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Solid State

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Q 1. A metal has a {tex}fcc{/tex} lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72{tex} \mathrm { g } \mathrm { cm } ^ { - 3 }{/tex}.The molar mass of the metal is {tex} \left( N _ { A } \text { Avogadro's constant } = 6.02 \times 10 ^ { 23 } \mathrm { mol } ^ { - 1 } \right) {/tex}

27{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}

20{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}

40{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}

30{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}

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Q 2. The number of carbon atoms per unit cell of diamond unit cell is

6

1

4

8

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Q 3. A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 {tex}\mathrm { pm }{/tex}.The diameter of the metal atom is

288 {tex} \mathrm { pm } {/tex}

408 {tex} \mathrm { pm } {/tex}

144 {tex} \mathrm { pm } {/tex}

204 {tex} \mathrm { pm } {/tex}

For FCC,

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Q 4. The number of octahedral void(s) per atom present in a cubic close-packed structure is

1

3

2

4

Number of octahedral voids in ccp, is equal to effective number of atoms, in ccp, effective number of atoms are 4 so, 4 octahedral voids. So, 1 octahedral voids per atom

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Q 5. Structure of a mixed oxide is cubic close packed {tex}(ccp){/tex}. The cubic unit cell of mixed oxide ions is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

{tex} A B O _ { 2 } {/tex}

{tex} A _ { 2 } B O _ { 2 } {/tex}

{tex} A _ { 2 } B _ { 3 } \mathrm { O } _ { 4 } {/tex}

{tex} A B _ { 2 } \mathrm { O } _ { 2 } {/tex}

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Q 6. A solid compound {tex} X Y {/tex} has {tex} \mathrm { NaCl } {/tex} structure. If the radius of the cation is 100 {tex}\mathrm { pm } , {/tex} the radius of the anion {tex} \left( Y ^ { - } \right) {/tex} will be

275.1 {tex} \mathrm { pm } {/tex}

322.5 {tex} \mathrm { pm } {/tex}

241.5 {tex} \mathrm { pm } {/tex}

165.7 {tex} \mathrm { pm } {/tex}

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Q 7. {tex} A B {/tex} crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositely charged ions in the lattice is

335 {tex} \mathrm { pm } {/tex}

250 {tex} \mathrm { pm } {/tex}

200 {tex} \mathrm { pm } {/tex}

300 {tex} \mathrm { pm } {/tex}

Distance between two oppositely charge ions for BCC

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Q 8. Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351{tex} \mathrm { pm } {/tex} , the atomic radius of lithium will be

151.8 {tex} \mathrm { pm } {/tex}

75.5 {tex} \mathrm { pm } {/tex}

300.5 {tex} \mathrm { pm } {/tex}

240.8 {tex} \mathrm { pm } {/tex}

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Q 9. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 {tex} \mathrm { pm } {/tex} . What is the radius of copper atom in {tex} \mathrm { pm } {/tex} ?

157

181

108

128

As given that copper crystallizes is FCC lattice (Face centred cubic). In FCC atoms are present on the corners of the cubic, unit cell as well as on the face centres of each face. The atoms on the face diagonal will be touching each other. Let, the radius of the atom be r and edge length of the cube be a.

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Q 10. Which of the following statements is not correct?

The number of carbon atoms in a unit cell of diamond is {tex} 8. {/tex}

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

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

Molecular solids are generally volatile.

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Q 11. With which one of the following elements silicon should be doped so as to give {tex} p {/tex}-type of semiconductor?

Selenium

Boron

Gemanium

Arsenic

The n-type semiconductors are obtained when Si or Ge are doped with elements of group 15, eg, Arsenic (As), while p-type semiconductors are obtained when Si or Ge are doped with traces of elements of group 13, ie indium (In), Boron (B).

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Q 12. If {tex} \mathrm { NaCl } {/tex} is doped with {tex} 10 ^ { - 4 } \mathrm { mol } \% {/tex} of {tex} \mathrm { SrCl } _ { 2 } {/tex} , the concentration of cation vacancies will be {tex} \left( N _ { A } = 6.02 \times 10 ^ { 23 } \mathrm { mol } ^ { - 1 } \right) {/tex}

{tex} 6.02 \times 10 ^ { 16 } \mathrm { mol } ^ { - 1 } {/tex}

{tex} 6.02 \times 10 ^ { 17 } \mathrm { mol } ^ { - 1 } {/tex}

{tex} 6.02 \times 10 ^ { 14 } \mathrm { mol } ^ { - 1 } {/tex}

{tex} 6.02 \times 10 ^ { 15 } \mathrm { mol } ^ { - 1 } {/tex}

One Sr^{+2} creates one cationic vacancies

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Q 13. The appearance of colour in solid alkali metal halides is generally due to

Interstitial positions

{tex} F {/tex} -centres

Schottky defect.

Frenkel defect.

The appearance in colour in solid alkali metal halides is generally due to F− centres. When electrons occupy the vacant space of anions due to the presence of excess of metal ions, are called F-centres.

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Q 14. CsBr crystallises in a body centred cubic lattice. The unit cell length is **436.6 pm**. Given that the atomic mass of **Cs = 133 amu** and that of **Br = 80 amu** and Avogadro number being **6.02 x 10 ^{23} mol^{-1}** the density of CsBr is

4.25{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}

42.5{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}

0.425{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}

8.25{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}

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Q 15. In a face-centered cubic lattice, a unit cell is shared equally by how many unit cells?

2

4

6

8

A cubic unti cell has six faces. Therefore, in a cubic lattice (irrespective of its nature) cubic unit cell is shared equally by 6 unit cells.

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Q 16. A compound formed by elements {tex} X {/tex} and {tex} Y {/tex} crystallises in a cubic structure in which the {tex} X {/tex} atoms are at the corners of a cube and the {tex} Y {/tex} atoms are at the face-centres. The formula of the compound is

{tex} X Y _ { 3 } {/tex}

{tex} X _ { 3 } Y {/tex}

{tex} X Y {/tex}

{tex} X Y _ { 2 } {/tex}

No of atoms in the X in the unit cell {tex}{ 8}\times{(1/8)}{/tex} = 1 No of atoms in the X in the unit cell {tex}{ 6}\times{(1/2)}{/tex} = 3 Hence the formula of compund is {tex} X Y _ { 3 } {/tex}

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Q 17. The pyknometric density of sodium chloride crystal is {tex} 2.165 \times 10 ^ { 3 } \mathrm { kg } \mathrm { m } ^ { - 3 } {/tex} while its X-ray density is {tex} 2.178 \times 10 ^ { 3 } \mathrm { kg } \mathrm { m } ^ { - 3 } {/tex} . The fraction of unoccupied sites in sodium chloride crystal is

5.96

{tex} 5.96 \times 10 ^ { - 2 } {/tex}

{tex} 5.96 \times 10 ^ { - 1 } {/tex}

{tex} 5.96 \times 10 ^ { - 3 } {/tex}

X ray density is higher because corresponds to occupied space only as X-rays are diffracted by constituet particles.On the hand ,pyknometric density is lower as it correspondence to occupied as well as unoccupied space. Thus, the fraction of unoccupied site in NaCl crystal is given as

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Q 18. When {tex} \mathrm { Zn } {/tex} converts from melted state to its solid state, it has {tex} h c p {/tex} structure, then find the number of nearest atoms.

6

8

12

4

When Zinc is converted from it's melted state to solid state ,it has HCP structure. In HCP structure , the coordination number is 12. Since Zn is in HCP structure the number of nearest neighbours is 12, six are in the plane ,3 above the plane and 3 below the plane. So correct option is C.

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Q 19. In cube of any crystal {tex} A {/tex} -atom placed at every corners and {tex} B {/tex} -atom placed at every centre of face. The formula of compound is

{tex} A B {/tex}

{tex} A B _ { 3 } {/tex}

{tex} A _ { 2 } B _ { 2 } {/tex}

{tex} A _ { 2 } B _ { 3 } {/tex}

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Q 20. In crystals of which one of the following ionic compounds would you expect maximum distance between centres of cations and anions?

{tex} \mathrm { CsI } {/tex}

{tex} \mathrm {CsF} {/tex}

{tex} \mathrm { LiF } {/tex}

{tex} \mathrm {LiI} {/tex}

The ionic radii of {tex} \mathrm { Cs } {/tex}^{+} is highest among the given cations and ionic radii of {tex} \mathrm { I } {/tex}^{-} is highest among the given anions. Therefore the maximum distance between the centers of cations and anions occurs in {tex} \mathrm { CsI } {/tex}

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Q 21. Schottky defect in crystals is observed when

Density of the crystal is increased

Unequal number of cations and anions are missing from the lattice

An ion leaves its normal site and occupies an interstitial site

Equal number of cations and anions are missing from the lattice.

In ionic crystals, the defect forms when oppositely charged ions leave their lattice sites, creating vacancies. These vacancies are formed in stoichiometric units, to maintain an overall neutral charge in the ionic solid. The surrounding atoms then move to fill these vacancies, causing new vacancies to form. Normally these defects will lead to a decrease in the density of the crystal.

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