Chemistry 111 Week 9 Lecture Notes
Chemistry 111 Week 9 Lecture Notes 111/40551
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This 10 page Class Notes was uploaded by email@example.com Notetaker on Friday November 6, 2015. The Class Notes belongs to 111/40551 at University of St. Thomas taught by Uzcategui-White in Summer 2015. Since its upload, it has received 14 views.
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Date Created: 11/06/15
Chemistry Week 9 Lecture Notes: 7.10 An X Ray has a a wavelength of 1.3 Angstroms. Calculate the energy (in J) of one photon of this radiation. Y = ? c= y x v à c/y E= hv E= h x c/ Y à 6.626 x 10 ^ -‐34 J s x 3.00 x 10^ 8 m/s / 1.3 A x 10 ^-‐10 m/ 1 A = 1.5 x 10^-‐15 J/ photon 7.13 A gamma ray emitted by Cobalt 60 has an energy of 1.33 MeV (million electron volts, 1 eV= 1.60 x 10^-‐19 J). What is the frequency (in Hz) and the wavelength (in m) of the ray? E= 1.33 MeV V= ? Y (wavelength) = ? E= hv à v = E/ h -‐à v = 1.33 MeV (10^6 eV/ 1 MeV) (1.602 x 10^-‐19 J/ 1 eV)/ 6.626 x 10^-‐34 J s V= 3.22 x 10^20 s^-‐1 (Hz) Y = c/v -‐à 3.00 x 10^8 m/s/ 3.22 x 10^20 1/s = 9.32 x 10 ^-‐13 m Section 7.2: Atomic Spectra-‐ o Particle/ atom (energy) à “excited” (move electrons to higher energy level) à give away extra energy à atomic spectra (only see colored lines) o Atomic spectra: line spectrum of different wavelengths o Line emission spectra: series of lines with black in between (Balmer series) o Equation = Rydberg Equation: (predict different wavelengths, n2 > n1) 1 / Y = R (1/ n1^2 – 1/n2^2) R= Rydberg constant= 1.096776 x 10^7 m^-‐1 Bohr’s Model of the Hydrogen Atom: o H atom only has certain energy levels= stationary states o Higher the energy level, farther from the nucleus o H atom in first orbit= lowest energy= ground state o In stationary state-‐ does not radiate energy o Atoms changes states by emitting/ absorbing photons o Electrons fall from high energy à lower = emit light o Light emitted by excited atoms= specific lines at certain wavelengths o Get excited by giving energy through heat or electricity • Only true for Hydrogen atoms • Equation: E= -‐2.18 x 10^-‐18 J (z^2/n^2) z = 1 Change in E = -‐2.18 x 10^ -‐18 J (1/ n^2 final – 1/ n^2 intial) • Conditions: a) Emitting energy: electrons move closer to the nucleus change in E < 0 b) Absorbing energy: electrons move away from the nucleus change in E > 0 What do we use excited electrons for? 1. Visible light 2. Identify atoms (using wavelength) 3. Quantify color solutions (energy absorbing/emitting light) Beer’s law: A = E b c A= absorption of light, E = constant (in problem), b= 1 cm (constant), c = concentration 7.19 Use the Rydberg Equation to find Y (in nm) of photon emitted when an H atom undergoes a transition from n= 5 to n=2 Y= ? 1/Y = 1.096776 x 10^7 m^-‐1 x (1/ 2^2 – 1/ 5^2) = 2303229.6 m^-‐1 Y= 1/ 2303229.6 1/m x 1 x 10^-‐9 m/ 1 nm) = 434.17 nm 7.21 Calculate the energy difference for 1 mol of H atoms (using #19) E= -‐2.18 x 10^-‐18 J (1/ 2^2 – 1/5^2)= -‐4.58 x 10^-‐19 J/ photon -‐4.58x 10^-‐19 J/ photon x 6.022 x 10^23 photon/ 1 mol = -‐2.76 x 10^5 J/ mol 7.23 Arrange the H atom electron transitions in order of increasing frequency. a) N= 2 to n= 4 b) N= 2 to n=1 c) N=2 to n=5 d) N= 4 to n=3 • The bigger the difference= the more energy= higher frequency • Wavelength = opposite Answer: D < A < C < B 7.26 An electron in the n= 5 level of an H atom emits a photon of wavelength 1281 nm. To what energy level does it move? N= 5 N= ? Y = 1281 nm à 1281 nm x 1 m/ 1 x 10^-‐9 nm = 1.281 x 10^-‐6 m Emits = moves down; n2 = 5 1/ (1.281 x 10^-‐6 m) (1.099776 x 10^7 1/m) = .07118= 1/ n1^2 – 1/25 1/n1^2 = .1118 n1^2= 1/ .1118 à 8.994 n1= 3 Section 7.4: The wave-‐ particle duality of matter and energy Louis de Broglie: -‐ Matter is wave-‐ like -‐ Energy à wave à matter of particle Y= h/ m x U Y= wavelength (J x s), m= mass (in kg), U= velocity of particle (m/s) -‐ Wavelength is inversely proportional to mass -‐ We do have a wavelength in our movement à hard to see -‐ Mass increases, speed increases, wavelength decreases o Momentum: Y= h/P (P= momentum, P= m x U) de Broglie/ Einstein Equation -‐ Shorter wavelengths = higher energy= higher momentum -‐ Shorter wavelengths = more dangerous Heisenberg’s Uncertainty: o “It is impossible to know simultaneously the position and momentum (mass x speed) of a particle”. Change in X x mass x change in U > h/ 4π U = uncertainty X= position 7.31 A 232 lb fullback runs 40 yards at 19.8 +- .1 mi/h a) what is the de Broglie wavelength (in m)? b) What is the uncertainty in his position? a) Y= h/ m x U M= 232 lb x 1 kg/ 2.205 lbs = 105 kg U= 19.8 mi/ h x 1 km/ .620 mi x 1 hr/ 3600 s x 1000 m/ 1 km = 8.87 m/s Y = 6.626 x 10^ -34 J/s/ 105 kg x 8.87 m/s = 7.10 x 10^ -37 m b) X x m x U > h/ 4π U = .1 mi/ h x 1km/ .620 mi x 1000 m/ 1 km x 1 hr/ 3600 s = 5 x 10^2 m/s X = 6.626 x 10^-34 J / 105 kg x 5 x 10^2 m/s x 4 π = 1 x 10^-35 m The Quantum- Mechanical Model of the Atom: o The matter wave of the electron occupies the space near the nucleus and is continuously influenced by it. o The Schrodinger wave equation allows us to solve for the energy states associated with a particular atomical orbit HW = EW W^2 = probability of finding electrons (electron density) Orbitals à location of an electron (regions of space within an electron) Quantum Numbers and Atomic Orbitals: o Quantum number is used to identity energy states and orbitals o Orbital specified by 3 quantum numbers: 1) Principal quantum number (n): positive integers -‐ Indicates the relative size of the orbital and its distance from the nucleus -‐ Electron shell or energy level 2) Angular Momentum Quantum Number (l): integer # that goes from 0 to n-1 -‐ Value l indicates the shape of the orbital -‐ Subshell or sublevel L = 0 à ‘s’ sublevel L= 1 à ‘p’ sublevel L= 2 à ‘d’ sublevel L= 3 à ‘f’ sublevel 3) Magnetic Quantum Number (ml): value of –l, 0, +l -‐ Indicates the spatial orientation of the orbitals (# of orbitals) N= 1 L= 0 Ml = 0 N= 2 L= 0,1 Ml = 0, -1, 0, +1 (4 total orbitals) Shape of Orbitals o ‘s’ orbital = spherical o node region = 0 probability of finding electrons o 3s > 2s > 1s (in size) o angular number (l) -‐ l= 0 à spherical -‐ ‘p’ sublevel (dumbbell) à two regions of high electron density -‐ 1 node -‐ ml = -1,0,+1 o l= 0 à s sublevel o l= 1à p sublevel o l= 2 à d sublevel o l=3 à f sublevel o d sublevel: -‐ l= 2 à m1= -2, -1, 0, 1 , 2 (s orbitals/regions of high electron density) -‐ 4 clover leaf, 1 donut shaped -‐ both have 2 nodes o n= level of energy (size of orbital) o bigger n value = bigger orbital ex: 4p > 3p > 2p >1p 7.45 Give the sublevel designation, the ml values and the # of orbitals a) n= 2, l=0 sub: 2s ml= 0 orb # = 1 b) n=3 l=2 sub: 3d ml= -2, -1, 0, 1, 2 orb= 5 c) n=5 l=1 sub: 5p ml= -1, 0, 1 orb= 3 7.49 Are the following combinations allowed? If not, fix them. a) N=1, l= 0, ml=0 correct b) N=2, l=2, ml= +1 Wrong;; N=2, l=0, ml=0 N=2, l=1, ml=-1 c) N=7, l=1, ml= +2 Wrong;; Ml= -1 or 0 or +1 d) N=3, l=1, ml= -2 Wrong;; Ml= -1 or 0 or +1 Or N=3, l=2, ml= +2 Chapter 8: Electron Configuration and Chemical Periodicity: Spin quantum number: o Symbol= ms o Values= +1/2, -1/2 o Property: direction of spin Pauli Exclusion Principle: o no two electrons in the same atom can have the same 4 quantum numbers o an atomic orbital can hold a maximum of 2 electrons and must have opposing spin values Energy Level Splitting: o electrostatic interactions between electrons with atoms with more than one electron causes the energy levels to split into sublevels. o The electrostatic effects include: -‐ Effect of nuclear charge (Z) -‐ Effect of electron repulsion and shielding -‐ Effect of orbital shape on orbital energy (penetration) 1. Nuclear charge: -‐ Tendency of nucleus to attract electrons -‐ The closer the electrons are to the nucleus, the more stable they are -‐ The higher the nuclear charge à lower energy/ energy level 2. Electron Repulsion/ shielding: -‐ Shielding effect: preventing attraction to nucleus -‐ Attraction stronger when no other electrons are present -‐ Shielding by inner electrons greatly lowers the full nuclear charge (Z) to an effective nuclear charge (Zeff)