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Electronic Structure of Atoms Chapter 6 Light is a Wave • • • • Electromagnetic wave Wavelength (l), m Frequency (n), Hz or s-1 Travels at c (3.00 X 108 m/s) in a vacuum • Wavelength and frequency are inversely proportional • c = ln •How many complete waves are shown above? •What is the wavelength of light shown above? Electromagnetic Spectrum Blu-Ray = 405 nanometers (blue light) DVD = 650 nanometers (red light) • Visible light = 4 X 10-7 m to 7X 10-7 m (400 to 700 nm) Waves: Ex 1 Calculate the wavelength of a 60 Hz EM wave Waves: Ex 2 Calculate the wavelength of a 93.3 MHz FM radio station Waves: Ex 3 Calculate the frequency of 500 nm blue light. Planck’s Quantum Hypothesis: Light is a Particle • Photon –light particle • Photons emitted in “packets” (whole numbers) • Light is both a wave and a particle E = hn (for one photon) h = 6.63 X 10-34 J s (Planck’s constant) Photons: Ex 1 Calculate the energy of a photon of wavelength 600 nm. 600 nm 1 X 10-9m = 6 X10-7 m 1 nm Photons: Ex 2 Calculate the energy of a photon of wavelength 450 nm (blue light). Also, calculate the energy per mole. Ans: (4.42 X 10-19 J , 266 kJ/mol) Photons: Ex 3 Calculate the energy of laser light with a frequency of 4.69 X 1014 s-1 . Also, calculate the energy per mole. Ans: (3.11 X 10-19 J, 187 kJ/mol) Photons: Ex 3a If the laser emits 1.3 X 10-2 J per pulse, how many photons (quanta) are released during this pulse? Ans: 4.18 X 1016 photons Photoelectric Effect (Einstein) • When light shines on a metal, electrons are emitted • Can detect a current from the electrons • Used in light meter, scanners, digital cameras Three Key Points 1. Below a certain frequency, no electrons are emitted 2. Greater intensity light produces more electrons 3. Greater Frequency light produces no more electrons, but they come off with greater speed Low Frequency Not enough energy to eject electron High Frequency Can eject electron Energy of photon is greater than W (Work function) 2. More intensity – More photons – More electrons ejected with same KE 3. Greater Frequency – No more electrons ejected – Electrons ejected with greater speed (KE) Photon/Matter Interactions 1. Electron excitation (photon disappears) 2. Ionization/photoelectric effect (photon disappears) 3. Scattering by nucleus or electron 4. Pair production (photon disappears) Electron Excitation Ionization/Photoelectric Effect •Photon is absorbed (disappears) •Photon is absorbed (disappears) •Electron jumps to an excited state •Electron is propelled out of the atom Scattering •Photon collides with a nucleus or electron •Photon loses some energy •Speed does not change, but the wavelength increases Pair Production •Photon closely approaches a nucleus •Photon disappears •An electron and positron are created. Principle of Complimentarity • Any experiment can only observe light’s wave or particle properties, not both • Different “faces” that light shows Line Spectra • Discharge tube – Low density gas (acts like isolated atoms) – high voltage • Light emitted only at certain (discrete) wavelengths Hydrogen Helium Solar absorption spectrum • Electrons orbit in ground state (without radiating energy) • Jumps to excited state by absorbing a photon • Returns to ground state by emitted a photon hn = Ee - Eg 4. Ways to make something glow Bohr Model Photon Absorption -Glow in the dark Collision -Heat -Electricity -Chemical Reaction Photon Absorption Collision Bohr’s Equation • Works only for H and other 1electron atoms (He+, Li2+, Be3+, etc…) En = -RH n2 En RH n = Energy of an orbital = Rydberg constant (2.18 X 10-18 J) = Principal quantum number of orbital Bohr’s Equation: Ex 1 Calculate the energy of the first three orbitals of hydrogen En = - 2.18 X 10-18 J n2 E1 = - 2.18 X 10-18 J 12 E1 = - 2.18 X 10-18 J E2 = - 2.18 X 10-18 J 22 E2 = - 5.45 X 10-19 J E3 = - 2.18 X 10-18 J 32 E3 = - 2.42 X 10-19 J Bohr’s Equation: Ex 2 What wavelength of light is emitted if a hydrogen electron drops from the n=2 to the n=1 orbit? E1 = - 2.18 X 10-18 J E2 = - 5.45 X 10-19 J DE = (- 2.18 X 10-18 J - - 5.45 X 10-19 J) DE = -1.64 X 10-18 J DE = hn n = 2.48 X 1015 s-1 c = ln l = c/n = (3 X 108m/s)(2.48 X 1015 s-1 ) l = 122 nm DE = hn c = ln DE = hc/l l = hc DE l = (6.626 X 10-34 J s)(3.0 X 108 m/s) (1.63 X 10-18J) l = 1.22 X 10-7 m = 122 nm (UV) Bohr’s Equation: Ex 3 Calculate the wavelength of light emitted if a hydrogen electron drops from the n=6 to the n=2 orbit? (ANS: 410 nm (violet)) Bohr’s Equation: Ex 4 Calculate the wavelength of light that must be absorbed to exite a hydrogen electron from the n=1 to the n=3 orbit? (ANS: 103 nm (UV)) Bohr Model: Other Atoms En = (Z2)(- 2.18 X 10-18 J) n2 Z = Atomic number of the element (H=1, He=2, etc) The Batcave Intruder Alert Laser uses a beam that has a frequency of 3.53 X 1014 Hz. a) Calculate the wavelength in nanometers. (850) b) Calculate the energy per photon. (2.34X10-19 J) c) Calculate the energy per mole of photons. (1.41 X 105 J/mol) d) Calculate the number of photons in a 50.0 mJ pulse of this laser. (2.13 X 1017 photons) e) Calculate the moles of photons in that pulse. (3.54 X 10-7 moles) f) Identify the range of the electromagnetic spectrum of this laser. Wave Nature of Matter • Louis DeBroglie • All matter has wave and particle properties • “matter waves” l= h mv momentum • Everything has a wavelength •Wavelike properties only matter for small objects •Electrons, protons, light, etc… Diffraction pattern of electrons scattered off aluminum foil DeBroglie Wavelength: Ex 1 Calculate the wavelength of a baseball of mass 0.20 kg moving at 40.25 m/s (90 mph) l= h mv l = (6.626 X 10-34 J s) (0.20 kg)(40.25 m/s) l = 8.2 X 10-35 m DeBroglie Wavelength: Ex 2 Calculate the wavelength of an electron (9.109 X 10-31 kg) moving at 2.2 X 106 m/s l= h mv l = (6.626 X 10-34 J s) (9.11 X 10-31 kg)(2.2 X 106 m/s) l = 3.3 X 10-10 m or 0.33 nm DeBroglie Wavelength: Ex 3 What velocity must a neutron (1.67 X 10-27 kg) move to have a wavelength of 500 pm (1 pm = 1X10-12 m)? 500 pm l= h mv v= h ml 1X10-12 m 1 pm = 5.0 X 10-10 m = (6.626 X 10-34 J s) = 794 m/s (1.67 X 10-27 kg)(5.0 X 10-10 m/s) Quantum Mechanical Model 1. Electron is both wave and particle 2. Probabilistic view – Where does the electron “hang out” rather than where the electron “is Quantum Mechanical Model Electron as a particle • Heisenberg Uncertainty Principle – can never know both the position and velocity of an electron at the same time • Results a. Electron moves randomly (not like a planet) b. Electron cloud – 90% probability c. Important for transistors/chips nucleus Random electron cloud Quantum Mechanical Model Electron as Wave • Schrodinger Wave Equation (1926) – treats electron solely as a wave Quantum Mechanical Model Result One Explains the forbidden zone (waves do not match) Forbidden Zone Quantum Mechanical Model Result Two Orbital are not circular a. Orbital – region of space where there is a significant chance of finding an electron b. Does not move like a planet Three Major Discoveries Light is both a wave and a particle Electron Orbitals are Quantized Planck Einstein Bohr The electron (and all matter) is both a wave and a particle De Broglie (Later Schrodinger/ Heisenberg) Quantum Numbers 1. First (principal) QN (n)– how far the electron is from the nucleus (larger the number, farther away) – Level or shell n=2 n=1 2. Second (azimuthal) QN (l) – the shape of the orbital Value of l Letter used 0 s 1 p 2 d 3 f Quantum Mechanical Model 3. Third (magnetic) QN (ml)– the suborbital Orbital # suborbitals s p d f 0 3 (px,py,pz) 5 7 Link to Periodic table Total e- 2 6 10 14 • In a magnetic field, you can “see” the three p suborbitals in a line spectrum 4. Fourth (spin) QN (ms)– spin of the electron Pauli Exclusion Principle – No two electrons in an atom can have the same four quantum numbers – – Two electrons in the same suborbital (ex: py) must have opposite spins Can have values of +1/2 or -1/2 Shows the two electrons, spin +½ and spin -½ n Possible Values of l Possible Values of ml Subshell name 1 0 0 1s 2 0 1 0 -1, 0, +1 2s 2p 3 0 1 2 0 -1, 0, +1 -2, -1, 0, +1, +2 3s 3p 3d 4 0 1 2 3 0 -1, 0, +1 -2, -1, 0, +1, +2 -3, -2, -1, 0, +1, +2, +3 4s 4p 4d 4f Quantum Numbers: Ex 1 What is the designation for a subshell n=5 and l =1? How many suborbitals are in this subshell? What are the values of ml for each of the orbitals? Quantum Numbers: Ex 1 What is the designation for a subshell n=5 and l =1? 5p How many orbitals are in this subshell? 3 What are the values of ml for each of the orbitals? -1, 0, +1 Quantum Numbers: Ex 2 Which of the following sets of quantum numbers are not allowed in the hydrogen atom? A. B. C. D. n=2, l=0, ml =0, ms=1/2 n=1, l=0, ml=0, ms= -1/2 n=3, l=1, ml= 2, ms=1/2 n=4, l=2, ml= -2, ms=1/2 Quantum Numbers: Ex 2 Which of the following sets of quantum numbers are not allowed in the hydrogen atom? A. B. C. D. n=2, l=0, ml =0, ms=1/2 n=1, l=0, ml=0, ms= -1/2 n=3, l=1, ml= 2, ms=1/2 n=4, l=2, ml= -2, ms=1/2 Quantum Numbers: Ex 3 Which of the following sets of quantum numbers are not allowed in the carbon atom? A. B. C. D. n=2, l=0, ml =0, ms=1/2 n=1, l=1, ml=0, ms= -1/2 n=2, l=1, ml= -1, ms=1/2 n=4, l=2, ml= -3, ms=1/2 Quantum Numbers: Ex 3 Which of the following sets of quantum numbers are not allowed in the carbon atom? A. B. C. D. n=2, l=0, ml =0, ms=1/2 n=1, l=1, ml=0, ms= -1/2 n=2, l=1, ml= -1, ms=1/2 n=4, l=2, ml= -3, ms=1/2 Electron Configurations 1. Electron Configuration – shorthand notation to tell you the locations of all the electrons in an atom or ion 2. Notation 2p3 Orbit Shape # e- Electron Configuration Electron Configurations Why is “d” one less? • Energy is slightly above the next s orbital • Complete the orbital filling diagram for manganese Electron Configurations 1. Easy Examples H He O 2. Write e- configuration Li V N F Sr Ar P Se Mg Kr Fe S Electron Configurations 3. Which element is represented by the following electron configurations? 1s22s22p63s23p64s23d5 1s22s22p63s23p64s23d104p65s24d7 1s22s22p63s23p64s1 1s22s22p63s23p3 1s22s22p63s1 1s22s22p63s23p2 1s22s22p63s23p64s23d104p6 Noble Gas (Condensed) Shortcut 1. Rule – Use the noble gas in the previous row 2. Examples Ne and P Ru Kr You try: Br Ar S Ca I Xe Electron Configuration Exceptions • Mostly with transition metal elements • There is a special stability to filled and half-filled orbitals Element Cr Mo Cu Ag Actual configuration [Ar]4s13d5 [Kr]5s14d5 [Ar]4s13d10 [Kr]5s14d10 Instead of [Ar]4s23d4 [Kr]5s24d4 [Ar]4s23d9 [Kr]5s24d9 Ions p Sr Sr+ 2+ Sr 2+ Al Al3+ e e- configuration p S 1S S21Br Ba 2+ Ba B3+ e e- configuration Ions Transition Metal Ions Lose their “s” electrons first. Fe [Ar]4s23d6 Fe1+ Fe2+ Fe3+ Cu Cu2+ Ru Ru3+ Transition Metal Ions Zn2+ Co1+ Co2+ V3+ Sc2+ Sc3+ 4. a) Increase b) Decrease c) line spectrum(forbidden zone) 14. a) Gamma < (d)yellow < (e)red < (b)93.1 radio < (c)680 AM 16. a) 3.00 X 1017 Hz b) 3.94 X 10-3 m c) (a) is X-Ray and (b) is microwave d) 7.64 X 10-6 m 18. UV has a higher frequency and shorter l than IR. UV produces more energy 22.a) 5.09 X 1014Hz b) 20.3 kJ c) 3.37 X 10-19 J d) Na+ 24. AM: 6.69 X 10-28 J FM: 6.51 X 10-26 J 26.1.56 X 10-18 J/photon, 127 nm 28. a) microwave b) 6.4 X 10-11 J/hr 34.a) Absorbed b) Emitted c) Absorbed 36.a) 9.7 X10-8 m, emitted b) 434 nm, emitted c) 1.06 X 10-6m, absorbed 38. a) n=1 larger DE b) 121 nm, 103 nm, 97.2 nm 40. a)2.626X10-6m (IR) b) 6 4 transition 42. 3.97 X 10-10 m (3.97 A) 44. 7.75 X 10-11 m (0.775 A) 50. a) n = 3 (l=2,1,0) (9 ml values) b) n = 5 (l = 4,3,2,1,0) (25 ml values) 52. a) 2,1,1 2,1,0 2,1,-1 b) 5,2,2 5,2,1 5,2,0 5,2,-1 5,2,-2 54. 2p not allowed 1s 4d not allowed 5f 3d not allowed Solution to 41c 6.941 g X 1kg =0.006941kg/mol 1mol 1000g 0.006941 kg X 1mol = 1.15 X 10-26kg/atom 1mol 6.022X1023 atoms l= h (1.15 X 10-26kg)(2.5 X 105 m/s) = 2.3 X 10-13 m 66 C P Ne Config [He]2s2sp2 [Ne]3s23p3 [He]2s22p6 Core 2 10 2 Valence 4 5 8 Unpaired 2 3 0 68. a) [Ar]4s23d104p1 (1 unpaired) b) [Ar]4s2 (0 unpaired) c) [Ar]4s23d3 (3 unpaired) d) [Kr]5s24d105p5 (1 unpaired) e) [Kr]5s24d1 (1 unpaired) f) [Xe]6s14f145d9 (2 unpaired) g) [Xe]6s24f145d1 (1 unpaired) 70. a) [Ar]3d10 b) [Xe]4f145d8 c) [Ar]3d3 d) [Ne]3s23p6 72.a) 7A b) 4B c) 3A d) Sm and Pm (f-block) 74.a) [He]2s22p3 b) [Ar]4s23d104p4 c) [Kr]5s24d7 76.a) Ba Ca K Na b) Au (shortest) Na(longest) c) 455 nm, Ba 78 a) 9.37X1014 s-1 b) 374 kJ/mol c) UV-B Shorter l, higher energy d)UV-B 100. O3 O2 + O DH = 105.2 kJ/mol DH = 1.052 X 105 J/mol 1.052 X 105 J 1 mol = 1.75 X 10-19 J 1 mol 6.022X1023 molec. DE = hu l = c/ u u = 2.63 X 1014 Hz l = 1140 nm Extra Problem A certain biomolecule requires 598 kJ/mol to break one of its bonds. What wavelength of light (nm)would a single photon need to break this bond? What region of the EM spectrum is this? A photon of wavelength 550 nm will break a bond in a synthetic dye. Calculate the energy per mole of that bond (kJ/mol) (218 kJ/mol) a) Calculate the wavelength of light emitted from a 72 transition in a hydrogen atom. (398 nm) b) Calculate the speed an electron would need to have the wavelength. (1830 m/s) c) Calculate the wavelength of a proton moving at that same speed. (2.17 X 10-10 m or 0.217 nm) d) Why is the proton’s wavelength so much smaller? Which has more energy, e- or p+? me = 9.109 X 10-31 kg mp = 1.673 X 10-27 kg a) Calculate the wavelength of a proton that has been accelerated to 2 X 106 m/s. (1.98X10-13 m) b) Calculate the frequency. (1.51 X 1021 Hz) c) Calculate the energy. (1.00 X 10-12 J/photon) mp = 1.673 X 10-27 kg