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JEE Main

Explore popular questions from Solid State for JEE Main. This collection covers Solid State previous year JEE Main questions hand picked by experienced teachers.

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Q 1. A cubic solid is made up of two elements {tex} \mathrm { A } {/tex} and {tex} \mathrm { B } {/tex} . Atoms {tex} \mathrm { B } {/tex} are at the corners of the cube and {tex} \mathrm { A } {/tex} at the body centre. What is the formula of compound?

{tex} \mathrm { AB } {/tex}

B

{tex} \mathrm { AB } _ { 2 } {/tex}

C

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

D

None

Explanation

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Q 2. A compound alloy of gold and copper crystallises in a cubic lattice in which gold occupy that lattice point at corners of the cube and copper atom occupy the centres of each of the cube faces. What is the formula of this compound?

A

{tex} \mathrm { AuCu } _ { 6 } {/tex}

B

{tex} \mathrm { AuCu } {/tex}

{tex} \mathrm { AuCu } _ { 3 } {/tex}

D

None of these

Explanation

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Q 3. {tex} \mathrm {KF}{/tex} crystallises in the {tex}\mathrm {NaCl}{/tex} type structure. If the radius of {tex} \mathrm { K } ^ { + } {/tex} ions 132{tex} \mathrm { pm } {/tex} and that of {tex} \mathrm { F^- } {/tex} ion is {tex} 135 \mathrm { pm } , {/tex} what is the closet {tex} \mathrm { K } - \mathrm { K } {/tex} distance?

A

Cannot say

B

534{tex} \mathrm { pm } {/tex}

C

755{tex} \mathrm { pm } {/tex}

378{tex} \mathrm { pm } {/tex}

Explanation

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Q 4. AgCl has the same structure as that of NaOH. The edge length of unit cell of AgCl is found to be 555 pm and the density of AgCl is {tex} 5.561 \mathrm { g } \mathrm { cm } ^ { - 3 } . {/tex} Find the percentage of sites that are unoccupied.

0.24{tex} \% {/tex}

B

2.4{tex} \% {/tex}

C

24{tex} \% {/tex}

D

None

Explanation

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Q 5. The effective radius of the iron atom is 1.42 {tex}\overset{\circ}{\text{A}}{/tex}. It has fcc structure. Calculate its density (Fe = 56 amu).

A

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

B

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

C

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

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

Explanation


For fcc,

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Q 6. The density of {tex} \mathrm { CaF } _ { 2 } {/tex} (fluorite structure) is {tex}3.18 \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex} . The length of the side of the unit cell is

A

253{tex} \mathrm { pm } {/tex}

344{tex} \mathrm { pm } {/tex}

C

546{tex} \mathrm { pm } {/tex}

D

273{tex} \mathrm { pm } {/tex}

Explanation

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Q 7. If the anions (A) form hexagonal closest packing and cations (C) occupy only {tex}2 / 3 {/tex} octahedral voids in it, then the general formula of the compound is

A

{tex} \mathrm { CA } {/tex}

B

{tex} \mathrm { CA } _ { 2 } {/tex}

{tex} \mathrm { C } _ { 2 } \mathrm { A } _ { 3 } {/tex}

D

{tex} \mathrm { C } _ { 3 } \mathrm { A } _ { 2 } {/tex}

Explanation

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Q 8. A solid is formed and it has three types of atoms {tex} \mathrm { X } , \mathrm { Y }, \mathrm { Z }{/tex}. {tex} \mathrm { X } {/tex} forms a FCC lattice with Y atoms occupying all the tetrahedral voids and {tex} \mathrm { Z } {/tex} atoms occupying half the octrahedral voids. The formula of the solid is

{tex} \mathrm { X } _ { 2 } \mathrm { Y } _ { 4 } \mathrm { Z } {/tex}

B

{tex} \mathrm { XY } _ { 2 } \mathrm { Z } _ { 4 } {/tex}

C

{tex} \mathrm { X } _ { 4 } \mathrm { Y } _ { 2 } \mathrm { Z } {/tex}

D

{tex} \mathrm { X } _ { 4 } \mathrm { YZ } _ { 2 } {/tex}

Explanation

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Q 9. A compound XY crystallises in BCC lattice with unit cell edge length of 480{tex} \mathrm { pm } {/tex} . If the radius of {tex} \mathrm { Y } ^ { - } {/tex} is 225 {tex} \mathrm { pm } , {/tex} then the radius of {tex} \mathrm { X } ^ { + } {/tex} is:

A

127.5{tex} \mathrm { pm } {/tex}

190.68{tex} \mathrm { pm } {/tex}

C

225{tex} \mathrm { pm } {/tex}

D

255{tex} \mathrm { pm } {/tex}

Explanation

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Q 10. {tex} \mathrm {NH_4 Cl}{/tex} crystallises in a body centred cubic type lattice with a unit cell edge length of 387{tex} \mathrm { pm } {/tex} . The distance between the oppositely charged ions in the lattice is

335.1{tex} \mathrm { pm } {/tex}

B

83.77{tex} \mathrm { pm } {/tex}

C

274.46{tex} \mathrm { pm } {/tex}

D

137.23{tex} \mathrm { pm } {/tex}

Explanation

For bcc structure d=
Here d is the distance between the oppositely charged ions in the lattice and a is the edge length.

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Q 11. An element crystallises in a structure having fcc unit cell of an edge 200 pm. Calculate the density, if 200{tex} \mathrm { g } {/tex} of this element contains {tex} 5 \times 10 ^ { 24 } {/tex} atoms.

A

5{tex} \mathrm { m } / \mathrm { cm } ^ { 3 } {/tex}

B

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

C

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

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

Explanation

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Q 12. If the length of the body diagonal for {tex}CsCl\ {/tex}which crystallises into a cubic structure with {tex}Cl^-{/tex} ions at the corners and {tex}Cs^{+}{/tex}ions at the centre of the unit cells is 7{tex}{\text{Å}}{/tex} and the radius of the {tex}Cs^+{/tex} ion is 1.69 {tex}{\text{Å}}{/tex}. What is the radii of {tex}Cl^-{/tex} ion?

1.81{tex} {\text{Å}} {/tex}

B

5.31{tex} {\text{Å}} {/tex}

C

3.62{tex}{\text{Å}} {/tex}

D

None

Explanation

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Q 13. Which of the following formulas is consistent with the unit cell of the rhenium oxide compound shown below?

{tex} \mathrm { Re } _ { 2 } \mathrm { O } _ { 6 } {/tex}

B

{tex} \mathrm { Re } _ { 2 } \mathrm { O } _ { 3 } {/tex}

C

{tex} \mathrm { ReO } _ { 6 } {/tex}

D

{tex} \mathrm {ReO } {/tex}

Explanation

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Q 14. The figure below shows a unit cell of the mineral Perovskite (the titanium atom is at the centre of the cube). Which of the following is a correct chemical formula for this mineral?

A

{tex} \mathrm { Ca } _ { 8 } \mathrm { TiO } _ { 6 } {/tex}

B

{tex} \mathrm { CaTiO } {/tex}

C

{tex} \mathrm { Ca } _ { 2 } \mathrm { TiO } _ { 3 } {/tex}

{tex} \mathrm { CaTiO } _ { 3 } {/tex}

Explanation

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Q 15. An atomic solid has hexagonal arrangement of unit cell with height of hexagonal (in close packed arrangement) as "{tex}h{/tex}". The radius of atom in terms of height is

A

{tex} \sqrt [ 4 ] { \frac { 2 } { 3 } } h {/tex}

B

{tex} \frac { h } { 4 } \sqrt { \frac { 2 } { 3 } } {/tex}

C

{tex} \frac { h } { 2 } \sqrt { \frac { 3 } { 2 } } {/tex}

{tex} \frac { h } { 4 } \sqrt { \frac { 3 } { 2 } } {/tex}

Explanation