August 8 Field Trip

I got the following email from our President (Bruce, not Barack) ...

"Our field trip to the calcite location and Riley’s Canyon was a big success. We found some huge banded calcites that must be worth $300 apiece. We found the red horn coral pits that are well hidden in the trees … it took us a while and at least 6 miles of hiking to find them, so the price of red horn coral just went up! Too bad only Pat and I went on the trip. You couldn't ask for a better day."

Feldspar Follow-Up

Gerry Austin's excellent discussion of the feldspars at our July meeting provoked the question, "Why isn't there a series between pure potassium and pure calcium?" Is it the nature of the lattice, or something else? I asked Bill Wray, mineralogist extraordinaire, and got the following response (published with permission):

" … The answer has to do with differing ionic radii and ionic charge (and related thereto, Si to Al ratio within the minerals). Na(1+) and Ca(2+) have nearly the same ionic radii, 1.00 Angstroms for Ca(2+) and 1.02 Angstroms for Na(1+). Calcium has a higher atomic number (20) than sodium (11), but the divalent ionic state of calcium causes the ionic radius to be smaller [more 'drawn in'] thus compensating for what ordinarily would be a larger radius. Likewise, Si(4+) has an ionic radius of 0.40 Angstroms, smaller but fairly close to the 0.54 Angstroms of Al(3+). Thus, with relatively little crystal lattice distortion, plagioclase can form a complete solid solution series between the Na (albite) and Ca (anorthite) end members. Also, plagioclase, as an anhydrous Na-Ca-Si-Al mineral, crystallizes generally at relatively high temperatures which thermodynamically permits a greater distortion of the crystal lattice on formation, which lattice can (and does) persist down to 'room temperature'.

" In contrast, K(1+) is significantly larger an ion than either Na(1+) or Ca(2+), and has an ionic radius of 1.38 Angstroms. For this reason, even at high temperatures K-feldspar and plagioclase do not form a solid solution series. With any amount of Ca beyond only a small amount, the necessary Al substitution for Si (to maintain charge balance) would so further distort the crystal lattice that the mineral simply will not form, even at high temperatures (and pressures) which would otherwise favor formation. However, at high temperatures of formation a mixed (Na,K) feldspar can form, e.g. disordered monoclinic high-temperature sanidine or high-temperature plutonic alkali feldspar. These feldspars have little or no Ca, and so the further and probably fatal distortion of the crystal lattice due to the Al for Si substitution is avoided.

" But, these mixed (Na,K) feldspars are not true solid-solution minerals at lower temperatures, and are metastable (or unstable) at 'room temperature'. Ordinarily, given slow cooling conditions, the original (Na,K) feldspar mineral will partition through ionic diffusion into coexisting K-feldspar and Na-feldspar phases. K-feldspar crystals (microcline, orthoclase or sanidine) with exsolution blebs or lamellae of Na-feldspar (albite) are termed perthite, or microperthite if the perthitic structure is visible only under magnification. Na-feldspar crystals (albite) with blebs or lamellae of K-feldspar (microcline or orthoclase) are termed antiperthite, or microantiperthite."

The short answer is it's the size and charge of the atoms and how they fit (or don't) into the lattice, that determines whether or not species exist in a potential potassium-calcium series.