Exercise Booklet 2017-2018_be392c25d71ad447c78263dc2823f879.pdf

2017-2018

4TCH403

Inorganic Chemistry

Exercise Booklet

Corresponding author :

Graziella GOGLIO [email protected]

Exercise 1: The different families of materials After describing the following materials in terms of composition, specify whether they are a ceramic, metal or metal alloy, a polymer or a composite. Name an application for each of these materials. Alumina, Polyvinyl chloride, Teflon, Bronze, Steel, Talcum, Titanium Oxide, Quartz, Magnetite, Cermet, Nitinol, Barium titanate, Polyethylene, SiC/SiC, Polyethylene terephthalate. Exercise 2: Point defects in a copper crystal network The main point defects in the crystallographic network of a pure metal such as copper are vacancies and self-interstitials. Describe what is the nature of the disorder introduced into a crystal respectively by a vacancy and a self-interstitial,

1-

At any temperature T (K), a given number of vacancies or self interstitial point defects may exist in equilibrium within the crystal.

2-

2.1- Given that the enthalpy of formation of a vacancy in the copper crystal is 1 eV, calculate the concentration of vacancies in a copper crystal at a temperature of 25, 250, 500, 750 1080 and 1100 ° C (Tt of Cu = 1083 ° C) 2.2- Respond to the same question for the relative number of self-interstitials knowing that for this type of defects, Hf = 4 eV. Exercise 3: Energy of vacancies formation in a metal Calculate the energy required to form a mole of vacancies in a metal. The concentration in vacancies was measured experimentally as a function of temperature: Temperature (K) nt/N

300

600

800

2.62 10-15 5.18 10-8 3.4 10-6

1000 4.22 10-5

1200

1300

1350

2.3 10-4

4.3 10-4

5.8 10-4

Calculate the enthalpy of formation of one mole of vacancies. Deduce the enthalpy of formation of 1 vacancy. Exercise 4: Determination of the nature of defects in a material To metals A and B, with molar masses MA and MB, where MA < MB, form a single phase AmBn = γ over a composition domain limited by the molar fractions XB min=X1=0.5 and XB max=X2=0.8. Density measurements p of various composites show that the density of γ decreases as XB increases. 1- Determine the range of chemical composition for this solid solution γ. 2- Within this solid solution, the continuous transition from x1 to x2 results in the presence of a single type of defect. 2.1- Determine the three possible types of defects. 2.2- Establish using the composition corresponding to x 1 the expression that gives the chemical formula for each case for each pattern. Specify for each the range for the quantities of defects. 3- Express, assuming a constant lattice volume, the density of each of the three solid solutions as a function of p(x1). 4- Deduce the real nature of the defect in γ.

Exercise 5: Substitutions in covalent materials Given: Element Carbon Silicon Germanium 12-

3-

Molar mass (g mol-1) 12.011 28.086 72.59

Electronegativity (Pauling scale) 2.5 1.8 1.8

Covalent radius (pm) 118 122

What is the common feature between C, Si and Ge? Silicon and germanium both crystallize as carbon does, in the diamond structure. 2.1- Project the diamond structure in the plane (xOy) and label the z coordinate 2.2- Indicate the number of nearest neighbors and determine the shortest atom-atom distance as a function of the lattice parameter a. 2.3- Knowing that silicon density is 2.3 g cm-3, deduce the lattice parameter of silicon and the shortest Si-Si distance. Compare this result with the radius of covalently bound silicon. Ge and Si are totally miscible in the solid state. 3.1- Give the reasons allowing this total miscibility 3.2- A 27.9 wt% mixture of silicon in germanium yields a substitutional solid solution with a lattice parameter of 5.54 Å. Calculate the lattice parameter of germanium.

Exercise 6: Substitutions in metallic materials Calculate the lattice parameter of a substitutional solid solution containing 72 wt% iron and 28 wt% chromium. Given: M(Fe) = 55.847 g mol-1 M(Cr) = 51.996 g mol-1 R(Fe) = 1.24 Å R(Cr) = 1.29 Å Fe and Cr structure : body-centered Exercise 7: Point defects in ionic crystals The ionic crystal of NaCl crystallizes in a face-centered cubic unit cell. 1The lattice parameter of NaCl is a=0.552 nm. The ionic radii are respectively 96 pm for the cation and 180 pm for the anion. Show and justify the relationship between the lattice parameter and the ionic radii. 2Calculate the density of NaCl considering the following data: M(Na)=22.99 g mol-1 and M(Cl)=35.5 g mol-1 3In these crystals, some Schottky defects can form. Give the definition of a Schottky defect in an ionic crystal. 4In real NaCl crystals we determine there exist 5% Schottky defects and in another 5% Frenkel defects. Calculate the solid density for both cases. 5Justify the following data considering the nature of the chemical bond in the materials listed in the table below: Material Sodium Sodium chloride

Young’s modulus (GPa) 5 37

Melting point (K) 370 1073

Exercise 8: A lacunar oxide As many transition oxides, niobium oxide Nb2O5 can form a nonstoichimetric compound at high temperature, resulting in the formation of doubly ionized oxygen vacancies at high temperature. 1Write the equation describing the equilibrium between this oxide and dioxygen. 2The electronic charges liberated during this reaction induce valency defects which interact with cations in normal positions in the crystalline network. 2.1- What is the nature of these valency defects? 2.2- Write the charge transfer mechanism responsible for electronic conduction. 3Demonstrate that during the conditions defined above, the concentration of ionized O2vacancies scales with the partial oxygen pressure, PO2. 4The difference in stoichiometry of the Nb2O5-x oxide, always proportional to the vacancy concentration x, also varies with temperature. One measures, under PO2= 10-13 bar: T(°C) - ln X

927 2.54

977 2.30

1027 2.05

1077 1.82

1127 1.59

1177 1.41

1227 1.21

Determine the standard enthalpy of formation of the vacancies, ΔfH°.

Exercise 9: Nonstoichiometry in NiO Let’s consider NiO:

Structure type NaCI a=4.159 Å M(Ni)= 58.7 g mol-1

M(O)=16 g mol-1

1- Calculate the density, ρ0, of the stoichiometric material. 2- Chemical analyses reveal a non-stoichiometry in NiO. The material can be described by the formula NiOx with x having an interval of [1.0; 1.1]. 2.1- What are the three possible types of defects? For each defect type, write the chemical formula of the material, introducing a variable (such as l, s or i) that describes the defect formation appearing in the ideal, stoichiometric compound. 2.2- Establish the expression of x as a function of l, s and i. Determine the minimum and maximum values of x forming the limits for l, s and i. 2.3- For each type of defect, establish an expression for the density ρ as a function of ρ0 for l, s and i. Calculate the density of the NiOx material assuming the largest nonstoichiometry (x=1.1). 2.4- The density of the Ni01.1 material is characterized by pycnometry, and determined to be 6.41 g cm-3. What is the nature of the defect? 2.5- Considering the nature of the defect identified in the previous question, determine the kind of conduction expected for NiOx. 2.5.1- Explain the formation of this defect, and specifically indicate whether it is favoured by large or small oxygen partial pressures (a text is expected rather than an equation). 2.5.2- Establish the equation of the defect formation in the Kröger Vink notation. 2.5.3- Does this defect induce a change to the cation valency? Justify your answer. 2.5.4- Deduce from the nature of the charge carriers the kind of conduction, then the relationship between the charge carrier concentration and the oxygen partial pressure. 2.5.5- Conclude with by expressing the conductivity as a function of the oxygen partial pressure.

Exercise 10: Non-stoichiometry of the pyrrhotites Pyrrhotite, FeS, is a covalent compound that crystallizes in a hexagonal structure, in which the iron atoms occupy the octahedral sites of a compact network of sulfur atoms. Its cell parameters are: a=3.45 Å and c=5.70 Å. Pyrrhotite is actually nonstoichiometric, with a formula Fe uS, where u belongings between the [0.875; 1.0] interval. The variation of cell parameters within the solid solution should be considered to be negligible. Given: M(Fe)=55.87 g mol-1 and M(S)=32.06 g mol-1 12-

Calculate the density of the stoichiometric FeS crystal. Establish as a function of u the expression of the density of FeuS pyrrhotites when: a) It is an interstitial solid solution b) It is a lacunar solid solution

3- The density of Fe0,875S is about 4.5 g.cm-3. Deduce the nature of the nonstoichiometry in pyrrhotites. Clarify the chemical composition of Fe0,875S by taking into account electron transfers.

Exercise 11: Linear defects Characterize the defect below from the layout of the Burgers ‘s circuit:

0000000 000

Exercise 12 : Dislocation motion 1- A linear defect moves when a sufficient level of shear stress  is applied to it. Is the stress required for the dislocation motion higher in Si or Fe? Suppose the same dislocation in Si crystal and Fe crystal. Please, justify your answer. 2- Show that the figure below displays two dislocations of opposite sign. Represent the final shape of the crystal when a stress field  is applied. The two possible orientations for will be considered.

Exercise 13 : Microstructure of pure zirconium

Describe this image. What are the defects highlighted? What is the mean size of grains?

Exercise 14 : Austenite microstructure The microstructure below corresponds to austenite, a solid solution of face-centered cubic crystal structure.

(magnification : X 325)

1234-

Identify the defects in the austenite in this micrograph. Determine approximately the mean size (in µm ) of the austenite grains. How are the 2D defects highlighted? What are the other possible defects in austenite?

Exercise 15 : 3D defects Copper (FCC) has a radius of 0.128 nm. The incorporation of tin of radius 0.151 nm gives bronze having 3 at% tin and 97 at% copper. 1- Bronze retains the fcc structure to form a solid solution. Its lattice parameter increases, its value at 3 at% tin is 0.364 nm. Justify the tin substitution and calculate the density of this bronze. 2- In realty, tin-saturated copper leads to Cu3Sn precipitation in the copper matrix. This compound has an orthorhombic structure with lattice parameters: a=0,434 nm, b=0,556 nm and c=3,81 nm Calculate the number of tin and copper atoms in a cell knowing that the density of this compound is 8.94 g/cm3 3- Assuming that precipitates are distributed in the matrix in the cell shape defined above, their 26 3 volume density would be 10 /m . In fact, they are parallelepiped needles such as a'= 100a, 19 3 b'=100b, c'= 1000c and their volume density then becomes 10 /m . Show that the growth of precipitates leads to a decrease of internal energy. The copper matrix/Cu3Sn interface energy is 0,8J/m2. MCu=63,55 g mol-1, MSn=118,69 g mol-1

Exercise 16: Properties and microstructure of a copper-aluminium system Let us consider the copper-aluminium system Given:

R(Al)=1.431 Å

M(Al)=26.981 g mol-1

FCC structure

R(Cu)=1.278 Å

M(Cu)=63.546 g mol-1

FCC structure

The experimental lattice parameters for both copper and aluminium were determined using X-ray diffraction. The following values were obtained: aexp(Al)=4.04901 Å aexp(Cu)=3.61503 Å 1. Are these lattice parameter values in accordance with theory? Explain your response. 2. Solid solution a) Explain why Cu and Al cannot form a total solid solution via substitution. b) The phase SSmax is the maximum solubility of copper in aluminium, having a lattice parameter of 4.04864 Å. Calculate the molar fractions of copper and aluminium in SSmax. Hint: conserve 5 digits after the decimal point in your calculations. 3. Precipitation within the Cu matrix. If the quantity of introduced copper is greater than that defined in the previous question, the obtained material contains AlyCuz precipitates dispersed in the SSmax matrix. A elemental chemical analysis shows that AlyCuz contains 54.08% copper, by mass. a) Determine the composition of AlyCuz. b) Explain the impact of the AlyCuz precipitates on the mechanical properties of the SSmax matrix in comparison to the properties of pure Al and AlyCuz-SSmax. Hint: First compare Al and SSmax.

Exercise 17: Thermal and crystallographic analysis of coral Coral is a natural ceramic mainly composed of calcium carbonate (CaCO3) (M = 100.08 g/mol). During the stage of biomineralization, it crystallizes as aragonite. This ceramic also possesses another crystalline variety, which is calcite (See the table below).

Crystal System Lattice parameters (Å)

Density ρ (g/cm3)

Aragonite Orthorhombic a=4.9623 b=7.968 c=5.7439 2.93

Calcite Hexagonal a=4.990 c=17.002 2.72

1- Calculate the number of patterns in each lattice. 2- During heat treatment, aragonite is transformed into calcite. Calculate the volume variation (ΔV/V) associated with this transformation. 3- Explain the transformations that occur during heating and cooling from the dilatometric analysis. Comment on the stability of both aragonite and calcite. 4- Calculate the thermal coefficient of expansion of calcium carbonate.

5- Assuming that the properties of coral are isotropic, show that the dilatometry curve is coherent with the calculation made in question 2. 6- Coral is used medically as a bone substitute. Before its implantation, coral must be heattreated at 500 °C to remove the organic compounds. Determine the volume of coral to use so that the implant will accommodate a cubic cavity of 1 cm3.

heating

cooling

Temperature (°C)

Exercise 18: Problem of thin layer deposition A technician meets the following problem: he or she tries to deposit a layer of diamond by PVD (Physical Vapor Deposition) on a copper substrate. For this, the substrate must be treated at high temperature. Under the deposition conditions, we observe a diamond layer on Cu (see Figure a) below). This layer contains no defect and adheres to the substrate. However, after cooling the material to ambient temperature, the technician observes that the layer cracks and then bursts (see Figure b below). Diamond layer Copper substrate a) Formation of the diamond layer on Cu substrate during deposition (T > Tambient) Explain why the layer cracks and then bursts.

diamond copper b) The diamond layer on the Cu substrate after the deposition and cooling to ambient conditions (T = Tambient)

Exercise 19: Intrinsic properties 1- We give the following values of Young’s modulus E: 1) 0.7 GPa 2) 40 GPa 3) 250 GPa 4) 400 GPa 5) 450 GPa Attribute to each of the following materials the corresponding E value: a) Silicon carbide

c) Alumina

b) Magnesia (MgO) d) Magnesium

e) Polyethylene

f) Diamond

2- We give the following values of melting point Tm: 4) 2053 °C 1) 113.7 °C 2) 660 °C 3) 800.7 °C Attribute to each of the following materials the corresponding Tm value: a) Aluminium b) Diiodide (I2) c) Alumina

d) Sodium chloride

6) 1000 GPa