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JEE • CHEMISTRY • SOLID STATE

Solid state practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 16

Analyzing +4 | -1

A compound of formula

{A_2}{B_3}

has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms?

View answer and explanation

Correct option: C

hcp lattice-

B

\frac{1}{3}

tetrahedral voids-

A

The correct answer is C (hcp lattice-

B

\frac{1}{3}

tetrahedral voids-

A

) because in an hcp lattice, the number of tetrahedral voids is twice the number of atoms, and the formula

{A_2}{B_3}

implies that A occupies 1/3 of the tetrahedral voids.

2 Hard

2 of 16

Application +4 | -1

For a given crystal, the lattice parameter ‘

a

’ is 318 pm. The

d

-spacing for a (III) plane is

View answer and explanation

Correct option: B

184\mathrm{ }pm

Option B (184 pm) is correct because it is calculated using the formula for d-spacing in a cubic crystal system, given the lattice parameter and Miller indices.

3 Hard

3 of 16

Application +4 | -1

How many unit cells are present in a cube shaped ideal crystal of

\mathrm{NaCl}

of mass 1.00g?

\left[ {\mathrm{Atomic}\;\mathrm{ masses}:\mathrm{Na} = 23,\;\mathrm{Cl} = 35.5} \right]

View answer and explanation

Correct option: A

2.57 \times {10^{21}}

Option A (

2.57 \times {10^{21}}

) is correct because it is calculated by dividing the total mass by the mass of one unit cell.

4 Hard

4 of 16

Application +4 | -1

An element (atomic mass

100 {~g} / {mol}

) having

b c c

structure has unit cell edge

400 {pm}

. The density of element is (no. of atoms in

b c c, Z=2

)

View answer and explanation

Correct option: C

5.188 {~g} / {cm}^3

Option C (

5.188 {~g} / {cm}^3

) is correct because the density is calculated using the formula

\rho = \frac{{Z \times M}}{{N_A \times a^3}}

, where Z=2 for bcc, M is the molar mass,

N_A

is Avogadro's number, and a is the edge length.

5 Hard

5 of 16

Application +4 | -1

Silver (atomic weight

= 108\;\mathrm{g}\;\mathrm{mo}{\mathrm{l}^{ - 1}}

) has a density of

10.5\;g\;\mathrm{c}{\mathrm{m}^{ - 3}}.

The number of silver atoms on a surface of area

{10^{ - 12}}\;{\mathrm{m}^2}

can be expressed in scientific notation as

y \times {10^x}.

The value of

x

View answer and explanation

Correct option: C

Option C is correct because the calculation involves finding the volume of a silver atom, then its area, and finally dividing the given surface area by the area of each silver atom.

6 Hard

6 of 16

Analyzing +4 | -1

In a fcc lattice

A, B, C

and

D

atoms are arranged at corners, face centres octahedral voids and half of tetrahedral voids respectively and if the crystal deposited is defected such that all the particles at one body diagonal of each unit cell are missing, correspondingly then what is the resulted formula of the compound?

View answer and explanation

Correct option: B

{AB}_4 {C}_4 {D}_4

Option B is correct. After removing atoms along the body diagonal, the formula becomes AB₄C₄D₄, calculated by considering the contributions of atoms at corners, faces, octahedral voids, and tetrahedral voids.

7 Hard

7 of 16

Application +4 | -1

Silver (atomic weight

= 108\;\mathrm{g}\;\mathrm{mo}{\mathrm{l}^{ - 1}}

) has a density of

10.5\;g\;\mathrm{c}{\mathrm{m}^{ - 3}}.

The number of silver atoms on a surface of area

{10^{ - 12}}\;{\mathrm{m}^2}

can be expressed in scientific notation as

y \times {10^x}.

The value of

x

View answer and explanation

Correct option: C

Option C is correct because the calculation involves finding the volume of a silver atom, then its area, and finally dividing the given surface area by the area of each silver atom.

8 Hard

8 of 16

Understanding +4 | -1

If the radius of an octahedral void and that of atom in close packing is

{r} \,\&\,{R}

respectively then relation between

{r} \,\&\, {R}

is:

View answer and explanation

Correct option: C

{r}=0.414 {R}

Option C (

{r}=0.414 {R}

) is correct because the relationship between the radius of an octahedral void (r) and the radius of the atom in close packing (R) is given by r = 0.414R.

9 Hard

9 of 16

Application +4 | -1

Sodium metal crystallizes as a body centred cubic lattice with the cell edge

4.29{A^{^{^ \circ }}}

What is the radius sodium atom?

View answer and explanation

Correct option: A

1.857\mathrm{ \times }{10^{( - 8)}}cm

Option A (

1.857\mathrm{ \times }{10^{( - 8)}}cm

) is correct because the radius of the sodium atom in a BCC lattice is calculated using the formula

r = \frac{{\sqrt 3 a}}{4}

, where a is the cell edge.

10 Hard

10 of 16

Application +4 | -1

An element occurring in bcc structure has

12.08 \times {10^{23}}

unit cells. The total number of atoms of the elements in these cells will be:

View answer and explanation

Correct option: A

24.16 \times {10^{23}}

Option A (24.16 × 10²³) is correct because in a bcc structure, each unit cell contains 2 atoms, so the total number of atoms is twice the number of unit cells.

11 Hard

11 of 16

Remembering +4 | -1

If the radius ratio is in the range of

0.414 - 0.732

, then the coordination number will be

View answer and explanation

Correct option: C

Option C (6) is correct because a radius ratio in the range of 0.414 - 0.732 corresponds to a coordination number of 6, which typically results in an octahedral geometry.

12 Hard

12 of 16

Application +4 | -1

A metal crystallises with a bcc structure with a density of 19.3 g.cm

{ }^{-3}

Calculate the length of the body diagonal of the unit cell. (Atomic weight of metal = 183.9)

View answer and explanation

Correct option: D

5.48 Å

Option D (5.48 Å) is correct because it is calculated using the bcc structure formula, density, and atomic weight to find the edge length and then the body diagonal.

13 Hard

13 of 16

Application +4 | -1

CsBr

has BCC type structure with edge length 4.3 pm. The shortest inter ionic distance in between

C{s^ + }\,\& \,B{r^ - }

View answer and explanation

Correct option: A

3.72 pm

Option A (3.72 pm) is correct because it is calculated using the formula for the shortest interionic distance in a BCC structure, which is

\frac{{\sqrt{3}a}}{{2}}

, where a is the edge length.

14 Hard

14 of 16

Remembering +4 | -1

In diamond, the coordination number of carbon is

View answer and explanation

Correct option: A

four and its unit cell has eight carbon atoms

Option A (four and its unit cell has eight carbon atoms) is correct because in a diamond structure, each carbon atom is tetrahedrally bonded to four other carbon atoms, and the unit cell contains eight carbon atoms.

15 Hard

15 of 16

Application +4 | -1

A metallic element exists as simple cubic lattice. Each edge of the unit cell is 3Å. The density of the metal is

9 \mathrm{~g} \mathrm{~cm}^{-3}

. How many number of unit cells will be present in

100 \mathrm{~g}

of the metal :-

View answer and explanation

Correct option: B

4.12 \times 10^{23}

Option B (

4.12 \times 10^{23}

) is correct because it correctly calculates the number of unit cells using the density, edge length, and the fact that it's a simple cubic lattice.

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16 Hard

16 of 16

Analyzing +4 | -1

Assertion (A): If the length of the unit cell of

LiCl

having

{NaCl}

structure is

5.14\,\,\mathop {\text{A}}\limits^ \circ ,

the ionic radius of ion is

4.82 \,\,\mathop {\text{A}}\limits^ \circ.

Reason (R): Action-anion contact is retained in

{LiCl}

structure because anions constitute the lattice.

View answer and explanation

Correct option: A

Both (A) and (R) are correct and (R) is the correct explanation of (A)

Option A (Both (A) and (R) are correct and (R) is the correct explanation of (A)) is correct because the assertion is derived from the anion-anion contact in the LiCl structure.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 16 single-correct-answer MCQs on SOLID STATE.
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