CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 13 - MCQExams.com

$$z =4 \text { (for } f C C) $$

$$M =197 \text { (for gold) } $$

$$a =4.07 \text { A( given) } $$

$$=407 \mathrm{pm}$$

$$P =\frac{2 \times M}{N a \times a^{3} \times 10^{-30}} g \mathrm{~cm}^{-3} $$

$$=\frac{4 \times 191}{6.022 \times 10^{23} \times(401)^{3} \times 10^{-30}} $$

$$=\frac{7899}{401}=19.4 \mathrm{~g} \mathrm{~cm}^{-3}$$

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. 

Face centred cubic (fcc) lattice has a packing efficiency of $$74$$%.

Body centred cubic (bcc) lattice has a packing efficiency of $$68$$%.

Simple/ Primitive cubic (pc) lattice has a packing efficiency of $$52.4$$%.

Hence, the correct order is $$\text {fcc > bcc > pc}$$.

$$CsBr$$ has body centered cubic (BCC) structure. The edge length of unit cell is given by $$4.4 \ \mathring A$$

Now, for a cube of edge $$a$$ units, the length of the diagonal is $$\sqrt 3 a$$ units.

The shortest ionic distance between $$Cs^+$$ and $$Br^-= \cfrac {\sqrt 3a}{2}$$

$$= \cfrac {\sqrt 3 \times 4.4}{2}= 3.81 \ \mathring A$$ 

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 $$= \cfrac 12 \ moles = \cfrac 12 \times 6.022 \times 10^{23}= 3.0125 \times 10^{23}$$

We know that,

$$density \ = \cfrac {Mass}{Volume} \Rightarrow Volume = \cfrac {Mass}{Density}$$

$$\Rightarrow Mass = density \times volume$$

Now, given densities of gold and quartz are respectively $$d_1$$ and $$d_2$$

Masses of gold and quartz are $$x \ g $$ and $$y \ g$$ respectively.

So,

$$Volume \ of \ gold = \cfrac {x}{d_1}$$.....(i)

$$Volume \ of \ quartz = \cfrac {y}{d_2}$$.....(ii)

$$ Volume \ of \ Nugget = \cfrac {Mass \ of \ nugget}{density \ of \ Nugget}$$

$$=\cfrac {x+y}{d}$$.....(iii)

Now, $$volume \ of \ nugget = volume \ of \ gold + volume \ of \ quartz$$

$$\Rightarrow \cfrac {x+y}{d}= \cfrac {x}{d_1} + \cfrac {y}{d_2}$$

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.

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

In $$NaCl$$ crystal, there are $$4$$ formula units per unit cell. So, $$n=4$$

$$\Rightarrow a^3 = \cfrac {nM}{N_o \rho}= \cfrac {4 \times 58.5}{6.022 \times 10^{23} \times 3.165}$$

$$\Rightarrow a= (123)^{(1/3)} \times (10^{-24})^{1/3} \ cm$$

$$\Rightarrow a = 4.97 \times 10^{-8} \ cm = 497 \ pm$$

Now, in $$NaCl$$, $$Cl^-$$ ions are present at corners and at face centers and $$Na^+$$ ions are present at edge centers and body centers.

So, distance between $$Na^+$$ and $$Cl^-= \cfrac a2= \cfrac {497}{2}=248.5 \ pm$$

Since, cube close packing is analogous to face-centered cubic (fcc) structure and number of atoms in a fcc crystal is $$4 \ per \ unit \ cell$$

So, number of tetrahedral voids $$= 2 \times Number \ of \ atoms \ in \ unit \ cell $$

$$= 2 \times 4 = 8$$

Hence, Option "D" is the correct answer.

Square close packing in 2D: In this kind of arrangement the AAA type of pattern of stacking is followed. The coordination number of each sphere in this arrangement will be $$4$$.

Hexagonal close packing in 2D has triangular voids as shown in the figure.

$$Na_2O$$ structure is antifluorite type structure in which $$O^{2-}$$ ions forms ccp (cubic close packing) and $$Na^{+}$$ occupy all tetrahedral voids.