Match the defects given in column I with statements given in column II and mark the appropriate choice. Column I Column II (A) Simple vacancy defect (i) Shown by non-ionic solids and increases the density of the solid. (B) Simple interstitial defect (ii) Shown by ionic solution and decreases the density of the solid (C) Frenkel defect (iii) Shown by non-ionic solids and decreases the density of the solid (D) Schottky defect (iv) Shown by ionic solids and density of the solid remains the same

(A)$$\rightarrow$$(iii), (B)$$\rightarrow$$(i), (C)$$\rightarrow$$(iv), (D)$$\rightarrow$$(ii)

(A)$$\rightarrow$$(iv), (B)$$\rightarrow$$(iii), (C)$$\rightarrow$$(ii), (D)$$\rightarrow$$(i)

(A)$$\rightarrow$$(iii), (B)$$\rightarrow$$(iv), (C)$$\rightarrow$$(i), (D)$$\rightarrow$$(ii)

(A)$$\rightarrow$$(i), (B)$$\rightarrow$$(iii), (C)$$\rightarrow$$(iv), (D)$$\rightarrow$$(ii)

MCQ Questions for Class 12 Engineering Chemistry The Solid State Quiz 14

Match the defects given in column I with statements given in column II and mark the appropriate choice.

	Column I		Column II
(A)	Simple vacancy defect	(i)	Shown by non-ionic solids and increases the density of the solid.
(B)	Simple interstitial defect	(ii)	Shown by ionic solution and decreases the density of the solid
(C)	Frenkel defect	(iii)	Shown by non-ionic solids and decreases the density of the solid
(D)	Schottky defect	(iv)	Shown by ionic solids and density of the solid remains the same

0%
(A) $\rightarrow$ (iv), (B) $\rightarrow$ (iii), (C) $\rightarrow$ (ii), (D) $\rightarrow$ (i)
0%
(A) $\rightarrow$ (iii), (B) $\rightarrow$ (iv), (C) $\rightarrow$ (i), (D) $\rightarrow$ (ii)
0%
(A) $\rightarrow$ (iii), (B) $\rightarrow$ (i), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (ii)
0%
(A) $\rightarrow$ (i), (B) $\rightarrow$ (iii), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (ii)

Packing fraction in diamond is (approximately)

0%
$74\%$
0%
$78\%$
0%
$68\%$
0%
$34\%$

Packing fraction in

$2D-$ hexagonal arrangement of identical sphere is

0%
$\dfrac{\pi}{3\sqrt {2}}$
0%
$\dfrac{\pi}{3\sqrt {3}}$
0%
$\dfrac{\pi}{2\sqrt {3}}$
0%
$\dfrac{\pi}{6}$

If oxidation states of A , B and C respectively are

$+2, +3$ and

$-2$ then the possible formula of the compound is

0%
$A_3(B_2C)_2$
0%
$A_2(BC)_3$
0%
$B_3(AC_3)_2$
0%
$A_3(BC_3)_2$

The pink colour of lithium chloride crystal is due to:

0%
frenkel defect
0%
metal excess defect
0%
metal deficiency defect
0%
impurity defect

Explanation

$LiCl$ has non-stochiometric metal excess defect due to anion vacancies. The negative ions $(Cl^-)$ are missing from their lattice sites leaving the holes in which electrons are entrapped so that electrical neutrality is maintained.

When

$LiCl$ is heated,

$Li$ atoms gets deposited on the surface of the crystal. The

$Cl^-$ ions diffuse into the surface and combine with

$Li$ atoms to give

$LiCl$ . This is so because of loss of electrons by

$Li$ atoms to form

$Li^+$ . The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of

$Li$ . The anionic sites occupied by unpaired electrons are F-centers which imparts a pink color to

$LiCl$ crystals. The color is observed as a result of excitation of these electrons when they absorb energy from visible light falling on crystals.

Select correct for given arrangement.

0%
It has $3$ tetrahedron
0%
Each tetrahedron share its two vertex with other tetrahedron
0%
It is observed in $\gamma -SO_3$
0%
It has a six member ring

Molybdenum (At. wt.

$= 96 \ g/mol$ ) crystallizes as

$bcc$ crystal. If density of crystal is

$10.3 \ g/cm^3$ , then radius of

$Mo$ atom is: [Use :

$\,N_A = 6 \times 10^{23}$ ]

0%
$111$ pm
0%
$314$ pm
0%
$135.96$ pm
0%
none of these

Explanation

Since, Molybladenum crystallizes as

$bcc$ rystal. So,

$N = 2$ .

Expression for density is:

$\rho \quad =\quad \dfrac { N\times M }{ { N }_{ A }\times { a }^{ 3 } }$

${ a }^{ 3 }\quad =\quad \dfrac { N\times M }{ { N }_{ A }\times \rho }$

${ a }^{ 3 }\quad =\quad \dfrac { 2\times 96 }{ 6\times { 10 }^{ 23 }\times 10.3 }$

${ a }^{ 3 }\quad =\quad 3.1\times { 10 }^{ -23 }\quad { cm }^{ 3 }$

$a\quad =\quad 3.14\times { 10 }^{ -8 }\ cm\quad =\quad 314\quad pm$ .

For,

$bcc$ crystal,

$4r\quad =\quad \sqrt { 3 } a$

$r\quad =\quad \dfrac { \sqrt { 3 } }{ 4 } a$

$r\quad =\quad \dfrac { \sqrt { 3 } }{ 4 } \times 314$

$r\quad =\quad 136\quad pm$ .

In fcc, a tetrahedral void is formed by atoms at

0%
3 corners and 1 face centre
0%
3 face centres and 1 corner
0%
2 face centres and 2 corners
0%
2 face centres, 1 corner and 1 body centre

The unit cell with the structure above refer to which crystal system?

0%
Cubic
0%
Orthorhombic
0%
Tetragonal
0%
Trigonal

Explanation

For orthorhombic crystal structure all three sides are not equal i.e.

$a\neq b\neq c$ and

$\alpha=\beta=\gamma=90^o$

Therefore, it is an orthorhombic crystal system.

Option

$B$ is correct.

$KF$ has

$NaCl$ structure. What is the distance (in

$A^o$ ) between

$K^+$ and

$F^-$ in

$KF$ , if density of

$KF$ is

$2.48$

$gm/cm^3$ ?

$[K=39, F=19]$

0%
$2.86$
0%
$2.68$
0%
$3.16$
0%
$3.68$

Explanation

Effective number of KF molecules is 4.

$\rho=\cfrac{Z\times M_A}{N_A\times a^3}$

$\implies 2.48=\cfrac{4\times 58}{6.023\times 10^{23}\times a^3}$

$\implies a^3=155.3\times 10^{-24}$

$\implies a=5.36\times 10^{-8}$

$cm=5.36\mathring A$

$NaCl$ type stucture distance between cation and anion

$=\cfrac{a}{2}$

$\implies$ Distance between cation and anion

$=\cfrac{a}{2}=2.68\mathring A$

987599_1026340_ans_6522e52d1e214d56bb9c5e7820b2561e.PNG

Radii of

$A^+$ and that of

$X^-$ and

$Y^-$ have been given as

$A^+=1.00$ pm

$X^-=1.00$ pm

$Y^-=2.00$ pm
Determine the volume of unit cells of AX and AY crystals.

0%
$123.2\times10^{-36} m^3, 2160\times10^{-36} m^3$
0%
$12.32\times10^{-36} m^3, 216\times10^{-36} m^3$
0%
$22.62\times10^{-36} m^3, 76.36\times10^{-36} m^3$
0%
$93.23\times10^{-36} m^3, 142\times10^{-36} m^3$

Explanation

$i)$ For

$A^+X^-$

$r_{A+}=1.00$

$pm$

$r_{A-}=1.00$

$pm$

$\cfrac{r_{A^+}}{r_{X^-}}=1$ , then it is in CsCl type structure

In CsCl type structure,

$A$ is present at body centre and

$B$ is present at corner.

$2(r_{A^+}+r_{X^-})=a\sqrt{3}$

$a=\cfrac{4}{\sqrt{3}}$

$pm$

Volume

$=a^3=\cfrac{64}{3\sqrt{3}}\times 10^{-36}$

$m^3$ =

$12.32\times 10^{-36}$

$m^3$

$ii)$ For

$A^+Y^-$

$r_{A^+}=1.00$

$pm$

$r_{Y^-}=2.00$

$pm$

$\cfrac{r_{A^+}}{r_{Y^-}}=\cfrac{1.00}{2.00}=0.5$ . Then the crystal will have octahedral structure. (

$NaCl$ type)

$2(r_{A^+}+r_{Y^-})=a$

$a=6$

$pm$

Volume

$=(6)^3=216\times 10^{-36}$

$m^3$

Packing fraction in a body-centered cubic cell of crystals is:

0%
$\frac{{\sqrt 3 }}{8}\pi$
0%
$\frac{\pi }{6}$
0%
$\frac{{\sqrt 2 }}{6}\pi$
0%
$\frac{1}{{3\sqrt 2 }}\pi$

Explanation

We know that,

$\begin{array}{l}APF = \dfrac{{{N_{atoms}}{V_{atom}}}}{{{V_{unit\_cell}}}} = \frac{{6.\dfrac{4}{3}\pi {r^3}}}{{\dfrac{{3\sqrt 3 }}{2}{a^2}c}}\\ = \dfrac{{6.\frac{4}{3}\pi {r^3}}}{{\dfrac{{3\sqrt 3 }}{2}{{(2r)}^2}\sqrt {\dfrac{2}{3}} .4r}} = \dfrac{{6.\frac{4}{3}\pi {r^3}}}{{\dfrac{{3\sqrt 3 }}{2}.\sqrt {\dfrac{2}{3}} .16{r^3}}}\\ = \dfrac{\pi }{{\sqrt {18} }} = \dfrac{\pi }{{3\sqrt 2 }} \approx 0.74048048\end{array}$

hence option (D) is correct

Aluminium crystallizes in a cubic close packed structure. Its metallic radius is

$125pm$ .
i) Calculate the edge length of unit cell
ii) How many unit cells are there in

$1.00cm^3$ of aluminium?

0%
$354 pm, 2.41\times 10^{22}$
0%
$355 pm, 3.41\times 10^{20}$
0%
$356 pm, 4.41\times 10^{21}$
0%
$357 pm, 5.41\times 10^{23}$

Explanation

For a

$FCC$ or

$CCP$ unit cell,

$4r = \sqrt2 \times a$ or

$a = \dfrac{4r}{\sqrt2} = 354pm$

$\because$ Volume of one unit crystal

$= a^3 = 44361864\times 10^{-30}cm^3$

$\therefore$ No. of unit cells in

$1cm^3$ aluminium =

$\dfrac{1\times 10^{30}}{44361864}= 2.41\times 10^{22}$

$AB$ has

$NaCl$ structure while

$XB$ has

$CsCl$ structure if

${ r }_{ A }|{ r }_{ B }=\ 5$ and

${ r }_{ X }|{ r }_{ B }= 0.75$ then what is the ratio of edge length of unit cell of

$AB$ and

$XB$ .

0%
$1.52$
0%
$1.50$
0%
$1.40$
0%
$0.85$

Explanation

$AB\dfrac{r_{A}}{r_{B}}=0.5=\dfrac{1}{2}$

$XB\dfrac{r_{X}}{r_{B}}=0.75=\dfrac{1.5}{2}$

$r_{B=2}$

$r_{X=1.5}$

$r_{A=1}$

$XB$ form

$C_{5}Cl$ structure

$50\ B^{-}$

takes

$CCP$ porition

$4r=\sqrt{2}a$

$a=\dfrac{4r_{B}}{\sqrt{2}}=\dfrac{4\times 2}{\sqrt{2}}=\dfrac{8}{\sqrt{2}}$

As form

$NaCl$ structure

$50\ B^{-}$ takes simple cable structure

$a=2r_{B}=2_{X2}=4$

eape length of

$A-B$ : edge length of

$XB$

$\dfrac{8}{\sqrt{2}}:4=1.4$

The density of

$Ca{F_2}$ (fluorine structure) is

$3.18$ g /

${cm^{3}}$ . The length of the side of the unit cell is:

0%
$253$ pm
0%
$344$ pm
0%
$546$ pm
0%
$273$ pm

In a solid lattice, the cation has left a lattice site and is located at interstitial position, the lattice defect is ____

0%
Interstitial defect
0%
Vacancy defect
0%
Frenkel defect
0%
Schottky defect

The number of octahedral voids present in 234 g of NaCl is:

0%
$N_{A}$
0%
$4N_{A}$
0%
$16N_{A}$
0%
$\cfrac{N_{A}}{4}$

Crystal

$AB$ shows

$ZnS$ (zinc blend) type structure if the radius of

$B^{-}$ in

$50\ pm$ then the radius of

$A^{+}$ , if edge length of unit cell is

$160\ pm$ .

0%
$11.25\ pm$
0%
$27.94\ pm$
0%
$15.25\ pm$
0%
$19.28\ pm$

In a solid

$AB$ having

$NaCl$ structure atoms, occupy the corners of the unit cell. If all the face centred atoms along one of the axis are removed, then the resultant stoichiometry of the solid is:

0%
$AB_2$
0%
$A_2B$
0%
$A_4B_3$
0%
$A_3B_4$

Explanation

A occupies

$8$ corners which contributes

$\cfrac{1}{8}\Rightarrow \cfrac{1}{8}\times 8=1$

and

$6$ face centered contributes

$\cfrac{1}{2}\Rightarrow \cfrac{1}{2}\times 6=3$

But face centered atom along any one axis is removed which will be

$2$ .

Now, face centre have

$4\times \cfrac{1}{2}=2$

$Z_{effective}$ of

$A$ after removing face centres at any axis

$=4\times \cfrac{1}{2}+\cfrac{1}{8}\times 8=3$

$B$ occupies octahedral voids which contributes

$\cfrac{1}{4}$

Therefore,

$1+12\times \cfrac{1}{4}=4=Z_{effective}$ of

$B$

Formula

$=A_3B_4$

Which of the following has face-centred Bravais lattice?

0%
Hexagonal
0%
Monoclinic
0%
Cubic
0%
Orthorhombic

The ratio of a number of rectangular planes and diagonal plane in a cubic unit cell is ____.

0%
$1 : 2$
0%
$3 : 1$
0%
$2 : 3$
0%
$3 : 4$

Explanation

The ratio of rectangular plane and diagonal plane in cubic unit cell is

$3:6$ i.e,

$1:2$

There are 3 planes that lie parallel to the side of the cube,that these are rectangular planes.while 6 planes goes through opposite edges.And also there are diagonal planes.

And Hence, the ratio of rectangular plane and diagonal plane in cubic unit cell is

$3:6$ i.e,

$1:2$

A CCP of

$n$ spheres of radius

$r$ ,

$n$ spheres of radius

$0.414\, r$ and

$2n$ spheres of radius

$0.225\, r$ are introduced in the octahedral and tetrahedral void respectively. Therefore packing of the lattice is:

0%
$0.74$
0%
$0.81$
0%
$0.86$
0%
$0.79$

Explanation

$\begin{array}{l} \text { Given: - CCP or FCC packing radius=r } \\ \text { Zeffective }=\frac{1}{8} \times 8+\frac{1}{2} \times 6=4 \end{array}$

$\begin{array}{l} \text { Let radius of sphere of atom of Fcc be r } \\ \text { and side of } F c c \text { be a } \end{array}$

$\begin{aligned} \text { Then } &: a \sqrt{2}=4 \mathrm{~r} \\ & \Rightarrow a=2 \sqrt{2} \mathrm{r} \end{aligned}$

$\begin{array}{l} \because \text { There are n spheres, } 80 \text { total number of } \\ \text { unit cells = } \frac{n}{4} \\ \text { volume of Lattice }=\frac{n}{4} \times(a)^{3}=\frac{n}{4}(2 \sqrt{2} r)^{3} \\ =4 \sqrt{2} n r^{3} \end{array}$

$\begin{aligned} \text { volume of atoms }=& n \times \frac{4}{3} \times \pi r^{3}+n \times \frac{4}{3} \times \pi \times(0.414 \mathrm{~r})^{9} \\ +2 n \times \frac{4}{3} \times \pi(0.225 \mathrm{})^{3} \\ =& \frac{n \times 4}{3} \times r^{3} \times \pi\left[1+(0.414)^{3}+(0.225)^{9}\right] \\ =& 4.534 \mathrm{nr}^{3} \end{aligned}$

$\begin{array}{l} \therefore \text { packing of lattice }=\dfrac{4.534 \mathrm{nr}^{9}}{4 \sqrt{2} \mathrm{nr}^3}=0.81 \\ \text { So, option (B) is correct. } \end{array}$

Select right expression for determining Packing fraction (P.F.) of NaCl unit cell (assume ideal), if ions along an edge diagonal are absent ?

0%
$P.F. = \dfrac{\dfrac{4}{3}\pi(r_+^3 +r_-^3)}{16\sqrt{2}r_-^3}$
0%
$P.F. = \dfrac{\dfrac{4}{3}\pi(\dfrac{5}{2}r_+^3 +4r_-^3)}{16\sqrt{2}r_-^3}$
0%
$P.F. = \dfrac{\dfrac{4}{3}\pi(\dfrac{5}{2}r_+^3 +r_-^3)}{16\sqrt{2}r_-^3}$
0%
$P.F. = \dfrac{\dfrac{4}{3}\pi(\dfrac{7}{2}r_+^3 +r_-^3)}{16\sqrt{2}r_-^3}$

Arrange the following in decreasing order of melting point:

$NaCl,SiO_{2},Dry Ice, Ice$

0%
$NaCl>SiO_{2}>Dry Ice>Ice$
0%
$SiO_{2}>NaCl>Dry Ice>Ice$
0%
$NaCl>SiO_{2}>Ice>Dry Ice$
0%
$SiO_{2}>NaCl>Ice>Dry Ice$

The number of parameters that defines a unit cell is

0%
2
0%
3
0%
4
0%
6

Match the column 1 correctly with the column 2 given below

0%
A=2 B=1 C=3 D=4
0%
A=3 B=1 C=4 D=2
0%
A=3 B=2 C=1 D=4
0%
A=1 B=2 C=3 D=4

Given length of side of hexagonal unit cell is

$\dfrac { 100 }{ \sqrt { 2 } }$ pm. The volume of hexagonal unit cell is (in pm

$^{ 3 }$ ):

0%
${ 8\times 10 }^{ 6 }$
0%
${ 1.5\times 10 }^{ 6 }$
0%
${ 64\times 10 }^{ 6 }$
0%
${ 36\times 10 }^{ 6 }$

Amorphous solids are those which have:

0%
fixed arrangement of atoms and molecules
0%
no fixed melting point
0%
regular geometry
0%
no of the above

The ratio of the volume of tetragonal lattice to that of a hexagonal lattice is?

0%
$\frac { \sqrt { 3 } }{ 2 } abc$
0%
$\frac { 2 }{ \sqrt { 3 } }$
0%
$\frac { 2a^ 2 }{ \sqrt { 3b } } c$
0%
$\frac { \sqrt { 3a^ 2 } }{ 2bc }$

The incorrect statement among the following is:

0%
The fraction of total volume occupied by the atoms in a primitive cell is 0.48.
0%
Molecular solids are generally volatile.
0%
The number of carbon atoms in an unit cell of diamond is 8.
0%
The number of Bravais lattices in which a crystal can be categorized is 14.

CsBr crystallises in a body centred cubic lattice.The until cell length is 436.6pm.Given that the atomic mass of Cs=133 and that of Br=80 amu and Avogadro number being

$6.02\times { 10 }^{ 23 } { mol }^{ -1 }$ ,the density of

$CsBr$ is:

0%
$8.25{ g }/{ { cm }^{ 3 } }$
0%
$4.25{ g }/{ { cm }^{ 3 } }$
0%
$42.5{ g }/{ { cm }^{ 3 } }$
0%
$0.425{ g }/{ { cm }^{ 3 } }$

The close-packing sequence ABAB____represents:

0%
SC packing
0%
FCC packing
0%
Hexagonal packing
0%
BCC packing

An element crystallizing in the three Bravais Cubic lattices, the ratio of densities

$\rho_{P}: \rho_{1}: \rho_{F}$ is:

0%
0.125 : 0.163 : 0.177
0%
0.125 : 0.177 : 0.163
0%
0.177 :0.163 : 0.125
0%
0.163 :0.177 : 0.125

Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of

$390$ pm. If the size of chloride ion is

$180$ pm, the size of ammonium ion would be:

0%
$158$ pm
0%
$174$ pm
0%
$142$ pm
0%
$126$ pm

Schottky defect is observed in the crystal of:

0%
$AgCl$
0%
$NaCl$
0%
$MgCl_2$
0%
$TiCl$

Explanation

Schottky defect is a type of point defect in crystal lattice named after walter H scottky. In non-iodic crystals, it means a lattice vacany defect. In ionic-crystals, the defect forms when oppositely charged ions leave their lattice sites, creating vacancies.

Eg.

$AgBr,NaCl$

$\Rightarrow$ Schottky defect abserved in

$NaCl$ crystal.

An element exists as hexagonal close packed structure as well as cubic closed packed structure. In which case the element would have higher density?

0%
CCP
0%
HCP
0%
Same in both
0%
Unpredictable

Explanation

The packing efficiencies of both

$hcp$ and

$ccp$ is approximately same that is

$74$ % density of the element will be same in both the cases as two structures will have the same coordination number and hence, the same packing fraction and density.

Which of the following is /are found in the solid state?

0%
$\mathrm { LiHCO } _ { 3 }$
0%
$\mathrm { KHCO } _ { 3 }$
0%
$\mathrm { NaHCO } _ { 3 }$
0%
$\mathrm { NH } _ { 4 } \mathrm { HCO } _ { 3 }$

Maximum possible numbers of two dimensional and three dimensional lattices are respectively __________.

0%
5 and 14
0%
7 and 14
0%
14 and 4
0%
5 and 13

In FCC crystal, which of the following shaded planes the following type of arrangement of atoms

Crystalline allotrope of silicon has:

0%
Diamond like structure
0%
Graphite like structure
0%
fullerene like structure
0%
$s{p^2}$ hybridization

What is the packing efficiency of arrangement in a body centred unit cell?

0%
$53.26\%$
0%
$74.00\%$
0%
$68.00\%$
0%
$64.00\%$

Explanation

Packing efficiency is defined as the percentage of space occupied by constituent particles packed inside the lattice.

Consider edge length is

$a$

In BCC,

$AC=\sqrt{AB^2+BC^2}=\sqrt{a^2+a^2}=\sqrt2 a$

$AC=FD$

and

$AF=\sqrt{AD^2+FD^2}=\sqrt{2a^2+a^2}=\sqrt{3}a$

Thus

$\sqrt3 a=4r$ i.e.,

$a=\dfrac{4r}{\sqrt3}$

Volume of the unit cell=

$a^3=\left(\dfrac{4r}{\sqrt3}\right)^3=\dfrac{64r^3}{3\sqrt3}$

Now, No. of spheres per unit cell=

$8\times \dfrac18+1=1+1=2$

Thus volume of two spheres=

$2\times \dfrac43 \pi r^3=\dfrac83 \pi r^3$

$\therefore \text{Packing efficiency}=\dfrac{\dfrac83 \pi r^3}{\dfrac{64}{3\sqrt3}r^3}=0.68$

Thus,

$\%\text{Packing efficiency}=68\%$

1340237_1646223_ans_3f772657a4cd472e9971855668bfc836.png

Sodium crystallises in bcc arrangement with the interfacial sepration between the atoms of the edge

$53 pm$ . The density of the solid is :

0%
$1.23 g/cc$
0%
$485 g/cc$
0%
$4.85 g/cc$
0%
$123 g/cc$

Consider the bcc unit cells of the solids

$1$ and

$2$ with the position of atoms as shown below. The radius of atom

$B$ is twice that of atom

$A$ . The unit cell edge length is

$50 \%$ more in solid

$2$ than in

$1$ . What is the approximate packing efficiency in solid

$2$ ?

0%
$45 \%$
0%
$65 \%$
0%
$90 \%$
0%
$75 \%$

Explanation

Solution:- Given

$r_B=2r_A$

Packing fraction for Solid 2 is :

$p.f. = \dfrac{\left(z_{eff} \times \dfrac{4}{3}\pi r^3_A\right)_A + \left(z_{eff} \times \dfrac{4}{3} \pi r^3_B\right)_B}{a^3}$

$2(r_A + r_B) = \sqrt{3}a$

$\Rightarrow 2(r_A + 2r_A) = \sqrt{3}a$

$\Rightarrow 2\sqrt{3} r_A = a$

$\Rightarrow p.f. = \dfrac{1 \times \dfrac{4}{3}\pi r^3_A + \dfrac{4}{3}\pi (8r^3_A)}{8 \times 3\sqrt{3}r^3_A} = \dfrac{9 \times \dfrac{4}{3}\pi}{8 \times 3\sqrt{3}} = \dfrac{\pi}{2\sqrt{3}}$

p. efficiency

$= \dfrac{\pi}{2\sqrt{3}} \times 100 \simeq 90 \%$

Option C is correct.

Select the correct statement related to hexagonal close packing of identical spheres in three dimensions.

0%
In one unit cell, there are 12 octahedral voids and all are completely inside the unit cell.
0%
In one unit cell, there are six octahedral voids and all are completely inside the unit cell.
0%
The number of atoms per unit cell in hcp are 12
0%
Coordination number of every sphere is 8

Select the correct statement(s) related to hexagonal close packing of identical spheres in three dimensions.

0%
In one unit cell, there are 12 octahedral voids and all are completely inside the unit cell
0%
In one unit cell, there are six octahedral voids and all are completely inside the unit cell
0%
In one unit cell, there are six octahedral void and of which three are completely inside the unit cell and other three are partially inside the unit cell
0%
Co-ordination number of every sphere is 12 in hcp lattice.

In a cubic lattice if atoms are present only at all the corners and alternate face centres then the packing efficiency will be:

0%
$74 \%$
0%
$37\%$
0%
$68\%$
0%
Cannot be determined

Crystals of different ionic compounds having similar arrangement of ions as well as geometry are known:

0%
Isomorphs
0%
Isomers
0%
Anomers
0%
Isotopes

A lattice is defined as

0%
the amount of energy required, per mole, to separate the ions from their lattice positions to an infinte distance in the gas phase
0%
the distance separating the cations and anions
0%
a set of all points with identical environments within crystal
0%
the arrangement of electrons in various energy levels

Packing fraction in FCC lattice is

0%
$\cfrac { 1 }{ 6 } \pi \quad$
0%
$\cfrac { \sqrt { 2 } }{ 6 } \pi$
0%
$\cfrac { \sqrt { 3 } }{ 8 } \pi \quad$
0%
$\cfrac { \sqrt { 2 } }{ 3 } \pi$

Explanation

For a face-centered cubic unit cell, the number of atoms is four. A line can be drawn from the top corner of a cube diagonally to the bottom corner on the same side of the cube, which is equal to 4r. Using geometry, and the side length, a can be related to r as:

$a = 2r\sqrt{2}$

Packing fraction =

$\dfrac{Number \ of\ atom \times volume \ of \ atom}{volume \ of \ unit \ cell}$

$\Rightarrow \dfrac{4 \times \frac{4}{3} \pi r^{3}}{(2\sqrt{2}r)^{3}}$

$\therefore$ Packing fraction =

$\dfrac{\pi \sqrt{2}}{6}$

Sodium oxide as anti-fluorite structure. The percentage of the tetrahedral voids occupied by the sodium ions is

0%
$12$ %
0%
$25$ %
0%
$50$ %
0%
$100$ %

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 14 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 14 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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