Solid State Test

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Solid State Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

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)

A
$$1 Fe^{2+}$$ and $$1 S^{2-}_2$$

B
$$2 Fe^{2+}$$ and $$2 S^{2-}_2$$

C
$$4 Fe^{2+}$$ and $$4 S^{2-}_2$$

D
$$6 Fe^{2+}$$ and $$6 S^{2-}_2$$

Solution

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

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

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

$$Z=4 \, i.e., 4Fe^{2+}$$ and $$4S^{2-}_2$$ ions are present in a single unit cell.
Question 2
1 / -0

The unit cell in a body centred cubic lattice is given in the figure. Each sphere has a radius, $$r$$ and the cube has a side, $$a$$.
What fraction of the total cube volume of empty?

A
$$1 - \dfrac {8}{3} \pi \dfrac {r^{3}}{a^{3}}$$

B
$$\dfrac {4}{3}\cdot \pi \dfrac {r^{3}}{a^{3}}$$

C
$$\dfrac {r}{a}$$

D
$$2 - \dfrac {4}{3} \pi \dfrac {r^{3}}{a^{3}}$$

Solution

$$Z_{eff}$$ of BCC=$$2$$
Volume occupied by atoms= $$2\times \cfrac {4}{3}\pi r^3$$
Volume of unit cell= $$a^3$$
Packing fraction= $$\cfrac {8}{3}\pi \cfrac {r^3}{a^3}$$
Fraction of empty unit cell= $$1-\cfrac {8}{3}\pi \cfrac{r^3}{a^3}$$
Answer: (A) $$1-\cfrac {8}{3}\pi \cfrac{r^3}{a^3}$$
Question 3
1 / -0

The closest-packing sequence $$ABAB .....$$ represents

A
primitive cubic packing

B
body-centred cubic packing

C
face-centred cubic packing

D
hexagonal packing

Solution

The closest-packing sequence ABAB..... represents hexagonal packing.
The closest-packing sequence AAAA..... represents primitive cubic packing.
The closest-packing sequence ABCABC..... (with spheres of third layer placed into octahedral voids) represents face-centred cubic packing.
The closest-packing sequence ABCABC..... (with spheres of third layer placed into tetrahedral voids) represents hexagonal packing.
Question 4
1 / -0

The crystal system of a compound with unit cell dimensions $$a = 0.387, b = 0.387, c = 0.504\ nm$$ and $$\alpha = \beta = 90^{\circ}$$ and $$\gamma = 120^{\circ}$$ is:

A
cubic

B
hexagonal

C
orthorhombic

D
rhombohedral

Solution

The crystal system of a compound with unit cell dimensions:

$$\displaystyle a = 0.387, b = 0.387, c = 0.504\ nm$$, $$\displaystyle \alpha = \beta = 90^{\circ}$$ and $$\displaystyle \gamma = 120^{\circ}$$ is hexagonal.
The dimensions are $$\displaystyle a =   b \neq c $$ and $$\displaystyle \alpha = \beta =$$ $$\displaystyle \gamma $$.

Note :
For cubic crystal, the dimensions are $$\displaystyle a =   b = c $$ and $$\displaystyle \alpha = \beta = \gamma= 90^{\circ}$$
For orthorhombic crystal, the dimensions are $$\displaystyle a \neq   b \neq c $$ and $$\displaystyle \alpha = \beta = \gamma = 90^{\circ}$$
For rhombohedral crystal, the dimensions are $$\displaystyle a =   b = c $$ and $$\displaystyle \alpha = \beta = \gamma \neq 90^{\circ}$$
Question 5
1 / -0

Which of the following statements is not correct?

A
The number of carbon atoms in a unit cell of diamond is $$4$$

B
The number of Bravis lattices in which a crystal can be categorized in $$14$$

C
The fraction of the total volume occupied by the atoms in a primitive cell is $$0.48$$

D
Molecular solids are generally volatile

Solution

Contribution of each atom at the corners = $$\cfrac{1}{8}$$
Contribution of each atom on the face center is $$\cfrac{1}{2}$$.
Total contribution is = $$\left(\cfrac{1}{8} \times 8\right) + \left(\cfrac{1}{2} \times 6\right) = 1+3 = 4$$
Now 4 Carbon atoms are entirely inside the crystal in 4 alternate tetrahedral voids in staggered form.
$$\therefore$$ Total no. of atoms = 4 + 4 = 8
Hence, Statement (A) is wrong.
Question 6
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Example of unit cell with crystallographic dimensions $$a\neq b \neq c, \alpha = \gamma = 90^{\circ}, \beta \neq 90^{\circ}$$ is:

A
Calcite

B
Graphite

C
Rhombic sulphur

D
Monoclinic sulphur

Solution

The unit cell with crystallographic dimensions,
$$\displaystyle a\neq b\neq c, \alpha = \gamma = 90^{\circ}$$ and $$\displaystyle \beta \neq 90^{\circ}$$ is monoclinic.
Triclinic unit cell has crystallographic dimensions $$\displaystyle a\neq b\neq c, \alpha \neq \beta \neq \gamma \neq 90^{\circ}$$
Tetragonal unit cell has crystallographic dimensions $$\displaystyle a= b\neq c, \alpha = \beta = \gamma = 90^{\circ}$$
Orthorhombic unit cell has crystallographic dimensions $$\displaystyle a\neq b\neq c, \alpha = \beta = \gamma = 90^{\circ}$$
Question 7
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In which of the following type of lattices, the coordination number is 4?

A
BCC

B
Simple cubic

C
HCP

D
FCC

Solution

In Face centered cubic structure octahedral voids are located at:

(i) Body Center
(ii) Edge Centers
Question 8
1 / -0

Close packing is maximum in the crystal lattice of

A
face-centred cubic

B
body-centred cubic

C
simple-centred cubic

D
none of these

Solution

Close packing is maximum in the crystal lattice of face-centered cubic (fcc) or cubic close packing (ccp) (packing efficiency 74 %).
It is minimum in the crystal lattice of simple cubic unit cell (packing efficiency 52.4 %).
Note: Body-centred cubic unit cell has packing efficiency of 68%.
Question 9
1 / -0

A match box exhibits:

A
Cubic geometry

B
Monoclinic geometry

C
Orthorhombic geometry

D
Tetragonal geometry

Solution

A match box exhibits orthorhombic geometry.

Orthorhombic crystal system has the following unit cell dimensions

$$\displaystyle a \neq b \neq c$$ and $$\displaystyle \alpha = \beta = \gamma = 90^{\circ}$$

The lengths of all sides are different from each other. All the three angles are equal to $$\displaystyle 90^{\circ} $$.

Thus, the option (C) is the correct answer.
Question 10
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The formula for determination of density of a cubic unit cell is:

A
$$\dfrac {a^{3}\times N_{0}}{Z \times M}g\ cm^{-3}$$

B
$$\dfrac {Z\times N_{0}}{M \times a^{3}}g\ cm^{-3}$$

C
$$\dfrac {a^{3}\times M}{Z \times N_{0}}g\ cm^{-3}$$

D
$$\dfrac {Z\times M}{N_{0}\times a^{3}}g\ cm^{3}$$

Solution

The formula for determination of density of cubic unit cell is $$\displaystyle \rho =\dfrac {Z\times M}{N_{0}\times a^{3}}g\ cm^{3} $$
$$\displaystyle \rho = $$ density
$$\displaystyle Z =$$ number of atoms per unit cell
$$\displaystyle M=$$ molecular weight
$$\displaystyle N_0=$$ Avogadro's number
$$\displaystyle a=$$ edge length

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Solid State Test - 37

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Solid State Test - 37

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Question 1

$$1 Fe^{2+}$$ and $$1 S^{2-}_2$$

$$2 Fe^{2+}$$ and $$2 S^{2-}_2$$

$$4 Fe^{2+}$$ and $$4 S^{2-}_2$$

$$6 Fe^{2+}$$ and $$6 S^{2-}_2$$

Solution

Question 2

$$1 - \dfrac {8}{3} \pi \dfrac {r^{3}}{a^{3}}$$

$$\dfrac {4}{3}\cdot \pi \dfrac {r^{3}}{a^{3}}$$

$$\dfrac {r}{a}$$

$$2 - \dfrac {4}{3} \pi \dfrac {r^{3}}{a^{3}}$$

Solution

Question 3

primitive cubic packing

body-centred cubic packing

face-centred cubic packing

hexagonal packing

Solution

Question 4

cubic

hexagonal

orthorhombic

rhombohedral

Solution

Question 5

The number of carbon atoms in a unit cell of diamond is $$4$$

The number of Bravis lattices in which a crystal can be categorized in $$14$$

The fraction of the total volume occupied by the atoms in a primitive cell is $$0.48$$

Molecular solids are generally volatile

Solution

Question 6

Calcite

Graphite

Rhombic sulphur

Monoclinic sulphur

Solution

Question 7

BCC

Simple cubic

HCP

FCC

Solution

Question 8

face-centred cubic

body-centred cubic

simple-centred cubic

none of these

Solution

Question 9

Cubic geometry

Monoclinic geometry

Orthorhombic geometry

Tetragonal geometry

Solution

Question 10

$$\dfrac {a^{3}\times N_{0}}{Z \times M}g\ cm^{-3}$$

$$\dfrac {Z\times N_{0}}{M \times a^{3}}g\ cm^{-3}$$

$$\dfrac {a^{3}\times M}{Z \times N_{0}}g\ cm^{-3}$$

$$\dfrac {Z\times M}{N_{0}\times a^{3}}g\ cm^{3}$$

Solution

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