1. Chapter 7 Basic Mineralogy

2. Mineral: a naturally occurring, inorganic substance with a characteristic internal structure and a chemical composition that is either fixed or varies within certain limits. http://www.oum.ox.ac.uk/thezone/minerals/in dex.htm

3. Ionization potential: a measure of the energy required to remove an electron from an atom and place it at an infinite distance from the nucleus. Electronegativity: a measure of the ability of an atom to attract electrons. (The smaller the electronegativity, the less likely the atom will attract electrons—it will most likely donate them instead.)

4. A Measure of electronegativity of elements as seen in the periodic table.

7. Example 7-1 The mineral fluorite has the chemical composition CaF 2. Calculate the ionic character of the bond between Ca-F. From Table 7-2, the difference in electronegativity = 3.98 (F - ) -1.00(Ca 2+ ) = 2.98 From table 7-3, the bond is ~89% ionic. http://skywalker.cochise.edu/wellerr/mineral/fluorite/fluoriteL.htm

8. Coordination number: the number of anions that surround a cation in an ionic crystal. Radius ratio: the radius of the cation divided by the radius of the anion. http://web.arc.losrios.edu/~borougt/MineralogyDiagrams.htm

9. So, we seem to think that silica (SiO 4 4- ) has a coordination number of 4. Let’s test this. From appendix III, the ionic radius of Si 4+ = 0.48 & O 2- = 1.32. Then R c /R a = 0.48/1.32 = 0.36. If we were to check the corresponding radius ratios from figure 7-2, we would see that it fits nicely in the tetrahedral arrangement with a coordination number of 4. Of course, we already knew that one! Forsterite: Mg 2 SiO 4 http://www.minerals.net/Image/5/97/Olivine.aspx

10. The Unit cell is the basic building block for a crystal. In order to understand this concept, think of the unit cell as being like a brick in a wall (if the wall is built by stacking bricks directly upon one another).

11. X-ray Crystallography: the science of determining the arrangement of atoms within a crystal from the manner in which a beam of X-rays is scattered from the electrons within the crystal. The method produces a three-dimensional picture of the density of electrons within the crystal, from which the mean atomic positions, their chemical bonds, their disorder and sundry other information can be derived.

12. Bragg’s Law describes the relationship between the angle of the incident monochromatic x-ray beam and the diffracted ray as a result of the crystalline structure and interplanar spacing. n = 2dsin  A-C is the interplanar spacing and is equal to d. is the wavelength of the x-ray and  is the angle of incidence and diffraction.

13. http://www.mindat.org/min-1584.html X-ray Diffraction Pattern for Forsterite Mg 2 SiO 4. http://www.ccp14.ac.uk/poster-talks/phase-id-1999/html/phaseid.htm & Fluorite CaF 2

14. Si O O O O Silica (SiO 4 )

15. Examples of silicate minerals olivine epidote beryl augite hornblende muscovite quartz Mineral pictures from: mindat.org

16. Pink (Rose) : due to traces of iron, manganese or titanium. Amethyst : Maybe be manganese but some believe it could be organic, iron or even aluminum. Citrine : iron Aventurine : inclusion of green mica (fushite) Tiger's eye : inclusion of fiber of silicified crocidolite (variety of asbestos) Prasiolite : Iron or copper Milk quartz : gas and liquid inclusions Smoky : Radioactivity on quartz containing aluminium Blue : pressure. Chalcedony is a variety of quartz with micro-crystals. Agate is a multicolor variety of chalcedony and onyx is a variety of agate with parallel strips of various nuances of black. Quartz Varieties

17. Ionic Substitutions When minerals crystallize, certain minor or trace elements that are present in the environment can enter the structure of the crystallizing mineral. There are four rules that predict, with many exceptions, the uptake of trace elements by crystallizing minerals. 1.Ions of one element can substitute for those of another in a crystal structure if their radii differ by less than ~15%. 2.Ions that differ by one charge unit substitute readily for each other as long as charge neutrality is maintained. 3.When two ions occupy the same site in a crystal structure, the ion with the higher ionic potential preferentially enters the site. 4.Even if the size and charge of the minor and major ion are similar, substitution may be limited for the minor ion if it has a very different electronegativity and forms a bond of very different character from that of the major ion.

19. General clay types Kaolinite, illites, smectites, vermiculite Kaolinite – 1 tetrahedral and 1 octachedral layer (1:1) -Limited subsitution of Al in the basic formula (Al 2 Si 2 O 5 (OH) 4 ) -net surface charge minimal, negligible CEC Illite – 2 tetrahedral and 1 octachedral layer (2:1) ….the octahedral sandwich -Al substitution for Si in tetrahedral layer -marginal net surface charge minimal, low CEC Smectites – also a 2:1 clay -lots of Fe and Mg substitutions for Al in octahedral layer -lots of Al substitution for Si in the tetrahedral layer -swelling clay -significant net surface charge, high CEC Vermiculites – also 2:1 clay -higher net surface charge -high CEC

20. 1:1 Clays: consist of tetrahedral layer and an octahedral layer; substitutions are limited and the net charge is minimal (have a low CEC.) kaolinite

21. 2:1 clays: consists of two tetrahedral layers with an intervening octahedral layer. The octahedral layer can be either di- or tri- octahedral and a large variety of substitutions are possible. 2:1 clays have a greater variation with net charge possibilities and generally have a greater C.E.C. montmorillonite

22. The octahedral and tetrahedral layers are arranged in different ways with different amounts of elemental substitutions to produce different clay minerals.

23. Clays will have negative net surface charge caused by: 1)Subsitutions ….Al 3+ for Si 4+ in tetrahedral layer, Mg 2+ for Al 3+ in octahedral layer 2)Imperfections in crystal structure (e.g. missing cations) 3)Broken bonds at edges of crystals (exposing O 2- or OH - ions) For the 2:1 clays surface charge arises mostly from substitutions and imperfections For 1:1 clays surface charge arises mostly from broken bonds at crystal edges

24. What is Cation Exchange Capacity (CEC) and why is it important? http://www.finerminds.com/health-fitness/vitaminwater-not-healthy/ http://www.finesttreeserviceaz.com/

25. http://fernroadfarm.blogspot.com/

26. Cation Exchange Capacity The net negative surface charge will attract (adsorb) ambient dissolved cations If other cations are introduced, these adsorbed cations will be replaced by the new cations to varying degrees ….cation exchange The cation exchange capacity (CEC) will vary from clay to clay depending on clay structure, amount and type of substitution, pH, and particle surface area CEC – specifically defined as the uptake or release of ammonium (NH 4 + ) ions when the clay is exposed to a 1 M ammonium acetate solution at pH 7.0 Units for CEC = meq / 100g CEC(meq/100g) = N SM (mole sites/g)x10 5 If surface has a net positive charge then the AEC (anion exchange capacity) is measured

27. Surface area effects The number of negative surface sites per area (N SA ) is related to the surface area (S A ),and the mole sites per unit mass (N SM ) N SA = N SM / (1.66 E -6 ) (S A ) And the N SM is used to calc CEC CEC = N SM (1.0 E 5 ) units N SA – sites nm -2 N SM – mole sites g -1 S A – m 2 g -1 CEC – meq 100g -1 clay particles have a large surface to mass ratio

28. Example 7-6: Given a smectite with a negative surface charge of 0.8 mole sites kg -1, what is the CEC? CEC= (0.8 mole sites kg -1 ) (.001 kg g -1 ) (1 E 5 ) = 80meq 100g -1 Determining ion-exchange properties Batch method Start with solution of known cation conc., ionic strength, and pH Add clay Compare cation conc. before and after addition of clay Repeat at different pHs, ionic strengths, cation concentrations Use data to construct adsorption isotherms

29. Adsorption Isotherms Represent partitioning of a particular species between an aqueous phase and solid particles (sorbate) Figure 7-8. Representation of a typical adsorption isotherm showing the distribution of a species between an aqueous phase and a solid (sorbent). At very low concentrations, the distribution behaves ideally and can be represented by a unique value, K d. At higher concentrations, the partitioning deviates from ideality. If precipitation occurs, the concentration of the species in solution will remain constant; i.e., the solution is saturated with respect to the particular species. Kd is the tangent to the isotherm found at the origin At high concentrations, precipitation keeps the aqueous concentration constant

30. Column Test Method In this case, the sorbent is packed into a column and a volume of solution is passed through the column. The concentration of the ion of interest in the original solution is compared to that in the eluent and the K d is calculated using the following formula. K d = ((C i – C f )/C f ))(V/M) http://www.cresp.org/cresp-projects/waste-processing-special-nuclear- materials/leaching-assessment-for-alternative-waste-forms/

31. Example 7-7 Ten grams of montmorillonite are placed in a column and 100ml of solution are passed through the column. The initial solution has a zinc concentration of 20 mg L -1 and the eluent has a zinc concentration of 14.1 mg L -1. Calculate the K d for zinc between the solution and the montmorillonite. K d = ((20 – 14.1)/14.1))(100 cm 3 /10g) = 4.18 cm 3 g -1

32. Zeolites: a crystalline structure characterized by a framework of linked tetrahedra, each consisting of four O atoms surrounded by a cation. This framework open cavities in the form of channels and cages. These channels are usually occupied by H 2 O, but large enough to allow the passage of guest species. Zeolites have relatively large CEC and are useful for a variety of environmental remediation processes.

33. Asbestos minerals: a group of silicate minerals that occur as long, thin fibers. They have high tensile strength, flexibility, and heat and chemical resistance. Asbestos minerals can be described by two different structures: chrysotile and amphibole.

34. Chrysotile structure: consists of a layer of silica tetrahedra bonded to a layer of octahedrally coordinated Mg ions. Each Mg 2+ is surrounded by four hydroxyl molecules and two oxygens. The distance between the oxygens in the octahedral layer is slightly greater than the distance between the oxygens in the tetrahedral layer. This results in the octahedral layer curling around the tetrahedral layer forming a scrolled tube.

35. Amphibole structure: consists of a strip of octahedrally coordinated cations sandwiched between two double silica chains. The chains extend for an infinite distance. The cations can be Na, Li, Ca, Mn, Fe, Mg, Al, and Ti.

36. Health Effects of Asbestos Exposure Asbestosis: a lung disease caused by asbestos particles deposited in the lungs through inhalation. Over time, the lung encapsulates these fibers and hardens leading to a decrease in the efficiency of the O 2 /CO 2 exchange. Mesothelioma: a rare, diffuse malignant cancer of the lining of the lung and stomach. It has a long latency period of 35 to 40 years. Lung cancer: usually linked to smoking, however, some cases have been attributed to radon, second-hand smoke, or exposure to asbestos.

37. Crystalline and Amorphous Silica There are six polymorphs (same chemical composition, but different crystalline structure) of silica composition with a chemical formula of SiO 2.

38. Amorphous silica (opal, SiO 2 ·nH 2 O) is found in siliceous oozes in the seafloor sediments and on land as preserved deposits of marine sediments or precipitated from geyer fluids that contain high amounts of dissolved silica.

39. Dissolution of Silica Minerals SiO 2(s) + 2H 2 O  H 4 SiO 4(aq) For quartz: logK sp = 1.8814 – 2.028 x 10 -3 T – 1560.46 / T For amorphous silica: logK sp = 0.338037 – 7.8896 x 10 -4 T – 840.075 / T Where T is the temperature in Kelvin. When figuring the solubility in ppm, remember to multiply the K sp (in units of moles/L) by the gram-molecular weight of silica (SiO 2 ) regardless if you are calculating quartz or amorphous silica.

40. Example 7-8 Calculate the K sp for quartz and amorphous silica at a pH of 5.5 and T = 25 ⁰ C. Which form of silica is more soluble? For Quartz: Log K sp = 1.8814 – (2.028 x 10 -3 )(298.15) – 1560.46/298.15 = -3.96 K sp = 10 -3.96 For Amorphous silica: Log K sp = 0.338037 – (7.8896 x 10 -4 )(298.15) – (840.075/298.15) = -2.71 K sp = 10 -2.71 Amorphous silica has the larger Ksp and is the more soluble form of solid silica.

41. Chapter 7 Problem set due November 26: #s: 1, 9, 10, 14, 36, 49, 55, 57 http://agwired.com/2010/11/12/the-cost-of-thanksgiving-dinner/