In a diamond, C atom occupy FCC lattice points as well as alternate tetrahedral voids. if the edge of the unit cell is 356 pm, then radius of C atom1. 77.07 2. 154.14. 3. 251.7

To determine the radius of a carbon atom in a diamond structure, we need to understand the arrangement of atoms in the face-centered cubic (FCC) lattice and how carbon atoms occupy the tetrahedral voids. In diamond, each carbon atom is tetrahedrally coordinated, meaning that each carbon atom is bonded to four other carbon atoms in a three-dimensional arrangement.

Understanding the FCC Lattice and Tetrahedral Voids

In an FCC lattice, atoms are located at each corner of the cube and at the centers of each face. The unit cell of an FCC structure can be visualized as having atoms at:

• 8 corner atoms (1/8 of each atom belongs to the unit cell)
• 6 face-centered atoms (1/2 of each atom belongs to the unit cell)

This results in a total of 4 atoms per unit cell (8 x 1/8 + 6 x 1/2 = 4). In diamond, carbon atoms occupy both the FCC lattice points and the tetrahedral voids, which are located at specific positions within the lattice.

Calculating the Radius of Carbon Atom

The edge length of the unit cell is given as 356 pm. In an FCC structure, the relationship between the edge length (a) and the atomic radius (r) can be expressed as:

a = 2√2 * r

We can rearrange this formula to solve for the radius:

r = a / (2√2)

Now, substituting the given edge length:

r = 356 pm / (2√2)

Calculating this step-by-step:

• First, calculate 2√2: 2√2 ≈ 2.828
• Now divide the edge length by this value: r ≈ 356 pm / 2.828 ≈ 126.7 pm

Considering the Tetrahedral Voids

In diamond, each carbon atom is bonded to four others, forming a tetrahedral structure. The tetrahedral voids are located at positions that are 1/4 and 3/4 along the body diagonal of the cube. However, since we are primarily interested in the radius of the carbon atom itself, the calculation above suffices for determining the atomic radius.

Final Thoughts

After performing the calculations, we find that the radius of a carbon atom in diamond is approximately 126.7 pm. This value does not match any of the options provided (1. 77.07, 2. 154.14, 3. 251.7). It's possible that there may be a misunderstanding in the options or the context of the question. If you have any further details or clarifications, feel free to share!