WA 8 Dearing

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Apr 3, 2024

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Name: Eric Dearing College ID: 0550796 Thomas Edison State College General Chemistry I with Labs (CHE-121) Section no.: Semester and year: May 2014 Written Assignment 8: The Electronic Structure of Atoms Answer all assigned questions and problems, and show all work. 1. The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5 × 10 14 Hz. (a) Calculate the wavelength, in nm, associated with this radiation, and (b) calculate the energy, in joules, of a single photon associated with this frequency. (8 points) (Reference: Chang 7.16) Wavelength = c – 3.00x10 8 m/s / v – 7.5 × 10 14 Hz /10 -9 = 4.0 x 10 2 nm Energy = hv, (6.63x10 -34 )( 7.5 × 10 14 Hz) = 5.0x10 -19 J 2. (a) What is an energy level? Explain the difference between ground state and excited state. (b) What are emission spectra: How do line spectra differ from continuous spectra? (6 points) (Reference: Chang 7.23) (a) An energy level is the possible locations around an atom where electrons of specific values may be found. (b) Emission spectra is the range of electromagnetic radiation emitted by a substance when an atom jumps from a higher energy state to a lower one. Line spectra is light released at a specific wavelength where as Continuous spectra comes from white light and contains all colors and frequencies. 3. Explain the statement, Matter and radiation have “dual nature.” (3 points) (Reference: Chang 7.35) Dual nature refers to the fact that matter can exhibit wave or radiation like properties and radiation can behave like particles or matter. 4. Which of the four quantum numbers ( n , ℓ, m , m s ) determine (4 points) (a) the energy of an electron in a hydrogen atom and in a many-electron atom = n (b) the size of an orbital = n (c) the shape of an orbital = (d) the orientation of an orbital in space. m (Reference: Chang 7.54) 5. Determine the maximum number of electrons that can be found in each of the following subshells: 3 s , 3 d , 4 p , 4 f . (2 points) (Reference: Chang 7.65) 3 s = 2 electrons , 3 d = 10 electrons , 4 p = 6 electrons , 4 f = 14 electrons 6. How many electrons would fill the third energy level ( n = 3)? (2 points) 2+6+10 = 18 electrons max 1
7. Explain the meaning of the symbol 4 d 6 . (2 points) (Reference: Chang 7.72) 4 = n or energy level, d = suborbital, 6 = number of electrons in d suborbital 8. For each of the following pairs of subshells, indicate which is higher in energy: (a) 2 s or 2 p , (b) 3 d or 4 p , (c) 4 s or 4 p , (d) 3 d or 4 s . (4 points) (a) 2 s = 2+0=2, 2 p = 2+1=3, 2p higher (b) 3 d = 3+2=5, 4 p =4+1= 5 , 3d & 4p have same energy level (c) 4 s = 4+0=4, 4 p =4+1=5, 4p higher (d) 3 d = 3+2=5, 4 s =4+0=4, 3d higher 9. Explain the meaning of diamagnetic and paramagnetic. Give an example of an element that is diamagnetic and one that is paramagnetic. What does it mean when we say that electrons are paired? (3 points) (Reference: Chang 7.73) Paramagnetic substances contain net unpaired spins and are attracted by a magnet. ↑↑ or↓↓ Diamagnetic substances do not contain net unpaired spins and are slightly repelled by a magnet. ↑↓ or ↓↑ 10. Write the ground-state electron configurations for the following elements: (12 points) a. B = 1 s 2 2 s 2 2 p 1 b. V = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4s 2 3 d 3 c. Ni = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4s 2 3 d 8 d. I = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4s 2 3 d 10 4 p 6 5s 2 4d 10 5p 5 (Reference: Chang 7.89) 11. The electron configuration of a neutral atom is 1 s 2 2 s 2 2 p 6 3s 2 . Name the element. (2 points) Na Sodium (Reference: Chang 7.91) 12. A 368 g sample of water absorbs infrared radiation at 1.06 × 10 4 nm from a carbon dioxide laser. Suppose all the absorbed radiation is converted to heat. Calculate the number of photons at this wavelength required to raise the temperature of the water by 5.00ºC. (4 points) (Reference: Chang 7.107) (1.06 × 10 4 nm) (1x10 -9 ) = 1.06 × 10 -5 m ms∆T = (368g)(4.184 J)(5C) = 7.70 x 10 3 J hc/wavelength = (6.63x10 -34 )(3.00x10 8 m/s ) / (1.06 × 10 -5 nm) = 1.88 x 10 -20 J 7.70 x 10 3 J / 1.88 x 10 -20 J = 4.10 x 10 23 photons 2
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