Solid State Test

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Solid State Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

If silver iodide crystallizes in a zinc blendek structure with $$I^-$$ ions forming the lattice, then the fraction of the tetrahedral voids occupied by $$Ag^+$$ ions will be:

A
10%

B
25%

C
50%

D
75%

Solution

In $$Agl,$$ $$I^-$$ ions form the lattice.

Let the number of $$I^-$$ ion $$= 1$$

There are two tetrahedral voids for 1 ion (sphere).

As there are one $$Ag^+$$ ion in $$AgI$$ so fraction of tetrahedral voids occupied by$$Ag^+$$ ions $$=\cfrac {1}{2}$$ or $$50\%$$

Option C is correct.
Question 2
1 / -0

Crystals are formed by ions, atoms, and molecules having an internal pattern that is ___________ and ___________.

A
irregular, non-repeating

B
regular, repeating

C
regular, non-repeating

D
none of the above

Solution

In crystalline solids, the atoms, ions or molecules are arranged in an ordered and symmetrical pattern that is repeated over the entire crystal. The smallest repeating structure of a solid is called a unit cell, which is periodically repeated in three dimensions on a lattice.
Question 3
1 / -0

Which one of the following has a different crystal lattice from those of the rest?

A
$$Ag$$

B
$$V$$

C
$$Cu$$

D
$$Pt$$

E
$$Au$$

Solution

Vanadium crystallizes in bcc structures while $$Ag, Cu, Pt$$ and $$Au$$ crystallize in ccp structures.
Question 4
1 / -0

In a crystal, atoms are located at the position of:

A
Maximum potential energy

B
Minimum potential energy

C
Zero potential energy

D
Infinite potential energy

Solution

In a crystal, atoms are located at the position of minimum potential energy.
As two atoms approach each other, the potential energy gradually decreases, reaches a minimum value and then sharply increases.

Hence, option C is correct.
Question 5
1 / -0

An element occurs in two crystalline forms $$\alpha$$ and $$\beta$$. The $$\alpha$$-form a fcc arrangement with $$a=3.68$$ $$\mathring{A}$$ and the $$\beta$$-form has a bcc arrangement with $$a=2.92 \mathring { A}$$. Calculate the ratio of their densities.

A
2 : 1

B
1 : 2

C
1 : 3

D
1 : 1

Solution

$$Unit\; cell\; length\; in\; Fcc\; =\; 3.68{ A }^{ 0 }\; ({ a }_{ 1 })\\ Unit\; cell\; length\; in\; Bcc\; =\; 2.92{ A }^{ 0 }\; ({ a }_{ 2 })\\ Density\; \alpha \; (Fcc)\; =\; \cfrac { M\times { Z }_{ 1 } }{ { N }_{ 0 }\times { a }_{ 1 }^{ 3 } } \\ Density\; \beta \; (Bcc)\; =\; \cfrac { M\times { Z }_{ 2 } }{ { N }_{ 0 }\times { a }_{ 2 }^{ 3 } } \\ \cfrac { { d\alpha } }{ d\beta } =\cfrac { { Z }_{ 1 }\times ({ { a }_{ 2 }) }^{ 3 } }{ { Z }_{ 2 }\times { ({ a }_{ 1 } })^{ 3 } } =\cfrac { 4\times ({ 2.92) }^{ 3 } }{ 2\times { (3.68) }^{ 3 } } =\; 0.999\sim 1\\ Hence\; ratio\; of\; density\; =\; 1:1$$
Question 6
1 / -0

In a face centred cubic arrangement of $$A$$ and $$B$$ atoms when $$A$$ atoms are at the corner of the unit cell and $$B$$ atoms of the face centres. One of the $$A$$ atom is missing from one corner in unit cell. The simplest formula of compound is:

A
$$A_{7}B_{3}$$

B
$$AB_{3}$$

C
$$A_{7}B_{24}$$

D
$$A_{7}B_{8}$$

Solution

Hint: Consider the arrangement of atoms A and B in the lattice.

Explanation:
Step 1: Atoms calculation in the lattice.
Atoms are present at the face center of the lattice.
As there are $$8$$ corners in a face-centered cubic arrangement, each contributes to $$\frac {1}{8}$$.
Since $$1$$ atom is missing from lattice, then total atoms = $$7$$ x $$\frac {1}{8}$$ = $$\frac {7} {8}$$
Therefore $$\frac{7}{8}$$ atoms are present of atom A.
Now B atoms are present at the face center and there are $$6$$ faces.
Each face center contributes half an atom to the cell.
Therefore total contribution of B = $$6$$ x $$\frac {1} {2}$$ = $$3$$.

Step 2: Calculation of ratio.
$$A:B$$ = $$\frac{7}{8}$$ : $$3$$.
Therefore $$A:B$$ is $$A_7B_{24}$$

Hence B is the correct option.
Question 7
1 / -0

The pyknometer density of $$NaCl$$ crystal is $$2.165\times { 10 }^{ 3 }kg\quad { m }^{ -3 }$$ while its X-rays density is $$2.178\times { 10 }^{ 3 }kg\quad { m }^{ -3 }$$. The fraction of the unoccupied sites in $$NaCl$$ crystal is:

A
$$5.968$$

B
$$5.96\times { 10 }^{ -2 }$$

C
$$5.96\times { 10 }^{ -3 }$$

D
$$ 5.96\times { 10 }^{ -4 }$$

Solution

Fraction occupied = $$\dfrac{\text{X- ray density - pyknometer density}}{\text{X- ray density}}$$

Fraction unoccupied $$ = \dfrac{2.178\times 10^3 - 2.165\times 10^3}{2.178\times 10^3}$$

$$ = 5.96\times 10^{-3}$$

Hence, option C is correct.
Question 8
1 / -0

When heated above $$ 916^o C , $$ iron changes its crystal structure from bcc to ccp structure without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is :

A
$$ 1. 089 $$

B
$$ 0.918 $$

C
$$ 0.725 $$

D
$$ 1.231$$

Solution

I) for BCC , $$Z=2$$, $$4r=a\sqrt{3}$$, $$M_A=56$$ edge length=$$a_1$$
$$\rho_{BCC}=\cfrac {2\times 56}{N_A\times {a_1}^{3}}=\cfrac {112\times 3\sqrt {3}}{N_A\times 64r^{3}}=K\times 3\sqrt{3}$$

II) for FCC, $$Z=4$$, $$4r=a_2\sqrt{2}$$, $$M_A=56$$ edge length=$$a_2$$

$$\rho_{FCC}=\cfrac {4\times 56}{N_A\times {a_2}^{3}}=\cfrac {2\times 112\times 2\sqrt{2}}{N_A\times {64r}^{3}}=\cfrac {224\times 2\sqrt{2}}{N_A\times {64r}^{3}}=K\times 4\sqrt{2}$$

$$\cfrac {\rho_{BCC}}{\rho_{FCC}}=\cfrac {3\sqrt{3}}{4\sqrt{2}}=0.918$$
Question 9
1 / -0

Which among the following metals crystallise as a simple cube?

A
Polonium

B
Iron

C
Copper

D
Gold

Solution

Polonium crystallizes as simple cube. Iron crystallizes as BCC unit cell. Copper and gold crystallizes as FCC unit cell.
Question 10
1 / -0

Which of the following defect is seen in $$FeO$$?

A
Metal excess defect

B
Metal deficiency defect

C
Displacement defect

D
Impurity defect

Solution

$$FeO$$ has metal deficiency defect. Metal deficiency defect is supposed to arise when there are lesser number of positive ions than negative ions. In case of $$FeO$$, the positive ions are missing from their lattice sites.
The additional negative charge is balanced by some nearby metal ion by acquiring one more positive charge. It happens in $$FeO$$ because $$Fe$$ has capacity of showing variable oxidation states.

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Solid State Test - 40

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Solid State Test - 40

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Question 1

10%

25%

50%

75%

Solution

Question 2

irregular, non-repeating

regular, repeating

regular, non-repeating

none of the above

Solution

Question 3

$$Ag$$

$$V$$

$$Cu$$

$$Pt$$

$$Au$$

Solution

Question 4

Maximum potential energy

Minimum potential energy

Zero potential energy

Infinite potential energy

Solution

Question 5

2 : 1

1 : 2

1 : 3

1 : 1

Solution

Question 6

$$A_{7}B_{3}$$

$$AB_{3}$$

$$A_{7}B_{24}$$

$$A_{7}B_{8}$$

Solution

Question 7

$$5.968$$

$$5.96\times { 10 }^{ -2 }$$

$$5.96\times { 10 }^{ -3 }$$

$$ 5.96\times { 10 }^{ -4 }$$

Solution

Question 8

$$ 1. 089 $$

$$ 0.918 $$

$$ 0.725 $$

$$ 1.231$$

Solution

Question 9

Polonium

Iron

Copper

Gold

Solution

Question 10

Metal excess defect

Metal deficiency defect

Displacement defect

Impurity defect

Solution

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