Question

Describe the relative hardnesses of CaAl2Si2O8 (similar to orthoclase), Mg3Si4O10(OH)2, and corundum. Justify your reasoning, and for each case, specify which bond is broken.

Answer 1

The hardness of a mineral is mainly controlled by the bonding between the atoms and slightly by the size of the atoms.

CalAl2SiO8 - Anorthite has an hardness of 6 and an absolute hardness of 72 in Mohs Scale . All oxygen atoms are shared between two SiO4 4- tetrahedron. Consisting of an infinite network of tetrahedra inter-connected via bridging oxygen atoms. Here the Si and O bonds are broken.

Mg3Si4O10( OH)2 has Ferromagnesiun minerals . This is hydrated magnesium silicate more commonly known a Talc .Talc has an hardness of 1 in Mohs Scale and an absloute hardness of 1 .It is composed of a layer of magnesium-oxygen/hydroxyl octahedra and this is sandwiched between two layers of tetrahedral silica. Talc both hydrophobic and inert as the main surfaces, known as basal surfaces, of the elementary sheet contain neither hydroxyl groups nor active ions. It has a planar structure and the silicate sheets are held together by vander waals forces .These are really weak forces. The Si and Si bonds are broken.

Corundum - Al2O3 - has an hardness of 9 and an abosolute hardness of 400 Mohs Scale .  It has an hexagonal (rhombohedral) crystal system which has each aluminum atom surrounded by six oxygen atoms. The oxygen atoms are not bonded at the corners of a regular octahedron.The hardness of corundum is kind of related to the strong and short oxygen-aluminum bonds. It is these that bonds pull the oxygen and aluminum atoms close together and making it hard. Here the Aluminium and oxygen bonds get broken.

Al2O3 + H20 -à 2 Al( OH) 3 which is gibb site which is relatively softer