Exam 1 Cheat Sheet.docx - SURVEY OF ELEMENTS Most elements...

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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 l-lyll'ogen 2 i: . ll , lime: :lll atoms in the animatinn an! identical: the center amt-LL 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° 453-153“ (31 able) "'"."'.“." I'rmI-uc. :52. Hits um... r." Inlet.- Lu .l-LIII'I .IInmLL-‘u '3“ “‘5‘" I‘ *‘ “""L'fi “"- ' Why not stable? Valence (outer) shell usually not WIN-9'11JIM-W" 3 “mm-mu m1}; 1 Emmy + 3 comm-5 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 ' Close-packed 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 Face-Centered Cubic Structure (FCC) int-uni 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 Ill-l 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.' hurl-1g. :Li, Lufu'u'srw .1 = ' ' ' ' ' - um um: 'III2L mu. .mmHL-u _ . mum-mu: .u. _ y" weight. fraCtlon of Chilll'ls 1" stze range I "L'mILc-Jgr 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 body-centered wbic structure . 0.68 for Simple Cubic 4R . J5 a volume atoms .r’ atom 45 a a unit cell ”‘1 inwma . . J— 3 - For close-packed directions =0.5e APF . — = 0.52 R= ii an: l volume volume ' . - atom close-packed ”'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?=2.406x10-23 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 1r-c 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 face-centered cubic structure = 0.74 maximum achievable APF For close-packed 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 -a-a__2: b_b/2-1I2; ;-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. (Di-D}. (DO?) ...
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