Using the data given below, find the type of cubic lattice to which the crystal belongs.
`{:(,Fe,V,Pd),("a in pm",286,301,388),(rho" in gm cm"^(-3),7.86,5.96,12.16):}`

Answer

To determine the type of cubic lattice for the given metals (Fe, V, Pd) using their edge lengths and densities, we will use the formula relating density, molar mass, and the number of atoms in the unit cell.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The density (ρ) of a cubic lattice can be expressed as:
\[ \rho = \frac{n \cdot M}{V \cdot N_A} \]
where:
- \( n \) = number of atoms in the unit cell
- \( M \) = molar mass (g/mol)
- \( V \) = volume of the unit cell (in cm³)
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol⁻¹)

The volume \( V \) of the cubic unit cell can be calculated as:
\[ V = a^3 \]
where \( a \) is the edge length in cm.

2. **Convert Edge Lengths from pm to cm**:
- For Fe: \( a = 286 \) pm = \( 286 \times 10^{-10} \) cm
- For V: \( a = 301 \) pm = \( 301 \times 10^{-10} \) cm
- For Pd: \( a = 388 \) pm = \( 388 \times 10^{-10} \) cm

3. **Calculate the Number of Atoms (n) for Each Metal**:
Rearranging the density formula gives:
\[ n = \frac{\rho \cdot V \cdot N_A}{M} \]

- **For Fe**:
- Density \( \rho = 7.86 \) g/cm³
- Molar mass \( M = 56 \) g/mol
\[ V = (286 \times 10^{-10})^3 \text{ cm}^3 \]
\[ n_{Fe} = \frac{7.86 \cdot (286 \times 10^{-10})^3 \cdot 6.022 \times 10^{23}}{56} \]
After calculation, \( n_{Fe} \approx 2 \) (BCC).

- **For V**:
- Density \( \rho = 5.96 \) g/cm³
- Molar mass \( M = 51 \) g/mol
\[ V = (301 \times 10^{-10})^3 \text{ cm}^3 \]
\[ n_{V} = \frac{5.96 \cdot (301 \times 10^{-10})^3 \cdot 6.022 \times 10^{23}}{51} \]
After calculation, \( n_{V} \approx 2 \) (BCC).

- **For Pd**:
- Density \( \rho = 12.16 \) g/cm³
- Molar mass \( M = 106 \) g/mol
\[ V = (388 \times 10^{-10})^3 \text{ cm}^3 \]
\[ n_{Pd} = \frac{12.16 \cdot (388 \times 10^{-10})^3 \cdot 6.022 \times 10^{23}}{106} \]
After calculation, \( n_{Pd} \approx 4 \) (FCC).

4. **Conclusion**:
- Fe (Iron) has \( n = 2 \) → Body-Centered Cubic (BCC)
- V (Vanadium) has \( n = 2 \) → Body-Centered Cubic (BCC)
- Pd (Palladium) has \( n = 4 \) → Face-Centered Cubic (FCC)

### Final Answer:
- Fe: BCC
- V: BCC
- Pd: FCC