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Part i

  1. 1. Inorganic Chemistry Bonding and Coordination Chemistry Books to followInorganic Chemistry by Shriver & Atkins Physical Chemistry: Atkins
  2. 2. Bonding in s,p,d systems: Molecular orbitals of diatomics,d-orbital splitting in crystal field (Oh, Td).Oxidation reduction: Metal Oxidation states, redoxpotential, diagrammatic presentation of potential data.Chemistry of Metals: Coordination compounds (Ligands &Chelate effect), Metal carbonyls – preparation stability andapplication.Wilkinson’s catalyst – alkene hydrogenationHemoglobin, myoglobin & oxygen transport
  4. 4. Failure of Classical Mechanics• Total energy, E = ½ mv2 + V(x)• p = mv ( p = momentum )• E = p2/2m + V(x) ……… . . Eq.1 “ A moving ball I know it all ”
  5. 5. • Newton’s second law is a relation between the acceleration d2x/dt2 of a particle and the force F(x) it experiences. • Therefore, v = p/m • Or, p• = F(x) “ Hit the ball hard, it will move fast Hit it soft, it will move slow” • Continuous variation of energy is possible.Macroscopic World: “Classical Mechanics - the God”
  6. 6. Max Planck E = hν 1900 German physicist A young Max Planck was to give a lecture on radiant heat. When he arrived he inquired as to the room number for the Planck lecture. He was told, "You are much too young to be attending the lecture of the esteemed professor Planck."“Each electromagnetic oscillator is limited to discretevalues and cannot be varied arbitrarily”
  7. 7. Plank had applied energy quantizationto the oscillators in the blackbody buthad considered the electromagneticradiation to be wave.
  8. 8. Hertz J.J. Thomson PHOTOELECTRIC EFFECTWhen UV light is shone on a metal plate in a vacuum, it emitscharged particles (Hertz 1887), which were later shown to beelectrons by J.J. Thomson (1899). Classical expectations Vacuum Light, frequency ν As intensity of light increases, force chamber Collecting increases, so KE of ejected electrons Metal should increase. plate plate Electrons should be emitted whatever the frequency ν of the light. Actual results: I Maximum KE of ejected electrons is Ammeter independent of intensity, but dependent on νPotentiostat For ν<ν0 (i.e. for frequencies below a cut- off frequency) no electrons are emitted
  9. 9. Photoelectric Effect.
  10. 10. (i) No electrons are ejected, regardless of the intensity of the radiation, unless its frequency exceeds a threshold value characteristic of the metal.(ii) The kinetic energy of the electron increases linearly with the frequency of the incident radiation but is independent of the intensity of the radiation.(iii) Even at low intensities, electrons are ejected immediately if the frequency is above the threshold.
  11. 11. Major objections to the Rutherford-Bohr model • We are able to define the position and velocity of each electron precisely. • In principle we can follow the motion of each individual electron precisely like planet.• Neither is valid.
  12. 12. Werner HeisenbergHeisenbergs name will always be associated withhis theory of quantum mechanics, published in1925, when he was only 23 years. • It is impossible to specify the exact position and momentum of a particle simultaneously. • Uncertainty Principle. ∀ ∆x ∆p ≥ h/4π where h is Plank’s Constant, a fundamental constant with the value 6.626×10-34 J s.
  13. 13. Einstein h ν = ½ mv2 + φ• KE 1/2mv2 = hν- φ∀ φ is the work function• hν is the energy of the incident light.• Light can be thought of as a bunch of particles which have energy E = hν. The light particles are called photons.
  14. 14. If light can behave as particles,why not particles behave as wave?Louis de BroglieThe Nobel Prize in Physics 1929French physicist (1892-1987)
  15. 15. Wave Particle Duality• E = mc2 = hν• mc2 = hν• p = h /λ { since ν = c/λ}∀ λ = h/p = h/mv• This is known as wave particle duality
  16. 16. Flaws of classical mechanicsPhotoelectric effectHeisenberg uncertainty principle limitssimultaneous knowledge of conjugate variablesLight and matter exhibit wave-particle dualityRelation between wave and particle propertiesgiven by the de Broglie relationsThe state of a system in classical mechanics is defined byspecifying all the forces acting and all the position andvelocity of the particles.
  17. 17. Wave equation? Schrödinger Equation.• Energy Levels• Most significant feature of the Quantum Mechanics: Limits the energies to discrete values.• Quantization. 1887-1961
  18. 18. The wave functionFor every dynamical system, there exists a wave function Ψthat is a continuous, square-integrable, single-valued functionof the coordinates of all the particles and of time, and fromwhich all possible predictions about the physical properties ofthe system can be obtained.Square-integrable means that the normalization integral is finiteIf we know the wavefunction we know everything it is possible to know.
  19. 19. Derivation of wave equationTime period = T, Velocity = v, v = λ/T, Frequency, ν = 1/T, v = ν λ
  20. 20. y xIf the wave is moving to the right with velocity ‘v’ at time ‘t’y(x,t) = A sin 2π/λ(x-vt)λ= v/ ν• y = A sin 2πν(x/v - t)• Differentiating y W.R.T x, keeping t constant A wave eqn.• δ 2y/δx2 + (4π 2 / λ 2 ) y = 0 is born
  21. 21. • In three dimension the wave equation becomes:• δ 2 ψ/δx2 + δ 2 ψ/δy2 + δ 2 ψ/δz2 + (4π 2 /λ 2 )ψ = 0· It can be written as ∇ 2ψ + (4π 2 /λ 2 )ψ = 0· We have λ = h/mv• ∇ 2ψ + (4π 2 m2v2/h2 ) ψ = 0· E = T + V or T = (E-V) (E = total energy)· V = Potential energy, T = Kinetic energy· T = 1/2 mv2 = m2v2/2m· m2v2 = 2m(E-V)
  22. 22. ∇ 2ψ + (8π 2 m/ h2 )(E - V) ψ = 0• This can be rearranged as• {(− h2/8π 2 m) ∇ 2 + V}ψ = Ε ψ· Hψ = Ε ψ• Η = [(− h2/8π 2 m)∇ 2 + V) Hamiltonian operator{(-h2/8π 2m)(∂ 2/∂x2 + ∂ 2/∂y2 + ∂ 2/∂z2) + V} Ψ = E Ψ δ 2y/δx2 + (4π 2 / λ 2 ) y = 0
  23. 23. How to write Hamiltonian for different systems? {(-h2/8π 2m)∇ 2 + V} Ψ = E Ψ -e • Hydrogen atom: r• KE = ½ m (vx2 + vy2 + vz2) +Ze• PE = -e /r, (r = distance between the 2 electron and the nucleus.)• H = {(-h2/8π 2m) ∇ 2 –e2/r}∀ ∇ 2 Ψ + (8π 2 m/h2)(E+e2/r) Ψ = 0• If the effective nuclear charge is Ze • H = {(-h2/8π 2m )∇ 2 –Ze2/r}
  24. 24. H2+ Molecule e (x,y,z) ra rb A RAB B the wave function depends on the coordinates of the two nuclei,represented by RA and RB, and of the single electron, represented by
  25. 25. e (x,y,z) H2 + ra{(-h /8π m)∇ + V} Ψ = E Ψ 2 2 2 rb A Rab B• PE = V = -e2/ra – e2/rb + e2/Rab• H = (-h2/8π 2m)∇ 2 + ( – e2/ra - e2/rb + e2/Rab)• The Wave equation is∀ ∇ 2 Ψ + (8π 2 m/h2) (E+ e2/ra + e2/rb – e2/Rab) Ψ = 0 Born-Oppenheimer approximation
  26. 26. V = -e2/4πε 0[1/ra+1/rb-1/Rab]
  27. 27. e1 (x1, y1, z1)He Atom r12 r1 e2 (x2, y2, z2) r2 Nucleus (+2e) {(-h2/8π 2m)∇ 2 + V} Ψ = E Ψ• V = -2e2/r1 – 2e2/r2 + e2/r12• H = (-h2/8π 2m) (∇ 12 + ∇ 22) + V• The Wave equation is• (∇ 12 + ∇ 22 )Ψ + (8π 2 m/h2)(E-V) Ψ = 0
  28. 28. e1 (x1, y1, z1) r12 e2 (x2, y2, z2) ra2 H2 ra1 rb2 rb1 A Rab B• PE = V = ?• H = (-h2/8π 2m)(∇ 12 + ∇ 22) + V• The Wave equation is• (∇ 12 + ∇ 22 )Ψ + (8π 2 m/h2)(E-V) Ψ = 0
  29. 29. V = -e2/4πε 0[1/ra1+1/rb1 + 1/ra2 +1/rb2 -1/r12 -1/Rab]attractive potential energy Electron-electron repulsion Internuclear repulsion
  30. 30. Particle in a boxAn electron moving along x-axis in a field V(x) a V=0 x =0 x =a
  31. 31. d2 Ψ /dx2 + 8π 2 m/h2 (E-V) Ψ = 0 a Assume V=0 between x=0 & x=a V=0 d2Ψ/dx2 Ψ [8π 2at x = 20 Ψ a 0 Also + = 0 mE/h ] & = x =0 x =a d2Ψ/dx2 + k2Ψ = 0 where k2 = 8π 2mE/h2 Solution is: Ψ = C cos kx + D sin kx• Applying Boundary conditions:∀ Ψ = 0 at x = 0 ⇒ C = 0 ∴ Ψ = D sin kx
  32. 32. ∀ Ψ = D sin kx a• Applying Boundary Condition:∀ Ψ = 0 at x = a, ∴ D sin ka = 0 V=0• sin ka = 0 or ka = nπ, x =0 x =a• k = nπ/a• n = 0, 1, 2, 3, 4 . . .∀ Ψ n = D sin (nπ/a)x• k2 = 8π 2m/h2[E] or E = k2h2/ 8π 2m• E = n2 h2/ 8ma2 k2= n2 π 2/a2• n = 0 not acceptable: Ψ n = 0 at all x• Lowest kinetic Energy = E = h2/8ma2
  33. 33. An Electron in One Dimensional Box aV=∝ V=∝ ∀ Ψ n = D sin (nπ/a)x • En = n2 h2/ 8ma2 • n = 1, 2, 3, . . . • E = h2/8ma2 , n=1 • E = 4h2/8ma2 , n=2 • E = 9h2/8ma2 , n=3 x=0 x=a Energy is quantized
  34. 34. Characteristics of Wave Function: What Prof. Born Said• Heisenberg’s Uncertainty principle: We can never know exactly where the particle is.• Our knowledge of the position of a particle can never be absolute.• In Classical mechanics, square of wave amplitude is a measure of radiation intensity• In a similar way, ψ 2 or ψ ψ* may be related to density or appropriately the probability of finding the electron in the space.
  35. 35. The wave function Ψ is the probability amplitude 2 ψ = ψ *ψ Probability density
  36. 36. The sign of the wave function has not direct physical significance: thepositive and negative regions of this wave function both correspondsto the same probability distribution. Positive and negative regions ofthe wave function may corresponds to a high probability of finding aparticle in a region.
  37. 37. Characteristics of Wave Function: What Prof. Born Said• Let ρ (x, y, z) be the probability function, ∀ ∫ρ dτ = 1 Let Ψ (x, y, z) be the solution of the wave equation for the wave function of an electron. Then we may anticipate that ρ (x, y, z) ∝ Ψ 2 (x, y, z)• choosing a constant in such a way that ∝ is converted to =∀ ρ (x, y, z) = Ψ 2 (x, y, z)∴ ∫Ψ 2 dτ = 1The total probability of finding the particle is 1. Forcing this condition onthe wave function is called normalization.
  38. 38. ∀ ∫Ψ 2 dτ = 1 Normalized wave function• If Ψ is complex then replace Ψ 2 by ΨΨ *• If the function is not normalized, it can be done by multiplication of the wave function by a constant N such that• N2 ∫Ψ 2 dτ = 1• N is termed as Normalization Constant
  39. 39. Acceptable wave functionsThe wave equation has infinite number of solutions, all of whichdo not corresponds to any physical or chemical reality.• For electron bound to an atom/molecule, the wave function must be every where finite, and it must vanish in the boundaries• Single valued• Continuous• Gradient (dΨ/dr) must be continuous• Ψ Ψ*dτ is finite, so that Ψ can be normalized• Stationary States• E = Eigen Value ; Ψ is Eigen Function
  40. 40. Need for Effective ApproximateMethod of Solving the Wave Equation• Born Oppenheimer Principle.• How can we get the most suitable approximate wave function?• How can we use this approximate wave function to calculate energy E?
  41. 41. Operators“For every dynamical variables there is a corresponding operator” Energy, momentum, angular momentum and position coordinates Operators Symbols for mathematical operation
  42. 42. Eigen valuesThe permissible values that a dynamical variablemay have are those given by αφ = aφφ- eigen function of the operator α thatcorresponds to the observable whose permissiblevalues are a α -operator φ - wave function a - eigen value
  43. 43. αφ = aφIf performing the operation on the wave function yieldsoriginal function multiplied by a constant, then φ is an eigenfunction of the operator α φ = e2x and the operator α = d/dx Operating on the function with the operator d φ /dx = 2e2x = constant.e2x e2x is an eigen function of the operator α
  44. 44. For a given system, there may be various possiblevalues.As most of the properties may vary, we desire todetermine the average or expectation value.We know αφ = aφMultiply both side of the equation by φ *φ *αφ = φ *aφTo get the sum of the probability over all space∫ φ *αφ dτ = ∫ φ *aφ dτa – constant, not affected by the order of operation
  45. 45. Removing ‘a’ from the integral and solving for ‘a’ a = ∫ φ *αφ dτ/ ∫ φ *φ dτα cannot be removed from the integral.a = <φ  α  φ >/ <φ  φ >
  46. 46. Variation Method: Quick way to get E• HΨ = EΨ∀ Ψ HΨ = Ψ EΨ = EΨ Ψ• If Ψ is complex,• E = ∫ Ψ *H Ψ dτ/ ∫ Ψ * Ψdτ• E=<Ψ H Ψ> /<Ψ Ψ> …… (4)• Bra-Ket notation
  47. 47. What does E = <Ψ H Ψ> /<Ψ Ψ> tell us ?• Given any Ψ, E can be calculated.• If the wave function is not known, we can begin by educated guess and use Variation Theorem.Ψ 1 ⇒ E1Ψ 2 ⇒ E2“If a trial wave function is used to calculate the energy,the value calculated is never less than the true energy”– Variation Theorem.
  48. 48. ∀ Ψ 1 ⇒ E1 ∀ Ψ 2⇒ E2 The Variation Theorem tells that • E1 , E2> Eg, Eg true energy of the ground state • IF, E1 > E2,• Then E2 and Ψ 2 is better approximation to the energy and corresponding wave function Ψ 2 to the true wave function
  49. 49. Variation Method: The First Few Steps• We can chose a whole family of wave function at the same time, like trial function with one or more variable parameters C1, C2, C3,….• Then E is function of C1, C2, C3 …….etc.• C1, C2, C3 …. etc. are such that E is minimized with respect to them.• We will utilize this method in explaining chemical bonding.
  50. 50. Chemical Bonding• Two existing theories,• Molecular Orbital Theory (MOT)• Valence Bond Theory (VBT) Molecular Orbital Theory• MOT starts with the idea that the quantum mechanical principles applied to atoms may be applied equally well to the molecules.
  51. 51. H-C≡C-H
  52. 52. MOT: We can write the following principles Describe Each electron in a molecule by a certain wave function Ψ - Molecular Orbital (MO). Each Ψ is defined by certain quantum numbers, which govern its energy and its shape. Each Ψ is associated with a definite energy value. Each electron has a spin, ± ½ and labeled by its spin quantum number ms. When building the molecule- Aufbau Principle (Building Principle) - Pauli Exclusion Principle.
  53. 53. Simplest possible molecule: H2+ : 2 nuclei and 1 electron.• Let the two nuclei be labeled as A and B & wave functions as Ψ A & Ψ B.• Since the complete MO has characteristics separately possessed by Ψ A and Ψ B, ∀ Ψ = CAΨ A + CBΨ B • or Ψ = N(Ψ A +λ Ψ B)∀ λ = CB/CA, and N - normalization constant
  54. 54. This method is known as Linear Combination of Atomic Orbitals or LCAO ∀ Ψ A and Ψ B are same atomic orbitals except for their different origin. • By symmetry Ψ A and Ψ B must appear with equal weight and we can therefore write • λ 2 = 1, or λ = ±1 • Therefore, the two allowed MO’s are ∀ Ψ = Ψ A± Ψ B
  55. 55. For Ψ A+ Ψ B we can now calculate the energy• From Variation Theorem we can write the energy function as• E = <Ψ A+Ψ B H Ψ A+Ψ B> /<Ψ A+Ψ B Ψ A+Ψ B>
  56. 56. Looking at the numerator: E = <Ψ A+Ψ B H Ψ A+Ψ B> /<Ψ A+Ψ B Ψ A+Ψ B>∀ <Ψ A+Ψ B H  Ψ A+Ψ B> = <Ψ A H Ψ A> +• <Ψ B H Ψ B> +• <Ψ A H Ψ B> +• <Ψ B H Ψ A>• = <Ψ A H  Ψ A> + <Ψ B H Ψ B> +2<Ψ AH Ψ B>
  57. 57. = <Ψ A H  Ψ A> + <Ψ B H Ψ B> + 2<Ψ AH Ψ B>ground state energy of a hydrogen atom. let us call this as EA <Ψ A H  Ψ B> = <Ψ B H Ψ A> = β β = resonance integral∴ Numerator = 2EA + 2 β
  58. 58. Physical Chemistry class test answer scripts will be shown tothe students on 3rd March (Tuesday) at 5:30 pm inRoom C-306: Sections 11 and 12
  59. 59. Looking at the denominator: E = <Ψ A+Ψ B H Ψ A+Ψ B> /<Ψ A+Ψ B Ψ A+Ψ B>• <Ψ A+Ψ B  Ψ A+Ψ B> = <Ψ A  Ψ A> +• <Ψ B  Ψ B> +• <Ψ A  Ψ B> +• <Ψ B  Ψ A>• = <Ψ A  Ψ A> + <Ψ B  Ψ B> + 2<Ψ A  Ψ B>
  60. 60. = <Ψ A Ψ A> + <Ψ B Ψ B> + 2<Ψ A Ψ B>Ψ A and Ψ B are normalized,so <Ψ A  Ψ A> = <Ψ B  Ψ B> =1 <Ψ A Ψ B> = <Ψ B Ψ A> = S, S = Overlap integral. ∴ Denominator = 2(1 + S)
  61. 61. Summing Up . . . E = <Ψ A+Ψ B H Ψ A+Ψ B> /<Ψ A+Ψ B Ψ A+Ψ B>Numerator = 2EA + 2 β Denominator = 2(1 + S) E+ = (EA + β)/ (1 + S)Also E- = (EA - β)/ (1 – S) E± = E A ± βS is very small ∴ Neglect S
  62. 62. Energy level diagram EA - β βψA ψB β EA + β
  63. 63. Linear combination of atomic orbitalsRules for linear combination1. Atomic orbitals must be roughly of the same energy.2. The orbital must overlap one another as much aspossible- atoms must be close enough for effectiveoverlap.3. In order to produce bonding and antibonding MOs,either the symmetry of two atomic orbital must remainunchanged when rotated about the internuclear line orboth atomic orbitals must change symmetry in identicalmanner.
  64. 64. Rules for the use of MOs* When two AOs mix, two MOs will be produced* Each orbital can have a total of two electrons (Pauliprinciple)* Lowest energy orbitals are filled first (Aufbau principle)* Unpaired electrons have parallel spin (Hund’s rule)Bond order = ½ (bonding electrons – antibondingelectrons)
  65. 65. Linear Combination of Atomic Orbitals (LCAO)The wave function for the molecular orbitals can be approximatedby taking linear combinations of atomic orbitals. A B ψA ψB ψ AB = N(cA ψ A + cBψ B) c – extent to which each AO contributes to the MO ψ 2AB = (cA2 ψ A2 + 2cAcB ψ A ψ B + cB2 ψ B 2)Probability density Overlap integral
  66. 66. Constructive interference . + . +. +. ψg bonding cA = cB = 1 ψ g = N [ψ A + ψ B]Amplitudes of wave functions added
  67. 67. ψ 2AB = (cA2 ψ A2 + 2cAcB ψ A ψ B + cB2 ψ B 2) density between atoms electron density on original atoms,
  68. 68. The accumulation of electron density between the nuclei put theelectron in a position where it interacts strongly with both nuclei.Nuclei are shielded from each otherThe energy of the molecule is lower
  69. 69. node +. -. +. .- cA = +1, cB = -1 ψu antibondingψu = N [ψA - ψB] Destructive interference Nodal plane perpendicular to the H-H bond axis (en density = 0) Energy of the en in this orbital is higher. Ψ A-Ψ B Amplitudes of wave functions subtracted.
  70. 70. The electron is excluded from internuclear region  destabilizing Antibonding
  71. 71. When 2 atomic orbitals combine there are 2 resultant orbitals. orbitals. Eg. Eg. s orbitals σ*s 1E high energy antibonding orbital 1sb 1sa σ1s Molecular orbitals low energy bonding orbital
  72. 72. Molecular potential energy curve shows the variationof the molecular energy with internuclear separation.
  73. 73. Looking at the Energy Profile• Bonding orbital• called 1s orbital• s electron• The energy of 1s orbital decreases as R decreases• However at small separation, repulsion becomes large• There is a minimum in potential energy curve
  74. 74. H2 11.4 eVLCAO of n A.O ⇒ n M.O. 109 nm Location of Bonding orbital 4.5 eV
  75. 75. The overlap integralThe extent to which two atomic orbitals on different atomoverlaps : the overlap integral S = ∫ ψ A ψ B dτ *
  76. 76. S > 0 Bonding S < 0 antiBond strength depends on the S = 0 nonbonding degree of overlap
  77. 77. Homonuclear Diatomics• MOs may be classified according to:(i) Their symmetry around the molecular axis.(ii) Their bonding and antibonding character.∀ σ 1s< σ 1s*< σ 2s< σ 2s*< σ 2p< π y(2p) = π z(2p) <π y*(2p) =π z*(2p)<σ 2p*.
  78. 78. dx2-dy2 and dxy 2-Cl4Re ReCl4
  79. 79. g- identical B under inversion Au- not identical
  80. 80. Place labels g or u in this diagram σ∗u π∗g πu σg
  81. 81. First period diatomic molecules H H2 H σ1s2 σ u* Bond order: 1Energy 1s 1s σg Bond order = ½ (bonding electrons – antibonding electrons)
  82. 82. Diatomic molecules: The bonding in He2 He He2 He σ1s2, σ *1s2 σu* Bond order: 0 Energy 1s 1s σgMolecular Orbital theory is powerful because it allows us to predict whethermolecules should exist or not and it gives us a clear picture of the of theelectronic structure of any hypothetical molecule that we can imagine.
  83. 83. Second period diatomic molecules Li Li2 Li σ1s2, σ *1s2, σ2s2 2σu* Bond order: 1 2s 2sEnergy 2σg 1σu* 1s 1s 1σg
  84. 84. Diatomic molecules: Homonuclear Molecules of the Second Period Be Be2 Be 2σu* σ1s2, σ *1s2, σ2s2, 2s 2s σ *2s2Energy 2σg Bond order: 0 1σu* 1s 1s 1σg
  85. 85. Simplified
  86. 86. Simplified
  87. 87. MO diagram for B2 3σ u* 1π g* 1π u 3σ g Diamagnetic?? 2σ u* 2σ g
  88. 88. Li : 200 kJ/molF: 2500 kJ/mol
  89. 89. Same symmetry, energy mix-the one with higher energy moves higher and the one with lower energy moves lower
  90. 90. MO diagram for B2 B B2 B 3σu* 3σ u*1π g* 1πg* 2p (px,py)1π u 3σ g 2p LUMO 3σg 2σ u* HOMO 1πu 2σu* 2s 2s 2σg 2σ g Paramagnetic
  91. 91. C2 1π g 1π g 1π u 1π u 1σ g 1σ g X Paramagnetic ? Diamagnetic
  92. 92. General MO diagrams 1π 1π g g 1π u 1π u 1σ g 1σ gLi2 to N2 O2 and F2
  93. 93. Orbital mixing Li2 to N2
  94. 94. Bond lengths in diatomic molecules Filling bonding orbitals Filling antibonding orbitals
  95. 95. SummaryFrom a basis set of N atomic orbitals, N molecular orbitals areconstructed. In Period 2, N=8.The eight orbitals can be classified by symmetry into two sets: 4 σand 4 π orbitals.The four π orbitals from one doubly degenerate pair of bondingorbitals and one doubly degenerate pair of antibonding orbitals.The four σ orbitals span a range of energies, one being stronglybonding and another strongly antibonding, with the remainingtwo σ orbitals lying between these extremes.To establish the actual location of the energy levels, it is necessaryto use absorption spectroscopy or photoelectron spectroscopy.
  96. 96. Distance between b-MO and AO
  97. 97. Heteronuclear Diatomics…. The energy level diagram is not symmetrical. The bonding MOs are closer to the atomic orbitals which are lower in energy. The antibonding MOs are closer to those higher in energy.c – extent to which each atomicorbitals contribute to MOIf cA>cB the MO is composed principally of φ A
  98. 98. HF
  99. 99. HF 1s 1 2s, 2p 7ψ =c1 φ H1s + c2 φ F2s + c3 φ F2pz Largely 2px and 2py nonbonding 1σ 2 2σ 21π 4 Polar