Chemistry 101 ANSWER KEY REVIEW QUESTIONS Chapter 7 1. Calculate the wavelength and energy of a photon of radiation with frequency of 2.85x10 12 s –1 . c
3.00x108 m/ s λ = = = 1.05x10­4 m ­1 12 υ 2.85x10 s ­1 E = hυ = (6.634x10­34 J s )(2.85x1012 s ) = 1.89x10­21 J 2. What is the wavelength of radiation with energy of 8.23x10 –19 J? In what region of the electromagnetic spectrum would this radiation be found? E
8.23x10­19 J υ = = = 1.24x1015 s ­1 ­34 h 6.634x10 J s c 3.00x108 m/s λ = = = 2.42x10­7 m = 242 nm (UV) υ 1.24x1015 s ­1 3. For each of the following transitions in the hydrogen atom, calculate the energy, wavelength and frequency of the associated radiation and determine whether radiation is absorbed or emitted during the transition. a) from n=5 to n=1 1
1 1
1 1 = 1.097x107 ( 2 ­ 2 ) m ­1 = 1.097x107 ( ­ ) λ
n1 n 2 1 25 λ = 9.50x10­8 m c 3.00x108 m /s υ = = = 3.16x1015 s ­1 λ 9.50x10­8 m ­1 E = hυ = (6.634x10­34 J s )(3.16x1015 s ) = 2.10x10­18 J Energy is emitted by this transition b) from n=2 to n=7 1
1 1
1 1 = 1.097x107 ( 2 ­ 2 ) m ­1 = 1.097x107 ( ­ ) λ
n1 n 2 4 49 λ = 3.97x10­7 m c 3.00x108 m /s υ = = = 7.56x1014 s ­1 ­7 λ 3.97x10 m ­1 E = hυ = (6.634x10­34 J s )(7.56x1014 s ) = 5.02x10­19 J Energy is absorbed by this transition
1 4. Use the diagrams below to determine the wavelength and frequency of each wave shown. (Assume the same time and distance scale for both waves) l
For wave (a)
l = 0.75 m υ = x (m) 1 wave = 0.125 s ­1 8.0 s For wave (b)
l = 0.60 m υ = x (m) l
1 wave = 0.166 s ­1 6.0 s 5. The energy needed to remove an electron completely from an atom is called its ionization energy. In terms of Bohr’s model, ionization can be considered a process in which the electron moves to an “orbit” of infinite radius. The ionization of a ground­state hydrogen atom can therefore be calculated by assuming that the electron undergoes a transition from ni=1 state to nf=¥ state. Calculate this energy in kJ/mol.
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1 1
1 1
1 = 1.097x107 ( 2 ­ 2 ) m ­1 = 1.097x107 ( ­ ) =0 λ
n1 n 2 1 ¥
¥ λ = 9.12x10­8 m c 3.00x108 m /s υ = = = 3.29x1015 s ­1 ­8 λ 9.12x10 m ­1 E = hυ = (6.634x10­34 J s )(3.29x1015 s ) = 2.18x10­18 J E = 2.18x10­18 J 6.02x10 23 atoms 1 kJ x x 3 = 1.31x103 kJ/mol 1 atom 1 mol
10 J
2 6. Determine if each set of quantum numbers below is permissible or not. If yes, write the orbital designation for each. a) n= 2 l=1 ml= +1 Yes b) n= 1 l=0 ml= –1 No (ml cannot be greater than l) c) n= 4 l= 2 ml= +1 Yes d) n= 3 l= 3 ml= 0 No (l values cannot exceed n–1) 2p 4d 7. Write the quantum numbers associated with each of the following orbitals: a) 4p n= 4 l= 1 ml= +1, 0 or –1 b) 3d n= 3 l= 2 ml= +2, +1, 0, –1 or –2 c) 7s n= 7 l= 0 ml= 0 d) 5f n= 5 l= 3 ml= +3, +2, +1, 0, –1, –2 or –3 8. The quantum numbers listed below are for four different orbitals. List them in order of increasing energy. Indicate whether any two have the same energy. a) n= 4 l= 0 ml= 0 4s b) n= 3 l= 2 ml= +1 3d c) n= 3 l= 1 ml= –1 3p d) n= 3 l= 2 ml= 0 3d c < a < b = d
3 9. Rank the following orbitals in increasing order of energy: 4d 3p 3p < 4d < 6s 5p 5p < 6s 4f < 4f 10. How many orbitals are possible with each of the following designations: a) 2p 3 b) 5s 1 c) 3d 5 d) 4f 7 11. When a compound containing cesium is heated in a Bunsen burner flame, photons with energy of 4.30 x10 –19 J are emitted. What color is the cesium flame? E
4.30x10­19 J = = 6.48x1014 s ­1 h 6.634x10­34 J s c 3.00x108 m/s λ= = = 4.63x10­7 m = 463 nm (Blue­green color) 14 ­1 υ 6.48x10 s υ =
12. The He + ion contains one electron and is therefore a hydrogen­like ion. Calculate the wavelength of the first four electron transitions (6®2, 5®2, 4®2 and 3®2) for this ion and compare with the same transitions in the H atom. Comment on the differences. (RH for He + =8.72x10 –18 J) 1 R H 1 1
1 1 = ( 2 ­ 2 ) = 4.38x107 ( 2 ­ 2 ) m ­1 λ hc n1 n 2
n1 n 2 R H 8.72x10­18 J = = 4.38x107 m ­1 hc (6.634x10­34 Js )(3.00x108 m /s) For transition 6 ® 2 λ = 1.03x10­7 m = 103 nm (H= 411 nm) For transition 5 ® 2 λ = 1.09x10­7 m = 109 nm (H= 434 nm) For transition 4 ® 2 λ = 1.22x10­7 m = 122 nm (H= 487 nm) For transition 3 ® 2 λ = 1.64x10­7 m = 164 nm (H= 657 nm) All H transitions are in the visible region, while He+ transitions are in the UV region
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