Lecture 15 Chemistry 362 M. Darensbourg 2017 Spring term Molecular orbitals for heteronuclear diatomics (HHe, HF and CO) Delocalized MO’sSlide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Electron Configurations: Assign electrons to energy levels but put in horizontal scheme:Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Lecture 15 Chemistry 362 M. Darensbourg 2017 Spring term Molecular orbitals for heteronuclear diatomics (HHe, HF and CO) Delocalized MO’sThe earth’s atmosphereIt is both beautiful and precious.Mixing of s and p orbitals No mixing of s and p orbitalsConsequence of no mixing on energy level diagram: Homonuclear Diatomics No mixing of sigma orbitals Derived from 2s and 2pz Core (closed shell) electrons; little effect.Copyright © 2014 Pearson Education, Inc. Homonuclear Diatomics No mixing vs. mixing of sigma orbitals Derived from 2s and 2pz No mixing (O2 and F2) Mixing (B2, C2, N2)A AA22p2s2p2sσ2sσ*2sσ2pzπ2px,π2pyπ*2px,π*2pyσ*2pzM.O.Energy Level Diagram for A2 (A = O )O2Electronic configuration: σs2σ*s2σpz2πpx2πpy2π*px1π*py1Note Hund's rule again! Bond order = (8 - 4)/2 = 2(double bond) and PARAMAGNETIC.V.B. theory could not explain paramagnetism.Electron Configurations: Assign electrons to energy levels but put in horizontal scheme: [KK] < σ2s < σ*2s<s2pz<(π2px, π2py) < (π*2px, π*2py) < σ*2pz [KK] < σ2s < σ*2s< (π2px, π2py) <σ2pz < (π*2px, π*2py) < σ*2pz For no mixing: O2, F2, Ne2 Assign electrons and determine bond order, magnetism, and Term Symbols! B.O. = (# bonding electrons - # anti-bonding electrons)/2 For mixing: B2, C2, N2 But what about heteronuclear diatomics: CN-, NO, CO?HETERONUCLEAR DIATOMIC MOLECULESSimplest would be HHe. Differs from H2 in two ways:(1) A.O. energies for H, He different. He - greater nuclear charge, electrons more tightly bound.(2) Now three electrons to feed into m.o.'s.Energy level diagram is now:1sH1sHeψbψaaverage energyof a.o.'sH HHe HeFor heteronuclear diatomics, m.o.'s formed symmetrically above and below AVERAGE energy of constituent a.o.'sΨb = c1(1sH) + c2 (1sHe) where c2 > c1 Ψa= c1(1sH) - c2 (1sHe) where c1 > c2For HHe, bond order = (2 - 1)/2 = 1/2 i.e. v. wk. "1/2" bond - not formed under normal conditions - v. unstable.Unpaired electron, PARAMAGNETIC.Note for "He2" - extra electron in antibonding m.o. - therefore bond order = 0. Molecule does not exist - no force to hold atoms together.He is monatomic gas.More Dramatic for HF: Must use H1s + F 2pzs/pz gives bonding and anti-bonding pair.+++++++––––s/pzσspzσ*spz. ...s/px or py and pz/px or py all non-bonding (positive and negative overlaps cancel. No overlap at all for px/py.++–s/pxors/pynon-bonding++–pz/pxorpz/pynon-bonding–Before moving on to show the energy level diagram for A2 molecules - we need to be clear about the labels for m.o.’sMO Energy Level Diagrams for HF and CO from Shriver, 2.21Carbon Monoxide VSIE O2s = 32.3 eV O2p = 15.8 eV C2s = 19.4 eV C2pz = 10.6 eVCarbon Monoxide Consequence: CO binds to metals, such as iron, Through Carbon rather than Oxygen. Can produce dative bond by donating Electron density from the 3σ to empty orbital On metal, and will accept electron density From the metal via π-backbonding.Molecular Orbitals in Polyatomic Molecules In the vast majority of cases, bonds can be described as LOCALISED between pairs of nuclei - therefore can use VB approach. At this stage we only need to use molecular orbitals (m.o’s) when DELOCALISATION of electrons occurs - i.e. when several Lewis structures have to be drawn. Even in these cases we can use the VB approach for the σ-framework, and construct m.o.’s only for the π-bonds. Just look at TWO examples - one inorganic (CO32-), one organic (benzene, C6H6).Carbonate ion, CO32-.Regular trigonal planar, equal bonds, all angles 1200.σ-framework:use sp2 hybrids on both C and O's.2 electrons in each bond + lone pairs in remaining sp2 hybrids on O's = 18 electrons+++– ––C++–––O++–––O+++–––OTotal number of valence electrons = 4 (C) + 18 (3 O's) + 2 (negative charges) = 24Therefore have 6 electrons to put into π-type orbitalsWhich orbitals are available for π-bonding?On each atom, sp2 hybrid formation uses px and py orbitals. Therefore one pz orbital on each atom isavailable for π-bonding - 4 orbitals in total.Remember general rule:n a.o's n m.o.'sTherefore we will form 4 m.o.'sDetails of how to calculate what these look likeleft until later.O COz++++----By analogy with diatomic m.o.'s - most strongly bonding m.o. will be the one which increases electron density in all of the bonds:O CO++++----The most antibonding m.o. will have NODES on all of the bonds:O CO++++----nodesThe remaining two turn out to be non-bonding, giving the energy level for the π-bonding in CO32- as follows:C 3OCO32-pzpzππ0π∗The 6 electrons just fill the bonding and non-bonding m.o.'s.One π-bond shared out equally in a delocalised m.o.Each bond therefore 1σ + 1/3πAll electrons paired - therefore