0
$\begingroup$

"diamond structure. Every carbon atom is bonded to four other carbon atoms forming a three-dimensional lattice of chair conformations."

SENTENCE FROM A BOOK

What does this mean? Where I am struck is that I knew carbons occupy zinc blende like lattice so, how can this line be true.

$\endgroup$
2
$\begingroup$

Where I am struck is that I knew carbons occupy zinc sulfide like lattice so, how can this line be true.

Zinc sulfide can exist in two different crystalline forms, Zincblende (sphalerite) and wurtzite. Zincblende is face-centered cubic (also known as diamond cubic), each ion is tetracoordinate and has local tetrahedral geometry - just like the carbons in diamond. The wurtzite structure is hexagonal close-packed with interconnected 6-membered rings. You may have been thinking of the wurtzite structure, while it is the zincblende structure that is related to diamond. You can see pictures of the two structures here.

$\endgroup$
  • $\begingroup$ YES I TO KNOW IT IS ZINC BLENDE STRUCTURE . BUT, THE BOOK SAYS "Every carbon atom is bonded to four other carbon atoms forming a three-dimensional lattice of chair conformations" $\endgroup$ – DSinghvi Aug 14 '14 at 9:58
  • 1
    $\begingroup$ If you know the fcc structure, maybe you should be more specific what you cannot understand. $\endgroup$ – Greg Aug 14 '14 at 10:10
  • $\begingroup$ well what is this "chair conformation" related to to diamond structure. I know chair conformation are for hexane . But how does diamond form this structure $\endgroup$ – DSinghvi Sep 6 '14 at 10:40
  • $\begingroup$ Take a look at the link in my answer and look at the picture for the wurtzite structure - do you see the chair arrangement of the carbon atoms? If not, here is another link with a good picture of the chair arrangement link.springer.com/article/10.1007/BF00721087#page-1 When 6 sp3 carbons are placed in a ring, the chair arrangement is a low energy arrangement, no matter whether those carbons are bonded to hydrogen atoms (cyclohexane) or other carbon atoms (diamond). $\endgroup$ – ron Sep 6 '14 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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