Problem 90 - Chemistry The Central Science In Si Units

Question

The diameter of a rubidium atom is \(495 \mathrm{pm}\) We will consider two different ways of placing the atoms on a surface. In arrangement \(\mathrm{A},\) all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the "depressions" formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side? (b) How many \(\mathrm{Rb}\) atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side, using arrangement B? (c) By what factor has the number of atoms on the surface increased in going to arrangement \(\mathrm{B}\) from arrangement A? If extended to three dimensions, which arrangement would lead to a greater density for Rb metal?

Step-by-Step Solution

Answer

Arrangement B fits about 1.16 times more atoms than Arrangement A, and it leads to greater density in 3D.

Step 1: Convert Diameter to Radius

First, convert the diameter of a rubidium atom into its radius because the radius is used in calculation. The given diameter is \(495 \, \text{pm}\), so the radius \(r\) is \(\frac{495}{2} = 247.5 \, \text{pm}\).

Step 2: Convert Radius from Picometers to Centimeters

To align the unit with the given surface, convert the radius from picometers to centimeters. Recall that \(1\, \text{pm} = 1 \times 10^{-12}\, \text{m} = 1 \times 10^{-10}\, \text{cm}\). Thus, \(r = 247.5 \times 10^{-10} \, \text{cm}\).

Step 3: Calculate the Number of Atoms in Arrangement A

In arrangement A, atoms are placed in a square grid. Thus, the side length of the square each atom occupies is \(2r = 495 \times 10^{-10} \, \text{cm}\). The number of atoms along one edge of the square is \(\frac{1.0 \text{cm}}{495 \times 10^{-10} \text{cm}}\). Since the grid is two-dimensional, the total number of atoms is this value squared.

Step 4: Calculate the Number of Atoms Along One Edge for Arrangement A

Calculate the number of atoms along one edge: \[\frac{1.0}{495 \times 10^{-10}} = 2.02 \times 10^7\]

Step 5: Calculate Total Atoms for Arrangement A

Square the result from Step 4 to get the total number of atoms: \[(2.02 \times 10^7)^2 = 4.08 \times 10^{14}\] So, about \(4.08 \times 10^{14}\) atoms fit on the surface in arrangement A.

Step 6: Calculate Atoms in Close-Packed Arrangement B

In arrangement B, atoms occupy a close-packed hexagonal arrangement. The average area occupied per atom decreases by a factor of \(\frac{\sqrt{3}}{2}\). Multiply the number of atoms from arrangement A by this factor to find the result for arrangement B.

Step 7: Calculate Factor Increase for Arrangement B

Compute the close-packed arrangement: \[4.08 \times 10^{14} \times \frac{2}{\sqrt{3}} = 4.72 \times 10^{14}\]

Step 8: Compare Arrangement B to Arrangement A

Divide the result for Arrangement B by Arrangement A to determine the increase factor: \[\frac{4.72 \times 10^{14}}{4.08 \times 10^{14}} = 1.16\] Thus, Arrangement B fits about 1.16 times more atoms than Arrangement A.

Step 9: Analyze for 3D Packing Density

In 3D, arrangement B involves face-centered cubic (FCC) packing, which is denser than body-centered cubic (BCC) or simple cubic packing. Therefore, 3D close-packed arrangement B leads to greater density for rubidium metal.

Key Concepts

Square grid arrangement

When considering atomic placement on a surface, a square grid arrangement involves lining up atoms in neat, perpendicular rows and columns. This pattern resembles a checkerboard where each atom sits at each corner of squares of identical size. To understand this arrangement, picture atoms as circles that create a square when put side by side.

In these arrangements, the atoms' centers form a perfect square, and each side of this square is equal to twice the atomic radius. For rubidium with an atomic diameter of 495 pm, each atom effectively occupies a space of 495 pm x 495 pm. When translated to cm (converting picometers to centimeters using the conversion factor of 1 pm = 1x10^-10 cm), this occupancy length turns out to be crucial for calculations.

This setup is straightforward but not the most space-efficient because there's empty space between atoms, reducing how many can fit on a surface. In our exercise, the number of atoms placed on a 1 cm by 1 cm surface following this arrangement is around 4.08 x 10^14 atoms, highlighting the setup’s inefficiency.

Close-packed arrangement

The close-packed arrangement maximizes the number of atoms placed on a surface by minimizing wasted space. Unlike the square grid, here each row of atoms finds itself in the 'depressions' or interstitial spaces created by the row preceding it. This forms a tighter configuration, sometimes called a hexagonal grid, where atoms snugly fill the voids left by their neighbors.

Visualize this by imagining a set of marbles squished together. Each subsequent row of marbles nestles perfectly between the valleys created by the overlapping marbles of the previous line, effectively using the gaps left in the square arrangement. This method reduces the area occupied per atom because it allows extra atoms to be fitted into the existing spaces, thereby using the available surface more effectively.

In the context of our exercise, this close-packed method allows for more atoms than the square arrangement. Specifically, using this type of packing, the atom count increases by a factor of approximately 1.16, or 4.72 x 10^14 atoms on the same 1 cm surface. This increase demonstrates how efficiently space is utilized in close-packed configurations.

Face-centered cubic packing

Taking the close-packed concept into three dimensions gives us face-centered cubic (FCC) packing. In the three-dimensional world of atoms inside solids, FCC is one of the most efficient ways to arrange atoms, maximizing density and stability.

The FCC structure involves atoms situated at each face and corner of a cube. This means in every small cube unit, atoms touch each other along the face diagonals. Each corner atom is shared among eight surrounding cubes, and each face atom is shared between two, forming an extremely tight and dense configuration.

For materials like rubidium, such an arrangement has significant implications. Compared to other cubic arrangements like body-centered cubic, FCC packing fills more space with atoms, enhancing the material's overall density. In our exercise, extending close-packed arrangements to 3D with FCC transitions from a simple surface examination to understanding the natural compactness and which leads to the denser, more robust structure of rubidium as a metal.