In the table given below, dimensions and angles of various crystals are given. Complete the table by filling the blanks. Types of crystal Dimensions Angles 1 Cubic $$a = b = c$$ $$\alpha =\beta = \gamma =$$____(P) 2 Tetragonal ______(q) $$\alpha =\beta = \gamma=90^o$$ 3 Orthorhombic $$a \neq b \neq c ______(r) 4 Hexagonal ______(s) $$\alpha = \beta = 90^o,\gamma = $$______(t)

$$(p) 90^o,\quad (q) a=b\neq c, \quad (r) \alpha = \beta=\gamma=90^o, \quad (s) a = b\neq c, \quad (t) - 120^o$$

$$(p)120^o, \quad (q) a=b=c, \quad (r) \alpha = 90^ o, \beta = \gamma =120^0, \quad (s) a \neq b \neq c, \quad (t) 90^o$$

$$(p)90^o, \quad (q) a\neq b=c, \quad (r) \alpha = \beta=\gamma=120^o, \quad (s) a \neq b \neq c, \quad (t) 90^o$$

$$(p)120^o,\quad (q) a \neq b \neq c, \quad (r) \alpha \neq \beta \neq \gamma \neq90^o, \quad (s) a \neq b = c, \quad (t) 120^o$$

Match the types of packing given in column I with the items given in column II. Column I Column II (i) Square close packing in two dimension (p) Triangular voids (ii) Hexagonal close packing in two dimensions (q) Pattern of spheres is repeated in every fourth layer (iii) Hexagonal close packing in three dimension (r) Coordination number=4 (iv) Cubic close packing in three dimension (s) Pattern of spheres is repeated in alternate layers

(i)- (r), (ii)- (p), (iii)- (s), (iv)- (q)

(i)- (p), (ii)- (r), (iii)- (q), (iv)- (s)

(i)- (q), (ii)- (s), (iii)- (p), (iv)- (r)

(i)- (r), (ii)- (p), (iii)- (q), (iv)- (s)

MCQ Questions for Class 12 Engineering Chemistry The Solid State Quiz 13

Absorption of photons by crystal

0%
has no effect on imperfection
0%
produce atomic displacement leading to imperfections
0%
decreases number of defects
0%
none of the above

Metallic gold crystallises in fcc lattice. The edge length of the cubic unit cell is

$4.07$

$\mathring A$ . What is the density (in g cm

$^{-3}$ ) of gold?

0%
$10.2$
0%
$19.4$
0%
$7.4$
0%
$13.6$

Explanation

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

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

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

$=407 \mathrm{pm}$

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

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

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

The density of solid argon is

$1.65$ g mL

$^{-1}$ at

$-233^\circ C$ . If the argon atom is assumed to be the sphere of radius

$1.54\times10^{-8}$ cm, what percentage of solid argon is apparently empty space?

[Atomic mass of Ar is

$40$ amu.]

0%
$62\%$
0%
$76\%$
0%
$68\%$
0%
$52\%$

Explanation

Volume of one atom of

$\displaystyle Ar=\frac{4}{3}\pi r^3$

Also, number of atoms in

$1.65$ g or in

$\displaystyle1$ mL

$=\dfrac{1.65}{40}\times6.023\times10^{23}$

$\therefore$ Total volume of all atoms of

$Ar$ in solid state

$\displaystyle=\frac{4}{3}\pi r^3\times\frac{1.65}{40}\times6.023\times10^{23}$

$\displaystyle=\frac{4}{3}\times\frac{22}{7}\times(1.54\times10^{-8})^3\times\frac{1.65}{40}\times6.023\times10^{23}$

$=0.380$ cm

$^3$

Volume of solid argon

$=1$ cm

$^3$

$\therefore\%$ Empty space

$\displaystyle=\frac{[1-0.380]}{1}\times100=62\%$

Packing factors for a body centred cubic and simple cubic structure are, respectively :

0%
$0.68$ and $0.524$
0%
$0.524$ and $0.68$
0%
$0.48$ and $0.52$
0%
$0.52$ and $0.48$

Explanation

$\text{Body centred cubic}:$

Body centred cubic lattice has radius

$=r=\displaystyle\cfrac{\sqrt3a}{4}$

The unit cell of body centred cubic has

$2$ atoms on the average.

$\therefore$ Volume of atoms

$=\displaystyle2\times \cfrac{4}{3}\pi r^3$

$=\displaystyle2\times\cfrac{4}{3}\times\pi\times\begin{bmatrix}\cfrac{\sqrt3a}{4}\end{bmatrix}^3=\cfrac{\sqrt3}{8}\pi{a^3}$

The volume of unit cell

$=a^3$

$\therefore$ Packing factor

$\displaystyle=\cfrac{\sqrt3}{8}\cfrac{\pi a^3}{a^3}=0.68$

$\text{Simple Cubic}:$

Simple cubic has radius

$=\displaystyle r=\cfrac{a}{2}$

The unit cell of sc has one atom.

$\therefore$ Volume of atom

$=\displaystyle\cfrac{4}{3}\pi\begin{pmatrix}\cfrac{a}{2}\end{pmatrix}^3=\cfrac{4\pi a^3}{24}$

The volume of unit cell

$=a^3$

$\therefore$ Packing factor

$=\displaystyle\cfrac{4}{24}\cfrac{\pi a^3}{a^3}=0.524$

Hence, the correct option is

$\text{A}$

An element (atomic mass

$= 100\ g / mol$ ) having bcc structure has unit cell edge

$400\ pm$ . Then, density of the element is:

0%
$10.376\ g/cm^{3}$
0%
$5.188\ g/cm^{3}$
0%
$7.289\ g/cm^{3}$
0%
$2.144\ g/cm^{3}$

Explanation

$\rho = \dfrac {Z\times M}{N_{A} \times a^{3}} = \dfrac {2\times 100}{6.023\times 10^{23} \times (400 \times 10^{-10})^{3}}$

$= 5.188\ g/cm^{3}$

The density of solid argon is

$1.65 g$ per cc at

$-233$ . If the argon atom is assumed to be a sphere of radius

$1.54\times { 10 }^{ -8 } cm$ , what percent of solid argon is apparently empty space?

$\left( Ar=40 \right)$

0%
$16.5$ %
0%
$38$ %
0%
$50$ %
0%
$62$ %

Explanation

Volume of one molecule

$=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }=\dfrac { 4 }{ 3 } \pi { \left( 1.54\times { 10 }^{ -8 } \right) }^{ 3 }{ cm }^{ 3 }$

$=1.53\times { 10 }^{ -23 }{ cm }^{ 3 }$
Volume of molecules in

$1.65 g$

$Ar$

$=\dfrac { 1.65 }{ 40 } \times { N }_{ 0 }\times 1.53\times { 10 }^{ -23 }=0.380{ cm }^{ 3 }$
Volume of solid containing

$1.65 g$

$Ar$

$=1{ cm }^{ 3 }$

$\therefore$ Empty space

$=1 - 0.380=0.620$

$\therefore$ Percent of empty space

$=62$ %

What is not a type of crystal?

0%
Ionic
0%
Covalent network
0%
Metallic
0%
Covalent molecular
0%
Amorphous

The percentage packing efficiency of the two dimensional arrangement of sphere for plane ABCDEF shown below is :

0%
$90.64\%$
0%
$74.05\%$
0%
$68.02\%$
0%
$78.54\%$

Select the correct statement(s) about cubic structure of diamond.
Effective number of atoms present in a diamond cubic cell is

$8$ .
Each carbon is surrounded by four more carbon atoms.
Atomic radius of carbon is

$\displaystyle \frac{\sqrt{3}\times a}{8}$ , where a is edge length of cube.
Approximate packing fraction of unit cell is

$\displaystyle \frac{\sqrt{3}\times \pi }{16}$ .

0%
1, 2 and 4
0%
2, 3 and 4
0%
2 and 3
0%
All 1, 2, 3 and 4

Explanation

All the four statements about cubic structure of diamond are correct.

1. Effective number of atoms present in a diamond cubic cell is

$8$ .

2. Each carbon is surrounded by four more carbon atoms. Thus each C atom is present at the centre of tetrahedron and four other C atoms are present at the corners of tetrahedron.

3. Atomic radius of carbon is

$\displaystyle \frac{\sqrt{3}\times a}{8}$ , where a is edge length of cube.

4. Approximate packing fraction of unit cell is

$\displaystyle \frac{\sqrt{3}\times \pi }{16}$ which approximates to 34%. In other words, in the unit cell of diamond, 34% of volume is occupied and 66% of volume is empty.

The hcp and ccp structure for a given element would be expected to have_______.

0%
same coordination number
0%
same density if made up of same element
0%
same packing fraction
0%
same number of atoms in single unit cell

An element (atomic mass

$100\ g/mol$ ) having

$BCC$ structure has unit cell edge

$400\ pm$ . The density of element is:

(No. of atoms in

$BCC, Z = 2$ )

0%
$2.144\ g/cm^{3}$
0%
$7.289\ g/cm^{3}$
0%
$5.188\ g/cm^{3}$
0%
$10.376\ g/cm^{3}$

Explanation

Mass of two atoms

$= \dfrac {100}{6.02\times 10^{23}} \times 2\ g$

$= \dfrac {2}{6.02} \times \dfrac {10^{-21}}{10^{3}} kg$

volume of cell

$= (4\times 10^{-10})^{3} = 64\times 10^{-30}\ m^3$

$Density = \dfrac {mass}{volume} = \dfrac {2\times 10^{24}}{6\times 10^{-30} \times 6.02}$

$Density = \dfrac {2}{6.02\times 64} \times 10^{6} kg/ m^{3}$

$= \dfrac {2\times 10^{6} \times 10^{3}}{6.02\times 64 \times 10^{6}} gm/ cc$

$= \dfrac {2}{6.02\times 64} \times 10^{3} gm/ cc = 5.188/ gm/cc$

The crystal system of a compound with unit cell dimension

$"a = 0.387, b = 0.387$ and

$c = 0.504\ nm$ and

$\alpha = \beta = 90^{\circ}$ and

$\gamma = 120^{\circ}"$ is :

0%
cubic
0%
hexagonal
0%
orthohombic
0%
rhombohedral

Explanation

Thus it is hexagonal crystal system as satisfies the criterion

$a = b\neq c, \alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$
These are the characteristics of a hexagonal system.

The vacant space in a

$bcc$ unit cell is:

0%
$23\%$
0%
$32\%$
0%
$26\%$
0%
$48\%$

Explanation

Packing fraction of BCC unit cell is 68%. Therefore, the vacant space in BCC unit cell is

$32%$ .

The efficiency of packing is

$68\%$ in:

0%
hcp structure
0%
ccp structure
0%
fcc structure
0%
bcc structure

Explanation

$\begin{array}{l}\text{ FCC or CCP packing efficiency }=74\%\\\text{ HCP Packing Efficiency }=74\%\\\text{ BCC Packing Efficiency }=68\%\end{array}$

If the distance between

$Na^+$ and

$Cl^-$ ions in

$NaCl$ crystal is

$X$ -pm, the length of edge of the unit cells (in pm) is:

0%
$\displaystyle\frac{X}{2}$
0%
$X$
0%
$2X$
0%
$4X$

Explanation

$NaCl$ ,

Edge length

$=2(r^+_{Na}+r^-_{Cl})$

$=2\times$ distance between

$Na^+$ and

$Cl^-$ ions

$=2X$

Edge length

$=2X$ .

Certain crystals produce electric signals on application of pressure. This phenomenon is called:

0%
Ferroelectricity
0%
Ferrielectricity
0%
Pyroelectricity
0%
Piezoelectricity

Which of the following compounds represents an inverse

$2 : 3$ spinel structure?

0%
$Fe^{III} [Fe^{II} Fe^{III}]O_{4}$
0%
$PbO_{2}$
0%
$Al_{2}O_{3}$
0%
$Mn_{3}O_{4}$

Explanation

Inverse spinel is generally denoted as

$A^{+2} {B_2}^{+3}O_4$

$A^{+2}$

$\cfrac {1}{8}$ tetrahedral voids

$A^{+3}$

$\cfrac {1}{8}$ tetrahedral voids +

$\cfrac {1}{4}$ octahedral voids

$O^{2-}$ at FCC lattice points.

$Fe^{+3}[Fe^{+3}Fe^{+2}]O_4$ or

${Fe}_3O_4$

$Fe^{+3}$ present at

$\cfrac {1}{8}$ tetrahedral void and

$\cfrac {1}{4}$ octahedral void

${Z_{Fe}}^{+3}=2$

$Fe^{+2}$ present at

$\cfrac {1}{8}$ tetrahedral void

${Z_{Fe}}^{+2}=8\times \cfrac {1}{8}=1$

$O^{2-}$ at FCC lattice points

$Z_{eff}=4$

Formula:

${Fe}^{+2} {Fe}_2^{+3}O_4\Rightarrow {Fe}_{3}O_4$

Hence, the correct option is

$\text{A}$

At room temperature, sodium crystallises in body-centred cubic lattice with

$a = 4.24\overset {\circ}{A}$ .

Calculate the theoretical density of sodium. (Atomic mass of

$Na = 23.0$ )

0%
$1.002 \ g \ cm^{-3}$
0%
$2.002 \ g \ cm^{-3}$
0%
$3.002 \ g \ cm^{-3}$
0%
$None$

Explanation

A body-centred cubic unit cell contains

$8$ atoms at the

$8$ corners and

$1$ in the centre.
Hence,
Total number of atoms in a unit cell

$= 8\times \dfrac {1}{8} + 1 = 2$

Volume of unit cell

$= a^{3} = (4.24 \times 10^{-8})^{3} cm^{3}$

So,

$Density = \dfrac {Z\times M}{N_{0}\times V}$

$= \dfrac {2\times 23}{(6.023\times 10^{23})(4.24\times 10^{-8})^{3}}$

$= 1.002\ g\ cm^{-3}$ .

Option A is correct.

Given an alloy of Cu, Ag, and Au in which Cu atoms constitute the CCP arrangement. If the hypothetical formula of the alloy is

$Cu_{4}Ag_{3}Au$ . What are the probable locations of Ag and Au atoms?

0%
Ag - all Tetrahedral voids; Au - all Octahedral voids
0%
Ag - 3/8th Tetrahedral voids; Au - 1/4th Octahedral voids
0%
Ag - 1/2 Octahedral voids; Au - 1/2 Tetrahedral voids
0%
Ag - all Octahedral voids; Au - all tetrahedral voids

Explanation

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.

Ice crystallises in a hexagonal lattice. At the, low temperature, the lattice constants were

$a = 4.53 A^{\circ}$ and

$b = 7.41A^{\circ}$ .

How many

$H_2O$ molecules are contained in a unit cell?

$[d(ice) = 0.92 g /cm^3]$

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

Explanation

The volume of the unit cell,

$\displaystyle V = a^2bsin60^o$

$\displaystyle V= (4.53 \times 10^{-8})^2 \times 7.41 \times 10^{-8} \times 0.866$

$(1 \mathring {A} = 10^{-8} cm)$

$\displaystyle V=1.317 \times 10^{-22} \: cm^3$

The density is

$\rho= \displaystyle 0.92 \: g/cm^3$

$N_A = 6.023 \times 10^{23} /mol, M_{H_2O}= 18 g/mol$

$\rho =\cfrac{ZM}{VN_A} \Rightarrow Z= \cfrac {\rho V N_A}{M}$

Putting all the given values in the above equation we get,

No. of molecules per unit cell,

$Z=4$

Argon crystallises in fcc arrangement and the density of solid and liquid

$Ar$ is 1.59

$g /$

$cm^3$ and 1.42

$g /$

$cm^3$ , respectively . The percentage of empty space in liquid

$Ar$ is:

0%
34.84%
0%
43.8%
0%
23.4%
0%
21.6%

Explanation

Let the volume of solid

$Ar = 100$ mL

Mass of solid

$Ar = V$

$\times \rho = 100 \times 1.59 g / cm^3 = 159 g$
Volume of liquid

$Ar$

$= \dfrac {Mass}{\rho} = \dfrac {159 g}{1.4 gcm^{-3}}$
= 113.57 ml

$\because$

$Ar$ crystallises in fcc type lattice, the packing fraction = 0.74

Actual volume occupied by

$Ar$

= Packing fraction

$\times$ volume of solid

$Ar$

= 0.74 x 100 = 74 mL

$\therefore$ % empty in liquid Ar

$= \dfrac {(113.57-74) \times 100}{113.57}$

= 34.84%

The packing efficiency of the two-dimensional square unit cell shown below is:

0%
$39.27$ %
0%
$68.02$ %
0%
$74.05$ %
0%
$78.54$ %

Explanation

$L\sqrt {2} = 4r; L = 2\sqrt {2}r$

Packing fraction

$= \dfrac {Occupied\ area}{Total\ area} \times 100$

$= \dfrac {2\pi r^{2}}{(2\sqrt {2}r)^{2}}\times 100 = 78.5$ %.

The packing efficiency of the face centered cubic (fcc), body centered cubic (bcc) and simple primitive cubic (pc) lattices follows the order:

0%
$fcc > bcc > pc$
0%
$bcc > fcc > pc$
0%
$pc > bcc > fcc$
0%
$bcc > pc > fcc$

Explanation

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 $\text {fcc > bcc > pc}$ .

Which one of the following is purely a crystalline compound?

0%
Rock Salt
0%
Ice
0%
Quartz silica
0%
Dry Ice

How many defects exists in the arrangement of constituent particles of

$7.45\ g$

$KCl$ ?

$[K=39,Cl=35.5gm/mole]$

0%
$10\times { 10 }^{ 23 }$
0%
$1\times { 10 }^{ 6 }$
0%
$1.0\times { 10 }^{ -6 }$
0%
$1.0\times { 10 }^{ 4 }$

CsBr ha cubic structure with edge length

$4.4\overset{o}{A}$ . The shortest interionic distance between

$Cs^{+}$ and

$Br^-$ is:

0%
$3.82\overset{o}{A}$
0%
$1.86\overset{o}{A}$
0%
$7.44\overset{o}{A}$
0%
$4.3\overset{o}{A}$

Explanation

$CsBr$ 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 $Br^-= \cfrac {\sqrt 3a}{2}$

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

Percentage of free space in a body centred cubic unit cell is

0%
30%
0%
32%
0%
34%
0%
28%

Explanation

Packing fraction

$\displaystyle =\frac { volume\, occupied \,by\, atoms\, in \,a \,unit\, cell } {volume \,of\, the\, unit\, cell }$

For bcc,

Packing fraction

$\displaystyle =\dfrac {2\times \dfrac{4}{3}\pi r^3}{\left ( \dfrac{4r}{\sqrt{3}} \right )^3}=\dfrac{\sqrt{3}\pi}{8}=0.68$

Volume occupied

$= 68\%$

Volume vacant

$=100 - 68 = 32\%$

$AB$ has

$ZnS$ type structure. What will be interionic distance between

${A}^{+}$ and

${B}^{-}$ if lattice constant of

$AB$ is

$200pm$ ?

0%
$50\sqrt {3}$
0%
$100\sqrt{3}$
0%
$100$
0%
$\cfrac{100}{\sqrt{2}}$

Explanation

$ZnS$ has cubic packing (ccp) crystal lattice structure which is analogous to the face-centered cubic lattice (FCC)

$ZnS$ , anions form FCC and cations occupy alternate tetrahedral voids.

Given, lattice constant

$=$ edge length of a cube

$=a=200 \ pm$

Now, tetrahedral voids are present on the body diagonal at

$\left (\cfrac 14 \right)^{th}$ of the distance from each corner.

So, distance between

$A^+$ and

$B^-= \cfrac 14 \times \sqrt 3 a$

$= \cfrac {\sqrt 3}4 \times 200 \ pm = 50 \sqrt 3 \ pm$

A compound formed by elements X and Y crystallizes in the cubic structure, where X atoms are at the corners of a cube and Y atoms are at the centre of the body. The formula of the compound is?

0%
XY
0%
$XY_2$
0%
$X_2Y_2$
0%
$XY_3$

Explanation

$X=8\times \displaystyle\frac{1}{8}=1$ (at corner)

$Y=1\times 1=1$ (at body center)

So the formula of compound is

$=$ XY.

${ Al }^{ 3+ }$ ions replace

${ Na }^{ + }$ ions at the edge centres of

$NaCl$ lattice, then the number of vacancies in 1 mole of

$NaCl$ will be:

0%
$3.01{ \times 10 }^{ 23 }$
0%
$6.02{ \times 10 }^{ 23 }$
0%
$12.04{ \times 10 }^{ 23 }$
0%
$12.04{ \times 10 }^{ 21 }$

Explanation

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

A metallic element exist as cubic lattice Each edge of the unit cell is

$4.0 \mathring{A}$ .The density of the metal is

$6.25 g/cm^3$ .How many unit cells will be present in

$100 g$ of the metal?

0%
$1.0\times 10^{22}$
0%
$2.5\times10^{23}$
0%
$5.0\times10^{23}$
0%
$2.0\times10^{23}$

Choose the correct statements.

0%
equivalent points in unit cells of a periodic lattice lie on a Bravais lattice
0%
equivalent points in unit cells of a periodic lattice do not lie on a Bravais lattice
0%
There are four Bravais lattices in two dimensions
0%
There are five Bravais lattices in three dimensions

There are three ionic compounds, first compound has fcc lattice of anions and cations are present in all tetrahedral voids, second compound has fcc lattice of anions and cations are present in all octahedral voids and the third compound has simple cubical lattice in which cubical void is occupied by cation, then the order of packing efficiency of these will be :

0%
$I > II > III$
0%
$II > I > III$
0%
$III > II > I$
0%
$III > I > II$

Explanation

Since packing efficiency of various crystal systems are

Crystal System	Packing Efficency
FCC	$74$ %
Simple Cubic	$52.4$ %

Now, number of tetrahedral voids

$= 2 \times$ Number of octahedral voids.

Since the crystal system (III) is a simple cubic lattice, it will have the least packing efficiency.

Now, both crystal systems

$I$ and

$II$ are FCC but the number of voids is filled in crystal system I than II because of a greater number of tetrahedral voids than octahedral voids. So, crystal system I have the most packing efficiency.

A nugget of gold and quartz was found to contain x g of gold and y g of quartz and has density d if the densities of gold and quartz are

$d_{1}$ and

$d_{2}$ respectively then the correct relation is :

0%
$\dfrac{x}{d_{1}} + \dfrac{y}{d_{2}} = \dfrac{x + y}{d}$
0%
$xd_{1} + yd_{2} = (x + y) d$
0%
$\dfrac{x}{d_{2}} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d}$
0%
$\dfrac{x + y}{d} + \dfrac{x}{d_{1}} + \dfrac{x}{d_{2}} = 0$

Explanation

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

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

If x is the length of body diagonal, then the distance between two nearest cations in rock salt structure is:

0%
$\dfrac { X }{ \sqrt { 6 } }$
0%
$\dfrac { X }{ \sqrt { 5 } }$
0%
$\dfrac { X }{ \sqrt { 3 } }$
0%
$\dfrac { X }{ \sqrt { 2 } }$

Explanation

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)

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

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

The packing efficiency of the two- dimensional square unit cell shown below is :

0%
39.27%
0%
68.2 %
0%
74.05%
0%
78.54%

Explanation

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 $= \cfrac {Area \ occupied}{Total area} \times 100$ %

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

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

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

$\Rightarrow Packing Efficiency = 78.54$ %

An element crystallizes in a structure having a

$fcc$ unit cell of an edge length

$200\ pm$ . If

$200\ g$ of this element contains

$24\times10^{23}$ atoms then its density is:

0%
41.66 g cm $^{-3}$
0%
313.9 g cm $^{-3}$
0%
8.117 g cm $^{-3}$
0%
400 g cm $^{-3}$

Explanation

Theoretical density of crystal,

$\rho = \cfrac {z\times M}{N_o\times a^3} \ g/cm^3$

Given that,

Edge length,

$a= 200 \times 10^{-10} \ cm$ and

$24 \times 10^{23}$ atoms weigh

$200 \ g.$

So, weight of

$6.02 \times 10^{23}$ atoms (Avogadro’s Number)

$= \cfrac {6.02 \times 10^{23}}{24 \times 10^{23}} \times 200 \approx 50.16 \ g$

So, molar mas of element,

$M= 50.16 \ g/mol$

Now, for fcc,

$z=4 \$

$\therefore \text{Density, } \rho = \cfrac {4 \times 50.16}{6.02 \times 10^{23} \times (200)^3 \times 10^{-30}}= 41.66 \ g/cm^3$

Beautiful crystals arise from various chemical substances. A single chemical can, sometimes, give rise to different crystals because of different:

0%
angles of vision
0%
arrangement of atoms in space
0%
composition
0%
weight of atoms

In the table given below, dimensions and angles of various crystals are given. Complete the table by filling the blanks.

	Types of crystal	Dimensions	Angles
1	Cubic	$a = b = c$	$\alpha =\beta = \gamma =$ ____(P)
2	Tetragonal	______(q)	$\alpha =\beta = \gamma=90^o$
3	Orthorhombic	$$a \neq b \neq c	______(r)
4	Hexagonal	______(s)	$\alpha = \beta = 90^o,\gamma =$ ______(t)

0%
$(p) 90^o,\quad (q) a=b\neq c, \quad (r) \alpha = \beta=\gamma=90^o, \quad (s) a = b\neq c, \quad (t) - 120^o$
0%
$(p)120^o, \quad (q) a=b=c, \quad (r) \alpha = 90^ o, \beta = \gamma =120^0, \quad (s) a \neq b \neq c, \quad (t) 90^o$
0%
$(p)90^o, \quad (q) a\neq b=c, \quad (r) \alpha = \beta=\gamma=120^o, \quad (s) a \neq b \neq c, \quad (t) 90^o$
0%
$(p)120^o,\quad (q) a \neq b \neq c, \quad (r) \alpha \neq \beta \neq \gamma \neq90^o, \quad (s) a \neq b = c, \quad (t) 120^o$

Explanation

For, cubic crystal,

$a=b=c\quad \& \quad \alpha =\beta =\gamma =90^o$ . So, P is

$90^o$ .

For tetragonal crystal,

$a=b\neq c\quad \& \quad \alpha =\beta =\gamma =90^o$ . So, q is

$a=b\neq c$ .

For orthorhombic crystal,

$a=b=c\quad \& \quad \alpha =\beta =\gamma =90^o$ . so, r is

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

For Hexagonal crystal,

$a=b\neq c\quad \& \quad \alpha =\beta =90^o\quad \& \quad \gamma =120^o$ . So, s is

$a=b\neq c$ and t is

$120^o$ .

Which of the following sets of axial angles and axial lengths represent the maximum number in Bravais lattices?

0%
$\alpha = \beta = \gamma = 90^0$ and $a=b\neq c$
0%
$\alpha = \beta = \gamma = 90^0$ and $a\neq b\neq c$
0%
$\alpha = \beta = \gamma \neq 90^0$ and $a=b= c$
0%
$\alpha = \beta = \gamma = 90^0$ and $a=b= c$

$R$ is the radius of the sphere in the close packed arrangement and

$r$ is the radius of the tetrahedral void then :

0%
$R=0.224r$
0%
$r=0.225R$
0%
$r=0.414R$
0%
$R=0.414r$

Which one is called pseudo solid?

0%
$CaF_2$
0%
$Glass$
0%
$NaCl$
0%
All of these

The distance between

$Na^+$ and

$Cl^-$ ions in

$NaCl$ with a density

$3.165$ g

$cm^{-3}$ is:

0%
$300$ pm
0%
$248.5$ pm
0%
$248.5$ cm
0%
$24.85$ pm

Explanation

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 $Cl^-= \cfrac a2= \cfrac {497}{2}=248.5 \ pm$

Experimentally it was found that a metal oxide has formula

$M_{0.98}O.$ Metal

$M$ , is present as

$M^{2+}$ and

$M^{3+}$ in its oxide. Fraction of the metal which exists as

$M^{3+}$ would be:

0%
5.08%
0%
7.01%
0%
4.08%
0%
6.05%

Explanation

Since the oxidation state of oxygen is

$-2$ . So, for

${ M }_{ 0.98 }0$ to be neutral, the total oxidation state of

${ M }_{ 0.98 }$ has to be

$+2$ .

Let the fraction of

${ M }^{ 3+ }$ be

$x$ .

Then fraction of

${ M }^{ 2+ }$ will be

$(0.98-x)$ .

Now for the compound to be neutal,

$3x+2(0.98-x)=2\\ 3x+1.96-2x=2\\ x=2-1.96\\ x=.04$

So, fraction of

${ M }^{ 3+ }$ will be 4%.

So, the correct answer is option

$C$ .

In the cubic close packing, this unit cell has:

0%
4 tetrahedral voids each of which is shared by four adjacent unit cells.
0%
4 tetrahedral voids within the unit cell.
0%
8 tetrahedral voids each of which is shared by four adjacent unit cell.
0%
8 tetrahedral voids within the unit cells.

Explanation

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.

Which of the following is an example for interstitial solid solution?

0%
$SiC$
0%
$TiC$
0%
$Li_{2}C_{2}$
0%
$Al_{4}C_{3}$

Explanation

$\begin{array}{l} \text { Interstitial solid solution - Interstitial solid solution } \\ \text { forms when the solute atom is small enough to } \\ \text { fit at interstitial sites between the solvent atoms. } \\ \text { example- TiC } \end{array}$

Which of the following points defects are shown by

$AgBr_{(s)}$ crystals?
(I) Schottky defect
(II) Frenkel defect
(III) Metal defect
(IV) Metal deficiency defect

0%
(I) and (II)
0%
(III) and (IV)
0%
(I) and (III)
0%
(II) and (IV)

Explanation

$AgBr$ shows both Schottky and Frenkel defects. In

$AgBr$ , both

$Ag^+$ and

$Br^-$ ions are absent from the lattice causing a Schottky defect. However,

$Ag^+$ ions are mobile so they have a tendency to move aside the lattice and trapped in the interstitial site, hence cause Frenkel defect.

Hence, Option "A" is the correct answer.

Match the types of packing given in column I with the items given in column II.

	Column I		Column II
(i)	Square close packing in two dimension	(p)	Triangular voids
(ii)	Hexagonal close packing in two dimensions	(q)	Pattern of spheres is repeated in every fourth layer
(iii)	Hexagonal close packing in three dimension	(r)	Coordination number=4
(iv)	Cubic close packing in three dimension	(s)	Pattern of spheres is repeated in alternate layers

0%
(i)- (p), (ii)- (r), (iii)- (q), (iv)- (s)
0%
(i)- (q), (ii)- (s), (iii)- (p), (iv)- (r)
0%
(i)- (r), (ii)- (p), (iii)- (s), (iv)- (q)
0%
(i)- (r), (ii)- (p), (iii)- (q), (iv)- (s)

Explanation

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.

The pattern of arrangement in hcp and ccp in 3D is shown in the figure.

The unit cell with the given structure and a = b = c, represents ____________ crystal system.

0%
cubic
0%
orthorhombic
0%
tetragonal
0%
trigonal

Explanation

$a=b=c$ and

$\alpha = \beta = \gamma = 90^o$ , the given structure will be a cubic crystal system.

Image I :

$a = b = c$

all angles

$= 90^o$ (Isometric ) cubic

Image II :

$a \neq b \neq c$

all angles

$= 90^o$

Orthorhombic

1171359_936234_ans_f23667f85d954800b18c6f71e66dc014.png

$N{a_2}O$ has:

0%
NaCl structure
0%
ZnS structure
0%
fluorite structure
0%
antifluorite structure

Explanation

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

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

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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