Question

A metal X of atomic mass 75 u forms a cubic lattice of edge length 5 Å. If the density of the lattice is 2 g cm$^{-3}$, the radius (in Å) of the metal atom is ($N=6\times10^{23} \text{ mol}^{-1}$)

• A. 1.083
• B. 4.330
• C. 2.165
• D. 6.495

Hint:

First compute $Z$ to determine the cubic class: $Z=1$ (SCC), $Z=2$ (BCC), $Z=4$ (FCC). Then apply the correct geometric radius equation.

Answer 1

The Correct Option is C

Step 1: Concept
This problem requires relating the unit cell dimension and bulk density of a crystal to identify its lattice type ($Z$), which subsequently allows calculating the atomic radius using the appropriate geometric formula.

Step 2: Meaning
The density ($\rho$) formula for a cubic system is: $$\rho = \frac{Z \cdot M}{a^{3} \cdot N_{A}}$$ where $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $a$ is the edge length, and $N_{A}$ is Avogadro's number.

Step 3: Analysis
Given parameters: $M = 75 \text{ g mol}^{-1}$ $a = 5 \text{ Å} = 5 \times 10^{-8} \text{ cm}$ $\rho = 2 \text{ g cm}^{-3}$ $N_{A} = 6 \times 10^{23} \text{ mol}^{-1}$ Rearranging to find $Z$: $$Z = \frac{\rho \cdot a^{3} \cdot N_{A}}{M} = \frac{2 \cdot (5 \times 10^{-8})^{3} \cdot 6 \times 10^{23}}{75}$$ $$Z = \frac{2 \cdot 125 \times 10^{-24} \cdot 6 \times 10^{23}}{75} = \frac{1500 \times 10^{-1}}{75} = \frac{150}{75} = 2$$ Since $Z = 2$, the metal crystallizes in a Body-Centered Cubic (BCC) lattice structure. For a BCC unit cell, the relationship connecting radius ($r$) and edge length ($a$) is: $$4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4}a$$ Substituting $a = 5 \text{ Å}$ and $\sqrt{3} \approx 1.732$: $$r = \frac{1.732 \cdot 5}{4} = \frac{8.66}{4} = 2.165 \text{ Å}$$

Step 4: Conclusion
The atomic radius of metal X matches 2.165 Å precisely.

Final Answer: (C)