7. Arrange the molecular orbitals in order of increasing energy and add the electrons.

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1 Molecular Orbital Theory I. Introduction. A. Ideas. 1. Start with nuclei at their equilibrium positions. 2. onstruct a set of orbitals that cover the complete nuclear framework, called molecular orbitals (MO's).3. Use the rules of quantum mechanics to arrange the molecular orbitals in order of increasing energy and add the electrons. 4. Fill the MO's with the molecule's electrons. a. Lowest energy MO's filled first. b. No more than two electrons in the same MO with their spins paired. c. alf fill a degenerate set with spins parallel before pairing up in the same MO. 5. Start with the atomic orbitals (AO s). an construct the MO's by taking Linear ombinations of the Atomic Orbitals using at least one AO from each atom (LAO method). 6. Most of the time the only electrons that need be considered are the valence electrons and the AO's used in the linear combinations are the valence orbitals 7. Arrange the molecular orbitals in order of increasing energy and add the electrons. a. Lowest energy MO's filled first. b. No more than two electrons in the same MO with their spins paired. c. alf fill a degenerate set with spins parallel before pairing up in the same MO. B. Lowest energy states of homonuclear diatomic molecules. 1. Get the two lowest energy MO's by taking linear combinations of the lowest energy atomic orbitals, the 1s atomic orbitals. There are two ways to combine, can add them together or subtract them. The resulting MO's are sketched below. 1

2 Atom A Atom B! 1sA! 1sB "# 1s! 1sA! 1sB! 1sA +! 1sB " 1s a. Ψ 1s is a spherical function that is positive in all regions in space. When two 1s functions are added the resulting wave function, labeled σ 1s, is also positive in all regions. (A sketch of σ 1s is shown above.) On the other hand, when the two 1s atomic orbital functions are subtracted, the resulting function, σ * 1s, is positive around nucleus A, negative around nucleus B, and is equal to zero halfway between the two nuclei (see above). The wave function is said to have a nodal point between the two nuclei. b. The σ designation indicates that the wave function is symmetric with respect to the inter-nuclear axis. That is, the wave function does not change sign as one goes above or below the internuclear line. The subscript, 1s, tells the atomic orbitals that are involved in the molecular orbital. 3. The relative energies of the orbitals. a. The molecular orbital energy diagram. 2

3 !" 1s Energy $ # 1sA # 1sB $ Atom A! 1s AB Molecule Atom B b. The σ 1s orbital is lower in energy than the atomic orbitals by an amount β. Occupation of this orbital will tend to stabilize the molecule and promote bonding between the two atoms. The molecular orbital is said to be a bonding molecular orbital. c. The σ * 1s orbital is higher in energy (less stable) than the atomic orbitals by an amount β. Occupation of this orbital will tend to destabilize the molecule and detract from bonding. It is called an antibonding molecular orbital. Antibonding molecular orbitals are identified by a right-hand superscript asterisk ( * ). d. In more complex molecules it is also possible to have nonbonding molecular orbitals. These orbitals have the same energy at their input atomic orbitals. Occupation of the orbitals should neither promote nor detract from bonding. e. In general, the more electrons one has in bonding compared to antibonding molecular orbitals, the more stable will be the molecule. 3. Filling of the two lowest energy orbitals. a. onsider + 2 molecule-ion. This is a one electron system. 1) The electron will occupy the σ 1s MO and + 2 will have a σ 1 1s configuration and be stabilized by an amount β compared to the separated atoms. Therefore, + 2 should be stable. 3

4 2) This is a known substance whose bond dissociation energy is 255 kj and bond distance is 106 pm. b. onsider the 2 molecule. This is a two electron system. 1) The two electrons will occupy the σ 1s MO and will have their spins paired. Since both electrons are stabilized by β, to give an over-all stability of 2β, the molecule is stable. 2) 2 is the form of elemental hydrogen. It is diamagnetic, has a bond energy of 431 kj, and a bond distance of 74 pm. c. onsider the e + 2 molecule-ion. This is a three electron system. 1) Two of the three electrons will occupy the σ 1s MO while the third will be in the σ * 1s MO. Since two electrons are stabilized and only one is destabilized, the ion should be stable. 2) e + 2 is a known substance whose bond energy is 241 kj. d. onsider e 2 molecule. This is a four electron system. 1) Two electrons will occupy the σ 1s MO and two will be in the σ * 1s MO. Since two electrons are stabilized by β and two are destabilized by β, there should be no net stability and the molecule should not be stable. 2) The e 2 molecule has never been observed. 4. Molecular orbital electron configurations and bond order. a. Molecular orbital electron configurations can be written in the same manner as atomic electron configurations. 1) The molecular orbitals are written in order of increasing energy and the population of each MO is given as a right-hand superscript. 2) Examples. + 2 σ 1 1s 2 σ 2 1s e + 2 σ 2 1s σ * 1s 1 e 2 σ 2 1s σ * 1s 2 b. The stability will depend on the relative occupation of bonding MO's versus antibondingmo's. This can be measured by the bond order (BO). BO = 1 2 ( number of electrons in bonding MO's - number of electrons in antibonding MO's) 4

5 Examples: BO in + 2 = 1 2 ( 1-0) = 1 2 BO in e + 2 = 1 2 ( 2-1) = 1 2 BO in 2 = 1 2 ( 2-0 ) = 1 BO in e 2 = 1 2 ( 2-2 ) = 0 c. Note that the higher the BO, 1) the higher the bond energy. 2) the shorter the bond distance for the same atoms. d. A substance with a BO = 0 should not be stable.. igher energy MO's. 1. The next set of MO's can be obtained by taking linear combinations of the next higher energy atomic orbitals. These are the 2s and 2p atomic orbitals. Recall that the 2s energy is lower than that of the 2p atomic orbital energies. 2. The 2s AO's can be combined to give a bonding σ 2s MO ( Ψ 2sA + Ψ 2sB ) and an antibonding σ * 2s MO ( Ψ 2sA - Ψ 2sB ). These will generally have the same shapes as do the corresponding MO's obtained from the 1s AO's. 3. The 2p orbitals. a. We must now worry about the orientations of the AO's. Assume that the bonding axis is the x axis. The 2p x will have σ symmetry and the 2p y and 2p z will have π symmetry with respect to this axis. b. The 2p x orbitals combine to form σ MO's. subtract! 2px A! 2px B "# 2p x! 2px A! 2px B add " 2p x! 2px A +! 2px B 5

6 c. The 2p y orbitals will combine to form π MO's as will the 2p z orbitals. The 2p z orbitals could be combined similarly.! 2py A! 2p y B! 2pz A! 2p z B "# 2p y or "# 2pz! 2py A +! 2py B! 2py A! 2pz A! 2py B! 2pz B! 2pz A +! 2pz B " 2p y or " 2pz * * The π 2 py and the π 2 pz will have the same energy and the π 2py and the π 2pz will also have the same energies. 4. Energies. a. The energies change with atomic number as shown below. 6

7 a. At N 2 and below, the relative energies of these MO's are σ 2s < σ * 2s < π 2py = π2p z < σ2p x < π * 2p y = π * 2p z < σ * 2p x b. Above N 2 the relative energies are σ 2s < σ * 2s < σ 2p x < π 2py = π2p z < π * 2p y = π * 2p z < σ * 2p x c. Note that the π2p y and π2p z orbitals have the same energy, that is, are degenerate, as are the π * 2p y and π * 2p z orbitals. 5. Orbital filling. a. A total of 16 more electrons can be accommodated in these orbitals for a total of 20 electrons. b. Examples: Substance N Electron onfiguration B.E rx-x Magnetism BO N2 14 σ 2 1s σ* 2 1s σ2 2s σ* 2 2s π2 2py π2 2pz σ2 2px O2 16 σ 2 1s σ* 2 1s σ2 2s σ* 2 2s σ2 2px π2 2py π2 2pz π* 1 2py π* 1 2pz F2 18 σ 2 1s σ* 2 1s σ2 2s σ 2 2px σ2 2px π2 2py π2 2pz π* 2 2py π* 2 2pz diamagnetic paramagnetic diamagnetic 1 II. Other molecular systems. A. eteronuclear diatomic molecules. 1. The energies of the input atomic orbitals will not be the same. This will give rise to unequal splitting as shown below. Ψ a * = 3 Φ A 4 Φ B Φ B ENERGY Φ A Ψ b = 1 Φ A + 2 Φ B Atom A Molecule AB Atom B 7

8 a. The bonding MO will have more of the character of ψ A than ψ B, that is, the weighting factor 1 will be greater than 2. onversely, the antibonding MO will have more of ψ B character than ψ A character, that is, 4 > 3. b. The greater the energy difference between the two atomic orbitals, the greater will be the disparity in the weighting factors. If the energy difference is too much, then Ψ b will become identical with ψ A and Ψ a will become identical with ψ B. This means that two AO's do not interact. Therefore, only AO's of comparable energies can be combined to form MO's. 2. This is the case in a molecule such as F. X 1s ENERGY XX XX 2p X XX XX XX XX XX 2S F F The 1s orbital of is too high in energy interact with either the 1s or the 2s F orbitals. It can interact with the F 2p orbital having σ symmetry, to form MO's as in the above diagram. In this case the 1s orbital would be ψ B (the higher energy AO) and the F 2p σ would be ψ A. a. The F molecule has a total of 8 valence electrons, but only two are in delocalized MO's. 8

9 The 1s and 2s F AO's do not interact because of energy and only one of the three F 2p AO has the correct symmetry. b. There are only two electrons to be distributed in the MO's (one from and one from F). The electrons will occupy Ψ b and the molecule should be stable. Note that the electron density will be polarized towards the F ( 1 > 2 ). The bond will be a polar bond. 3. In general, valence orbitals are of comparable energies and are the ones that can be combined to form MO's. Therefore, for most molecules the inner core electrons can be ignored and valence orbitals used to form the MO's that are filled by the valence electrons of the constituent atoms. 2. onsider the SO 3 molecule. It is a trigonal planar molecule containing 24 valence electrons. a. Assume that the atoms are sp 2 hybridized and that the molecular plane is the xy plane. Therefore, the p z orbital on each atom will have π symmetry. Of the 24 electrons, 6 will be involved in π bonding. b. The qualitative π energy level diagrams and rough sketches of the MO's, in terms of the input AO's, are shown on the following page. c Note that two electrons occupy the bonding MO, Ψ 1, while four are in the nonbonding MO's, Ψ 2 and Ψ 3. Therefore, π interactions should stabilize the molecule. d. Also note that, considering the sum total of electron density from all the occupied MO's, electron density is evenly distributed between the sulfur and the three oxygens. Thus, the oxygens should be equivalent. In the valence bond theory such equivalency could be obtained only by using resonance. In molecular orbital theory the electrons are naturally delocalized and resonance is not necessary. Molecular orbital theory is preferred in treating molecules with delocalized π electrons 9

10 ENERGY O S O O " MO = 1! 1 + 2! 2 + 3! 3 + 4!! 1! 4! 2! 3! 4! 4! 3 " 4! 3! 2 " 1 " 2 " 3! 2 S! 1 SO 3 O's! 1 QUALITATIVE ENERGY DIAGRAM AND ORBITAL SKETES OF TE SO 3 Π MOLEULAR ORBITALS. 10

e. Overall electron density in the molecule. 3. onsider benzene, 6 6. It is a planar molecule consisting of six sp 2 hybridized carbons that form a prefect hexagonal ring.

The net effect of the resonance structures is that the π electrons are evenly distributed about the ring. b.

11 e. Overall electron density in the molecule. 3. onsider benzene, 6 6. It is a planar molecule consisting of six sp 2 hybridized carbons that form a prefect hexagonal ring. A hydrogen is bonded to each carbon, as shown below. a. The molecule exhibits resonance. The two major resonance structures are those shown above. The net effect of the resonance structures is that the π electrons are evenly distributed about the ring. b. To show this symmetric distribution of electron density, benzene is written as shown below. The circle signifies that the π electrons are delocalized about the planar ring. Benzene is a very stable molecule and forms the basis of aromatic organic compounds. The bonding is best treated using MO theory. 11

12 ENERGY c. In the molecular orbital approach to describing the π electronic structure, linear combinations of the following six p π orbitals on benzene are taken to construct six MO's.!! 2 3! 1! 4! 6! 5 " MO = 1! 1 + 2! 2 + 3! 3 + 4! 4 + 5! 5 + 6! 6 d. There are six π electrons which can then be placed in the MO's. The following diagram shows the relative energies, atomic orbital contributions, and filling of these MO's. Note that the net effect is that the π electron density is evenly distributed around the six member ring. 2# B! B! A # E 2! E2(a)! E2(b)! E2(a)! E2(b) 0 "# E 1! E1(a)! E1(b)! E1(a)!E1(b) "2# A! A! A Qualitative Energy Level Diagram and Orbital Sketches for the! Orbitals of Benzene 12

13 e. π Molecular Orbitals. f. Overall π electron density 13

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Chemistry 6 (9 am section) Spring Covalent Bonding

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