Transcript Presentation 1

Minerals
Ionic Solids
Types of bonds
Covalent
bonding 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+
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 difference
in electronegativities of A and B.
Thus, the electronegativity of either is ½ (IP + EA)
Ionization Potential
IP = energy required for A  A+ + e-
Electron Affinity
EA = energy released in A + e  A-
What is meant by the statements that Si – O has 50 % ionic character or
that Al – O has 60 % ionic character?
F
Cs
= 3.98, most electronegative
= 0.79, least
O
Si
Al
= 3.44
= 1.90
= 1.61
(3.44 – 1.90) / (3.98 – 0.79) = 0.48
(3.44 – 1.61) / (3.98 – 0.79) = 0.57
Derive minimum radius ratio, r+ / r- for coordination 6
If r- = 1, r+ + r- = 21/2. So, r+ / r- = 0.414.
Approach for coordination #s 8 and 12 is
similar. That for #4 is trickier. Maybe think,
equilateral (tetrahedral) pyramids.
Derive minimum radius ratios for coordination 4, 8 and 12
Isomorphic substitution of Na+ by Ca2+ is much more common than
Na+ for K+. Similarly, Li+ replaces Mg2+ more often than it replaces
Na+. Use ionic radii to explain.
See Table 2.1. This is just a matter of size compatibility.
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.
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 O2-.
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.
SiO4 held together with bivalent cations
Si / O = 0.25, which is low
So little covalency, easily weathered
Si2O6 held together with bivalent cations
in octahedral coordination
Single chains
Si4O11 held together with bivalent cations
in octahedral coordination
Isomorphic substitution of Al for Si occurs
Double chains
Si2O5 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 O2- or OHIsomorphic substitution of Al for Si
and Al or Fe for Mg
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.
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
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
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
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)
AlSi3O8- or Al2Si2O82- in 3-D framework
with mono- or divalent cations balancing
negative charge
Isomorphic substitution, Al for Si
NaAlSi3O8(s) + 8H2O(l) =
Al(OH)3(s) + Na+(aq) + 3Si(OH)4(aq) + OH-(aq)
Weathering of albite to gibbsite involves hydrolysis
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 involves
acidic hydrolysis
Generally, weathering of primary silicates involves
Loss of tetrahedrally coordinated Al
Oxidation of Fe2+
Consumption of H+
Release of silicic acid and cations
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
1:1 kaolin and serpentine
Dioctahedral, kaolin (common in soil)
Trioctahedral, serpentine (rare)
Little isomorphic substitution
Kaolin
Kaolinite crystalline units
H-bonded together
Halloysite interlayer includes
structural water
but will dehydrate
Morphology usually tubular
Ormsby et al. (1962) fractionated kaolinites and found
Sample #
Particle-size fraction
(micrometer)
44-10
10-5
2-1
1-0.5
________________________________________________________
A
------------------------------ m2 / g --------------------------------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)
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.
Which serpentine is
suspected of causing
cancer?
2:1 pyrophyllite and talc
Dioctahedral, pyrophyllite
Trioctahedral, talc
Negligible isomorphic substitution
 Essentially ideal structures
2:1 smectite and saponite
Dioctahedral, smectite
Trioctahedral, saponite
Smecites differentiated
Based on site of
isomorpthic substitution
Montmorillonite, substitution
predominantly in octahedral
Beidellite, substitution in
tetrahedral
Nontronite, substitution in
tetrahedral and
Fe3+ dominant in octahedral
Saponites differentiated based on site of isomorphic substitution
Saponite, substitution in tetrahedral sheet
Hectorite, substitution in octahedral sheet
Also, include presence of Li+
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
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)
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+
Oxides, oxyhydroxides and hydroxides
Al
Al(OH)3
AlOOH
Fe
FeOOH
FeOOH
Fe3O4
Fe2O3
Fe2O3
Problems
7, 10, 11, 12 and 14