Solid State — Page 8 | NEET Chemistry

Q71

A compound is formed by two elements

\mathrm{X}

and

\mathrm{Y}

. The element

\mathrm{Y}

forms cubic close packed arrangement and those of element

\mathrm{X}

occupy one third of the tetrahedral voids. What is the formula of the compound?

A

\mathrm{XY}_{3}

B

\mathrm{X_3Y}_{2}

C

\mathrm{X_3Y}

D

\mathrm{X_2Y}_{3}

Correct Answer

Option D

Solution

A compound is formed by two elements

\mathrm{X}

and

\mathrm{Y}

. The element

\mathrm{Y}

forms a cubic close-packed (CCP) arrangement, and element

\mathrm{X}

occupies one third of the tetrahedral voids. In a CCP structure, there are 4 atoms of

\mathrm{Y}

in a unit cell. This means there are 8 tetrahedral voids in the CCP structure. Since

\mathrm{X}

occupies one third of the tetrahedral voids, there are

\frac{1}{3} \times 8 = \frac{8}{3}

atoms of

\mathrm{X}

in the formula unit. The ratio of

\mathrm{X}

to

\mathrm{Y}

in the formula unit is

\frac{8}{3} : 4 = \frac{2}{3} : 1

. Multiplying both parts of the ratio by 3, we get

2 : 3

. So, the formula of the compound is

\mathrm{X_2Y_{3}}

Q72

Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom?

A 108 pm

B 127 pm

C 157 pm

D 181 pm

Correct Answer

Option B

Solution

For

fcc

unit cell,

4r = \sqrt 2 \,\,a;\,r = {{\sqrt 2 \times 361} \over 4} = 127\,\,pm

Q73

CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct?

A

{r_{C{s^ + }}} + {r_{C{l^ - }}} = {{\sqrt 3 } \over 2}a

B

{r_{C{s^ + }}} + {r_{C{l^ - }}} = {3 \over 2}a

C

{r_{C{s^ + }}} + {r_{C{l^ - }}} = \sqrt 3 a

D

{r_{C{s^ + }}} + {r_{C{l^ - }}} = 3a

Correct Answer

Option A

Solution

Relation between radius of cation, anion and edge length of the cube

2{r_{C{s^ + }}} + 2{r_{C{l^ - }}} = \sqrt 3 a

\Rightarrow \,\,\,\,{r_{C{s^ + }}} + {r_{C{l^ - }}} = {{\sqrt 3 a} \over 2}

Q74

In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is :

A M2A3

B M4A3

C M4A

D MA3

Correct Answer

Option B

Solution

In HCP unit cell, Z = 6 so A = 6 Also we know, in HCP Tetrahedral voids = 2Z = 12 $\therefore$ No. of M =

{2 \over 3}

[TV] =

{2 \over 3}

$\times$ 12 = 8 $\therefore$ Formula = M8A6 = M4A3

Q75

An element X has a body centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm

-

3. The number of atoms present in 300 g of the element X is _______________. Given : Avogadro constant, NA = 6.0

\times

1023 mol

-

1.

A 5 NA

B 6 NA

C 15 NA

D 25 NA

Correct Answer

Option D

Solution

$\rho=\dfrac{Z \times M}{a^{3} \times N_{\mathrm{A}}}$

Z=2 \text{ for } b c c

\begin{aligned} & 5 \mathrm{~g} / \mathrm{cm}^{3}=\frac{2 \times M}{\left(200 \times 10^{-10} \mathrm{~cm}\right)^{3} \times 6.0 \times 10^{23}} \Rightarrow M=12 \mathrm{~g} \end{aligned}

$12 \mathrm{~g}$ of element contain $=N_{\mathrm{A}}$ atoms $300 \mathrm{~g}$ of element contains $=N_{\mathrm{A}} \times \dfrac{300}{12}=25 N_{\mathrm{A}}$

Q76

A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms :

A hcp lattice - A,

{1 \over 3}

Tetrahedral voids-B

B hcp lattice - B,

{1 \over 3}

Tetrahedral voids - A

C hcp lattice - A,

{2 \over 3}

Tetrahedral voids - B

D hcp lattice - B,

{2 \over 3}

Tetrahedral voids - A

Correct Answer

Option B

Solution

A2B3 has HCP lattice If A form HCP, then

{{{3^{th}}} \over 4}

of THV must occupied by B to form A2B3 If B form HCP, then

{{{1^{th}}} \over 3}

of THV must occupied by A to form A2B3

Q77

Which type of ‘defect’ has the presence of cations in the interstitial sites?

A Metal deficiency defect

B Schottky defect

C Vacancy defect

D Frenkel defect

Correct Answer

Option D

Solution

In Frenkel defect in a molecule an atom or ion (normally the cation) leave their original site and places itself in the interstitial site (area between all other cations and anions).

Which is shown below

Q78

The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively :

A 8 : 1 : 6

B 4 : 2 : 1

C 1 : 2 : 4

D 4 : 2 : 3

Correct Answer

Option C

Solution

ZSC = 1 ZBCC = 2 ZFCC = 4 $\therefore$ Ratio = ZSC : ZBCC : ZFCC = 1 : 2 : 4

Q79

A diatomic molecule X2 has a body-centred cubic (bcc) structure with a cell edge of 300 pm. The density of the molecule is 6.17 g cm–3. The number of molecules present in 200 g of X2 is : (Avogadro constant (N A) = 6

\times

1023 mol–1 )

A 8 NA

B 40 NA

C 4 NA

D 2 NA

Correct Answer

Option C

Solution

d =

{{Z \times M} \over {{a^3} \times {N_A}}}

$\Rightarrow$ 6.17 =

{{2 \times M} \over {{{\left( {3 \times {{10}^{ - 8}}} \right)}^3} \times 6 \times {{10}^{23}}}}

[For BCC Z = 2] $\Rightarrow$ M = 50 g/mol Number of moles in 200 gm =

{{{200} \over {50}}}

= 4 $\therefore$ Number of molecules = 4 $\times$ NA

Q80

At 100oC, copper (Cu) has FCC unit cell structure with cell edge length of x

\mathop A\limits^o

. What is the approximate density of Cu (in g cm

-

3) at this temperature? [Atomic Mass of Cu = 63.55 u]

A

{{205} \over {{x^3}}}

B

{{105} \over {{x^3}}}

C

{{211} \over {{x^3}}}

D

{{422} \over {{x^3}}}

Correct Answer

Option D

Solution

FCC unit cell Z $=$ 4 We know, d =

Z \times {M \over {{N_A} \times {a^3}}}

d = {{63.5 \times 4} \over {6 \times {{10}^{23}} \times {x^3} \times {{10}^{ - 24}}}}g/c{m^3}

d = {{63.5 \times 4 \times 10} \over 6}g/c{m^3}

d = {{423.33} \over {{x^3}}} \simeq \left( {{{422} \over {{x^3}}}} \right)