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AP Review: Unit 6A

AP Review: Unit 6A. By Harrison Alch and Karen Sittig May 1, 2008. Quantum Mechanics: Constants. c = the speed of light = 2.998*10 8 m/s λ = wavelength (generally in nm) h = Planck’s constant = 6.63*10 -34 J*s E = energy (generally in J) m = mass mass of an electron = 9.11*10-31 kg .

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AP Review: Unit 6A

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  1. AP Review: Unit 6A By Harrison Alch and Karen Sittig May 1, 2008

  2. Quantum Mechanics: Constants • c = the speed of light = 2.998*108 m/s • λ = wavelength (generally in nm) • h = Planck’s constant = 6.63*10-34 J*s • E = energy (generally in J) • m = mass • mass of an electron = 9.11*10-31 kg

  3. Quantum Mechanics: Equations • c=λv • Wavelength of a photon (massless) • E = hv • Energy of a photon • λ = h / mv • deBroglie wavelength – takes mass of a particle into account

  4. Example problem • It requires a photon with a minimum energy of 4.41 x 10-19 J to emit electrons from sodium metal. • What is the minimum frequency needed to emit electrons from sodium via the photoelectric effect? • E = hv • 4.41 x 10-19 J = (6.626 x 10-34 J*s) * v • v = 6.66 x 1014 Hz • Calculate the wavelength of one photon of this light. What color would it be? • λ = c/v • Λ = (2.998 x 108 m/s)/(6.66 x 1014 Hz) • λ = 450 nm • This light would be blue

  5. Example problem cont. • If sodium is irradiated with light of 439 nm, what is the maximum possible kinetic energy of the emitted electrons? • E439 = hc/ λ • E439 = (6.626 x 10-34 J*s)(2.998 x 108 m/s)/(439 x 10-9 m) • E439 = 4.52 x 10-19 J • Ek = E439 – Emin • Ek = (4.52 x 10-19 J) – (4.41 x 10-19 J) • Ek = 1.1 x 10-20 J/e-

  6. IMFAs • “Intermolecular forces of attraction” • General types: • London Dispersion Forces • Dipole-Dipole Forces • Ion-Dipole Forces • Hydrogen Bonds

  7. London Dispersion Forces • Exist between all atoms and molecules • Are the only forces between nonpolar atoms and molecules • The weakest type of intermolecular forces, caused by instantaneous dipoles • Dispersion forces tend to increase in strength with increasing molecular weight.

  8. Dipole-Dipole and Ion-Dipole • Dipole-Dipole Forces • Exist between polar molecules when the net positive end of one attracts the net negative end of another • Only exist when molecules are close together. • Ion-Dipole Forces • exist between ions and polar molecules • A cation attracts the negative ends of polar molecules, an anion attracts the positive ends.

  9. Hydrogen Bonds • Exist only between an H bonded to an O, F, or N in a very polar bond and usually another O, F, or N, or other very electronegative atom. • The strongest type of intermolecular forces. from http://kentsimmons.uwinnipeg.ca/

  10. Comparative Strength of IMFAs • When molecules of 2 substances are relatively the same mass, the different strengths of attractive forces are due to differences in the length of the dipole moment. • When molecules of 2 substances are different masses, the different strengths of attractive forces are due to the strength of dispersion forces (the more massive one generally has stronger attractive forces).

  11. Some Properties of Liquids • Viscosity • The resistance of liquid to flow • Liquids that have stronger IMFAs are more viscous • Viscosity also increases with the increasing ability of molecules to become tangled (some isomers are more viscous than others) • Surface Tension • The amount of energy required to increase the surface area of a liquid by a given amount • Liquids that have stronger IMFAs tend to have more surface tension • The surface tension of water at 293 K is 0.0729 J/m2 • The surface tension of mercury at 239 K is 0.46 J/m2

  12. Example Problems • Why is the surface tension of CHBr3 greater than CHCL3? • CHBr3 has a higher molar mass, is more polarizable, and has stronger dispersion forces, so the surface tension is greater. • Why does as temperature increases, oil flow faster through a narrow tube? • The viscosity of the oil decreases because the average kinetic energy of the molecules increase.

  13. Periodic Trends • Size • Generally, size decreases across a period and increases down a family • As the nuclear charge increases but no new electron orbitals are added, the valence electrons are held more strongly by the nucleus, decreasing the size • In transition metals, the electrons in filled d orbitals tend to repel one another strongly enough to result in a deviation from this trend • Electron Affinity • Generally, electron affinity increases across a period and decreases down a family • Across a period, the effective nuclear charge increases, more readily attracting electrons • Down a family, the distance from the nucleus decreases, making it more difficult to attract electrons

  14. Period Trends cont. • Ionization Energy • Ionization energy tends to increase across a period and up a family • As more protons are added to the nucleus, they increase the effective nuclear charge and hold the valence electrons more tightly, so removing an electron requires more energy • As more orbitals are added, the valence electrons are farther away from the nucleus, reducing the pull of the protons on the electrons and allowing them to be removed more easily. • Ionic Radii • Cations are smaller than their parent atoms due to the remaining electrons being held more tightly than before • Anions are larger than their parent atoms due to increased electron-electron repulsions • For a series of isoelectronic ions and atoms, atomic radius decreases with increasing atomic number (and increasing nuclear charge)

  15. Example Problems • Put these atoms in order of increasing size: Na, Ca, K, Be, & S • Be, S, Na, Ca, K • Which gas has a lower boiling point, Cl2 or Kr? • Kr would have a lower boiling point, despite its higher molar mass. • Which has stronger IMFAs, CH4 or CH3OH? • CH3OH would have stronger IMFAs due to its polar –OH group.

  16. Drawing Lewis Structures • Pick a central atom. This should be of the element with the lowest electronegativity (generally from families 3, 4, or 5, or a noble gas) • Determine how many bonds are needed by subtracting the valence electrons (number of family) from the total electrons needed to complete octets (or duets etc) and dividing by two. • Connect each atom with a single bond. • Add more bonds if needed. Otherwise, complete the octets on the outer atoms. • If there are extra electrons, add them to the central atom.

  17. Example Problems • XeF4 • CH4 • NH3 • H3O+

  18. Molecular Shapes • Can be classified by electron geometry or molecular geometry • Electron geometry doesn’t take lone pairs into account when assigning names, molecular geometry does • The hybridization of the central atom is determined by the electron geometry

  19. Electron Geometry Linear 2 bonding domains No hybrid orbitals Tetrahedral 4 bonding domains sp2 hybridized Trigonal Planar 3 bonding domains sp hybridized Pictures from www.wikipedia.org

  20. Electron Geometry cont. Octahedral 6 bonding domains sp2d2 hybridized Trigonal Bipyramidal 5 bonding domains sp2d hybridized Pictures from www.wikipedia.org

  21. Molecular Geometry Bent 1 or 2 lone pairs sp or sp2 hybridized See-Saw 6 bonding domains sp2d hybridized Trigonal Pyramidal 1 lone pair sp2 hybridized Pictures from www.wikipedia.org

  22. Molecular Geometry cont. Linear 3 lone pairs sp2d hybridized T-Shaped 2 lone pairs sp2d hybridized Square Pyramidal 1 lone pair sp2d2 hybridized Square Planar 2 lone pairs sp2d2 hybridized Pictures from www.wikipedia.org

  23. Sample Problems • Name the electron geometry: • XeF4 • CH4 • SF6 • Name the molecular geometry: • NH3 • H2O • PCl5 • State the hybridization of the central atom: • XeF2 • BrF5 • SO2

  24. Solutions to Sample Problems • Name the electron geometry: • XeF4 - Octahedral • CH4 - Tetrahedral • ClF3 – Trigonal Bipyramidal • Name the molecular geometry: • NH3 – Trigonal Pyramidal • H2O- Bent • PCl5 - Octahedral • State the hybridization of the central atom: • XeF2 -sp2d2 • BeCl2 - unhybridized • SO2 – sp2

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