Assertion (A) Graphite is a good conductor of electricity however diamond belongs to the category of insulators.
Reason (R) Graphite is soft in nature on the other hand diamond is very hard and brittle.
Assertion (A) Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four.
Reason (R) Besides the body centre there is one octahedral void present at the centre of each of the six faces of the unit cell and each of which is shared between two adjacent unit cells.
Assertion (A) The packing efficiency is maximum for the fcc structure.
Reason (R) The coordination number is 12 in fcc structures.
Assertion (A) Semiconductors are solids with conductivities in the intermediate range from $10^{-6}-10^4 \mathrm{ohm}^{-1} \mathrm{~m}^{-1}$.
Reason (R) Intermediate conductivity in semiconductor is due to partially filled valence band.
With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
Cubic close packed structure contains one atom at each of eight corners of a unit cell and one atom at each of six faces which can be represented below
As we know any atom surrounded by six atoms (hard sphere) creates an octahedral void. In case of fcc body centre is surrounded by six identical atoms present at face centre hence, there is a octahedral void at body centre of each unit cell.
Beside the body centre there is one octahedral void at centre of each of 12 edge as shown below
Since, each void is shared by 4 unit cell. Therefore, contribution of octahedral void to each edge of a unit cell is $\frac{1}{4}$.
Number of octahedral void at centre of 12 edge $=\frac{1}{4} \times 12=3$
Number of octahedral void at body centre $=1$
Therefore, total number of octahedral void at each $\operatorname{ccp}$ lattice $=3+1=4$