EPSC501 Crystal Chemistry

WEEK 5

Oxidation states of transition elements(many more in aqueous solutions than in the

common rock-forming minerals)

Notice that almost every transition metal has a +2 oxidation state. This is because these elements lose 2 valence electrons from their outermost “s” orbital before losing any electrons from the d orbital.

How we fill up the electronic configurations of neutral elements…

Cr (Z=24) is an example of apparent “irregularities” in the pattern of filling orbitals...

Cr (instead of [Ar]3d44s2) is [Ar]3d54s1

... because this preferred configuration is half filling of the five outer “3d” orbitals, and of the “4s” orbital.

For the same reason, Cu (Z=29) is NOT [Ar]3d94s2

because the configuration [Ar]3d104s1 is more stable. All transition elements that readily lose their 2 “s”

electrons before giving up any d electrons can therefore adopt the +2 valence state.

The enhanced stability of half-filled orbitalsexplains why some transition metals have more than one possible oxidation state.

Take Fe (Z=26), rarely found in its native state on Earth (except in meteorites). Iron occurs as ferrous and/or ferric ions in minerals.

Fe: [Ar]3d64s2 Fe2+ : 3d6 (rather than d44s2)Fe3+ : 3d5 (half-filled d subshell)

In presence of O2- (electron acceptor), Fe readily gives up either 2 or 3 electrons.

Oxidation states of transition elements

Other common valence states are explained by the filling of d orbitals that experience energy splitting when a cation is coordinated to anions in certain geometric configurations.

In octahedral coordination, the the five d orbitalsof the central cation interact with those of the surrounding anions.

(a) An octahedral array of negative charges approaching a metal ion. (b-f) The orientations of the dorbitals relative to the negatively charged ligands. Notice that the lobes of the dz2 and dx2-y2 orbitals (b and c) point toward the charges. The lobes of the dxy, dyz, and dxz orbitals (d-f) point between the charges.

These d orbitals of the central cation experience the greatest repulsion from the nearest neighbour anions.

As a result, they have a higher potential energy relative to the other three d orbitals (shown below).

If the cation is surrounded by six anions (usually O2-) in a regular octahedron…

Filling these orbitals is energetically unfavorable…

Filling these orbitals is energetically favorable…

d0 Ca2+,Sc+ 0 0 0 0d1 Ti3+ 1 0.4 1 0.6d2 V3+ 2 0.8 2 1.2d3 Cr3+,V2+ 3 1.2 3 0.8

low-spin state high-spin stated4 Cr2+,Mn3+ 2 1.6 4 0.6 4 0.4d5 Mn2+, Fe3+ 1 2.0 5 0 5 0d6 Fe2+, Co3+ 0 2.4 4 0.4 4 0.6d7 Co2+ 1 1.8 3 0.8 3 1.2d8 Ni2+ 2 1.2 2 1.2d9 Cu2+ 1 0.6 1 0.6d10 Cu+, Zn2+ 0 0 0 0

dn example Octahedral field Tetrahedral------------------------------- ---------------

lone e- CFSE lone e- CFSE

CFSE = crystal field stabilization energy (also called “ligand field stabilization energy”)

Polyhedra of CN=6 but of lower symmetry give different energy splitting of the d orbitals…

4 nearest anions along the x, y axes

2 nearest anions along

the x axis

Why Cu2+ (9 “d” electrons) and Zn2+ (10 “d” electrons) don’t readily substitute for Fe2+…

Despite very similar crystal radii (Cu2+ = 0.77Å, Zn2+=0.74Å) Cu and Zn rarely substitute for Fe2+ (0.78Å) and Mg2+ (0.72Å) in octahedral sites, because their many d electrons must fill the higher-energy “eg” orbitalsrepulsed by the surrounding anions.

Because of their 4 “d”electrons, Cr2+ (0.62Å) and Mn3+ (0.65Å) get a larger CFSE than Mn2+ or Fe3+

(0.65Å) in tetrahedral sites or in distorted octahedral sites.

Why should the effect of Cr on colour be different from mineral to mineral?

In ruby (corundum) Al2O3, Cr3+ substitutes for Al3+ (CN=6)

The Al-O bond length = 1.8389; 1.9953 (why two values?)

- each oxygen is satisfied by 4 Al-O bonds: 4* 3/6 e.v.

- relatively ionic (especially where Al-O is longer)

- crystal-field splitting effect is strong

In beryl Be3Al2(SiO3)6

Cr3+ for Al3+ (also in CN=6), but bond length Al-O= 1.9096

“O2” is satisfied by Be-O 2/4, Al-O 3/6 and Si-O bonds (4/4)

- crystal field splitting is weaker than in corundum

Absorption spectrum for 2 different garnet samplesSolid line: garnet with 5.5 wt% Cr2O3Dotted line: garnet with 17 wt. % Cr2O3

Problem Set #2 (for next week)

Why should Fe2+

... induce a dark red colour in almandine (a garnet)

Fe3Al2(SiO4)3

... but induce a green colour in olivine (Mg, Fe)2SiO4

Answer by describing how the bonding environment of Fe2+

(its CN in each mineral, and the distortion of the polyhedron) and the bonding character (Fe-O bond length(s), what else bonds to O, and the degree of ionic vs. covalent character) differ in these two minerals.

Are the differences you note likely to result in different energy splitting of the d orbitals?

In some minerals, color may be caused instead by

- electron transfer (“electron hopping”) among molecular orbitals

This happens if ions of different charges occupy adjacent sites and share orbitals (with some degree of covalent bonding)

- beryl (aquamarine), an electron jumps from Fe2+ to Fe3+

among adjacent FeO6 octahedra...

- in corundum, Al2O3 (sapphire), electron hopping between Fe2+ and Ti4+ is made possible by edge-sharing Metal-O6 octahedra...

A question from Yumi about what else we should be able to read from the crystallographic data files…

“How can one determine the molar proportions of ions/atoms from a crystallographic data file?”

Fluorite CaF2

Can one determine the molar proportions of ions/atoms from this crystallographic data file?

Wyckoff R W G Crystal Structures 1 (1963) 239-444 Second edition. Interscience Publishers, New York, New YorkFluorite structure

5.46295 5.46295 5.46295 90 90 90 Fm3m

atom x y z

Ca 0 0 0

F .25 .25 .25

Each special position in a space group has a letter label known as a Wyckoff site...

Use http://www.cryst.ehu.es/

Select “WYCKPOS” and choose the appropriate space group

Fluorite CaF2

Wyckoff R W G Crystal Structures 1 (1963) 239-444Second edition. Interscience Publishers, New York, New YorkFluorite structure

5.46295 5.46295 5.46295 90 90 90 Fm3m

atom x y z

Ca 0 0 0

F .25 .25 .25

matches Wyckoff letter a, mutiplicity = 4

matches Wyckoff letter c, multiplicity = 8

Therefore, there are 4 * (CaF2) in the unit cell of this crystal structure.

If the formula is CaF2, Z=4

Tribaudino M, Nestola F, Ohashi H European Journal of Mineralogy 17 (2005) 297-304 High temperature single crystal investigation in a clinopyroxene of composition (Na0.5Ca0.5)(Cr0.5Mg0.5)Si2O69.656 8.833 5.262 90 106.5281 90 C2/catom x y z occ Uiso U(1,1) U(2,2) U(3,3)...NaM2 0 .3015 .25 .5 .0137 .0173 .011 .011 ...CaM2 0 .3015 .25 .5 .0137 .0173 .011 .011 ...CrM1 0 .9065 .25 .5 .0065 .0060 .0077 .0056 ...MgM1 0 .9065 .25 .5 .0065 .0060 .0077 .0056 ...Si .2889 .0924 .2301 .0070 .0060 .0087 .0065 ...O1 .1141 .0818 .1383 .0093 .006 .012 .011...O2 .3602 .2542 .3102 .0108 .014 .011 .008 ...O3 .3520 .0138 .0006 .0082 .007 .007 .008 ...

Cr-diopside M2M1Z2O6

The M2, M1 ions are in similar special positions “0 y .25” (Wyckoff site 4e), for a total of 4*M2, 4*M1 ions per cell.

Each Si is in a general position, Wyckoff site 8f, for a total of 8 Si+4

ions per cell.

Each oxygen ion is in a general position “x y z”, Wyckoff site 8f, for a total of 3*8 = 24 oxygen ions.

Z = 4 formula units/cell

Zircon, ZrSiO4

Robinson K, Gibbs G V, Ribbe P HAmerican Mineralogist 56 (1971) 782-790 The structure of zircon: A comparison with garnet 6.607 6.607 5.982 90 90 90 *I4_1/amd0 -.25 .125atom x y z B(1,1) B(2,2) B(3,3)...Zr 0 .75 .125 .00096 .00096 .0012Si 0 .75 .625 .0014 .0014 .00270 0 .0661 .1953 .0037 .0031 .0029

Note that this is a special setting, with origin at 0 -.25 .125

The corresponding Wyckoff sites are:

Zr on Wyckoff position 4a

Si on Wyckoff position 4b

O on Wyckoff position 16h

Zr4Si4O16 or 4*ZrSiO4

Z = 4 formula units/cell

Types of ionic substitutions

1) Homogeneous (or one-for-one)

2) Coupled (for charge balance)

3) Interstitial

4) Omission

Read the article on zircon chemistry by Finch and Hanchar (2003) posted (Week 4 Class Notes and Readings). Note the various types of substitution mentioned, and how they are accommodated by the crystal structure.

In introductory mineralogy courses, the ionic size and valence of ions are stressed as the factors that predict the degree of compositional variation in minerals.

The “radii within 15%” rule of thumb, as the condition leading to complete solid solution is a very broad generalization.

Electronic configuration (including crystal-field effects) and bonding character (covalency vs. ionic) are often more important than similar radii and valences.

Consider the differences in valences of elements forming all these zircon-group minerals and compounds. Which ones will form solid solutions with each other?

Take the time to examine (with Xtaldraw) the structure of zircon.

Are its coordination polyhedra distorted?

What might be the reason for this?

How does this compare with the structure of garnet?

We will look at these diagrams in some detail next week (week 6).