328 ELEMENTARY CHEMICAL MICROSCOPY 



Since the crystals usually lie on one of the faces of the octa- 

 hedron, there is apt to result an abnormal development of this 

 face and the diagonally opposite and parallel face; the resulting 

 crystal will thus exhibit an hexagonal outline when seen through 

 the microscope, i.e., viewed from above. Combinations of cube 

 and octahedron may lead to a somewhat similar appearance. 



Not infrequently preparations are obtained in which twinning 

 is very marked, and others in which there is a grouping of crys- 

 tals in threes or fours. Of the twin crystals, one form seems to 

 predominate; it results from the union, in reversed position, of 

 two halves of an octahedron where the dividing plane is parallel 

 to the two opposite faces. 



The size and rate of development of the crystals formed will 

 depend largely upon the concentration of the test drop. In 

 very concentrated solutions, minute crystalline grains or the 

 skeletons of octahedra are produced. In very dilute solutions 

 the crystals appear only after some time. In case the test drop 

 proves to be of the latter sort, heat it gently to cause slight 

 evaporation, or expose to alcohol vapor, see Method VI, page 



3°5- 



Thin crystals are lemon yellow in color, but those which attain 



a considerable thickness are of a decided orange tint. 



The best results are obtained from neutral solutions or those 

 which are very slightly acid with hydrochloric acid. Excess of 

 mineral acids is to be avoided, sulphuric acid in particular. 

 Either evaporate and remove them, or mitigate their action by 

 adding sodium acetate or sodium carbonate. If the latter salt 

 is used, care should be taken to avoid making an alkaline solu- 

 tion and a large excess of the chloroplatinic acid must always 

 be used. 



Ammonium, rubidium, cesium and thallous-thallium also give 

 octahedral crystals with chloroplatinic acid, the composition of 

 the salts being similar to that of the potassium salt. The 

 solubility of these compounds, and consequently the size of 

 the crystals produced, decreases rapidly in the order in which 

 the elements are named. Ammonium will give octahedra of the 

 same size as those of potassium, hence its absence must be* 



