Solid State Test - 64

In $$NaCl$$ lattice, $$Cl^-$$ ions forms FCC and $$Na^+$$ ions are present at edge centers and body center.

Number of $$Na^+$$ ions present $$= \cfrac 14 \times 12 = 3$$

Now, charge on $$Al^{3+} = +3$$ and charge on $$Na^+ =1$$.

So, each $$Al^{3+}$$ ion will replace $$3  Na^+$$ ions and two vacant sites will be created.

Now, $$3/4$$ moles of $$Na^+$$ ions will be replaced by $$1/4$$ moles of $$Al^{3+}$$ ions to maintain electrical neutrality. Now, remaining $$1/4$$ moles of $$Na^+$$ ions and $$1/4$$ moles of $$Al^{3+}$$ ion will occupy $$1/2$$ moles of lattice sites.

So, Number of vacancies $$= \cfrac 12 \ moles = \cfrac 12 \times 6.022 \times 10^{23}= 3.0125 \times 10^{23}$$

Given that, $$Cu$$ atoms forms CCP arrangement which is analogous to FCC type arrangement.So, 

Number of atoms of $$Cu$$ present in unit cell $$=4$$

So,

 Number of octahedral voids $$=$$ Number of atoms in FCC arrangement $$=4$$

Number of tetrahedral voids $$= 2 \times $$ Number of octahedral voids

$$= 2 \times 4 = 8$$

The given hypothetical formula is $$Cu_4Ag_3Au$$

This is possible if $$\left (\cfrac 14 \right)^{th}$$ of the octahedral voids and $$\left (\cfrac 32 \right)^{th}$$ of the tetrahedral voids are occupied. 

In $$NaCl$$ structure, $$Cl^-$$ ions form FCC and $$Na^+$$ are present at body center and edge center.

Now, if $$a$$ is the edge length of the cube, then body diagonal is given by

$$d= \sqrt3 a$$

Given, $$d=x$$

$$\Rightarrow x = \sqrt 3 a$$.......(i)

Since, atoms at face centere and corners are touching each other, so distance between two nearest cation $$= \cfrac {\sqrt 2 a}2$$........(ii)

(i) and (ii)

$$\Rightarrow \cfrac {x}{\sqrt 3}= \sqrt 2 d’$$

$$\Rightarrow d’ = \cfrac {x}{\sqrt 6}$$

Area od square of side length $$l=l^2$$

Number of atoms in unit cell $$= 1+ 4 \times \cfrac 14 = 2$$

Let, radius of each circle be $$r$$

Area occupied $$= \pi r^2 \times 2$$

Now, diagonal of square $$= l \sqrt 2$$

$$\Rightarrow 4r = l \sqrt 2$$

$$r= \cfrac l{2 \sqrt 2}$$

Now, Packing efficiency $$ = \cfrac {Area \ occupied}{Total area} \times 100$$%

$$= \cfrac {2 \pi r^2}{l^2} \times 100$$%

$$=\cfrac {2 \pi l^2}{(2 \sqrt 2)^2 \times l^2} \times 100$$%

$$= \cfrac {100 \times 2 \pi}{8}$$%

$$\Rightarrow Packing Efficiency = 78.54$$%

Different types of crystal system arised due to difference in the arrangement of atoms in the space.

When light strikes a photographer $$(AgBr)$$ paper, it gives energy to the electrons present in the film. These energetic electrons when strike silver ions turn them to silver atoms. So eventually,  ions leave their lattice site and occupy interstitial sites. Since silver atoms are black in color so whenever light strikes a silver ion the photographic film will turn black.

Theoretical density of crystal, $$\rho = \cfrac {nM}{N_o a^3} \ g/cm^3$$

For NaCl, $$n=4, \ M=58.5 \ g, \ a=0.564 \times 10^{-7} \ cm$$

$$\therefore \rho = \cfrac {4 \times 58.5}{6.022 \times 10^{23} \times (0.564)^3 \times 10^{-21}}=2.16 \ g/cm^3$$

Hence, the correct option is $$\text{B}$$