Solid State Test

Result

Solid State Test - 6

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A solid has three types of atoms X, Y and Z. X forms an FCC lattice with Y atoms occupying all tetrahedral voids, and Z atoms occupying half of the octahedral voids. The formula of the solid is

A
X₄YZ₂

B
X₄Y₂Z

C
XY₂Z₄

D
X₂Y₄Z

Solution

We know that, X – FCC = 4,Y – Tetrahedral void = 8,Z – ½ of Octahedral void = 2.
Hence, formula becomes X₄Y₈Z₂or X₂Y₄Z.
Question 2
1 / -0

Which of the following statements is correct?

A
Ionic crystal of AgBr does not have Frenkel defect.

B
The unit cell having lattice parameter, i.e. a = b ≠ c and crystal angles, i.e. α = β = 90^o and γ = 120^o, is hexagonal.

C
In an ionic compound having Schottky defect, the ratio r⁺/ r^- is very high.

D
The coordination number of Na⁺ in NaCl is 4.

Solution

The coordination of the cation in NaCl is six as each Na⁺ ion is surrounded by six Cl^- ions.
Question 3
1 / -0

Titanium crystallises in a face-centred cubic lattice. It reacts with hydrogen interstitially allowing the atoms of this element to occupy holes in the host lattice. If hydrogen occupies the tetrahedral holes, how many hydrogen atoms are present per titanium atom in the compound?

A
1

B
2

C
3

D
4

Solution

There are 4 atoms per unit cell and 8 tetrahedral sites per unit cell. So, there will be 8/4 = 2 hydrogen atoms per titanium atom.
Question 4
1 / -0

Which of the following options is not correctly matched?

A
Monoclinic: a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90^o

B
Tetragonal: a = b ≠ c and α = β = γ = 90^o

C
Orthorhombic: a ≠ b ≠ c and α = β = γ = 90^o

D
Rhombohedral: a = b = c and α = β = γ ≠ 90^o

Solution

Monoclinic: a ≠ b ≠ c and α = γ = 90^oβ ≠ γ ≠ 90^o
Hence, the option (1) is not correctly matched.

The given characteristics are for the triclinic unit cell: a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90^o
Question 5
1 / -0

Which of the following lattice structures have the same packing efficiency?

A
Simple cubic and bcc

B
Simple cubic, bcc, and ccp

C
bcc and ccp

D
hcp and ccp

Solution

Packing efficiency in both hexagonal close packed (hcp) and cubic close packed (ccp) lattices is 74%.

Question 6

1 / -0

Match the columns:

Column – I	Column – II
A. Frenkel defect	1. Group 15 element doped in group 14 element
B. Schottky defect	2. Density remains the same
C. n-type semiconductor	3. Density decreases

A - 1, B - 2, C - 3

A - 1, B - 3, C - 2

A - 2, B - 1, C - 3

A - 2, B - 3, C - 1

Solution

Frenkel defect leads to no change in density but Schottky defect leads to decrease in density.

Question 7
1 / -0

Which of the following statements is correct about the structure of zinc blende?

A
Sulphide ions are present at the corner of the cube and at the centre of the cube.

B
Zinc ions occupy all the tetrahedral sites.

C
There are two zinc ions for every sulphide ion.

D
Each zinc ion is surrounded by four sulphide ions.

Solution

Each zinc ion is surrounded by four sulphide ions, which are disposed towards the corner of the regular tetrahedron.
Question 8
1 / -0

In the crystals of which of the following ionic compounds would you expect maximum distance between the centres of the cations and the anions?

A
LiF

B
CsF

C
CsI

D
LiI

Solution

CsI compound would have the maximum distance between the centres of the cations and the anions due to its cubic structure and large size of cation and anion.
Question 9
1 / -0

Which of the following sets contains only hexagonal crystal system?

A
Graphite, ZnO and CdS

B
NaCl, HgS and graphite

C
KNO₃, BaSO₄ and CaCO₃

D
ZnO, HgS and CdS

Solution

Hexagonal crystal system is present in graphite, ZnO and CdS.
Question 10
1 / -0

An element has a body-centred cubic (bcc) structure with a cell edge of 210 pm. The density of the element is 4.2 g/cm³.
The number of atoms present in 210 g of the element is

A
5.03 x 10²⁴

B
10.8 x 10²⁴

C
20.12 x 10²⁴

D
30.18 x 10²⁴

Solution

Volume of the unit cell = (210 pm)³= (210 × 10^-12 m)³ = (210 × 10^-10cm)³= 9.26 × 10^-24 cm³
Volume of 210 g of the element = Mass/Density = 210 g/4.2 g cm^-3 = 50 cm³
Number of unit cells in this volume = 50 cm³/9.26 x 10^-24cm³ = 5.4 x 10²⁴
Since each bcc cubic unit cell contains 2 atoms, so the total number of atoms in 210 g of the element = 2 (atoms/unit cell) × 5.4 x 10²⁴ unit cells = 10.8 × 10²⁴ atoms

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Solid State Test - 6

Result

Solid State Test - 6

Score

-

Rank

-

SHARING IS CARING

Question 1

X4YZ2

X4Y2Z

XY2Z4

X2Y4Z

Solution

Question 2

Ionic crystal of AgBr does not have Frenkel defect.

The unit cell having lattice parameter, i.e. a = b ≠ c and crystal angles, i.e. α = β = 90o and γ = 120o, is hexagonal.

In an ionic compound having Schottky defect, the ratio r+/ r- is very high.

The coordination number of Na+ in NaCl is 4.

Solution

Question 3

1

2

3

4

Solution

Question 4

Monoclinic: a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90o

Tetragonal: a = b ≠ c and α = β = γ = 90o

Orthorhombic: a ≠ b ≠ c and α = β = γ = 90o

Rhombohedral: a = b = c and α = β = γ ≠ 90o

Solution

Question 5

Simple cubic and bcc

Simple cubic, bcc, and ccp

bcc and ccp

hcp and ccp

Solution

Question 6

A - 1, B - 2, C - 3

A - 1, B - 3, C - 2

A - 2, B - 1, C - 3

A - 2, B - 3, C - 1

Solution

Question 7

Sulphide ions are present at the corner of the cube and at the centre of the cube.

Zinc ions occupy all the tetrahedral sites.

There are two zinc ions for every sulphide ion.

Each zinc ion is surrounded by four sulphide ions.

Solution

Question 8

LiF

CsF

CsI

LiI

Solution

Question 9

Graphite, ZnO and CdS

NaCl, HgS and graphite

KNO3, BaSO4 and CaCO3

ZnO, HgS and CdS

Solution

Question 10

5.03 x 1024

10.8 x 1024

20.12 x 1024

30.18 x 1024

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

X₄YZ₂

X₄Y₂Z

XY₂Z₄

X₂Y₄Z

The unit cell having lattice parameter, i.e. a = b ≠ c and crystal angles, i.e. α = β = 90^o and γ = 120^o, is hexagonal.

In an ionic compound having Schottky defect, the ratio r⁺/ r^- is very high.

The coordination number of Na⁺ in NaCl is 4.

Monoclinic: a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90^o

Tetragonal: a = b ≠ c and α = β = γ = 90^o

Orthorhombic: a ≠ b ≠ c and α = β = γ = 90^o

Rhombohedral: a = b = c and α = β = γ ≠ 90^o

KNO₃, BaSO₄ and CaCO₃

5.03 x 10²⁴

10.8 x 10²⁴

20.12 x 10²⁴

30.18 x 10²⁴