MOLECULAR STRUCTURE. Molecular Structure - B. Molecular Structure - B. Molecular Structure - B. Molecular Structure - B. Molecular Structure - B

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1 MOLECULAR STRUCTURE Molecular Orbital all orbitals of the appropriate symmetry contribute to a molecular orbital. Bundet Boekfa Chem Div, Faculty Lib Arts & Sci Kasetsart University Kamphaeng Saen Campus 1 2 π orbitals The two 2p x orbitals overlap to give a bonding and antibonding π x orbital, and the two 2p y orbitals overlap to give two π y orbitals. The overlap integral The extent to which two atomic orbitals on different atoms overlap is measured by the overlap integral, S: 3 4 The electronic structures of homonuclear diatomic molecules The electronic structures of homonuclear diatomic molecules The molecular orbital energy level diagram for homonuclear diatomic molecules. The lines in the middle are an indication of the energies of the molecular orbitals that can be formed by overlap of atomic orbitals

2 N 2 molecule The ground-state configuration of N 2 is 1σ g2 1σ u2 1π u4 2σ g2. A measure of the net bonding in a diatomic molecule is its bond order, b: b = 1/2 (n n*) b = 1/2 (8 2) = 3 7 Bond Order The bond order is a useful parameter for discussing the characteristics of bonds, because it correlates with bond length and bond strength: 1 The greater the bond order, the shorter the bond. 2 The greater the bond order, the greater the bond strength. 8 Heteronuclear diatomic molecules HF molecule The electron distribution in the covalent bond between the atoms in a heteronuclear diatomic molecule is not shared evenly because it is energetically favourable for the electron pair to be found closer to one atom than the other. A polar bond, a covalent bond in which the electron pair is shared unequally by the two atoms. 9 The accumulation of the electron pair near the F atom results in that atom having a net negative charge, which is called a partial negative charge and denoted δ. There is a matching partial positive charge, δ +, on the H atom. The general form of the molecular orbitals is ψ = c H χ H + c F χ F 10 Electronegativity The electronegativity is a parameter introduced by Linus Pauling as a measure of the power of an atom to attract electrons to itself when it is part of a compound. χ A χ B = 0.102{D(A-B) ½ [D(A-A) + D(B-B)]} ½ Robert Mulliken present other definition: χ M = ½(I + Eea) where I is the ionization energy of the element and Eea is its electron affinity 11 The variation principle If an arbitrary wavefunction is used to calculate the energy, the value calculated is never less than the true energy. The arbitrary wavefunction is called the trial wavefunction. 12 2

3 (αa E)cA + (β ES)cB = 0 (β ES)cA + (αb E)cB = 0 α is Coulomb integral β is Resonance integral The first step is to express the two integrals in terms of the coefficients To solve the secular equations for the coefficients we need to know the energy E of the orbital. As for any set of simultaneous equations, the secular equations have a solution if the secular determinant, the determinant of the coefficients, is zero To find the energies E of the bonding and antibonding orbitals of a homonuclear diatomic molecule set with αa =αb =α The solutions of this equation are

4 when the two atoms are the same, and we can write αa =αb =α, the solutions are The second simple case is for a heteronuclear diatomic molecule but with S = 0 (a common approximation). The secular determinant is then In this case, the bonding orbital has the form the corresponding antibonding orbital is 19 The solutions can be expressed in terms of the parameter ζ (zeta), with 20 Molecular orbitals for polyatomic systems The π molecular orbital energy level diagrams of conjugated molecules can be constructed using a set of approximations suggested by Erich Hückel in the π orbitals are treated separately from the σ orbitals, and the latter form a rigid framework that determines the general shape of the molecule. All the C atoms are treated identically, so all the Coulomb integrals α for the atomic orbitals that contribute to the π orbitals are set equal. For example, in ethene, we take the σ bonds as fixed, and concentrate on finding the energies of the single π bond and its companion antibond. 22 Ethene We express the π orbitals as LCAOs of the C2p orbitals 23 1 All overlap integrals are set equal to zero. 2 All resonance integrals between nonneighbours are set equal to zero. 3 All remaining resonance integrals are set equal (to β ). The assumptions result: 1 All diagonal elements: α E. 2 Off-diagonal elements between neighbouring atoms: β. 3 All other elements:

5 These approximations lead to The roots of the equation are 25 Two possible energy levels are corresponding to the bonding (+ ) and antibonding (- ). The ground state configuration is 1 2. C2p A 2 * The excitation energy from 1 to 2 * is 2. Frontier orbitals HOMO: The highest occupied molecular orbital LUMO: The lowest unoccupied molecular orbital C2p B 26 The building-up principle leads to the configuration 1π 2, because each carbon atom supplies one electron to the π system. The highest occupied molecular orbital in ethene, its HOMO, is the 1π orbital; the lowest unfilled molecular orbital, its LUMO, is the 2π orbital (or, the 2π* orbital). These two orbitals jointly form the frontier orbitals of the molecule. Butadiene electron binding energy: The sum of the energies of each electron. Ethene (1 -bond) Energy levels -electron binding energy Butadiene (2 -bonds) Energy levels -electron binding energy The energy of the butadiene molecule is 0.48 more stable than the sum of two individual bonds. Comparison between the -electron binding energies of ethene and butadiene shows the extra stabilization of a conjugated system called the delocalization energy

6 Benzene Aromatic Stability 1 2 b 2u e 2u e 1g a 2u b 2u e 2u e 2u The -bond formation energy: 8 The delocalization energy: 2 a 2u e 1g e 1g Homework 1. Give the description in the chemical bond in terms of valence-bond theory, overlap integral 2. Give the ground state electron configuration and bond order of (a) CO, (b) NO, (c) O 2 and (d)cn - 3. From the ground-state electron configuration of NO and O 2 predict which molecule should have the shorter bond length? 4. Write the secular equation and energy for homonuclear diatomic molecule. 5. Find the roots of the butadiene secular determinant by Hückel approximation (write E in term of α and β) 6. Predict the delocalization energies of ethene, butadiene and benzene from Hückel approximation 33 References 1. Peter Atkin et. al, Physical Chemistry, Oxford Univ press. 2. David O Hayward, Quantum Mechanics for Chemists, RSC, 184 p. 3. Jack Barrett, Structure and Bonding, RSC press, 178 p. 34 6