minerals ionic solids types of bonds covalentbonding e - s shared equally
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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. Electronegativity Calculate stabilities of A + B - and A - B + - PowerPoint PPT PresentationTRANSCRIPT
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Minerals
Ionic Solids
Types of bonds
Covalent bonding e-s shared equally
Ionic coulombic attraction between anion and catione-s localized
Ionic / covalent character depends on difference in electronegativity
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Electronegativity
Calculate stabilities of A+B- and A-B+
Difference in energies given by
E(A+B-) – E(A-B+) = (IPA – EAB) – (IPB – EAA)
= (IPA + EAA) – (IPB + EAB)
According to Mulliken half the above difference is the differencein electronegativities of A and B.
Thus, the electronegativity of either is ½ (IP + EA)
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Ionization Potential
IP = energy required for A A+ + e-
Electron Affinity
EA = energy released in A + e A-
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What is meant by the statements that Si – O has 50 % ionic character orthat Al – O has 60 % ionic character?
F = 3.98, most electronegativeCs = 0.79, least
O = 3.44Si = 1.90Al = 1.61
(3.44 – 1.90) / (3.98 – 0.79) = 0.48
(3.44 – 1.61) / (3.98 – 0.79) = 0.57
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Derive minimum radius ratio, r+ / r- for coordination 6
Derive minimum radius ratios for coordination 4, 8 and 12
If r- = 1, r+ + r- = 21/2. So, r+ / r- = 0.414.
Approach for coordination #s 8 and 12 issimilar. That for #4 is trickier. Maybe think,equilateral (tetrahedral) pyramids.
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Isomorphic substitution of Na+ by Ca2+ is much more common than Na+ for K+. Similarly, Li+ replaces Mg2+ more often than it replacesNa+. Use ionic radii to explain.
See Table 2.1. This is just a matter of size compatibility.
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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.
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Use the equation s = Z / CN (where s is bond strength, Z is cationvalance 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 O2-.
However, in the second case Σ s = 1 + ¾ = 1 ¾ so that additionalbond(s) are needed to satisfy Rule 2. Conceivably, this might involveeither 1) 1/8 + 1/8 = 1/4 or 2) 2/8 = 1/4.
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SiO4 held together with bivalent cations
Si / O = 0.25, which is lowSo little covalency, easily weathered
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Si2O6 held together with bivalent cationsin octahedral coordination
Single chains
Si4O11 held together with bivalent cations in octahedral coordination
Isomorphic substitution of Al for Si occurs
Double chains
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Si2O5 as Si tetrahedral sheet fused toM octahedral sheet
Bonding via apical O of Si tetrahedral sheet
M = Al, Fe or Mg, typically coordinated to O2- or OH-
Isomorphic substitution of Al for Si and Al or Fe for Mg
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Biotite and muscovite common
K+ balances excess negative charge arising from substitution
Located in holes of opposing Si tetrahedral sheets
What is the coordination numberfor K+?
It fits here, 6Os fromthe 2 adjacent Sitetrahedral sheets.Therefore, CN = 12.
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Weathering of muscovite, congruent dissolution
K2[Si6Al2]Al4O20(OH)4(s) + 6C2O4H2(aq) + 4H2O =
2K+ + 6C2O4Al+(aq) + 6Si(OH)4(aq) + 8OH-(aq)
Involves complexation, hydrolysis and loss of silicic acid
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Weathering of muscovite, incongruent dissolution
K2[Si6Al2]Al4O20(OH)4(s) + 0.8Ca2+(aq) + 1.3Si(OH)4(aq) =
2K+(aq) + 0.4OH-(aq) + 1.6H2O +
1.1Ca0.7[Si6.6Al1.4]Al4O20(OH)4(s)
Involves reduction of interlayer charge and cation exchange
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Weathering of biotite to vermiculite
K2[Si6Al2]Mg4Fe(II)2O20(OH)4(s) + 3Mg2+(aq) + 2Si(OH)4(aq) =
1.25Mg0.4[Si6.4Al1.6]Mg5.2Fe(III)0.8O20(OH)4(s) +
FeO(OH)(s) + 2K+(aq) + 4H+(aq)
Involves Fe oxidation, also reducing interlayer charge
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What is the interlayer charge of muscovite, cmol(+) / kg?
K2[Si6Al2]Al4O20(OH)4
2 moles of – charge per unit formula due to substitution of Al3+ for Si4+.Therefore, 200 cmol(+) / mass of unit formula (kg)
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AlSi3O8- or Al2Si2O8
2- in 3-D frameworkwith mono- or divalent cations balancing negative charge
Isomorphic substitution, Al for Si
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NaAlSi3O8(s) + 8H2O(l) =
Al(OH)3(s) + Na+(aq) + 3Si(OH)4(aq) + OH-(aq)
Weathering of albite to gibbsite involves hydrolysis
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4KAlSi3O8(s) + 0.5Mg2+(aq) + 2H+(aq) + 10H2O(l) =
K[Si7.5Al0.5]Al3.5Mg0.5O20(OH)4(s) +
4.5Si(OH)4(aq) + 3K+(aq)
Weathering of orthoclase to montmorillonite involvesacidic hydrolysis
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Generally, weathering of primary silicates involves
Loss of tetrahedrally coordinated AlOxidation of Fe2+
Consumption of H+
Release of silicic acid and cations
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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
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1:1 kaolin and serpentine
Dioctahedral, kaolin (common in soil)
Trioctahedral, serpentine (rare)
Little isomorphic substitution
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Kaolin
Kaolinite crystalline units H-bonded together
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Halloysite interlayer includes structural water but will dehydrate
Morphology usually tubular
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Ormsby et al. (1962) fractionated kaolinites and found
Sample # Particle-size fraction (micrometer)
44-10 10-5 2-1 1-0.5________________________________________________________
------------------------------ m2 / g ---------------------------------A 5.02 5.86 8.60 8.80
H 6.09 6.59 8.86 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 / cm3 (Deeds and van Olphen, 1963)
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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 cm2 / g or 0.226 m2 / g
and 2.26 m2 /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.
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Which serpentine issuspected of causingcancer?
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2:1 pyrophyllite and talc
Dioctahedral, pyrophyllite
Trioctahedral, talc
Negligible isomorphic substitution
Essentially ideal structures
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2:1 smectite and saponite
Dioctahedral, smectiteTrioctahedral, saponite
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Smecites differentiated Based on site of isomorpthic substitution
Montmorillonite, substitution predominantly in octahedral
Beidellite, substitution in tetrahedral
Nontronite, substitution intetrahedral and Fe3+ dominant in octahedral
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Saponites differentiated based on site of isomorphic substitution
Saponite, substitution in tetrahedral sheet
Hectorite, substitution in octahedral sheet Also, include presence of Li+
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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 Mg2+ is the octahedral cation
Typically form from micas
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Weathering of biotite to vermiculite
K2[Si6Al2]Mg4Fe(II)2O20(OH)4(s) + 3Mg2+(aq) + 2Si(OH)4(aq) =
1.25Mg0.4[Si6.4Al1.6]Mg5.2Fe(III)0.8O20(OH)4(s) +
FeO(OH)(s) + 2K+(aq) + 4H+(aq)
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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+
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Oxides, oxyhydroxides and hydroxides
Al
Al(OH)3 AlOOH
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Fe
FeOOH
Fe2O3
FeOOH
Fe3O4
Fe2O3
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Problems
7, 10, 11, 12 and 14