# Chapter 9 Molecular Geometry and Bonding Theories

SO32- Molecular Geometry / Shape and Bond Angles (note precise angle is 106 degrees)
SO32- Molecular Geometry / Shape and Bond Angles (note precise angle is 106 degrees)

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

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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.

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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).

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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.

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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? Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral Answer: d

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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 repulsions.)

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The VSPER Model 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 # of electron domains in the Lewis diagram

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Suppose a particular AB3 molecule has the resonance structure
Does this structure follow the octet rule? No, there are 10 electrons around A. Yes, there are 8 electrons around A. Yes, there are three bonds and one electron pair around A. Yes, there are four bonds around A. Answer: a

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Suppose a particular AB3 molecule has the resonance structure
How many electron domains are there around the A atom? One electron domain Two electron domains Three electron domains Four electron domains Answer: d

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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 on the central atom.

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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 to see what molecular geometries are possible.

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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, not the nonbonding pairs.

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From the standpoint of the VSEPR model, what do nonbonding electron pairs, single bonds, and multiple bonds have in common? There are no common features. Each occurs about the central atom only. Each represents a single electron domain. All exist when a particular Lewis structure is drawn. Answer: c

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With each electron domain, then, there might be more than one molecular geometry.

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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) H- Be – H

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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. SO3

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Tetrahedral Electron Domain
There are three molecular geometries: tetrahedral, if all are bonding pairs, trigonal pyramidal, if one is a nonbonding pair, and bent, if there are two nonbonding pairs.

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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 O3:

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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).

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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

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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).

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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–.

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Nonbonding Pairs and Bond Angle
Nonbonding pairs are physically larger than bonding pairs. Therefore, their repulsions are greater; this tends to compress bond angles.

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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? Answer: b 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°. Yes, the existence of resonance with three resonance structures equalizes repulsions between electron domains resulting in all bond angles equaling 120°.

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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.

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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).

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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? Bond angles are not determined by a particular arrangement of electron domains. A tetrahedral arrangement of electron domains results in greater electron repulsions and a less favorable geometry than electron domains in a square planar geometry. A tetrahedral arrangement of electron domains results in smaller electron repulsions and a more favorable geometry than electron domains in a square planar geometry. A tetrahedral arrangement of electron domains results in no greater electron repulsions and no greater favorable geometry than electron domains in a square planar geometry. Answer: c

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Trigonal Bipyramidal Electron Domain
There are two distinct positions in this geometry: Axial Equatorial Lone pairs occupy equatorial positions.

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Trigonal Bipyramidal Electron Domain
Lower-energy conformations result from having nonbonding electron pairs in equatorial, rather than axial, positions in this geometry.

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Trigonal Bipyramidal Electron Domain
There are four distinct molecular geometries in this domain: Trigonal bipyramidal Seesaw T-shaped Linear

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Octahedral Electron Domain
All positions are equivalent in the octahedral domain. There are three molecular geometries: Octahedral Square pyramidal Square planar

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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

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Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells
Continued The sulfur has five electron domains around it: four from the S—F 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:

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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 S—F 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

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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):

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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 B—A—B 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+.

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Shapes of Larger Molecules
For larger molecules, look at the geometry about each atom rather than the molecule as a whole. <120 , <107.5

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Shapes of Larger Molecules
This approach makes sense, especially because larger molecules tend to react at a particular site in the molecule. <120 , <107.5

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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 H—O—C and O—C—C 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 H—O—C, the O atom has four electron domains (two bonding, two nonbonding). The electron-domain geometry around O is therefore tetrahedral, which gives an ideal angle of 109.5°. The H—O—C angle is compressed somewhat by the nonbonding pairs, so we expect this angle to be slightly less than 109.5°. To predict the O—C—C 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°.

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Sample Exercise 9.3 Predicting Bond Angles
Continued Practice Exercise 1 The atoms of the compound methylhydrazine, CH6N2, 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 C—N—N and H—N—H 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 H—C—H and C—C—C bond angles in propyne:

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9.3 Molecular Shape & Polarity
Ask yourself: COVALENT or IONIC? If COVALENT: Are the BONDS polar? NO: The molecule is NONPOLAR! YES: Continue—Do the AVERAGE position of δ+ and δ– coincide? YES: The molecule is NONPOLAR. NO: The molecule is POLAR. NOTE: Different atoms attached to the central atom have different polarity of bonds.

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Comparison of the Polarity of Two Molecules
A NONPOLAR molecule 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

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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.

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Molecule Polarity

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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? Yes, because COS has different elements than in CO2. No, because COS is linear. Yes, because O and S have different electronegativities, and CO and CS bond dipoles do not cancel each other. No, because O and S have similar electronegativities, and CO and CS bond dipoles cancel each other. Answer: c

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Polarity

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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 μ = 0.57 D.

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Sample Exercise 9.4 Polarity of Molecules
Continued (b) Because oxygen is more electronegative than sulfur, SO2 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:

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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 S—F dipole is shown. The six S—F 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 μ = 0.

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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.

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Intermolecular Force Intermolecular Force – forces of attraction between molecules. Dipole-dipole: strong force of attraction between polar molecules. 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 dipoles of constantly moving electrons. – Larger molecules have stronger momentary dipoles.

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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. Start here 6/14/10

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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) of opposite spin in the orbital overlap. Start here 6/14/10

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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. Start here 6/14/10

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Overlap and Bonding Increased overlap brings the electrons and nuclei closer together while simultaneously minimizing electron-electron repulsion. However, if atoms get too close, the internuclear repulsion greatly raises the energy. Atoms can’t get too close because the internuclear repulsions get too great.

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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 (σ) bond.

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Sigma bonding orbitals
Form s orbitals on separate atoms.

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Sigma bonding orbitals
From px orbitals (complete overlap of lobes) on separate atoms.

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There are two different ways to overlap!
Sigma bonds (σ) (used in single bonds) Pi bonds (π) (2nd and 3rd bond in multiple bonds)

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Pi bonding orbitals p orbitals (py and pz are a sideways overlap) on separate atoms. Pi bonding molecular orbital

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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.

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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… From VSPER: bent, angle between H atoms! It is not 900

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9.5 VSEPR and Hybrid Orbitals
If overlapping orbitals don’t always seem to work, then how can we explain other geometries? Valence Bond Theory (Hybrids Orbitals)

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9.5 VSEPR and Hybrid Orbitals
VSEPR predicts shapes of molecules very well. How does that fit with orbitals? Let’s 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. Wouldn’t that make the angle 90°? Why is it 104.5°?

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Hybridization … the blending of orbitals Valence bond theory is based on two assumptions:
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. An atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded atoms. (this is hybridization)

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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?

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Hybrid Orbitals You can see why they must be equal in energy if we look at methane.

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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! How does carbon overcome this problem so that it may form four bonds?

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Carbon’s Empty Orbital
The first thought that chemists had was that carbon promotes one of its 2s electrons… …to the empty 2p orbital.

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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 electron was a 2p, matched with the lone 1s electron from a hydrogen atom.

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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.

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But is this what they observe?
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?

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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.

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These new orbitals have slightly MORE energy than the 2s orbital…
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. Sp3 Hybrid Orbitals

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Carbon’s 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 and are equal in length and bond energy.

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Anytime you have four electron domains, sp3 hybridization can be used to explain the tetrahedral electron domain geometry.

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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!

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Be—sp 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)

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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.

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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.

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Hybrid Orbitals With hybrid orbitals the orbital diagram for Be would look like this.

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What is the orientation of the two unhybridized p orbitals on Be with respect to the two Be—F bonds?
Both p orbitals are parallel to the Be—F bonds. Both p orbitals intersect the Be—F bonds. Both p orbitals are at an angle of 120° to the Be—F bonds. Both p orbitals are perpendicular to the Be—F bonds. Answer: d

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Boron—Three Electron Domains Gives sp2 Hybridization BF3
Using a similar model for boron leads to three degenerate (equal) sp2 orbitals.

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Hybrid orbitals Using a similar model for boron leads to…

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Hybrid Orbitals …three degenerate sp2 orbitals.

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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? One Two Three None Answer: d

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Hybridization How about expanded octets like PF5! (Hypervalent Molecules)

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Hybrid Orbitals For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids.

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Hybridization Involving d Orbitals

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Hybrid Orbitals

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Hybrid Orbital Summary
Draw the Lewis structure. Use VSEPR to determine the electron-domain geometry. Specify the hybrid orbitals needed to accommodate these electron pairs.

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Sample Exercise 9.5 Describing the Hybridization of a Central Atom
Describe the orbital hybridization around the central atom in NH2–. 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

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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.

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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) CO (b) H2S (c) O (d) CO32– (e) More than one of the above Practice Exercise 2 Predict the electron-domain geometry and hybridization of the central atom in SO32–.

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9.6 Multiple Bonds How does a double or triple bond form?
What happened to sigma and pi bonds! It can’t, 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 (σ) bond Pi (π) bond

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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?

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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 a Pi () Bonds. – Double bond (1 and 1 ) triple bond (1 and 2 )

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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.

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Bonding in Molecules Single bonds are always σ-bonds.
Multiple bonds have one σ-bond, all other bonds are π-bonds.

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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)? The molecule is both linear and planar. The molecule is not linear, but is planar. The molecule is linear, but not planar. The molecule is neither linear nor planar. Answer: b

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Localized or Delocalized Electrons: Resonance
Bonding electrons (σ or π) that are specifically shared between two atoms are called localized electrons. In many molecules, we can’t describe all electrons that way (resonance); the other electrons (shared by multiple atoms) are called delocalized electrons.

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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.

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Sample Exercise 9.6 Describing σ and Π 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 𝜎 bonds, and double bonds consist of one σ bond and one π bond. The ways in which these bonds form can be deduced from the molecular geometry, which we predict using the VSEPR model.

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Sample Exercise 9.6 Describing σ and Π 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 C—H and one C—O σ bonds to C. There remains an unhybridized 2p orbital (a pπ orbital) on carbon, perpendicular to the plane of the three sp2 hybrids.

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Sample Exercise 9.6 Describing σ and Π 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 C—O σ 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 pπ orbital that is perpendicular to the plane of the molecule. The two pπ orbitals overlap to form a C—O π bond (Figure 9.25).

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Sample Exercise 9.6 Describing σ and Π 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 π bond in the molecule. (ii) Six of the electrons in the molecule are used to make the σ bonds in the molecule. (iii) The C—O bond length in formaldehyde should be shorter than that in methanol, H3COH. (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 σ and π bonds in the molecule.

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Sample Exercise 9.7 Delocalized Bonding
Describe the bonding in the nitrate ion, NO3–. Does this ion have delocalized π 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 π 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 π component of the double bonds is delocalized. Solve In Section 8.6 we saw that NO3– has three resonance structures:

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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 π 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 pπ 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 N—O σ 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: Notice that we have accounted for a total of 18 electrons—six in the three N—O σ bonds, and 12 as nonbonding pairs on the O atoms. The remaining six electrons will reside in the π system of the ion.

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Sample Exercise 9.7 Delocalized Bonding
Continued The four pπ orbitals—one 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 N—O π bond formed by the overlap of the pπ orbital on N and a pπ orbital on one of the O atoms. The remaining two O atoms have nonbonding pairs in their pπ orbitals. Thus, for each of the resonance structures, we have the situation shown in Figure Because each resonance structure contributes equally to the observed structure of NO3–, however, we represent the π bonding as delocalized over the three N—O bonds, as shown in the figure. We see that the NO3– ion has a six-electron π system delocalized among the four atoms in the ion. Practice Exercise 1 How many electrons are in the π 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: SO2, SO3, SO32–, H2CO, NH4+?

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When two atoms are bonded by a triple bond, what is the hybridization of the orbitals that make up the Σ-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 Answer: a

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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

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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.

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Molecular Orbital (MO) Theory
Whenever there is direct overlap of orbitals, forming a bonding and an antibonding orbital, they are called sigma (σ) molecular orbitals. The antibonding orbital is distinguished with an asterisk as σ*. Here is an example for the formation of a hydrogen molecule from two atoms. Start here 6/16/10

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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 = ½(# of bonding electrons – # of antibonding electrons) = ½(2 – 0) = 1 bond

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Can He2 Form? Use MO Diagram and Bond Order to Decide!
Bond Order = ½(2 – 2) = 0 bonds Therefore, He2 does not exist.

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Would you expect Be2+ to be a stable ion?
No, because the bond order is 0. Yes, because the bond order is 0.5. Yes, because the bond order is 1. Yes, because the bond order is 1.5. Answer: b

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Suppose one electron in H2 is excited from the σ1s MO to the σ. 1s MO
Suppose one electron in H2 is excited from the σ1s MO to the σ*1s MO. Would you expect the H atoms to remain bonded to each other, or would the molecule fall apart? Remain bonded, as the bond order remains the same. Fall apart, as the bond order changes from 1 to 1.5. Remain bonded, as the bond order changes from 1 to 1.5. Fall apart, as the bond order changes from 1 to zero. Answer: d

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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 He2+ 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 H2 or He2 (Figure 9.33). If the bond order is greater than 0, we expect a bond to exist, and the ion is stable.

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Sample Exercise 9.8 Bond Order
Continued Solve The energy-level diagram for the He2+ ion is shown in Figure 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 He2+ ion to be stable relative to the separated He and He+. Formation of He2+ in the gas phase has been demonstrated in laboratory experiments.

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Sample Exercise 9.8 Bond Order
Continued Practice Exercise 1 How many of the following molecules and ions have a bond order of : H2, H2+, H2–, and He22+? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 Practice Exercise 2 What are the electron configuration and the bond order of the H2– ion?

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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 orbitals.

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MO Theory The resulting MO diagram:
There are σ and σ* orbitals from s and p atomic orbitals. There are π and π* orbitals from p atomic orbitals. Since direct overlap is stronger, the effect of raising and lowering energy is greater for σ and σ*.

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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. The px and py orbitals are degenerate orbitals.

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MO Diagrams for Diatomic Molecules of 2nd Period Elements

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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 diagram! It is paramagnetic.

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Figure 9. 43 indicates that C2 is diamagnetic
Figure 9.43 indicates that C2 is diamagnetic. Would that be expected if the σ2p MO were lower in energy than the π2p MOs? Yes, because the reversal of MOs does not change the electron occupation of each MO. No, because the reversal of MOs results in the two π2p MOs containing one electron each. Answer: b

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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).

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Sample Exercise 9.9 Molecular Orbitals of a Period 2 Diatomic Ion
For the O2+ ion predict (a) number of unpaired electrons, (b) bond order, (c) bond enthalpy and bond length. Solution Analyze Our task is to predict several properties of the cation O2+. 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.

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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 O2. The electron removed from O2 to form O2+ is one of the two unpaired π*2p electrons (see Figure 9.43). Therefore, O2+ has one unpaired electron. (b) The molecule has eight bonding electrons (the same as O2) 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 O2 and N2, approximately 700 kJ ⁄ mol and 1.15 Å. (The experimentally measured values are 625 kJ ⁄ mol and Å.)

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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: C22+, 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–.

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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.

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