CHEMISTRY 103 – Help Sheet #20 Phase Diagrams and Solids-Part II Do the topics appropriate for your course

Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page)

Nuggets: Crystalline versus Amorphous Solids; Types of Solids; Unit Cells; Packing; Packing Efficiency; Counting Atoms; Coordination Number; Density Calculations; Not included: Semiconductors, Polymers SOLIDS Crystalline substance: A regular, repeating, ordered structure. Amorphous substance: Non-repeating micro-structure; e.g., glass. TYPES OF CRYSTALLINE SOLIDS

Metallic Solids are solids that contain only metals held together with metallic bonds. Metals have fixed atom positions surrounded by a “sea of electrons” that give metals their high electrical and thermal conductivity. Examples: Fe(s), etc. Properties: ductile, malleable, high electrical and thermal conductivity, shiny, wide range of mp/bp.

Ionic Solids solids comprised of ions and held together by electrostatic interactions/ionic bonds. Examples: NaCl, Ca3P2, MgO, etc. Properties: brittle, cleaved along internal atom planes, high mp/bp, electrical conductivity in molten (liquid) state.

Network Solids made from nonmetal atoms that are held together with covalent bonds. Examples: Cgraphite, Cdiamond, SiO2 (sand), Si. Properties: very high mp/bp.

Molecular Solids compounds that are made from nonmetal atoms, have covalent bonds between atoms within the molecule but individual molecules are held together with weak intermolecular forces (H-bonding, dipole-dipole, LDF). Examples: sugar (C12H22O11), all the typical non-ionic Lewis dot structure molecules, etc. Properties: very low mp/bp.

UNIT CELLS OF CRYSTALLINE SOLIDS - The structure in the crystalline solid that when repeated generates the entire substance. There are 4 common unit cells for Chem 103 (memorize):

Primitive Cubic (PC) Body-Centered Cubic (BCC) Face-Centered Cubic (FCC) (also called Cubic Closest Packing, CCP) Hexagonal Closest Packing (HCP or HEX)

PACKING OF SOLIDS: Two ways to pack atoms: Cubic packing and Closest packing

CUBIC PACKING (one layer): atoms are aligned or in registry

• Primitive Cubic (PC): Atoms sit on the corners of a cube. 1 atom is within the unit cell. CN = 6; packing efficiency = 52%

• Body-Centered Cubic (BCC): Atoms sit on the corners of a cube and one atom sits in the center. 2 atoms are within the unit cell. CN = 8; packing efficiency = 68%

= body-centered atom= corner atom

CLOSEST PACKING (one layer): Atoms offset or out of registry Atoms are packed as closely as possible

• Face-Centered Cubic (FCC) also called Cubic Closest Packing (CCP): Atoms sit on the corners of a cube and one atom sits on each face of the cube. 4 atoms are within the unit cell for cells that contain only one type of atom. CN = 12; packing efficiency = 74% has repeating layers abcabcabc

• Hexagonal Closest Packing (HCP or HEX): Not a cube structure. 6 atoms are within the unit cell for cells that contain only 1 type of atom. CN = 12; packing efficiency = 74%; repeating layers abababa.

COUNTING ATOMS IN A UNIT CELL: Add up all the atoms: 1) corner atoms = 1/8 of an atom

2) edge atoms = 1/4 of an atom 3) face atoms = 1/2 of an atom 4) atoms in the unit cell = an entire atom = 1

Corner atom shown in 2 images. Note how the atom is located within 8 unit cells so only 1/8 of the sphere is in any one unit cell.

Example 1: Determine the unit cell and empirical formulas given the unit cell structure:

Answer 1 Step 1. Take each element separately; will start with metal B. (Atoms are darkened and have an arrow pointing at them.) 8 B atoms; each is a corner atom worth 1/8. B: 8(1/8) = 1 B atom

Step 2. Now work with metal A atoms. 1 A atom; this atom is inside of unit cell and worth 1. A: 1(1) = 1 A atom

Step 3. Now focus on the O atoms. There are 2 kinds of O atoms in the unit cell; work with edge O atoms first. There are 8 O edge atoms; each worth 1/4. O: 8(1/4) = 2 O atoms

Step 4. Continue with O atoms. Work with face O atoms. There are 4 O face atoms; each worth 1/2. O: 4(1/2) = 2 O atoms Total O atoms = 2 + 2 = 4 O atoms

Step 5. Unit Cell Formula: ABO4 Since the unit cell formula can not be simplified the empirical formula is the same as the unit cell formula. Empirical Formula: ABO4

abab

= metal A= oxygen

= metal B

abca

= face atom= corner atom

COORDINATION NUMBER (CN): In a solid with only one type of atom: number of atoms that touch (nearest neighbors) the atom of interest. If a solid contains anions and cations: the number of cations that touch the anions, and the number of anions that touch the cations (these may not be the same!).

HOLES (skip if not covered!): Spaces within solids created by atoms (metallic solids) or anions (ionic solids)

Metals: the holes are usually empty though they may be filled with other metal/nonmetal atoms Ionic compounds: holes are usually created by anions and are completely or partially filled by cations.

Closest packing: Tetrahedral holes (Td) and Octahedral holes (Oh). • Td has CN = 4; made from 1 atom sitting on top of 3 atoms.

• Oh has CN = 6; made from 2 sets of 3 atoms – one set of 3 atoms sits on top of the other

set of 3 atoms offset by 120˚. In closest packing systems, there are always three holes: 1 Oh and 2 Td for every anion, regardless if there is anything inside of these holes.

IONIC COMPOUNDS – Coordination Number (skip if not covered!): The CN for an ionic compound refers to the number of cations that touch anions and vice-versa. The CNanions and the CNcations may or may not be the same but the total are equal; i.e., CNall anions = CNall cations; for example: In NaCl, the Cl- anions create a FCC structure in which the Na+ sit in the Oh created by the Cl-. The CNCl- does not refer to the number of Cl- that touch other Cl-

ions (it would be 12 for a FCC) but rather how many Na+ touch the Cl-. Since the Na+ sit in Oh holes the CNNa+ = 6 (all ions sitting in a Oh have a CN = 6).

Since there is a 1:1 ratio between Na+ and Cl- from the chemical formula, the CNCl- must also equal 6. CNCl- = 6; CNNa+ = 6.

In AlCl3, the Cl- anions create a FCC structure in which the Al+3 sit in the Oh created by the Cl-. Since the Al+3 sit in the Oh the CNAl+3 = 6 (all ions sitting in a Oh

have a CN = 6). Since there is a 1:3 ratio between Al+3 and Cl- from the chemical formula, the CNCl- equals 2 (i.e., each Cl- must touch 2 Al+3); the sum of all the CNCl- must equal the sum of all the CNAl+3; total CNCl- = (#anions)(CNfor each Cl-) = 3(2) = 6;

total CNAl+3 = (#cations)(CNfor each Al+3) = 1(6) = 6. Therefore: CNall Cl- = 6; CNall Al+3 = 6.

PACKING EFFICIENCY: How efficient do the atoms in the solid pack:

Area of circle = pr2 Volume of sphere = 4/3pr3

Example 2: What is the packing efficiency for this 2-D unit cell? Answer 2: 78.5% Step 1. Count #circles in unit cell. 4 circles each worth 1/4. Circles = 4(1/4) = 1 circle in unit cell

Step 2. Determine side length and area of square. side = 2r area = side x side area = (2r)(2r) = 4r2

Step 3. Packing Efficiency

PE = 78.5%

€

Packing Efficiency (2D) = (#circles)(area of circles)area of square

x 100% = (# circles)(πr2)(side)2 x 100%

€

Packing Efficiency (3D) = (#atoms)(volume of atom)volume of cube

x 100%= (# atoms)(4

3πr3)

(side)3 x 100%

PE = areaof circlesareaof square

x100%

PE = (# circles)(areaof a circle)areaof square

x100%

PE = (1)(πr2 )

4r2x100% = π

4x100%

rr

DENSITY OF SOLIDS

mass = #atoms in unit cell x mass of each atom mass of each atom = (atomic mass) x 1/(6.022 x 1023)

volume = l x w x h (where l = length, w = width, and h = height; in a cube: l = w = h = side) when an atomic radius is given and no cube dimensions, calculate the side length as follows:

Unit Cell Side Dimensions

Primitive cubic

Body centered cubic

Face centered cubic

Example 3: If the radius of a Cu atom is 128pm and Cu packs in the face-centered cubic (FCC) arrangement, what is the density of Cu in g/cm3? Answer 3: 8.895g/cm3 solve for mass: mass = #atoms x mass of each atom; from the FCC structure ® 4 atoms;

mass of each Cu atom = (63.55g/mol)/(6.022 x 1023atoms/mol) = 1.055 x 10-22g/atom; mass in unit cell = (4 atoms)(1.055 x 10-22g/atom) = 4.221 x 10-22g;

solve for volume: volume = (side)3; ; ; side = 3.620x10-8cm;

volume = (side)3 = (3.620 x 10-8cm)3 = 4.745 x 10-23cm3;

solve for D:

UNIT CELL SUMMARY unit cell PC BCC FCC HEX #atoms 1 2 4 6 CN (metals) 6 8 12 12 packing eff. (%) 52 68 74 74

side length 2R ---

STRUCTURAL ANALYSIS X-ray crystallography is used to study and determine the structural arrangement of atoms in crystals. X-rays are used because the size of the x-rays (0.01 to 10nm) are approximately the same size as the atoms and layers. X-rays are diffracted off the crystal and a diffraction pattern is observed.

Density = massvolume

= (# atoms)(massof eachatom)side3

=(# atoms) atomicmass

6.022x1023⎛⎝⎜

⎞⎠⎟

side3

R R

side

R

RR R

side

R

R

RR

side

side = 2Rside = 4 3

3R side = (2 2)R

side = 2 2 side = 2 2(128pm) 1m

1x1012pm

⎛

⎝⎜⎜

⎞

⎠⎟⎟100cm1m

⎛⎝⎜

⎞⎠⎟

D = massvolume

= 4.221 x 10−22g

4.745 x 10−23cm3= 8.895g / cm3

€

4 33

R

€

2 2R

dd(sinθ)

θ

X-rays

X-rays

1. What type of solid (molecular, ionic, metallic, or network) will each of the following substances form? a. Ru b. KBr c. H2O d. NaOH e. U f. CO2 g. SiO2 h. Si i. CH4 j. I2 k. CaCO3 l. PH3 2. If there were two samples of the same metal, one with the body-center cubic structure and the other with the face-centered cubic structure, which is more dense? 3. Given that Zn packs in hexagonal closest packing, how many nearest neighbors will it have? 4. Determine the unit cell formula and empirical formula for each structure below.

a.

b. 5. An ionic solid contains a metal, M, and oxygen anions. The O ions form a face-centered cubic structure. The M atoms lie on the center of all the edges of the unit cell. What is the empirical formula of this compound? a. MO b. MO2 c. M2O d. M3O2 e. M3O4 6. In an ionic solid the metal ions (M) occupy two face positions and four edge positions in the cubic unit cell. The anions (X) occupy all the corner positions in the cubic unit cell. What is the empirical formula of the ionic solid? 7. Determine the empirical formula for the structure below. a. MX b. MX2 c. M2X3 d. M3X2 e. M5X2

8. If a solid packs in a body-centered cubic structure and has a cell unit length of 286.65pm, what is the radius (in pm) of the atom?

= copper= gold

= copper (inside unit cell)

= oxygen

= M

= X

9. In an ionic solid the metal ions (M) occupy all of the edge positions in the cubic unit cell. The anions (X) occupy all corner positions and one inner position in the cubic unit cell. What is the empirical formula of the ionic solid? 10. Tungsten, W, packs in the face centered cubic arrangement and has a density of 19.35 g/cm3. What is the radius (in cm) of a tungsten atom? 11. If lead packs in a face centered cubic arrangement and the volume of the unit cell is 1.21 x 10-22 cm3, what is the density of lead? 12. Pt has a density of 21.45 g/cm3 and takes on the face centered cubic unit cell. What is the side dimension of the unit cell in cm? a. 1.39 x 10-8 cm b. 2.47 x 10-8 cm c. 3.21 x 10-8 cm d. 3.92 x 10-8 cm e. 5.22 x 10-8 cm 13. Potassium iodide (KI) unit cell structure contains 4 potassium and 4 iodide ions. The unit cell side is shown at the right. The radius of I- = 206pm and the radius of K+ = 152pm. What is the density of KI? 14. A unit cell of CaF2(s) has the calcium ions creating a FCC structure. The fluoride ions sit in tetrahedral holes. What is the coordination number for the fluoride ion and the calcium ion? 15. What is the packing efficiency for the 2-dimensional unit cell shown?

16. What is the packing efficiency for the 2-dimensional unit cell shown?

ANSWERS 1. a. metallic b. ionic c. molecular d. ionic e. metallic f. molecular g. network h. network i. molecular

j. molecular k. ionic l. molecular 2. Face-centered cubic because the more efficient the packing, the more dense the substance. 3. 12 {memorized} 4. a. unit cell = Cu3Au; empirical formula = Cu3Au {Cu = 6(1/2) = 3Cu; Au = 8(1/8) = 1Au; Cu3Au1} b. unit cell = Cu4O2; empirical formula = Cu2O {Cu = 4(1) = 4Cu; O = 8(1/8) + 1(1) = 2O; Cu4O2 = unit cell formula;

empirical formula is simplest ratio = Cu2O} 5. e {O: 4 O atoms since FCC; M: 12 edges x ¼ = 3M; formula = M3O4} 6. M2X {M: 2(1/2) + 4(1/4) = 2 M atoms; X: 8(1/8) = 1 X atom; M2X}

I- I-K+

7. d {M: 8(1/4) + 1(1) = 3 M atoms; X: 8(1/8) + 2(1/2) = 2 X atoms; M3X2 = unit cell formula = empirical formula}

8. 124.12pm {in BCC: ; ; solve for R; R = 124.12pm}

9. M3X2 {M: 12(1/4) = 3 M atoms; X: 8(1/8) + 1(1) = 2 X atom; M3X2}

10. 1.408 x 10-8cm { ; solve for side; =

= 3.9813 x 10-8 cm; in FCC: side = ; ; solve for R;

R = 1.4076 x 10-8cm}

11. 11.4g/cm3 { ; ; D = 11.374g/cm3}

12. d { ; solve for side; ;

; side = 3.924 x 10-8 cm}

13. 3.00g/cm3 {D = mass/volume; solve for mass: mass = #KI x mass of each KI;

mass of each KI = (39.10 + 126.90g/mol)/(6.022 x 1023atoms/mol) = 2.757 x 10-22g/KI; mass in unit cell = (4 KI)(2.757 x 10-22g/KI) = 1.103 x 10-21g; solve for volume: volume = (side)3; side = 2(K+ radius) + 2(I- radius) = 2(152) + 2(206) = 716pm; side = 716pm(1m/1 x 1012pm)(100cm/1m) = 7.16 x 10-8cm; volume = (side)3 = (7.16 x 10-8cm)3 = 3.671 x 10-22cm3; solve for D: D = mass/volume = (1.103 x 10-21g)/(3.671 x 10-22cm3) = 3.00g/cm3}

14. CNF- = 4; CNCa+2 = 8 {Since the F- are in tetrahedral holes, they will have a CNF- = 4 since all ions in tetrahedral holes

have CN = 4. The CNall anions = (#anions)(CNeach anion) = (2)(4) = 8. Since CNall anions = CNall cations,

CNall cations = 8. Therefore, CNall cations = (#Ca+2)(CNCa+2); CNall cations = (1)(CNCa+2) = 8; CNCa+2 = 8} 15. 78.5% {There is 1 circle in the unit cell with an area of pR2. The side dimension = 2R so the area of unit cell = (2R)2 = 4R2;

; ; }

16. 78.5% {There are 2 circles in the unit cell; each circle has an area of pR2. The diagonal = 4R and from this,

the from the Pythagorean theorem or by recognizing that the unit cell is a face-centered cubic unit cell. Area of

unit cell = 8R2. ; ; }

side = 4 33R 286.65 = 4 3

3R

D =#atoms atomicmass

6.022x1023⎛

⎝⎜

⎞

⎠⎟

(side)3side =

#atoms atomicmass6.022x1023

⎛

⎝⎜

⎞

⎠⎟3

D

side =

4 183.846.022x1023

⎛

⎝⎜

⎞

⎠⎟3

19.35side = 2 2( )R 3.9813x10−8 = 2 2( )R

D =#atoms atomicmass

6.022x1023⎛

⎝⎜

⎞

⎠⎟

volumeD =

4 207.26.022x1023

⎛

⎝⎜

⎞

⎠⎟

1.21x10−22

D =#atoms atomicmass

6.022x1023⎛

⎝⎜

⎞

⎠⎟

(side)3side =

#atoms atomicmass6.022x1023

⎛

⎝⎜

⎞

⎠⎟3

D

side =

4 195.086.022x1023

⎛

⎝⎜

⎞

⎠⎟3

21.45

PE = (#circles)(area of circle)area of unitcell

x100% PE = (1)(πR2)

4R2x100% PE = π

4x100% = 78.5%

side = 2 2( )RPE = (#circles)(area of circle)

area of unitcellx100% PE = (2)(πR

2)

8R2x100% PE = 2π

8x100% = π

4x100% = 78.5%