The structure of diamond follows, with each sphere...

The structure of diamond follows, with each sphere representing a carbon atom. (a) How many carbon atoms are there per unit cell in the diamond structure? (b) Verify, by extending the drawing if necessary, that each carbon atom has four nearest neighbors. What is the arrangement of these nearest neighbors? (c) What is the distance (center to center) from any carbon atom to its nearest neighbor, expressed in terms of $a$, the unit cell edge? (d) The observed unit cell edge length in diamond is $3.567 \AA$. What is the $C-C$ single bond length in diamond? (e) Calculate the density of diamond.

The structure of diamond follows, with each sphere representing a carbon atom. (a) How many carbon atoms are there per unit cell in the diamond structure? (b) Verify, by extending the drawing if necessary, that each carbon atom has four nearest neighbors. What is the arrangement of these nearest neighbors? (c) What is the distance (center to center) from any carbon atom to its nearest neighbor, expressed in terms of $a$, the unit cell edge?
(d) The observed unit cell edge length in diamond is $3.567 \AA$. What is the $C-C$ single bond length in diamond? (e) Calculate the density of diamond.

Key Concepts

Crystal Density Calculation

The density of a crystalline solid is calculated by dividing the mass of atoms in the unit cell by the volume of the unit cell. This calculation incorporates the number of atoms per unit cell, the atomic mass (with the aid of Avogadro's number), and the unit cell’s dimensions. This process connects the atomic-scale structure to the bulk material property of density.

Bond Length Determination

Determining bond length in a crystal involves relating the interatomic distances to the dimensions of the unit cell. For the diamond structure, the shortest distance between the centers of two neighboring carbon atoms is expressed in terms of the unit cell edge length. This geometrical relationship is key to understanding how the microscopic arrangement of atoms defines the macroscopic bond lengths observed in the material.

Unit Cell

A unit cell is the smallest, repeating segment of a crystal lattice that encapsulates the full symmetry and structure of the material. In the context of the diamond structure, careful counting of the carbon atoms, including those at corners, faces, edges, and inside the cell, provides the number of atoms per unit cell. This count is fundamental for calculating various properties, including density and bonding relationships.

Diamond Cubic Structure

The diamond cubic structure is a type of crystal lattice where a face?centered cubic (FCC) lattice is adorned with a basis of two atoms, resulting in a unique arrangement. In this structure, each atom is tetrahedrally coordinated, meaning it is bonded to four other atoms arranged at the vertices of a tetrahedron. This configuration is responsible for many of diamond's remarkable physical properties, such as its hardness and high thermal conductivity.

Coordination Number

The coordination number refers to the number of nearest-neighbor atoms surrounding an atom within a crystal lattice. In diamond, every carbon atom is surrounded by four nearest neighbors in a tetrahedral geometry. This tetrahedral coordination is essential for understanding the spatial arrangement and bonding nature in diamond, influencing its overall structural and mechanical characteristics.

Transcript

00:01 So in this question in diamond, the baseless is a face -centric cubic unit cell, and it contains c atoms or carbon atoms, and half of the tetrahedral sites are filled with more carbon atoms.

00:15 So normally, in a face -centre cubic cell, we have four atoms.

00:21 However, in diamond, there are empty sites that are filled by carbon, and the unit cell ends up having eight total carbon atoms.

00:32 In it.

00:37 Carbon is tetrahedrally bonded to neighboring carbons in the unit cell.

00:42 That means that they each have four neighbors.

00:46 The arrangement is tetrahedral.

00:47 So carbon is bonded like this in the unit cell.

00:53 We need to solve for the distance between two carbons.

00:57 So here i've drawn part of the unit cell.

01:01 So this is one atom of carbon and this is the bond to the next carbon.

01:08 And there would be more carbons that are arranged tetrategially and it would expand because they're a covalent network they would just repeat like that but to simplify let's just look at this one bond because all of the bonds are the same length due to it being tetrahedral so if we close up on this this is what we will see so the edge length is a and this length is fourth of that and then we have two triangles being formed here we and and 2r is equal to the distance between the two carbon atoms.

01:49 So we need to find the base of this top triangle using the bottom triangle in order to find 2r in terms of a.

01:58 So let's solve for x.

02:01 So we're going to use the pythagrin theorem because this is the right triangle.

02:06 So first let's look at the bottom triangle...

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