CdO is cubic with a lattice constant a=4.695 A˚ and a density ρ of 8.15 g/cm3. Determine the crystal structure of CdO.
Simple Cubic
Face Centered Cubic
Body Centered Cubic
Face Centered Cubic
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To determine the crystal structure of CdO, we need to analyze the given information about the lattice constant and density and then compare it with the characteristics of different crystal structures.
1. Simple Cubic (SC):
In a simple cubic structure, each lattice point is occupied by a single atom/ion, and each atom/ion has eight nearest neighbors. The coordination number for a simple cubic lattice is 6. The unit cell consists of only one lattice point.
2. Face Centered Cubic (FCC):
In a face-centered cubic structure, each lattice point is occupied by an atom/ion, and each atom/ion has twelve nearest neighbors. The coordination number for a face-centered cubic lattice is 12. The unit cell contains additional atoms/ions at the face centers in addition to the lattice points.
3. Body Centered Cubic (BCC):
In a body-centered cubic structure, each lattice point is occupied by an atom/ion, and each atom/ion has eight nearest neighbors. The coordination number for a body-centered cubic lattice is 8. The unit cell contains an additional atom/ion at the center of the cube, in addition to the lattice points.
To determine the crystal structure, we can calculate the number of atoms/ions per unit cell using the given lattice constant and density.
Step 1: Calculate the unit cell volume (V):
The unit cell volume can be calculated using the formula:
V = a^3
where 'a' is the lattice constant.
V = (4.695 Å)^3
V ≈ 101.238 Å^3
Step 2: Convert the unit cell volume to cubic centimeters (cm^3):
To convert the unit cell volume from Å^3 to cm^3, we can use the conversion factor:
1 cm^3 = 1 x 10^-24 Å^3
V = 101.238 Å^3 x (1 cm^3 / 1 x 10^-24 Å^3)
V ≈ 101.238 x 10^24 cm^3
V ≈ 1.01238 x 10^26 cm^3
Step 3: Calculate the number of atoms/ions per unit cell (N):
The number of atoms/ions per unit cell can be determined using Avogadro's number (N_A):
N = (ρ x V) / (M x N_A)
where ρ is the density, V is the unit cell volume, M is the molar mass, and N_A is Avogadro's number (6.022 x 10^23 mol^-1).
The molar mass of CdO can be found from the atomic masses of Cd and O:
M(Cd) = 112.41 g/mol
M(O) = 16.00 g/mol
M(CdO) = M(Cd) + M(O)
M(CdO) ≈ 112.41 g/mol + 16.00 g/mol
M(CdO) ≈ 128.41 g/mol
N = (8.15 g/cm^3 x 1.01238 x 10^26 cm^3) / (128.41 g/mol x 6.022 x 10^23 mol^-1)
N ≈ 1.020
Based on the calculated number of atoms/ions per unit cell (N ≈ 1.020), we can determine the crystal structure:
- If N is close to 1, the crystal structure is simple cubic (SC).
- If N is close to 4, the crystal structure is face-centered cubic (FCC).
- If N is close to 2, the crystal structure is body-centered cubic (BCC).
In this case, since N ≈ 1.020 is closest to 1, the crystal structure of CdO is Simple Cubic (SC).