Solid State — Page 5 | NEET Chemistry

Q41

In cube of any crystal A -atom placed at every corners and B -atom placed at every centre of face. The formula of compound is

A AB

B AB 3

C A 2 B 2

D A 2 B 3

Correct Answer

Option B

Solution

A atoms are at 8 corners of the cube $\therefore$ No. of A atoms per unit cell = 8 $\times$

{1 \over 8}

= 1 B atoms are at the face centre of six faces. $\therefore$ No. of B atoms per unit cell = 6 $\times$

{1 \over 2}

= 3 Thus, formula of compound = AB 3

Q42

Which is the incorrect statement?

A Density decreases in case of crystals with Schottky defect.

B NaCl (s) is insulator, silicon is semi conductor, silver is conductor, quartz is piezoelectric crystal.

C Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal

D FeO 0.98 has no-stoichimetric metal deficiency defect.

Correct Answer

Option C

Solution

Frenkel defect is favoured in those ionic compounds in which there is large difference in the size of cations and anions.

Non-stoichiometric defects due to metal deficiency is shown by FexO where x = 0.93 to 0.96.

Q43

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}

Q44

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.

Q45

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.

Q46

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

Q47

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

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

Q48

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}

Q49

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

Q50

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.