Solid State Test

Result

Solid State Test - 20

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Question 1
1 / -0

Only One Option Correct Type
This section contains 9 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.

A
Schotty defect

B
Frenkel defect

C
metal excess defect

D
none of these

Solution

. (a) A Schottky defect consists of a pair of holes in the crystal lattice. One positive ion and one negative ion are absent.
Note This sort of defect occurs mainly in highly ionic compounds where cation and anion are of similar size, hence the coordination number is high (8.6) as in NaCI, CsCI, KCI, KBr
Question 2
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Structure shown in the figure represents

A
metal excess defect because of absence of anion

B
metal excess defect because of absence of cation

C
Frenkel defect

D
Schottky defect

Solution

The picture is shown as cation excess that means metal excess with the absence of anion.
Question 3
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Which of the following defects is also known as dislocation defect?

A
Frenkel

B
Schottky

C
Non-Stoichiometric

D
Simple interstitial

Solution

(a) If there is no deficie ncy of cation or anion but cation migrates to interstitial site, it is called dislocation. It is observed in the crystals with smaller cation and larger anion, so that cation can dislocate easily. It is called Frenkel defect.
Question 4
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Cations are present in the interstitial sites in

A
Frenkel defect

B
Schottky defect

C
vacancy defect

D
metal deficiency defect

Solution

. (a) Frenkel defect is observed in the crystals in which cation is much smaller than the anion so that cation can migrate to interstitial position.
Question 5
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Select the correct statement about non-stoichiometric compounds

A
They are called Berthollide compounds

B
They do not obey the law of constant composition

C
Electrical neutrality is maintained either by having extra electrons in the structure or by changing the charge on some of the metal ions

D
All of the above

Solution

(a) Non-stoichiometric compounds are also called Berthollide compounds thus, correct.
(b) Since, composition of the constituent elements is not fixed, thus law of constant composition is not followed as is Fe 0,98 O thus, correct.
(c) If cation (say A⁺) is missing then, charge is balanced by the cation A²⁺. If anion (say B^-) is missing, then charge is balanced by inserting electron (e^-)thus, correct.
Question 6
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In a face-centred cubic lattice, atom A occupies the corners of the cube and atom B occupies the face-centred positions. If one atom of B is missing from one of the face-centred points, the formula of the compound is

A
AB

B
AB₂

C
A₂B

D
A₂B₅

Solution
Question 7
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An element crystallises in fcc lattice having edge length 40 0 pm. Maximum radius of the atom which can be placed in the interstitial site without distorting the structure is

A
58.55pm

B
117pm

C
166pm

D
83pm

Solution

(a) For fcc structure
= 0.414 for octahedral void
= 0.225 for tetrahedral void
Thus, maximum packing can be for octahedral void
Question 8
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Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, is present as M²⁺ and M³⁺ in its oxide. The fraction of the metal which exists as M³⁺ would be:

A
7.01%

B
4.08%

C
6.05%

D
5.08%

Solution

Metal oxide = M_0.98O
If ‘x’ ions of M are in +3 state, then
3x + (0.98 – x) × 2 = 2
x = 0.04
So the percentage of metal in +3 state would be
Question 9
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If NaCI is doped with 10^-3 mole % of SrCI₂ then , number of cation ic vacancies is

A
10^-5 mol ^-1

B
6.02 x 10²⁰ mol ^-1

C
6.02 x 10¹⁸mol ^-1

D
6.02 x 10^-18 mol^-1

Solution

(c)
Due to the addition of SrCI₂, each Sr²⁺ ion replaces two Na⁺ ions, but occupies one Na⁺ lattice point. Thus, this exchange of Na+ ion by Sr²⁺ ion makes one cationic vacancy.
SrCI² doped = 10^-3 mol per 100 mol = 10^-5mol per 1 mol
∴ Cation vacancies = 10^-5mol per 1 mol
= 10^-5 x N₀ mol^-1
= 10^-5 x 6.02 x 10²³
Total = 6.02 x 10¹⁸cationic vacancies mol^-1
Question 10
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Matching List Type
Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

A
a

B
b

C
c

D
d

Solution

(i) When cation or anion is absent, molar mass decreases hence, density decreases.
Thus, (i) - » (q)
(ii) When cation is packed in the interstitial site, molar mass increases. Hence, density increases.
Thus, (ii) - » (s)
(iii) When smaller cation migrates into the interstitial site, Frenkel defect is observed. There is no change in molar mass and thus, no change in density.
Thus, (iii) -> (p)
(iv) As in NaCI, when cation and anion are missing (Schottky defect), molar mass and thus, density decreases.
Thus, (iv) - 0