Thallium chloride (TICI) crystallizes in a cubic lattice whose edge length is found to be 385 pm. If the density of the solid is found to be ` 7.0 " g cm"^(-3)` , predict the type of lattice to which the crystals of TICI belong .
(Atomic mass of TI = 204, Cl = 35.5)

Explanation

To determine the type of lattice to which thallium chloride (TlCl) belongs, we can use the formula for density in a crystalline solid:

\[ D = \frac{Z \times M}{N_A \times A^3} \]

Where:
- \( D \) = density of the solid (in g/cm³)
- \( Z \) = number of formula units per unit cell
- \( M \) = molar mass of the compound (in g/mol)
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol⁻¹)
- \( A \) = edge length of the unit cell (in cm)

### Step 1: Convert the edge length from picometers to centimeters
Given edge length \( A = 385 \) pm, we convert this to centimeters:
\[ A = 385 \, \text{pm} = 385 \times 10^{-12} \, \text{m} = 385 \times 10^{-10} \, \text{cm} = 3.85 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the molar mass of TlCl
The molar mass \( M \) can be calculated as follows:
\[ M = \text{Atomic mass of Tl} + \text{Atomic mass of Cl} = 204 \, \text{g/mol} + 35.5 \, \text{g/mol} = 239.5 \, \text{g/mol} \]

### Step 3: Substitute the values into the density formula
We know:
- Density \( D = 7.0 \, \text{g/cm}^3 \)
- Molar mass \( M = 239.5 \, \text{g/mol} \)
- Edge length \( A = 3.85 \times 10^{-8} \, \text{cm} \)

Now substituting these values into the density formula:
\[ 7.0 = \frac{Z \times 239.5}{6.022 \times 10^{23} \times (3.85 \times 10^{-8})^3} \]

### Step 4: Calculate \( A^3 \)
Calculating \( A^3 \):
\[ A^3 = (3.85 \times 10^{-8})^3 = 5.73 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Rearranging to find \( Z \)
Rearranging the density formula to solve for \( Z \):
\[ Z = \frac{D \times N_A \times A^3}{M} \]
Substituting the values:
\[ Z = \frac{7.0 \times 6.022 \times 10^{23} \times 5.73 \times 10^{-23}}{239.5} \]

### Step 6: Calculate \( Z \)
Calculating \( Z \):
\[ Z = \frac{7.0 \times 6.022 \times 5.73}{239.5} \approx \frac{7.0 \times 34.38}{239.5} \approx \frac{241.66}{239.5} \approx 1.01 \]

### Step 7: Determine the type of lattice
Since \( Z \approx 1 \), this indicates that there is approximately one formula unit of TlCl per unit cell. This suggests that TlCl crystallizes in a simple cubic lattice.

### Conclusion
Thallium chloride (TlCl) belongs to a simple cubic lattice structure.

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