E nuc rep = Z AZ B r AB m =0.499 hartree =13.60 ev. E nuc rep (H 2 )= m/a m =0.714 hartree =19.43 ev.

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1 Chemistr 31 Phsical Chemistr Homework Assignment # 7 1. Sketch qualitative contour maps of the following molecular orbitals for a diatomic molecule AB. Identif each as a bonding or an anti-bonding molecular orbital. The notation is the same as the one we have been using for H + and other eamples. Unless otherwise specified, the Z-ais is the molecular ais. Ignore the radial node(s) in the atomic orbitals. 1 (a) s A +p zb (b) s A p zb (c) 1 [s A +p za ]+1s B (d) [s A p za ]+1s B Answer (a) See the answer to Question below (antibonding orbital). (b) See the answer to Question below (bonding orbital). (c) Discussed in class (an sp hbrid orbital on atom A forms a bonding orbital with the 1s orbital on atom B). (d) Discussed in class (an sp hbrid orbital on atom A forms an atni-bonding orbital with the 1s orbital on atom B).. Given that the equilibrium internuclear distance of H + is 16 pm, and that of H is 7.1 pm, calculate the internuclear repulsion energ in each case. Given that the bond dissociation energ D e is.79 ev for H + and.78evforh, calculate the electronic energ in each case. Answer: The internuclear repulsion energ, in atomic units, is given b E nuc rep = Z AZ B r AB. For H + and H the nuclear charges are the same. Therefore, E nuc rep (H + )= m/a m =.99 hartree =13.6 ev. E nuc rep (H )= m/a m =.71 hartree =19.3 ev. In these calculations, we have used the conversion factors 1a = m, and 1 hartree = 7.11 ev. The bond dissociation energ D e is the energ difference between the molecule at its equilibrium geometr and the two atoms at infinite separatation. The energ of the molecule is the sum of the electronic energ and the nuclear repulsion energ. Therefore, Therefore, D e = E el + E nuc rep E el (H + )= = 1.81 ev. E el (H )= = 1.65 ev. 1

2 3. The overlap integral S for can be evaluated as ³ S = 1+r AB r AB e r AB, where r AB is the internuclear distance in atomic units (i.e., a ). (a) At what internuclear separation is this function a maimum? (b) At what value of r AB is it a minimum? (c) What is the value of S at the equilibrium internuclear separation of 16 pm? Answer: (a) To find the maimum of the function, we differentiate the function with respect to r AB and set the result equal to zero: µ ds = = 1+ ³ dr AB 3 r AB e r AB 1+r AB r AB e r AB, or 1+ 3 r AB=1 + r AB r AB. This result simplifies to ield = 1 3 (1 + r AB) r AB. The onl phsicall reasonable (since r AB cannot be negative!) solution is r AB =. (b) The eponential deca of the function ensures that its minimum value is at r AB =. (c) At r AB =16pm, we get (in atomic units, 16 pm = m/( m/a )=.3a ) S = ³ e.3 =.5856 Note that each term in the epression as well as the eponent is implicitl being multiplied b a coefficient with appropriate units to make the overalp integral have units of volume.. In a linear variational treatment of a diatomic molecule AB, the following trial function is used: ϕ = c 1 (s A )+c (p z,b ), where the Z-ais is the internuclear ais. The following matri elements are given for the Hamiltonian and overlap matrices at a particular value of the internuclear separation r AB : H AA = H BB = 3; H AB = H BA = 5; S AB =1/ 5. (a) Find the variational energies for the bonding and antibonding orbitals. (b) Assuming that the linear variational parameters c 1 and c are of comparable magnitude, sketch the bonding and antibonding orbtitals and clearl mark the positions of the nodes. (You ma ignore the presence of the radial node in the s orbital although I would recommend that ou tr to sketch the molecular orbitals with the radial node present.) Answer (a) The variational energies are W ± = H AA H AB 1 S = / 5.

3 This ields W + = / 5 =.15 W = / 5 = (b) The bonding orbital is the one corresponding to the lower energ or c 1 = c. The antibonding orbitals correspond to the higher energ, or c 1 = c. The sketches of the orbitals are shown on the following pages. 3

4 Fig. 1. The atomic orbitals under consideration. Red contours represent the positive parts of the wave function and the blue contours the negative parts. The magenta line is the contour where the wave function passes through zero Fig.. The bonding orbital (s +p z ). The angular node of the p z orbital is distorted b the negative part of the s orbital. Note that there is no node between the two nuclei.

5 Fig. 3. The antibonding orbital (s p z ). There is a node between the two nuclei. 5