The structures available through this page have been created to give you a better view of the 3D spatial arrangement of atoms in the solid cubic lattices examined during lecture. Clicking on the thumbnail pictures below will launch the RasMol program for viewing each structure. Remember, you must select Spacefill from the Display menu in RasMol in order to view the lattice. Rotating each of the structures will give you a better sense of the three dimensionality of these lattices.
Before you begin, please follow these instructions to configure RasMol
as a helper application!
Simple Cubic Unit Cell
The simple cubic unit cell is delineated by eight lattice points, which mark the actual cube.
These are corner atoms, so each one only contributes one eighth of an atom to the unit cell,
thus giving us only one net atom.
Body Centered Cubic Unit Cell
This unit cell uses nine lattice points, eight of which are corner atoms (forming the cube)
and one more in the center of the cube. The corners contribute only one net atom and the
center atom contributes another for a total of two net atoms. Please, notice that the
corner atoms are not touching one another. Instead, the atoms touch through the center atom along
the diagonal of the cube. Since the center atom is in contact with eight other atoms, we say
that it has a coordination number of eight.
Face Centered Cubic Unit Cell
This unit cell uses 14 lattice points, eight of which are corner atoms (forming the cube) with
the other six in the center of each of the faces. Since the face atoms are shared by two cubes,
they only contribute three atoms to the unit cell. These three atoms plus the one from the corners
yield the observed four net atoms in the unit cell. Please, notice that once more, the
corner atoms do not touch one another, touching instead through the diagonal of the face.
Look at the structure carefully and corroborate that we have a coordination number of 12.
This single layer is the precursor of the simple cubic (SC) unit cell. The lattice points are
touching one another but not in the most efficient way. In other words, we are creating a
lot of empty space between the atoms.
In order to generate the simple cubic structure, we need to place subsequent layers in such a
way that they eclipse one another. Try to see the unit cell from this two-layer lattice.
This single layer is the precursor of the body centered cubic (BCC) unit cell. The lattice
points are in the same general form as the ones above for the SC unit cell, with the exception
that the atoms are not touching one another.
In order to generate the BCC unit cell, we need to place a second layer in the indentations of
the first layer, as shown in this figure. This can be seen if you rotate it until it is on its
side. The second layer is presented in blue in order to differentiate it from the first layer.
To complete the BCC unit cell, the third layer is placed in such a way that it eclipses the first
one (pattern is therefore ABA). Can you see the BCC unit cell from the lattice? Rotate it until
you see it clearly.
The remaining cubic unit cell (Face Centered Cubic - FCC) begins with a totally different layer,
one that is packed much more efficiently. This figure shows the basic layer that will be repeated
throughout the entire lattice. Let's think about what we are about to do. We want to place the
second layer in the indentations of the first layer. Notice that the indentations are too close
to one another for us to fill them all. In other words, if we were to place an atom in each
indentation, they would not all fit. Because of this, the second layer will only cover half of
the holes in the first layer.
Notice that having placed the second layer on top of the first still leaves a lot of holes. You
can see all the way through them in this figure. There is no way that you could have
placed another atom (in this same layer) to cover even one of these holes because it simply does
not fit. The question then becomes, where do we place the third layer? There are two options,
both of which are equally efficient. We will show both of them, but will concentrate on only one.
The first option places the third layer such that it eclipses the first one, that is, the ABA
pattern. This leads to a different unit cell: the Hexagonal Unit Cell, and this type of packing
is referred to as hexagonal closest packing (hcp). Although it may be difficult to see, demonstrate
to yourself that it has a coordination number of 12 by picking one atom in the center of a hexagon
and seeing that it touches 12 other atoms.
The other option places the third layer in the indentations of the second layer (the holes that
were still clearly visible in the two-layer (AB) figure). This generates an ABC pattern.
The fourth layer, eclipsing the first, creates the ABCA pattern shown here.
Since I realize that it is difficult to pin point the actual unit cell from the lattice, let's
cut some of the atoms from the end of the lattice in order to facilitate this. This figure
will allow you to see the FCC unit cell. First, rotate the lattice straight up until you align
the blue, green, and pink atoms at the far left. One of the faces of the unit cell should be
clearly visible. Now slowly rotate the lattice diagonally to the right (NE) until you see another
face. This face is part of the same unit cell.
To make it even easier, let's cut most of the atoms out of the lattice and just leave enough to
see the FCC unit cell. Rotate straight up until the two blue atoms are aligned north and south.
These are the only two atoms remaining from the original layer (A). Continue rotating and you
will see that you have only the 14 lattice points that form the unit cell. This figure clearly
shows the atoms from each layer that make up the FCC unit cell.
NaCl Unit Cell
The radii of the ions is: rNa+= 0.95 Å and rCl-= 1.81 Å. Since the sodium
ion is 52.5% the size of the chloride ion, we can predict that the chloride ions will occupy the
lattice points that define the face centered cubic unit cell, with the sodium ions occupying the
octahedral holes. Notice that the octahedral holes are along the edges (for a net of three cations)
and in the center of the cube (one more cation). Thus we have four anions (FCC) and four cations,
which corresponds with the stoichiometry of the salt (1:1).
CsCl Unit Cell
With a radius of 1.60 Å, the cesium ion is 88.4% the size of the chloride ion. Based on this we can
predict that the chloride ions will occupy the lattice points that define a simple cubic unit cell,
while the cesium ions will occupy the cubic hole at the center of the SC unit cell.
Li2Se Unit Cell
The radii of the ions is: rLi+= 0.60 Å and rSe2-= 1.98 Å. Since the lithium
ion is 30.3% of the selenide ion, we can predict that the selenide ions will occupy the lattice points
that define the face centered cubic unit cell, with the lithium ions occupying the tetrahedral holes.
Remember that the tetrahedral holes are found between each corner and the three adjacent faces.
Therefore, there are eight tetrahedral holes inside each cube. Since the stoichiometry of the salt is
2:1, we must fill all of the tetrahedral holes (the FCC structure gives four selenides, thus we
need eight lithiums).
ZnS Unit Cell
The radius of the zinc ion is 35.0% that of the sulfide ion. Based on this, we can predict that the
sulfide ion will occupy the latice points that define the face centered cubic unit cell, with the zinc
ions occupying the tetrahedral holes. Since there are four sulfide ions (FCC), we need only four zinc
ions to satisfy the stoichiometry. Therefore, only half of the eight tetrahedral holes inside of the
unit cell will be filled. Notice that the filled holes form a tetrahedron themselves.