0
$\begingroup$

Since lots of materials with some remarkable properties are some form of carbon structures:

Incredible strength of graphene is often explained by it having a hexagonal atomic lattice. Hardness of a diamond is also attributed to it's atomic structure geometry.

My question is - Aren't there other elements in the periodic table that can form rigid structures, either 2D or 3D? Not just hexagon lattice, but also triangular, or other? And if non-carbon materials with similar rigid structures do exist, why don't they have such outstanding properties?

Being new to this topic, I'd be extremely grateful, if someone could name a discipline or a subject that focuses on atomic structures and their characteristics, or recommend a further reading or research direction.

$\endgroup$
1
  • 1
    $\begingroup$ Cubic boron nitride is similar to diamond. $\endgroup$
    – Jon Custer
    Commented Mar 26, 2022 at 21:48

1 Answer 1

2
$\begingroup$

Carbon is more or less unique among the elements in that it has four bond sites available to link it up with other atoms (including carbon, if desired); in fact, two carbons can share one bond, two bonds, or three bonds, and it's also possible to fit a 3-dimensional, solid network of carbons together where each individual carbon is bound to four nearest-neighbor carbons (yielding diamond).

This furnishes the possibility of an almost endless variety of carbon-based molecules that resemble round globs, loops, kinked strings, lumpy sheets or solid networks, any of which can be bonded to another glob, loop, sheet, etc.

The area of technical specialization that deals with carbon-based molecular structures is called organic chemistry.

$\endgroup$
2
  • $\begingroup$ thanks for your answer, but is there a technical specialization area that deals with designing solid molecular structures, not specifically of carbon? $\endgroup$ Commented Mar 27, 2022 at 23:56
  • $\begingroup$ @eyeballpaul, perhaps materials science. $\endgroup$ Commented Mar 28, 2022 at 4:52

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.