Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

Effective number of 'A’ atoms $$ = \left( {8 \times \frac{1}{8}} \right) + \left( {4 \times \frac{1}{2}} \right) = 3$$
Effective number of ‘B’ atoms $$ = \left( {12 \times \frac{1}{4}} \right) + 1 = 4$$
∴ Formula of the solid = $${A_3}{B_4}.$$

The electrical resistance of metals depends upon temperature. Electrical resistance decreases with decrease in temperature and becomes zero near the absolute temperature. Material in this state is said to possess super conductivity.

Paramagnetism occurs due to the presence of one or more unpaired electrons. They have permanent magnetic dipoles and are weakly attracted by the magnetic field.

Solid $$AB$$  crystallizes as $$NaCl$$  structure, so it has coordination number 6 and its $$\frac{{{r^ + }}}{{{r^ - }}}$$  ranges from 0.414 – 0.732.
For maximum radius of anion, we have to take the lower limit of the range 0.414 – 0.732.
$$\eqalign{ & {\text{So,}}\,\frac{{{r^ + }}}{{{r^ - }}} = 0.414 \cr & \Leftrightarrow {r^ - } = \frac{{0.100}}{{0.414}}nm = 0.241\,nm \cr} $$

Ferrimagnetic substance become paramagnetic on heating. This is due to randomisation of spins on heating.

$$\frac{{{r_c}}}{{{r_a}}} = \frac{{0.066}}{{0.140}} = 0.47$$
So, cation $$\left( {M{g^{2 + }}} \right)$$  fitted in octahedral holes of $${O^{2 - }}$$ are present at $$fcc$$  lattice point.

$$\eqalign{ & {\text{Volume of the unit cell}} \cr & = {a^3} = {\left( {400\,pm} \right)^3} \cr & = {\left( {400 \times {{10}^{ - 10}}\,cm} \right)^3} \cr & = 64 \times {10^{ - 24}}\,c{m^3} \cr & {\text{Density of the unit cell}} = 10\,g/c{m^3} \cr & {\text{Mass of unit cell = Volume}} \times {\text{density}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 64 \times {10^{ - 24}} \times 10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 640 \times {10^{ - 24}}\,g \cr & {\text{Now}},640 \times {10^{ - 24}}\,g \equiv {\text{1}}\,{\text{unit cell}} \cr} $$
$$\therefore 32\,g \equiv \frac{1}{{640 \times {{10}^{ - 24}}}} \times 32$$      $$ = 0.05 \times {10^{24}} = 5 \times {10^{22}}$$

In Frenkel defect, ions get displaced from their original position and move to interstitial sites. Hence, there is no change in the density of the crystal.

Since in $$NaCl$$  type of structure 4 formula units form a cell. Number of formulas in cube shaped crystals
$$\frac{{1.0}}{{58.5}} \times 6.02 \times {10^{23}}$$
No. of unit cells present in a cubic crystal $$ = \frac{{P \times {a^3} \times {N_A}}}{{M \times Z}} = \frac{{m \times {N_A}}}{{M \times Z}}$$
∴ units cells $$ = \frac{{1.0 \times 6.02 \times {{10}^{23}}}}{{58.5 \times 4}} = 2.57 \times {10^{21}}$$       unit cells.

Since each $$S{r^{ + + }}\,ion$$   provides one cation vacancy, hence
Concentration of cation vacancies
$$\eqalign{ & = mole\,\% \,{\text{of}}\,SrC{l_2}\,{\text{added}} \cr & {\text{ = 1}}{{\text{0}}^{ - 4}}\,mole\,\% \cr & = \frac{{{{10}^{ - 4}}}}{{100}} \times 6.023 \times {10^{23}} \cr & = 6.023 \times {10^{17}} \cr} $$