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Question about coordination number of face centered cubic.

  1. Oct 31, 2009 #1
    May i know how to count the coordination number of fcc?
    i know coordination number is count by the number in contact.
    i can understand the coordination number of bcc which is 8 , becoz that is 8 number in contact .
    but why fcc is 12 , not 13?
    i counted it as 13 if we take one face atom as 'reference atom'..
    am i count wrongly?
    actually how count the coordination number?
     
  2. jcsd
  3. Oct 31, 2009 #2

    tiny-tim

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    Hi aiklone1314! :wink:

    You count the number of atoms "touching" the reference atom, so that doesn't include the reference atom itself. :smile:

    See http://en.wikipedia.org/wiki/Coordination_number" [Broken] …
     
    Last edited by a moderator: May 4, 2017
  4. Nov 4, 2009 #3
    Both fcc and hcp have a co-ordination number 12. The highest co-ordination number you can achieve in any packing is 12. Try it out with a set of ping-pong balls/marbles.
     
  5. Nov 21, 2009 #4
    First of all, i am assure you that the answer for coordination number of FCC is 12. Here i am to explain to you how am i calculate it. First, now we are dealing with the central atom of any of the face(out of 6, choose 1). The distance from this reference atom(central atom) to the four end of the atom is same, which is a/square root of 2. Thus we have 4 now. Next, consider that the reference atom again, from this atom to the another 4 faces of the cubic( top, bottom, left, right), the distance from the central atom of the 4 faces and our reference atom is a/square root of 2 again! I have calculate it, u may go and check out again in case i wrong. The reason we not consider the opposite face of the reference atom is because the distance from that will be differ, which is a(so that is not in consideration). At this stage, we have 8. Finally, remember we are dealing with 3-D problem, if we attach another cubic( 2 cubic connected side by side now), we can have another 4, making it a total of 12! Remember, from the reference atom, the ends of the new cubic attach is not consider because they are sharing the same node, which means we have already calculate that at the very beginning, which is the 1st 4. We only deal with again from the reference atom to the 4 faces of the new cubic( top, bottom, left and right). Thats it. Hopefully my explanation is correct and can help you all to understand that. Remember, you need imagination of 3-D image when solving problem.
     
  6. Nov 21, 2009 #5
    thanks all the guys above....i got the meaning now!!
     
  7. Jan 6, 2011 #6
    Hy.I got the idea till the point that we wont consider the opposite face. But after that how 8 still remain and then what about next 4 ?
     
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