Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value. Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g/cm^3, and an atomic weight of 192.2 g/mol. Calculate the radius of a vanadium atom, given vanadium has a BCC crystal structure, a density of 5.96 g/cm^3, and an atomic weight of 50.9 g/mol. Description: Iron has a BCC crystal structure, an atomic radius
The Correct Answer and Explanation is:
1. Theoretical Density of Iron (Fe)
Given:
- Structure: BCC (Body-Centered Cubic)
- Atomic radius (r) = 0.124 nm = 1.24 × 10⁻⁸ cm
- Atomic weight (A) = 55.85 g/mol
- Avogadro’s number (Nₐ) = 6.022 × 10²³ atoms/mol
BCC Unit Cell Characteristics:
- Number of atoms/unit cell (n) = 2
- In BCC: a=4r3a = \frac{4r}{\sqrt{3}}
Step 1: Calculate unit cell volume VcV_c: a=4×1.24×10−8 cm3=2.864×10−8 cma = \frac{4 \times 1.24 \times 10^{-8} \text{ cm}}{\sqrt{3}} = 2.864 \times 10^{-8} \text{ cm} Vc=a3=(2.864×10−8)3=2.35×10−23 cm3V_c = a^3 = (2.864 \times 10^{-8})^3 = 2.35 \times 10^{-23} \text{ cm}^3
Step 2: Theoretical Density ρ\rho: ρ=n×AVc×Na=2×55.852.35×10−23×6.022×1023=7.87 g/cm3\rho = \frac{n \times A}{V_c \times N_a} = \frac{2 \times 55.85}{2.35 \times 10^{-23} \times 6.022 \times 10^{23}} = 7.87 \text{ g/cm}^3
✅ Result: Theoretical density of Fe = 7.87 g/cm³, which matches well with the experimental value of 7.87 g/cm³.
2. Radius of Iridium (Ir)
Given:
- Structure: FCC (Face-Centered Cubic)
- Density (ρ) = 22.4 g/cm³
- Atomic weight (A) = 192.2 g/mol
- n = 4 atoms/unit cell (FCC)
Step 1: Volume of the unit cell: Vc=n⋅Aρ⋅Na=4⋅192.222.4⋅6.022×1023=5.714×10−23 cm3V_c = \frac{n \cdot A}{\rho \cdot N_a} = \frac{4 \cdot 192.2}{22.4 \cdot 6.022 \times 10^{23}} = 5.714 \times 10^{-23} \text{ cm}^3
Step 2: Lattice parameter: a=Vc1/3=(5.714×10−23)1/3=3.84×10−8 cma = V_c^{1/3} = (5.714 \times 10^{-23})^{1/3} = 3.84 \times 10^{-8} \text{ cm}
In FCC: a=22ra = 2\sqrt{2}r → r=a22r = \frac{a}{2\sqrt{2}}: r=3.84×10−822=1.36×10−8 cm=0.136 nmr = \frac{3.84 \times 10^{-8}}{2\sqrt{2}} = 1.36 \times 10^{-8} \text{ cm} = 0.136 \text{ nm}
✅ Radius of Ir atom = 0.136 nm
3. Radius of Vanadium (V)
Given:
- Structure: BCC
- ρ = 5.96 g/cm³
- A = 50.9 g/mol
- n = 2 atoms/unit cell
Step 1: Volume of the unit cell: Vc=2⋅50.95.96⋅6.022×1023=2.83×10−23 cm3V_c = \frac{2 \cdot 50.9}{5.96 \cdot 6.022 \times 10^{23}} = 2.83 \times 10^{-23} \text{ cm}^3
Step 2: Lattice parameter a=Vc1/3a = V_c^{1/3}: a=(2.83×10−23)1/3=3.05×10−8 cma = (2.83 \times 10^{-23})^{1/3} = 3.05 \times 10^{-8} \text{ cm}
In BCC: a=4r3a = \frac{4r}{\sqrt{3}} → r=a34r = \frac{a \sqrt{3}}{4}: r=3.05×10−8⋅34=1.32×10−8 cm=0.132 nmr = \frac{3.05 \times 10^{-8} \cdot \sqrt{3}}{4} = 1.32 \times 10^{-8} \text{ cm} = 0.132 \text{ nm}
✅ Radius of V atom = 0.132 nm
Conclusion (Summary Table)
| Element | Structure | Atomic Radius (nm) | Theoretical Density (g/cm³) |
|---|---|---|---|
| Iron (Fe) | BCC | 0.124 | 7.87 |
| Iridium (Ir) | FCC | 0.136 | 22.4 (given) |
| Vanadium (V) | BCC | 0.132 | 5.96 (given) |
These calculations show how atomic structure, radius, and density interrelate in crystalline materials, helping us understand their physical properties.
