Number of $$A$$ cations per unit cell = 4
Number of $$B$$  anions per unit cell must be = 2 × 4 = 8
Number of tetrahedral voids per unit cell = 2 × 4 = 8
Number of octahedral voids per unit all = 1 × 4 = 4.
Hence, occupancy of tetrahedral voids = 100%

In a $$p$$ - type semiconductor, trivalent impurity from the III group elements is added as the impurity. trivalent impurities like Aluminium, Indium and Gallium are added to the intrinsic semiconductor. The Trivalent impurities added provides extra holes known as the acceptor atom. The majority carries in a $$p$$ - type semiconductor are holes.
In an $$n$$ - type semiconductor, pentavalent impurity from the V group is added to the pure semiconductor . Examples of pentavalent impurities are Arsenic, Antimony, Bismuth etc. The pentavalent impurities provide extra electrons and are termed as donor atoms. Electrons are the majority charge carries in $$n$$ - type semiconductors.

Intrinsic semiconductors are insulators at room temperature and become semiconductors when temperature is raised.

In Frenkel defect some of ion ( usually cation due to their small size ) missing from their normal position and occupies position in interstitial position.

The radius ratio range of crystalline structure is 0.225 - 0.414. This range indicates 4 coordination number and tetrahedral arrangement.

The coordination number is 12.

Lattice sites in a pure crystal are occupied by the constituent units like atoms, molecules, ions but not occupied by electrons.

No. of atoms per unit cell in face centred unit cell = 4
∴ No. of tetrahedral voids = 2 × 4 = 8

Out of the given substances, only $$Li$$  has high electrical and thermal conductivity as $$Li$$  is a metallic solid.

Suppose no. of atoms of $$Z = a$$
No. of tetrahedral voids $$ = 2a$$
No. of atoms of $$Y = \frac{1}{3} \times 2a = \frac{{2a}}{3}$$
Ratio of atoms of $$Y$$ and $$Z = \frac{{2a}}{3}:a = 2:3$$
Hence, the formula of the compound $$ = {Y_2}{Z_3}$$