MCQ Questions for Class 12 Engineering Chemistry The Solid State Quiz 7

Which of the following statements is/are correct about tetrahedral voids (TVs) in an fcc unit cell ?

0%
Number of TVs per atom in fcc unit cell is $2$ .
0%
Number of TVs per unit cell is $8$ .
0%
Number of TVs is twice the number of atoms in the fcc unit cell.
0%
Number of TVs is equal to the number of atoms in the fcc unit cell.

Explanation

Tetrahedral interstices : We have seen that in hexagonal close packing (hcp) and cubic close packing (ccp), each sphere of second layer touches with three spheres of first layer. Thus, they leave a small space in between which is known as tetrahedral site or interstices or the vacant space between

$4$ touching spheres is called as tetrahedral void. Since a sphere touches three spheres in the below layer and three spheres in the above layer, there are two tetrahedral sites associated with one sphere. It may by noted that a tetrahedral site does not mean that the site is tetrahedral in geometry but it means that this site is surrounded by four spheres and the centres of these four spheres lie at the apices of a regular tetrahedron.
In FCC, one corner and its face centres form a tetrahedral void,
In FCC, two tetrahedral voids are obtained along one cube diagonal. So, in FCC,

$8$ tetrahedral voids are present.
In FCC, total number of atoms

$= 4$ .
In FCC, total number of tetrahedral voids

$= 8$ .
So, we can say that in 3D close packing,

$2$ tetrahedral voids are attached with one atom.

Statement: The maximum number of Bravais lattices is shown by tetragonal type crystals.

State whether the given statement is true or false.

0%
True
0%
False

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	$a\neq b\neq c$	$\alpha\neq \beta \neq \gamma \neq 90^o$	Primitive $=$ $1$
Hexagonal	$a= b\neq c$	$\gamma=120^o$	Primitive $=$ $1$

The maximum number of bravais lattices are shown by orthorhomic crystals. Hence, the above statement is false.

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%
$ABCBCABCBC$

The empty space left in hcp in three dimensions is :

0%
$26\%$
0%
$74\%$
0%
$52.4\%$
0%
$80\%$

Explanation

Packing efficiency is the ratio of the volume occupied by spheres in unit cell to the volume of hcp unit cell multiplied by 100.

For hcp unit cell, packing efficiency (or volume occupied) is 74%.

The empty space left in hcp in three dimensions is

$100-74=26$ %.

Chromium metal crystallizes as a body-centered cubic lattice. The length of the unit cell edge is found to be

$287$ pm. What would be the density (in g cm

$^{-3}$ ) of chromium?

0%
$3.40$
0%
$5.20$
0%
$12.10$
0%
$7.30$

Explanation

$a$ is the edge length, then for bcc lattice, we have

$r=\dfrac {\sqrt3 a}{4}=\dfrac{\sqrt 3 \times 287}{4} =124.27 \mathring {A}$
Density

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

$=\dfrac{2\times 51.99}{(287 \times 10^{-10})^3 \times 6.023 \times 10^{23}}$

$=7.30$ g cm

$^{-3}$

Z for BCC is 2.

Hence, the correct option is

$\text{D}$

Assertion (A): Triclinic system is the most unsymmetrical system.

Reason (R): No axial angle is equal to

$90^o$ in the triclinic system.

Choose the correct option.

0%
Both (A) and (R) are correct and (R) is the correct explanation of (A)
0%
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
0%
(A) is correct but (R) is incorrect
0%
(A) is incorrect but (R) is correct
0%
Both (A) and (R) are incorrect

Explanation

The edges of the triclinic system are not equal as

$a \neq b \neq c$ . Hence, it is known as the most unsymmetrical system.

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 $=14$

Statement: Packing fraction of a lattice structure depends on the radius of the atom crystallizing in it.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

Packing efficiency is the ratio of the total volume occupied by atoms to the total volume of the cube.

Packing efficiency

$=\dfrac{n\times \dfrac 43 \pi r^3}{V}$

For FCC,

$n=4$ ,

$r=\dfrac{a}{2\sqrt2}$ ,

$V=a^3$

Packing fraction

$=\dfrac{4\times \dfrac 43\pi \times \left(\dfrac {a}{3\sqrt 2} \right)^3}{a^3} =\dfrac{\pi}{3\sqrt 2} =0.74$

Hence, packing efficiency depends on the radius of an atom.

Hence, the given statement is

$\text{true}$

Statement: It is much more difficult to describe the crystal structure of compounds than those of elements.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

It is much more difficult to describe the crystal structure of compounds than those of elements. In compounds, different constituents may occupy different lattice points. Hence, each constituent has a different position. This increases complexity.

Hence, the given statement is

$\text{true}$

Which of the following is not a crystalline solid?

0%
Common salt
0%
Sugar
0%
Iron
0%
Rubber

Explanation

Common salt, sugar and iron are crystalline solids whereas rubber is an amorphous solid as it does not have well developed perfectly ordered crystalline structure. Other amorphous solids include tar, glass, plastic, butter etc.

Hence, the correct option is

$\text{D}$

KF has a NaCl structure. What is the distance (in pm) between

$K^+$ and

$F^-$ in KF if the density is

$2.48$ g cm$$^{-3}

0%
$220$
0%
$135$
0%
$540$
0%
$269$

Explanation

As we know,
for such structure

$2r^+ + 2r^ - = a$
Also, as we know,
density

$= nM/VN_o$
here, n= 4

$2.48 = 4*58/(a)^3(N_o)$
distance between

$K^+\, and\, F^-$ in KF = a/2 = 269 pm

Which of the following is an amorphous substance?

0%
Zinc
0%
Clay
0%
Salt
0%
Sugar

Ice crystallizes in a hexagonal lattice. At low temperature, at which the structure was determined, the lattice constants were

$a\, =\, 4.53\ \mathring A$ and

$b = 7.60\ \mathring A$ (see figure). How many molecules of water are contained in a given unit cell? [Density of ice is

$0.92$ g cm

$^{-3}$ .]

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

As we know, density

$= \dfrac{nM}{VN_o}$ , where n is the number of atoms per unit cell.
Density

$=0.92 = \dfrac{n\times 18}{(a^2b)\times (6\times 10^{23})}$

$\implies n =4$
So, one unit of this structure contains

$4$ molecules of water.

Potassium crystallizes in a body-centered cubic lattice with edge length aequal to 5.2

$\mathring A$ , what is the calculated density (in g cm

$^{-3}$ ) of crystalline potassium ?

0%
$2.2$
0%
$1.9$
0%
$0.92$
0%
None of the above

Explanation

As we know, density

$=d= \dfrac{nM}{VN_o}$
Here,

$n= 2$

$d = \dfrac{2\times 39}{(a)^3(N_o)}=0.92$ g cm

$^{-3}$

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

$124$ pm, compute the density (in g cm

$^{-3}$ ) of iron in bcc and fcc respectively :

0%
$8.4,\ 9.8$
0%
$9.3,\ 4.6$
0%
$8.6,\ 7.9$
0%
$7.9,\ 8.6$

Explanation

The expression for density is as follows:

Density

$=d = \dfrac{zM}{VN_o}$

For bcc,

$z =2$

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

So, density

$=\dfrac{ 2\times 56}{a^3(6\times 10^{23})} = 7.9$ g cm

$^{-3}$

For fcc,

$n =4$

$a = 2\sqrt {2} r$

So, density

$= \dfrac{4\times 56}{a^3(6\times 10^{23})} = 8.6$ g cm

$^{-3}$

Hence, the density of iron in bcc and fcc are

$7.9$ g cm

$^{-3}$ and

$8.6$ g cm

$^{-3}$ respectively.

Hence, the correct option is

$\text{D}$

A unit cell of sodium chloride has four formula units. The edge of length of the unit cell is

$0.564$ nm. What is the density (in g cm

$^{-3}$ ) of sodium chloride?

0%
$1.5$
0%
$2.0$
0%
$2.2$
0%
$3.2$

Explanation

As we know, the expression for density is as follows:
Density

$=d = \dfrac{nM}{VN_o}$ , where

$n$ is the number of atoms and

$M$ is the molecular mass.
For NaCl,

$n =4$
Density

$= \dfrac{4\times 58.5}{(0.564\times 10^{-9})^3(6\times 10^{23})} = 2.2$ g cm

$^{-3}$

Pyrite,

$FeS_2$ crystallises in a cubic unit cell with a cell edge of 0.54 nm. The density of pyrite is

$5.02 g/cm^3$ . How many

$Fe^{2+}$ and

$S^{2-}_2$ ions are present in a single unit cell?
(S = 32, Fe =56u)

0%
$1 Fe^{2+}$ and $1 S^{2-}_2$
0%
$2 Fe^{2+}$ and $2 S^{2-}_2$
0%
$4 Fe^{2+}$ and $4 S^{2-}_2$
0%
$6 Fe^{2+}$ and $6 S^{2-}_2$

Explanation

$d=\dfrac{ZM}{a^3N_A}$

$M= 2 \times 32 + 56=120$

$5.02=\dfrac {Z\times 120}{6.023 \times 10^{23}\times (5.4 \times 10^{-8})^3}$

$Z=4 \, i.e., 4Fe^{2+}$ and

$4S^{2-}_2$ ions are present in a single unit cell.

A compound having bcc geometry has atomic mass

$50$ amu. Calculate the density (in g cm

$^{-3}$ ) of the unit cell, if its edge length is

$290$ pm.

0%
$6.81$
0%
$3.40$
0%
$13.62$
0%
None of the above

Explanation

The expression for the density of the unit cell is given below.

$d=\displaystyle\frac{n\times M}{a^3\times N_A}$

Here,

$M=50$

For bcc,

$n=2$

$a=290$ pm

$=290\times10^{-10}$ cm

Substituting values in the expression for density, we get

$d=\displaystyle\frac{2\times50}{(290\times10^{-10})^3\times6.023\times10^{23}}=6.81$ g cm

$^{-3}$

Hence, the density of the unit cell is

$6.81$ g cm

$^{-3}$ .

The most unsymmetrical crystal system is:

0%
cubic
0%
hexagonal
0%
triclinic
0%
orthorhombic

Explanation

The most unsymmetrical crystal system is triclinic while the most symmetrical crystal system is cubic.

The axial distances and the axial angles,

$a\neq b\neq c$ and

$\alpha\neq\beta\neq\gamma\neq 90^{\circ}$ represents triclinic system.

None of the edge lengths is equal and none of the angles is equal to each other. Also, none of the angles is equal to

$90^0$ .

Examples of triclinic crystal systems include

$K_2Cr_2O_7$ and

$H_3BO_3$ .

Hence, the correct option is

$\text{C}$

A pseudo solid is:

0%
glass
0%
pitch
0%
KCl
0%
glass and pitch both

Explanation

Pseudo solids (such as glass & pitch) lack a regular three-dimensional arrangement of atoms.

Pitch is a viscoelastic polymer which can be natural or manufactured, derived from petroleum.

Solids are most stable in crystalline form. However, if a solid is formed rapidly (for example, when a liquid is cooled suddenly), its atoms or molecules do not have time to align themselves and may become locked in positions other than those of a regular crystal. Such resulting solids are called Pseudo solids.

Hence, the correct option is

$\text{D}$

The lattice parameters,

$a\neq b\neq c$ and

$\alpha\neq\beta\neq\gamma\neq 90^{\circ}$ represents a__________.

0%
tetragonal system
0%
orthorhombic system
0%
monoclinic system
0%
triclinic system

Explanation

The axial distances and the axial angles,

$a\neq b\neq c$ and

$\alpha\neq\beta\neq\gamma\neq 90^{\circ}$ represents triclinic system.
None of the edge lengths are equal and none of the angles are equal to each other. Also, none of the angles are equal to

$90^0$ .
Examples of triclinic crystal systems include

$K_2Cr_2O_7$ and

$H_3BO_3$ .

In a simple cubic cell, each point on a corner is shared by:

0%
2 unit cells
0%
1 unit cell
0%
8 unit cells
0%
4 unit cells

Explanation

In a simple cubic cell, each point on a corner is shared by

$8$ unit cells.
Thus, each point on a corner contributes one-eight to the unit cell.

Hence, the correct option is

$\text{C}$

Which of the following are the correct axial distances and axial angles for rhombohedral system?

0%
$\;a=b=c,\,\alpha=\beta=\gamma\neq 90^{\circ}$
0%
$\;a=b\neq c,\;\alpha=\beta=\gamma=90^{\circ}$
0%
$\;a\neq b=c,\;\alpha=\beta=\gamma=90^{\circ}$
0%
$\;a\neq b\neq c,\;\alpha\neq\beta\neq\gamma\neq 90^{\circ}$

Explanation

The correct axial distances and axial angles for rhombohedral system are

$\;a=b=c,\,\alpha=\beta=\gamma\neq 90^{\circ}$ .

All the three edge lengths are equal and all the three angles are equal but none of the angles is equal to

$90^o$ .

Examples of rhombohedral crystal include calcite,

$NaNO_3$ and

$CaCO_3$ .

Hence, the correct option is

$\text{A}$

Close packing is maximum in the crystal lattice of:

0%
simple cubic
0%
face-centred
0%
body-centred
0%
hexagonal closed packing

Explanation

The space occupied in sc, bcc and fcc and hcp structures are $52\%$ , $68\%, 74\%$ and 74% respectively. Hence, the close packing is maximum in the crystal lattice of face centred cubic and hexagonal close packing.

Hence, the correct options are $\text{B}$ and $\text{D}$

The empty space in hcp and ccp is about :

0%
$26\%$
0%
$30\%$
0%
$35\%$
0%
$40\%$

Explanation

The space occupied by spheres of equal size in three dimensions in both hcp and ccp arrangements is

$74\%$ . The vacant space is

$100-74=26\%$ .

For an orthorhombic system, axial ratios are

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

0%
$\;\alpha=\beta=\gamma\neq 90^{\circ}$
0%
$\;\alpha=\beta=\gamma=90^{\circ}$
0%
$\;\alpha\neq \beta=\gamma=90^{\circ}$
0%
$\;\alpha\neq\beta\neq\gamma=90^{\circ}$

Explanation

For orthorhombic system axial ratios are

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

$\;\alpha=\beta=\gamma=90^{\circ}$ . Thus, all the three edge lengths are unequal but all the angles are equal. They are equal to

$90^0$ .

Which of the following is a primitive unit cell?

0%
Simple cubic
0%
Body-centred cubic
0%
Face-centred cubic
0%
Both body-centred and face-centred cubic

A tetrahedral void in a crystal implies that:

0%
the shape of the void is tetrahedral
0%
molecules forming the void are tetrahedral in shape
0%
the void is surrounded tetrahedrally by four spheres
0%
the void is surrounded by six spheres

Which one of the following is not a close packing?

0%
HCP
0%
CCP
0%
BCC
0%
FCC

Explanation

The space occupied in hcp, ccp, bcc and fcc structures are $74\%$ , $74\%$ , $68\%$ and $74\%$ respectively. Thus, bcc is not a close packing.

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

If the co-ordination number of an element in its crystal lattice is 8, then packing is :

0%
fcc
0%
hcp
0%
bcc
0%
none of the above

Explanation

The co-ordination numbers of an element in fcc, hcp and bcc crystal lattice are $12$ , $12$ and $8$ respectively. Hence, if the coordination number of an element in its crystal lattice is $8$ , then packing is bcc.

The number of octahedral sites in a cubical close packed array of

$N$ spheres is :

0%
$\dfrac N2$
0%
2N
0%
N
0%
4N

Explanation

The ratio of spheres and octahedral void in ccp is

$1:1$ . Therefore, for N spheres, the number of octahedral voids will be N.

The number of octahedral sites in a cubical close packed array of N spheres is N.

For example, a ccp lattice contains

$4$ atoms per unit cell. It has

$4$ octahedral voids and

$8$ tetrahedral voids.

At zero kelvin, most of the ionic crystals possess :

0%
Frenkel defect
0%
Schottky defect
0%
metal excess defect
0%
No defect

An element (atomic mass

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

$400$ pm. The density (in g cm

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

0%
$10.376$
0%
$5.188$
0%
$7.289$
0%
$2.144$

Explanation

Given, M

$=$ 100 g/mol, a

$=$ 400pm

For BCC -

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

As we know,

$\text{Density}\displaystyle=\frac{n\times M}{a^3\times N_A}$

$\displaystyle=\frac{2\times100}{(400\times10^{-10})^3\times6.02\times10^{23}}$

$=5.188$ g cm

$^{-3}$

$(\because 400pm= 400\times 10^{-10} cm)$

Hence, the correct option is B.

The available space occupied by spheres of equal size in three dimensions in both hcp and ccp arrangement is:

0%
74 %
0%
70 %
0%
60.4 %
0%
52.4 %

Explanation

$\text { Solution:}-\quad(z=\text { no. of atoms per unit cell) } \\$

$\text { [Percentage of occupied volume }=z \times \dfrac{\text { Volume of sphere }}{\text { volume of cube /hexagon}}]$

$z=4,$

$4 \pi=a \sqrt{2} \\$

$r=a \sqrt{2} \\$

$\text { percentage }=4\times \dfrac{4 / 3 \pi \gamma^{3}}{a^{3}} \times 100 \%$

$=\dfrac{16}{3} \times \dfrac{\pi \gamma^{3}}{a^{3}} \\$

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

$\text { (%of occupied Vol. } =74 \% \text { ) }$

$\text { for hcp, }$

$z=6 \quad 2 r=a \\$

$r=9 / 2 \\$

$\text { percentage }=6 \times \dfrac{4 / 3 \pi \gamma^{3}}{3 \sqrt{2} a^{3}}$

$\text { (Volume of hexagon } \left.=3 \sqrt{2} \mathrm{a}^{3}\right) \\$

$\quad=\dfrac{\pi \gamma^{3}}{3 \sqrt{2} a^{3}} \times 100 \\$

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

$\text { (%of occupied Vol. } =74 \% \text { ) }$

$\text { Hence correct ans is }(A)$

A truncated tetrahedron has_______________.

0%
$12$ vertices
0%
$8$ faces
0%
$18$ edges
0%
$3$ hexagonal faces

Solid sodium amide,

$NaNH_2$ has a cubic closed packed arrangement of

$N{ H }_2^-$ ions with _____ voids occupied by

$Na^+$ ions.

0%
half the number of octahedral
0%
all tetrahedral
0%
half the number of tetrahedral
0%
double the number of octahedral

Explanation

$Z_{effective}\quad of\quad {NH_2}^{-}=6\times \cfrac {1}{2}+8\times \cfrac {1}{8}=4$ (CCP arrangement)

To maintain the

$NaNH_2$ formula

$Z_{effective}$ of

$Na^{+}$ must be

$4$ .

So,

$Na^{+}$ will be present at alternate tetrahedral voids.

$Z_{effective}$ of

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

Answer: (C) half the number of tetrahedral

An element crystallizes in a structure having a fcc unit cell with edge length

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

$^{-3}$ ) if

$200$ g of this element contain

$24\times10^{23}$ atoms.

0%
$10.4$
0%
$2.6$
0%
$41.6$
0%
$83.2$

Explanation

Weight of

$24\times10^{23}$ atoms

$=200$ g

$\therefore\,$ Weight of

$4$ atoms

$=\displaystyle\frac{200}{24\times10^{23}}\times4$ g

The fcc structure contains

$4$ atoms in a unit cell.

Density

$=\displaystyle\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$

$=\displaystyle\frac{200\times4}{24\times10^{23}\times(200\times10^{-10})^3}$

$=41.6$ g cm

$^{-3}$

Hence, the density of the unit cell is

$41.6$ g cm

$^{-3}$ .

The more efficient mode of packing of identical atoms in one layer is :

0%
square close packing pattern
0%
hexagonal close packing pattern
0%
trigonal closed packing
0%
none of the above

A metallic crystal crystallizes into a lattice containing a sequence of layers AB AB AB.... . Any packing of spheres leaves out voids in the lattice. What percentage of the volume of this lattice is empty space?

0%
$\;74\%$
0%
$\;26\%$
0%
$\;50\%$
0%
$\;33\%$

Explanation

AB AB AB describes the hexagonal close-packed arrangement of spheres.

The available space occupied by spheres of equal size in three dimensions in the hexagonal close packed arrangement is 74%.

The vacant space is

$100-74=26$ %.

Hence, the correct option is

$\text{B}$

Crystal system in which maximum number of Bravais lattices are possible is:

0%
cubic
0%
triclinic
0%
orthorhombic
0%
rhombohedral

The maximum proportion of available volume that can be filled by hard spheres in diamond is:

0%
$0.52$
0%
$0.34$
0%
$0.32$
0%
$0.68$

Explanation

The maximum packing or the maximum proportion of volume filled by hard spheres in various arrangements are as follows:

(a) Simple cubic

$\displaystyle=\frac{\pi}{6}=0.52$

(b) Body-centered cubic

$\displaystyle=\frac{\pi\sqrt{3}}{8}=0.68$

(c) Face-centered cubic

$\displaystyle=\frac{\pi\sqrt{2}}{6}=0.74$

(d) Hexagonal closed packing

$\displaystyle=\frac{\pi\sqrt{2}}{6}=0.74$

(e) Diamond

$\displaystyle=\frac{\pi\sqrt{3}}{6}=0.34$

Which crystal system does the unit cell dimensions,

$a\neq b\neq c$ and

$\alpha= \beta= \gamma$ represent?

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

Explanation

$a \neq b \neq c$

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

3 types:-

1. Body centred orthorhombic: lattice point in middle of unit cell.

2. Base centerd: lattice point in middle of each of 2 ends.

3. Face centred = middle of each side.

1998681_316703_ans_8079165f4ff1466f9e76c8209026ef3b.jpg

The ABAB packing and ABCABC packing are respectively called as:

0%
hexagonal close packing (hcp) and cubic close packing (ccp)
0%
cubic closed packing (ccp) and hexagonal closed packing (hcp)
0%
body centered cubic (bcc) and hexagonal close packing (hcp)
0%
hexagonal closed packing (hcp) and body-centered cubic (bcc)

Explanation

$\text { Solution:- Packing of lattice points in a crystal is of 2-types:- } \\$

$\text { 1.) Cubic dose packing (CCP) } \\$

$\text { 2.) hexagonal close packing (HCP) } \\$

$\text { layers of packing sphere or lattice points } \\$

$\text { in ccp & hcp are } 3 \text { . }$

$\text { - CCP is ABCABC ... type } \\$

$\text { and } \\$

$\text { - HCP is } A B \triangle B A B \text { .... type. }$

$\text { Examples:- } \\$

$\qquad \text { - Zinc blende }(Z n S)(4: 4)-c c p \text { . } \\$

$\qquad \text { - Wurtzite }(Z n S)(4: 4)-h c p \text { . } \\$

$\text { hence correct option is (A) }$

1998226_278670_ans_c6ccee84d7e14113b133a6f267a08b07.png

Tetragonal crystal system has which of the following unit cell dimensions?

0%
$a = b = c$ and $\alpha = \beta = \gamma = 90^{\circ}$
0%
$a = b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$
0%
$a \neq b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$
0%
$a = b \neq c$ and $\alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$

$ZnS$ can be existing in the ________ structure other than zing blende structure.

0%
Bcc
0%
Wurtzite
0%
Simple cubic
0%
Rock salt

Average composition of ordinary glass is:

0%
$Na_2O\cdot CaO\cdot 6SiO_2$
0%
$Na_2O\cdot 2CaO\cdot 3SiO_2$
0%
$Na_2O\cdot CaO\cdot 6SiO_2\cdot Cr_2O_3$
0%
$Na_2O\cdot CaO\cdot PbO\cdot 6SiO_2$

Explanation

$Na_2CO_3+xSiO_2\rightarrow Na_2O\cdot xSiO_2+CO_2$

$Na_2SO_4+C\rightarrow Na_2O+CO+SO_2$

$Na_2O+ySiO_2\rightarrow Na_2O\cdot ySiO_2$

$CaCO_3+zSiO_2\rightarrow CaO\cdot zSiO_2+CO_2\uparrow$

The overall reaction on simultaneous fusion may be represented as:-

$Na_2CO_3+CaCO_3+6SiO_2\rightarrow Na_2O\cdot CaO\cdot 6SiO_2+CO_2\uparrow$

Specifically called as soda lime glass composed of approximately

$75$ %

$SiO_2$ ,

$Na_2O$ and

$CaO$ (lime).

Option A is the answer.

A compound

$AB$ has a rock type structure with

$A:B =1:1$ . The formula mass of

$AB$ is

$6.023$

$y^{1/3}$ y amu and the closest

$A-B$ distance is

$y^{1/3}$ nm.Calculate the density of lattice.

0%
$5.0 kg/m^3$
0%
$10 kg/m^3$
0%
$50kg/m^3$
0%
$20 kg/m^3$

Explanation

$AB$ has rock-salt

$(A:B: :1:1)$ structure, i.e., fcc structure

$(n=4)$ and formula mass of

$AB$ is 6.023 y g having closest distance

$A-B\:y^{1/3}$ nm. Therefore, edge length of unit cell

$(a)=2(A^++B^-)=2\times y^{1/3}\times10^{-9}m$

$\therefore$

$Density of AB=\displaystyle\frac{n\times\:formula\:mass} {Av.no.\times V}=\frac{n\times\:formula\:mass}{6.023\times10^{23}\times a^3}$

$=\displaystyle\frac{4\times6.023\times y\times10^{-3}}{6.023\times10^{23}\times(2y1/3\times10^{-9})^3}$ (formula mass in kg)

$=5.0kg/m^3$

Pick out the crystalline substance from the following.

0%
Coal
0%
Wax
0%
Clay
0%
Sugar

Explanation

In sugar, the constituent particles are arranged in a specific manner. Hence, it is a crystalline solid.

In coal, wax, and clay the constituent particles are arranged in a random manner. Hence, it is an amorphous solid.

Hence, the correct option is

$\text{D}$

Which of the following is an amorphous substance?

0%
Zinc
0%
Clay
0%
Salt
0%
Sugar

The space lattice of graphite is:

0%
cubic
0%
tetragonal
0%
rhombic
0%
hexagonal

Identify the crystalline solid among the following :

0%
Glass
0%
Carbon black
0%
Resin
0%
Silver

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 7 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 7 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.