Solid State Test

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Solid State Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

A metal crystallizes in two cubic phases, fcc and bcc whose unit cell lengths are $$3.5 \mathring A$$ and $$3.0 \mathring A$$ respectively. The ratio of density of fcc and bcc is :

A
$$2$$

B
$$1.26$$

C
$$3.34$$

D
$$1.8$$

Solution

The formula for the calculation of density is given below:

$$\text{Density}= \displaystyle\frac { Z \times M} {N_0 \times a^3 \times 10^{-30}}$$ g cm$$^{-3}$$

$$Z$$ is the number of formula units in the unit cell.

$$M$$ is the formula mass of the substance.

$$a$$ is the length (in pm) of the unit cell.

$$N_0$$ is Avogadro's number.

For FCC, $$Z= 4, \ a=3.5 \mathring A=350$$ pm

$$D_{fcc}= \displaystyle\frac { 4 \times M} {N_0 \times (350)^3 \times 10^{-30}}$$ g cm$$^{-3}$$ $$......(1)$$

For BCC, $$Z= 2,\ a=3.0 \mathring A=300$$ pm

$$D_{bcc}= \displaystyle\frac { 2 \times M} {N_0 \times (300)^3 \times 10^{-30}}$$ g cm$$^{-3}$$ $$......(2)$$

Dividing equation $$(1)$$ by equation $$(2)$$, we get

$$\displaystyle\frac {D_{fcc}} {{D_{bcc}}}=\displaystyle\frac {\displaystyle\frac { 4 \times M} {N_0 \times (350)^3 \times 10^{-30}}} {\displaystyle\frac { 2 \times M} {N_0 \times (300)^3 \times 10^{-30}}}=\displaystyle\frac {\displaystyle\frac {4}{350^3}}{\displaystyle\frac {2}{300^3}}=1.26$$
Question 2
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$$TiO_{2}$$ is a well known example of :

A
triclinic system

B
tetragonal system

C
monoclinic System

D
none of the above

Solution

Tetragonal crystal lattice result from stretching a cubic lattice along with one of its lattice vectors so that the cube becomes a rectangular prism with a square base and a height.

The titanium cations have a coordination number of $$6$$ meaning they are surrounded by an octahedron of $$6$$ oxygen atoms. The oxygen anions have a co-ordination number of $$3$$ resulting in trigonal planar coordination. Rutile also shows a screw axis when its octahedral are viewed sequentially.
Question 3
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Hydrogen occupies the _________ holes, while carbon and nitrogen occupy __________ holes.

A
tetrahedral and octahedral

B
octahedral and tetrahedral

C
octahedral and octahedral

D
tetrahedral and tetrahedral

Solution

The size of the hole in tetrahedral voids is smaller than the size of the hole in octahedral voids. The size of the H atom is smaller than the size of C and N atoms. Hence, hydrogen occupies tetrahedral voids and carbon and nitrogen occupy octahedral voids.

Hence, the correct option is $$A$$
Question 4
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The fraction of total volume occupied by the atoms present in a simple cubic lattice is:

A
$$\dfrac{\pi }{4}$$

B
$$\dfrac{\pi }{6}$$

C
$$\dfrac{\pi }{3\sqrt{2}}$$

D
$$\dfrac{\pi }{4\sqrt{2}}$$

Solution

For simple cubic lattice,

Radius (r) = $$\dfrac {a}{2}$$

Volume of the atom = $$ \dfrac{4}{3}\times \pi \times \left ( \dfrac{a}{2} \right )^3$$

Packing fraction = $$ \dfrac{\dfrac{4}{3}\times \pi \times \left ( \dfrac{a}{2} \right )^3}{a^3} = \dfrac{\pi }{6}$$

Hence, the correct answer is option $$\text{B}$$.
Question 5
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An element crystallises in a bcc lattice. Nearest neighbours and percentage of volume occupied by spheres in the unit cell are :

A
$$8,\ 74\%$$

B
$$8,\ 68\%$$

C
$$6,\ 74\%$$

D
$$6,\ 68\%$$

Solution

There are $$8$$ nearest neighbours in BCC. Percentage of volume occupied by spheres in the unit cell of BCC is $$68\%$$.
Question 6
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Aluminium has fcc structure. The edge length of the unit cell is $$404$$ pm. If the density of the metal is $$2.7$$ g cm$$^{-3}$$, the molar mass (in gmol$$^{-1}$$) of Al atom is :

A
$$28.20$$

B
$$30.40$$

C
$$26.80$$

D
$$25.20$$

Solution

The expression for density is as follows:

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

where $$Z = 4 $$ for fcc

$$d=\dfrac { 4\times M}{ { 6.023\times 10 }^{ 23 }\times { (404\times { 10 }^{ -10 }) }^{ 3 } } = 2.7$$ g cm$$^{-3}$$

$$\implies M=26.80$$ g mol$$^{-1}$$

Hence , the molecular mass is $$26.80$$ gmol$$^{-1}$$.
Question 7
1 / -0

Edge length of a cube is $$400$$ pm. Its body diagonal (in pm) would be:

A
$$566$$

B
$$600$$

C
$$500$$

D
$$693$$

Solution

Edge length of a cube is 400pm. Hence, the body diagonal $$= \sqrt{3}a = \sqrt{3} \times 400 = 693$$ pm
Question 8
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A metal crystallises as body centred cubic lattice with the edge length of unit cell equal to $$0.304$$ nm. If the molar mass of the metal is $$50.3$$ g mol$$^{-1}$$, its density (in g cm$$^{-3}$$) is :

A
$$5.945$$

B
$$2.9725$$

C
$$8.9175$$

D
$$4.458$$

Solution

The expression for density is given below:

Z=2 for BCC, M= 50.3 g/ mol, $$a= 0.304\ nm = 0.304 \times 10^{-7} cm$$ and $$N_A= 6.0 \times 10^{23}$$

$$d=\dfrac { zM }{ { N }_{ A }{ a }^{ 3 } } $$ $$=\dfrac { 2\times 50.3 }{ { 6\times 10 }^{ 23 }\times { (0.304\times { 10 }^{ -7 }) }^{ 3 } } =\quad 5.945$$ g cm$$^{-3}$$

Option A is correct.
Question 9
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Percentage of free space in a body-centred cubic unit cell is:

A
$$34%$$

B
$$28%$$

C
$$30%$$

D
$$32%$$

Solution

Packing fraction = $$ \dfrac{ \text{Volume occupied by atoms in a unit cell}}{\text {Volume of the unit cell}}$$

For Body centered cubic cell , Packing fraction = $$ \dfrac{ 2 \times \dfrac{4}{3}\pi \times r^3} {\left({\dfrac {4r}{\sqrt 3}} \right )^2}$$ = $$\dfrac {\sqrt3 \pi }{8} = 0.68$$

where ,edge length a= $$ \dfrac{4r}{\sqrt 3} $$

Therefore, volume occupied = $$68\%$$

Hence vacant volume = $$32\%$$

So correct answer is option $$D$$.
Question 10
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An element crystallizes as a face-centred cubic lattice with edge length equal to $$460$$ pm. The density (in gcm$$^{-3}$$) of the element X when molar mass of X atom is $$60$$ gmol$$^{-1}$$ is:

A
$$4.096$$

B
$$2.048$$

C
$$6.144$$

D
$$3.072$$

Solution

The formula for calculating density is given below:

$$d=\dfrac { z\times M }{ { N }_{ A }\times { a }^{ 3 } } $$

where,
$$d=$$ density of the unit cell
$$z=$$ effective number of atoms in the fcc unit cell $$=4$$
$$M=$$ molar mass $$=60\ g/mol$$
$$N_A=$$ Avogadro number
$$a^3=$$ volume of the unit cell $$=(460\times { 10 }^{ -10 })^3cm ^{ 3 }$$

Substituting the value, we get

$$d=\dfrac { z\times M }{ { N }_{ A }\times { a }^{ 3 } } $$ $$=\dfrac { 4\times 60}{ { 6.023\times 10 }^{ 23 }\times { (460\times { 10 }^{ -10 }) }^{ 3 } } = 4.096$$ g cm$$^{-3}$$

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Solid State Test - 22

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Solid State Test - 22

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Question 1

$$2$$

$$1.26$$

$$3.34$$

$$1.8$$

Solution

Question 2

triclinic system

tetragonal system

monoclinic System

none of the above

Solution

Question 3

tetrahedral and octahedral

octahedral and tetrahedral

octahedral and octahedral

tetrahedral and tetrahedral

Solution

Question 4

$$\dfrac{\pi }{4}$$

$$\dfrac{\pi }{6}$$

$$\dfrac{\pi }{3\sqrt{2}}$$

$$\dfrac{\pi }{4\sqrt{2}}$$

Solution

Question 5

$$8,\ 74\%$$

$$8,\ 68\%$$

$$6,\ 74\%$$

$$6,\ 68\%$$

Solution

Question 6

$$28.20$$

$$30.40$$

$$26.80$$

$$25.20$$

Solution

Question 7

$$566$$

$$600$$

$$500$$

$$693$$

Solution

Question 8

$$5.945$$

$$2.9725$$

$$8.9175$$

$$4.458$$

Solution

Question 9

$$34%$$

$$28%$$

$$30%$$

$$32%$$

Solution

Question 10

$$4.096$$

$$2.048$$

$$6.144$$

$$3.072$$

Solution

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