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Unformatted text preview: SURVEY OF ELEMENTS ' Most elements: Electron configurations not stable. BedrCentered Cubic Structure (BCC) Elgment Mimi“ 1* “(13‘3er configuration * Arenas located at 8 cube corners with a single aunt! at cube
llyll'ogen 2 i: . ll , lime: :lll atoms in the animatinn an! identical: the center amtLL is sh
“ml—w— different I); [11: easu or viewing.
thhlum 3 15225]
Beryllium 4 1522.53 ex: Cr, 'll‘, Fe (( ll. Ta, .‘rlo
Boron 5 15225323] . . _ a _
Carbon 5 15225329: ' Coordination n — 8
Neon 10 152253 2g“ (stable)
Sodium 11 is2 252 25:15 3:21
Magnesium 12 is2 253 2p“ 353
Aluminum 13 is2 253 2515 353 Sp]
Argon 18 152 252 2963533536 (stable)
Kll‘EtOH 36 1532532253523253d1° 453153“ (31 able) "'"."'.“." I'rmIuc. :52.
Hits um... r." Inlet. Lu .lLIII'I .IInmLL‘u '3“ “‘5‘" I‘ *‘ “""L'ﬁ “"
' Why not stable? Valence (outer) shell usually not WIN9'11JIMW" 3 “mmmu m1}; 1 Emmy + 3 comm5 x 1 completely filled. Simple Cubic (SC) Crystal Degree of Polymerization, DP Structure
 Centers at atoms located at the eight comers etawbe DP = average number of repeat units per chain ' Rare due to low packing density (only Pu has Lhis SET'LIC
H i'l H H H H H
e+e++++
H H H H ' Closepacked directions are cube edges.
H H H H
' CiHJT‘IJlnHLlun i‘r‘ = ﬁ H—¢
H (# n earesl n ei gh hers) H H we H
H I'IHH
“—3:
at BP where rﬁ  average molecular weight of repeat unit
for copolymers this is calculated as felons: “it litigi‘. '  81m.
Chain frmtim—J L sol. I: of :11qu
mi: r‘
MOLECULAR [EIGHT DISTRIBUTION FaceCentered Cubic Structure (FCC) intuni rn Fi‘ I.I. (ﬂiixl'cr d  Atoms located at 8 cube corners and at the centers of the 6 faces. lime: All ﬂ'lﬂtllS in the animation are identical: thrI face centered Illl
are 911anqu :li ['l'ei'ent ly l'ns' ease of viewing. ﬁ _ total wtet polymer
"—— tota] #ot molecules
ex: h}. Cu. ALL. Pb. Si. Pt. Ag 4 ' C(mrdinarien n = 12 is", = mean (middle) molecular weight of size range r'
Jr, = number fraction of chains in size range 1'
Rd.l.n.c1.' hurl1g. :Li, Lufu'u'srw .1
= ' ' ' ' '  um um: 'III2L mu. .mmHLu _ . mummu: .u. _
y" weight. fraCtlon of Chilll'ls 1" stze range I "L'mILcJgr Is.“ .u..s_..r.4.. ’1 ﬂLFIHSlel'liL cell: 6 Face x 1.52 + 3 corners x Atomic Packing Factor (APF) Atomic Packing Factor: BCC
 APF tor the bodycentered wbic structure . 0.68 for Simple Cubic
4R . J5 a
volume
atoms .r’
atom 45 a
a unit cell ”‘1 inwma . .
J— 3  For closepacked directions
=0.5e APF . — = 0.52 R= ii an:
l volume volume
' .  atom
closepacked ”'1" ca“ H“ 2 — 11651103 H 3 J5 3
directions . 5.11: —] . 0 68
Unit cell contains 1 atom . a x we .. 1 atomiunit cell Volume
unit cell Theoretical Density Computation for Chromium  Cr has 360 crystal structure
A = 52.00 g/mol
Fl = 0.125 nm
n = 2 atomslunit cell
a = 41%? = 0.2887 nm /8/ v0: a?=2.406x1023 cm3
atoms
. 9
n A unitcell\.‘2 52.00 1/ mo!
)= — = —— =7.199.icm5
I.  atoms
volume / mol
unit cell plural 7.1Bglcm3 Theoretical Density for Metals, ) . Mass of Atoms in Unit Cell (Ml NA}
Densw' ) ' Total Volume otUnitCell ' Vc
n A
) = VCNA
where n = number of atoms/unit ceil A = atomic weight
Vc = Volume oi unit cell = a3 for cubic NA = Avogadro’s number
= 6.022 x 1023 atoms/moi Crystallographic Planes Example Problem I X Z Y 1. Relocate origin — not needed
2. Intercepts a h c
3. Reciprocajs we 11b 1rc
4. Normalize e‘a bib due
a
1 1 0
5. Reduction 1 1 o X
6. Miller Indices (110) Atomic Packing Factor: FCC APF for the facecentered cubic structure = 0.74
maximum achievable APF For closepacked directions: 4R= J58 [ J54! i.e., R —
4 1 Unit cell contains: 6 x12 + 8 x 118
 4 atomsfunit cell volume  0.74 3.
.\ volume unit cell Crystallographic Planes Algorithm ior determining the Miller Indices oi a plane
1. It plane passes through selected origin, establisl a
new origin in another unit cell
2. Read all values of intercepts ol plane (designated
A, B, C) with x, y, and z axes in terms or a. b, c
3. Take reciprocals of intercepts
4. Normalize reciprocals oi intercepts by multiplying
by lattice parameters a, b, and c
4. Reduce to smallest integer values
5. Enclose resulting Miller lndices in parentheses, no
commas i.e., (hid) Crystallographic Directions II:
Example Problem 2 1. Point coordinates oi tail and head
tail pt. 1 x. a a. .11 =
Head apt; a 43. = a. 2 3 i Qubtract and normalize aa__2: b_b/21I2; ;1
a b c l: => —2. 1/2. 1 4W5. Multiply by 2 to eliminate the traction, then place in square brackets (no commas) —4, 1, :) [—412] where the overbar represents a
2 negative index Family 0! directions — all directions that are crystallographically equivalent
(have the same atomic spacing) — indicated by indices in angle brackets Ex: <1oo> = [1001010],[001],[_100],[0_10],[m_l] r; = II. X Coordination Number andr Ionic Ftadii
 Coordination Number increases with w r anion
To term a stable structure, how many anions can
surround around a cation?
'cation Guard. 4... '7
ZnS
a (zinc blends}
Q mpmm Fina r ' Nurrtler
< 0.155 2 linear “
0.155  0.225 3 triangular 0.2254414 4 tetrahedral
(141470.732 6 octahedral 0.732  1.0 3 MMTﬂaﬁl
Mimic. ethic Crystallographic Planes Example Problem "I
x y z Z 1. Relocate orig‘n — not needed 2. Intercepts a2 it 3:14 3. Recipmcals 2J'a tip «as 4. Normalize zeta bib 4113:: y
2 1 #3 a b 5. Heductbnth) s s a
6. Miller lndioes {see} X Familyr of planes  all panes that are orystallographicaily eqtiualent {have
the same atomic packing)  indicated by inches in braces Ex! [1001 (100). (010). {901). (loot. (DiD}. (DO?) ...
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 Winter '12
 DaRocha
 Crystallography, Cubic crystal system, Diamond cubic, Atomic packing factor

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