Solid State Questions — JEE Chemistry

Q1

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is :

A AB2

B A2B3

C A2B5

D A2B

Correct Answer

Option C

Solution

No. of atoms in the corners

\left( A \right) = 8 \times {1 \over 8} = 1

No. of atoms at face centers

\left( B \right) = 5 \times {1 \over 2} = 2.5

$\therefore$ Formula is

= A{B_{2.5}}

or

{A_2}{B_5}

Q2

Percentages of free space in cubic close packed structure and in body centred packed structure are respectively :

A 30% and 26%

B 26% and 32%

C 32% and 48%

D 48% and 26%

Correct Answer

Option B

Solution

Packing fraction is defined as the ratio of the volume of the unit cell that is occupied by the spheres to the volume of the unit cell.

P.F. for cpp and bcc are

0.74

and

0.68

respectively. So, the free space in ccp and bcc are

26\%

&

32\%

respectively.

Q3

Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?

A CsCl

B AgBr

C ZnS

D KBr

Correct Answer

Option B

Solution

Since AgBr has intermediate radius ratio.

AgBr shows both, Frenkel as well as Schottky defects.

ZnS only shows Frenkel defects.

KBr, CsCl only shows Schottky defects.

Q4

10 mL of 1mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head is approximated as cube, what is its edge length?

A 1.0 pm

B 2.0 nm

C 0.1 nm

D 2.0 pm

Correct Answer

Option D

Solution

No of moles formed = 10-3 $\times$

{{10} \over {1000}}

= 10-5 $\therefore$ No of molecules formed = 10-5 $\times$ NA In unimolecular layer formation each cube occupy an area = a2 $\therefore$ Total area occupied = 10-5 $\times$ NA $\times$ a2 According to the question, 10-5 $\times$ NA $\times$ a2 = 0.24 $\Rightarrow$ 10-5 $\times$ 6 $\times$ 1023 $\times$ a2 = 0.24 $\Rightarrow$ a2 = 4 $\times$ 10-20 $\Rightarrow$ a = 2 $\times$ 10-10 cm = 2 pm

Q5

A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at

\dfrac{1}{3}^{\mathrm{rd}}

of the total faces. The empirical formula of the compound is :

A

\mathrm{{X_2}{Y_{1.5}}}

B

\mathrm{{X_{3}}{Y_2}}

C

\mathrm{X{Y_{2.5}}}

D

\mathrm{{X_{2.5}}Y}

Correct Answer

Option B

Solution

$\begin{aligned} & \text{ Number of } X \text{ particles }=4 \times \dfrac{1}{8}+1=1.5 \\\\ & \text{ Number of } Y \text{ particles }=6 \times \dfrac{1}{3} \times \dfrac{1}{2}=1 \\\\ & \therefore \text{ Empirical formula }=X_{1.5} Y_1=X_3 Y_2\end{aligned}$

Q6

Which of the following expressions is correct in case of a

\mathrm{CsCl}

unit cell (edge length 'a')?

A

\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\dfrac{\sqrt{3}}{2} \mathrm{a}

B

\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\dfrac{\mathrm{a}}{\sqrt{2}}

C

\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\mathrm{a}

D

\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}^{-}}=\dfrac{\mathrm{a}}{2}

Correct Answer

Option A

Solution

$\mathrm{CsCl}$ has body centered type structure in which $\mathrm{Cs}^{+}$ occupies at corner of a cube and $\mathrm{Cl}^{-}$ occupies the centre of the cube. $2 \mathrm{r}_{\mathrm{Cs}^{+}}+2 \mathrm{r}_{\mathrm{Cl}^{-}}=\sqrt{3} \mathrm{a}$ (where a is the edge length of the cube)

\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl^-}}=\frac{\sqrt{3}}{2} \mathrm{a}

Q7

The correct relationships between unit cell edge length '

a

' and radius of sphere '

r

' for face-centred and body-centred cubic structures respectively are :

A

r=2 \sqrt{2} a

and

\sqrt{3} r=4 a

B

r=2 \sqrt{2} a

and

4 r=\sqrt{3} a

C

2 \sqrt{2} r=a

and

\sqrt{3} r=4 a

D

2 \sqrt{2} r=a

and

4 r=\sqrt{3} a

Correct Answer

Option D

Solution

In a face-centered cubic (FCC) unit cell, atoms are present at the corners as well as at the centers of the faces.

Hence, the diagonal of the face of the unit cell is equal to 4 times the radius of an atom.

This gives us the equation:

\sqrt{2} a = 4r

Which simplifies to:

a = 2\sqrt{2}r

In a body-centered cubic (BCC) unit cell, atoms are present at the corners and at the center of the unit cell.

The body diagonal of the unit cell is equal to 4 times the radius of an atom.

This gives us the equation:

\sqrt{3} a = 4r

Which simplifies to:

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

Therefore, Option D is correct:

2\sqrt{2}r = a

and

4r = \sqrt{3}a

Q8

How many unit cells are present in a cubeshaped ideal crystal of NaCl of mass 1.00 g? [Atomic masses: Na = 23, Cl = 35.5]

A 5.14

\times

1021 unit cells

B 1.28

\times

1021 unit cells

C 1.71

\times

1021 unit cells

D 2.57

\times

1021 unit cells

Correct Answer

Option D

Solution

Since in

NaCl

type of structure

4

formula units form a cell. Number of formulas in cube shaped crystals

= {{1.0} \over {58.5}} \times 6.02 \times {10^{23}}

No. of unit cells present in a cubic crystal

= {{P \times {a^3} \times {N_A}} \over {M \times Z}} = {{m \times {N_A}} \over {m \times Z}}

$\therefore$ units cells

= {{1.0 \times 6.02 \times {{10}^{23}}} \over {58.5 \times 4}}

= 2.57 \times {10^{21}}

unit cels.

Q9

Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius) :

A 20/3

\pi r^3

B 24/3

\pi r^3

C 16/3

\pi r^3

D 12/3

\pi r^3

Correct Answer

Option C

Solution

The face centered cubic unit cell contains

4

atom $\therefore$ Total volume of atoms

= 4 \times {4 \over 3}\pi {r^3} = {{16} \over 3}\pi {r^3}

Q10

All of the following share the same crystal structure except :

A LiCl

B NaCl

C RbCl

D CsCl

Correct Answer

Option D

Solution

RbCl, LiCl, and NaCl have face centered cubic structure and CsCl body centered cubic structure.