If in diamond, there is a unit cell of carbon atoms as foc and if carbon atom is `sp^3` hybridised, what fractions of voids are occupied by carbon atom?

Options

A. 25% tetrahedral
B. 50% tetrahedral
C. 25% octahedral
D. 50% octahedral

Answer

B

Explanation

`sp^3` C-atom suggests a tetrahedral arrangement of 4 C-atoms. For fcc number of atoms (N) per unit cell = 4
`:.` Total available tetrahedral voids `= 2N=2 xx 4 = 8`
Occupied tetrahedral void = 4 i.e., 50% of the total.