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Hand in Assignment 2 Solutions

Hand in Assignment 2 Solutions - CHEM 212(DE HAND-IN...

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CHEM 212 (DE) HAND-IN ASSIGNMENT #1 (Fall 2008) page 1 of 4 1. (6 marks) Consider a Li 2+ ion with its electron in a 4 s orbital. What is the shortest wavelength of light that can be emitted by this ion? What is the third ionization energy for Li? Give your answer for the ionization energy in kJ mol 1 . (Note: For a hydrogen-like ion, E n = R H Z 2 / n 2 .) Since the energy of a photon is inversely proportional to the wavelength of the light (i.e. E photon = hc / λ ), the shortest wavelength of light is emitted for the transition of highest energy. Starting from n = 4, the highest energy transition is to n = 1. Therefore: E photon = (2.179×10 18 J)(3) 2 [ 1/16 1/1 ] λ = (6.626×10 34 Js)(2.998×10 8 m s 1 ) / E photon To calculate the third ionization energy, use the energy formula above to calculate E 1 . E 1 = (2.179×10 18 J)(3) 2 [ 1/1 2 ] = 1.961×10 17 J IE(3) = N A × |E 1 | = 1.181×10 7 J mol 1 = 11,810 kJ mol 1 2. (12 marks) Sketch plots of R ( r ) versus r , R ( r ) 2 versus r and 4 π r 2 R ( r ) 2 versus r for a 5 s orbital. Repeat the process for a 4 d orbital. Accurate plots are not required (i.e. no calculations are required).
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CHEM 212 (DE) HAND-IN ASSIGNMENT #1 (Fall 2008) page 2 of 4 3. (6 marks) Use Slater’s rules to calculate the effective nuclear charge for a 4s electron in a ground- state zinc atom. Repeat the process for a 3d electron. The electron configuration of Zn is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 . The “grouped” electron configuration is (1s) 2 (2s,2p) 8 (3s,3p) 8 (3d) 10 (4s) 2 .
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