1)  Construction/Atomic Arrangement –

Diamond is made up of carbon atoms. Carbon is tetravalent.

∴ Each carbon atom forms 4 covalent bonds.

Diamond cubic structure is combination of two interpenetrating FCC sub-lattices along the body diagonal by ¼th  cube edge. One sub-lattice has its origin at (0,0,0) & other sub-lattices has its origin 1/4 of way along body diagonal i.e. at point (a/4, a/4, a/4). The basic diamond lattice and the atomic position of cubic cell of diamond are as shown in above fig.

The fractions e.g. $\dfrac{3}{4},\dfrac{1}{2},\dfrac{1}{4}$ denote the height above the base in units of cube edge. The atomic basis is 2 atoms/lattice point. One atom at (0,0,0) & the other at (a/4, a/4, a/4) as stated above.

2)  Effective number of atoms/unit cell (n) –

a)  Each corner atoms is shared by 8 adjacent unit cells, therefore it contributes 1/8th of its volume & mass of 1 unit cell. There are such 4 corner atoms in one square face. Further, there are 2 square faces: One at bottom & the other at the top.

$\therefore$ Contribution of corner atoms to unit cell $=2\times 4\times \dfrac{1}{8}=1$

b)  There is one atom at centre of each square face which is shared by 2 adjacent cells. Therefore it contributes 1/2 of its volume & mas to 1 unit cell. There is one atom in one square face. Furthermore, there are 6 faces.

∴ Contribution of such one centre atom to unit cell $=6\times 1\times \dfrac {1}{2} =3$

c) There are 4 atoms, located inside the unit cell & they are not shared by adjacent cells.

∴ Contribution of such 4 inside atoms to unit cell = 4

$∴ n=1+3+4=8$

3)  Atomic Radius (r) –

$XZ^{2}=XY^{2}+YZ^{2} \\ \ \\ = (XW^{2}+WY^{2})+YZ^{2} \\ \ \\ =\left [ \left ( \dfrac {a}{4}\right )^{2}+\left ( \dfrac {a}{4}\right )^{2} \right ] +\left ( \dfrac {a}{4} \right )^{2} \\ \ \\ =\dfrac {3a^{2}}{16} \\ \ \\ \therefore XZ=\dfrac {\sqrt{3}a}{4} $

But $ XZ=2r$

$\therefore 2r=\dfrac {\sqrt{3}}{4}a \\ \therefore r=\dfrac {\sqrt {3}}{8}a$

4)  Atomic Packing Factor (APF) –

APF=$ \dfrac {(Number\ of\ atoms/unit\ cell) \times (Volume\ of\ 1\ atoms)}{Volume\ of\ unit\ cell} \\$ $=\dfrac {8\times \dfrac {4}{3} \Pi r^{3}}{a^{3}}\$

$=8\times \dfrac {4}{3} \Pi \times \left (\dfrac {\sqrt {3}}{8}a \right )^{3} =0.34\$

∴ It is loosely packed structure but covalent bonds being the strongest.

∴ It is hard.