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Solid State

JEE Chemistry · 42 questions · Page 4 of 5 · Click an option or "Show Solution" to reveal answer

Q31
CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct?
A rCs++rCl−=32a{r_{C{s^ + }}} + {r_{C{l^ - }}} = {{\sqrt 3 } \over 2}a
B rCs++rCl−=32a{r_{C{s^ + }}} + {r_{C{l^ - }}} = {3 \over 2}a
C rCs++rCl−=3a{r_{C{s^ + }}} + {r_{C{l^ - }}} = \sqrt 3 a
D rCs++rCl−=3a{r_{C{s^ + }}} + {r_{C{l^ - }}} = 3a
Correct Answer
Option A
Solution

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

2rCs++2rCl−=3a2{r_{C{s^ + }}} + 2{r_{C{l^ - }}} = \sqrt 3 a
⇒    rCs++rCl−=3a2\Rightarrow \,\,\,\,{r_{C{s^ + }}} + {r_{C{l^ - }}} = {{\sqrt 3 a} \over 2}
Q32
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 =

23{2 \over 3}

[TV] =

23{2 \over 3}

×\times 12 = 8 ∴\therefore Formula = M8A6 = M4A3

Q33
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

ρ=Z×Ma3×NA\rho=\dfrac{Z \times M}{a^{3} \times N_{\mathrm{A}}}

Z=2 for bccZ=2 \text{ for } b c c
5 g/cm3=2×M(200×10−10 cm)3×6.0×1023⇒M=12 g\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 g12 \mathrm{~g} of element contain =NA=N_{\mathrm{A}} atoms 300 g300 \mathrm{~g} of element contains =NA×30012=25NA=N_{\mathrm{A}} \times \dfrac{300}{12}=25 N_{\mathrm{A}}

Q34
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, 13{1 \over 3} Tetrahedral voids-B
B hcp lattice - B, 13{1 \over 3} Tetrahedral voids - A
C hcp lattice - A, 23{2 \over 3} Tetrahedral voids - B
D hcp lattice - B, 23{2 \over 3} Tetrahedral voids - A
Correct Answer
Option B
Solution

A2B3 has HCP lattice If A form HCP, then

3th4{{{3^{th}}} \over 4}

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

1th3{{{1^{th}}} \over 3}

of THV must occupied by A to form A2B3

Q35
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

Q36
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

Q37
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×Ma3×NA{{Z \times M} \over {{a^3} \times {N_A}}}

⇒\Rightarrow 6.17 =

2×M(3×10−8)3×6×1023{{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 =

20050{{{200} \over {50}}}

= 4 ∴\therefore Number of molecules = 4×\timesNA

Q38
At 100oC, copper (Cu) has FCC unit cell structure with cell edge length of x Ao\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 205x3{{205} \over {{x^3}}}
B 105x3{{105} \over {{x^3}}}
C 211x3{{211} \over {{x^3}}}
D 422x3{{422} \over {{x^3}}}
Correct Answer
Option D
Solution

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

Z×MNA×a3Z \times {M \over {{N_A} \times {a^3}}}
d=63.5×46×1023×x3×10−24g/cm3d = {{63.5 \times 4} \over {6 \times {{10}^{23}} \times {x^3} \times {{10}^{ - 24}}}}g/c{m^3}
d=63.5×4×106g/cm3d = {{63.5 \times 4 \times 10} \over 6}g/c{m^3}
d=423.33x3≃(422x3)d = {{423.33} \over {{x^3}}} \simeq \left( {{{422} \over {{x^3}}}} \right)
Q39
Na and Mg crystallize in BCC and FCC type crystals respectively, then the number of atoms of Na and Mg present in the unit cell of their respective crystal is :
A 4 and 2
B 9 and 14
C 14 and 9
D 2 and 4
Correct Answer
Option D
Solution

In bcc - points are at corners and one in the center of the unit cell. Number of atoms per unit cell

=8×18+1=2.= 8 \times {1 \over 8} + 1 = 2.

In fcc - points are at the corners and also centre of the six faces of each cell. Number of atoms per unit cell

=8×18+6×12=4.= 8 \times {1 \over 8} + 6 \times {1 \over 2} = 4.
Q40
Which of the following exists as covalent crystals in the solid state?
A Silicon
B Sulphur
C Phosphorous
D Iodine
Correct Answer
Option A
Solution

Among the given crystals only silicon is as a covalent solid.

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