Earth and Planetary Materials

Transcription

1 Earth and Planetary Materials Spring 2013 Lecture

2 Example Beryl Be 3 Al 2 (SiO 3 ) 6 Goshenite Aquamarine Emerald Heliodor Red beryl Morganite pure Fe 2+ & Fe 3+ Cr 3+ Fe 3+ Mn 3+ Mn 2+

3 Rules for ionic (atomic) substitution Size and charge matter!

4 Victor M. Goldschmidt Swiss born Norwegian mineralogist and petrologist who laid the foundation of inorganic crystal chemistry and founded modern geochemistry

5 Goldschmidt s Rules Size Atomic substitution is controlled by size (radii) of the ion If size difference is < 15%: free substitution 15 ~30%: limited substitution > 30%: little to no substitution If there is a small difference of ionic radius, the smaller ion enters, the crystal preferentially

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7 Goldschmidt s Rules Charge Atomic substitution is controlled by charge of the ions If charge difference is: = 1: substitute readily if electrical neutrality is maintained >1 : substitution is minimal For ions of similar radius but different charges, the ion with the higher charge enters the crystal preferentially Forms a stronger bond with the anions surrounding the site

8 Ringwood s Modifications (1955) Substitutions may be limited, even when the size and charge criteria are satisfied, when the competing ions have different electronegativities and form bonds of different ionic character Explains discrepancies i with respect to the Goldschmidt rules Example: Na + and Cu + have the same radius and charge, but do not substitute for one another

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10 Other factors affecting substitution Temperature Minerals expand at higher T Greater tolerance for ionic substitution at higher T Pressure Increasing pressure causes compression Less tolerance for ionic substitution at higher P Availability of ions ions must be readily available for substitution to occur

11 Types of substitution Simple substitution Interstitial substitution Vacancy substitution

12 (1) Simple substitution Involves the same sites in the crystal structure Type 1: Complete binary solid solution: ions have same charge and nearly same size, substitute freely Example: Fe 2+ Mg 2+ Ni 2+ (0.86 Å) (0.80 Å) (0.77 Å)

13 Example Olivine group End members are forsterite (Mg 2 SiO 4 ; Fo ) and fayalite (Fe 2 SiO 4 ; Fa ) Any composition in between is possible: (Mg x Fe 1 x ) 2 SiO 4 x =0.7 would be 70 mole percent forsterite ; or Fo 70

14 (1) Simple substitution Type 2: Coupled substitution: ions have nearly same size, but different charge; balance charge by simultaneous substitution on a second site.

15 Example Si 4+ and Al 3+ (CN = 4): balanced by also substituting Ca 2+ for Na + on a different site write as a charge balanced reaction: Si 4+ + Na + = Al 3+ + Ca 2+ Example Plagioclase feldspars: Albite (NaAlSi 3 O 8 ) Anorthite (CaAl 2 Si 2 O 8 )

16 Giving isomorphic derivative structures

17 (2) Interstitial substitution Coupled substitution where the charge balancing substitution is accommodated on a site normally vacant in the mineral Many minerals have large channels or space between layers that can accommodate cations

18 Example Beryl Be 3 Al 2 (SiO 3 ) 6 Channels in beryl accommodating charged ions

19 (3) Vacancy (Omission) substitution Charge is balanced by leaving sites vacant Example: pyrrhotite: Fe (1 x) S contains both Fe 2+ and Fe 3+ For charge balance 3(Fe 2+ ) = 2(Fe 3+ ) + where represents and Fe site that is vacant. Another way to represent the formula is: (Fe 2+ ) 1 3x (Fe 3+ ) 2x x x S

20 Point defects

21 Point defects under microscope

22 Summary on substitution

23 Summary on substitution Simple substitution Complete binary substitution Coupled substitution Interstitial substitution Vacancy substitution Varied composition

24 Chemical composition

25 Isomorphs and derivative structures Isomorphs: minerals that have same crystal structure but different chemical composition Example: berlinite is AlPO 4 with the quartz structure (substitute t Si 4+ + Si 4+ = Al 3+ + P 5+ )

26 Isomorphs Examples KMg 3 (AlSi 3 O 10 )(OH) 2 phlogopite K(Li,Al) 2 3 (AlSi 3O 10)( )(OH)) 2 lepidolite KAl 2 (AlSi 3 O 10 )(OH) 2 muscovite Amphiboles: Ca 2 Mg 5 Si 8 O 22 (OH) 2 tremolite Ca 2 (Mg,Fe) 5 Si 8 O 22 (OH) 2 actinolite Actinolite series minerals (K,Na) 0 1 (Ca,Na,Fe,Mg) 2 (Mg,Fe,Al) 5 (Si,Al) 8 O 22 (OH) 2 Hornblende

27 From chemical composition to formula Or vice versa

28 Reporting a mineral/rock composition Element weight %: e.g. 5 g Fe in 100 g sample Oxide weight %: Majority of minerals contain large quantities of oxygen, e.g. silicates, oxides, carbonates Number of atoms mineral chemical formula Relating all elements to one: normalization

29 From element wt% to formula Example A wet chemical analysis: Na = 32.8 % Al = 12.8 % F = 54.4 % What is the unknown mineral? wt Atomic weight wt / atomic weight mole normalization mole ratio Na / / ~3 Al / / ~1 F / / ~ 6 Na 3 AlF 6

30 Report as oxides Mineral composition reported as oxides, e.g. Al as Al 2 O 3 Fe 2+ and Fe 3+ often reported together as FeO (i.e., all Fe 2+ ) Conventional order SiO 2, Al 2 O 3, Fe 2 O 3, TiO 2, Cr 2 O 3, FeO MnO, MgO, CaO, BaO, Na 2 O, K 2 O, F, Cl

31 Between oxide and element wt% Example Mg MgO Molecular weight: Mg = 24.3 MgO = = wt % Mg?? wt% MgO Step 1: Assuming 100 g of sample, mole of Mg: 4.55 / 24.3 = Step 2: weight of MnO = = 7.54 g in 100g sample wt% = 7.54%

32 From oxide wt% to formula Example Again, assuming 100 g of sample wt% mw m.w. Moles Moles Moles oxide cation anion SiO / MgO / Atomic ratio Mg : Si : O = : : = 2 : 1: 4 Mg 2 SiO 4

33 Step 1 Moles of oxide oxide A B C weight percent Molecular weight A / B (g oxide / 100 g (g oxide / mol oxide) (mol oxide / 100 g mineral) mineral) SiO Al 2 O CaO Na 2 O K 2 O Total 99.7

34 Step 2 Normalize Oxygens C D E = Number of oxygens per oxide A / B Oxygen C * D 100 g mineral (mol s / oxide (Oxygens / We know one formula unit of oxide / 100 g 100 g mineral) mineral contains 8 oxygens (for a SiO 2 mineral) feldspar) So we renormalize column E to Al 2 O eight total oxygens, giving us one CaO formula unit of mineral. Na 2 O Multiply each value by the ratio: K 2 O oxygens per formula unit Total / oxygens per 100 g mineral

35 Step 2 Normalize Oxygens oxide C D E F A / B (mol oxide / 100 g mineral) Oxygens / oxide C * D (Oxygens / 100 g mineral) E * (8/2.908) (Oxygens / formula unit) SiO Al 2 O CaO Na 2 O K 2 O Total

36 Step 3 Calculate Cations oxide Ox / oxide D F G H E * (8/2.908) (Oxygens / formula unit) Cations / oxide (F/D)*G (Cations / formula unit) SiO Al 2 O CaO Na 2 O K 2 O Total

37 Step 4 Allocate cations to sites K Na Ca Al Si O 8 H oxide (F/D)*G (Cations / formula unit) SiO Al 2 O CaO Na 2 O K 2 O 0.056