Solid State Test

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Solid State Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of a number of rectangular planes and diagonal plane in a cubic unit cell is ____.

A
$$1 : 2$$

B
$$3 : 1$$

C
$$2 : 3$$

D
$$3 : 4$$

Solution

The ratio of rectangular plane and diagonal plane in cubic unit cell is $$3:6$$ i.e, $$1:2$$

There are 3 planes that lie parallel to the side of the cube,that these are rectangular planes.while 6 planes goes through opposite edges.And also there are diagonal planes.

And Hence, the ratio of rectangular plane and diagonal plane in cubic unit cell is $$3:6$$ i.e, $$1:2$$
Question 2
1 / -0

A CCP of $$n$$ spheres of radius $$r$$, $$n$$ spheres of radius $$0.414\, r$$ and $$2n$$ spheres of radius $$0.225\, r$$ are introduced in the octahedral and tetrahedral void respectively. Therefore packing of the lattice is:

A
$$0.74$$

B
$$0.81$$

C
$$0.86$$

D
$$0.79$$

Solution

$$ \begin{array}{l} \text { Given: - CCP or FCC packing radius=r } \\ \text { Zeffective }=\frac{1}{8} \times 8+\frac{1}{2} \times 6=4 \end{array} $$

$$ \begin{array}{l} \text { Let radius of sphere of atom of Fcc be r } \\ \text { and side of } F c c \text { be a } \end{array} $$

$$ \begin{aligned} \text { Then } &: a \sqrt{2}=4 \mathrm{~r} \\ & \Rightarrow a=2 \sqrt{2} \mathrm{r} \end{aligned} $$

$$ \begin{array}{l} \because \text { There are n spheres, } 80 \text { total number of } \\ \text { unit cells = } \frac{n}{4} \\ \text { volume of Lattice }=\frac{n}{4} \times(a)^{3}=\frac{n}{4}(2 \sqrt{2} r)^{3} \\ =4 \sqrt{2} n r^{3} \end{array} $$

$$ \begin{aligned} \text { volume of atoms }=& n \times \frac{4}{3} \times \pi r^{3}+n \times \frac{4}{3} \times \pi \times(0.414 \mathrm{~r})^{9} \\ +2 n \times \frac{4}{3} \times \pi(0.225 \mathrm{})^{3} \\ =& \frac{n \times 4}{3} \times r^{3} \times \pi\left[1+(0.414)^{3}+(0.225)^{9}\right] \\ =& 4.534 \mathrm{nr}^{3} \end{aligned} $$

$$ \begin{array}{l} \therefore \text { packing of lattice }=\dfrac{4.534 \mathrm{nr}^{9}}{4 \sqrt{2} \mathrm{nr}^3}=0.81 \\ \text { So, option (B) is correct. } \end{array} $$
Question 3
1 / -0

Molybdenum (At. wt. $$= 96 \ g/mol$$) crystallizes as $$bcc$$ crystal. If density of crystal is $$10.3 \ g/cm^3$$, then radius of $$Mo$$ atom is: [Use : $$\,N_A = 6 \times 10^{23}$$]

A
$$111$$pm

B
$$314$$pm

C
$$135.96$$pm

D
none of these

Solution

Since, Molybladenum crystallizes as $$bcc$$ rystal. So, $$N = 2$$.
Expression for density is:

$$\rho \quad =\quad \dfrac { N\times M }{ { N }_{ A }\times { a }^{ 3 } } $$

$$ { a }^{ 3 }\quad =\quad \dfrac { N\times M }{ { N }_{ A }\times \rho } $$

$$ { a }^{ 3 }\quad =\quad \dfrac { 2\times 96 }{ 6\times { 10 }^{ 23 }\times 10.3 } $$

$$ { a }^{ 3 }\quad =\quad 3.1\times { 10 }^{ -23 }\quad { cm }^{ 3 }$$

$$a\quad =\quad 3.14\times { 10 }^{ -8 }\ cm\quad =\quad 314\quad pm$$.

For, $$bcc$$ crystal,

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

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

$$ r\quad =\quad \dfrac { \sqrt { 3 } }{ 4 } \times 314$$

$$r\quad =\quad 136\quad pm$$.
Question 4
1 / -0

The unit cell with the structure above refer to which crystal system?

A
Cubic

B
Orthorhombic

C
Tetragonal

D
Trigonal

Solution

For orthorhombic crystal structure all three sides are not equal i.e. $$a\neq b\neq c$$ and $$\alpha=\beta=\gamma=90^o$$
Therefore, it is an orthorhombic crystal system.

Option $$B$$ is correct.
Question 5
1 / -0

Arrange the following in decreasing order of melting point: $$NaCl,SiO_{2},Dry Ice, Ice$$

A
$$NaCl>SiO_{2}>Dry Ice>Ice$$

B
$$SiO_{2}>NaCl>Dry
Ice>Ice$$

C
$$NaCl>SiO_{2}>Ice>Dry
Ice$$

D
$$SiO_{2}>NaCl>Ice>Dry
Ice$$
Question 6
1 / -0

The number of parameters that defines a unit cell is

A
2

B
3

C
4

D
6
Question 7
1 / -0

Match the column 1 correctly with the column 2 given below

A
A=3 B=1 C=4 D=2

B
A=3 B=2 C=1 D=4

C
A=1 B=2 C=3 D=4

D
A=2 B=1 C=3 D=4

Solution
Question 8
1 / -0

The incorrect statement among the following is:

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

B
Molecular solids are generally volatile.

C
The number of carbon atoms in an unit cell of diamond is 8.

D
The number of Bravais lattices in which a crystal can be categorized is 14.
Question 9
1 / -0

$$AB$$ has $$NaCl$$ structure while $$XB$$ has $$CsCl$$ structure if $${ r }_{ A }|{ r }_{ B }=\ 5$$ and $${ r }_{ X }|{ r }_{ B }= 0.75$$ then what is the ratio of edge length of unit cell of $$AB$$ and $$XB$$.

A
$$1.52$$

B
$$1.50$$

C
$$1.40$$

D
$$0.85$$

Solution

$$AB\dfrac{r_{A}}{r_{B}}=0.5=\dfrac{1}{2}$$
$$XB\dfrac{r_{X}}{r_{B}}=0.75=\dfrac{1.5}{2}$$
$$r_{B=2}$$ $$r_{X=1.5}$$ $$r_{A=1}$$
$$XB$$ form $$C_{5}Cl$$ structure $$50\ B^{-}$$
takes $$CCP$$ porition
$$4r=\sqrt{2}a$$
$$a=\dfrac{4r_{B}}{\sqrt{2}}=\dfrac{4\times 2}{\sqrt{2}}=\dfrac{8}{\sqrt{2}}$$
As form $$NaCl$$ structure $$50\ B^{-}$$ takes simple cable structure
$$a=2r_{B}=2_{X2}=4$$
eape length of $$A-B$$: edge length of $$XB$$
$$\dfrac{8}{\sqrt{2}}:4=1.4$$
Question 10
1 / -0

The ratio of the volume of tetragonal lattice to that of a hexagonal lattice is?

A
$$\frac { \sqrt { 3 } }{ 2 } abc$$

B
$$\frac { 2 }{ \sqrt { 3 } } $$

C
$$\frac { 2a^ 2 }{ \sqrt { 3b } } c$$

D
$$\frac { \sqrt { 3a^ 2 } }{ 2bc } $$

Solution

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Solid State Test - 67

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Solid State Test - 67

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Question 1

$$1 : 2$$

$$3 : 1$$

$$2 : 3$$

$$3 : 4$$

Solution

Question 2

$$0.74$$

$$0.81$$

$$0.86$$

$$0.79$$

Solution

Question 3

$$111$$pm

$$314$$pm

$$135.96$$pm

none of these

Solution

Question 4

Cubic

Orthorhombic

Tetragonal

Trigonal

Solution

Question 5

$$NaCl>SiO_{2}>Dry Ice>Ice$$

$$SiO_{2}>NaCl>Dry Ice>Ice$$

$$NaCl>SiO_{2}>Ice>Dry Ice$$

$$SiO_{2}>NaCl>Ice>Dry Ice$$

Question 6

2

3

4

6

Question 7

A=3 B=1 C=4 D=2

A=3 B=2 C=1 D=4

A=1 B=2 C=3 D=4

A=2 B=1 C=3 D=4

Solution

Question 8

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

Molecular solids are generally volatile.

The number of carbon atoms in an unit cell of diamond is 8.

The number of Bravais lattices in which a crystal can be categorized is 14.

Question 9

$$1.52$$

$$1.50$$

$$1.40$$

$$0.85$$

Solution

Question 10

$$\frac { \sqrt { 3 } }{ 2 } abc$$

$$\frac { 2 }{ \sqrt { 3 } } $$

$$\frac { 2a^ 2 }{ \sqrt { 3b } } c$$

$$\frac { \sqrt { 3a^ 2 } }{ 2bc } $$

Solution

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$$SiO_{2}>NaCl>Dry
Ice>Ice$$

$$NaCl>SiO_{2}>Ice>Dry
Ice$$

$$SiO_{2}>NaCl>Ice>Dry
Ice$$