NEET Practice Questions (MCQs) with Answers & Solutions

For (fcc) unit cell, atoms touch each other along the face diagonal.

Hence, Atomic radius (R) $=\frac{a\sqrt{2}}{4}$

$a=\frac{4R}{\sqrt{2}}=\frac{4}{\sqrt{2}}\times 144.5pm=408.70pm=408.70\times {10}^{-10}cm$

Density (D) $=\frac{ZM}{V{N}_0},$ V = a³

D = $\frac{ZM}{a^3{N}_0}$; where Z for (fcc) unit cell = 4 , Avagadro’s number $\left(N_0\right)=6.023\times {10}^{23}$, Volume of cube (V) $={(408.70\times {10}^{-10})}^{3}c{m}^3$ and M (Mol. wt.) of silver = 108,

D $=\frac{4\times 108}{{(408.70\times {10}^{-10})}^3\times 6.023\times {10}^{23}}$$=10.50g/c{m}^3$  

We know that, $\rho =\frac{ZM}{N_{0}V}=\frac{4\times (21.76gmo{l}^{-1})}{(6.023\times {10}^{23}mo{l}^{-1})(6.81\times 4.43\times 7.17\times {10}^{-24}c{m}^3)}$$=0.668gc{m}^{-3}$  

The volume of unit cell (V) = a³ = (2.88Å)³ = $23.9\times {10}^{-24}c{m}^3$

Volume of 100 g of the metal = $\frac{\text{Mass}}{\text{Density}}=\frac{100}{7.20}=13.9c{m}^2$

Number of unit cells in this volume = $\frac{13.9c{m}^3}{23.9\times {10}^{-24}c{m}^3}=5.82\times {10}^{23}$  

The unit cell has a volume of (0.408 $\times {10}^{-9}m{)}^3=6.79\times {10}^{-29}{m}^3$ per unit cell and contains four atoms. The volume of 1 mole of silver is,

$107.9g/mol\left[\frac{{\displaystyle {(1\times {10}^{-2}m)}^3}}{{\displaystyle 10.6g}}\right]=1.02\times {10}^{-5}{m}^3/mol$; where 107.9 g/mol is the molecular mass of the silver

The number of unit cells per mol. is,

$1.02\times {10}^{-5}{m}^3/mol\left(\frac{{\displaystyle 1\text{unit cell}}}{{\displaystyle 6.79\times {10}^{-29}{m}^3}}\right)=1.50\times {10}^{23}$ unit cells per mol.

and the number of atoms per mol. is, $\left(\frac{{\displaystyle 4\text{atoms}}}{{\displaystyle \text{unit cell}}}\right)\left(\frac{{\displaystyle 1.50\times {10}^{23}\text{unit cell}}}{{\displaystyle \text{mol}}}\right)=6.00\times {10}^{23}$ atom/mol.  

In a simple cubic system, number of atoms a = 2r

∴ Packing fraction $=\frac{\text{Volume occupied by one atom}}{\text{Volume of unit cell}}=\frac{\frac{4}{3}\pi r^3}{a^3}=\frac{\frac{4}{3}\pi r^3}{{\left(2r\right)}^3}=\frac{\pi }{6}$  

We know that for the NaCl structure

radius of cation/radius of anion = 0.414; $\frac{r_{A^{+}}}{r_{B^{-}}}=0.414$; $r_{B^{-}}=\frac{r_{A^{+}}}{0.414}=\frac{120}{0.414}=290pm$  

From Bragg’s equation, $n\lambda =2d\sin \theta$; $d=\frac{n\times \lambda }{2\sin \theta }=\frac{1\times 0.154}{2(\sin 16.3)}=\frac{0.154nm}{2\times 0.281}=0.274nm$  

Using Bragg's equation $2d\sin \theta =n\lambda$

$d=\frac{n\lambda }{2\sin \theta }$, where d is the distance between two diffracted planes, $\theta$the angle to have maximum intensity of diffracted X-ray beam, n the order of reflection and $\lambda$ is the wavelength

∴ $d=\frac{1\times 2.29\AA }{2\times \sin {30}^o}=2.29\AA$     $\left(\sin {30}^o=\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)$  

$2d\sin \theta =n\lambda$

$\lambda =2d\sin \theta =2\times 2.64\times {10}^{-10}\times \sin 7{.75}^o=0.7123\times {10}^{-10}m$

$E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{0.7123\times {10}^{-10}}$$=27.88\times {10}^{-16}J$  

⇒ expansion, electrical and thermal conductivities etc. is different in different directions, with in the crystal

(a)  FeO_0.98 has non-stoichiometric metal excess defect. 

(b) In an ionic crystal of A⁺B^- type, if equal number of cations and anions are missing from their lattice sites, the defect is called Schottky defect. Due to such defect, density of solid decreases. Thus, statement (b) is correct.

(c) NaCl-insulator Silicon (Si) -semiconductor. Silver (Ag) - conductor; Quartz - piezoelectric crystal.

Thus, statement (c) is correct.

(d) In an ionic crystal when an j£n is missing from its lattice site and occupies interstitial site, the defect is called Frenkel defect. This type of defect is seen in those crystals where the difference in the size of cations and anions is very large and their coordination number is low. Thus, statement (d) is incorrect.