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Question
If in diamond, there is unit cell of carbon atoms as fcc and if carbon atom is sp3 hybridized what fractions of void occupied by carbon atom?
If in diamond, there is unit cell of carbon atoms as fcc and if carbon atom is sp3 hybridized what fractions of void occupied by carbon atom?
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Solution
The correct option is A 25% tetrahedral
Diamond has a FCC lattice, means the number of atoms present in the lattice is 4. FCC lattice has atoms at the corners of the cube as well as in the center of faces.So, packing fraction for the diamond atom is 74%. The fraction of voids occupied by the carbon atoms is 25% tetrahedral because the 4 atoms of tetrahedral are placed at the opposite corners which rise into a tetrahedral voids.
Diamond has a FCC lattice, means the number of atoms present in the lattice is 4. FCC lattice has atoms at the corners of the cube as well as in the center of faces.
So, packing fraction for the diamond atom is 74%. The fraction of voids occupied by the carbon atoms is 25% tetrahedral because the 4 atoms of tetrahedral are placed at the opposite corners which rise into a tetrahedral voids.
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