Question

A metallic element, Mk, with molar mass 40 crystallizes in a face-centered cubic structure, but can transform to hexagonal close-packed and body-centered cubic polymorphs at high temperature. The FCC structure has a density of 1.523 g/cm-3.
​​​​​​1. For the BCC phase, draw the packing of spheres one of the 2D layers (>10 atoms). Indicate with
a * the positions of the spheres in the layer below.

Answer 1

Answer 2

(BCC Structure)

Given:

• Molar mass of Mk (M) = 40 g/mol

• FCC density (ρ) = 1.523 g/cm³

• FCC has 4 atoms per unit cell.

Step 1: Find the FCC Lattice Parameter (a)

Density formula:

$$\rho =\frac{Z\cdot M}{a^3\cdot N_A}$$

For FCC, $Z=4$:

$$1.523=\frac{4\times 40}{a^3\times 6.022\times 10^{23}}$$$$a^3=\frac{160}{1.523\times 6.022\times 10^{23}}=1.742\times 10^{-22}{\text{cm}}^3$$$$a=\sqrt[3]{1.742\times 10^{-22}}=5.59\times 10^{-8}\text{cm}=559\text{pm}$$

Step 2: Find the Atomic Radius (r) in FCC

In FCC:

$$4r=\sqrt{2}a⟹r=\frac{\sqrt{2}\times 559}{4}=197.6\text{pm}$$

Step 3: Find the BCC Lattice Parameter (a')

In BCC:

$$4r=\sqrt{3}{a}^{\mathrm{\prime }}⟹a^{\mathrm{\prime }}=\frac{4\times 197.6}{\sqrt{3}}=456.4\text{pm}$$

Step 4: Density of BCC Phase

BCC has 2 atoms per unit cell ($Z=2$):

$${\rho }^{\mathrm{\prime }}=\frac{2\times 40}{(456.4\times 10^{-10}{)}^3\times 6.022\times 10^{23}}=1.39{\text{g/cm}}^3$$

Final Results:

• BCC Lattice Parameter (a') = 456.4 pm

• BCC Density (ρ') = 1.39 g/cm³

• Atomic Radius (r) = 197.6 pm (same for FCC and BCC