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JEE Advanced

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

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Q 1. A metal of density {tex} 7.5 \times 10 ^ { 3 } \mathrm { kg } \mathrm { m } ^ { - 3 } {/tex} has an fcc crystal structure with lattice parameter {tex} a = 400 \mathrm { pm } {/tex}. Calculate the number of unit cells present in {tex} 0.015 \mathrm { kg } {/tex} of the metal

A

{tex} 6.250 \times 10 ^ { 22 } {/tex}

B

{tex} 3.125 \times 10 ^ { 23 } {/tex}

{tex} 3.125 \times 10 ^ { 22 } {/tex}

D

{tex} 1.563 \times 10 ^ { 22 } {/tex}

Explanation

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Q 2. The material used in solar cells contains

A

{tex} \mathrm { Cs } {/tex}

{tex} \mathrm { Si } {/tex}

C

{tex} \mathrm { Sn } {/tex}

D

{tex} \mathrm {T i} {/tex}

Explanation

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Q 3. Two ionic solids {tex} \mathrm { AB } {/tex} and {tex} \mathrm { CB } {/tex} crystallize in the same lattice. If {tex} r _ { A ^ \oplus }/ r _ { \mathrm { B ^\ominus} } {/tex} and {tex} r _ { \mathrm { C ^\oplus } } / r _ { \mathrm { B ^ \ominus} } {/tex} are 0.50 and 0.70 respectively, then the ratio of edge length of {tex} \mathrm { AB } {/tex} and {tex} \mathrm { CB } {/tex} is

A

0.68

B

0.78

0.88

D

0.98

Explanation


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Q 4. The edge length of unit cell of a metal {tex} ( M w = 24 ) {/tex} having cubic structure is {tex} 4.53 {\text{Å}}{/tex}. If the density of metal is {tex} 1.74 \mathrm { g } \mathrm { cm } ^ { - 3 } , {/tex} the radius of metal is {tex} \left( N _ { \mathrm { A } } = 6 \times 10 ^ { 23 } \right) {/tex}

A

{tex} 180 \mathrm { pm } {/tex}

{tex} 160 \mathrm { pm } {/tex}

C

{tex} 140 \mathrm { pm } {/tex}

D

{tex} 190 \mathrm { pm } {/tex}

Explanation

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Q 5. The number of atoms in {tex} 100 \mathrm { g } {/tex} of an fcc crystal with density {tex} = 10.0 \mathrm { g } \mathrm { cm } ^ { - 3 } {/tex} and cell edge equal to {tex} 200 \mathrm { pm } {/tex} is equal to

{tex} 5 \times 10 ^ { 24 } {/tex}

B

{tex} 5 \times 10 ^ { 25 } {/tex}

C

{tex} 6 \times 10 ^ { 23 } {/tex}

D

{tex} 2 \times 10 ^ { 25 } {/tex}

Explanation


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Q 6.

In the crystals of which of the following ionic compounds would you expect maximum distance between the centers of the cations and anions?

A

B

D

Explanation

and ions have largest sizes

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Q 7.

The edge length of a face-centered cubic unit cell is 508 pin. If the radius of the cation is 110 pm, the radius of the anion is

144 pm

B

288 pm

C

618 pm

D

398 pm

Explanation

For face-centered cubic unit cell (e.g., that of ), edge length = Distance between cation and anion pm (given). Putting 110 pm, we get pm

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Q 8.

A TV in fcc is formed by atoms at

A

3 corners + 1 face center

3 face centers + 1 corner

C

2 face centers + 2 corners

D

2 face centers + 1 corner + 1 body center

Explanation

A TV in fcc is formed by 3 face center atoms and one corner atom. In total, 8 such tetrahedrons are possible

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Q 9.

Analysis show that nickel oxide consists of nickel ion with 96% ions having configuration and 4% having configuration. Which amongst the following best represents the formula of the oxide?

A

B

C

Explanation

(At no.=28)

and

Hence, 96% ion of and 4% ions of are present

Let number of ion present in the crystal =

Applying electroneutrality rule,

Total positive charge = Total negative charge

So formula of crystal or

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Q 10.

The -form of iron has fcc-structure (edge length 386 pm) and -form has bcc structure (edge length 290 pm). The ratio of density in -form and -form is

0.9788

B

1.02

C

1.57

D

0.6344

Explanation

for and for /unit cell

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Q 11.

What is the maximum number of layers of atoms in close packed planes that will lie within two imaginary parallel planes having a distance between them of (where is the radius of atom) in the copper crystal (fcc)?

(Consider the atoms to be within the parallel planes if their centers are on or with in the two parallel planes)

A

5

B

6

7

D

8

Explanation

Distance between two layer (A and B)

Let is the number of imaginary planes

Hence,

Thus, maximum number of layers = 7 (see figures below)

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Q 12.

The packing fraction for a body-centered cube is

A

0.42

B

0.53

0.68

D

0.82

Explanation

Factual statement

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Q 13.

The coordination number of Al in the crystalline state of is

A

2

B

4

6

D

8

Explanation

Coordination number of Al in in (solid) crystalline state is 6.

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Q 14.

When molten zinc is cooled to solid state, it assumes hcpstructure. Then the number of nearest neighbours of zinc atom will be

A

4

B

6

C

8

12

Explanation

hcp has 12 nearest neighbours

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Q 15.

A molecule occupies triclinic lattice with and . If the density of is , the number of molecules present in one unit cell is

A

2

3

C

4

D

5

Explanation

Volume of unit cell

Mass of unit cell g

Number of molecules in one unit cell

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Q 16.

What type of crystal defect is indicated in the diagram given below

A

Both Frenkel and Schottky defects

Schottky defect

C

Interstitial defect

D

Frenkel defect

Explanation

Since equal number of cations and anions are missing in the figure given, so it Schottky defect

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Q 17.

The number of hexagonal faces that are present in a truncated octahedron is

A

2

B

4

C

6

8

Explanation

The truncated octahedron is the 14-faced Archimedean solid, with 14 total faces : 6 squares and 8 regular hexagons.

The truncated octahedron is formed by removing the six right square pyramids one from each point of a regular octahedron as :

Truncated Octachedron

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Q 18.

What are types of following semiconductors I and II

A

-type -type

-type -type

C

Both -type

D

Both -type

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Q 19.

Which of the following is not a ferroelectric compound?

A

Rochelle salt

C

D

Explanation

Factual statement

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Q 20.

The number of unit cells in 58.5 g of is nearly

A

B

D

Explanation

units

One unit cell contains 4 units. Hence, number of the unit cell present =

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Q 21.

In the closest packing of atoms

A

The size of TV is greater than that of OV

The size of TV is smaller than that of OV

C

The size of TV is equal to that of OV

D

The size of TV may be greater or smaller or equal to that of OV depending upon the size of atoms

Explanation

For a given radius of anion

Radius ratio for OV and TV is as follows:

Hence, size of OV is larger than that of TV

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Q 22.

The density of an ionic compound ( ) is 2.165 and the edge length of unit cell is 562 pm, then the closest distance between and of unit cell is

281 pm,4

B

562 pm,2

C

562 pm,4

D

281 pm,4

Explanation

-type structure)

For fcc,

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Q 23.

In a hypothetical solid, C atoms are found to form cubical close-packed lattice. A atoms occupy all tetrahedral voids and B atoms occupy all octahedral voids

A and B atoms are of appropriate size, so that there is no distortion in the ccp lattice of C atoms. Now if a plane as shown in the following figure is cut, then the cross section of this plane will look like

A

B

D

Explanation

From figure, it is clear 4 corners and 2 face centers lie on the shaded plane. Therefore, there will be six C atoms, and atoms (marked A) in TVs do not touch other.

Fig. (a) is not possible; four atom marked C

Fig. (b) is not possible, atoms A in TVs are not shown in figure

Fig. (c) is possible, since atoms A in TVs are not touching each other

Fig. (d) is not possible, since atoms A in TVs are touching each other

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Q 24.

The interionic distance for cesium chloride crystal will be

A

B

D

Explanation

As is body-centered,

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Q 25.

The number of nearest neighbours and next nearest neighbours of an ion in a crystal of are, respectively,

A

C

D

Explanation

lies in OVs formed by ( touches six ions)

are not shown touching in the figure. Each atom shown is present at the face center of each cube

Try to visualize two cubes exactly above these two cubes.Each atom shown is present at the edge center and body center of each cube

Distance between two nearest

  • Thus, the number of nearest neighbours of ion = 6 ions

  • The number of next nearest neighbours of ion = 12 ions