Chemistry - Solid State Question with Solution | TestHub

In diamond, carbon atoms form a face-centered cubic (FCC) lattice with additional atoms in alternate tetrahedral voids. The atoms in the tetrahedral voids touch the atoms at the FCC corners.

The distance between an atom at the corner and an atom in a tetrahedral void is $\frac{\sqrt{3}a}{4}$.

Since these atoms are touching, this distance is equal to $2r$, where $r$ is the radius of the carbon atom.

So, $2r=\frac{\sqrt{3}a}{4}$

Given $a=356$ pm.

$r=\frac{\sqrt{3}\times 356pm}{8}$

$r=\frac{1.732\times 356pm}{8}$

$r=\frac{616.912pm}{8}$

$r=77.114pm$

The closest option is 77.07 pm.

The final answer is $\boxed{77.07pm}$.