NEET Practice Questions (MCQs) with Answers & Solutions

(B). For bcc, $r=\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{4}\times 4.29=1.86$$\stackrel{0}{A}$

(B).  The volume of the unit cell

                         $={\left(2.88\stackrel{0}{A}\right)}^3=23.9\times {10}^{-24}c{m}^3$

The volume of 100 g of the metal = $\frac{m}{\rho }=\frac{100}{7.20}=13.9 c{m}^3$

Number of unit cells in this volume

                           $=\frac{13.9 c{m}^3}{23.9\times {10}^{-24}c{m}^3}=5.82\times {10}^{23}$

(C). ccp lattice = fcc lattice

$\therefore Z_{cu}=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

$Z_{Ag}=12\times \frac{1}{4}=3$

(Each edge centered atom is contributed by 4 unit cell)

$Z_{Au}=1$

$\therefore Formula of alloy=C{u}_{4}A{g}_{3}Au$

Frenkel defect is not found in pure alkali metal halides. This is because of the larger size of the cations of the alkali metals, due to the large size the cations cannot fit into the interstitial sites.

(C). Molecular solids are the substances having molecules as constituent units having interparticle forces such as Vander Waal’s forces or hydrogen bonds.

High temperature changes 8 : 8 co-ordination to 6 : 6 whereas high pressure changes 6 : 6
co-ordination to 8 : 8.

(C). The value of Z for a bcc unit cell is 2.

Volume V=${\left(4.24A\right)}^3$

$\therefore \rho =\frac{ZM}{NV}=\frac{2\times 23}{\left(6.023\times {10}^{23}\right)\times {\left(4.24\times {10}^{-8}\right)}^3}=1.00g/c{m}^3$

(D). Amorphous solids neither have ordered arrangement (i.e. no definite shape) nor have sharp melting point like cyrstals, but when heated, they become soft until they assume the properties usually related to liquid. It is therefore they are regarded as super cooled liquids.

(A). It is based on the definition of the cubic lattice.

(C). The hexagonal close packing and the face-centred cubic system have the closest packing as their packing efficiency is 74 %.