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

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)

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 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

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.

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.

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 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.

SiO4 held together with bivalent cations

Si / O = 0.25, which is lowSo little covalency, easily weathered

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

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

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.

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 Al2Si2O8

2- in 3-D frameworkwith 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 involvesacidic hydrolysis

Generally, weathering of primary silicates involves

Loss of tetrahedrally coordinated AlOxidation 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________________________________________________________

------------------------------ 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)

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 issuspected of causingcancer?

2:1 pyrophyllite and talc

Dioctahedral, pyrophyllite

Trioctahedral, talc

Negligible isomorphic substitution

Essentially ideal structures

2:1 smectite and saponite

Dioctahedral, smectiteTrioctahedral, saponite

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

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

Fe2O3

FeOOH

Fe3O4

Fe2O3

Problems

7, 10, 11, 12 and 14