Final Exam Review
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Applicable concepts/equations (6.626π₯10−34)(2.9979π₯108) πΈπβππ‘ππ = 422π₯10−9π Etotal = Ephoton* # photons πΈπβππ‘ππ = 1.99π₯10−25π½ βπ πΈ= l π½ #πβππ‘πππ = 40.0π ππ∗2π ππ πΈπ‘ππ‘ππ = πΈ πβππ‘ππ 1.99π₯10−25π½ = 4.03x1026photons Can a wave do this? ****Things we discussed in this course. Can a Particle do this? yes yes yes no yes no yes no no yes https://www.youtube.com/watch?v=DfPeprQ7oGc Double Split Experiment Does this show that light has wavelike or particle like properties? Why? Wave-like: It creates an interference pattern. What would the results look like if they had particle like properties? Two bright lines behind the splits no interference pattern. Photoelectric Effect Does this show that light has wavelike or particle like properties? Why? What would the results look like if they had only wave-like properties? Particle: 1 photon=1 electron, which must be of high enough energy. High total energy still won’t eject an electron if each isn’t of sufficient energy. What is the effect of increasing the intensity of a laser of a frequency less than the threshold frequency? None: it must be OVER the threshold frequency to eject electrons. What is the effect of increasing the intensity of a laser of a frequency greater the threshold frequency? The rate of the ejected electrons is increased. http://phet.colorado.edu/en/simulation/photoelectric Ke Ek= βπ£ − π€ Slope=h 1.3x10−19 = (6.626x10−34)π£ − (7.53x10−19) Ek= βπ£ − π€ y=mx+b π£ = 1.33π₯10−15Hz frequency Y-intercept = work function Threshold Frequency Ψ Ψ2 Schrodinger equation solutions Particle in a box (1d) Particle in a box (2/3D) Didn’t cover Hydrogen Atom Multi electron atoms= many types of approximation, no exact solutions Subject of Current research: We saw result, of appoximations For each of the previous neutral electron configurations, give an excited state. Keep same number of electrons, move at least one up in energy: Many many many correct answers. Sb: [Kr]5s24d105p3 ο [Kr]5s14d105p4 or [Kr]4d105p5 ect….. Cu: [Ar]4s13d10 ο [Ar]4s23d9 or [Ar]3d104p1 ect….. -1,0,1 -1,0,1 9+4=13 orbitals 2 electrons= 13*2=26 electrons Effective Nuclear Charge Electronegativity Ionization Energy Electron Affinity Atomic Radius Electron Affinity Ionization Energy Electronegativity Atomic Radius Effective Nuclear Charge Electronegativity Ionization Energy Electron Affinity Atomic Radius Electron Affinity Ionization Energy Electronegativity Atomic Radius Write the electron configurations for C, N, and O. Place in order of increasing ionization energy, increasing electron affinity and increasing electronegativity. For each characteristic, do these follow the trend? Why or why not? Wrong Answer 1: All: C, N, O Wrong Answer 2: Variety of wrong/right answers Yes No Because the trend goes up and to the right. But hydrogen is half filled. Because the trend goes up and to the right. Correct: I.E. C, O, N : No, while the effective nuclear charge increases as you go to the right, nitrogen’s half filled shell has increased stability making it unlikely to ionize. E.A. N, C, O: No, while the effective nuclear charge increases as you go to the right, nitrogen’s half filled shell has increased stability making it unlikely to ionize. E.N. C, N, O: Yes. Because it is speaking of atoms in a stable bond, electronegativity is only based on effective nuclear charge which increases as you go to the right. Hint: what is the p block exceptions to electron affinity? Half filled subshells are more stable and therefore have a lower electron affinity. Hint: What would be the equivalent of that with the dblock? Half filled subshells are more stable and therefore have a lower electron affinity. ns2np5 and also ns1np5 Which group do these elements belong to? First is always smallest, The “jump” indicates stable configuration I1 Element 1 I2 0.605 1.110 Element 2 0.203 3.215 I3 I4 1.45 5.10 3.89 4.42 1st, 2nd , 3rd come off, then big jump: so group 3 Big jump after 1st electron, so group 1 Why do we see the sun as different colors? Which lights depicted in this diagram are emitted, which are scattered. What causes the events depicted by the blue arrows in the atmosphere. CN C N 2p E Bonding orbitals are a lower energy than antibonding and the original atomic orbitals. Most of the electron density in bonding orbitals are between the nuclei rather than outside of it. This adds to the bond order, rather than subtracts. [He]s2s2s2s*2p2p4 diamagnetic Bond order= ½ (6-2)=2 Yes! +0 Except sp3 Tetrahedral 109.5 sp3 Trig. Planar 120 p s sp3 s sp3 p sp3 sp2 sp2 p sp3 s p sp2 p p OH HO O CH3OH can be synthesized by the reaction shown. What volume of H2 gas (in L), at 748 mmHg and 86 oC, is required to synthesize 25.8 g CH3OH? πΆπ π + 2π»2 π → πΆπ»3ππ» Grams product moles product conversion 25.8π moles reactants conversion Volume Reactant Ideal gas law 1πππ πΆπ»3ππ» 2 πππ π»2 πΆπ»3ππ» = 1.610 mol H2 32.04 π πΆπ»3ππ» 1πππ πΆπ»3ππ» π= ππ π π = π∗ππ‘π ∗ 86+273 πππ∗π πππ»π 748πππ»π/760 ππ‘π (1.610 πππ∗0.0821 πΎ =48.2 L H2 Part 2: If the reaction is known to only be 70.0% efficient (aka has a 70% yield) what volume is required (at the same conditions)? πππ‘π’ππ 48.2 ∗ 100% πππ£ππ : = 0.700 π€βππβ πππ£ππ π₯ = 68.8 πΏ π‘βπππππ‘ππππ π₯
