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Lecture 27 Molecular orbital theory III Applications of MO theory Previously, we learned the bonding in H2+. We also learned how to obtain the energies and expansion coefficients of LCAO MO’s, which amounts to solving a matrix eigenvalue equation. We will apply these to homonuclear diatomic molecules, heteronuclear diatomic molecules, and conjugated πelectron molecules. MO theory for H2+ (review) j±k E± = E1s + 4pe 0 R 1± S e2 j - k φ– = N–(A–B) E1s + 4pe 0 R 1- S anti-bonding e2 E1s + e2 4pe 0 R - j + k φ+ = N+(A+B) 1+ S bonding E1s φ– is more anti-bonding than φ+ is bonding R MO theory for H2+ and H2 MO diagram for H2+ and H2 (analogous to aufbau principle for atomic configurations) H2+ H2 Reflecting: anti-bonding orbital is more anti-bonding than bonding orbital is bonding Matrix eigenvalue eqn. (review) Y = c A A + cB B L= E-l { ¶ E-l ( ) * Y ò Y dt -1 ¶L ¶L ¶L = = =0 ¶c A ¶cB ¶l ( ò Y Y dt -1)} = ¶{ E - l ( ò Y Y dt -1)} = 0 * * ¶c A ¶cB æ a b A ç çè b a B öæ c ö æ c ö æ ö ÷ ç A ÷ = Eç 1 S ÷ ç A ÷ ÷ø çè cB ÷ø è S 1 ø çè cB ÷ø (a A - E)(a B - E) - (b - ES)(b - ES) = 0 MO theory for H2 A B (a - E)(a - E) - ( b - ES)( b - ES) = 0 (a - E) = ±( b - ES) a + b = (1+ S)E or a - b = (1- S)E a +b a -b E= 1+ S or E = 1- S α is the 1s orbital energy. β is negative. anti-bonding orbital is more anti-bonding than bonding orbital is bonding. MO theory for H2 æ c ö æ a b öæ c ö æ ö A 1 S A ÷ = Eç ç ÷ ç ÷ç ÷ çè b a ÷ø ç cB ÷ ç ÷ S 1 c è ø B è ø è ø E= æ c A c A = cB ® ç çè cB ö æ ÷ =ç ÷ø ç è a +b 1+ S 1 2 1 2 ö ÷ ÷ ø or E = a -b 1- S æ c A c A = -cB ® ç çè cB ö æ 1 2 ÷ =ç ÷ø ç - 12 è ö ÷ ÷ ø MO theory for He2 and He2+ He2 has no covalent bond (but has an extremely weak dispersion or van der Waals attractive interaction). He2+ is expected to be bound. He2 He2+ σ and π bonds A π bond is weaker than σ bond because of a less orbital overlap in π. σ bond π bond MO theory for Ne2, F2 and O2 Hund’s rule O2 is magnetic Ne2 F2 O2 MO theory for N2, C2, and B2 Hund’s rule B2 is magnetic N2 C2 B2 Z E µ- 2 n 2 Polar bond in HF The bond in hydrogen fluoride is covalent but also ionic (Hδ+Fδ–). H 1s and F 2p form the bond, but they have uneven weights in LCAO MO’s . Y = CH1sy H1s + CF2 py F2 p 2 2 2 2 2 Y = CH1s y H1s + CF2 p y F2 p +… Hδ+Fδ– Polar bond in HF Calculate the LCAO MO’s and energies of the σ orbitals in the HF molecule, taking β = –1.0 eV and the following ionization energies (α’s): H1s 13.6 eV, F2p 18.6 eV. Assume S = 0. Matrix eigenvalue eqn. (review) With S = 0, æ a b A ç çè b a B öæ c ö æ c ö A A ÷ç ÷ = Eç ÷ ÷ø çè cB ÷ø çè cB ÷ø ( E - a )( E - a ) - b A a ( E= (a = +aB) ± A B (a 2 =0 2 a + 4 b ) A B 2 2 A + a B ) ± (a A - a B ) 2 æ 2b ö 1+ ç è a A - a B ÷ø 2 Polar bond in HF Ionization energies give us the depth of AO’s, which correspond to −αH1s and −αF2p. ( E= ) ( a H1s + a F2 p ± ) 2 a H1s - a F2 p + 4b 2 2 = -18.8 eV and -13.4 eV æ c H1s ç çè cF2 p ö æ ö 0.19 ÷ =ç ÷ ÷ø è 0.98 ø æ c H1s ç çè cF2 p ö æ ö 0.98 ÷ =ç ÷ ÷ø è -0.19 ø Hückel approximation We consider LCAO MO’s constructed from just the π orbitals of planar sp2 hybridized hydrocarbons (σ orbitals not considered) We analyze the effect of π electron conjugation. ˆ dt . Each pz orbital has the same a = ò pz Hp z Only the nearest neighbor pz orbitals have nonzero b = ò pz Hˆ pz¢ dt . Centered on the nearest neighbor carbon atoms Ethylene (isolated π bond) Coulomb integral of 2pz Resonance integral (negative) c1 c1 E c2 c2 α α β Ethylene (isolated π bond) ( = a + b ® ( c ,c ) = ( Ep * = a - b ® ( c1 ,c2 ) = Ep 1 2 1 2 1 2 ,, E 2 2 2 1 2 1 2 ) ) Butadiene 2 β 2 1 1 1 4 3 3 4 β β æ ç ç ç ç çè 2 3 4 a b 0 0 ö ÷ b a b 0 ÷ 0 b a b ÷ ÷ 0 0 b a ÷ø Butadiene Two conjugated π bonds E 2 1.62 2 0.62 4 4.48 extra 0.48β Two isolated π bonds stabilization = E = 2 2a + 2b = 4a + 4b π delocalization ( ) Cyclobutadiene β 1 1 β β 3 4 4 2 2 3 1 β æ ç ç ç ç çè 2 3 4 a b 0 b ö ÷ b a b 0 ÷ 0 b a b ÷ ÷ b 0 b a ÷ø Cyclobutadiene E 2 2 2 4 4 No delocalization energy; no aromaticity β2 β2 4 3 β1 1 1 2 3 a b1 0 2 1 2 b1 a b2 3 β1 0 b2 a 4 Cyclobutadiene b2 0 b1 æ ç ç ç ç ç è 4 b2 ö ÷ 0 ÷ ÷ b1 ÷ a ÷ø b1 < b < b2 < 0 Cyclobutadiene E = 2 (a + b1 + b2 ) + 2 (a + b1 - b2 ) = 4a + 4b1 Spontaneous distortion from square to rectangle? Homework challenge #8 Is cyclobutadiene square or rectangular? Is it planar or buckled? Is its ground state singlet or triplet? Find experimental and computational research literature on these questions and report. Perform MO calculations yourself (use the NWCHEM software freely distributed by Pacific Northwest National Laboratory). Summary We have applied numerical techniques of MO theory to homonuclear diatomic molecules, heteronuclear diatomic molecules, and conjugated π electron systems. These applications have explained molecular electronic configurations, polar bonds, added stability due to π electron delocalization in butadiene, and the lack thereof in cyclobutadiene. Acknowledgment: Mathematica (Wolfram Research) & NWCHEM (Pacific Northwest National Laboratory)