Name______Date______
Chapter 7 Review Sheet
1. Draw a diagram showing the classification of electromagnetic radiation, including the colors ofthe visible spectrum. Label each end as to low/high energy, short/long wavelength.
radio microwaves infrared visible ultraviolet x-rays gamma rays
waves
R O Y G B I V
low Ehigh E
long short
2. a) The primary visible emissions from mercury occur at 404.7 and 435.8 nm. Calculate the
frequencies of these emissions. Then calculate the energy of a single photon and a mole of
photons of light with each of these wavelengths.
404.7 nm: = c/ = (3.00 x 108 m/s)/(4.047 x 10–7 m) = 7.41 x 1014 s–1
E = hv = (6.626 x 10–34 J s)(7.41 x 1014 s–1) = 4.91 x 10–19 J / photon
4.91 x 10–19 J x 6.02 x 1023 photons = 2.96 x 105 J / mol
photon1 mol
435.8 nm: = c/ = (3.00 x 108 m/s)/(4.358 x 10–7 m) = 6.88 x 1014 s–1
E = hv = (6.626 x 10–34 J s)(6.88 x 1014 s–1) = 4.56 x 10–19 J / photon
4.56 x 10–19 J x 6.02 x 1023 photons = 2.75 x 105 J / mol
photon1 mol
b) Calculate the de Broglie wavelength of a proton moving at 90% of the speed of light. The
mass of a proton is 1.6726 x 10–27 kg.
0.9 (3.00 x 108 m/s) = 2.70 x 108 m/s
= h = 6.626 x 10–34 (kg m2/s2) s = 1.47 x 10–15 m
m v(1.6726 x 10–27kg)(2.70 x 108 m/s)
3. The outermost energy level configurations for the elements A – E are listed below. Use the
symbols A through E to answer each of the questions that follow:
A = 3s2B = 3s1C = 2p6D = 2p5E = 2p3
a. Which has the lowest first ionization energy?B
b. Which is a noble gas?C
c. Which has the highest electronegativity?D
d. Which is the largest atom?B
e. What is the identity of each element?
A = MgB = NaC = NeD = FE = N
4. For each of the following groups, arrange the atoms/ions in order of:
a. increasing radius c. increasing electron affinity
b. increasing first IEd. decreasing electronegativity
group 1:As, N, F
a. F N Asb. As N Fc. As N F d. F N As
group 2:S, Cl, F
a. F Cl Sb. S Cl Fc. S Cl F d. F Cl S
group 3:Cs, Li, K
a. Li K Csb. Cs K Lic. Cs K Lid. Li K Cs
group 4:N3–, O2–, F–(omit parts c and d)
a. F– O2– N3–b. N3– O2– F–
group 5:O, O–, O2–(omit parts c and d)
a. O O– O2–b. O2– O– O
5. Write the electron configuration for cobalt, and then write a set of 4 quantum numbers for eachelectron in a cobalt atom.
Co: 1s2 2s2 2p6 3s2 3p6 4s2 3d7
1s(1, 0, 0, ½)(1, 0, 0, – ½)
2s(2, 0, 0, ½)(2, 0, 0, – ½)
2p(2, 1, –1, ½) (2, 1, 0, ½) (2, 1, 1, ½) (2, 1, –1, –½) (2, 1, 0, –½) (2, 1, 1, –½)
3s(3, 0, 0, ½)(3, 0, 0, – ½)
3p(3, 1, –1, ½) (3, 1, 0, ½) (3, 1, 1, ½) (3, 1, –1, –½) (3, 1, 0, –½) (3, 1, 1, –½)
4s(4, 0, 0, ½)(4, 0, 0, – ½)
3d(3, 2, –2, ½) (3, 2, –1, ½) (3, 2, 0, ½) (3, 2, 1, ½) (3, 2, 2, ½)
(3, 2, –2, –½) (3, 2, –1, –½)
6. A ground-state hydrogen atom absorbs a photon of wavelength 94.91 nm. The atom then emitstwo photons: one of wavelength 1282 nm to reach an intermediate level and a second to returnto the ground state.
a. What higher level did the electron reach?
E = h c = (6.626 x 10–34 J s)(3.00 x 108 m/s) = 2.09 x 10–18 J
9.491 x 10–8 m
E = –2.178 x 10–18 J1 – 1 k
nf2ni2
2.09 x 10–18 J = –2.178 x 10–18 J [(1/n2) – 1]
–0.962 = 1/n2 – 10.0384 = 1/n2n2 = 26.0n = 5
b. What intermediate level did the electron reach?
E = h c = (6.626 x 10–34 J s)(3.00 x 108 m/s) = 1.55 x 10–19 J
1.282 x 10–6 m
–1.55 x 10–19 J = –2.178 x 10–18 J [(1/n2) – (1/25)]
7.12 x 10–2 = 1/n2 – 1/25
0.111 = 1/n2n2 = 8.99n = 3
c. What was the wavelength of the second photon emitted?
E = –2.178 x 10–18 J [ 1 – (1/9)] = –1.936 x 10–18 J
E = h c = h c = (6.626 x 10–34 J s)(3.00 x 108 m/s) =1.03 x 10–7 m =103nm
E1.936 x 10–18 J
7. You are walking across campus one night when a strange-looking (but seemingly friendly!) creature approaches you. He explains that he is a scientist from planet Rainhard in a far corner of the universe. Only 12 elements are known to exist on Rainhard. He has been trying unsuccessfully to organize a periodic table for these elements, and in desperation has come to the far-off planet Earth to seek the help of someone knowledgeable in periodic trends. Since you fulfill this criterion, you agree to help the friendly alien.
He tells you that the 12 elements on Rainhard are completely distinct from those found on Earth, and are symbolized: Az, By, Cx, Dw, Ev, Fu, Gt, Hs, Ir, Jq, Kp, Lo. He then relates to you the only known facts about these elements:
1. Hs is a noble gas.
2. Gt has the highest electronegativity of any element.
3. Fu is the most likely to gain an electron of any element in its period.
4. Ir has the lowest atomic mass.
5. It would require more energy to remove an electron from Az than from any other element.
6. Jq has one more valence electron than Dw but one less valence electron than Gt.
7. The sizes of the 3 isoelectronic species are: Cx+ < Hs < By–.
8. Ev and Kp both lose the same number of electrons when they form an ion, but Kp has a principal quantum number 2 less than Ev.
9. The third ionization energy of Jq would be especially high.
10. Element Lo is in the same period as the element with the largest atomic radius.
With these pieces of information the alien asks you to use your knowledge of general trends of the periodic table to fill in the periodic table below.
Ir / Kp / Gt / AzDw / Jq / By / Hs
Cx / Ev / Fu / Lo