Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

Among the three options $$KCl, NaCl$$   and $$MgC{l_2},$$  the size of anion is same. So larger the cation, larger will be the cation/anion ratio i.e., $$KCl$$  will have larger cation/anion ratio among the three. So, we left with two options $$KCl$$  and $$Ca{F_2}.$$  Among these two $$Ca{F_2}$$  will have maximum value of cation/ anion ratio because decrease in ionic radii of anion from $$C{l^ - }$$ to $${F^ - }$$ does not overcome the effect of decrease in ionic radii of cation from $${K^ + }$$  to $$C{a^{2 + }}.$$

$$\eqalign{ & {\text{Number of unit cells}} \cr & = \frac{{{N_A} \times Wt.\,{\text{of substance}}\,}}{{M.\,Wt. \times Z}} \cr & = \frac{{6.023 \times {{10}^{23}} \times 58.8}}{{58.8 \times 4}} \cr & = 1.5 \times {10^{23}} \cr} $$

For $$n$$ - type, impurity added to silicon should have more than 4 valence electrons.

Since in $$NaCl$$  type of structure 4 formula units form a cell.
$$\eqalign{ & 58.5\,gm.\,{\text{of}}\,NaCl = 6.023 \times {10^{23}}{\text{atoms}} \cr & 1\,gm\,\,{\text{of}}\,NaCl = \frac{{6.023 \times {{10}^{23}}}}{{58.5}}{\text{atoms}} \cr & {\text{4 atoms constitute 1 unit cell}} \cr & \therefore \,\,\frac{{6.023 \times {{10}^{23}}}}{{58.5}}{\text{atoms constitute}} \cr & = \frac{{6.023 \times {{10}^{23}}}}{{58.5 \times 4}} \cr & = 2.57 \times {10^{21}}\,{\text{unit cells}}{\text{.}} \cr} $$

In $$fcc$$  or $$ccp$$  number of oxide ions $$\left( O \right)$$  per unit cell = 4
Number of tetrahedral voids per ion in lattice = 8
Number of divalent cation $$\left( A \right) = \frac{1}{8} \times 8 = 1$$
Number of octahedral voids per ion in lattice = 4
Number of trivalent cations $$\left( B \right) = \frac{1}{2} \times 4 = 2$$
Formula $$ = A{B_2}{O_4}$$

$$AB$$  is just like $$NaCl.$$  Thus twelve $${A^ + }$$  are at edges and 1 within body of $$fcc$$  i.e. in octahedral voids and six $${B^ - }$$ at faces and 8 at corner.

Ferrimagnetism occurs when magnetic moments of domains are aligned in parallel and anti parallel directions in unequal numbers resulting in net magnetic moment.

In crystalline solid there is perfect arrangement of the constituent particles only at $$0\,K.$$  They have sharp $$M.P.$$

In antifluorite structure the oxide ions (anions) form a face-centered cubic array and the metal ions (cations) occupy all the tetrahedral voids. Example : $$N{a_2}O$$

8 $$X$$ atoms present at the corners.
Atoms contribute to 1 unit cell $$ = \frac{1}{8} \times 8 = 1$$
6 $$X$$ atoms present at the face centres.
Atoms contribute to 1 unit cell $$ = 6 \times \frac{1}{2} = 3$$
Total $$X$$ atoms = 3 + 1 = 4
4 $$M$$ atoms present at edge centres.
Atoms present in 1 unit cell $$ = 4 \times \frac{1}{4} = 1$$
1$$M$$ atom present at body centre and it contributes completely to 1 unit cell.
Thus, total $$M$$ atoms in one unit cell = 1 + 1 = 2
Ratio is $$M:X\,\,:\,:\,\,2:4:\,:\,\,1:2$$
Thus, empirical formula is $$M{X_2}.$$