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CHEMISTRY IS A MIRACLE ONE
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Senin, 08 Maret 2010

KRISTAL Cl YANG MENGELILINGI Na PADA NaCl, BERAPA????

Sodium chloride also crystallizes in a cubic lattice, but with a different unit cell.


However, the tightly-packed structures make it difficult to view the interior relationships.
The same structure, but with the ions moved further apart allows the interior to be viewed.
If we take the NaCl unit cell and remove all the red Cl ions, we are left with only the blue Na. If we compare this with the fcc / ccp unit cell, it is clear that they are identical. Thus, the Na is in a fcc sublattice.


Since the repetition patterns of Na and Cl are the same in the lattice, the Cl sublattice must also be fcc / ccp. Recall that there may be more than one way to divide a lattice into unit cells. Although it may not be obvious, the red Cl's represent a different fcc / ccp unit cell.
Notice that there are 6 Cl surrounding the Na, and 6 Na around each Cl. We can look at the NaCl as made up of fcc / ccp lattices interpenetrating. The Na's occupy the octahedral sites in the Cl sublattice, and the Cl's occupy the octahedral sites in the Na sublattice.

Sabtu, 06 Maret 2010

Simple Cubic Unit Cell





The simple cubic unit cell is a cube (all sides of the same length and all face perpendicular to each other) with an atom at each corner of the unit cell.

The unit cell completely describes the structure of the solid, which can be regarded as an almost endless repetition of the unit cell.

The volume of the unit cell is readily calculated from its shape and dimensions. This calculation is particularly easy for a unit cell that is cubic. In this example, the atoms are in contact with each other along the edges of the unit cell. Thus the side of the unit cell has a length of 2 r, where r is the radius of an atom.

Atoms, of course, do not have well-defined bounds, and the radius of an atom is somewhat ambiguous. In the context of crystal structures, the diameter (2 r) of an atom can be defined as the center-to-center distance between two atoms packed as tightly together as possible. This provides an effective radius for the atom and is sometime called the atomic radius.

A more challenging task is to determine the number of atoms that lie in the unit cell. As described above, an atom is centered on each corner. In this case, however, none of these atoms lies completely within the cell. Part of each atom lies within the unit cell and the remainder lies outside the unit cell. In determining the number of atoms inside the unit cell, one must count only that portion of an atom that actually lies within the unit cell.

The density of a solid is the mass of all the atoms in the unit cell divided by the volume of the unit cell.

Packing Efficiency of Face Centered Cubic (Efisiensi Kemas Kubus Pusat Muka)


Packing Efficiency of Body Centered Cubic (Efisiensi Kemas Kubus Pusat Badan)


Packing Efficiency of Simple Cubic (Efisiensi Kemas Kubus Sederhana)

Body Centred Cubic Structure
At room temperatures, elements Li, Na, K, Rb, Ba, V, Cr and Fe have structures that can be described as body centre cubic (bcc) packing of spheres. The other two common ones are face centred cubic (fcc) and hexagonal closest (hcp) packing. This type of structure is shown by the diagram below.

In a crystal structure, the arrangement extends over millions and millions of atoms, and the above diagram shows the unit cell, the smallest unit that, when repeatedly stacked together, will generate the entire structure.
Actually, the unit we draw is more than a unit cell. We use the centre of the atoms (or spheres) to represent the corners of the unit cell, and each of these atoms are shared by 8 unit cells. There is a whole atom located in the centre of the unit cell.
Usually, the length of the cell edge is represented by a. The direction from a corner of a cube to the farthest corner is called body diagonal (bd). The face diagonal (fd) is a line drawn from one vertex to the opposite corner of the same face. If the edge is a, then we have:
fd2 = a2 + a2 = 2 a2
bd2 = fd2 + a2
= a2 + a2 + a2
= 3 a2



Atoms along the body diagonal (bd) touch each other. Thus, the body diagonal has a length that is four times the radius of the atom, R.
bd = 4 R
The relationship between a and R can be worked out by the Pythagorean theorem:
(4 R)2 = 3 a2
Thus,
4 R = sqrt(3) a
or
a = 4/sqrt(3) R

Recognizing these relationships enable you to calculate parameters for this type of crystal. For example, one of the parameter is the packing fraction, the fraction of volume occupied by the spheres in the structure.


DIAGRAM TINGKAT ENERGI HCL

DIAGRAM TINGKAT ENERGI KF

DIAGRAM TINGKAT ENERGI C2

DIAGRAM TINGKAT ENERGI F2

DIAGRAM TINGKAT ENERGI Be2