MCQ Questions for Class 12 Engineering Chemistry The Solid State Quiz 6

The interstital hole is called tetrahedral because ?

0%
it is formed by four spheres
0%
partly same and partly different
0%
it is formed by four spheres, the centres of which form a regular tetrahedron
0%
none of the above

The lattices of Na and Al are bcc and fcc respectively. Presuming them to be closed packed, their packing fractions are, respectively :

0%
$0.68$ and $0.34$
0%
$0.68$ and $0.74$
0%
$0.34$ and $0.34$
0%
$0.52$ and $0.52$

Explanation

Packing fraction is known as the fraction of volume occupied by the constituent particles in crystal structure.

GIVEN

Lattices of

$Na$ is Body centred cubic

$BCC$

Lattices of

$Al$ is Face centred cubic

$FCC$

SOLUTION

$Packing\hspace{0.1cm}fraction$ =

$\dfrac{N_{atom}\hspace{0.1cm}V_{atom}}{V_{unit\hspace{0.05cm}cell}}$

Packing fraction of BCC ( $Na$ )

Number of atoms ( $N_{atom}$ ) = $2$

Volume of atoms (

$V_{atom}$ ) =

$\dfrac{4}{3}$

$\pi$

$r^{2}$

Volume of unit cell (

$V_{unit\hspace{0.05cm}cell}$ ) =

$($

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

$)^{3}$

Particle fraction $_{BCC}$ = $\dfrac{N_{atom}\hspace{0.1cm}V_{atom}}{V_{unit\hspace{0.05cm}cell}}$

$\dfrac{2\times\dfrac{4}{3}\pi r^{2}}{(\dfrac{4r}{\sqrt3})^{3}}$

$\dfrac{\pi \sqrt{3}}{8}$

$\mathbf Particle fraction$ $_{BCC}$ $\cong$ $\mathbf 0.6802$

Packing fraction of FCC ( $Al$ )

Number of atoms (

$N_{atom}$ ) =

$4$

Volume of atoms (

$V_{atom}$ ) =

$\dfrac{4}{3}$

$\pi$

$r^{30}$

Volume of unit cell (

$V_{unit\hspace{0.05cm}cell}$ ) =

$($

$\dfrac{4r}{\sqrt2}$

$)^{3}$

Particle fraction

$_{FCC}$ =

$\dfrac{4\times\dfrac{4}{3}\pi r^{3}}{(\dfrac{4r}{\sqrt2})^{3}}$

$\dfrac{\pi \sqrt{2}}{12}$

Particle fraction

$_{FCC}$ = $0.74$

The particle fractions for

$Na$ and

$Al$ are $0.68$ and $0.74$ .

The correct option : B

The packing factor for a simple cubic lattice is :

0%
$\displaystyle \frac{\sqrt2 \pi}{6}$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{\sqrt3 \pi}{8}$
0%
$\pi$

In FCC unit cell, the percentage of unit cell not occupied by atoms is :

0%
$33\%$
0%
$29\%$
0%
$26\%$
0%
$74\%$

Explanation

In FCC,

$26\%$ of the total volume is not be recovered by atoms.

If A is the molecular mass of an element, a is the edge length and

$N$ is the Avogadro's number, then the density for face-centered cubic lattice is :

0%
$\displaystyle\frac{4A}{Na^3}$
0%
$\displaystyle\frac{2A}{Na^3}$
0%
$\displaystyle\frac{A}{Na^3}$
0%
$1$

Explanation

$\text{Density}$

$=$

$\dfrac{\text{Number of molecules present in lattice (Z)} \times \text{Mass of each molecule (m)}}{\text{Volume of lattice (V)}}$

In face-centered cubic lattice, each lattice has four molecules and mass of each molecule is

$\dfrac{\text{Molar mass (A)}}{\text{Avogadro's number (N)}}$

Hence,

$\text{Density} = \dfrac{Z\times A}{N\times a^{3}}=\dfrac{4\times A}{N\times a^3}$ , where

$a$ is the edge length.

Which one of the following schemes of ordering closed packed sheets of equal sized spheres do not generate close packed lattice?

0%
ABCABC
0%
ABACABAC
0%
ABBAABBA
0%
ABCBCACBC

The volume of a crystal in BCC structure having N atoms is :

[a = edge length]

0%
$N_Aa^3$
0%
$0.5N_Aa^3$
0%
$0.33N_Aa^3$
0%
$0.25N_Aa^3$

Explanation

The body-centred cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell.

$V = \dfrac{N_A\times a^3}{Z}$

$Z = 2\because \text{BCC structure}$

$V = \dfrac{N_A\times a^3}{2}$

$V = 0.5N_Aa^3$

$a$ is the edge length, first three nearest neighbour distances for primitive cubic lattice are, respectively:

0%
$a,$ ${\sqrt 2a},$ ${\sqrt 3a}$
0%
${\sqrt3a},$ ${\sqrt2a},$ $a$
0%
$a,$ ${\sqrt2a},$ $2a$
0%
$a,$ ${\sqrt 3a},$ $2a$

Explanation

First three nearest neighbor distances for primitive cubic lattice are, respectively, (edge length of unit cell

$=a$ )

$a,$

${\sqrt 2a},$ and

${\sqrt 3a}$ .

An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by

$60.3$ pm. If the atomic weight of A is

$48$ , then the density of the solid is nearly:

0%
$2.7$ g/cc
0%
$5.07$ g/cc
0%
$3.5$ g/cc
0%
$1.75$ g/cc

Explanation

According to the given diagram,

The separation between the inner surface of the atoms at adjacent corners is given by-

$a - 2r =60.3$

For BCC,

$\sqrt3 a = 4r$

So, density

$= \dfrac{nM}{VN_A}$

$= \dfrac{2 \times 48 \times 10^{30}}{(463.8)^3 \times 6.023 \times 10^{23}} = 1.7$

$g/cc$

The density of a pure substance A whose atoms are in cubic closed packed arrangement is

$1$ g/cc. If all the tetrahedral voids are occupied by B atoms, what is the density (in g cm

$^{-3}$ ) of resulting solid?[Atomic mass of

$A=30$ g/mol and atomic mass of

$B=50$ g/mol]

0%
$3.33$
0%
$4.33$
0%
$2.33$
0%
$5.33$

Explanation

The formula for density is given below:

$d=\dfrac{nM}{VN_A}$ and therefore,

$V = \dfrac{4\times 30}{1\times N_A}$

After adding B,

$d= \dfrac{(4\times 30+8\times 50)}{VN_A} = \dfrac{520}{120} = 4.33$ (CCP has

$8$ tetrahedral voids.)

Therefore, the density is

$4.33$ g cm

$^{-3}$ .

Option B is correct.

The density of solid argon (atomic mass

$=40$ amu) is

$1.68$ g/ml at

$40$ K. If the argon atom is assumed to be a sphere of radius

${1.50 \times 10^{-8}}$ cm, then the percentage of solid Ar which is space will be:

[use

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

0%
$35.64$
0%
$64.36$
0%
$73$
0%
None of the above

Explanation

As we know, density

$=1.68 = \dfrac{nM}{VN_A}=\dfrac{n\times 40}{V\times6\times 10^{23}}$

$\implies V = 39.53 \times 10^{-24}\ n$
The radius of argon is

$1.5 \times 10^{-8}$ cm and so, the empty space is

$\left (1-\dfrac{4}{3} \right ) \times n\times \pi r^3 \times \dfrac{100}{V} = 64.3\%$

The packing efficiency of a simple cubic crystal with an interstitial atom exactly fitting at the body center is:

0%
$0.48$
0%
$0.52$
0%
$0.73$
0%
$0.91$

Explanation

For a simple cubic,

$a=2r$
Therefore, the radius of the interstitial atom exactly fitting at the centre of the cube is

$R= \dfrac{1}{2} \times (a\sqrt 3 -2r)$

$R= \dfrac{1}{2} \times (a\sqrt 3 -a)$

$R= 0.366a$

Packing fraction

$= \dfrac{\text{volume occupied by spheres}}{a^3}$

Packing fraction

$= \dfrac{ (\dfrac{1}{8} \times 8 \times \dfrac{4 \pi r^3}{3}) + ( 1 \times \dfrac{4 \pi R^3}{3} ) }{a^3}$

Packing fraction

$= \dfrac{4\pi ({(\dfrac{a}{2})}^3 + (0.366a)^3)}{3a^3}$

Packing fraction

$= 0.7285$ .

Sodium (atomic mass

$=23$ amu) crystallizes in a bcc arrangement with the interfacial separation between the atoms at the edge

$53.6$ pm. The density (in g cm

$^{-3}$ ) of sodium crystal is:

0%
$2.07$
0%
$2.46$
0%
$1.19$
0%
none of the above

Explanation

According to given condition,

${a-2R} = 53.6$ pm

$a(1 - {2(\sqrt 3 )\over 4}) = 53.6$ pm

$\implies a=400$ pm

So, density

$= \dfrac{nM}{VN_A}$

$= \dfrac{2 \times 23 }{(400)^3 \times 6.023 \times 10^{23}\times 10^{-24}} = 1.19$ g/cc

Graphite has hcp arrangement of carbon atoms and the parallel planes are

$3.35$

$\mathring A$ apart. The density (in g cm

$^{-3}$ ) of graphite is :

0%
$2.46$
0%
$2.12$
0%
$1.20$
0%
$1.41$

Explanation

One unit cell of graphite contains

$4$ C atoms. Thus,

$Z=4$
The formula mass of graphite is

$12$ g/mol.
Avogadro's number is

$N_o = 6.023 \times 10^{23}$
The edge length is

$a=3.35\ A^{o}$
The density of graphite

$=\dfrac {Z \times \text{formula mass}} {N_o \times a^3}=\dfrac {4 \times 12} {6.023 \times 10^{23} \times (3.35 \times 10^{-8})^3}=2.12$ g/cc

Which of the following is a crystalline solid ?

0%
Graphite
0%
Calcite
0%
Gallium
0%
Oxygen

Select the correct statement(s).

0%
A cubic system possesses a total of $23$ elements of symmetry
0%
A cubic system contains a center of symmetry, planes of symmetry as well as axes of symmetry
0%
For triclinic system, ${a\neq b\neq c}$ and ${\alpha \neq \beta \neq \gamma \neq 90^{\circ}}$
0%
The total number of Bravais space lattices belonging to all the seven crystals is $14$

Explanation

A cubic crystal all 23 elements of symmetry
(i) Plane of symmetry = 3 + 6 = 9
(ii) axis of symmetry = 13
(iii) center of symmetry = 1
In triclinic system,

$a \neq b \neq c$ and

$\alpha \neq \beta \neq \gamma$
The total no.of Bravais space lattice belonging to all the seven crystals are 14.

Packing fraction of an identical solid sphere is

$74\%$ in:

0%
simple cubic structure
0%
fcc structure
0%
hcp structure
0%
bcc structure

Explanation

$4R=\sqrt{2}a\Rightarrow R=\dfrac{\sqrt{2}}{4}a$

Packing factor=

$\dfrac{V_s}{V_e}=\dfrac{4\times \dfrac{4\pi}{3}R^3}{a^3}=\frac{4\times \dfrac{4\pi}{3}\times (\dfrac{\sqrt{2}}{4}a)^3}{a^3}$

$=\dfrac{\sqrt{2}\pi}{6}=74.05\%$

$e=\sqrt{\dfrac{8}{3}}a,R=\dfrac{a}{2}$

Packing factor

$=\dfrac{V_s}{V_3}=\dfrac{6\times \dfrac{4\pi}{3}R^3}{6\times \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}a^2\times c }=\dfrac{6\times \dfrac{4\pi}{3}\times (\dfrac{a}{2})^2}{6\times \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}a^2\times \dfrac{\sqrt{8}}{3}a}$

$=\dfrac{\sqrt{2\pi}}{6}=74.05\%$

Hence, both option

$B$ and

$C$ are correct.

Select the correct statement(s) :

0%
The co-ordination number of each type of ion in a CsCl crystal is twelve.
0%
A metal that crystallizes in a bcc structure has a co-ordination number of twelve.
0%
A unit cell of an ionic crystal shares some of its ions with other unit cells.
0%
The length of the unit cell in NaCl is $552$ pm (given that ${r_{Na+}= 85}$ pm and ${r_{Cl-}= 181}$ pm).

Explanation

The co-ordination number of each type of ion in a CsCl crystal is

$8$ .

A metal that crystallizes in a bcc structure has a co-ordination number of

$8$ .

The length of the unit cell in NaCl is

$532$ pm (given that

${r_{Na+}= 85}$ pm and

${r_{Cl-}= 181}$ pm).

Therefore, edge length

$=2r+2R = 2\times 85+2\times 181 = 532$ pm

A unit cell of an ionic crystal shares some of its ions with other unit cells like face molecules are shared by

$2$ unit cells and corner molecules are shared by

$8$ unit cells.

Option C is correct.

Which are amorphous solids?

0%
Sodium chloride
0%
Calcium flouride
0%
Glass
0%
Plastics

The density for an FCC lattice is:

[A = atomic wt, N = Avogadro's number, a = lattice parameter]

0%
$\displaystyle \frac{4A}{Na^3}$
0%
$\displaystyle \frac{2A}{Na^3}$
0%
$\displaystyle \frac{A}{Na^3}$
0%
$\displaystyle \frac{A}{Na^2}$

Explanation

$FCC$ lattice, the no. of atoms per unit cell

$= 4$

Density

$=\displaystyle \frac{mass\ of\ unit\ cell}{volume \ of\ unit\ cell} = \frac{4 A/N}{a^3} = \frac{4A}{Na^3}$

Stacking of hexagonal close packed layer gives rise to:

0%
bcc structure
0%
hcp structure
0%
fcc structure
0%
sc structure

Which of the following contains only one Bravais lattice?

0%
Hexagonal
0%
Monoclinic
0%
Rhombohedral
0%
Triclinic

Correct order for lattice energy of the given crystals is :

0%
$KCl < NaCl < NaF$
0%
$NaF > KCl > NaCl$
0%
$KCl > NaCl > NaF$
0%
$NaF < NaCl < KCl$

Explanation

Lattice energy is directly proportional to the charge of the ion and inversely proportional to the size of ion.

$F^{-}$ is smaller in size than

$Cl^-$ .

$NaF$ has greater lattice energy than

$NaCl$ and as

$Na^{+}$ is smaller in size than

$K^+$ .

Therefore,

$NaCl$ has greater lattice energy than

$KCl$ .

$KCl<NaCl<NaF$ .

Hence, option B is correct.

The elements of symmetry in a crystal are:

0%
plane of symmetry
0%
axis of symmetry
0%
centre of symmetry
0%
none of these

Which of the following is not an amorphous substance?

0%
Copper
0%
Rubber
0%
Glass
0%
Polymers

Explanation

The amorphous substance is defined as any noncrystalline solid in which the atoms and molecules are not organized in a definite lattice pattern. As Rubber, Glass and Polymers of high molecular mass are examples of amorphous solids and Copper is not an amorphous substance.

Copper is an example of a crystalline solid.

Hence, the correct option is $A$

Crystalline solids do not have :

0%
a definite latent heat of fusion.
0%
irregular atomic arrangement.
0%
incompressibility.
0%
free movement of atoms.

$A$ is the atomic mass number of an element,

$N$ is the Avogadro number and

$a$ is the lattice parameter, then the density of the element, if it has bcc crystal structure, is_________.

0%
$\displaystyle \dfrac{A}{Na^3}$
0%
$\displaystyle \dfrac{2A}{Na^3}$
0%
$\displaystyle \dfrac{\sqrt 3 A}{Na^3}$
0%
$\displaystyle \dfrac{2\sqrt 2A}{Na^3}$

Explanation

$\displaystyle d = \dfrac{Z \times A}{N \times a^3}$

$For\ bcc, Z = 2$

$\displaystyle d = \dfrac{2A}{Na^3}$

Stacking (a pile) of square close packed layer gives rise to:

0%
BCC structure
0%
FCC strncture
0%
SC structure
0%
none of the above

Statement: The unit cell of many metals is one of the three cubic unit cells.

State whether the given statement is true or false.

0%
True
0%
False

Statement: Face-centred cubic structure possesses the most close packing of atoms in it.

State whether the given statement is true or false

0%
True
0%
False

Explanation

Unit cell	Number of atoms at			No. of atoms per unit cell	Volume occupied by particles $(\%)$
Unit cell	Corners	Centres	Faces	No. of atoms per unit cell	Volume occupied by particles $(\%)$
Simple cubic	$8\times \dfrac{1}{8}=1$	0	0	1	52.4
Body centred cubic (BCC)	$8\times \dfrac{1}{8}=1$	1	0	2	68
Face centred cubic (BCC)	$8\times \dfrac{1}{8}=1$	0	$6\times \dfrac{1}{2}=3$	4	74

State whether the given statement is true or false:
Diamond is the hardest compound.

0%
True
0%
False

Explanation

Diamond is no more the hardest known material. Wurtzite boron nitride and lonsdaleite exist naturally or have been made in the lab. So, until now, no one had realised their superior strength. The simulation showed that wurtzite boron nitride would withstand $18$ per cent more stress than diamond, and lonsdaleite $58$ per cent more. If the results are confirmed with physical experiments, both materials would be far harder than any substance ever measured.

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

Statement: Glass is supercooled liquid.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

Glass is supercooled liquid or pseudo solid. Glass is an amorphous solid in which the regular arrangement of constituent particles is present only up to a short distance. The structure of glass is similar to the structure of liquids. Glass behaves like a fluid and very slowly falls under gravity. Due to this, the lower sides of glass panes of windows, photo frames, cupboards and showcases become slightly thicker and upper ends become thinner.

Hence, the correct option is

$\text{true}$

Lattice energy of NaCl is

$-186$ kcal/mol. If solution enthalpies of Na

$+$ and Cl

$^-$ are

$-97$ kcal/mol and

$-85$ kcal/mol respectively. The enthalpy NaCl solution in kcal will be:

0%
$-4$
0%
$182$
0%
$368$
0%
$4$

Explanation

The enthalpy of solution of NaCl in kcal is:

$\displaystyle \Delta H_{solution} = \Delta H_{\text {solution of cation}} + \Delta H_{\text {solution of anion}} - \Delta H_{\text {Lattice energy of NaCl}}$

$\Delta H_{solution}= (-97) + (-85) - (-186)=4$ kcal/mol

The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This enthalpy of solution (ΔHsolution) can either be positive (endothermic) or negative (exothermic).

Hence, the correct option is

$D$

Statement: The imperfectness in crystal increases with increase in temperature.

State whether the given statement is true or false

0%
True
0%
False

Statement: At

$0$ Kelvin, crystals have a perfect orderly arrangement.

State whether the given statement is true or false

0%
True
0%
False

If the radius of

$B ^-$ ion is

$0.180$ nm, the size (in nm) of the cation that can fit in the tetrahedral void is:

0%
$0.125$
0%
$0.04$
0%
$0.08$
0%
$0.180$

Explanation

For tetrahedral void, we have

$\dfrac {r_+ }{r_-}=0.225$

$\implies r_+=0.225\times r_-= 0.225 \times 0.180 =0.04$ nm

Hence, the size of cation that can fit in the tetrahedral void is

$0.04$ nm.

If a diamond has a face-centered cubic structure with two atoms per lattice point and a unit cell of the edge of

$3.569\:\mathring{A}$ . The density (in g cm

$^{-3}$ ) of the diamond is :

0%
$3.569$
0%
$2.569$
0%
$1.899$
0%
$7.946$

Explanation

Diamond is a carbon allotrope. We can use density formula to calculate the density as follows:

Density

$=\rho =\dfrac { ZM }{ N_{ 0 }a^{ 3 } }$

Substituting values in the above expression, we get

$\rho=\dfrac { 2\times 12 }{ (6.022\times { 10 }^{ 23 })\times (3.569\times { 10 }^{ -8 })^{ 3 } }$

On solving the above equation, we get

$\rho$ as

$3.569$ g cm

$^{-3}$ .

Hence, the correct option is

$\text{A}$

MgO has NaCl type structure and its unit cell edge length is

$421$ pm. Its density (in g cm

$^{-3}$ ) will be:

0%
$3.54$
0%
$1.46$
0%
$2.14$
0%
None of the above

Explanation

The expression for density is as follows:

$\rho =\dfrac{ZM}{N_{0}a^{3}}$

As MgO has the same structure as NaCl, we can say that MgO has FCC structure.

So, putting values accordingly, we get

$\rho =\dfrac { 4\times 40 }{ (6.022\times { 10 }^{ 23 })\times (4.21\times { 10 }^{ -8 })^{ 3 } } =3.54$ g cm

$^{-3}$

Select the correct order for covalent radii :

0%
Octahedral radii > linear radii > tetrahedral radii
0%
Octahedral radii > tetrahedral radii > linear radii
0%
Linear radii > tetrahedral radii > octahedral radii
0%
Tetrahedral radii > octahedral radii > linear radii

Metallic copper crystallises in bcc lattice. The length of the cubic unit cell is

$362$ pm. The density (in g cm

$^{-3}$ ) of crystalline copper is:

0%
$2.78$
0%
$3.51$
0%
$4.45$
0%
none of the above

Explanation

For a BCC lattice,

$z =2$
density =

$\dfrac{zM}{N_a \times (a \times 10^{-10})^3}$
density =

$\dfrac{2 \times 65}{6 \times 10^{23}\times (362 \times 10^{-10})^3}$
density =

$4.45 \text{ gcm}^{-3}$

Truncated octahedral (cube-octahedra) is called 'case' or 'sodalite cage'.

0%
True
0%
False

The ratio of packing density in fcc, bcc and cubic structure is respectively_________.

0%
$1:0.92 :0.70$
0%
$0.70 :0.92 :1$
0%
$1:0.70:0.92$
0%
$0.92:0.70:1$

Explanation

As we know,
Packing fraction in fcc

$= 74\%$ , bcc

$= 68\%$ and sc

$= 52\%$ .
Hence, their ratio is

$74:68:52=1:0.92:0.70$ .

Lithium borohydride

$(LiBH_4)$ crystallizes in an orthorhombic system with

$4$ molecules per unit cell. The unit cell dimensions are

$a = 6.8$

$\mathring {A }$ ,

$b = 4.4$

$\mathring {A}$ and

$c = 7.2$

$\mathring {A}$ . If the molar mass of the compound is

$21.76$ g/mol, the density (in g cm

$^{-3}$ ) of the compound will be:

0%
$0.67$
0%
$0.33$
0%
$1.72$
0%
none of the above

Explanation

Density

$= \dfrac{zM}{N_A \times (abc \times 10^{-24})}$

Density

$= \dfrac{4 \times 21.76}{6.023 \times 10^{23} \times (6.8 \times 4.4 \times 7.2 \times 10^{-24})}$

Density

$= 0.67 \text{ gcm}^{-3}$

Hence, the correct option is

$\text{A}$

If the effective radius of iron (Fe) atom is

$141.4$ pm which has rock salt type structure, then its density (in g cm

$^{-3}$ ) will be:

[

$Fe = 56$ amu]

0%
$5.81$
0%
$2.63$
0%
$1.08$
0%
None of the above

Explanation

In rock salt type structure its form fcc type unit cell where edge legth,

$a = 2\sqrt 2r$

$\implies a \approx 400 \text{ pm}$

Density,

$d = \dfrac{zM}{N_A \times (a)^3}$

For FCC structure,

$z=4$

Now,

Density,

$d = \dfrac{4 \times 56}{6.02 \times 10^{23} \times (400 \times 10^{-10})^3} \\$

Density,

$d =5.81 \text{ g cm}^{-3}$

Hence, the correct option is

$\text{A}.$

In which of the following systems, interfacial angles

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

$\beta \neq 90^o$ ?

0%
Monoclinic
0%
Rhombohedral
0%
Triclinic
0%
Hexagonal

Explanation

Name of system	Axes	Angles	Bravais Lattices
cubic	$a= b=c$	$\alpha=\beta=\gamma=90^o$	Primitive, Face-centred, Body centred $=$ $3$
Tetragonal	$a=b\neq c$	$\alpha=\beta=\gamma=90^o$	Primitive, Face-centred $=$ $2$
Rhombohedral or Trigonal	$a= b=c$	$\alpha=\beta=\gamma\neq 90^o$	Primitive $=$ $1$
Orthorhombic or Rhombic	$a= b=c$	$\alpha=\beta=\gamma=90^o$	Primitive, Face-centred, Body centred, End centred $=$ $4$
Monoclinic	$a\neq b\neq c$	$\alpha = \gamma = 90^o$ ; $\beta \neq 90^0$	Primitive, End - centred $=$ $2$
Triclinic	$a\neq b\neq c$	$\alpha\neq \beta \neq \gamma \ = 90^o$	Primitive $=$ $1$
Hexagonal	$a= b\neq c$	$\gamma=120^o$	Primitive $=$ $1$

Hence, the lattices are monoclinic and rhombohedral.

The total fraction of space occupied by voids in simple cubic structure is:

0%
$0.58$
0%
$0.25$
0%
$0.48$
0%
$0.86$

The compound CuCl has zinc blende structure and the edge length of its unit cell is

$500$ pm, its density (in g cm

$^{-3}$ ) is :

(Given that the atomic weight of

$Cu$ is

$63.5$ amu and of

$Cl$ is

$35.5$ amu.)

0%
$5.24$
0%
$1.14$
0%
$2.99$
0%
None of the above

Explanation

density

$= \dfrac{zM}{N_a \times (a \times 10^{-10})^3}$
density

$= \dfrac{4 \times 99}{6 \times 10^{23} \times (500 \times 10^{-10})^3}$
density

$= 5.24 \text{ gcm}^{-3}$

The packing fraction for a body-centered cube is________.

0%
$0.42$
0%
$0.53$
0%
$0.68$
0%
$0.82$

Explanation

Unit cell	Number of atoms at			No. of atoms per unit cell	Volume occupied by particles (%)
Unit cell	Corners	Centres	Faces	No. of atoms per unit cell	Volume occupied by particles (%)
Simple cube	$8\times \dfrac{1}{8}=1$	0	0	1	52.4
Body centred cube (BCC)	$8\times \dfrac{1}{8}=1$	1	0	2	68
Face centred cube (FCC)	$8\times \dfrac{1}{8}=1$	0	$6\times \dfrac{1}{2}=3$	4	74

Which of the following systems have lattice parameters as

$a\neq b \neq c$ ?

0%
Orthorhombic
0%
Monoclinic
0%
Triclinic
0%
Hexagonal

Explanation

Name of System	Axes	Angles	Bravais Lattices
Cubic	$a=b=c$	$\alpha = \beta = \gamma = 90^o$	Primitive, Face -centred, Body centred $=3$
Tetragonal	$a=b\neq c$	$\alpha = \beta = \gamma = 90^o$	Primitive, Face -centred, Body centred $=2$
Rhombohombic or trigonal	$a=b= c$	$\alpha = \beta = \gamma \neq 90^o$	Primitive $=1$
Orthorhombic or Rhombic	$a\neq b\neq c$	$\alpha = \beta = \gamma = 90^o$	Primitive, Face -centred, Body centred End centred $=4$
Monoclinic	$a\neq b\neq c$	$\alpha =\gamma=90^o\, ;\beta\neq 90^o$	Primitive, End-centred $=2$
Triclinic	$a\neq b\neq c$	$\alpha \neq \beta \neq \gamma = 90^o$	Primitive $=1$
Hexagonal	$a=b\neq c$	$\alpha = \beta =90^o\, \gamma = 90^o$	Primitive $=1$ total $=14$

The space in which atoms are not present in unit cell is :

0%
$48\%$ in simple cubic
0%
$32\%$ in body-centred cubic
0%
$26\%$ in face-centred cubic
0%
$26\%$ in hexagonal

Explanation

Unit cell	Number of atoms at			No. of atoms per unit cell	Volume occupied by particles (%)
Unit cell	Corners	Centres	Faces	No. of atoms per unit cell	Volume occupied by particles (%)
Simple cube	$8\times \dfrac{1}{8}=1$	0	0	1	52.4
Body centred cube (BCC)	$8\times \dfrac{1}{8}=1$	1	0	2	68
Face centred cube (FCC)	$8\times \dfrac{1}{8}=1$	0	$6\times \dfrac{1}{2}=3$	4	74

The adjoined table gives information about the cubic lattices where the empty spaces in simple cube, BCC and FCC are 48% , 32% and 26% respectively.
Hexagonal closest packing structure also has a packing efficiency of 74% and therefore the empty space in HCP is 26%

Therefore, all options are correct.

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 6 - MCQExams.com

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CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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