if in diamond,there is a unit cell of carbon atoms as fcc and if carbon atom is sp3 what fraction of void are occupied by c-atom?plzz..give answer in details.

To understand the fraction of voids occupied by carbon atoms in a diamond structure, we first need to delve into the arrangement of atoms in the diamond lattice and the nature of the carbon atoms themselves. Diamond crystallizes in a face-centered cubic (FCC) lattice, but with a unique arrangement that leads to its characteristic properties.

Understanding the Diamond Structure

In a diamond, each carbon atom is tetrahedrally coordinated, meaning that each carbon atom forms four covalent bonds with neighboring carbon atoms. This tetrahedral arrangement is a result of the sp³ hybridization of carbon, where one s orbital and three p orbitals combine to form four equivalent sp³ hybrid orbitals.

Unit Cell Composition

The diamond structure can be visualized as an FCC lattice with two carbon atoms per unit cell. In the FCC arrangement, there are atoms located at each corner of the cube and at the center of each face. However, in diamond, the additional carbon atoms occupy specific tetrahedral sites within the FCC lattice.

Calculating the Volume of the Unit Cell

The volume of the unit cell can be calculated using the formula for the volume of a cube:

• Volume of unit cell (V) = a³, where 'a' is the edge length of the cube.

For diamond, the edge length can be derived from the atomic radius of carbon. The radius of a carbon atom in diamond is approximately 0.77 Å. The relationship between the atomic radius and the edge length in an FCC structure can be expressed as:

• a = 2√2 * r, where r is the atomic radius.

Substituting the radius:

• a = 2√2 * 0.77 Å ≈ 2.18 Å.

Volume of Carbon Atoms in the Unit Cell

Next, we need to calculate the volume occupied by the carbon atoms in the unit cell. The volume of a single carbon atom can be approximated using the formula for the volume of a sphere:

• Volume of a single carbon atom (V_atom) = (4/3)πr³.

Substituting the radius:

• V_atom = (4/3)π(0.77 Å)³ ≈ 1.54 ų.

Since there are two carbon atoms in the unit cell, the total volume occupied by carbon atoms is:

• Total volume of carbon atoms = 2 * V_atom ≈ 2 * 1.54 ų ≈ 3.08 ų.

Calculating the Fraction of Voids Occupied

Now, we can find the fraction of the volume of the unit cell that is occupied by the carbon atoms. This is done by dividing the total volume occupied by the carbon atoms by the volume of the unit cell:

• Fraction occupied = (Total volume of carbon atoms) / (Volume of unit cell).

Substituting the values:

• Volume of unit cell = (2.18 Å)³ ≈ 10.43 ų.
• Fraction occupied = 3.08 ų / 10.43 ų ≈ 0.295.

Conclusion on Voids Occupied

This means that approximately 29.5% of the volume of the unit cell is occupied by carbon atoms. The remaining volume represents the voids or empty spaces within the lattice structure. This efficient packing and the strong covalent bonds between the carbon atoms contribute to the exceptional hardness and other unique properties of diamond.

In summary, in a diamond structure with an FCC arrangement of carbon atoms that are sp³ hybridized, about 29.5% of the voids are occupied by carbon atoms, illustrating the dense and organized nature of this remarkable material.