LECTURE #05. Learning Objectives. How does atomic packing factor change with different atom types? How do you calculate the density of a material?

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1 LECTURE #05 Chpter : Pcking Densities nd Coordintion Lerning Objectives es How does tomic pcking fctor chnge with different tom types? How do you clculte the density of mteril? 2 Relevnt Reding for this Lecture... Pges Re cp: Atomic Pcking Fctor (APF) APF = olume of toms in * olume of *ssume hrd spheres Unit cell contins: 6 x1/2 + 8 x1/8 = toms/ Describes How efficiently toms fill spce within given 2 Adpted from Fig..1(), Cllister 7e. APF = Close pcked directions: length = R = 2 ( 2/) ( volume tom volume APF for fce centered cubic structure = 0.7 1

2 If the hs different toms wtch out! Cermic Crystl Structures! Wtch Out! Different Atoms. Must Modify APF Equtions. R blue 2 Adpted from Fig..2(), Cllister 7e. Close pcked directions: digonl = 2R blue + 2R red = 5 toms APF = 2 Must put in correct vlue for ech tom type (/) volume volume tom 2

3 Wht if ions from the crystl structure? Cs + Cl Sme rules s metls w/respect to crystl symmetry. Structure: CsCl type Brvis lttice: simple cubic Ions/: 1 Cs Cl Ionic pcking fctor (IPF), similr definition t APF. 6 Cesium chloride (CsCl) showing () ion positions nd the two ions per lttice point nd (b) full size ions. Note tht the Cs + Cl pir ssocited with given lttice point is not molecule becuse the ionic bonding is non directionl nd becuse given Cs + is eqully bonded to eight djcent Cl, nd vice vers. Clculte the following: (1) The CN of CsCl (2) Ionic Pcking Fctor of CsCl Recll from the lst lecture CN r/r =r + /r Wht re the r s for Cs + nd Cl -? r Cs+ = nm r Cl- = nm These numbers cme from inside cover of text book! r + /r =

4 Now clculte the Ionic Pcking Fctor of CsCl For these ions, they touch long the body digonl (mke sure you know which direction the hrd spheres re touching) IPF = [# of toms/] [volume of sphere] volume of One tom of Cs + One tom of Cl /π(0.170nm) + /π(0.181nm) r Cs+ = nm r Cl- = nm Cs = nm Cl 2 = 2r Cs+ + 2r Cl = 0.05 nm = = 0.066nm 8 IPF = 0.68 For kicks, wht if the ions touched long the edge length (not the body digonl)? IPF = [# of toms/] [volume of sphere] volume of One tom of Cs + One tom of Cl /π(0.170nm) + /π(0.181nm) Cs r Cs+ = nm r Cl- = nm which r? = 0.05 nm verge them? Cl = 2r Cs+ + 2r Cl = 0.05 nm = = 0.066nm = 2r = 0.05 nm 0.52nm = = 0.06nm r vg = IPF = Not possible! >100% pcking!!! 9 It is criticl tht you identify the correct hrd sphere touching directions!

5 Let s do nother exmple: Consider NCl Wht is the CN? Wht is the IPF? How do we determine CN? Correct, CN r + /r Wht re the r s for N + nd Cl -? r N+ = nm r Cl- = nm Agin, got these numbers from inside cover of text book! r + /r = Not very cler representtion let s expnd it! Motif 2 ions per lttice point Do you see how the green spheres form FCC structure?! The blue spheres do the sme thing, form FCC structure We hve two interpenetrting FCC lttices Commonly clled Rock Slt structure structure seen in other mterils including TiC, TC, MgO, CO, etc. 11 5

6 Clculte the IPF of NCl IPF = [# of toms/] [volume of sphere] volume of Wht type of do we hve? Two FCC lttices one for N nd one for Cl toms/n FCC + toms/cl FCC = 8 toms/ totl Four toms of N + Four toms of Cl x /π(0.102 nm) + x /π(0.181 nm) = 0.117nm 12 Wht is the volume of the? = 2r N + 2r Cl = 0.566nm = = 0.181nm IPF = 0.117nm (previous slide) 0.181nm = WthO Wtch Out! t!diff Different Ions. CORRECTION! Though it looks like FCC symmetry, the fce digonl toms don t touch; but the edge toms do touch! 1 Why? Ctions nd nions do not hve the sme size! Before we considered toms of the sme size 6

7 Atoms cn occupy interstitil sites Unoccupied spce in center cn ccommodte smll toms, e.g. He in UO 2 fuel rods Fluorite (CF 2 ) showing () ion positions (b) full size ions. FCC interstitil sites Note: useful info on tom plcement Structure: fluorite (CF 2 ) type Brvis lttice: FCC Ions/: C F - Typicl cermics: UO 2, ThO 2, TeO 2 1 CLASS ROOM EXAMPLE: Clculte the ionic pcking fctor for UO 2, which hs the CF 2 structure U nd O ions touch long portion of the body digonl 15 7

8 CLASS ROOM EXAMPLE: SOLUTION Clculte the ionic pcking fctor for UO 2, which hs the CF 2 structure This problem is tricky! The fce digonl hs length of 2. The body digonl hs length of. U 2+ O - 2 Along with the cell edge, they form right tringle within the. The C 2+ nd F - ions touch short distnce long the body digonl. By the principle of similitude, this smller tringle, mesures s ¼ the size of the lrge one. Becuse of this, the length of the bond becomes: 1 R R nm nm U O 16 Now you cn clculte nd the corresponding volume. CLASS ROOM EXAMPLE: SOLUTION continued R R nm 2 U O IPF ions in Solving for we get: 0.58 nm 0.58 nm 0.16 nm single ion R 16 2 ions R 8 R nm U O ions nm IPF nm 17 8

9 THEORETICAL DENSITY, Density = mss/volume mss = number of toms per mss of ech tom mss of ech tom = tomic weight / Avogdro s number # toms/ Atomic weight (g/mol) olume/ (cm /) n A c N A Avogdro's number (6.02 x 10 2 toms/mol) 18 THEORETICAL DENSITY, # toms/ Atomic weight (g/mol) olume/ (cm /) n A c N A Exmple: Copper crystl structure = FCC: toms/ tomic weight = 6.55 g/mol (1 mu = 1 g/mol) tomic rdius R = nm (1 nm = 10 7 cm) Avogdro's number (6.02 x 10 2 toms/mol) c = ; For FCC, = R/ 2 ; c =.75 x 10-2 cm Result: theoreticl Cu = 8.89 g/cm Compre to ctul: Cu = 8.9 g/cm Why the difference? 19 9

10 Theoreticl Density, toms = volume 20 R x10 2 Ex: Cr (BCC) A = g/mol R = nm n = 2 = R/ = nm g mol toms mol theoreticl = 7.18 g/cm ctul = 7.19 g/cm 20 Densities of Mteril Clsses In generl metls > cermics > polymers Why? Metls hve... close pcking (metllic bonding) often lrge tomic msses Cermics hve... less dense pcking often lighter elements Polymers hve... low pcking density (often morphous) lighter elements (C,H,O) Composites hve... intermedite vlues 21 (g/cm ) Metls/ Alloys Pltinum Gold, W Tntlum Silver, Mo Cu,Ni Steels Tin, Zinc Titnium Aluminum Mgnesium Grphite/ Cermics/ Semicond Polymers Composites/ fibers Bsed on dt in Tble B1, Cllister *GFRE, CFRE, & AFRE re Glss, Crbon, & Armid Fiber Reinforced Epoxy composites (vlues bsed on 60% volume frction of ligned fibers in n epoxy mtrix). Zirconi Al oxide Dimond Si nitride Glss sod Concrete Silicon G rphite Dt from Tble B1, Cllister 7e. PTFE Silicone PC PET PC HDPE, PS PP, LDPE Glss fibers GFRE* Crbon fibers CFRE * Armid fibers AFRE * Wood 10

11 Summry If there re different ions in the, APF/IPF equtions must be modified! Find out which direction the hrd spheres touch The ionic pcking fctor cn be clculted by IPF ions in Theoreticl Density cn be clculted by # toms/ Atomic weight (g/mol) olume/ (cm /) n A c N A Avogdro's number (6.02 x 10 2 toms/mol) 22 11