Chemistry Test

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Chemistry Test - 3

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Weekly Quiz Competition

Question 1
3 / -1

In a crystal of diamond, the maximum proportion of free volume that is available is:

A
0.52

B
0.34

C
0.32

D
0.66

Solution

Solution

Diamond has a ZnS structure with 'c' atoms occupying alternate tetrahedral voids.

Proportion of the free Volume = 1-0.34 = 0.66.
Question 2
3 / -1

From among the following, the incorrect statement is

A
In sodium chloride crystal, the number of next nearest neighbours of each Na⁺ ion is 12 Na⁺ ions.

B
Diamond follows ZnS structure and the number of carbon atoms per unit cell is 8 .

C
When AgCl is dopped with CdCl ₂ , the number of cation vacancies is equal to number of cadmium ions that get inducted.

D
In case of metal excess defect having F - centres, the crystal is diamagnetic.

Solution

Solution

In sodium chloride crystal, the number of next nearest neighbours of each Na + ion is 12 Na + ions.

Diamond follows ZnS structure and the number of carbon atoms per unit cell is 8 .

When AgCl is dopped with CdCl 2 , the number of cation vacancies is equal to number of cadmium ions that get inducted.

Crystals having F - centres are paramagnetic and not diamagnetic.
Question 3
3 / -1

An element occupying a BCC structure has 12.08 x 10²³ unit cells. The total number of atoms of the element in the cells will be:

A
24.16 x 10²³

B
36.18 x 10²³

C
6.04 x 10²³

D
12.08 x 10²³

Solution

Solution

BCC unit cell contains 2 atoms

No. of atoms in 12.08 x 10²³ unit cells = 2 x 12.08 x 10²³ = 24.16 x 10²³
Question 4
3 / -1

The no. of atoms in 100 g m of an f c c crystal with density d = 10 g / c m 3 and cell edges as 200 p m is equal to -

A
3×10²⁵

B
5×10²⁴

C
1×10²⁵

D
5.96×10^–3

Solution

Solution

where, a is the edge of a unit cell

N_A = Avogadro's number (6.022 × 10²³)

M = Molar mass

Z = number of atoms per unit cell

For FCC, Z=4

From the given data, we can calculate the molar mass of the given substance as:
Question 5
3 / -1

If x = radius of Na⁺and y = radius of Cl - and a is the unit cell edge length for NaCl crystal, then which of the given relation is correct?

A
x+y=a

B
2x+2y=a

C
x+y=2a

D
x+y=√2a

Solution

Solution

In NaCl , chlorides Cl - having the CCP / FCC type of arrangement. i.e., Ions of Cl - are present in a corner site and center of each face of cubic lattice. Sodium ions (Na ⁺) are present at the octahedral voids. i.e., Located around six chloride ions.

Edge length = 2 × r_Cl ^- + 2 × r _Na^₊

Where,

r_Cl^- = Radius of chlorine ion = yr _Na ⁺ = Radius of sodium ion = x

Hence, edge length of NaCl crystal, a = 2 x + 2 y
Question 6
3 / -1

In 3 D close-packed structures, for every 100 atom, it contains

A
50 octahedral voids.

B
100 tetrahedral voids.

C
200 octahedral voids.

D
100 octahedral voids.

Solution

Solution

In a closed packed structure:

The number of octahedral voids = Effective number of atoms

The number of tetrahedral voids = 2 × Effective number of atoms

Given number of atoms n = 100

Hence, the number of tetrahedral voids = 2 × n = 2 × 100 = 200

The number of octahedral voids = n = 100
Question 7
3 / -1

Tetragonal crystal system has the following unit cell dimensions

A
a = b = c and γ = 120°

B
a = b ≠ c and α = β = γ = 90°

C
a = ≠ b ≠ candα = β = γ = 90°

D
a = b ≠ c and α = β = 90°,γ = 120°

Solution

Solution

Tetragonal type crystal system has the unit cell dimension a = b ≠ c and interfacial angles α = β = γ = 90 °.
Question 8
3 / -1

Which of the following are isomorphous with KNO₃

A
NaNO₃

B
CaCO₃

C
KClO₃

D
All of these

Solution

Solution

Isomorphous compounds are those which have a closely similar shape.

CO₃^2- , NO_3^- are sp² hybridised whereas ClO^-_3^- is sp³ hybridised

Since the size of Na⁺ and K⁺ are different therefore NaNO₃ and KNO₃ are not isomorphous.
Question 9
3 / -1

Which of the following is an example of a crystal in which some ions are totally missed from lattice sites?

A
F -centres

B
Interstitial defect

C
Frenkel defect

D
Schottky defect

Solution

Solution

Schottky defects are arise when one positive ion and one negative ion are missing from their respective positions leaving behind a pair of holes. These are more common in ionic compound with high co-ordination number and having almost similar size of cations and anions.
Question 10
3 / -1

Oxygen atoms form FCC lattice with 'A' atoms occupying all tetrahedral voids and 'B' atoms occupying all octahedral voids. Determine the formula of resultant compound formed when atoms are removed from two of the body diagonals?

A
A₄B₄O₇

B
A₈B₆O₇

C
A₈B₈O₇

D
A₆B₈O₆

Solution

Solution

This problem is based on determination of molecular formula of solid which can be determined by using following steps.
Since O -atoms form FCC lattice, the number of

Number of B atoms at octahedral voids = Number of O -atoms = 4

Number of A atoms at tetrahedral voids

= 2 x Number of O-atoms

= 2 x 4 = 8

Initial formula of compound is A₈ B₄ O₄

After removal atoms from the two body diagonals:

A atoms = 8 - 4 = 4 (Removed from 2 body diagonals)

B atoms = 4 - 1 = 3 (Removed from body centre)

So, formula will be A₄ B₃ O_3.5 i.e. A₈ B₆ O₇

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Chemistry Test - 3

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Question 1

0.52

0.34

0.32

0.66

Solution

Question 2

In sodium chloride crystal, the number of next nearest neighbours of each Na+ ion is 12 Na+ ions.

Diamond follows ZnS structure and the number of carbon atoms per unit cell is 8 .

When AgCl is dopped with CdCl 2 , the number of cation vacancies is equal to number of cadmium ions that get inducted.

In case of metal excess defect having F - centres, the crystal is diamagnetic.

Solution

Question 3

24.16 x 1023

36.18 x 1023

6.04 x 1023

12.08 x 1023

Solution

Question 4

3×1025

5×1024

1×1025

5.96×10–3

Solution

Question 5

x+y=a

2x+2y=a

x+y=2a

x+y=√2a

Solution

Question 6

50 octahedral voids.

100 tetrahedral voids.

200 octahedral voids.

100 octahedral voids.

Solution

Question 7

a = b = c and γ = 120°

a = b ≠ c and α = β = γ = 90°

a = ≠ b ≠ candα = β = γ = 90°

a = b ≠ c and α = β = 90°,γ = 120°

Solution

Question 8

NaNO3

CaCO3

KClO3

All of these

Solution

Question 9

F -centres

Interstitial defect

Frenkel defect

Schottky defect

Solution

Question 10

A4B4O7

A8B6O7

A8B8O7

A6B8O6

Solution

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In sodium chloride crystal, the number of next nearest neighbours of each Na⁺ ion is 12 Na⁺ ions.

When AgCl is dopped with CdCl ₂ , the number of cation vacancies is equal to number of cadmium ions that get inducted.

24.16 x 10²³

36.18 x 10²³

6.04 x 10²³

12.08 x 10²³

3×10²⁵

5×10²⁴

1×10²⁵

5.96×10^–3

NaNO₃

CaCO₃

KClO₃

A₄B₄O₇

A₈B₆O₇

A₈B₈O₇

A₆B₈O₆