Tungsten possesses a body-centered cubic structure. If its density is 19.28 g/cm3, what is its radius in pm?

• A. 910 pm
• B. 316 pm
• C. 137 pm
• D. 3.18 pm

Answer: (C)

Verified step by step guidance

1. Identify the type of crystal structure: Tungsten has a body-centered cubic (BCC) structure.
2. Recall the relationship between the edge length (a) of a BCC unit cell and the atomic radius (r): In a BCC structure, the body diagonal is equal to 4 times the atomic radius, and the body diagonal can also be expressed as \( \sqrt{3}a \). Therefore, \( 4r = \sqrt{3}a \).
3. Use the formula for density \( \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \), where \( \rho \) is the density, \( Z \) is the number of atoms per unit cell (2 for BCC), \( M \) is the molar mass of tungsten, \( a \) is the edge length, and \( N_A \) is Avogadro's number.
4. Rearrange the density formula to solve for the edge length \( a \): \( a = \left( \frac{Z \cdot M}{\rho \cdot N_A} \right)^{1/3} \). Substitute the known values: \( Z = 2 \), \( M = 183.84 \text{ g/mol} \), \( \rho = 19.28 \text{ g/cm}^3 \), and \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \).
5. Once you have calculated the edge length \( a \), use the relationship \( 4r = \sqrt{3}a \) to solve for the atomic radius \( r \). Convert the radius from cm to pm by multiplying by \( 10^{10} \).