Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

Number of $$Cu$$ -atoms per unit cell
$$ = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4.$$
Number of $$Ag$$ -atoms per unit cell
$$ = \frac{1}{4} \times 12 = 3$$
Number of $$Au$$ -atoms per unit cell $$= 1$$  ( at body centre )
Formula $$:C{u_4}A{g_3}Au$$

Iodine $$\left( {{I_2}} \right)$$  is a molecular solid and non polar in nature. Their molecules are held by weak dispersion forces or London forces.

$$\eqalign{ & {\left( {2R} \right)^2} = {\left( {R + r} \right)^2} + {\left( {R + r} \right)^2} \cr & \sqrt 2 R = R + r\,\,;\left( {\sqrt 2 - 1} \right)R = r \cr & 0.414R = r \cr} $$

$$AgBr$$  shows Frenkel and Schottky defects because radius ratio of $$AgBr$$  is intermediate. It has rock salt structure i.e., $$fcc$$  or $$ccp$$  lattice $$fo$$ $$Br$$  and $$Ag$$  occupies all the octahedral voids.

A sphere in square close packing is in contact with four spheres in two dimensions, so coordination number will be 4.

For $$bcc$$  structure of ionic compounds like $$CsCl,Z = 1$$
$$\eqalign{ & d = \frac{{Z \times M}}{{{N_A} \times {a^3}}} \Rightarrow {a^3} = \frac{{Z \times M}}{{d \times {N_A}}} \cr & = \frac{{1 \times 168.4}}{{3.988 \times 6.023 \times {{10}^{23}}}} \cr & = 7.014 \times {10^{ - 23}}\,c{m^3} \cr} $$

The face centered cubic unit cell contains 4 atom
∴ Total volume of atoms $$ = 4 \times \frac{4}{3}\pi {r^3} = \frac{{16}}{3}\pi {r^3}$$

$$\eqalign{ & {N_A} = 6.023 \times {10^{23}} \cr & {\text{Atomic mass of mercury}} = 200 \cr & {\text{Number of atoms present in }}200g{\text{ of }}Hg \cr & = 6.023 \times {10^{23}} \cr & {\text{So, number of atoms present in }}1g{\text{ of }}Hg \cr & = \frac{{6.023 \times {{10}^{23}}}}{{200}} \cr & = 3.0115 \times {10^{21}} \cr & {\text{Moss of 1 atom}} = \frac{1}{{3.011 \times {{10}^{21}}}}g \cr & {\text{Density of}}\,\,Hg = 13.6\,g/mL \cr & \because {\text{Density = }}\frac{{{\text{Mass}}}}{{{\text{Volume}}}} \cr & \therefore {\text{Volume}} = \frac{{{\text{Mass}}}}{{{\text{density}}}} \cr & {\text{Volume of 1 atom of mercury }}\left( {Hg} \right) \cr & = \frac{1}{{3.0115 \times {{10}^{21}} \times 13.6}}mL \cr & = 2.44 \times {10^{ - 23}}\,mL \cr} $$
As each mercury atom occupies a cube of edge length equal to its diameter, therefore
$$\eqalign{ & {\text{Diameter of 1 mercury atom}} \cr & = {\left( {2.44 \times {{10}^{ - 23}}} \right)^{\frac{1}{3}}}\,\,cm \cr & = 2.905 \times {10^{ - 8}}\,cm \cr & = 2.91\mathop {\text{A}}\limits^{\text{o}} \cr} $$

$$\eqalign{ & {\text{As per formula,}} \cr & {\text{radius ratio}} = \frac{{{\text{radius of cation}}}}{{{\text{radius of anion}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{94}}{{146}} = 0.643 \cr} $$
Since the value is between 0.414 - 0.732 hence the coordination no. will be 6 and geometry will be octahedral.

$$FeO$$  is mostly found with a composition of $$F{e_{0.95}}O.$$   In crystals of $$FeO,$$  some $$F{e^{2 + }}$$  cations are missing and the loss of positive charge is made up by the presence of the required number of $$F{e^{3 + }}\,ions.$$