This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 11I Problem 7 Using Fig. 11.59. calculate the critical radius ratio for octahedral coordination. Cut in n Figure 11.59 Chapter 11, Solution 7
Cation 2R (b) Figure 11.59 (a) Octahedral coordination of six anions (radii = R) around a central cation of radius r. (1'))
Horizontal section through center of (a). From the triangle shown above in Pig. 11.59 (3)), [204—3)]2 =(2R)2+(2R)2
(I‘+R)2 2 2R2
r+R:J§R
r :JER MR : 0.414]? i=0.414
R Chapter 11, Problem 8 Predict the coordination number for (a) BaO and (b) LiF. Ionic radii are Ba} = 0.143 11111. 03' = 0.132 11111. Li+
001811111. F" = 0.133 11111. Chapter 11, Solution 8
(a) Using Appendix IV. the radius ratio for BaO is . . 2+
Ication 2&2 swim my) 0.132 This ratio is greater than 0.732 and greater than 1.0. This is an unusual case in which the cation is larger than the anion because the elements respectively lie in Periods 6 and 2 of the periodic table. But if we inye1t this ratio to represent an anion surrounded by cations. we obtain 0.93. BaO should thus show cubic coordination (CN = 8). The solid is actually octahedral. CN = 6. (b) Using Appendix IV. the radius ratio for Lil: is I ‘ I+
Ication : 'I(L1 ): = 11.10.. Rf) “133 This ratio is greater than 0.414 and less than 0.132. thus Lil: should have an octahedron coordination number. C‘N = 6. and it does. Chapter 11, Problem 9 Calculate the density in grams per cubic centimeter of Csl. which has the CsCl structure. Ionic radii are Cs+ = 0.165 11111 and I' = 022011111. Chapter 11, Solution 9 For the CsCl st1uct'ure. £0 = 2(1‘ + R) . Thus. the lattice constant measures. a' 2110165111n+0220 11111) = 0.445 11111 = '—1.45><10_8 c111 .5 . . . ‘ . ‘ + — . . .
Slnce the unlt cell 01 C. sI contains one C s and one I 1011. its mass is _ (1Cs+ X1329 gr"1nol)+(11_ x 126.9 g.."'1nol) _ 23. —4.32x10—33g
6.02x10 lens11101 “lunit cell The density is thus calculated as. 4.32x10‘22 g
(4.45x10—E cm)3 I” V IN a3 p — — 4.90 grcm3 Chapter 11, Problem 14 Calculate the ionic packing factor for (a) M110 and (b) 810. Ionic radii are M113 = 0.091 11111. SI;— = 0.127 11111. and
02' = 0.132 11111. Chapter 11, Solution 14 Both MnO and 810 have the NaCl st111cture. Thus there are four anions and four cations per unit cell and the ionic
packing factor is calculated using the equation: Vol. ofions per unit cell _ FT?‘3)+ 4(%3R3) _ 1% 21(1'3 +213) Vol. of unit cell (13 (13 IPF = (a) For MnO. (1 = 2(0091 11111 + 0.132 11111) = 0.446 11111 a11dtl1e IPF is 1%“? +213) _1611[(0.09111111)3 + (0.132 111103]
(:3 3(0446 11111)J IPF— —0.577 (b) For 810. a = 2(0127 11111+ 0.132 11111) = 0.518 11111 and the IPF is 1%203 +213) 21621'[(0.12? 1111113 +(0132 11111)3:_ PF 2 3 3 — 0.524
a 3(0518 11111) Chapter 11, Problem 20 Calculate the linear density in ions per nanometer in the [l l l] and [110] directions for C‘eOg. which has the ﬂuorite
structure. Ionic radii are Ce4+ = 0.102 11111 and O" = 0.132 11111. Chapter 11, Solution 20 For CeOg in the [l l 1] direction. there are two 02— and one . 4+ . . . .
Ce along the cube diagonal. whrch is £11 111 length +21 ._):i(0.102 11111+0.132 11111) a 2:01
J3? Ce 0 J3 = 0234011111 p _ 1Ce4+ + 202‘ _ 1024+ + 202‘
L 60 ﬁ(0.540 11111)
= (1.07Ce“+ + 2.14 02—)1’n111 For the [110] direction. 4+
C e i 2Ce4+ 2Ce4+ p e e 2.62 Ce4+rnm
L JEO‘ J2 (0.540 11111) [110] Chapter 11, Problem 26 Calculate the ionic packing factor for CaTiO3 . which has the perovskite structure. Ionic radii are
Ca 2+ = 0.106 11111. T14+ = 0.064 11111. and 02— = 0.132 11111. Assume the lattice constant
a = 2014+ + ) Chapter 11, Solution 26 In the perovskite structure. there are a total of one calcium ion. one titanium ion. and three oxygen atoms. The ionic
packing factor is therefore calculated as. . .3 .3 ,3 '
Vol. ofions per unit cell i (%JT):(ICa3+ )+ (In4+ l+ (3:02' Vol. of unit cell a IPF * where (1 = 2(1‘Ti4+ +162 ) = 2(0.064 11111 + 0.132 nnl) = 0.392 11111. Substituting. 4K[(0.10611111)3 +(0.064 mu)3 +3(0.132)3] _ IPF — 3
3(0.392 11111) 0.581 Chapter 11, Problem 88
What is the purpose of (a) MgO and (15) A1103 additions to sodalime glass?J Chapter 11, Solution 88 (c) MgO is added to sodalime glass in small quantities of l to 4 percent to prevent devitriﬁcation. (d) Al203 is added in quantities of 0.5 to 1.5. percent to increase durability. ...
View
Full
Document
 Spring '08
 wholedepartment

Click to edit the document details