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FOCUS 1 Online Cumulative Example 
1
 You are working in a laboratory investigating the properties of semiconductor nanomateri-
als. In your research you must synthesize CdSe nanocrystals by the reaction of CdO with 
Se in high-temperature solutions. A solution of Se is prepared by dissolving 152.6 mg of 
selenium metal in 25.0 mL of 1- octadecene solvent. In a separate flask, 64.2 mg of CdO is 
dissolved in 3.00 mL of oleic acid and 50.0 mL of 1-octadecene at 225 8 C. 
 (a) Write the electron configurations of Cd and Se. 
 (b) Based on the data in Appendix 2D, which element has (i) the higher ionization 
energy ( I 1 ); (ii) the greater electron affinity ( E ea ); (iii) the larger atomic radius ( r )? 
 (c) CdSe can be considered a binary ionic compound. Based on the electron affinities 
you found in (b), which element is more likely to form an anion in this compound? 
Predict the charge on the ions each element is likely to form. 
 (d) The final product requires a 1:1 mole ratio of Cd:Se. What volume of the selenium 
solution must be added to the CdO solution? 
 (e) A sample of the material you prepared was found to emit light at 546 nm when it 
is excited with UV radiation. An electron in a nanocrystal can be treated as an electron 
trapped in a one-dimensional box and the emitted light is due to an electronic transi-
tion from the n 5 2 to the n 5 1 energy level. What is the diameter of the nanocrystals 
in your sample? Note that the effective mass ( m e *) of an electron in CdSe (the mass to 
use in the energy expression) is m e * 5 0.090 m e . The mass of the electron is given in 
the table inside the back cover of your textbook. 
 (f) What is the wavelength of radiation required to excite an electron in the nanocrys-
tals from n 5 1 to n 5 3? 
 ANTICIPATE For part (b), you should anticipate that Se has the higher electron affinity 
because it lies higher and more to the right than cadmium in the periodic table. For part (c) 
you should predict that Se forms 2 2 ions because it is in the same group as oxygen and sul-
fur. There is no way to predict the answers for parts (d)–(f), but you might know that CdSe 
nanocrystals that emit visible radiation have diameters in the range 1–10 nm. 
 PLAN For parts (a)–(c), use the tables in Appendix 2D and the periodic table. For part 
(d), determine the amount of Se required to react with Cd and then find the molar con-
centration of Se in solution to determine the volume of solution required. For parts (e) 
and (f), use Eq. 4 in Topic 1C in the form E n 1 1 2 E n 5 (2 n 1 1) h 2 /8 m e * L 2 . 
 SOLVE 
 (a) Cd is in Group 12, Period 5 and Se is in Group 16, Period 4. The electron configurations 
are therefore: 
 Cd: 3Kr 44d105s2 
 Se: 3Ar 43d104s24p4 
 (b) From Appendix 2D, 
I1/(kJ?mol21) Eea/(kJ?mol21) r/pm
Cd:
Se:
868
941
,0
1195
149
117
(i) Se has the greater first ionization energy; (ii) Se has the greater electron affinity; 
(iii) Cd has the greater atomic radius.
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2 2 Focus 1 Online Cumulative Example
 (c) Use the periodic table and Appendix 2D. 
 Selenium is in Group 16 and has a positive electron affinity, so it is 
likely to form anions with a charge of 2 2. Cadmium has a negative 
electron affinity and is likely to form cations. To maintain electri-
cal neutrality in CdSe, Cd will form 1 2 ions: 
 Se 2 2 , Cd 2 1 
 (d) Determine the amount of Cd present in 64.2 mg of CdO. 
64.2 mg 3
1 mmol CdO
128.41 mg
3
1 mmol Cd
1 mmol CdO
5 0.499p mmol Cd
 The concentration of Se in solution is: 
 156.2 mg 3
1 mmol
78.96 mg
3
1
0.025 L
5 79.12p mmol?L21
 Because 1 mmol Cd ≏ 1 mmol Se, you need 0.500 mmol Se. The 
volume of solution required is: 
0.499 p mmol Se 3
0.499 mmol
79.12 mmol?L21 5 6.32 3 1023 L 5 6.32 mL
 (e) First, convert the wavelength of the photons to energy by using E � hc / � : 
 
h
6447448 
c
6447448
E 5
6.626 07 3 10234 J?s 3 2.997 92 3 108 m?s21
546 3 1029 m
 
1442443
546 nm
5 3.63p 3 10219 J 
 and equate this energy to the energy difference D E 5 E n 1 1 2 E n , between states of the 
electron. 
 Use Eq. 4 from Topic 1C and solve for the length of the box, L with 
 n 5 1. 
From Eq. 4 in Topic 1C,
L 5 a (2n 1 1)h2
8me*¢E
b
1/2
Set n�1
}
5 a 3 3 (6.626 07 3 10234 J?s)2
8 3 (0.090 3 9.109 38 3 10231 kg) 3 (3.63p 3 10219 J)
b
1/2
 
1 4444444244444443
me*
5 2.34p 3 1029 m, which rounds to 2.35 nm
S
Se
Te
Kr
 79.12… mmol
 L
×
0.499… mmol
6.32 mL
1 mmol
0.499… mmol
64.2 mg 128.4 mg
2.0 mmol
 25 mL
×
156.2 mg
1 mmol
78.96 mg
0
L = 2.34… nm
x
n = 1
n = 2
546 nm
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3Focus 1 Online Cumulative Example
(f) Using Eq. 4 in Topic 1C for En 2 En12 for n 5 1 gives
 
energy of 
level n�2
6447448 
energy of 
level n
6447448 Set n�1
}
En12 2 En 5
(n 1 2)2h2
8me* L2 2
n2h2
8me*L2 5 (4n 1 4)
h2
8me*L2 5
h2
me*L2
¢E 5
h2
me*L2 5
(6.626 07 3 10234 J?s)2
0.090 3 9.109 38 3 10231 kg 3 (2.34p 3 1029 m)2
 
14444444244444443
me*
5 9.70p 3 10219 J
 Convert the energy to wavelength by using � 5 hc / E : 
l 5
(6.626 07 3 10234 J?s) 3 (2.997 92 3 108 m?s21)
9.70p 3 10219 J
5 2.05 3 1027 m
or 205 nm
 EVALUATE As predicted, Se forms anions of charge 2 2 and Cd forms cations of 
charge  1 2. The diameter of the nanocrystal is in the range 1–10 nm, as anticipated. 
205 nm
0
L
x
n = 1
n = 3
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