EPSC501 Crystal Chemistry WEEK 5

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]3d 4 4s 2 ) is [Ar]3d 5 4s 1... 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]3d 9 4s 2 because the configuration [Ar]3d 10 4s 1 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 orbitals explains 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]3d 6 4s 2 Fe 2+ : 3d 6 (rather than d 4 4s 2 ) Fe 3+ : 3d 5 (half-filled d subshell) In presence of O 2- (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 orbitals of 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 dz 2 and dx 2 -y 2 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).

Filling these orbitals is energetically unfavorable If the cation is surrounded by six anions (usually O 2- ) in a regular octahedron Filling these orbitals is energetically favorable

d n example Octahedral field Tetrahedral ------------------------------- --------------- lone e- CFSE lone e- CFSE d 0 Ca 2+,Sc + 0 0 0 0 d 1 Ti 3+ 1 0.4 1 0.6 d 2 V 3+ 2 0.8 2 1.2 d 3 Cr 3+,V 2+ 3 1.2 3 0.8 low-spin state high-spin state d 4 Cr 2+,Mn 3+ 2 1.6 4 0.6 4 0.4 d 5 Mn 2+, Fe 3+ 1 2.0 5 0 5 0 d 6 Fe 2+, Co 3+ 0 2.4 4 0.4 4 0.6 d 7 Co 2+ 1 1.8 3 0.8 3 1.2 d 8 Ni 2+ 2 1.2 2 1.2 d 9 Cu 2+ 1 0.6 1 0.6 CFSE = crystal field d 10 Cu +, Zn 2+ 0 0 0 0 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 Cu 2+ (9 d electrons) and Zn 2+ (10 d electrons) don t readily substitute for Fe 2+ Despite very similar crystal radii (Cu 2+ = 0.77Å, Zn 2+ =0.74Å) Cu and Zn rarely substitute for Fe 2+ (0.78Å) and Mg 2+ (0.72Å) in octahedral sites, because their many d electrons must fill the higher-energy eg orbitals repulsed by the surrounding anions. Because of their 4 d electrons, Cr 2+ (0.62Å) and Mn 3+ (0.65Å) get a larger CFSE than Mn 2+ or Fe 3+ (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) Al 2 O 3, Cr 3+ substitutes for Al 3+ (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 Be 3 Al 2 (SiO 3 ) 6 Cr 3+ for Al 3+ (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 samples Solid line: garnet with 5.5 wt% Cr 2 O 3 Dotted line: garnet with 17 wt. % Cr 2 O 3

Problem Set #2 (for next week) Why should Fe 2+... induce a dark red colour in almandine (a garnet) Fe 3 Al 2 (SiO 4 ) 3... but induce a green colour in olivine (Mg, Fe) 2 SiO 4 Answer by describing how the bonding environment of Fe 2+ (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 Fe 2+ to Fe 3+ among adjacent FeO 6 octahedra... - in corundum, Al 2 O 3 (sapphire), electron hopping between Fe 2+ and Ti 4+ 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 CaF 2 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 CaF 2 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 * (CaF 2 ) in the unit cell of this crystal structure. If the formula is CaF 2, Z=4

Cr-diopside M2M1Z 2 O 6 Z = 4 formula units/cell Tribaudino M, Nestola F, Ohashi H European Journal of Mineralogy 17 (2005) 297-304 High temperature single crystal investigation in a clinopyroxene of composition (Na 0.5 Ca 0.5 )(Cr 0.5 Mg 0.5 )Si 2 O 6 9.656 8.833 5.262 90 106.5281 90 C2/c atom x y z occ The UisoM2, U(1,1) M1 ions U(2,2) are in U(3,3)... similar NaM2 0.3015.25.5 special.0137.0173 positions.011 0 y.25.011 (Wyckoff... CaM2 0.3015.25.5 site.0137 4e),.0173 for a total.011 of 4*M2,.0114*M1... CrM1 0.9065.25.5 ions.0065 per.0060 cell..0077.0056... MgM1 0.9065.25.5 Each.0065 Si.0060 is in a general.0077 position,.0056... Si.2889.0924.2301 Wyckoff.0070.0060 site 8f, for.0087 a total.0065 of 8 Si... +4 O1.1141.0818.1383 ions.0093 per.006 cell..012.011... O2.3602.2542.3102.0108.014.011.008... Each oxygen ion is in a general O3.3520.0138.0006 position.0082 x.007 y z, Wyckoff.007.008 site 8f,... for a total of 3*8 = 24 oxygen ions.

Zircon, ZrSiO 4 Z = 4 formula units/cell Robinson K, Gibbs G V, Ribbe P H American Mineralogist 56 (1971) 782-790 The structure of zircon: A comparison with garnet 6.607 6.607 5.982 90 90 90 *I4_1/amd 0 -.25.125 atom x y z B(1,1) B(2,2) B(3,3)... Zr 0.75.125.00096.00096.0012 Si 0.75.625.0014.0014.0027 0 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 Zr 4 Si 4 O 16 or 4*ZrSiO 4

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 crystalfield 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 zircongroup 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).