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

Silver crystallizes in a fcc lattice and has a density of

$10.6 \,g/cm^3$ . What is the length of an edge of the unit cell ?

0%
$0.407 \,nm$
0%
$0.2035 \,nm$
0%
$0.101 \,nm$
0%
$4.07 \,nm$

Explanation

$10.6 = \dfrac{4 \times 108}{a^{3} \times 6.023 \times 10^{23}}$

$\therefore a = 4.07 \,nm$

An element crystallizes in a structure having FCC unit cell of an edge length

$200$ pm. If

$200$ g of this element contains

$24\times 10$

$^{23}$ atoms, the density (in g/cc) of the element is :

0%
$50.3$
0%
$63.4$
0%
$41.6$
0%
$34.8$

Explanation

It is given that an element crystallizes in a structure having FCC unit cell of an edge length

$200$ pm.

$200$ gm of the element contains

$24 \times 10^{23}$ atoms, which implies

$6.022 \times 10^{23}$ atoms are present in

$50$ gm, which is its molecular weight.

Since, the lattice is FCC,

$z=4$

$d=\dfrac { zM }{ { N }_{ A }{ a }^{ 3 } }$

$d=\dfrac { 4\times50}{ { 6\times10 }^{ 23 }\times{ (200\times{ 10 }^{ -10 }) }^{ 3 } } = 41.6$ g/cc

The number of hexagonal faces that are present in a truncated octahedron is:

0%
$8$
0%
$9$
0%
$10$
0%
$11$

Explanation

Correct Answer: Option A

Explanation:

When an object is cut through a cutting plane inclining towards its base at a certain angle, the remaining part after deleting the top part is Truncated Solid.

It is made from a regular octahedron.

The dimensions of octahedron comprise of side length $3l$ . After the removal of $6$ right square pyramids from each point, a truncated octahedron is obtained.

faces. ( $8$ regular hexagons and $6$ squares)

It has $36$ edges.

It has $24$ vertices.

Thus, there are $8$ hexagonal faces in a truncated octahedron.

Structure:

The packing efficiency of the two-dimensional square unit cell shown in the given figure is :

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

Explanation

Let

$a=2\sqrt2r$
Let us consider the square plane and

$L$ be the length of square (given).

$\text{Area} = L^{2}$
By applying pythagoras theorem,

$L^{2}$ +

$L^{2}$ =

$(2r)^{2}$

$4r^{2}=$

$2L$

$\dfrac{\sqrt{2L}}{2}=r$

$\text{Area} =$

$2 \pi \times \dfrac{\sqrt{(2L)}}{4}^{2}$
Packing efficiency

$=\displaystyle \frac{2\times\pi \mathrm{r}^{2}}{(2\sqrt{2}\mathrm{r})^{2}}=\frac{2\pi \mathrm{r}^{2}}{8\mathrm{r}^{2}}=\frac{\pi}{4}= 78.5\%$

An element crystallizes in a structure having FCC unit cell of an edge length

$200$ pm. Calculate the density (in g cm

$^{-3}$ ) if

$200$ grams of it contains

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

0%
$41.6$
0%
$42.6$
0%
$43.6$
0%
$44.6$

Explanation

The formula of density is as follows:

$d=\dfrac { z\times w }{ { N }\times { a }^{ 3 } }$

For FCC unit cell,

$z = 4$

$N = 24\times 10^{23}$ atoms

$a=200 pm= 2\times 10^{−8} cm$

Now,

$d = \dfrac{4\times 200}{24\times 10^{23}\times 8\times 10^{−24} }$

$= \dfrac{1000}{24}$

$d = 41.67\ g.cm^{−3}$

So, the correct option is A.

A solid made up of ions of A and B possesses an edge length of unit cell equal to

$0.564$ nm has four formula units. Among the two ions, the smaller one occupies the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass as

$9.712\times10^{-23}$ g. The density (in g cm

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

0%
$2.16$
0%
$0.54$
0%
$1.08$
0%
$1.562$

The empty space in this HCP unit cell is :

0%
$74\%$
0%
$47.6\%$
0%
$32\%$
0%
$26\%$

Assertion: The close packing of atoms in cubic structure is in the order: scp > bcc > ccp.

Reason:

$\text{Packing density} = \dfrac{\text{Volume of unit cell}}{a^{3}}$

0%
Both Assertion and Reason are correct and Reason is correct explanation of Assertion.
0%
Both Assertion and Reason are correct but Reason is not a correct explanation of Assertion.
0%
Assertion is wrong but Reason is correct.
0%
Assertion is correct but Reason is not correct.

Explanation

Packing efficiency of SCP(simple cubic)

$= 52\%$

Packing efficiency for BCC(body centred)

$= 68\%$

Packing efficiency of CCP(cubic-closed/face-centred)

$= 74\%$

The order of close packing would be, FCC<BCC<SCP

$\text{Packing density} = \dfrac{\text{Volume of unit cell}}{a^{3}}$

Match the elements in list I with the shape of its crystal in list II.

List I	List II
A) $Be$	1) Body-centred cubic
B) $Ca$	2) Simple cubic
C) $Ba$	3) Face-centred cubic
D) $Po$	4) Hexagonal close-packed

0%
A - 4, B-3, C-1, D-2
0%
A - 4, B-3, C-2, D-1
0%
A-2, B-4, C-1, D-3
0%
A-4, B-1, C-3, D-2

Explanation

A. Beryllium is a steel gray and hard metal that is brittle at room temperature and has a close-packed hexagonal crystal structure.

B. Crystal Structure of Calcium. —The X-ray pattern obtained with powdered calcium shows that the atoms are arranged in a face-centered cubic lattice. The side of the elementary cube is 5.56 Å.

C. At room temperature and pressure, barium has a body-centered cubic structure, with a barium–barium distance of 503 picometers.

D. Polonium is a radioactive element that exists in two metallic allotropes. The alpha form is the only known example of a simple cubic crystal structure in a single atom basis at STP, with an edge length of 335.2 picometers.

$Be$ (Beryllium) - HCP packing

$Ca$ (Calcium) - FCC packing

$Ba$ (Barium)- BCC packing

$Po$ (Polonium) - Simple cubic arrangement

If the edge length of an NaH unit cell is

$488$ pm, what is the length of an Na-H bond if its crystal has fcc structure?

0%
$488$ pm
0%
$122$ pm
0%
$244$ pm
0%
$976$ pm

Explanation

Length of an

$Na-H$ bond

$= r_+ + r_-$

$\because 2({ r }_{ + }{ +r }_{ - })=a$ (edge length)

$\because 2({ r }_{ + }{ +r }_{ - })=488\ pm$

The length of Na-H bond is the interatomic separation between

$Na^+$ and

$H^-$ ions

$\therefore { r }_{ + } + {r }_{ - }=244\quad pm$

Calculate the density (in g cm

$^{-3}$ ) of diamond from the fact that it has face-centered cubic structure with two atoms per lattice point and unit cell edge length of

$3.569\, \mathring A$ .

0%
$7.5$
0%
$1.7$
0%
$3.5$
0%
None of the above

Explanation

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

For FCC, z=2

$density = \dfrac{(2 \times 4) \times 12}{6 \times 10^{23} \times (3.569 \times 10^{-8})^3} = 3.5 \text{ g cm}^{-3}$

Which of the following statements are true?

0%
The unit cell content of potassium which crystallizes in body centered cubic unit cell is two
0%
For the planes, $(100), (110),\ (111)$ in a cubic unit cell, the ratio of $d_{100},\ d_{110}$ and $d_{111}$ is $\sqrt{2} : 1 :\sqrt{\dfrac{2}{3}}$
0%
A cubic crystal has total $23$ elements of symmetry
0%
In tetragonal crystal system, the number of Bravais lattices is only two

Explanation

In body centred cubic structure, cubic unit cell =

$\dfrac{1}{8} \times 8 + 1 = 2$
For the planes

$(1,0,0) \\ d_{100} = \dfrac{1}{\sqrt{1^2 + 0^2 + 0^2}} = \dfrac{1}{1} \\ d_{110} = \dfrac{1}{\sqrt{1^2 + 1^2 + 0^2}} = \dfrac{1}{\sqrt{2}} \\ d_{111} = \dfrac{1}{\sqrt{1^2 + 1^2 + 1^2}} = \dfrac{1}{\sqrt{3}} \\$
Hence, the given ratio is

$\sqrt 2 : 1 : \sqrt{\dfrac{2}{3}}$
A cubic crystal has all 23 elements of symmetry:
Planes of symmetry - 9
Axes of symmetry - 13
Centre of symmetry - 1
In tetragonal crystal system, the number of bravais lattices is two namely primitive and body centred.

Lithium metal crystallizes in a body-centered cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of lithium will be:

0%
300.5 pm
0%
240.8 pm
0%
151.98 pm
0%
75.5 pm

Explanation

In case of BCC crystal ,

$a \times \sqrt 3 =4r$

Hence , atomic radius of lithium,

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

$\dfrac{ 351\times 1.732}{4}$

= 151.98 pm

So option C is correct.

Fraction of empty space in ABAB type arrangement in 3D is :

0%
$0.74$
0%
$0.26$
0%
$0.68$
0%
$0.32$

Explanation

Percentage of packing fraction of ABAB (HCP) arrangement in 3D

$= 74\%$
Percentage of vacant space

$= 26\%$

Fraction of vacant space

$=0.26$

The face centred cubic cell of platinum has a length of

$0.392 nm$ . Calculate the density of platinum

$(g/cm^3):$

(Atomic weight : Pt=195)

0%
20.9
0%
20.4
0%
19.6
0%
21

Explanation

$Density, d = \dfrac{Z{\times} M}{N_A{\times}a^3}$

Here, Z = No. of atoms in fcc crystal = 4

$N_a =$ Avagadro No.

$= 6\times 10^{23} mol^{-1}$ ,

$a=0.392\ nm = 0.392 \times 10^{-7} cm$

$M = 195\ g/mol$

$d = \dfrac{4{\times} 195}{N_a{\times}{(0.392\times 10^{-7})}^3}$

$d = 21\dfrac{g}{cm^3}$

Option D is correct.

The pyknometric density of sodium chloride crystal is

$2.165\times 10^{3}\ kg\ m^{-3}$ while its X-ray density is

$2.178\times 10^{3} kg\ m^{-3}$ . The fraction of unoccupied sites in sodium chloride crystal is:

0%
$5.96\times 10^{-3}$
0%
$5.96$
0%
$5.96\times 10^{-2}$
0%
$5.96\times 10^{-1}$

Explanation

Therefore volume unoccupied site in sodium chloride

$=5.96 \times 10^{-3}$
1468571_144091_ans_ca8d8eddcdb04b1596f4dbfb47a38ff6.PNG

Statement 1: Increasing temperature increases the density of point defects.
Statement 2: The process of formation of point defects in solids is endothermic and has

$\Delta S > 0$ .

0%
Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is false
0%
Statement-1 is false, Statement-2 is true

Explanation

The process of formation of point defects in solids is endothermic.

Hence,

$\Delta H > 0$ and also,

$\Delta S > 0$ .
When temperature is increased,

$\Delta G = \Delta H-T\Delta S < 0$ . The process of formation of defects become spontaneous. Hence, more and more defects are created per unit volume.
Thus, increasing temperature increases the density of point defects.

Which of the following can conduct electricity in a fused state?

0%
$MgCl_2$
0%
$CaCl_2$
0%
$BaCl_2$
0%
$BeCl_2$

Explanation

In fused state, ionic compounds have free cations and anions which can move easily to conduct electricity.

$MgCl_2, CaCl_2$ and

$BaCl_2$ are ionic in nature. So, they can conduct electricity.
However,

$BeCl_2$ has a covalent character due to small size of Beryllium. In a fused state,

$BeCl_2$ exists in polymeric state. So, it does not have free ions to conduct electricity.

Option A, B and C are correct.

If the edge length is

$3.60A^o$ , density of diamond crystal is:

0%
3.92 gm/cc
0%
2.40 gm/cc
0%
3.37 gm/cc
0%
2.58 gm/cc.

Explanation

$Z_{diamond}=8\times \cfrac{1}{8}+6\times\cfrac {1}{2}+4\times 1$

$=8$

$\rho=$

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

$M_A=12$ ,

$a=3.6\mathring A$

$\rho=$

$\cfrac {8\times 12}{6.023\times 10^{23}\times 46.656\times 10^{-24}}$

$=3.37$

$gm/cm^3$

In a monoclinic unit cell, the relation of sides and angles are, respectively :

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

Explanation

Monoclinic crystal is one of the 7 crystal system. In monoclinic system, the crystal is describe by vectors of unequal lengths,

So, we have

$a\neq b\neq \ c$ and

$\beta = \gamma = 90^0 \neq \alpha$

Option C is correct.

An element crystallizes in a structure having FCC unit cell of an edge length equal to

$200$ pm. Calculate the density (in g cm

$^{-3}$ ) if

$200$ g of this element contains

$24\times10^{23}$ atoms.

0%
$41.67$
0%
$83.34$
0%
$119.43$
0%
None of the above

Explanation

$M = \dfrac{N_A \times 200}{24 \times 10^{23}} = 50$ g
For fcc

$z = 4$

$d = \dfrac{ zM}{N_A \times V }$

$d = \dfrac{ 4 \times 50}{6.023 \times 10^{23} \times (200 \times 10^{-10})^3 }$

$d = \dfrac{ 200}{6.023 \times 8 \times 10^{-1} }$

$d = 41.67$

$gcm^{-3}$

Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is

$407$ pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold

$=197$ amu.

0%
$19.4\:g/cm^{3}$ , $143.9$ pm
0%
$38.8\:g/cm^{3}$ , $143.9$ pm
0%
$19.4\:g/cm^{3}$ , $287.8$ pm
0%
None of the above

Explanation

For an FCC crystal, atoms are in contact along the diagonal of the unit cell.

$4r= \sqrt{2}a$ , where

$a$ is the edge length of unit cell and

$r$ is the radius of atom.
Substituting values in the above expression, we get

$r= \dfrac{ \sqrt{2} \times 407}{4} =143.9$ pm
Number of atoms per unit cell (FCC)

$=4$
Mass of

$1$ unit cell

$=197 \times 4=788\:amu=1.66 \times 10^{-27} \times 788\:kg=1.30808 \times 10^{-24}\:kg$
Volume of

$1$ unit cell

$= (407 \times 10^{-12})^{3}\:m^{3}=6.741 \times 10^{-29}\:m^{3}$

$=2.824 \times 10^{-28}\:m^{3}$

$\text{Density}= \dfrac{1.30808 \times 10^{-24}}{6.741 \times 10^{-29}}=19404.83 \:kg/m^{3}=19.4\:g/cm^{3}$

The distance between an octahedral and tetrahedral void in fcc lattice would be

0%
${\sqrt 3a}$
0%
$\displaystyle\frac{\sqrt{3a}}{2}$
0%
$\displaystyle\frac{\sqrt{3a}}{3}$
0%
$\displaystyle\frac{\sqrt{3}a}{4}$

Explanation

Distance between an octahedral and tetrahedral void in fcc lattice is

$\dfrac{\text{body diagonal}}{4}$

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

The density of crystalline

$CsCl$ is

$3.988\ g/cc$ . The volume effectively occupied by a single

$CsCl$ ion pair in the crystal is

0%
$7.014\times 10^{-23} cc$
0%
$2.81\times 10^{-22} cc$
0%
$6.022\times 10^{-23} cc$
0%
$3.004\times 10^{-23} cc$

Explanation

$BCC$ only one ion pair will be present in unit cell

$d= \cfrac{ZM}{VN_A}$

$d = \dfrac {2\times 168.4}{6.023\times 10^{23} \times a^{3}} = 3.988\ g/cc$

$a^{3} = \dfrac {1\times 168.4}{6.023\times 3.988\times 10^{23}} = 7.011\times 10^{-23} cc$ .

Option A is correct.

An element X (atomic weight

$= 24$ amu) forms a face-centred cubic lattice. If the edge length of the lattice is

${4\times10^{- 8}}$ cm and the observed density is

$2.40 \times 10^{3}$ kg m

$^{-3}$ , then the percentage occupancy of lattice points by element X is: (use

${N_{A}= 6\times 10^{23}}$ )

0%
$96$
0%
$98$
0%
$99$
0%
none of the above

Explanation

As we know, density

$(d)= \dfrac{nM}{VN_A}$

Here,

$a$ is the edge length of the unit cell,

$M$ is the molecular mass and

$V$ is the volume of an unit cell.

$d = \dfrac{4\times24}{(4\times10^{-8})^3\times6\times10^{23}} = 2.5$ g/cc
The observed density is

$2.4$ g/cc.
So, the percentage occupancy of lattice points by element X is

$\dfrac{2.4\times100}{2.5} = 96\%$

Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is

$124$ pm, the density (in g cm

$^{-3}$ ) of iron in both these structures are, respectively :

0%
$7.887,\ 8.59$
0%
$8.59,\ 7.887$
0%
$14.287,\ 16.248$
0%
$16.248,\ 14.287$

Explanation

For bcc unit cell, the edge length,

$\displaystyle a = \frac {4r}{\sqrt{3}} = \frac {4 \times 124}{\sqrt{3}} = 286.4$ pm
The number of formula units per unit cell

$\displaystyle Z=2$
Density

$\displaystyle = \frac {\text {Z} \times \text {Molar mass}}{\text {Avogadro's number } \times a^3}$
Density

$\displaystyle = \frac {2 \times 55.85}{6.023 \times 10^{23} \times (286.4 \times 10^{-10})^3} = 7.887gcm^{-3}$
For fcc unit cell, the edge length,

$\displaystyle a = \frac {4r}{\sqrt{2}} = \frac {4 \times 124}{\sqrt{2}} = 350.8$ pm
The number of formula units per unit cell

$\displaystyle Z=4$
Density

$\displaystyle = \frac {\text {Z} \times \text {Molar mass}}{\text {Avogadro's number } \times a^3}$
Density

$\displaystyle = \frac {4 \times 55.85}{6.023 \times 10^{23} \times (350.8 \times 10^{-10})^3} = 8.59gcm^{-3}$

The arrangement of

$X^{-}$ ions around

$A^{-}$ ion in solid

$AX$ is given in the figure (not drawn to scale). If the radius of

$X^{-}$ is

$250\ pm$ , the radius of

$A^{+}$ is

0%
$104pm$
0%
$125pm$
0%
$183pm$
0%
$57pm$

Select the correct statement(s) about three-dimensional hcp system.

0%
The number of atoms in hcp unit cell is $6$ .
0%
The volume of hcp unit cell is $24 \sqrt 2 r^3$ .
0%
The empty space in hcp unit cell is $26\%$ .
0%
The base area of hcp unit cell is $6\sqrt 3 r^2$ .

Explanation

All the four statements are correct.
(A) The number of atoms in hcp unit cell is 6.
12 atoms are present at corners and each atom contributes one sixth to the unit cell as each corner atom is shared by 6 unit cells.
2 atoms are present at face centers and each face center atom contributes one half to the unit cell as each face centered atom is shared by 2 unit cells.
3 atoms are present in the body with total contribution of 3 as each atom in body contributes 1 to the unit cell.
Hence, the number of atoms in hcp unit cell

$\displaystyle 12 \times \frac {1}{6} (corners) + 2\times \frac {1}{2}(face \: centers) +3 \times \frac {1}{1}(in body) = 2 + 1 + 3 = 6$
(B) and (D) Volume of unit cell

$=$ Base area

$\times$ height
Base area of regular hexagon

$= 6 \times$ area of equilateral triangle

$\displaystyle = 6\sqrt {3}r^2$
Height of unit cell

$\displaystyle=4r.\sqrt {\frac {2}{3}}$
Volume of the unit cell

$\displaystyle= 6\sqrt {3}r^2 \times 4r.\sqrt {\frac {2}{3}} = 24 \sqrt {2}r^3$

Which of the following statements are correct?

0%
Dislocation of ion from lattice site to interstitial site is called Frenkel defect.
0%
Missing of +ve and -ve ions from their respective position producing a pair of holes is called Schottky defect.
0%
The presence of ions in the vacant interstital sites along with lattice point is called interstital defect.
0%
Non-stoichiometric $NaCl$ is yellow solid.

Explanation

Frenkel defect: This type of defect is created when an ion leaves its appropriate site in the lattice and occupies an interstitial site. A hole or vacancy is thus produced in the lattice.

Schottky defect: This type of defect is produced when one cation and anion are missing from their respective positions leaving behind a pair of holes.

The presence of ions in the vacant interstital sites along with lattice point is called interstital defect.

When alkali metal halides are heated in an atmosphere of vapours of the alkali metal, anion vacancies are created.

The anions (halide ions) diffuse to the surface of the crystal from their appropriate lattice sites to combine with the newly generated metal cations. The electron lost by the metal atom diffuse through the crystal is known as F-centres.

The main consequence of metal excess defect in the development of colour in the crystal. For example, when

$NaCl$ crystal is heated in an atmosphere of

$Na$ vapours, it becomes yellow.

Therefore, all statements are correct.

The density of KBr is

$2.75$ g cm

$^{-3}$ . The length of the unit cell is

$654$ pm. Atomic mass of K

$= 39$ amu, Br

$= 80$ amu. What is true about the predicted nature of the solid?

0%
The unit cell is fcc.
0%
The number of atoms per unit cell is $4$ .
0%
There are four constituents per unit cell.
0%
There are $8$ ions at the corners and $6$ at the centres of the faces.

Explanation

As we know,

$Z_{eff}=\dfrac{p \times a^3 \times 10^{-30}\times N_A}{M}$

$=\dfrac{2.75 \times (654)^3 \times 10^{-30}\times 6 \times 10^{23}}{119}=4$
It suggests fcc structure.

Thus, there are four KBr units.

There are

$8$ ions at the corners and

$6$ at the face centers.

For orthorhombic system axial ratios are

$a\neq b\neq c$ and the axial angles are

0%
$\alpha = \beta = \gamma \neq 90^o$
0%
$\alpha = \beta = \gamma = 90^o$
0%
$\alpha = \gamma= 90^o , \beta \neq 90^o$
0%
$\alpha \neq \beta \neq \gamma = 90^o$

A metallic crystal crystallizes into a lattice containing a sequence of layers

$ABABAB$ .... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ?

0%
$74$ %
0%
$26$ %
0%
$50$ %
0%
$None\ of\ these$

The

$\gamma$ -form of iron has fcc structure (edge length

$=386$ pm) and

$\beta$ -form has bcc structure (edge length

$=290$ pm). The ratio of density in

$\gamma$ -form to that in

$\beta$ -form is :

0%
$0.85$
0%
$1.02$
0%
$1.57$
0%
$0.6344$

Explanation

$\begin{array}{l} Z_{e f f} \text { for } \mathrm{fcc}=4 \text { and forbcc }=2 / \text { unit cell } \\ \dfrac{\rho_{\lambda}}{\rho_{\beta}}=\dfrac{\rho_{f c c}}{\rho_{f c c}}=\left(\dfrac{Z_{e f f(f c c)}}{Z_{e f f(b c c)}}\right) \times\left(\dfrac{a_{b c c}}{a_{f c c}}\right)^{3} \\ =\dfrac{4}{2} \times\left(\dfrac{290}{386}\right)^{3}=0.848 \end{array}$

Identical spheres are undergoing two-dimensional packing in square close packing and hexagonal close packing. Which of the following statements are correct regarding the spheres?

0%
The ratio of coordination number for a sphere in first case to that in second case is $2 : 3$ .
0%
Packing in second case is more effective.
0%
Packing in first case is more effective.
0%
The stacking of layer on first type packing produces simple cubic structure.

Explanation

In square close packing, coordination number

$= 8$ , whereas in hcp, coordination number

$= 12$ . So, the ratio is

$2 : 3$ .

If the lattice parameter of Si is

$5.43$

$\mathring {A}$ and the mass of Si atom is

$28.08 \times 1.66 \times 10^{-27}$ kg, the density of silicon in kg m

$^{-3}$ is:

[Given: Silicon has a diamond cubic structure.]

0%
$2330$
0%
$1115$
0%
$3445$
0%
$1673$

Explanation

SInce Si has diamond cubic structure, its unit cell contains

$8$ atoms.
Density

$\displaystyle = \dfrac {\text{Mass}}{\text{Volume}}$
Mass

$\displaystyle = 8 \times 28.08 \times 1.66 \times 10^{-27} = 3.73 \times 10^{-25}$ kg
Volume

$\displaystyle = (5.43 \times 10^{-10})^3 = 1.6 \times 10^{-28} \:$ m

$^3$
Density

$\displaystyle = \frac {3.73 \times 10^{-25} }{ 1.6 \times 10^{-28}} = 2330$ kg m

$^{-3}$

Which of the following characteristic features match(es) with the crystal systems: cubic and tetragonal

0%
Both have the cell parameters; $a=b=c$ and $\alpha =\beta =\gamma$
0%
Both belong to different crystal systems
0%
Both have only two crystallographic angles of $90^o$
0%
Both belong to same crystal system

If the radius of the spheres in the close packing is R and the radius of octahedral voids is r, then r

$= 0.414$ R.

0%
True
0%
False

Explanation

Limiting radius ratio = r/R	Coordination Number of cation	Structural Arrangement (Geometry of voids)	Example
0.155-0.225 0.225-0.414 0.414-0.732 0.414-0.732 0.732-1.000	3 4 4 6 8	Plane Trigonal Tetrahedral Square planar Octahedral Cubic	Boron Oxide $ZnS, SiO_{2}$ - $NaCI,MgO_{2}$ $CsCI$

A metal crystallizes in bcc lattice. The percent fraction of edge length not covered by atom is :

0%
$10.4\%$
0%
$13.4\%$
0%
$12.4\%$
0%
$11.4\%$

Explanation

In BCC lattice,

$4r = a\sqrt{3}$

Fraction of each edge covered by atoms is

$\dfrac{2r}{a} = \dfrac{a\sqrt{3}}{2a} =0.866$ "or"

$86.6 \%$

Therefore, the fraction of edge not covered by atom is

$13.4 \%$

A metal crystallizes into two cubic phases, face-centered cubic and body-centered cubic which have unit cell lengths as

$3.5\ \mathring A$ and

$3.0\ \mathring {A}$ , respectively. Calculate the ratio of densities of fcc and bcc.

0%
$1.26$
0%
$3.25$
0%
$\sqrt2$
0%
None of the above

Explanation

As we know,
Density

$(\rho )=\dfrac {Z_{eff}\times Mw}{a^3\times N_A}$

For fcc,

$Z_{eff}=4$

$\rho _{fcc}=\dfrac {4\times Mw}{(3.5\times 10^{-8})^3 \times N_A}$

For bcc,

$Z_{eff}=2$

$\rho _{fcc}=\dfrac {2\times Mw}{(3\times 10^{-8})^3 \times N_A}$

$\dfrac{\rho _{fcc}}{\rho_{bcc}}=\dfrac {4\times (3)^3}{(3.5)^3 \times 2}=1.26$

The packing efficiency of a two-dimensional square unit cell shown is __________.

0%
39.27%
0%
68.02%
0%
74.05%
0%
78.57%

Explanation

The diagonal is given as

$4r$ , where

$r$ is the radius of an atom.

Diagonal

$= \sqrt {L^2+L^2}=\sqrt {2L}$

$4r$

$=\sqrt {2L}$ or

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

Total area

$=L^2$

$\left ( \dfrac {4r}{\sqrt 2} \right )^2=8r^2$

Number of spheres inside the square is

$1 + 4 \left ( \dfrac {1}{4} \right )=2$

Area of each sphere

$= \pi r^2$

Total area of spheres

$=2 \times \pi r^2$

Packing fraction

$=\dfrac {\text{Total area of spheres}}{\text{Total area}}$

$= \dfrac{2 \times \pi r^2}{8 r^2}=\dfrac{\pi}{4}=0.785$

So, the percentage fraction is

$78.5\%$ .

An element crystallizes in a structure having FCC unit cell of an edge length

$200$ pm. Calculate the density (in g cm

$^{-3}$ ) if

$200$ g of this element contain

$5\, \times\, 10^{24}$ atoms.

0%
$5$
0%
$30$
0%
$10$
0%
$20$

Explanation

Molecular mass of the given element is

$= \dfrac{200 \times 6 \times 10^{23}}{5 \times 10^{24}} = 24$

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

$density = \dfrac{4 \times 24 }{6 \times 10^{23} \times (200 \times 10^{-10})^3}$

$density = 20 \text{ g cm}^{-3}$

The pattern of successive layers of ccp arrangement can be designated as:

0%
AB AB AB
0%
AB ABC AB ABC
0%
ABC ABC ABC
0%
AB BA AB BA

Explanation

The pattern of successive layers of cubic close packing (ccp) arrangement can be designated as ABC ABC ABC.

Simple cubic arrangement can be designated as AAAAAA.

Hexagonal close packing (hcp) arrangement can be designated as ABABAB.

Hence, the correct option is

$\text{C}$

Statement: Packing fraction of FCC and HCP units cells are the same.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

Packing fraction of FCC and HCP units cells are the same.

Both fcc and hcp have a packing fraction of

$0.74$ .

This means that

$74\%$ of both fcc and hcp unit cells are occupied by atoms and

$26\%$ remains empty.

Hence, the given statement is

$\text{true}.$

An element has a FCC structure with edge length 200 pm. Calculate density if 200 g of this element contains

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

0%
$4.16\ g cm^{-3}$
0%
$41.6\ g cm^{-3}$
0%
$4.16\ kg cm^{-3}$
0%
$41.6\ kg m^{-3}$

Explanation

A fcc unit cell contains 4 atoms per unit cell.

Mass of unit cell is equal to mass of 4 atoms. It is equal to

$\displaystyle 4 \times \frac {200}{24 \times 10^{23}} = 3.33 \times 10^{-22}$ g

Volume of the unit cell

$\displaystyle = (200 \times 10^{-10})^3 = 8 \times 10^{-24} \: cm^3$

$\displaystyle Density = \frac {Mass}{Volume} = \frac { 3.33 \times 10^{-22} }{8 \times 10^{-24}} = 41.6 \: g/cm^3$

Which of the following does not posses any plane of symmetry?

0%
Cubic
0%
Monoclinic
0%
Triclinic
0%
Rhombohedral

Explanation

Triclinic crystal lattice does not posses any plane of symmetry.

Plane of symmetry is that imaginary plane which passes through the centre of the crystal and divides it into two equal portions (just mirror images of each other).

For triclinic crystal system,

$\displaystyle a \neq b \neq c, \alpha \neq \beta \neq \gamma = 90^o$

Calculate the void fraction per unit volume of unit cell for such crystal.

0%
0.236
0%
0.67
0%
0.45
0%
0.82

Explanation

When the cation radii are greater or equal to 0.414 R, but less than 0.732 R, where R is the radius of an atom, the cations occupy the octahedral sites. Let us approximate it to 0.5
Thus, the radius of atom B is approximately equal to one half of the radius of atom A.
The packing fraction is

$\displaystyle \frac {\text {volume occupied by the atoms}}{\text {total volume of the unit cell}}=\frac {\displaystyle4 \times \frac {4}{3} \pi r^3+\frac {4}{3} \pi (\frac {r}{2})^3}{(\displaystyle\frac {4r}{\sqrt {2}})^3}=0.764$
The void fraction is

$1-0.764=0.236$

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

0%
$39.27\%$
0%
$68.02\%$
0%
$74.05\%$
0%
$78.57\%$

Explanation

Here the diagonal is 4r, where, r is the radius of atom

We know diagonal

$= \sqrt(L^{2} +L^{2})= \sqrt{2 \times L}$

$4 \times r= {\sqrt2 \times L}$

$L= \dfrac{ 4 \times r }{\sqrt{2}}$

Number of spheres inside the square are

$1+ 4 \times \dfrac{1}{4}$

Area of sphere

$= \pi \times r \times r$

Packing fraction

$= \dfrac{2 \times \pi \times r \times r}{8 \times r^{2} }=\dfrac{\pi}{4}=0.7857$

Percentage is

$78.57\%$

The crystal with

$a= b \neq c, \alpha = \beta =\gamma =90^o$ is:

0%
cubic
0%
orthorhombic
0%
tetragonal
0%
hexagonal

Lithium borohydride 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 is

$21.76$ g mol

$^{-1}$ , calculate the density (in g cm

$^{-3}$ ) of crystal.

0%
$1.34$
0%
$0.87$
0%
$0.94$
0%
$0.67$

Explanation

The expression for density is as follows:

$\displaystyle \text{Density}=\frac{n\times \text{Molar mass}}{V\times N_a}$
Here,

$n=4$

$M=21.76$ g mol

$^{-1}$

$N_a=6.023\times10^{23}$

$V=a\times b\times c$

$\therefore V=6.8\times10^{-8}\times4.4\times10^{-8}\times7.2\times10^{-8}$

$=2.154\times10^{-22}$ cm

$^3$

$\displaystyle\therefore \text{Density}=\frac{4\times21.76}{2.154\times10^{-22}\times6.023\times10^{23}}$

$=0.6709$ g cm

$^{-3}$

CsBr crystallizes in a body-centred cubic lattice of unit cell edge length

$436.6$ pm. Atomic masses of Cs and Br are

$133$ amu and

$80$ amu respectively. The density (in g cm

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

0%
$8.50$
0%
$4.25$
0%
$42.5$
0%
$0.425$

Explanation

For bcc, z=4

The formula for density is given below:

$\displaystyle \text{Density}=\frac{z\times M}{a^3\times N_0}$

$\displaystyle=\frac{2\times213}{(4.366\times10^{-8})^3\times6.02\times10^{23}}$

$=8.50$ g cm

$^{-3}$

Hence, the correct option is

$\text{A}$

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

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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