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 $=3+1=4$

Cubic close packed structure contains one atom at each corner of an unit cell and at face centre of each unit cell. Each unit cell consists of 8 small cubes.

Each small cube contains 4 atoms at its alternate corner when these atoms are joined to each other lead to creation of a tetrahedral void as shown below

Since, there are total 8 smaller cubes present at one unit cell and each smaller cube has one tetrahedral void hence, total number of tetrahedral void present in each unit cell is equal to eight.

As we know ccp structure has 4 atoms per unit cell. Thus, total number of tetrahedral void in one ccp unit cell is equal to 8.

Conductivity of a semiconductor is too low for practical use. The conductivity of a semiconductor can be increased by adding a suitable amount of impurity to perfect crystal. This process is known as doping. It can be done by adding either of two types of impurity to the crystal.

(A) By adding electron rich impurities i.e., group 15 elements to the silicon and germanium of group 14 elements. Out of 4 valence electrons of group 14 elements and 5 valence electrons of group 15 elements, four electrons of each element led to formation of four covalent bonds while the one extra electron of group 15 elements become delocalised.

Thus, increases conductivity of semiconductor. This type of semiconductor is known as $n$-type semiconductor.

(B) By adding electron deficient impurity i.e., group 14 to the perfect crystal of group 14 elements when group 13 element is doped to group 14 element it lead to create a hole in the ideal crystal which is known as electron hole or electron vacancy.

An electron from the neighbouring atom come and fill the electron hole in doing so an electron from the neighbour leaves an electron hole to its original position. Thus, it increases conductivity of semiconductor. This type of semiconductor is known as p-type semiconductor.

Let the formula of the sample be $\left(\mathrm{Fe}^{2+}\right)_x\left(\mathrm{Fe}^{3+}\right)_y \mathrm{O}$

On looking at the given formula of the compound

$$x+y=0.93\quad\text{.... (i)}$$

Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen

Therefore,

$$\begin{array}{r} 2 x+3 y=2 \quad\text{.... (ii)}\\ \Rightarrow\quad x+\frac{3}{2} y=1\quad\text{.... (iii)} \end{array}$$

On subtracting equation (i) from equation (iii) we have

$$\begin{array}{rlrl} & \frac{3}{2} y-y =1-0.93 \\ \Rightarrow & \frac{1}{2} y =0.07 \\ \Rightarrow & y =0.14 \end{array}$$

On putting the value of y in equation (i), we get

$$\begin{aligned} & x+0.14=0.93 \\ & \Rightarrow \quad x=0.93-0.14 \\ & \Rightarrow \quad x=0.79 \\ & \text { Fraction of } \mathrm{Fe}^{2+} \text { ions present in the sample }=\frac{0.79}{0.93}=0.849 \end{aligned}$$

Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.