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Notes One • • • • • • • • Electronic Structure Valence Electrons vs kernel Electrons Structure of the atom Electromagnetic Radiation Calculating Frequency for Electromagnetic Radiation Calculating Energy of Radiation Matter and Energy Calculate Smallest Increment of Energy Electronic Structure • Valence electrons • Kernel electrons • Nucleus • Protons • Neutrons Valence Electrons? kernel Electrons Kernel e-1 Valence e-1 Gold [Xe]6s26p45d94f14 Iodine [Kr]5s25p64d10 Structure of the atom Rutherford Nucleus -dense -heavy -small. Bohr -e-1 orbit nucleus -e-1 are Particles Planck h=6.626 x 10-34 Js Planck 1900 Quantum loss or gain E=h n Electrons De Broglie – Wave Nature. Wavelength() Lower Higher Energy ???? Higher! Electromagnetic Radiation amplitude Electromagnetic Spectrum gamma ray X-ray Short wavelength High Energy Visible UV Light IR radar FM TV Short wave AM Long wavelength Low Energy Electromagnetic Radiation Waves have a frequency n Is frequency. •n = c C= 3.00 x 108 m/sec Long low n Short high n Calculating Frequency for Electromagnetic Radiation Example: Red light has = 700 nm. Calculate the frequency, n. n • = C C n= 8 3.00x10 14 Hz = 4.29x10 n= 7.00x10-7 Calculating Energy of Radiation • Calculate the energy of 1.00 mol of photons of red light, when = 700 nm and n = 4.29 x 1014 sec-1. E=h n E = (6.63x10-34 J•s)(4.29x1014sec-1) E= 2.85 x 10-19 J per photon E =(2.85 x10-19J/ph)x(6.02x1023ph/mol) E per mol = 171.6 kJ/mol Matter and Energy Until 1900's matter particles energy waves. In 1901, Max Planck multiples of hn. h = 6.626 x 10-34Js E = nhn n = 1,2,3... It is quantized. quanta Calculate Smallest Increment of Energy Calculate the smallest increment of energy (the quantum of energy) that an object can absorb from yellow light whose wavelength is 589 nm. n • = C C n= 8m/s 2.998x10 14s-1 n= = 5.09x10 5.98x10-7m E=h n E = (6.626x 10-34J-s)(5.09x1014s-1) E = 3.37 x 10-19J The Bohr Model Energy quantized light quantized. Bohr showed that 1 E - 2.18 10 J 2 n -18 n = 1, 2, 3…. (principal quantum numbers) Limited to hydrogen The Bohr Model Hydrogen only. Electrons described as particles. E=h•n •n=c Ionization Energy • Calculate the ionization energy of the ground state hydrogen atom (n=1) per mole. E= -2.18x10-18J __ 1 -18J =-2.18x10 12 E=(-2.18x10-18J)x 6.022x1023 atoms/mol E=-1312KJ/mol Notes Two • • • • • • • • Photoelectric Effect Emission Spectra Structure of the atom Line Spectra (Energy) Limitations of the Bohr Model Energy Levels For Hydrogen Ingredients of a laser The Wave Behavior of Matter Quantum Mechanics and Atomic Orbitals Photoelectric Effect THE PHOTOELECTRIC EFFECT Different metals hold on to their electrons more strongly or weakly due to atomic structure. hn = Ek + W h is Planck’s constant. n is the frequency. Ek is the kinetic energy. W is the work function. Minimum ionization energy 1 electron volt = 1.60217646 × 10-19 joules Ek = 1/2mv2 1joule = 1Coul x 1 volt Ek = q x volts Momentum(photon) p = h / Momentum(electron) p = h / = 2m(Ek) THE PHOTOELECTRIC EFFECT hn = Ek Ek W hn = Ek + W Ek = 1/2mv2 n slope= h e-1 ↑nth W n eV 3.60x10-7 8.33x1014 4.00x10-7 7.50x1014 4.40x10-7 6.82x1014 4.90x10-7 6.12x1014 5.50x10-7 4.45x1014 5.80x10-7 5.17x1014 1.45 1.12 0.95 0.65 0.40 0.25 n= C hn = Ek + W Kinteic Energy eV m n +7 +6 +5 +4 +3 +2 +1 0 -1 -2 W ↑nth -3 -4 0 1 2 3 4 5 6 7 8 9 10 Frequency Hz(1014) If the threshold frequency required for emission of photoelectron is 4.00x1014 Hz, what is the work function of the surface in Joules and eV? W = h•n W = (6.63x10-34J-s) (4.00x1014 Hz) W = (2.65x10-19J) ÷(1.60x10-19J) W = 1.66eV If the work function of a metal surface is 3.00x10-19J, and it is illuminated by light with a wavelength of 500 nm, what is the speed of the escaping photoelectrons? hn = Ek + W hn - W = Ek h• C - W = Ek n = C 8m/s) -34 (3.00x10 (6.63x10 J-s)• (500x10-9m) -(3.00x10-19J) = Ek 1 2 m v Ek =2 1 (9.11x10-31Kg) v2 -20 = 9.78x10 J 2 V= 4.63x105m/s = 9.78x10-20J A metal surface has a work function of 2.06eV. If the fastest photoelectron emitted from is 6.00x105m/s, what is the frequency of the light source striking the surface? 1 Ek = m v 2 2 hn = Ek + W Ek + W ______ n= h 2 ÷h 1/2 (9.11x10-31Kg)(6.00x105m/s) + ( (2.06eV)x1.60×10-19j) (6.63x10-34J-s) n= 7.44x1014s-1 Light of wavelength 400 nm is incident on a metal surface, as shown above. Electrons are ejected from the metal surface with a maximum kinetic energy of 1.1 x 10-19 J. (a) Calculate the frequency of the incoming light. (b) Calculate the work function of the metal surface. (c) Calculate the stopping potential for the emitted electrons. (d) Calculate the momentum of emitted electrons. (a) Calculate the frequency of the incoming light. n • = C C n= 8m/s 2.998x10 14s-1 n= = 7.50x10 4.00x10-7m (b) Calculate the work function of the metal surface. hn = Ek + W hn - Ek = W (6.63x10-34J-s x7.50x1014s-1) - 1.1x10-19J = 3.87x10-19J (c) Calculate the stopping potential for the emitted electrons. Ek =q x V 1.1x10-19J = 1.60×10-19C x volts V = 0.687 volts (d) Calculate the momentum of emitted electrons. Momentum P= 2m(Ek) p = 2x 9.11x10-31kg (1.1x10-19J) =4.1x10-6kg-m/s Emission Spectra Emission Spectrum Slit Prism Hydrogen Lamp Photographic Plate Exciting Gas Atoms to Emit Light with Electrical Energy Hg He H Examples of Spectra Helium Barium Oxygen Neon Identifying Elements with Flame Tests Na K Li Ba Ingredients of a laser The Wave Behavior of Matter Light particle nature, wave property Einstein’s / Planck’s equations Broglie h/mv particle property effects on objects are small. Heisenberg’s Uncertainty Principle • The position, direction of motion, and speed simultaneously for an electron. Quantum Mechanics and Orbitals Schrödinger equation wave and particle terms. Solving wave functions. (wave function )2 electron density Notes Three • • • • • • • • • Quantum Numbers (QN) Pauli Exclusion Principle Orbital Shapes and Energies The History of the Periodic Table Mendeleev’s Periodic Table The Aufbau Principle and the Periodic Table Hund’s Rule Valence electrons Filling out the Periodic Table Orbital Filling Diagram Write the OFD for Cu. How many e-1 for Cu? 29 e-1 [Ar] 4s 3d 9 2 Short Hand Element? Ar How far up does Ar fill? Final Answer? 2 2 6 2 6 2 9 1s 2s 2p 3s 3p 4s 3d Short Hand Answer Electron Dot Diagram for Selenium ● Se ● ● ● ●● [Ar]4s 3d 4p 4 10 2 Quantum Number for Last e-1 in Cd ms = +1/2 or -1/2= l = 0(s), 1(p), 2(d), 3(f) = How many e-1 does Cd have? 48 e-1 Short hand element? Kr Many e-1 does Kr have? 36 e-1 How many e-1 to account for? 12 e-1 Short-hand fills to where? Fill-in Valence12 e-1 Quantum #’s for Last e-1 n=4 l= 2 ml=-2 ms=-1/2 Quantum Numbers n= 4 l= 3 ml = +2 ms = -1/2 Principal Quantum #(n) n= 1, 2, 3, . . Distance from 3d__ __ __ __ __ the nucleus Weaker Attraction -1 3p__ __ __ e more NRG 3s__ e-1 far Larger n higher energy Larger n less strongly bound to nucleus stronger Attraction 2p__ __ __ 2s__ e-1-1less NRG 1s__ e near Angular Momentum Quantum #(l ) relates to orbital shape. l = 0 to n - 1 n=1 l = 0 n=2 l = 0, 1 n=3 l = 0, 1, 2 n=4 l = 0, 1, 2, 3 n=5 l = 0, 1, 2, 3, 4 n=6 l = 0, 1, 2, 3, 4, 5 n=7 l = 0, 1, 2, 3, 4, 5, 6 l = 0 for s l = 1 for p l = 2 for d l = 3 for f 10-8 m 1s Probability density Ψ2 Probability density Ψ2 Probability density Ψ2 The s Orbital 10-8 m 2s l=0 ? ? ??? No Nodal Plane 10-8 m 3s The p Orbital Angular Momentum QN (Shape) ?? ? l=1 ?? One Nodal Plane The d Orbital Angular Momentum QN (Shape) l=2 ? ? TwoNodal Planes ?? The f Orbital Angular Momentum QN (Shape) l=3 ? ?? ? ? n 1 2 3 4 5 l Magnetic Quantum #(ml ) 0 1s 0 2s 1 2p 0 3s 1 3p 2 3d 0 4s 1 4p 2 4d 3 4f 0 5s 1 5p 2 5d 3 5f 4 5g ml 0 0 +1 0 -1 0 +1 0 -1 +2 +1 0 -1 -2 0 +1 0 -1 +2 +1 0 -1 -2 +3+2 +1 0 -1 -2 -3 0 +1 0 -1 +2 +1 0 -1 -2 +3+2 +1 0 -1 -2 -3 +4+3+2 +1 0 -1 -2 -3 -4 Spin Quantum #(ms) Relates to electron spin states. spinning charges. magnetic field Two spins north +1/2 south +1/2 Pauli Exclusion Principle No two electrons can have the same set of four quantum numbers (n, l, ml, ms). An orbital can hold only two electrons, Electrons can have the same n, l, and ml values If n = 3, l = 2 (d-orbital), ml = -2 (a single d-orbital) ms = +1/2, and one with ms =-1/2 Energy Order of Orbitals What determines the order of filling? Penetration 2p 2s 3s 3p 3d Distance from nucleus Distance from nucleus To find energy order between orbitals (Penetration). Determine the lowest n+l Determine the lowest n+l Which orbital has the greatest penetration? 3d or 4s? n+l 3d 3 + 2= 5 3d 3 + 2= 5 4f 4 + 3= 7 4s 4 + 0= 4 4 + 2= 6 2p 2 + 1= 3 6s 6 + 0= 6 Lowest n, if a tie. 3s 3 + 0= 3 7p 7 + 1= 8 4f 4 + 3= 7 5f 5 + 3= 8 5s 5 + 0= 5 4d end The History of the Periodic Table • • Patterns in element properties were recognized • Dobereiner (1780-1849) found “triads” of similar elements: Cl, Br, I • Newlands suggested in 1864 that elements should be arranged in “octaves” because similarities occurred every 8th element The Modern Periodic Table • The German Meyer (1830-1895) and Russian Mendeleev (18341907) independently developed the current arrangement of elements • Mendeleev predicted the properties of “missing” elements Aufbau Principle Lowest energy (penetration) orbital gets electron first. Hunds Rule Distribute electrons in multiple orbital energy levels C F Elect. Configuration 1s 2s 1s2 2s2 2p 2 1s2 2s2 2p 5 ↑↓ ↑↓ 2px 2py ↑↓ ↑ ↑↓ ↑↓ 2pz ↑ ↑↓ ↑ Elect. Configuration 4s 3d 3d 3d 3d 3d 2p 2p 2p 2 2 x -y 2 10 2 6 2 z xy xz 1 ↑↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Ga [Ar]4s 3d 4p Fe [Ar]4s 3d yz ↑↓ ↑ ↑↓ ↑ ↓ ↑ ↑ ↑ ↑ x y z •Mendeleev’s Periodic Table • • Left blank spaces for the elements that he predicted would be discovered Arranged elements by mass, instead of by atomic number Four Quantum Numbers Ionization Energy (I.E.) The quantity of energy required to remove an electron from the gaseous atom or ion. 1. X(g) X+ + e2. It is easiest to take away the first e-, more energy needed for others 3. Greater charge makes it harder to extract e4. First e- comes from the outermost (weakest bound) orbital 5. I. E. increases from left to right on periodic table because larger elements have larger +/- attraction for electrons 6. I. E. decreases down a group because outer e- becomes more weakly bound Ionization Energy (I.E.) Electron Affinity (EA) Energy change when e- is added to a gas atom X(g) + e- -------> X-(g) 1. 2. 3. EA generally become more negative (exothermic) from left to right a. Interactions between the +nucleus and the added electrons is favorable b. The larger the nucleus, the better the interaction Exceptions to the trend is a function of electron—electron interactions a. N- doesn’t form because N has 1s22s22p3 configuration b. The extra electron would have to go into an occupied 2p orbital c. C- does form because it has an empty orbital: 1s22s22p2 d. This phenomenon doesn’t always hold true. O- can form 1s22s22p4 EA diminishes down a group a. Added e- is farther from the nucleus, so less interaction is observed b. The difference in EA is small and has exceptions (small F 2p orbital) Atomic Radius: • C. half the distance between the nuclei • molecule consisting of identical atoms. 1. Easy to understand for diatomic X2 molecules 2. Can be calculated from known radii of XY molecules 3. Decreases from left to right on periodic table because +/attraction becomes greater 4. Increases down a group because more e- and larger shells filled Names for groups of the Periodic Table Metals: tend to give up electrons to form cations found on the lower left Nonmetals: tend to gain electrons to become anions found on the upper right Metalloids: have properties of both metals and nonmetals depending on the conditions