Physical Chemistry II Recommended Problems Chapter 12( 23)

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1 Physical Chemistry II Recommended Problems Chapter 1( 3) Chapter 1(3) Problem. Overlap of two 1s orbitals recprobchap1.odt 1

2 Physical Chemistry II Recommended Problems, Chapter 1(3), continued Chapter 1 Problem 3 MO diagram of CO. Energies of the molecular orbitals are not known. The energies graphed below are just guesses. Chapter 1 Problem 5 Greater bond order implies shorter bond. Calculate bond order. Greater bond order implies shorter bond. molecule or molecular ion configuration number of bonding electrons number of antibonding electrons bond order bond length order Li σ g 0 1 shorter Li + σ g / C g u 1 u 4 6 shorter C + g u 1 u 3 5 3/ O g u 1 u 4 3 g 1 g 8 4 O + g u 1 u 4 3 g 1 g / shorter F g u 1 u 4 3 g 1 g shorter F - g u 1 u 4 3 g 1 g 4 3 u / recprobchap1.odt

3 Physical Chemistry II Recommended Problems, Chapter 1(3), continued Chapter 1(3) Problem 6 sketch the HOMO for N +, Li +, O -, H -, and C +. Chapter 1(3) Problem 7 Why is IE of CO greater than IE of NO? The answer is based on the MO diagram for N because CO is isoelectronic to N and the electron configuration of NO is like that of N -. NO has one more electron than CO, and that extra electron is in an antibonding molecular orbital. Ionization of NO occurs from a higher-energy orbital than ionization of CO, so NO has the smaller ionization energy. recprobchap1.odt 3

4 Physical Chemistry II Recommended Problems, Chapter 1(3), continued Chapter 1(3) Problem 9 Calculate hydrogen-fluoride AO coefficients at S HF =0.45. ψ 1 = ϕ H1s + ϕ Fpz ψ = c H ϕ H1s + c F ϕ Fpz = 13.6eV H FF = 18.6 ev H HF = 1.75S HF H FF equations to use to find coefficients secular determinant H HF S HF H HF S HF H FF = 0 Expanding the determinant gives a quadratic equation for E. The first row of the secular equation gives this connection between and ( ) +(H HF S HF ) =0 which can be rearranged to normalization = H HF S HF + c 1F + S HF = 1 normalization can be written in terms of C 1F and the ratio. ( C ) c 1H 1F + C + c SHF 1F = 1 1F substitution of the ratio above reduces the normalization formula to just one variable [( C H HF S HF ) ( 1F + 1 S H HF S )] HF HF = 1 Division and taking the square root yields ±1 = ( H HF S ) HF + 1 ( H HF S ) HF S HF Here is a sample calculation for the case S HF = H HF = X 0.45 X (5.96) 1/ = ev The expanded determinant is (1 - S HF) E + ( H HF S HF - -H FF ) E + ( H FF - H HF ) = 0 For S HF =0.45, E E = 0 The lower-energy solution is E = ev. (The larger solution is ev.) Continuing with the case S HF =0.45, ( / ) = ±1 Then = = ( ) Finally (again, for S HF =0.45) = The determinant can be checked for the case S HF =0.45. recprobchap1.odt 4

5 secular determinant (without rounding) = = = Likewise, normalization and the secular equation can be used to check and. + + S HF = X 0.45 X X = ( - E) + (H HF - S HF E) = ( ) ( X 0.45) = = 0, as required. This table shows the results for S HF = 0.30 and S HF E ground (ev) = H HF S HF To better see the effect of S HF, Larger overlap increases interaction substitute 1.75 S HF H FF for H HF 1.75 = S H FF + E c HF 1F energy and electron sharing, increasing the ratio of H to F orbital coefficients. recprobchap1.odt 5

6 Physical Chemistry II Recommended Problems, Chapter 1(3), continued Chapter 1(3) Problem 14 bonding in O, O -, and O +. molecule or molecular ion configuration number of unpaired electrons O σg σu 1πu 4 3σg 1πg O - σ g σu 1πu 4 3σg 1πg 3 1 O + σ g σu 1πu 4 3σg 1πg 1 1 The HOMO for O, 1π g *, is antibonding. Adding an electron to make O - weakens the bond. Removing an electron to make O + strengthens the bond. Bond strength: O - < O < O +. Chapter 1(3) Problem 15. Calculate dipole moment from bond length and MO coefficients. R e =91.7pm. MO coefficients are c H =0.34 and c F =0.84. Overlap S HF =0.30. Electron density (from this MO) assigned to H and to F are: n e,h = ( c H + S HF c H c F ) = ( X 0.34 X 0.84) = ( ) = X 0.01 = 0.40 n e,f = ( c F + S HF c H c F ) = ( X 0.34 X 0.84) = ( ) = X = 1.58 which I rounded to 1.60 so n e,h +n e,f =. That number accounts for only one fluorine electron, the F electron that pairs with the H electron to make a bond in this MO. I will add the other 8 fluorine electrons to n e,f. n e.f = = Atomic charge is nuclear charge minus electron charge. The nuclear charge of H is 1, so q H = = 0.60 The nuclear charge of Fe is 9. q F = = (the opposite of the H charge, as it must be) Dipole moment = (q H - q F )R e = X0.60 (1.60X10-19 Coulomb/e) X 91.7X10-1 m μ = 1.0 X 1.60X10-19 C X 91.7X10-1 m = 1.76X10-9 C m Divide by 3.33X10-30 C m/debye. μ = 5.3 D. (This is approximately twice the answer given by Engel in the back of the book.) The experimental dipole moment is 1.91 Debye, a factor of.8 smaller. A Hartree-Fock STO-3G calculation gives charge separation, using the Mulliken method that is comparable to the method used above, of X0.0 = That is a factor of 3 smaller than the X0.60 = 1.0 calculated above. recprobchap1.odt 6

7 Chapter 1 Problem 19 sketch MO energy diagram for OH based on the diagram for FH The diagram for OH will be similar, but the Os and Op orbitals will be higher in energy than the Fs and Fp orbitals because O has a smaller nuclear charge than does F. OH has one fewer valence electron than has HF, so OH will have one unpaired electron and will have a doublet ground-state spin. As in HF, the HOMO will be nonbonding (oxygen p x and p y orbitals) and the LUMO will be antibonding (σ*). The valence MO diagram at right shows orbitals calculated using Hartree-Fock theory and the minimal STO-3G basis set. There is qualitative agreement with the HF MO diagram above. recprobchap1.odt 7

8 Physical Chemistry II Recommended Problems, Chapter 1(3), continued Chapter 1 Problem 1. Valence MO energies for HF When the overlap is small (equivalent to the atoms being far apart) then E 1 and E approximately equal the F and H orbital energies. Increasing S 1 causes greater interaction, lowering E 1 and raising E. recprobchap1.odt 8

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