Chemistry - Solid State Question with Solution | TestHub

Question Explanation: Calculate the density or volume for an FCC crystal lattice.

Concept: Density of crystal.

Solution: Density $D=\frac{Z\times M}{N_A\times V}$, so the Volume $V=\frac{Z\times M}{N_A\times D}$.

Given $Z=4$ (for FCC), $M=207$ g/mol, $N_A=6.02\times 10^{23}$ mol⁻¹, and $D=11.34$ g/cm³.

$$V=\frac{4\times 207}{6.02\times 10^{23}\times 11.34}=1.213\times 10^{-22}c{m}^3$$

Since $V=a^3$, we have:

$$a^3=1.213\times 10^{-22}c{m}^3$$

$$a=(1.213\times 10^{-22}{)}^{1/3}=(121.3\times 10^{-24}{)}^{1/3}$$

$$a=4.95\times 10^{-8}cm$$

For an FCC lattice, the relationship between the atomic radius $r$ and the edge length $a$ is $4r=\sqrt{2}a$.

$$r=\frac{\sqrt{2}a}{4}=\frac{4.95\times 10^{-8}cm\times \sqrt{2}}{4}$$

$$r=1.75\times 10^{-8}cm=175pm$$

Final answer: The correct options are (A) and (C).