Question

A metal has BCC structure. Atomic radius \(=173\,\text{pm}\) and molar mass \(M=56\,\text{g mol}^{-1}\). The density of the metal is:

• A. \(7.2\,\text{g cm}^{-3}\)
• B. \(5.6\,\text{g cm}^{-3}\)
• C. \(8.5\,\text{g cm}^{-3}\)
• D. \(3.2\,\text{g cm}^{-3}\)

Hint:

Key relations for cubic lattices:

• Density formula: \( \rho = \dfrac{Z M}{a^3 N_A} \)
• For BCC: \(Z=2\)
• Radius-edge relation: \( a = \dfrac{4r}{\sqrt{3}} \)

Remember: In BCC atoms touch along the body diagonal.

Answer 1

The Correct Option is A

Concept: The density of a crystal lattice is given by \[ \rho = \frac{Z M}{a^3 N_A} \] where

• \(Z\) = Number of atoms per unit cell
• \(M\) = Molar mass
• \(a\) = Edge length of the unit cell
• \(N_A\) = Avogadro's number

For a Body-Centered Cubic (BCC) structure: \[ Z = 2 \] Also, the relation between atomic radius \(r\) and edge length \(a\) in BCC is \[ a = \frac{4r}{\sqrt{3}} \]
Step 1: {Calculate the edge length of the unit cell.} \[ a = \frac{4r}{\sqrt{3}} \] Given \(r = 173\,\text{pm}\) \[ a = \frac{4 \times 173}{\sqrt{3}} \approx 400\,\text{pm} \] Convert to cm: \[ 400\,\text{pm} = 4 \times 10^{-8}\,\text{cm} \]
Step 2: {Substitute values into density formula.} \[ \rho = \frac{Z M}{a^3 N_A} \] \[ \rho = \frac{2 \times 56}{(4 \times 10^{-8})^3 \times 6.022 \times 10^{23}} \]
Step 3: {Simplify the expression.} \[ a^3 = 64 \times 10^{-24} \] \[ \rho \approx \frac{112}{6.022 \times 64 \times 10^{-1}} \] \[ \rho \approx 7.2\,\text{g cm}^{-3} \] Thus, the density of the metal is approximately \(7.2\,\text{g cm}^{-3}\). Hence, the correct option is (A).