Solid State Test

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Solid State Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Complete the following table:

A
(A)- Impurity defects, (B)- Stoichiometric defects, (C)- Non-stoichiomeric defects (D)- Anion excess defects

B
(A)- Stoichiometric defects, (B)- Non-stoichiomeric defects, (C)- Impurity defects, (D)- Metal excess defects

C
(A)- Non-stoichiomeric defects, (B)- Stoichiometric defects, (C)- Impurity defects, (D)- Cation vacancy

D
(A)- Impurity defects, (B)- Stoichiometric defects, (C)- Metal excess defects, (D)- Non-stoichiomeric defects

Solution

Points defects can be classified into three types:

1. Stoichiometric defects (stoichiometric ratios of cations and anions remains unchanged) (A).
2. Non-stoichiometric defects (ratios of anions and cations changes due to this type of defects) (B)
3. Impurity defects (foreign atoms replace some of the atoms in the lattice) (C)

The non-stoichiometric defects further classified into two types:

1. Metal Excess defect (due to anionic vacancies) (D)
2. Metal deficiency defect (due to cation vacancies)

Hence, the correct answer is option $$\text{B}$$.
Question 2
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Silver halides generally show

A
Schottky defect

B
Frenkel defect

C
both Frenkel and Schottky defects

D
cation excess defect

Solution

Silver halides show both Frenkel and Schotkky defects. For Frenkel defect the reason is that there is size difference between the sizes of silver and halide.
Schottky defect is possible because silver halides are highly ionic.

Hence, option "C" is the correct answer.
Question 3
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A compound is formed by two elements $$Y$$ and $$Z.$$ The elements $$Z$$ forms $$ccp$$ and atoms $$Y$$ at $$1/3$$ rd of tetrahedral voids. The formula of the compound is:

A
$$Y_2Z_3$$

B
$$YZ$$

C
$$YZ_3$$

D
$$Y_2Z$$

Solution

Since, in $$CCP$$, number of atoms = 4

So, the number of $$Z$$ atoms = 4

Number of tetrahedral voids = $$4\times 2=8$$

Number of $$Y$$ atoms = $$\cfrac { 1 }{ 3 } \times 8=\cfrac { 8 }{ 3 }$$

Ratio of $$Y$$ and $$Z$$ is = $$\cfrac { 8 }{ 3 } :4=2:3$$

So, the formula is $${ Y }_{ 2 }{ Z }_{ 3 }$$.

Hence, the correct answer is option "A".
Question 4
1 / -0

Iodine molecules are held in the crystals lattice by ______.

A
london forces

B
dipole-dipole interaction

C
covalent bonds

D
coulombic forces

Solution

The London dispersion force is the weakest intermolecular force. The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction.
Iodine molecules are a class of non- polar molecular solid in which constitute molecules are held together by London or dispersion forces. These solids are soft and non-conductor of electricity.
So, the correct answer is option $$A$$.
Question 5
1 / -0

The correct order of the packing efficiency in different types of the unit cell is:

A
fcc < bcc <simple cubic

B
fcc > bcc >simple cubic

C
fcc < bcc >simple cubic

D
bcc < fcc >simple cubic

Solution

Packing efficiencies of different cells are shown in the above table.
Hence, correct order is fcc (74%) > bcc(68%) > simple cubic (52%).

Hence, the correct option is $$\text{B}$$
Question 6
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Which of the following is true about the value of refractive index quartz glass?

A
Same in all direction

B
Different in different direction

C
Cannot be measured

D
Always zero

Solution

Since quartz glass is an amorphous solid having short range order of constituents. Hence, value of refractive index is same in all directions, can be measured and not be equal to zero always.

Hence, option "A" is the correct answer.
Question 7
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The sharp melting point of crystalline solids is due to______.

A
a regular arrangement of constituent particles observed over a short distance in crystal lattice.

B
a regular arrangement of constituent particles observed over a long distance in crystal lattice.

C
same arrangement of constituent particles in different direction

D
different arrangement of constituent particles in different directions

Solution

Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same.
So, option B is correct answer.
Question 8
1 / -0

The total number of tetrahedral voids in face centered unit cell is ______.

A
6

B
8

C
10

D
12

Solution

In face centered cubic unit cell (FCC), tetrahedral voids are present on body diagonals. There are two tetrahedral voids on each body diagonal at $$\left (\cfrac 14 \right)^{th}$$ of the distance from each corner.

So, total number of tetrahedral voids = (no. of tetrahedral voids per body diagonal) x (no. of body diagonals) = $$2 \times 4 = 8$$
Hence, Option "B" is the correct answer.
Question 9
1 / -0

In which pair most efficient packing in present?

A
$$hcp$$ and $$bcc$$

B
$$hcp$$ and $$ccp$$

C
$$bcc$$ and $$ccp$$

D
$$bcc$$ and simple cubic cell

Solution

The packing efficiency of the lattices are given below.
$$hcp$$ = $$74\%$$
$$ccp$$ = $$74\%$$
$$bcc$$ = $$68\%$$
Simple cubic = $$54\%$$.
So, the pair with most efficient packing is $$hcp$$ and $$ccp$$.

Hence, Option "B" is the correct answer.
Question 10
1 / -0

Mark the incorrect pair from the following.

A
Schottky defect- Equal number of cation and anions are missing.

B
Frenkel defect- Dislocation of cation from its normal site to interstitial site.

C
Impurity defect- $$CdCl_2$$ in $$AgCl$$ crystal to create cationic vacancy.

D
Metal excess defect- $$Fe_{0.93}O$$

Solution

In $${ Fe }_{ 0.93 }O$$, some of $${ Fe }^{ 2+ }$$ ions are absent from $$FeO$$ and some of $${ Fe }^{ 3+ }$$ are present.
So, it is due to metal deficiency defect.

Hence, option "D" is the correct answer.

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Solid State Test - 42

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Solid State Test - 42

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Question 1

(A)- Impurity defects, (B)- Stoichiometric defects, (C)- Non-stoichiomeric defects (D)- Anion excess defects

(A)- Stoichiometric defects, (B)- Non-stoichiomeric defects, (C)- Impurity defects, (D)- Metal excess defects

(A)- Non-stoichiomeric defects, (B)- Stoichiometric defects, (C)- Impurity defects, (D)- Cation vacancy

(A)- Impurity defects, (B)- Stoichiometric defects, (C)- Metal excess defects, (D)- Non-stoichiomeric defects

Solution

Question 2

Schottky defect

Frenkel defect

both Frenkel and Schottky defects

cation excess defect

Solution

Question 3

$$Y_2Z_3$$

$$YZ$$

$$YZ_3$$

$$Y_2Z$$

Solution

Question 4

london forces

dipole-dipole interaction

covalent bonds

coulombic forces

Solution

Question 5

fcc < bcc <simple cubic

fcc > bcc >simple cubic

fcc < bcc >simple cubic

bcc < fcc >simple cubic

Solution

Question 6

Same in all direction

Different in different direction

Cannot be measured

Always zero

Solution

Question 7

a regular arrangement of constituent particles observed over a short distance in crystal lattice.

a regular arrangement of constituent particles observed over a long distance in crystal lattice.

same arrangement of constituent particles in different direction

different arrangement of constituent particles in different directions

Solution

Question 8

6

8

10

12

Solution

Question 9

$$hcp$$ and $$bcc$$

$$hcp$$ and $$ccp$$

$$bcc$$ and $$ccp$$

$$bcc$$ and simple cubic cell

Solution

Question 10

Schottky defect- Equal number of cation and anions are missing.

Frenkel defect- Dislocation of cation from its normal site to interstitial site.

Impurity defect- $$CdCl_2$$ in $$AgCl$$ crystal to create cationic vacancy.

Metal excess defect- $$Fe_{0.93}O$$

Solution

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