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Honors Chemistry

Honors Chemistry. Chapter 7: Quantum Mechanics. 7.1 Wave Properties. Wavelength ( l ) = distance between two in-phase points Measured in meters Frequency ( n ) = number of waves per second Measured in Hertz (Hz) Amplitude ( y ) = distance of maximum displacement from rest position

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Honors Chemistry

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  1. Honors Chemistry Chapter 7: Quantum Mechanics

  2. 7.1 Wave Properties • Wavelength (l) = distance between two in-phase points • Measured in meters • Frequency (n) = number of waves per second • Measured in Hertz (Hz) • Amplitude (y) = distance of maximum displacement from rest position • Amplitude corresponds to wave energy

  3. 7.1 The Wave Equation • v = ln • Find the wavelength of a 256 HZ (middle C) sound wave traveling at 343 m/s. • v = ln • 343 m/s = l (256 Hz) • l = 1.34 m • Try this…. • Find the frequency of a 25.0 cm wave traveling at 0.75 m/s.

  4. 7.1 Electromagnetic Radiation • James Clerk Maxwell (1873) • Mathematical description of light waves • Light is an electromagnetic wave • Speed of light (c) is constant • c = 2.99792458 x 108 m/s • To 3 sig dig, 3.00 x 108 m/s is fine • Try this…. • Find the frequency of a 250 nm light wave. (Don’t forget about the “nano” prefix!)

  5. 7.1 Electromagnetic Spectrum • Radio, micro, IR, ROYGBIV, UV, X, g • long l ------------------------------- short l • low n -------------------------------- high n • Radio wave end of the spectrum is low energy radiation • Gamma ray end is high energy radiation • Black body radiation • Wave theory fails to account for this!

  6. 7.1 Quantum Theory • Max Planck (1900) • Energy is emitted and absorbed only in small, discrete packets called quanta • Energy of a quantum of energy given byE = hn • h = 6.626 x 10-34 Js (Planck’s constant) • Correctly accounts for blackbody curves • Planck has no idea why it works!

  7. 7.1 Quantum Theory • Find the energy of a 2.50 x 1014 Hz light wave. • E = hn • E = (6.626 x 10-34 Js)(2.50 x 1014 Hz) • E = 1.66 x 10-19 J • A quantum holds a tiny amount of energy! • Try this…. • Find the energy of a 475 nm light wave. • Hint: Use the wave equation first!

  8. 7.2 The Photoelectric Effect • Albert Einstein (1905) • Electrons ejected from surface of metal exposed to light • Depends on frequency of light • Electrons ejected at a certain cutoff frequency • Above cutoff n, electrons leave with more energy • Bright light ejects more electrons • Quantum theory explains results • Light is made of quanta called photons

  9. 7.3 Spectroscopy • Emission spectra – light given off by glowing objects • Can be continuous or discontinuous • Line spectra – series of bright lines emitted by gas phase atoms • Pattern of bright lines is characteristic of the element that is glowing • Absorption spectra – dark lines in spectrum as light passes through a gas

  10. 7.3 Bohr’s Model • Niels Bohr (1913) • Electron energies are quantized • Only certain orbits are allowed • - RHEn = ------ n2 • RH = 2.18 x 10-18 J (Rydberg constant) • n = 1, 2, 3, 4, ….

  11. 7.3 Bohr’s Model • DE = Ef – E0 • -RH -RHDE = ----- - ----- nf2 n02 • Factor out RH • 1 1 DE = RH (----- - ----- ) n02 nf2 Link to Hydrogen energy states

  12. 7.3 Bohr’s Model • Find the energy of a photon of light emitted by an electron jumping from level 5 down to level 2. • DE = RH (1/n52 – 1/n22) • DE = (2.18 x 10-18 J)(1/25 – 1/4) • DE = -4.58 x 10-19 J • Try this…. • Find the energy of the jump from level 1 to level 4. • Find the frequency of the light produced.

  13. 7.4 Duality • Louis de Broglie (1924) • Electrons can be treated as waves • Each orbit must contain a whole number of waves…explains orbit quantization! • hl = ---- mv • mv is momentum (p), so we can write l = h/p • Verified by Davisson, Germer, and Thomson Link to quantum atom model

  14. 7.4 Duality • Find the wavelength of a 3.00 kg duck flying at 5.00 m/s. • l = h/mv • l = (6.626 x 10-34 Js) / (3.00 kg)(5.00 m/s) • l = 4.42 x 10-35 m • Try this…. • Find the wavelength of an electron traveling at 500,000 m/s. (me = 9.11 x 10-31 kg)

  15. 7.5 Uncertainty Principle • Werner Heisenberg (1926) • Complementary variables cannot be known to arbitrary precision • dp dq ≥ħ/2 • Minimum limits to uncertainties in values are inversely proportional • Position and momentum are an important complementary pair • dx dpx≥ħ/2

  16. 7.5 Uncertainty Principle • Find the uncertainty in velocity of an electron confined to a hydrogen atom (dx = 0.037 nm). • dx dpx≥ħ/2 • (3.7 x 10-11 m) dp ≥ 5.27 x 10-35 Js • dp ≥ 1.4 x 10-24 kg m/s • p = mv • 1.4 x 10-24 kg m/s = (9.11 x 10-31 kg) dv • dv = 1.5 x 106 m/s

  17. 7.5 Uncertainty Principle • Try this… • Find the uncertainty in position of a 20.0 mg fly whose position is known to within ±0.5 mm. • Uncertainty limits are not significant for macroscopic objects, but they are significant to subatomic particles • Cannot know the position and momentum of an electron at the same time! • Concept of orbits will not work

  18. 7.5 Quantum Mechanics • Erwin Schrödinger (1926) • Schrödinger equation – treat electron as a standing wave surrounding the nucleus • Schrödinger equation is ugly! • Solve for amplitude function (y) • Remember – amplitude is energy • Produces an energy diagram like Bohr’s, but this one actually works • Wave function has no physical meaning

  19. 7.5 Copenhagen Interpretation • Max Born (1926) • y2 denotes probability • Electron is delocalized • Wave function collapses on observation • electron density refers to magnitude of the probability wave for the electron • Orbital = spatial probability distribution • Electron clouds • Objections: Schrödinger’s Cat • Other interpretations

  20. 7.6 Quantum Numbers • Set of numbers that describe the distribution of electrons in the atom • Principal Quantum Number (n) • n = 1, 2, 3, 4, … • Corresponds to the n value used by Bohr • Describes energy level of the shell • Defines the size of the electron cloud

  21. 7.6 Quantum Numbers • Angular Momentum Quantum Number (l) • l = 0, 1, 2, … , n – 1 • There are a total of n values • Sublevels of the energy level • Angular distribution of electron cloud • For hydrogen, sublevels are degenerate • Correspond to fine structure spectral lines • l = 0 is s orbital l = 2 is d orbital • l = 1 is p orbital l = 3 is f orbital

  22. 7.6 Quantum Numbers • Magnetic Quantum Number (ml) • ml = -l, … , 0, …, +l • There are a total of 2l + 1 values • Number of degenerate orbitals in sublevel • Spatial orientation of the orbital • Zeeman Effect • Electron Spin Quantum Number (ms) • ms = +½, -½ • Two possible electron spin states • Spin up, spin down

  23. 7.7 Atomic Orbitals • s, p, d, f orbitals • Radial probability distributions • distance from nucleus of high e- probability

  24. 7.7 Atomic Orbitals • Angular probability distributions • Show regions of high e- probability • Cool 3d pictures

  25. 7.7 Orbital Energies • n determines energy • For H, all subshells are degenerate • Multielectron atomseach subshell liesat a different energy • Shielding effect • Fill lowest energyorbitals first

  26. 7.7 The Diagonal Rule • Rule of thumb • Shows the orderin which orbitalsare filled • Paramagnetic • Unpaired e- • Attracted to mag • Diamagnetic • Paired e- • Not attracted

  27. 7.8 Electron Configurations • H has 1 electron • Put it in 1s • Write it 1s1 • Read “one-s-one” • What about He? • You got it…1s2 • Keep filling 1s until it is full • But when is it full?

  28. 7.8 Pauli Exclusion Principle • No two electrons may share the exact set of quantum numbers • Consider Helium’s 1s2 configuration • First electron: n = 1, l = 0, ml = 0, ms = +½ • Second electron: same n, l, ml • ms must be -½ • No room for more electrons in 1s orbital • Each orbital can hold only two electrons!

  29. 7.8 Electron Configurations • What is the electron configuration of Li? • 1s2 2s1 • What about N? • 1s2 2s2 2p3 • But how are p electrons organized? • Hund’s Rule – arrange electrons in such a way as to maximize total spin state • Put e-’s in separate orbitals, same spin

  30. 7.9 Aufbau Principle • Build up on previous e- configurations • Each atom adds one more e- • Express configuration with noble gas core • Al = 1s2 2s2 2p6 3s2 3p1 • First 3 terms are the same as Ne config. • Write it as [Ne] 3s2 3p1

  31. 7.9 Exceptions • Particularly stable configurations • Full sublevel • Half-full sublevel • Some transition metals rearrange • Cr = [Ar] 4s2 3d4 • Just missed the stable half-full 3d5 • Kick one e- up to d to get [Ar] 4s1 3d5 • What other family would do this? • Cu = [Ar] 4s2 3d9 Ar 4s1 3d10

  32. 7.9 Periodicity of Electron Configuration

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