CHEM 3340 Physical Chemistry II Molecular Structure Bond Theories PDF

CHEM 3340 Physical Chemistry II Molecular Structure Bond Theories When two atoms approach each other, there is a shift in the electron density and the energy decreases: chemical bond Chemical reactions: break and create bonds Ionic bond: transfer of electron from one atom to another:the consequent a...

CHEM 3340 Physical Chemistry II Molecular Structure Bond Theories When two atoms approach each other, there is a shift in the electron density and the energy decreases: chemical bond Chemical reactions: break and create bonds Ionic bond: transfer of electron from one atom to another:the consequent attraction between the ions keeps the atoms together Covalent bond: Two atoms share a pair electrons Old theory: Lewis, not based on quantum mechanics Oppenheimer Approximation The nuclei are heavy, and move relatively slowly We solve the Schrödinger equation for xed nuclei (easy to evaluate the potential energy) Molecular potential energy curve is obtained: energy vs. internuclear bond separation Equilibrium bond length: Re: minimum of the curve Energy Minimum: De fi Valence Bond (VB) Theory First quantum mechanical theory of bonding Interesting ideas (σ and π bonds) that are still widely used in chemistry 90 % of chemistry can be interpreted More descriptive than predictive Set the state for the development of Molecular Orbital (MO) Theory σ bond in H2 Two nucleus: A and B Two electrons: 1 and 2 Combine the 1s atomic orbitals: ψ=ψA(1)ψB(2) Implies no interactions between electrons However, the two electrons are indistinguishable: ψ=ψA(2)ψB(1) Full wavefunction: ψ=ψA(1)ψB(2)+ψA(2)ψB(1) Enhanced electron density between the two nuclei: σ bond σ bond in N2 Same approach but with pz orbitals Spin pairing: the spins needs to be opposite: two electrons in the same orbital cannot have same spins π bonds in N2 px and py orbitals cannot overlap like σ bonds: They overlap sideby-side Polyatomic molecules H2O: H: 1s1, O:1s2, 2s2, 2px2, 2py1, 2pz1 The 1s orbital of H overlaps with the two p orbitals of O: two σ bonds Predicted bond angle: 90o, experimental: 104.5o Promotion: CH4 For C: 1s2, 2s2, 2px1, 2py1 How do we get CH4? Promotion: with absorption of energy, one of the 2s electrons jumps to 2pz The energy required for promotion will be compensated by the bond energy 4 C atomic orbitals (2s1, 2px1, 2py1, 2pz1 are combined with the 1s orbitals of H: 1 ‘s’ type σ bond, and 3 ‘p’ type σ bonds Problems: the σ bonds in methane are equivalent with a bond angle of 109.5o Hybridization The electrons interact with each other -> new wavefunctions are created by combinations of the atomic orbitals of C sp3 hybridization: 4 new wavefunctions h1 = s + px + py + pz h2 = s - px - py + pz h3 = s - px + py - pz h4 = s + px - py - pz The sp3 hybrid orbitals are combined with the H 1s orbitals in the molecule Other hybridizations exist: for ethylene sp2, for the ethyne, sp (the remaining p orbitals form π bonds) More like: we know the structure and come up with the proper hybridization Resonance Forms Sometimes there are more then one possible combinations of structures The wavefunction (Resonance hybrid) is a combination of these: Benzene Ψ = c1 ΨI + c2 ΨII c1 and c2 are equal in this case Resonance decreases the energy ΨI ΨII VB theory has a lot of problems: too arbitrary, hybridization, resonances, + cannot interpret many experimental facts: O2 has overall spin moment (s=1) Molecular Orbital Theory Solves many of the problems of VB Theory Approach similar to atomic orbitals: 1. solve Schrödinger equation for one electron for the molecule -> energies and orbitals 2. place electrons on molecular orbitals The one-electron wavefunction (Molecular Orbital) spreads through the molecule Simplest to start with: H2+ Schrödinger Equation for H2+ 2 2me d d d + + dx2 dy 2 dz 2 2 2 2 ⇥ + V (x, y, z) =E V = 1 2 e 4⇥ 0 rA 1 Exact solutions are complicated -> approximate methods + 1 rB 1 1 R ⇥ Linear Combination of Atomic Orbitals LCAO We nd the solutions as linear combination of atomic orbitals Example: Ψ (MO) = Ψ(1s electron around nucleus A) + Ψ(1s electron around fi nucleus B) Atomic orbitals: for clarity use χ MO: Ψ = c1 χ(H1s, A) + c2 χ(H1s, B) Even simpler notation: Ψ = c1 A + c2 B C1 and C2 are weight, and adjusted to satisfy the Schrödinger equation Simpler method: Variational principle an approximate solution always has higher energy than a real solution -> vary c1 and c2 until energy is minimized + ollect your thoughts You need to find the factor N+ such that at the minimum value of E occurs σ whole of space. ψ *ψ dτ = 1, where the integration is over the ration To proceed, you should substitute the LCAO into this integral nd make use of the fact that the atomic orbitals are individully normalized. 49 been calculated using the two H1s orbitals The linear1 combination 1ψ− in eqn 9B.2 has higher ψ = e ψ = e π π ( ) ( a ) a than ψ+, and for now it is labelled σ* because it is and noting that orbital r and r has are nota independent (1). When orbital. This nodal plane perpendicula = 0.46 3 1/2 0 A1 3 1/2 0 − rB1 /a0 B1 expressed in Cartesian coordinates based on atom A (2), 2 2 2 1/2 2 2 these 2+ radii are given by rA1 = {x + y + z } and r B1 = {x + y + (z − R)2}1/2, where R is the bond length. A repeat of the analysis for ψ− gives the results shown in Fig. 9B.3. Ψ =1 N(A 1+ B) S # # $ !"$ !#"#$   !" c1=c2, N: normalization constant  = 0.39 ∫ψ *ψj0dτ/ =aN0  ∫ψ dτ + ∫ψ dτ + 2 ∫ψ ψ dτ  = 2(1+ S )N Ψ has cylindrical symmetry around the where τ and has a value that depends on the nucle29 j0S/=a∫ψ0 ψ dinternuclear axis: σ bond r separation (this ‘overlap integral’ will play a significant role 2 A 2 + 2 B A 2 + B B ater). For the integral to be equal to 1, B Lowest energy solution to H : he solution Substitution of the wavefunction gives A − rA1 /a0 Bonding Orbitals A It is also called 1σ, or 1σg Region of constructive interference e y rA1 0 A rB1 z x 1 =g: 7.21 eV, j 10.7 eV, and k = 7.9 eV. B gerade, German ‘even’ symmetry R N = z – R {2(1+ S )} ween the bonding MO and the H1s Probability density: tious with rounding) is SE≈σ0.59, − EsoH1sN ==0.56. or H at its equilibrium bond length 2 2 2 2 (A2+B2 + 2AB) Ψ = N (A+B) =N 2 356 9 Molecular structure elf-test 9B.1 Normalize the orbital ψ in eqn 9B.2 and evaluA2: probability density of an 1s atomic te N for S = 0.59. orbital around nucleus A Figure 9B.7 A representation of the constructive interfer Centre of B2: probability densitynotation of an 1s atomic 9B.2 Orbital center of inversion inversion of E σ against R relative tonucleus the energy that occurs when two H1s orbitals overlap and form a bon orbital around B + 1/2 + 2 + − − Answer: N− = 1/{2(1 − S)}1/2, so N− = 1.10 The energy of the σ orbital decreases For homonuclear diatomic orbital. molecules (molecules consist- Figure 9B.1 shows the contours of constant amplitude for Overlap density: AB, extra ing of two atoms of the same element, such as N2), it proves e molecular orbital ψ+ in eqn 9B.2. Plots like these are readcontribution, and B orbital overlap-> helpfulwhere to label aAmolecular according to its inversion obtained using commercially available software. The calcusymmetry, the behaviour the wavefunction when it is inelectron accumulates whereof the ion is quite straightforward, because all that it is necessary vertedforms through formally, the centre of indo is to feed inatomic the mathematical of thethe twocentre atomic(more orbital wavefunctions overlap: version, Topic 10A) of the molecule. Thus, any point on the bitals and then let the software do the rest. + + + σg Figure 9B.12 The parity of an orbital is eve wavefunction is unchanged under inversio constructive interference bonding σ orbital that is projected through the centre of the of symmetry of the molecule, but odd (u) i molecule and out an equal distance on9B.2 the other side leads toamplitude Figure Surfaces of constant of the wavefunction changes sign. Heteronuclear diatomic mol an identical value (and sign) of the wavefunction (Fig. 9B.12). ψ+ of the hydrogen molecule-ion. z 1 {2(1+ S )}1/2 x z–R B R at its equilibrium bond length S ≈ 0.59, so N+ = 0.56. Antibonding Orbitals 2 9B.1 Normalize the orbital ψ− in eqn 9B.2 and evalu- or S = 0.59. Answer: N− = 1/{2(1 − S)}1/2, so N− = 1.10 Higher energy solution to H2+: Ψ = N(A - B) 9B.1 shows the contours of constant amplitude for Ψ has cylindrical symmetry around cular orbital ψ+ in eqn 9B.2. Plots like these are readthe internuclear axis: σ bond ned using commercially available software. The calcu- quite straightforward, because all that it is necessary mathematical It is alsoforms called 1σ* , 1σu o feed in the of the two atomic nd then let the software do the rest. ‘odd’ symmetry u: ungerade, 56 9 Probability Molecular structure Region of destructive interference Figure 9B.2 Surfaces of constant amplitude of the wavefunction density: ψ of the hydrogen molecule-ion. Ψ2 = N2 (A-B)2=N2 (A2+B2 - 2AB) + Overlap density: AB, this time the Centre of electron will be positioned mostly B.2 Orbital notation inversion – of the destructive interferen + where the two two orbitals do notFigure 9B.8 A representation overlap and form or homonuclear diatomic molecules (molecules consist-occurs when two H1s orbitals + – an antibon overlap-> destructive interference ng of two atoms of the same element, such as N2), it provesorbital. + + aThe energy is higher the elpful to label molecular orbital accordingbecause to its inversion ymmetry, the behaviour of the each wavefunction nuclei repel other when it is inσg σu erted through the centre (more formally, the centre of inFigure 9B.12 The parity of an orbital is even (g) if its ersion, Topic 10A) of the molecule. Thus, any point on the (b) wavefunction is unchanged under inversion through the centre onding σ orbital that is projected through the centre of the of symmetry of theamplitude molecule,ofbut (u) if the wavefunction.1 (a) Theand amplitude the bonding molecular Figureto 9B.3 Surfaces of constant theodd wavefunction molecule out anofequal distance on theorbital other side leads gen molecule-ion in a plane containing the two nuclei changes sign. Heteronuclear diatomic molecules do not have a ψ of the hydrogen molecule-ion. Energies of bonding and antibonding MOs Energies depend on the distance between the nuclei Populate the electrons: ◊lowest energy available ◊consider the Pauli’s exclusion principle (two electrons can occupy one orbital with paired spins) ◊ Hund’s maximum multiplicity rule (degenerate orbitals are occupied by parallel spins) Molecular Orbital Energy Diagram Energies of the MO as a result of LCAO (note that the energy of antibonding MO is increased higher than the decrease of the bonding MO) From N atomic orbitals we can build N MOs For H2: energy gain MO Energy diagram of guration: 1σg2, 1σu2 fi The overall energy (at this distance) is larger then the separate atoms: He2 is unstable compared to He MO orbitals for n=2 In general, the solution to Schrödinger equation is expressed as a combination of atomic orbitals For chemical properties: valence orbitals are important Typically, the solutions show that combination of orbitals of same symmetry are dominant in certain solution For n=2, we have s, px, py, and pz atomic orbitals Three sets of solutions: 4 σ orbitals: combination of s and pz (two nuclei) 2 π orbital: combination of py 2 π orbital: combination of px σ orbitals for n=2 Solutions are sought by combining s and pz orbitals of nuclei A and B Ψ = c1 χ(H2s, A) + c2 χ(H2s, B) + c3 χ(H2pz, A) + c4 χ(H2pz, B) Energy The solutions are quite simple: ◊ Bonding σ orbital: 1σ = χ(H2s, A) + χ(H2s, B) ◊ Anti-bonding σ orbital 1σ*= χ(H2s, A) - χ(H2s, B) ◊ Bonding σ orbital: 2σ = χ(H2pz, A) + χ(H2pz, B) ◊ Anti-bonding σ orbital: 2σ*= χ(H2pz, A) - χ(H2pz, B) Energy Diagram for σ orbitals 2σ* 2pz 2pz 2σ 1σ* 2s 1σ 2s π Orbitals for n=2 Combining px orbitals of nuclei A and B Ψ = c1 χ(H2px, A) + c2 χ(H2px, B) Solution: ◊ Bonding π orbital: πx = χ(H2px, A) + χ(H2px, B) ◊ Anti-bonding π orbital: πx* = χ(H2px, A) - χ(H2px, B) Complication: symmetry is different: π = πu, π* = πg Similar solution for py orbitals: πy = χ(H2py, A) + χ(H2py, B) πy* = χ(H2py, A) - χ(H2py, B) πx and πy orbitals are different, but they have the same energies! Energy Diagram for π orbitals π* 2px, 2py 2px, 2py π The two π orbitals have the same energy! Energy Diagram for n=2 electrons Where are the energy of σ and π orbitals relative to each other? The electrons are not ‘shared’ equally: polar bond Example: HF, F has a partial negative, H has a partial positive charge MO interpretation: Ψ = c1 χ( A) + c2 χ(B) ◊ Nonpolar bond: c12=c22 , equal contribution ◊ Polar bond: c12< c22. ◊ Ionic bond: c12=0, c22=1 The atomic orbital with the lower energy (e.g., B) makes the larger contribution to the bonding orbital (c12< c22) The atomic orbital with the higher energy (e.g., A) makes the larger contribution to the anti-bonding orbital (c12> c22) MO orbitals for HF Electronegativity For two elements, χA, χB can be obtained from bond dissociation energies E(A-B) |χA - χB | = 0.102 x [∆E/(kJ/mol)]0.5 where: ∆E = E(A-B) - 0.5{E(A-A) + E(B-B)} Alternative de nition: χ = 0.5 (I + Ea) I: ionization energy [eV]: energy required to remove the highest energy electron from an atom (A -> A+ + e-) Ea: electron af nity [eV]: energy required to add one electron to the atom (A + e- -> A-) fi The extent bond polarity can be predicted from electronegativity numbers (χ): power of an atom to draw electrons to itself when it is part of a compound fi MO energy diagram for NO S=1/2, NO is paramagnetic 1 π: predominantly localized on O HOMO: highest occupied molecular orbital: 2π, one electron, predominantly localized on N LUMO: lowest unoccupied molecular orbital: 4σ Since NO is radical, it is reactive. Half life in cells: 1-5s Polyatomic molecules H2O More atomic orbitals are used to construct MOs The MOs spread over the whole molecule Example water: Nucleus A: χ(H 1s) Nucleus B: χ(O 2s), χ(O 2px) , χ(O 2py), χ(O 2pz) Nucleus C: χ(H 1s) General MO solution: ψ = c1 χ(H 1s, A) + c2 χ(O 2s, B) + c3 χ(O 2px, B) + c4 χ(O 2py, B) + c5 χ(O 2pz, B)+ c6 χ(H 1s,C) 6 atomic orbitals -> 6 solution with 6 energies Naming comes from symmetry HOMO: non-bonding, mostly on O cult General form for combination of N atomic orbitals (χ1, χ1,...., χN ) with coef cients c1, c2,..., cn k=1,...,N equations (Hk,1 - E Sk,1)c1 + (Hk,2 - E Sk,2)c2 +....+ (Hk,n - E Sk,n)cn=0 fi fi E: energy Only π orbitals are considered: the σ bonds are rigid and give a ‘frame’ for the molecule Ethene MO: ψ = c1 χ(C2pz, A)+ c2 χ(C2pz, B) (H1,1 - E S1,1)c1 + (H1,2 - E S1,2)c2 =0 (H2,1 - E S2,1)c1 + (H2,2 - E S2,2)c2 =0 Approximations: t’) to experimental data (spectrum, bond energies) Ab initio methods: Solve Schrödinger equation numerically Computation extensive -> simpli cations GTO: Gaussian type orbital: LCAO is too complex, we can use Gaussian-type orbitals (e-r) DFT: density functional theory: solves electron densities instead of orbitals (rewrites Schrödinger equation for electron densities) fi fi Problem: consider electron repulsion SCF: self consistent eld approximation Step 1. take an initial guess of MOs Step 2. calculate potential energy considering the electrons Step 3. solve the Schrödinger equation Step 4: With the new solution update the potential energy Step 5: Repeat steps 2-4 until convergence is reached fi Graphical Output of Computations Actual output: c1,c2,...,cn of atomic orbitals for each MO Graphical output: stylized shapes to represent solution: the boundary surfaces with 90% probability, colored with the sign of the wavefunctions Benzene Graphical Output of Computations Isodensity Surface Total electron density: sum of the squares of the wavefunctions Surface of constant electron density: isodensity surface Solid Transparent Mesh Ethanol Graphical Output of Computations Electrostatic potential surface diagram Net charge on each point on the isodensity surface: charge (potential) from nuclei - charge (potential) from electrons Electrostatic potential surface Many programs for these calculations/plots for catalysis and drug design