Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 Å.

Answer 1

Hi Precious W.!

I cannot be certain what question you're asking, but most often unit cell questions call for finding the radius of the atom and the density of the atom.

Part One: A Brief Introduction to Crystallography

In crystallography, the unit cell must meet the following criteria:

1. The unit cell is the simplest repeating unit in the crystal structure
2. The opposite faces of a unit cell are parallel to one another
3. The edge of the unit cell connects equivalent points within the structure

I am aware of fourteen unit cell types:

1. Cubic
2. Tetragonal
3. Monoclinic
4. Orthorhombic
5. Rhombohedral
6. Hexagonal
7. Triclinic

However, the unit cells with which you should have concern in general chemistry are all of the type cubic. Including:

1. Simple
2. Body-Centered (BCC)
3. Face-Centered (FCC)

Body-centered cubic unit cells are those in which the atoms are present at the corners and body center of the structure. This information will be crucial later.

Part Two: Finding the Radius of the W Atom

This next bit is going to be a bit tricky without a visual aid to assist. I'll do my best to describe how this looks, but you would do well to search for a diagram to follow along.

In the BCC structure, Tungsten (W) atoms are in contact with one another across the diagonal of the cubic shaped unit cell. It is often by convention to label the edge length as a or x. Here, we will use the convention a. Let's consider the bottom face of the cube. By use of the pythagorean theorem, we can show that the diagonal has a length of a√2. Keep in mind that for the moment, we are considering the bottom face only. So, we may think of it temporarily as a square. If all sides of the square have an edge length equal to a, then the length of the diagonal connected one corner to its opposite is:

a² + a² = c² => c = √a² + a² where c is the length of the diagonal

c = √a² + a² = √2a²

c = a√2

Suppose we connect the front right corner on the bottom of the cube with the back left corner on the bottom of the cube. We have just shown this length as equal to a√2. But we need to find a right triangle through the entire cubic structure. This is necessary to find the radius of the Tungsten atom. We will now connect this back bottom corner to the right front top corner. This right front top corner is already connected to the bottom right front corner (where we started originally) by the edge length a. We have now formed a right triangle. The hypotenuse (the length from the back bottom left corner to the front top right corner) has a length equal to four W radii, 4r. Again, this is where a diagram is useful.

According to the pythagorean theorem:

a² + b² = c²

Where:

1. The diagonal along the bottom face (which we could call the adjacent side) is represented by a
2. The edge (which we could call the opposite side) is represented by b
3. The hypotenuse is the diagonal cutting through one side of the cube to the other is represented by c

a² + b² = c²

=> a = a√2

=> b = a (edge length, not the letter a above)

=> c = 4r

(a√2)² + a² = (4r)²

2a² + a² = 16r²

r = √3a²/16

r = (a√3)/4

Now, let's recall that we were given the edge length in the information:

a = 3.165 A

r = (a√3)/4

=> a = 3.165 A

r = (3.165√3)/4

r = 1.37 A

Step Three: Find the Density of the W BCC

By ascertaining the density of its unit cell, we can determine the density of the Tungsten atom itself.

Density is measure of concentration found as the mass divided by the volume. Moreover, the density of the unit cell is found as the mass contained within the unit cell divided by the volume within it.

Step 3a: Find the Number of Atoms per Unit Cell

In any BCC unit cell structure, one atom is present at the center of the structure and one-eight of an atom is found at each of the eight corners of the cubic structure. The total number of atoms present in a BCC unit cell is:

(8 corners)(1/8 atom/corner) = 1 atom

1 atom present at the center

2 atoms total per BCC unit cell

Step 3b: Find the Mass Contained Within the Unit Cell

We will use the mass of Tungsten found from the periodic table and Avogadro's number through dimensional analysis to find the mass:

(1 W unit cell)*(2 W atoms/unit cell)*(1 mol W/6.022x10²³ atoms W)*(183.84g W/1 mol W)

mass contained in one W unit cell = 6.1056x10^-22 g W

Step 3c: Find the Volume of Tungsten Unit Cell

Volume is simply the length x width x height

In our case, we have a cube whose edges all have a length equal to 3.165 A

Because density of unit cells is often given as grams per cubic centimeter, we need to convert our edge length from Angstroms into centimeters:

(3.165A)(1cm/10⁸A)

a = 3.165x10^-8 cm

If this conversion was a bit tricky for you, an alternative could look like this:

(3.165A)*(1m/10¹⁰A)*(1cm/10^-2m)

This used meters as an intermediary and yields the same answer:

a = 3.165x10^-8 cm

Let's plug this into our equation for volume:

V = l*w*h = a³

<=> (3.165x10^-8 cm)³

V = 3.1705x10^-23 cm³

Step 3d: Calculate the Density

Density of W = mass of W contained within the unit cell/volume of the unit cell

<=> 6.1056x10^-22 g W/3.1705x10^-23 cm³

Density of W = 19.26 g*cm^-3

Hope this helps! Please feel free to follow-up with any questions!

Cheers