Final Exam 2009 Solutions

Final Exam 2009 Solutions - Name: _ Student Number: _ CHM...

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Name: ________________________ Student Number: ___________________ CHM 2311 Final Exam April 29 2009 Professor Darrin Richeson There are 15 pages in this exam (including supporting materials and tables). Please count the pages to make sure none are missing. You may carefully remove the last three pages of the exam and use them for scratch work. Please write legibly and show your work to receive credit for your answers. Partial marks may in some cases be awarded for partially correct work. Question Mark Question Mark 1 /5 8 /4 2 / 9 9 /13 3 / 7 10 /16 4 / 16 11 /5 5 / 10 12 /12 6 /9 7 /4 Total /110 Good luck and have a great summer!
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2 1. ( 5 points ) A hydrogen orbital has the following complete wave function (both angular and radial portions of the wave function are included and the angular part is written in Cartesian coordinate form): { } ( ) = = 2 2 2 3 / 2 2 / 3 15 4 / 1 2 15 81 1 ) , ( ) ( ) , , ( r y x e a Z Y r R r o π ρ ϕ θ ψ where o a Zr = and Z = 1 for a hydrogen atom a o = 52.9 pm a. For this orbital, determine the distance from the nucleus in picometers at which all radial nodes occur. The radial component becomes zero (has a node) when 2 = 0 or = 0 = Zr/a o This is only true at the r = 0 – the nucleus no radial nodes b. Determine the quantum numbers n and l for this orbital. Clearly show the process by which you determined these values. Identify the orbital (1s, 2s, 2p x , etc. ..). In part (a) we determined that there are no radial node s. The number of angular nodes can be determined from the angular part of the wavefunction. From the equation, = 0 for x 2 -y 2 = 0 (and all values for z). Rearranging gives a node for x = y and for x = -y There are 2 angular nodes. This corresponds to l = 2 or a d orbital. The equation for the number of radial nodes is given by n-l-1 = number of radial nodes. Since n-l-1= 0 then n = 2+1 = 3 Therefore this is a 3d orbital and specifically the 3d x2-y2 orbital.
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3 2. (9 points) a. Calculate Z eff for a 2p electron on each of the following species. O 2- , F - , Na + , Mg 2+ . Each of these ions has the same electronic configuration: (1s 2 ) (2s 2 , 2p 6 ) This means that they all have the same s value s =(2 x 0.85) + (7 x 0.35) = 4.15 Z eff = Z actual - s Ion Z Z eff O 2- 8 3.85 F - 9 4.85 Na + 11 6.85 Mg 2+ 12 7.85 Match the ion with the radii in the following table. Crystal (Ionic) Radius Ion 119pm 86pm 116pm 126pm Calculate the Z eff for a 4d electron of antimony (symbol Sb) (1s2) (2s2, 2p6) (3s2, 3p6) (3d10) (4s2, 4p6) (4d10) (5s2, 5p3) Z eff = Z-s = 51- 9x0.35 + 36x1 = 51-39.15 = 11.85
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4 3. (7 points) There are potentially three isomers for the anion NSO - . The two known species are NSO
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This note was uploaded on 01/22/2011 for the course BCH 2333 taught by Professor Mezl during the Winter '08 term at University of Ottawa.

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Final Exam 2009 Solutions - Name: _ Student Number: _ CHM...

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