NEET Practice Questions (MCQs) with Answers & Solutions

(a)

 $$\begin{gathered} \mathrm{Given}, \mathrm{Li} \mathrm{has} \mathrm{a} \mathrm{bcc} \mathrm{structure}. \\ \mathrm{Density} \left(\mathrm{\rho }\right) = 530 \mathrm{kg}.{\mathrm{m}}^{-3} \\ \mathrm{Atomic} \mathrm{mass} \left(\mathrm{M}\right) = 6.94 {\mathrm{gmol}}^{-1} \\ \mathrm{We} \mathrm{know} \mathrm{that}, \mathrm{number} \left({\mathrm{N}}_{\mathrm{A}}\right) = 6.02 \times {10}^{23}{\mathrm{mol}}^{-1} \\ \mathrm{We} \mathrm{know} \mathrm{that}, \mathrm{number} \mathrm{of} \mathrm{atoms} \mathrm{per} \mathrm{unit} \mathrm{cell} \mathrm{in} \mathrm{bcc} \left(\mathrm{Z}\right) = 2. \\ \therefore \mathrm{We} \mathrm{have} \mathrm{the} \mathrm{formula} \mathrm{for} \mathrm{density}, \\ \mathrm{\rho }=\frac{\mathrm{ZM}}{{\mathrm{N}}_{\mathrm{A}}{\mathrm{a}}^3} \\ \mathrm{Where} \mathrm{a}= \mathrm{edge}-\mathrm{length} \mathrm{of} \mathrm{a} \mathrm{unit} \mathrm{cell}. \\ \mathrm{or} \mathrm{a}=\sqrt[3]{\frac{\mathrm{ZM}}{{\mathrm{\rho N}}_{\mathrm{A}}}}= \sqrt[3]{\frac{2 \times 6.94 \mathrm{g} {\mathrm{mol}}^{-1}}{0.53 \mathrm{g} {\mathrm{cm}}^{-3}\times 6.02 \times {10}^{23}{\mathrm{mol}}^{-1}}} \\ = \sqrt[3]{4.35 \times {10}^{-23}{\mathrm{cm}}^{-3}} \\ = 3.52 \times {10}^{-8}\mathrm{cm} \\ \mathrm{a}= 352 \mathrm{pm} \end{gathered}$$

(b) Given, edge length = 361pm

Four metal atoms in one unit cell

i.e. effective number in unit cell (z) = 4 (given)
$\therefore$ It is a FCC Structure
$\therefore$ Face diagonal = 4r

                       $\sqrt{2}\mathrm{a} = 4\mathrm{r} \Rightarrow =\frac{\sqrt{2}\times 361}{4}=127 \mathrm{pm}$

In Frenkel defect, ions in solids dislocate from their positions. Hence, Frenkel delect is a dislocation defect.

(d)

$$\begin{gathered} \because \mathrm{Packing} \mathrm{efficiency} \mathrm{in} \mathrm{bcc} \mathrm{lattice} =68\% \\ \therefore \mathrm{Vacant} \mathrm{space} \mathrm{in} \mathrm{boc} \mathrm{lattice} = 100-68 =32\% \end{gathered}$$

For fcc lattice,

4r= √2a

r= √2 a = a   
      4       2√2

= 408  
   2√2

=144pm

diameter d=2r=2 x 144pm=288pm

In case of body centred  cubic (bcc) crystal, a$\sqrt{3}$=4r

Hence, atomic radius of lithium,r = a$\sqrt{3}$/4

=(351x1.732)/4

= 151.98 pm

Key Idea:

Packing fraction = Volume occupied by atoms in a unit cell

                                 Volume of the unit cell

For body centered cube,

Packing fraction = $\frac{2\mathrm{x}\frac{4}{3}{\mathrm{\pi r}}^3}{(4\mathrm{r}/\sqrt{3}{)}^3} = \frac{\sqrt{3}{\displaystyle \mathrm{\pi }}}{8}$ = 0.68

where, edge length a = 4r/$\sqrt{3}$

... Volume occupied = 68%

and volume vacant =32%

(d) The n-type semiconductors are obtained when Si or Ge are doped with elements of group 15, eg, arsenic (As), while p-type semiconductors are obtained when Si or Ge are doped witlh traces of elements of group 13, ie, indium (In), boron (B).

Note : The conductivity of semiconductors increases with rise in temperature

(C) Based on Fluorite structures of $Ca{F}_2$