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

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 B 50%
Diamond has $Z n S$ like structure, where each lattice points and tetrahedral
voids alternately occupied by $C - a t o m$
$∴$ Occupied tetrahedral voids $= 50 %$

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Each carbon atom is sp3 hybridized in diamond. The % of tetrahedral voids occupied in diamond is:

The correct option is B 50%Diamond has ZnS like structure, where each lattice points and tetrahedral voids alternately occupied by C−atom∴ Occupied tetrahedral voids =50%

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

The correct option is B 50%
Diamond has $Z n S$ like structure, where each lattice points and tetrahedral
voids alternately occupied by $C - a t o m$
$∴$ Occupied tetrahedral voids $= 50 %$