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Unformatted text preview: 41. nl=2dsi119,d: H?” zwi2313132313><10’10m2313pm
2sme 2><sm1422° 42. In a facecentered cubic unit cell (ccp structure), the atoms touch along the face diagonal: "u T (4r)2 :12 + 12 m. I 12rJ§
B' L chbe:l3:(r\/8_)3:22.63r3 There are four atoms in a facecentered cubic cell. Each atom has a volume of 4/3 7113. Vatoms : 4 X £7513 : 16.7612) 3
V 16.76 3 . . .
So mm = r3 = 0.7406, or 74.06% of the volume of each un1t cell 1s occupled by
chbe 22.63 r
atoms. A bodycentered cubic unit cell contains two net atoms (8 comer atoms x 1/ 8 atom per comer
+ 1 center atom : 2 atoms/unit cell). In a bodycentered unit cell (see Figure 16.18), the
atoms touch along the body diagonal of the unit cell. Therefore, body diagonal : 4r : J3 l.
The length of a cube edge (1?) is related to the radius of the atom (r) by the equation 1? = 4r/1/3 . Volume of unit cell : P : (4r/ J3 )3 : 12.32 r3 Volume of atoms in unit cell : 2 X 2719 : 8.378 r3
3
So Vamms = 8378 r3 : 0.6800 : 68.00% occupied.
V 12.32 r cube In a simple cubic unit cell, the atoms touch along the cube edge (I): T ‘ T 2(radius) : 2r : l
2 r I l vwbe= [3 = (203 = 8r3 There is one atom per simple cubic cell (8 corner atoms X 1/8 atom per corner : 1 atom/unit
cell). Each atom has an assumed volume of 4/3 m3 = volume of a sphere. 43. um: gas : 4.1891‘3 4. 189 1‘3
8r3 V atom V So : 0.5236, or 52.36% occupied. cube A cubic closest packed structure (facecentered cubic unit cell) packs the atoms more efficiently than a bodycentered cubic unit cell, which is more efficient than a simple cubic
unit cell. A cubic closest packed structure has a facecentered cubic unit cell. In a facecentered cubic
unit, there are: 1/8 t 1/2 t
8 comers >< ﬂ + 6 faces >< ﬂ : 4 atoms comer face The atoms in a facecentered cubic unit cell touch along the face diagonal of the cubic unit
cell. Using the Pythagorean formula, Where I : length of the face diagonal and r : radius of
the atom: £2 + :2 = (4r)2 '1“
aa l:r1/8_ I
l: rf: 197 Md12 m >< J§: 5.57 x 1010111 : 5.57 x108 cm Volume ofa unit cell : 13 : (5.57 X 10'8 cm)3 : 1.73 X 10'22 cm3 Mass ofa unit cell : 4 Ca atoms X % >< W : 2662 X 10*22 g Ca
6.022 x 10 atoms mol Ca
. 2.662 10'22
Den51ty : mass — x—ga : 1.54 g/cm3 volume — 1.73 >< 10'22 cm 46. A facecentered cubic unit cell contains four atoms. For a unit cell: mass ofX : volume >< density : (4.09 X 10'8 cm)3 X 10.5 g/em3 : 7.18 X 10'22 g mol X = 4 atoms X >< % = 6.642 >< 10*?14 mol X 6.022 x 1023 atoms 7.18 x 10*22 gX Molar mass : —
6.642 x 10’24 molX : 108 g/mol; the metal is silver (Ag). 50. If facecentered cubic: :2 “ls :(137pm)f :387pm23.87>< 10’8cm 4 atomsW >< —6 022 11:3: x —183.9g1 W
Density : ' X Satongs mo : 21.1 g/cm3
(3.87 >< 10— cm) If bodycentered cubic: z: i = M =316pm=3.16 X10’8cm
«5 J? 2 atoms W x —6 022 11:10:: x —183.9g1 W Density : ' >< atoms m0 : 19.4 gz’cm3 (3.16 >< 10'8 cm)3 The measured density of tungsten is consistent with a bodycentered unit cell. 52. To produce an n—type semiconductor, dope Ge with a substance that has more than four
valence electrons, e.g., a Group SA element. Phosphorus or arsenic are two substances that
will produce n—type semiconductors when they are doped into germanium. To produce a p
type semiconductor, dope Ge with a substance that has fewer than four valence electrons,
e.g., a Group 3A element. Gallium or indium are two substances that will produce ptype
semiconductors when they are doped into germanium. Doping germanium with phosphorus (or arsenic) produces an ntype semiconductor. The
phosphorus adds electrons at energies near the conduction band of germanium. Electrons do
not need as much energy to move from ﬁlled to unﬁlled energy levels, so conduction
increases. Doping germanium with gallium (or indium) produces a ptype semiconductor. Because gallium has fewer valence electrons than germanium, holes (unﬁlled energy levels)
at energies in the previously ﬁlled molecular orbitals are created, which induces greater
electron movement (greater conductivity). (6.626 x 1034 Js)(2.298 x 103 m/s) 2 730 10_9 2.72 X 10’19 J = energy ofband gap
. X m 58. E=E=
9t 64. From Figure 16.42, MgO has the NaCl structure containing 4 Mg2+ ions and 4 02— ions per
facecentered unit cell. lmol MgO x 40.31 g MgO — : 2.678 >< 10‘22 g MgO
6.022 x 1023 atoms lmolMgO 4 MgO formula units X 1 cm3 3.58g Volume ofunit cell : 2.678 >< 10’22 g MgO >< : 7.48 x 10’23 cm3 Volume ofunit cell : 13, l : cube edge length; l: (7.48 X 10'23 cm3)U3 : 4.21 X 10'8 cm
For a facecentered unit cell, the 02— ions touch along the face diagonal: _ J— x 4.21 x 10’“ cm
4 J21 = 4r0,_, r0,_ : 1.49 >< 10'8cm The cube edge length goes through two radii of the 02’ anions and the diameter of the Mg2+
cation. So: 1: 2r0,_ 4 2ng,., 4.21 >< 10—86111: 2(149 >< 10—86111) + 2ng,., ng,. : 6.15 >< 10‘9 cm 68. a. The NaCl unit cell has a facecentered cubic arrangement of the anions with cations in
the octahedral holes. There are four NaCl formula units per unit cell, and since there is a
1:1 ratio of cations to anions in MnO, then there would be four MnO formula units per
unit cell assuming an NaCltype structure. The CsCl unit cell has a simple cubic
structure of anions with the cations in the cubic holes. There is one CsCl formula unit per
unit cell, so there would be one lVInO formula unit per unit cell if a CsCl structure is observed.
Formula units of MnO = (4 47 X 104, cm)3 >< 5.28gMnO >< lmol MnO
Unit cell ' cm3 70.94 g MnO 6.022 x 1023 formula units lVInO
X
mol lVInO4 ’ 4.00 formula units MnO From the calculation, MnO crystallizes in the NaCl type structure. b. From the NaCl structure and assuming the ions touch each other, then 1 = cube edge
length : 2rMn2, + 2r02, . I: 4.47 ><10’8 cm : 2rMI12+ + 2(l.40 x10’8 cm), ran+ : 8.35 ><10’8 cm : 84 pm 114. TMH2+ _ 84pm :0.60 r02 — 140.pm From Table 16.6 of the text, octahedral holes should be ﬁlled when 0.414 < r+/r_ < 0.732. Because the calculated radius ratio falls within the prescribed limits, we would expect
lVIn2+ to occupy the octahedral holes formed by the cubic closest packed array of 02' ions
(as predicted in part a). The ionic radius is 148 pm for Rb+ and 181 pm for C1_. Using these values, let’s calculate the
density of the two stmctures. Normal pressure: Rb+ and C1‘ touch along cube edge (form NaCl structure). Cube edge =1: 2(148 +181): 658 pm = 6.58 X 10'8 cm; there are four RbCl units per unit
cell. 4(85.47) + 405.45) — =2.82 fcm3
6.022 x 1023(658 x 10*)3 g Density = d = High pressure: Rb+ and C1‘ touch along body diagonal (form CsCl structure).
2r. + 2r+ = 658 pm = body diagonal = [J3 , I: 658 pm/a/g= 380. pm . . . 85.47 +35.45
Each unit cell contams 1 RbCl unit: d = ﬂ = 3.66 g/cm3
6.022 ><10 (3.80 x 10 ) The highpressure form has the higher density. The density ratio is 366/282 : 1.30. We
would expect this because the effect of pressure is to push things closer together and thus
increase density. ...
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 Spring '09
 cabrera
 Chemistry

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