1. Minerals Ionic Solids Types of bonds Covalentbonding e - s shared equally Ionic coulombic attraction between anion and cation e - s localized Ionic / covalent character depends on difference in electronegativity

2. Electronegativity Calculate stabilities of A + B - and A - B + Difference in energies given by E(A + B - ) – E(A - B + ) = (IP A – EA B ) – (IP B – EA A ) = (IP A + EA A ) – (IP B + EA B ) According to Mulliken half the above difference is the difference in electronegativities of A and B. Thus, the electronegativity of either is ½ (IP + EA)

3. Ionization Potential IP = energy required for A  A+ + e - Electron Affinity EA = energy released in A + e  A -

4. What is meant by the statements that Si – O has 50 % ionic character or that Al – O has 60 % ionic character? F = 3.98, most electronegative Cs = 0.79, least O = 3.44 Si= 1.90 Al= 1.61 (3.44 – 1.90) / (3.98 – 0.79) = 0.48 (3.44 – 1.61) / (3.98 – 0.79) = 0.57

9. Derive minimum radius ratio, r + / r - for coordination 6 Derive minimum radius ratios for coordination 4, 8 and 12 If r - = 1, r + + r - = 2 1/2. So, r + / r - = 0.414. Approach for coordination #s 8 and 12 is similar. That for #4 is trickier. Maybe think, equilateral (tetrahedral) pyramids.

10. Isomorphic substitution of Na + by Ca 2+ is much more common than Na + for K +. Similarly, Li + replaces Mg 2+ more often than it replaces Na +. Use ionic radii to explain. See Table 2.1. This is just a matter of size compatibility.

11. Show that OH - in below structure for gibbsite satisfies Pauling Rule 2. s = Z / CN = 3 / 6 = 1 / 2. Σ s = ½ + ½ = ABS(-1), for OH -1 which is consistent with the above structure.

12. Use the equation s = Z / CN (where s is bond strength, Z is cation valance and CN is coordination number) and Pauling Rule 2 to show that a corner of a Si – O tetrahedron can be linked to one other Si – O tetrahedron but not solely to one other Al – O tetrahedron. In the latter case, show that either two monovalent cations or one bivalent cation with CN = 8 are needed to satisfy the rule. In the first case, s = 4 / 4 = 1, and Σ s = 1 + 1 = 2 = ABS(-2), for O 2-. However, in the second case Σ s = 1 + ¾ = 1 ¾ so that additional bond(s) are needed to satisfy Rule 2. Conceivably, this might involve either 1) 1/8 + 1/8 = 1/4 or 2) 2/8 = 1/4.

16. SiO 4 held together with bivalent cations Si / O = 0.25, which is low So little covalency, easily weathered

17. Si 2 O 6 held together with bivalent cations in octahedral coordination Single chains Si 4 O 11 held together with bivalent cations in octahedral coordination Isomorphic substitution of Al for Si occurs Double chains

18. Si 2 O 5 as Si tetrahedral sheet fused to M octahedral sheet Bonding via apical O of Si tetrahedral sheet M = Al, Fe or Mg, typically coordinated to O 2- or OH - Isomorphic substitution of Al for Si and Al or Fe for Mg

22. Biotite and muscovite common K + balances excess negative charge arising from substitution Located in holes of opposing Si tetrahedral sheets What is the coordination number for K + ? It fits here, 6Os from the 2 adjacent Si tetrahedral sheets. Therefore, CN = 12.

23. Weathering of muscovite, congruent dissolution  K 2 [Si 6 Al 2 ]Al 4 O 20 (OH) 4 (s) + 6C 2 O 4 H 2 (aq) + 4H 2 O = 2K + + 6C 2 O 4 Al + (aq) + 6Si(OH) 4 (aq) + 8OH - (aq) Involves complexation, hydrolysis and loss of silicic acid

24. Weathering of muscovite, incongruent dissolution  K 2 [Si 6 Al 2 ]Al 4 O 20 (OH) 4 (s) + 0.8Ca 2+ (aq) + 1.3Si(OH) 4 (aq) = 2K + (aq) + 0.4OH - (aq) + 1.6H 2 O + 1.1Ca 0.7 [Si 6.6 Al 1.4 ]Al 4 O 20 (OH) 4 (s) Involves reduction of interlayer charge and cation exchange

25. Weathering of biotite to vermiculite  K 2 [Si 6 Al 2 ]Mg 4 Fe(II) 2 O 20 (OH) 4 (s) + 3Mg 2+ (aq) + 2Si(OH) 4 (aq) = 1.25Mg 0.4 [Si 6.4 Al 1.6 ]Mg 5.2 Fe(III) 0.8 O 20 (OH) 4 (s) + FeO(OH)(s) + 2K + (aq) + 4H + (aq) Involves Fe oxidation, also reducing interlayer charge

26. What is the interlayer charge of muscovite, cmol(+) / kg? K 2 [Si 6 Al 2 ]Al 4 O 20 (OH) 4 2 moles of – charge per unit formula due to substitution of Al 3+ for Si 4+. Therefore, 200 cmol(+) / mass of unit formula (kg)

27. AlSi 3 O 8 - or Al 2 Si 2 O 8 2- in 3-D framework with mono- or divalent cations balancing negative charge Isomorphic substitution, Al for Si

28. NaAlSi 3 O 8 (s) + 8H 2 O(l) = Al(OH) 3 (s) + Na + (aq) + 3Si(OH) 4 (aq) + OH - (aq) Weathering of albite to gibbsite involves hydrolysis

29. 4KAlSi 3 O 8 (s) + 0.5Mg 2+ (aq) + 2H + (aq) + 10H 2 O(l) = K[Si 7.5 Al 0.5 ]Al 3.5 Mg 0.5 O 20 (OH) 4 (s) + 4.5Si(OH) 4 (aq) + 3K + (aq) Weathering of orthoclase to montmorillonite involves acidic hydrolysis

30. Generally, weathering of primary silicates involves Loss of tetrahedrally coordinated Al Oxidation of Fe 2+ Consumption of H + Release of silicic acid and cations

31. Phyllosilicates Dominate clay fraction for intermediate to advanced weathering stage Si tetrahedral and Al / Mg octahedral sheets Bonding via apical Os creates distortion –imperfect fit, corners of octahedra and hexangonal structure in Si tetrahedral sheet

34. 1:1 kaolin and serpentine Dioctahedral, kaolin (common in soil) Trioctahedral, serpentine (rare) Little isomorphic substitution

35. Kaolin Kaolinite crystalline units H-bonded together

36. Halloysite interlayer includes structural water but will dehydrate Morphology usually tubular

37. Ormsby et al. (1962) fractionated kaolinites and found Sample #Particle-size fraction (micrometer) 44-1010-52-11-0.5 ________________________________________________________ ------------------------------ m 2 / g --------------------------------- A5.025.868.608.80 H6.096.598.86 10.06 Calculate the surface area for kaolinite of 10 and 1 micrometer equivalent spherical diameter and compare to the above. Assume a density of 2.63 g / cm 3 (Deeds and van Olphen, 1963)

38. A / ρV = 3 / ρr which with r = 5 x 10 -4 and 5 x 10 -5 cm, respectively, gives 3 / 1.325 x 10 -3 = 2260 cm 2 / g or 0.226 m 2 / g and 2.26 m 2 /g, respectively, which are less than in Ormsby et al. (1962). See kaolinite figure (A and B). Deviation from minimum surface area (sphere) increases with increasing size fraction.

39. Which serpentine is suspected of causing cancer?

40. 2:1 pyrophyllite and talc Dioctahedral, pyrophyllite Trioctahedral, talc Negligible isomorphic substitution  Essentially ideal structures

41. 2:1 smectite and saponite Dioctahedral, smectite Trioctahedral, saponite

42. Smecites differentiated Based on site of isomorpthic substitution Montmorillonite, substitution predominantly in octahedral Beidellite, substitution in tetrahedral Nontronite, substitution in tetrahedral and Fe 3+ dominant in octahedral

43. Saponites differentiated based on site of isomorphic substitution Saponite, substitution in tetrahedral sheet Hectorite, substitution in octahedral sheet Also, include presence of Li +

44. 2:1 vermiculite Include dioctahedral and trioctahedral forms Dioctahedral forms exhibit isomorphic substitution in both tetrahedral and octahedral sheets whereas Trioctahedral forms exhibit substitution in tetrahedral sheet Mg 2+ is the octahedral cation Typically form from micas

45. Weathering of biotite to vermiculite K 2 [Si 6 Al 2 ]Mg 4 Fe(II) 2 O 20 (OH) 4 (s) + 3Mg 2+ (aq) + 2Si(OH) 4 (aq) = 1.25Mg 0.4 [Si 6.4 Al 1.6 ]Mg 5.2 Fe(III) 0.8 O 20 (OH) 4 (s) + FeO(OH)(s) + 2K + (aq) + 4H + (aq)

46. 2:1 illite (hydrous mica and other names) Dioctahedral mineral similar to and weathered from mica, including K + as the dominant interlayer cation but Less subsitution in tetrahedral sheet (more Si) Less K +

49. Oxides, oxyhydroxides and hydroxides Al Al(OH) 3 AlOOH

50. Fe FeOOH Fe 2 O 3 FeOOH Fe 3 O 4 Fe 2 O 3

52. Problems 7, 10, 11, 12 and 14