Each carbon atom is $s{p}^3$ hybridized in diamond. The % of tetrahedral voids occupied in diamond is:

Q. If in diamond, there is unit cell of carbon atoms as fcc and if carbon atom is $s{p}^3$ hybridized what fractions of void occupied by carbon atom?

Q. Diamond has a three dimensional structure of $C$ atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where $50\%$ of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is____________

Q. How many tetrahedral voids are occupied in diamond?

Q. In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356$ pm, then the radius (in pm) of carbon atom is:

Q. In the diamond cubic structure, carbon atoms form an FCC while the other carbon atoms occupy alternate tetrahedral voids, the atoms being connected by covalent bonds. Then the packing efficiency is approximately