Chemistry 200: General Chemistry I - LectureDr. Namphol SinkasetChapter 7 Study GuideConcepts1. Electromagnetic radiation consists of energy propagated by electric and magneticfields. These fields propagate through space in waves that are perpendicular to oneanother.2. EM waves have two important characteristics: frequency (ν) and wavelength (λ).These two characteristics are related by the speed of light (c).3. The electromagnetic spectrum is a continuum of all wavelengths of EM radiation.4. Since EM radiation can be pictured as a wave, it has the typical wave properties ofinterference and diffraction. Interference is the term used to describe how wavescan combine in different ways to make new waves. Diffraction describes how wavescan regenerate themselves after encountering obstacles or slits. When both of theseproperties are displayed, interference patterns are observed.5. The wave explanation of EM radiation underwent substantial change due to the ob-served phenomena of blackbody radiation, the photoelectric effect, and atomic spectra.6. When explaining blackbody radiation, Max Planck invoked the idea that atoms canonly possess and emit certain quantities of energy. These discrete packets of energywere called quanta (singular, quantum).7. To explain the photoelectric effect, Albert Einstein proposed that light had particlesof energy called photons.8. The Rydberg equation was formulated to predict lines in atomic spectra, but didnot explain why lines were observed.9. Niels Bohr combined Planck’s and Einstein’s ideas into a model of the hydrogen atom.(1) The H atom only has certain allowable energy levels. Each level is a fixed circularorbit of an e−around the nucleus. (2) The atom doesn’t radiate energy if the e−staysin its orbit. (3) When the e−changes orbit, it must emit or absorb energy to do so.The energy is exactly equal to the difference in energy between the orbits.10. Since it was observed that energy manifested as EM radiation can act as a particle,Louis de Broglie reasoned that matter must also be able to act like a wave.11. The Heisenberg Uncertainty Principle states that it’s impossible to know thelocation and momentum of a particle simultaneously. This phenomenon is on displayin the laser/double slit electron experiment. This applies specifically to the atom inthat it’s impossible to know the exact location of the e−.112. Quantum mechanics is the study of the wave nature of things at the atomic scale.13. Solution of the Schr¨odinger equation leads to a number of different wavefunctions(ψ). A wavefunction describes the state of the e−matter/wave.14. A wavefunction is often referred to as an atomic orbital.15. Electrons do not occupy atomic orbitals like orbits (i.e. they are not “in”orbitals); an atomic orbital is an electron.16. An atomic orbital is specified by 3 quantum numbers: principal quantum number,angular momentum quantum number, and magnetic quantum number.17. The principal quantum number (n) is a positive integer that indicates the relativesize of the orbital. It also gives the atom’s energy levels or shells.18. The angular momentum quantum number (`) is an integer from 0 to n − 1. Itis related to the shape of the orbital. This integer is also known as the sublevel orsubshell. Each sublevel has a letter designation: ` = 0, 1, 2, 3, 4 . . . corresponds tos, p, d, f, g . . ..19. The magnetic quantum number (m`) is an integer from −` to +`, including 0. Itindicates the orientation of the orbital.20. Sublevels of an atom are indicated by a combination of the n value and the letterdesignation of the ` value, e.g. for n = 3 and ` = 2, the designation is 3d.21. An orbital is specified by each allowed combination of n, `, and m`values.22. s orbitals are spherical in shape and are completely symmetric.23. 2p orbitals are dumbbell shaped. There are 3 possible orientations.24. Four of the 3d orbitals resemble 4-leaf clovers, and the third looks like a p orbital witha doughnut in the middle.25. The 4f orbitals look like the hilts of daggers and balloon animals.26. The energy of an e−in a hydrogen atom depends ONLY on the principal quantumnumber. That is, if the e−has enough energy to be in the 2nd energy level, it couldchoose to manifest itself as 2s or 2p because in either case, the energy is the same.Equations1. c = νλ2. E = nhν3. Ephoton= hν4.1λ= R1n21−1n2225. λ =hmvRepresentative Problems8. An FM station broadcasts classical music at 93.5 MHz (megahertz, or 106Hz). Find the wavelength (in m, nm, and˚Angstrom) of these radio waves.First, we convert MHz to Hz. Recall that Hz is the unit for 1/s.93.5 MHz ×106Hz1 MHz= 9.35 × 107HzNow we calculate the wavelength associated with this frequency.c = νλ3.00 × 108m/s = (9.35 × 1071/s)λλ = 3.208556mλ = 3.21 mThis corresponds to 3.21 × 109nm and 3.21 × 1010˚A.12. Rank the following photons in terms of decreasing energy: (a) IR (ν =6.5 × 1013s−1); (b) microwave (ν = 9.8 × 1011s−1); (c) UV (ν = 8.0 × 1015s−1).According to the equation Ephoton= hν, energy is directly related to the frequency, ν.Thus, the higher the frequency, the more energy the photon possesses. In order of decreasingenergy, we have: UV > IR > microwave.24. Use the Rydberg equation to calculate the wavelength (in˚A) of the photonabsorbed when a hydrogen atom undergoes a transition from n = 1 to n = 3.To solve this problem, we plug in values into the Rydberg equation.31λ= R1n21−1n22= (1.096776 × 107m−1)112−132= (1.096776 × 107m−1)11−19= (1.096776 × 107m−1)89= 974912¯0 m−1λ = 1.025733604674× 10−7m= 1.025734 × 10−7mNow we convert to˚Angstroms.1.025734 × 10−7m ×1 × 1010˚A1 m= 1025.734˚A48. What feature of an orbital is related to each of the following quantumnumbers? (a) Principal quantum number (n); (b) Angular momentum quantumnumber (`); (c) Magnetic quantum number (m`).(a) The principal quantum number gives an indication of the orbital’s size and energylevel. (b) The angular momentum quantum number gives an indication of the orbital’sshape. (c) The magnetic quantum number gives an indication of the orbital’s orientation.55. For each of the following, give the sublevel designation, the allowable m`values, and the number of orbitals: (a) n = 4, ` = 2; (b) n = 5, ` = 1; (c) n = 6,` = 3.(a) n = 4 and ` = 2 is a 4d orbital. The allowed m`values are -2, -1, 0, 1, 2. Thereare 5 4d orbitals. (b) n = 5 and ` = 1 is a 5p