Molecular Geometries Sp3 Hybrid Orbitals and Bonding Theories 2015 Pearson Education, Inc. Carbons original s, px, py and pz orbitals hybridize to 4 equal energy sp3 hybrid orbitals. This matches experimental evidence that the bonds in methane have a bond angle of 109.50 and are

equal in length and bond energy. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Anytime you have four electron domains, sp3 hybridization can be used to explain the tetrahedral electron domain geometry. Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. What Happens with Water? We started this discussion with H2O and the angle question: Why is it 104.5 instead of 90? Oxygen has two bonds and two lone pairs four electron domains. The result is sp3 hybridization! 2015 Pearson Education, Inc. Molecular

Geometries and Bonding Theories Besp hybridization BeF2 When we look at the orbital diagram for beryllium (Be), we see that there are only paired electrons in full sub-levels. Be makes electron deficient compounds with two bonds for Be. Why? sp hybridization (mixing of one s orbital and one p orbital) 2015 Pearson Education, Inc. sp Orbitals Mixing the s and p orbitals yields two degenerate

orbitals that are hybrids of the two orbitals. These sp hybrid orbitals have two lobes like a p orbital. One of the lobes is larger and more rounded, as is the s orbital. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Position of sp Orbitals These two degenerate orbitals would align themselves 180 from each other. This is consistent with the observed geometry of

Be compounds (like BeF2) and VSEPR: linear. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbitals With hybrid orbitals the orbital diagram for Be would look like this. Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. What is the orientation of the two unhybridized p orbitals on Be with respect to the two BeF bonds? a. Both p orbitals are parallel to the BeF bonds. b. Both p orbitals intersect the BeF bonds. c. Both p orbitals are at an angle of 120 to the BeF bonds. d. Both p orbitals are perpendicular to the BeF bonds. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

BoronThree Electron Domains Gives sp2 Hybridization BF3 Using a similar model for boron leads to three degenerate (equal) sp2 orbitals. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid orbitals Using a similar model for boron leads to Molecular

Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbitals three degenerate sp2 orbitals. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. In an sp2 hybridized atom, we saw that there was one unhybridized 2p orbital. How many

unhybridized 2p orbitals remain on an atom that has sp3 hybrid orbitals? a. One b. Two c. Three d. None Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybridization How about expanded octets like PF5! (Hypervalent Molecules)

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbitals For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids. Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. Hybridization Involving d Orbitals Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbitals Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Hybrid Orbital Summary 1) Draw the Lewis structure. 2) Use VSEPR to determine the electron-domain geometry. 3) Specify the hybrid

orbitals needed to accommodate these electron pairs. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories Sample Exercise 9.5 Describing the Hybridization of a Central Atom Describe the orbital hybridization around the central atom in NH . 2

Solution Analyze We are given the chemical formula for a polyatomic anion and asked to describe the type of hybrid orbitals surrounding the central atom. Plan To determine the central atom hybrid orbitals, we must know the electron-domain geometry around the atom. Thus, we draw the Lewis structure to determine the number of electron domains around the central atom. The hybridization conforms to the number and geometry of electron domains around the central atom as predicted by the VSEPR model. Solve The Lewis structure is Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Sample Exercise 9.5 Describing the Hybridization of a Central Atom Continued Because there are four electron domains around N, the electron-domain geometry is tetrahedral. The hybridization that gives a tetrahedral electron-domain geometry is sp3 (Table 9.4). Two of the sp3 hybrid orbitals contain nonbonding pairs of electrons, and the other two are used to make bonds with the hydrogen atoms. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.5 Describing the Hybridization of a Central Atom

Continued Practice Exercise 1 For which of the following molecules or ions does the following description apply? The bonding can be explained using a set of sp2 hybrid orbitals on the central atom, with one of the hybrid orbitals holding a nonbonding pair of electrons. (a) CO2 (b) H2S (c) O3 (d) CO32 (e) More than one of the above Practice Exercise 2

Predict the electron-domain geometry and hybridization of the central atom in SO 32. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. 9.6 Multiple Bonds How does a double or triple bond form? What happened to sigma and pi bonds! It cant, if we only use hybridized orbitals. However, if we use the orbitals which are not hybridized, we can have a side-ways overlap. Two types of bonds:

Sigma () bond Pi () bond Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. What does SO3 do? The sulfur needs to form 3 hybrid orbitals to attach and share with

the 3 oxygen! What happens to the unhybridized p orbital? Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sigma () and Pi () Bonds The bonds that form between the atoms (in the blue) are called Sigma () bonds.

- All single bonds and 1 of the bonds in a multiple bond are sigma bonds. The bond that forms when parallel p orbitals overlap is called Molecular a Pi () Bonds. Geometries Bonding - Double bond (1 and 1 ) triple bond (1 and 2 ) andTheories 2015 Pearson Education, Inc. Sigma () and Pi () Bonds Sigma bonds are characterized by head-to-head overlap. cylindrical symmetry of electron density about the internuclear axis.

Pi bonds are characterized by side-to-side overlap. electron density above and below the internuclear axis. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories Bonding in Molecules Single bonds are always -bonds. Multiple bonds have one -bond,

all other bonds are -bonds. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. The molecule called diazine has the formula N2H2 and the Lewis structure Do you expect diazine to be a linear molecule (all four atoms on the same line)? If not, do you expect the molecule to be planar (all four atoms in the same plane)? a. The molecule is both linear and planar.

b. The molecule is not linear, but is planar. c. The molecule is linear, but not planar. d. The molecule is neither linear nor planar. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories Localized or Delocalized Electrons: Resonance Bonding electrons ( or ) that are specifically shared between two atoms are called localized electrons. In many molecules, we cant describe

all electrons that way (resonance); the other electrons (shared by multiple atoms) are called delocalized electrons. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Resonance (Benzene) The organic molecule benzene (C6H6) has six -bonds and a p orbital on each C atom, which form delocalized bonds using one electron from each p orbital. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Sample Exercise 9.6 Describing and Bonds in a Molecule Formaldehyde has the Lewis structure Describe how the bonds in formaldehyde are formed in terms of overlaps of hybrid and unhybridized orbitals. Solution Analyze We are asked to describe the bonding in formaldehyde in terms of hybrid orbitals. Plan Single bonds are bonds, and double bonds consist of one bond and one bond. The ways in which these bonds form can be deduced from the molecular geometry, which we predict using the VSEPR model. Molecular Geometries

and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.6 Describing and Bonds in a Molecule Continued Solve The C atom has three electron domains around it, which suggests a trigonal-planar geometry with bond angles of about 120. This geometry implies sp2 hybrid orbitals on C (Table 9.4). These hybrids are used to make the two CH and one CO bonds to C. There remains an unhybridized 2p orbital (a p orbital) on carbon, perpendicular to the plane of the three sp2 hybrids. Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. Sample Exercise 9.6 Describing and Bonds in a Molecule Continued The O atom also has three electron domains around it, and so we assume it has sp2 hybridization as well. One of these hybrid orbitals participates in the CO bond, while the other two hold the two nonbonding electron pairs of the O atom. Like the C atom, therefore, the O atom has a p orbital that is perpendicular to the plane of the molecule. The two p orbitals overlap to form a CO bond (Figure 9.25). Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Sample Exercise 9.6 Describing and Bonds in a Molecule Continued Practice Exercise 1 We have just arrived at a bonding description for the formaldehyde molecule. Which of the following statements about the molecule is or are true? (i) Two of the electrons in the molecule are used to make the bond in the molecule. (ii) Six of the electrons in the molecule are used to make the bonds in the molecule. (iii) The CO bond length in formaldehyde should be shorter than that in methanol, H 3COH. (a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2

(a) Predict the bond angles around each carbon atom in acetonitrile: (b) Describe the hybridization at each carbon atom, and (c) determine the number of and bonds in the Molecular molecule. Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.7 Delocalized Bonding Describe the bonding in the nitrate ion, NO 3. Does this ion have delocalized bonds? Solution Analyze Given the chemical formula for a polyatomic anion, we are asked to describe the bonding and determine whether the ion has delocalized bonds.

Plan Our first step is to draw Lewis structures. Multiple resonance structures involving the placement of the double bonds in different locations would suggest that the component of the double bonds is delocalized. Solve In Section 8.6 we saw that NO3 has three resonance structures: Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.7 Delocalized Bonding Continued In each structure, the electron-domain geometry at nitrogen is trigonal planar, which implies sp2 hybridization of the N atom. It is helpful when considering delocalized bonding to consider atoms with lone pairs that are bonded to the central atom to be sp2 hybridized as well. Thus, we can envision that each of the O atoms in the anion has three sp2 hybrid orbitals in the plane of the ion. Each of the four atoms has

an unhybridized p orbital oriented perpendicular to the plane of the ion. The NO3 ion has 24 valence electrons. We can first use the sp2 hybrid orbitals on the four atoms to construct the three NO bonds. That uses all of the sp2 hybrids on the N atom and one sp2 hybrid on each O atom. Each of the two remaining sp2 hybrids on each O atom is used to hold a nonbonding pair of electrons. Thus, for any of the resonance structures, we have the following arrangement in the plane of the ion: Molecular Geometries Notice that we have accounted for a total of 18 electronssix in the three NO bonds, and 12and as Bonding nonbonding pairs on the O atoms. The remaining six electrons will reside in the system of the ion.Theories 2015 Pearson Education, Inc. Sample Exercise 9.7 Delocalized Bonding Continued

The four p orbitalsone on each of the four atoms are used to build the p system. For any one of the three resonance structures shown, we might imagine a single localized NO bond formed by the overlap of the p orbital on N and a p orbital on one of the O atoms. The remaining two O atoms have nonbonding pairs in their p orbitals. Thus, for each of the resonance structures, we have the situation shown in Figure 9.28. Because each resonance structure contributes equally to the observed structure of NO3, however, we represent the bonding as delocalized over the three NO bonds, as shown in the figure. We see that the NO3 ion has a six-electron system delocalized among the four atoms in the ion. Practice Exercise 1 How many electrons are in the system of the ozone molecule, O3 ?

(a) 2 (b) 4 (c) 6 (d) 14 (e) 18 Practice Exercise 2 Which of these species have delocalized bonding: SO 2, SO3, SO32, H2CO, NH4+? 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories When two atoms are bonded by a triple bond, what is the hybridization of the orbitals that make up the -bond component of the bond?

a. sp hybrid atomic orbitals b. sp2 hybrid atomic orbitals c. sp3 hybrid atomic orbitals d. sp3d hybrid atomic orbitals Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. 9.7 Molecular Orbital (MO) Theory Wave properties are used to describe the energy of the electrons in a molecule. Molecular orbitals have many characteristics like atomic orbitals:

maximum of two electrons per orbital Electrons in the same orbital have opposite spin. Definite energy of orbital Can visualize electron density by a contour diagram Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. More on MO Theory They differ from atomic orbitals because they represent the entire molecule, not a single atom. Whenever two atomic orbitals overlap, two molecular orbitals are formed: one bonding, one

antibonding. Bonding orbitals are constructive combinations of atomic orbitals. Antibonding orbitals are destructive combinations of atomic orbitals. They have a new feature unseen before: A nodal plane occurs where electron density equals zero. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Molecular Orbital (MO) Theory Whenever there is direct overlap of orbitals,

forming a bonding and an antibonding orbital, they are called sigma () molecular orbitals. The antibonding orbital is distinguished with an asterisk as *. Here is an example for the formation of a hydrogen molecule from two atoms. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories

MO Diagram An energy-level diagram, or MO diagram shows how orbitals from atoms combine to give the molecule. In H2 the two electrons go into the bonding molecular orbital (lower in energy). Bond order = (# of bonding electrons # of antibonding electrons)Molecular = Geometries (2 0) = 1 bond

and Bonding Theories 2015 Pearson Education, Inc. Can He2 Form? Use MO Diagram and Bond Order to Decide! Bond Order = (2 2) = 0 bonds Therefore, He2 does not exist. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Would you expect Be2+ to be a stable ion? a. No, because the bond order is 0. b. Yes, because the bond order is 0.5. c. Yes, because the bond order is 1. d. Yes, because the bond order is 1.5. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Suppose one electron in H2 is excited from the 1s MO to the *1s MO. Would you expect the H atoms

to remain bonded to each other, or would the molecule fall apart? a. Remain bonded, as the bond order remains the same. b. Fall apart, as the bond order changes from 1 to 1.5. c. Remain bonded, as the bond order changes from 1 to 1.5. d. Fall apart, as the bond order changes from 1 to zero. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.8 Bond Order What is the bond order of the He2+ ion? Would you expect this ion to be stable relative to the separated He atom and He+ ion?

Solution Analyze We will determine the bond order for the He 2+ ion and use it to predict whether the ion is stable. Plan To determine the bond order, we must determine the number of electrons in the molecule and how these electrons populate the available MOs. The valence electrons of He are in the 1s orbital, and the 1s orbitals combine to give an MO diagram like that for H 2 or He2 (Figure 9.33). If the bond order is greater than 0, we expect a bond to exist, and the ion is stable. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.8 Bond Order Continued

Solve The energy-level diagram for the He2+ ion is shown in Figure 9.34. This ion has three electrons. Two are placed in the bonding orbital and the third in the antibonding orbital. Thus, the bond order is Because the bond order is greater than 0, we predict the He 2+ ion to be stable relative to the separated He and He+. Formation of He2+ in the gas phase has been demonstrated in laboratory experiments. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.8 Bond Order Continued Practice Exercise 1

How many of the following molecules and ions have a bond order of (a) 0 (b) 1 (c) 2 (d) 3 : H 2, H2+, H2, and He22+? (e) 4 Practice Exercise 2 What are the electron configuration and the bond order of the H 2 ion?

Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. 9.8 s and p Orbitals Can Interact For atoms with both s and p orbitals, there are two types of interactions: The s and the p orbitals that face each other overlap in fashion. The other two sets of p orbitals overlap in fashion. These are, again, direct and

side-ways overlap of Molecular orbitals. Geometries and Bonding Theories 2015 Pearson Education, Inc. MO Theory The resulting MO diagram: There are and * orbitals from s and p atomic orbitals. There are and * orbitals from p atomic orbitals.

Since direct overlap is stronger, the effect of raising and lowering energy is greater for and *. 2015 Pearson Education, Inc. Molecular Geometries and Bonding Theories s and p Orbital Interactions In some cases, s orbitals can interact wit the pz orbitals more than the px and py orbitals.

It raises the energy of the pz orbital and lowers the energy of the s orbital. Molecular Geometries The px and py orbitals are degenerate orbitals. and Bonding Theories 2015 Pearson Education, Inc. MO Diagrams for Diatomic nd Molecules of 2 Period Elements Molecular Geometries and Bonding

Theories 2015 Pearson Education, Inc. MO Diagrams and Magnetism Diamagnetism is the result of all electrons in every orbital being spin paired. These substances are weakly repelled by a magnetic field. Paramagnetism is the result of the presence of one or more unpaired electrons in an orbital. Is oxygen (O2) paramagnetic or diamagnetic? Look back at the MO Molecular Geometries diagram! It is paramagnetic.

and Bonding Theories 2015 Pearson Education, Inc. Figure 9.43 indicates that C2 is diamagnetic. Would that be expected if the 2p MO were lower in energy than the 2p MOs? a. Yes, because the reversal of MOs does not change the electron occupation of each MO. b. No, because the reversal of MOs results in the two 2p MOs containing one electron each. Molecular Geometries and Bonding Theories

2015 Pearson Education, Inc. Paramagnetism of Oxygen Lewis structures would not predict that O2 is paramagnetic. The MO diagram clearly shows that O2 is paramagnetic. Both show a double bond (bond order = 2). Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc.

Sample Exercise 9.9 Molecular Orbitals of a Period 2 Diatomic Ion For the O ion predict (a) number of unpaired electrons, (b) bond order, (c) bond enthalpy and bond 2 + length. Solution Analyze Our task is to predict several properties of the cation O 2+. Plan We will use the MO description of O2+ to determine the desired properties. We must first determine the number of electrons in O2+ and then draw its MO energy diagram. The unpaired electrons are those without a partner of opposite spin. The bond order is one-half the difference between the number of bonding and antibonding electrons. After calculating the bond order, we can use Figure 9.43 to estimate

the bond enthalpy and bond length. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.9 Molecular Orbitals of a Period 2 Diatomic Ion Continued Solve (a) The O2+ ion has 11 valence electrons, one fewer than O 2. The electron removed from O2 to form O2+ is one of the two unpaired *2p electrons (see Figure 9.43). Therefore, O 2+ has one unpaired electron. (b) The molecule has eight bonding electrons (the same as O 2) and three antibonding electrons (one fewer than O2). Thus, its bond order is

(c) The bond order of O2+ is between that for O2 (bond order 2) and N2 (bond order 3). Thus, the bond enthalpy and bond length should be about midway between those for O 2 and N2, approximately 700 kJ mol and 1.15 . (The experimentally measured values are 625 kJ mol and 1.123 .) Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Sample Exercise 9.9 Molecular Orbitals of a Period 2 Diatomic Ion Continued Practice Exercise 1 Place the following molecular ions in order from smallest to largest bond order: C 22+, N2, O2, and F2.

(a) C22+ < N2 < O2 < F2 (b) F2 < O2 < N2 < C22+ (c) O2 < C22 + < F2 < N2 (d) C22+ < F2 < O2 < N2 (e) F2 < C22+ < O2 < N2 Practice Exercise 2 Predict the magnetic properties and bond orders of (a) the peroxide ion, O22; (b) the acetylide ion, C22. Molecular Geometries and Bonding Theories 2015 Pearson Education, Inc. Heteronuclear Diatomic Molecules

Diatomic molecules can consist of atoms from different elements. How does a MO diagram reflect differences? The atomic orbitals have different energy, so the interactions change slightly. The more electronegative atom has orbitals lower in energy, so the bonding orbitals will more resemble them in energy. 2015 Pearson Education, Inc. Molecular Geometries and Bonding

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