Chem 1211 1st Edition Lecture 17Outline of Last Lecture I. WavesII. Electromagnetic RadiationIII. Photoelectric EffectOutline of Current Lecture I. Atomic SpectraII. Quantization StepsIII. Wave Nature of ElectronsCurrent LectureI. Atomic Spectraa. ConcepTest: Which electronic transition would you expect to have the smallest energy gap? i. n1=1 to n2=2ii. n1=2 to n2=3iii. n1=1 to n2=3iv. n1=2 to n2=4v. n1=6 to n2=71. ANSWER: V (N1=6 to N2=7)2. REASONING: The gaps decrease as N increasesb. Rydberg Equationc. (1/wavelength) = R [(1/n12) – (1/n22)]i. R= 1.097 x 107 m-1ii. n1 < n2d. ConcepTest: Which electronic transition of hydrogen would you expect to occur in the visible range of electromagnetic radiation?i. n1=1 to n2=2ii. n1=1 to n2=3These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.iii. n1=2 to n2=3iv. n1=4 to n2=5v. depends on the value of Planck’s constant 1. ANSWER: iii (n1=2 to n2=3)2. REASONING: Anything that starts with 1 is too energetic; anything that starts with 3 is not energetic enough. Use a graph or calculateby handII. Quantization “Steps”a. Electrons: particles, wavesb. deBroglie wavelengthi. w= (h)/(mv)1. m=mass2. v=velocityc. ConcepTest: Usain Bold may be the fastest human with a speed of 23 mph (10.3 m/s). He weighs 94 kg. What is his deBroglie wavelength at that speed? [h=6.626 x 10-34 Js; J = kgm2s-2]i. ANSWER: 6.84 x 10-37m ii. REASONING: Use the equation w= (h)/(mv)III. Wave Nature of Electronsa. Louis de Broglie (1925)i. w= (h)/(mv)b. Many diagrams show a “Bohr” model in which the electrons orbit the nucleus in a perfect circle; THIS IS NOT RIGHT. If this were true, the electrons would fall into the nucleus when they decay, leading to the end of all matter. c. Quantum mechanics- orbitals probability i. You can find the probability of where an electron is, but you can never find out for sure. When you try to experiment, you change the orbital enough that the position of the electron changes as well. d. Quantum Numbers: Reviewi. Principal quantum number: n=1,2,3,4… “Shells” (rows in periodic table)ii. Angular momentum quantum number: I= 0,1,2,…, (n-1) OR l=s,p,d,f,g,h,…(n-1) [defines shapes]iii. Magnetic quantum number: mI = -I, (-I+1),…,0,…,(I-1), I1. I=0 (s orbital, mI=0 (only 1 value)2. I=1 (p orbitals), mI= -1,0,+1 (three p orbitals: x,y,z)3. I=2 (d orbitals), mI= -2,-1,0, +1, +2 (five shapes/directions4. I=3 (f orbitals), mI= -3,-2,-1,0,+1,+2,+3 (7 shapes/directions)iv. Spin quantum number, ms = +(1/2) or –(1/2)1. Means that there are no more than 2 electrons per orbital 2. Spin: parallel, antiparallelv. ConcepTest: What are the values of n, I, and ml for a 4s orbital?1. 0, 4, 12. 4, 0, 13. 4, 0, 04. Depends on the spin5. Requires more infoa. ANSWER: 3 (4, 0, 0)b. REASONING: For a 4s orbital, the n=4, and the only optionsfor I and mI are 0vi. Atomic Orbital Shapes1. s=sphere2. p=dumbbell3. d= multiple4. f= complexvii. Electron Spin and Electron Occupancy1. Spins describe direction of electon’s magnetic fielda. Simple description: spin up or spin down2. Paramagnetic atoms: unpaired elctrons a. Have net magnetic ions3. Paramagnetism: net field in external magnetic field4. Diamagnetism: no net magnetic field5. Ferromagnetism: permanent magnetic
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