Question

Post-lab question Part A 1. Metallic sodium forms a body centered cubic crystal. Why would the water displacement method used in part A not be suitable to determine the density of sodium? (Hint: watch the following video to help you answer this question: https://www.youtube.com/watch?v=18tOtZKpi04) 2. Calculate the density of sodium. (The radius of a sodium atom is 186pm) Post-lab Question Part B: Consider cubic array iodide ions (r =220pm) in which the anions are touching along the face of the cube. What is the maximum radius (in Angstroms) that a cation could have and still fit in the "octahedral" holes on the edges of the FCC unit cell? b) When sodium ions are fit into the octahedral holes along the edges, do the iodide ions still touch along the face? Draw a picture! c) Calculate the density of Nal.

Answer 1

to determine 1. water displacement method not density of sodium As in interaction with water. suitable gives explosive reaction ZM d = NA a3 pa 186 .pm 3 az 45 2x 239 - 4 3 6.023* 10 23 247 = 4 x 186 yrs 2 9.67X10-7 gree = 9.295810 m -8. For tetrahedral void, - 429x10 0.225 Y- 220pm I 0.225 X 220 pm 49.5 po 49.5x10-12 m d = ZH .495 NA a3 4 X 127 g ce 6-025*10***(252x2:23 03 сім Z 24 17 35006 glee 282 m 222 x 220 pm 262x220

1. density of the metal Volume 2 The cubic cell in conners and In Fee 6 atome I atom 3x6 = 3 atoms mass of metal = 91.670g 91.672 wolume of metail - (30-3-20) 10.3ml 01034 0-0103 L 8.900 FCC (Face-centred cubie tattice) lattice there ate 8 atoms at at face centres Fach canet combibutes to eight celle so per unit cell contribution Similarly each face ceulie contributes to two unit cells so contribution per unit cell by six face centres - i total number of atoms per unit cell of fee lattice = (1+3) = 4 atoms For cubic crystals, 2M density Do get M = molecular mast = 91.679 i volume of unit cell, a Na - Avogadro number length of edge of the cube 6.023x1023 x 8900ght of atoms per unit cell = 4 L 4. Length of the edge -26 6.84x10 -8 4.089 x 10 4.09x 10 3 - 8900 d = Na3 2 Н the NA d 4x91.679 a -26 6.84 X10 X 1000 de cen cm ein fce crystal 5. atomic radius of metal -8 1.44 X10 2/2 Foo Fee - AB²+ BC² = 144 pm AC2 a²ta² be 16 m² = 20 a- "B a covalent radius 76 % essas of metal (N) 124 pon (wikipedia) Experimental vele accepted value x 100% accepted value T4U-124 X 100%= 16.127 124

1. ce- ions occupies the comers Surround one are сре and to the half to the and face centres of the unit cell 6 cl ions Nat in There 8 cl ions at 8 commons of the cube and 6 at ions at 6 face centres. Each corner chloride ion contributes eighth unit cell each face centred ce-ion contributes one unit cell. Total number of cl ions 8x8 + 1 x6 = 1+3=4 Nations on edges and one in the centre of the unit cell. Each Nation present fourth to the unit cell and Nation centre contributes to the unit cell. Total number of sodium ions in the unit cell 12x+y + 1x + 3+1 = 4 Foro fce, A C²_ AB2 + BC² 2 There are the on the edge.contributes present at bord one body centre contubules 3. TO at 402 a B А 28 radius of ce in edge length ion or the space between the chloride along each edge a=222x 181pm 511.94 pm -