Use the data given below to find the type of cubic lattice to which the crystal of iron belongs
a/pm=286 , `rho//g cm^(-3)`=7.86

Explanation

To determine the type of cubic lattice to which the crystal of iron belongs, we can follow these steps:

### Step 1: Understand the given data
We have:
- Edge length (a) = 286 pm (picometers)
- Density (ρ) = 7.86 g/cm³

### Step 2: Convert the edge length to centimeters
Since the density is given in g/cm³, we need to convert the edge length from picometers to centimeters:
1 pm = 10⁻¹² m
1 cm = 10⁻² m
Thus,
\[ a = 286 \text{ pm} = 286 \times 10^{-12} \text{ m} = 286 \times 10^{-10} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of a cubic unit cell is given by:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (286 \times 10^{-10} \text{ cm})^3 \]

### Step 4: Calculate the mass of the unit cell using density
The mass (m) of the unit cell can be calculated using the formula:
\[ \rho = \frac{m}{V} \]
Rearranging gives us:
\[ m = \rho \times V \]
Substituting the values:
\[ m = 7.86 \text{ g/cm}^3 \times (286 \times 10^{-10} \text{ cm})^3 \]

### Step 5: Calculate the number of formula units (Z)
The mass of the unit cell can also be expressed in terms of the number of formula units (Z) and the molar mass (M) of iron:
\[ m = \frac{Z \times M}{N_a} \]
Where:
- M (molar mass of iron) = 56 g/mol
- \( N_a \) (Avogadro's number) = \( 6.022 \times 10^{23} \text{ mol}^{-1} \)

Equating the two expressions for mass:
\[ \rho \times V = \frac{Z \times M}{N_a} \]

### Step 6: Solve for Z
Rearranging gives:
\[ Z = \frac{\rho \times V \times N_a}{M} \]
Substituting the values we calculated:
1. Calculate \( V \)
2. Substitute \( V \) into the equation to find Z.

### Step 7: Determine the type of cubic lattice
- If \( Z = 1 \), the structure is Simple Cubic (SC).
- If \( Z = 2 \), the structure is Body-Centered Cubic (BCC).
- If \( Z = 4 \), the structure is Face-Centered Cubic (FCC).

After performing the calculations, we find that \( Z \) is approximately equal to 2, indicating that iron has a Body-Centered Cubic (BCC) structure.

### Final Answer
Iron belongs to the Body-Centered Cubic (BCC) lattice structure.

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