Why does quartz have no cleavage plane and halite does? What do I need to know about the specific mineral so that I can decide whether it has cleavage or not? I need some kind of rule.


1 Answer 1


I'm pretty sure a more rigorous answer deserves to come along, but I can give a simple overview of some of the important factors.

Cleavage planes have to do with bond strength and bond geometry. If there isn't a plane of bonds that can be cut through, then you won't get cleavage. When a mineral is fractured, the fracture "wants" to take the path where the total energy required to break all of the bonds along that path is at a minimum.

In Halite, you can "cleave" a perfect plane through weaker ionic bonds (the geometry works, and the bonds are weak). In quartz, all of the bonds are of the same strength (strong silicate bonds), and the geometry of the mineral is such that you can not find a plane that will cut through all of the bonds, therefore the fractures must "wander" creating the conchoidal structures. In mica, the cleavage planes form between the silicate sheets. The silica bonds are strong, the the sheets are held together by weaker ionic bonds.

  • 1
    $\begingroup$ It could still be mentioned that glass, which has short-range order, but no long-range order also only cleaves unevenly. So actually cleaving is a property all materials have, but depends on weak layers. Compare this to a stack of printer paper that has perfect cleavage between the sheets. $\endgroup$ Jun 6, 2014 at 10:57
  • $\begingroup$ So in minerals that are held together only by covalent bonds the cleavage tends to be nonexistent or weak. I that correct? I also noticed while learning about different mineral species that those with many covalent bonds in the crystal structure have greater hardness values. $\endgroup$
    – Tamás
    Jun 6, 2014 at 11:11
  • 1
    $\begingroup$ Diamond has perfect cleavage, and these are covalent bonds. "Weak" in the context of cleavage has to do with relative bond strength. $\endgroup$
    – equant
    Jun 7, 2014 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.