How many tetrahedral holes are occupied in diamond?

25%

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50%

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75%

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100%

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Solution

The correct option is D 50%
Diamond has a ZnS type structure. the carbon atoms occupy all the fcc lattice points along with alternative tetrahedral voids .
total no.of tetrahedral voids in fcc unit cell =8
Where carbon atom occupy only 4 of them . therefore percentage of tetrahedral voids occupied by carbon atom in diamond is 50 percent.

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In a normal spinel type structure, the oxide ions are arranged in ccp whereas $\frac{1}{8}$ tetrahedral holes are occupied by $Z n^{2 +}$ ions and 50% of octahedral holes are occupied by $F e^{3 +}$ ions. The formula of the compound is ?

C a S

exists in a cubic close packed arrangement of

S^{2 -}

ions in which

C a^{2 +}

ions occupy

1 / 2

of the available tetrahedral holes. How many

C a^{2 +}

and

S^{2 -}

ions are contained in the unit cell?

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