III. The Molecular Orbital Model

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III. The Molecular Orbital Model A. Problems with the L.E. Model 1. Assumes all e- are localized; this is not what is observed 2. Unpaired e- are not described very well at all 3. Gives us no clue to bond energies B. Molecular Orbital Model 1. Similar to atomic theory, but develops orbitals for a whole molecule 2. Gives molecular orbitals that electrons go into 3. Properties of the molecular orbitals (MO’s) a. Each MO can hold 2 e- of opposite spin b. Shape of the MO represents probability of where e- is C. H 2 Molecular Orbital Description 1. We can construct our MO’s from the AO’s we already know a. MO 1 = 1s a + 1s b b. MO 2 = 1s a - 1s b

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III. The Molecular Orbital Model. Problems with the L.E. Model Assumes all e- are localized; this is not what is observed Unpaired e- are not described very well at all Gives us no clue to bond energies Molecular Orbital Model - PowerPoint PPT Presentation

Transcript of III. The Molecular Orbital Model

  • III. The Molecular Orbital ModelProblems with the L.E. ModelAssumes all e- are localized; this is not what is observedUnpaired e- are not described very well at allGives us no clue to bond energies

    Molecular Orbital ModelSimilar to atomic theory, but develops orbitals for a whole moleculeGives molecular orbitals that electrons go intoProperties of the molecular orbitals (MOs)Each MO can hold 2 e- of opposite spinShape of the MO represents probability of where e- is

    H2 Molecular Orbital Description1.We can construct our MOs from the AOs we already knowMO1 = 1sa + 1sb MO2 = 1sa - 1sb

  • Size, shape, and energy of MOs

    MO1 = s-MOLarge probability between H nuclei = s interactionLower energy than H 1s orbitals = bonding MOThe 2 total valence e- pair up, giving a stabilization = bondMO2 = s*-MOLarge probability bracketing H nuclei = s interactionHigher energy than H 1s orbitals = antibonding MONo electrons in this MOs1ss*1s

  • 5.H2- has three electronsLess stable than H2 One e- in antibonding MO

    Bond Order = Difference between the number of bonding electrons and number of antibonding electrons divided by two.

    Example of B. O.H2 = (2 0)/2 = 1 (single bond strength)H2- = (2 1)/2 = 0.5 (less than single bond strength, but stable)He2 = (2 2)/2 = 0 (not stable, not a diatomic molecule)

  • I. Bonding in Homonuclear DiatomicsLi, Be, BLi2 MO descriptionLi 1s22s1, but we can only use valence e- (2s) in bonding1s orbital is core and cant be reached for bondingMOs for Li2

    Bond Order = (2 0)/2 = 1, so it is stable

    MO electron configuration would be s2s2 Li2 is stable, but only occurs as a gas

  • Be2 uses the same MOs as Li2but is s2s2 s*2s2 Bond Order = (2-2)/2 = 0Be2 is not a stable diatomicB2 must begin to use MOs generated from 2p AOsB = 1s22s22p1 p-orbital MO combinations

  • c.Head-on overlap gives s2p and s*2p interactions

    Side-on overlap gives p2p and p*2p interactions.There are 2 such interactions each, because of the 2Perpendicular p-orbitals not involved in the s-bond.

    2s--2s overlap is more favorable, at lower energy

    Bond Order = (4 2)/2 = 1, stable diatomic

    IV.MagnetismA.Magnetic Balance

  • Diamagnetic = repelled by a magnetic fieldPaired e- are diamagneticH2 is diamagnetic because all of its e- are pairedMost compounds are diamagnetic because paired e- are favorableRepulsion is very weak

    Paramagnetic = attracted by a magnetic fielda.unpaired electrons are paramagneticb.much stronger than diamagnetismc.Relatively few compounds have unpaired e-

    B.Magnetism and MO Theory1.We predicted from MO theory that B2 would be diamagnetic2.Experiment shows us that it is paramagnetic.3.We must adjust our model to explain this

  • We must allow our model to take into account 2s/2p interaction2s/2p mixing changes the ordering of some of our MOsSpacing is no longer symmetricPredicts paramagnetism of B2 d. Bond Order is still 1

  • Other homonuclear diatomicsC2 and N2 use same ordering and MO set as B2O2 and F2 have different ordering because s/p mixing changes

  • Trends in 2nd row diatomicsAs Bond Order increase, Bond Energy increases, Bond Length decreasesBond Order doesnt completely predict Bond EnergyB2 > F2 even though both have BO = 1F2 has much electron repulsion, weakening the bondN2 bond order = (8-2)/2 = 3. This triple bond is very strong and explains why N2 is very unreactive.O2 is paramagnetic.

    V.Heteronuclear Diatomics: Different elements have different A.O. energiesA.Adjacent elements are the most similar (NO, CN, etc)We can use same MOs as for homonuclear diatomics for adjacentsNO Bond Order = (8-3)/2 = 2.5

  • MO Theory better handles unpaired e-Paramagnetic

    5.NO+ and CN- have the same configurationBO = (8-2)/2 = 3 (triple bonds)We can use the same figureDiamagnetic

  • Elements far apart have very different energiesWe must devise new MO ordering for each caseHF (use H1s and F2p)F much lower energy (more tightly bound e-)F2p < H1s so the electrons will be closer to F (Predicts Electronegativity!!!)

  • VI. Combining LE and MO ModelsResonance StructuresL.E. model is not very accurate in some cases where more than one Lewis Structure can be drawnResonance structures = equivalent, but not exact representations of the bonding in a molecule

    O3

    NO3-

    Benzene

  • s-bonds can be localized without problems; Use LE Model p-bonds need to Delocalize in some molecules for stabilityStill use LE Model to describe the s-bondingUse MO model to describe p-bondingSimple as possible, but must be accurateBenzene

  • 4. NO3-