In the previous parts of this series (Part 1 & Part 2) , we presented an overview of what nanoparticles are and what they can do. We emphasized the importance of surface atoms as the only ones that can facilitate catalytic reactions. For our example, we outlined calculations on a nanoparticle of palladium with a radius of 1 nm. We calculated that this nanoparticle contained approximately 285 atoms total, 177 of them surface atoms (62%). Surface atoms are involved in catalytic reactions and therefore are most important. In this article, we shall build nanoparticles and explore close-packing. Close-packed lattices are associated with metals that have higher densities and are harder. Metals that exhibit body-centered unit cells such as the alkali metals are less dense and softer. So packing matters. The holes do not matter too much for metallic solids, but do with ionic solids and, as we will see, with nanoparticles.
The simplest close-packed structures can be made with tangerines, oranges or similar round fruit. Structures also can be made with Styrofoam balls or, as described here, with clear glass marbles. Making structures using PowerPoint is an exceptionally enriching activity and all of the drawings presented here were created in this way.
We are familiar with close packing from displays of oranges, grapefruits or even cannonballs. These stack as shown in the sequence shown in Fig. 1. Notice that the third layer (yellow) is positioned above the center hole of the first layer (green). This is an important characteristic of cubic close packing.
All of the atoms shown in Fig. 1 are surface atoms; there are no internal (subsurface) atoms. The smallest nanoparticle with an internal atom is shown in Fig. 2. The left figure shows just the middle layer consisting of seven atoms. The atom in the center is already internal in 2-dimensions, but we need to add top and bottom cap layers in order to make it internal in 3-dimensions. The other two figures show the top and side-view of the nanoparticle with just one internal atom. This “nanoparticle” has 13 atoms, 12 of which are surface atoms and one subsurface atom. Its diameter would be approximately 6 atomic radii.
Close packing options: cubic close packing and hexagonal close packing
When building a close-packed lattice of atoms, the first two layers are assembled as the blue and green layers in the examples above. We will refer to the blue layer as Layer A and the green layer as Layer B as shown in Fig. 3. The third layer could be positioned directly below the holes in the green layer forming another uniquely positioned layer, Layer C, and denoted as the red layer in Fig. 2 and Fig. 3(a). This arrangement is referred to as ABC packing or cubic close packing (ccp) and is preferred by palladium along with many other late transition metals including Rh, Ir, Ni, Pt, Cu, Ag, and Au. Alternatively, the third layer could be positioned directly below the blue layer — essentially forming another Layer A as shown in Fig. 3(b). This arrangement is referred to as ABAB packing, or hexagonal close packing (hcp). Both types of close packing exhibit the largest theoretical packing efficiency — there is no more efficient way to pack spheres. In both cases, 74% of the space is occupied by atoms and 26% of the space is from the holes between the atoms.
Cubic close packing is equivalent to the face-centered cubic unit cell
The face-centered cubic (fcc) unit cell is one of the three principle cubic unit cells. A typical view of the fcc unit cell is shown in Fig. 4(a). The atoms are shown with smaller diameters than normal so that atoms in the rear are not obstructed from view. In reality, the atoms would be touching across the diagonal of each face so that the diagonal distance across a face equals 4r. Fig. 4(b) is the same fcc lattice as Fig. 4(a) but with the color layer labels used in Figs. 1 – 3. Note how the layers are pitched at an angle from the perspective of Figs. 1 – 3. Fig. 4(c) gives a view from the perspective from one corner to the opposite corner — the red atom is directly below the yellow atom.
While building close-packed models, one cannot help but notice the holes between the spheres. There are two types of holes possible in the close packed lattice types: tetrahedral and octahedral. Tetrahedral holes are smaller and are formed whenever four spheres are arranged together as shown in Figs. 5(a) and (b). The space between these four spheres is a tetrahedral hole. In Figs. 5(c) and (d), six atoms come together to form a much larger octahedral hole. From geometry one can calculate that the radius of a tetrahedral hole is 0.225 that of the radius of the close-packed spheres and the radius of an octahedral hole is 0.414 as large as that of the spheres. In Part 1, we discussed how hydrogen atoms occupy holes within palladium. The atomic radius of hydrogen is too large for the tetrahedral holes, and so hydrogen atoms occupy the octahedral holes. In close-packed arrays, there are exactly as many octahedral holes as close packed spheres and twice as many tetrahedral holes as close-packed spheres. Both types of holes are important for ionic substances because they are occupied by the small ions.
Octahedral holes can be demonstrated by using one’s fingers: To form a model of an octahedral hole, hold the thumb, forefinger and middle finger together as shown in Fig. 6(a). One hand represents a layer of three green spheres as shown in Figs. 5(c) and (d). The other hand does the same to represent the three blue spheres. The two sets of fingers are brought together in a staggered position to represent the arrangement in Fig. 5(c). Lifting one forefinger reveals the octahedral hole.
Tetrahedral holes can also be modeled with one’s fingers. To make a tetrahedral hole, hold the thumb and forefinger together on both hands, representing two atoms per hand. Then position the two sets of fingers together in a staggered arrangement to create a finger model of Fig. 5(a), where two spheres (blue and green) are from one hand and the other two spheres are from the other hand. In this position, one can lift one forefinger to visualize the tetrahedral hole.
Building a nanoparticle model
Building a nanoparticle with an atom count of approximately 285, similar to the spherical nanoparticle discussed in Part 1 (1.0 nm radius) is both fun and informative — and we can check our estimate. We built our model using clear glass marbles, available from a hobby store, and joined together with an adhesive that works with glass — there are many to choose from. The model features the four layers shown in Fig. 7. The green layer contains 24 surface atoms (the ones around the perimeter) and 37 internal, subsurface atoms for a total of 61 atoms.
The circle in Fig. 7 shows how these atoms resemble the cross section of a sphere. The blue and red layers have smaller proportion of internal atoms to surface atoms. The yellow layer is the cap layer so all the atoms are surface atoms. Fig. 8 shows how the layers are assembled together to form the top half of the nanoparticle. There would be three more layers (similar to blue, red and yellow) below the green layer in order to form the entire “sphere”.
Fig. 8(a) shows the first two layers. Note that the blue spheres rest in half of the three-sided depressions created by the green layer. In Fig. 8(b), the third layer is positioned above the holes still visible through the first two layers. This creates an ABC arrangement — cubic close packed, rather than a hexagonal close packed model. Note the upper left atoms in the red layer in Fig. 8(b) is see-through so that the hole in the lower layers is seen to be directly below the red atoms. When the yellow layer is added, it is arranged the same as the green layer as shown by the see-through yellow atom in the Fig. 8(c). As more and more layers are added, they would continue this pattern of ABCABCABC…
The green layer is the largest layer in the model. A complete model of a nanoparticle would have a second set of layers similar to blue, red and yellow.* In total there are 152 surface atoms (54%) and 127 internal atoms for a total of 279 as summarized in the Table. This model features slightly fewer surface atoms than we estimated using our rough calculations based on a sphere.
Fig. 9 shows the individual layers of our actual model along with the assembled model of the top half of a nanoparticle. One may note that our model does not actually resemble a sphere. This is due to the very small number of atoms in this model. Rearranging a few atoms here and there does almost nothing to improve the sphericalness of the model. As the size of the nanoparticle increases, the shape could more closely resemble a sphere.
*The layer below the green layer has to occupy the same positions as the red layer does in cubic close packing – packing follows that pattern green – blue – red – green – blue – red, etc.
**The nanoparticle in its entirety would consist of two B and C layers and a second yellow A layer.
This ends our three-part series on nanoparticle metals. In Part 1 we discussed how chemicals interact with metal surfaces, and in the case of hydrogen, migrate down into the metal. The large surface area per mass of nanoparticles is the defining feature that sets them apart from other forms of the metal. We also discussed the various commercial forms of nanoparticles. In Part 2 we explored nanoparticle calculations involving surface area and particle size and how we can make estimations using simple assumptions. And here in Part 3 we investigated the construction of nanoparticles from the smallest possible groupings of atoms. Construction can be done using PowerPoint and using the “Arrange/Group” function to group and move layers. For maximum ease of visualization, actual models of nanoparticles also can be built with spheres and glue.
- How many tetrahedral and octahedral holes are created by adding the second layer in Fig. 1? And how many of each type of hole are in the three layer structure in Fig. 1? (Answer: 3 tetrahedral and 1 octahedral; one more tetrahedral hole is added.)
- How many of each type of hole are in the three-layer structure in Fig. 2? (Answer: 6 tetrahedral holes and no completely formed octahedral holes, but 6 partially formed octahedral holes — each needs one more atom.)
- How many interior and surface atoms are in Fig. 2? What is the degree of dispersion (fraction of atoms that are surface atoms as defined in Part 2) in Fig. 2? The percent surface atoms? (Answer: 1 interior and 12 exterior; 12/13 = 0.92 or 92%)
- The article stated “In close-packed arrays, there are exactly as many octahedral holes as close packed spheres and twice as many tetrahedral holes as close-packed spheres.” Confirm this by counting atoms, and holes for a fcc unit cell. Hint: Corner atoms are only 1/8 “inside” each unit cell (See Fig. 4), and there are eight of them. Face-centered atoms are only 1/2 inside each unit cell. The octahedral holes for fcc are located at edge centers and at the body-center positions. Tetrahedral holes are located directly below each of the eight corner atoms and along the diagonal line passing through the cube center and connecting opposite cube corners. (Answer: There are the equivalent of four atoms in each unit cell: 8 x 1/8 from the corners and 6 x 1/2 from the face centers. There are four octahedral holes from the 12 edge-centers, each ¼ within the unit cell + the body center. There are 8 tetrahedral holes, one under each of the eight corners.)
- Why does the hcp structure have the same number of holes and degree of dispersion as the ccp (fcc) structure? (Answer: Both hcp and ccp are close-packed with the same maximum efficiency. In fcc, the layering is ABC, and in ccp, it it ABAB — but both forms produce the same number and type of holes.)
- The article listed the transition metals that prefer cubic close packing (ccp). Others transition metals prefer hexagonal close packing (hcp). And still others do not use either close packing system — most of them preferring body-centered cubic packing which is less efficient than ccp and hcp. Use the internet to look up which transition metals prefer each of these types of solid structure.
- Imagine another layer to set on top of the yellow layer in Fig. 8. (a) Would the new layer be equivalent in terms of position to the green, blue or red layer? (b) What is the new count of surface and subsurface atoms and percent of surface atoms? (Answer: Another layer would be equivalent to blue. The new layer would be placed directly above the previous blue layer visible beneath the yellow and red layers and situated on the dimples created by three yellow atoms. This amounts to 15 atoms in the new layer, all surface atoms. By adding this new layer, the 15 peripheral atoms in the yellow layer remain surface atoms, but the 10 internal yellow atoms become sub-surface atoms. All together, there will now be 152 – 10 + 15 = 157 surface atoms and 127 + 10 = 137 subsurface atoms.)
- Geometry enthusiasts: You can confirm the 74% occupancy by atoms for close packed lattices using the fcc lattice as a convenient structure. The volume of any cubic lattice is given by V = e3, where e is the length of each edge of the cube. The fcc unit cell contains the equivalent of 4 atoms. Contact between atoms occurs along the face diagonal so that the diagonal equals 4r, where r is that atomic radius. (Do a search for images of face centered cubic to see this.) Show that √2 e = 4 r (Remember that the diagonal forms a right angle triangle with the edges.) Express the volume of the unit cell in terms of r. Assume each atom occupies a volume of a sphere. Show that the volume of four atoms divided by the volume of the unit cell equals 0.74048 or 74%.
- Repeat this calculation for the body centered unit cell that contains two atoms and the direct contact between atoms occurs along the body diagonal. Sketching a diagram of the inside body diagonal might be helpful. Confirm that this gives √3 e = 4 r. For percent occupancy by atoms, you should get 0.68017 or 68%.
- Geometry enthusiasts: Confirm the ratio of the radius of the octahedral holes to the radius of the atom is 0.414. To do this imagine the fcc lattice, use the relationship that e = 4r/√2 and e = 2ratom + 2rhole. If you want another challenge, determine this ratio for tetrahedral holes to the radius of the atom. (0.225)