2

Here 

      oh = octahedral 
      td = tetrahedral
      CN = co-ordination number

My problem is that, that why it is not possible to have a void with CN = 5 ?

My reasoning is:

According to the definition of CN i.e. the number of nearest neighbors of a particle (here a void because we are talking about it).

In the following figure if we treat silver sphere as the void then we say that it is a td void.

But if we place a sphere over this td void then we generate a void whose distance from center of void to the center of all the 5 spheres is equal. And so making the CN = 5

So kindly help me in understanding that why a void with CN = 5 not possible.

enter image description here

Singh
  • 276
  • 1
  • 8

1 Answers1

3

Trigonal bipyramidal complexes are abundant. However, if you want a periodic crystal, you must tesselate volume with translation - the unit cell of the crystal lattice. A five-fold symmetric figure cannot do this, for the angles do not add up without gaps. 2-, 3-, and 6-fold symmetries are allowed, offering 230 unique 3-D space groups overall.

If you allow two kinds of unit cell, you can get approximate periodicity, as in quasicrystals and gas-water clathrates. Both can have 5- or 10-fold symmetry overall and give sharp x-ray or neutron diffraction spots (the IUC definition of a crystal).

Uncle Al
  • 8,542
  • 17
  • 31