Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: N2 N2 + O2 O2+ Bond energy (kJ/mol) 945 841 498 623 Bond length (pm) 110 112 121 112 No. of valence electrons 10 9 12 11 Removing an electron from N2 decreases the bond energy of the resulting ion, whereas removing an electron from O2 increases the bond energy of the resulting ion. Explain these facts using M.O diagrams. Plan: We first draw the MO energy levels for the four species, recalling that they differ for N2 and O2. Then we determine the bond orders and compare them with the data: bond order is related directly to bond energy and inversely to bond length. Sample Problem - Continued Solution: The MO energy levels are: N2 N2 + O2 O2+ σ2p* σ2p* π2p* π2p* σ2p π2p π2p σ2p σ2s* σ2s* σ2s σ2s BO: 1/2(8-2) = 3 1/2[7-2] = 2.5 1/2(8-4) = 2 1/2(8-3) = 2.5 Bonding in Heteronuclear vs. Heteronuclear Diatomic Molecules (Zumdahl Section 14.4) Review from Chapter 13: Covalent Ionic Ionic Figure13.11 Covalent Figure 13.12 Homonuclear: H2 Electronegativity Nonpolar covalent bond (450kJ/mol bond) Figure 14.26 Heteronuclear: HF Electronegativity Polar covalent bond (565kJ/mol bond) Figure 14.45 Electrons are not equally shared in heteronuclear bonds HF Electronegativity Figure 14.45 Because F (χ = 4.0) is more electronegative than H (χ = 2.2), the electrons move closer to F. (Table 13.2 for electronegativities) This gives rise to a polar bond: H F M.O.s of a Polar Covalent Bond: HF σ Antibonding (σ*) Mostly H(1s) H F H F σ Bonding Mostly F(2p) When the electronegativities of the 2 atoms are more similar, the bonding becomes less polar. Example: NO 2p 2s N 2s NO O Electronegativity 2p χ(N) = 3.0 χ(O) = 3.4 . .. ..N=O.. Sample Problem: Use the M.O. model to predict the changes in magnetism and bond order upon oxidation of NO to make NO+ NO+ NO oxidation Bond Order = (8-3)/2 = 2.5 Paramagnetic Bond Order = (8-2)/2 = 3 Diamagnetic Combining the Localized Electron and Molecular Orbital Models (into a convenient working model) (Zumdahl Section 14.5) Figure 14.47 Only the π bonding changes between these resonance structures- The M.O. model describes this π bonding more effectively Atomic Orbitals Molecular Orbitals + 3 others at higher energies Figure 14.51 Another example: Benzene σ bonding: π bonding: p atomic orbitals π molecular orbital Molecular Spectroscopy (Zumdahl 14.7) Figure 14.52 Ozone – The Important Molecule z = O3 y 18 valence electrons 9 molecular orbitals x empty πxy πx πz∗ weak absorption hν UV strong absorption at 230 nm πz Electronic Spectroscopy Electronic Transitions Give Molecules Color! Ultraviolet Blue Green Yellow Red Wavenumber These experiments are used, for example, to determine the M.O. diagram experimentally Vibrational Spectroscopy For fun- see animated vibrations @ http://www.cem.msu.edu/~parrill/AIRS/ CH stretch HCH bend CN stretch Wavenumber ∝ 1/(Wavelength) 2000cm-1 = 1/(0.0005cm) = 1/(5000nm) These experiments are used, for example, to determine the identities of molecules in solution Molecular Vibrations z z N free atoms: 3N independent motions y x y z = 3N degrees of freedom x y x Molecule: N bound atoms z 3 translational 3 rotational (non-linear molecule) y 3N – 3 – 3 = 3N – 6 degrees of freedom x for vibrations Ozone: N=3 3 x 3 – 6 = 3 vibrational degrees of freedom Acetonitrile – An Important Solvent N=6 CH3CN 3 x 6 – 3 = 15 vibrational degrees of freedom Maximum of 15 vibrational absorptions in the infrared (IR) spectrum 5 bonds = 5 bond-stretching vibrational modes 15 - 5 = 10 bending (deformation) vibrational modes Rotational Spectroscopy