Solid State Test

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Solid State Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

Ice crystallises in a hexagonal lattice. At the, low temperature, the lattice constants were $$a = 4.53 A^{\circ}$$ and $$b = 7.41A^{\circ}$$.
How many $$H_2O$$ molecules are contained in a unit cell? $$[d(ice) = 0.92 g /cm^3]$$

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

The volume of the unit cell, $$\displaystyle V = a^2bsin60^o$$

$$\displaystyle V= (4.53 \times 10^{-8})^2 \times 7.41 \times 10^{-8} \times 0.866$$ $$(1 \mathring {A} = 10^{-8} cm)$$

$$\displaystyle V=1.317 \times 10^{-22} \: cm^3$$

The density is $$\rho= \displaystyle 0.92 \: g/cm^3$$

$$N_A = 6.023 \times 10^{23} /mol, M_{H_2O}= 18 g/mol$$

$$\rho =\cfrac{ZM}{VN_A} \Rightarrow Z= \cfrac {\rho V N_A}{M}$$

Putting all the given values in the above equation we get,

No. of molecules per unit cell, $$Z=4$$
Question 2
1 / -0

Argon crystallises in fcc arrangement and the density of solid and liquid $$Ar$$ is 1.59 $$g /$$ $$cm^3$$ and 1.42 $$g /$$ $$cm^3$$ , respectively . The percentage of empty space in liquid $$Ar$$ is:

A
34.84%

B
43.8%

C
23.4%

D
21.6%

Solution

Let the volume of solid $$Ar = 100$$ mL

Mass of solid $$Ar = V$$ $$ \times \rho = 100 \times 1.59 g / cm^3 = 159 g $$
Volume of liquid $$Ar$$

$$ = \dfrac {Mass}{\rho} = \dfrac {159 g}{1.4 gcm^{-3}} $$
= 113.57 ml

$$\because$$ $$Ar$$ crystallises in fcc type lattice, the packing fraction = 0.74

Actual volume occupied by $$Ar$$

= Packing fraction $$\times$$ volume of solid $$Ar$$
= 0.74 x 100 = 74 mL

$$\therefore$$ % empty in liquid Ar $$= \dfrac {(113.57-74) \times 100}{113.57}$$

= 34.84%
Question 3
1 / -0

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

A
$$39.27$$%

B
$$68.02$$%

C
$$74.05$$%

D
$$78.54$$%

Solution

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

Packing fraction $$= \dfrac {Occupied\ area}{Total\ area} \times 100$$

$$= \dfrac {2\pi r^{2}}{(2\sqrt {2}r)^{2}}\times 100 = 78.5$$%.
Question 4
1 / -0

Which one of the following is purely a crystalline compound?

A
Rock Salt

B
Ice

C
Quartz silica

D
Dry Ice
Question 5
1 / -0

How many defects exists in the arrangement of constituent particles of $$7.45\ g$$ $$KCl$$?
$$[K=39,Cl=35.5gm/mole]$$

A
$$10\times { 10 }^{ 23 }$$

B
$$1\times { 10 }^{ 6 }$$

C
$$1.0\times { 10 }^{ -6 }$$

D
$$1.0\times { 10 }^{ 4 }$$
Question 6
1 / -0

CsBr ha cubic structure with edge length $$4.4\overset{o}{A}$$. The shortest interionic distance between $$Cs^{+}$$ and $$Br^-$$ is:

A
$$3.82\overset{o}{A}$$

B
$$1.86\overset{o}{A}$$

C
$$7.44\overset{o}{A}$$

D
$$4.3\overset{o}{A}$$

Solution

$$CsBr$$ has body centered cubic (BCC) structure. The edge length of unit cell is given by $$4.4 \ \mathring A$$

Now, for a cube of edge $$a$$ units, the length of the diagonal is $$\sqrt 3 a$$ units.

The shortest ionic distance between $$Cs^+$$ and $$Br^-= \cfrac {\sqrt 3a}{2}$$

$$= \cfrac {\sqrt 3 \times 4.4}{2}= 3.81 \ \mathring A$$
Question 7
1 / -0

Percentage of free space in a body centred cubic unit cell is

A
30%

B
32%

C
34%

D
28%

Solution

Packing fraction $$\displaystyle =\frac { volume\, occupied \,by\, atoms\, in \,a \,unit\, cell } {volume \,of\, the\, unit\, cell }$$

For bcc,

Packing fraction $$ \displaystyle =\dfrac {2\times \dfrac{4}{3}\pi r^3}{\left ( \dfrac{4r}{\sqrt{3}} \right )^3}=\dfrac{\sqrt{3}\pi}{8}=0.68$$

Volume occupied $$= 68\%$$

Volume vacant $$=100 - 68 = 32\%$$
Question 8
1 / -0

The packing efficiency of the face centered cubic (fcc), body centered cubic (bcc) and simple primitive cubic (pc) lattices follows the order:

A
$$fcc > bcc > pc$$

B
$$bcc > fcc > pc$$

C
$$pc > bcc > fcc$$

D
$$bcc > pc > fcc$$

Solution

Face centred cubic (fcc) lattice has a packing efficiency of $$74$$%.

Body centred cubic (bcc) lattice has a packing efficiency of $$68$$%.

Simple/ Primitive cubic (pc) lattice has a packing efficiency of $$52.4$$%.
Hence, the correct order is $$\text {fcc > bcc > pc}$$.
Question 9
1 / -0

$$AB$$ has $$ZnS$$ type structure. What will be interionic distance between $${A}^{+}$$ and $${B}^{-}$$if lattice constant of $$AB$$ is $$200pm$$?

A
$$50\sqrt {3}$$

B
$$100\sqrt{3}$$

C
$$100$$

D
$$\cfrac{100}{\sqrt{2}}$$

Solution

$$ZnS$$ has cubic packing (ccp) crystal lattice structure which is analogous to the face-centered cubic lattice (FCC)
In $$ZnS$$, anions form FCC and cations occupy alternate tetrahedral voids.
Given, lattice constant $$=$$ edge length of a cube $$=a=200 \ pm$$
Now, tetrahedral voids are present on the body diagonal at $$\left (\cfrac 14 \right)^{th}$$ of the distance from each corner.
So, distance between $$A^+$$ and $$B^-= \cfrac 14 \times \sqrt 3 a$$
$$= \cfrac {\sqrt 3}4 \times 200 \ pm = 50 \sqrt 3 \ pm$$
Question 10
1 / -0

A compound formed by elements X and Y crystallizes in the cubic structure, where X atoms are at the corners of a cube and Y atoms are at the centre of the body. The formula of the compound is?

A
XY

B
$$XY_2$$

C
$$X_2Y_2$$

D
$$XY_3$$

Solution

$$X=8\times \displaystyle\frac{1}{8}=1$$(at corner)

$$Y=1\times 1=1$$(at body center)

So the formula of compound is $$=$$XY.

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Solid State Test - 63

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Solid State Test - 63

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Question 1

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 2

34.84%

43.8%

23.4%

21.6%

Solution

Question 3

$$39.27$$%

$$68.02$$%

$$74.05$$%

$$78.54$$%

Solution

Question 4

Rock Salt

Ice

Quartz silica

Dry Ice

Question 5

$$10\times { 10 }^{ 23 }$$

$$1\times { 10 }^{ 6 }$$

$$1.0\times { 10 }^{ -6 }$$

$$1.0\times { 10 }^{ 4 }$$

Question 6

$$3.82\overset{o}{A}$$

$$1.86\overset{o}{A}$$

$$7.44\overset{o}{A}$$

$$4.3\overset{o}{A}$$

Solution

Question 7

30%

32%

34%

28%

Solution

Question 8

$$fcc > bcc > pc$$

$$bcc > fcc > pc$$

$$pc > bcc > fcc$$

$$bcc > pc > fcc$$

Solution

Question 9

$$50\sqrt {3}$$

$$100\sqrt{3}$$

$$100$$

$$\cfrac{100}{\sqrt{2}}$$

Solution

Question 10

XY

$$XY_2$$

$$X_2Y_2$$

$$XY_3$$

Solution

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