6/12/24, 6:08 PM Question Answer Paper

Std.: 12th (Science) Marks: 30

Date: 30-05-24
SOHAM EDUCATORS Time: 1.5 hrs

SUBJECT : CHEMISTRY()

Q.1 Select and write the most appropriate answers from given alternatives: 5
1 Which of the following is n-type semiconductor?
(a) Pure Si
(b) Si doped with As
(c) Si doped with Ga
(d) Ge doped with In

Ans Si doped with As

2 The arrangement ABCABC……. Is referred as

a. Cubic close packing
b. Hexagonal close packing
c. Tetragonal close packing
d. Octahedral close packing

Ans Option (a)

3 Schottky defect in crystals is observed when

a. Unequal number of cations and anions are missing from the lattice
b. Equal number of cations and anions are missing
c. An ion leaves its normal site and occupies an interstitial site
d. Density of the crystal is increased

Ans Option (b)

4 Which of the following is not correct?

a. Four spheres are involved in the formation of tetrahedral void. |
b. The centres of spheres in octahedral voids are at the apices of a regular tetrahedron.
c. If the number of atoms is N the number of octahedral voids is 2N.
d. If the number of atoms is N/2, the number of tetrahedral voids is N.

Ans Option (d)

5 Frenkel defect is generally observed in

a. AgBr b. AgI c. ZnS d. All of these

Ans Option (d)

Q.2 Answer the following very short questions:(Any Three) 3

1 What are the types of particles in each of the four main classes of crystalline solids?

Ans The smallest constituents or particles of various solids are atoms, ions or molecules.

2 Mention two properties that are common to both hep and ccp lattices.

Ans In hcp and ccp crystal lattices coordination number is 12 and packing efficiency is 74%.

3 What is the cause of electrical conductivity in

(a) metals (b) ionic solids.

Ans (a) Due to flow of electrons.

(b) Due to flow of ions in solution or melt and defect in the solid state.

4 Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the
dimensions of unit cell of its crystal.

Ans Density,
n(no. of atoms per unit cell)M(molar mass)
P= NA (avogadro 's number) a3 (cell edge)

https://web.vizn.co.in/class-admin/generate-paper/view-model-answer/?d=NjcyNzI1 1/5
6/12/24, 6:08 PM Question Answer Paper

By knowing density, number of atoms per unit cell, cell edge and Avogadro's number,
we can calculate M, the atomic mass of metal

5 Let a small quantity of phosphorus be doped into pure silicon.

Will the resulting material contain the same number of total number of electrons as the original pure silicon ?

Ans Total number of electrons in doped silicon will be more than in original silicon.

Q.3 Answer the following Questions:(Any Three) 6

1 (a) Which of these two CdCl2 and NaCl will produce Schottky defect, if added to a AgCl crystal?
(b) Why common salt is sometimes yellow instead of being pure white?

Ans i. CdCl2, because each divalent Cd2+ ion will displace two Ag+ from their sites and occupy only one site,
thereby creating a cationic vacancy.
ii.This is due to the presence of electrons in some lattice sites in place of anions. These sites act as F-
centers.
2 In an ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one third of
tetrahedral voids. What is the formula of the compound?

Ans In the given hcp lattice, Y atoms are present at 12 corners and 2 face centres.
∴ Number of Y atoms = 12 × 12 + 2 × 12 = 3
There are 6 tetrahedral voids, the number of X atoms = 13 × 6 = 2
∴ Formula of the compound is X2Y3.

3 Classify the following solids in different categories based on the nature of intermolecular forces operating in
them:
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon
carbide.

Ans Ionic Molecular Molecular Covalent Metallic

(non - polar) (hydrogen
bonding)

Potassium Benzene Ammonia Graphite Tin

sulphide
Urea Water Silicon Rubidium
Zinc carbide
Argon
sulphide

4 What is a unit cell?

Ans Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice) which when
repeated in different directions produces the crystalline solid (lattice).
The crystal is considered to consist of an infinite number of unit cells.
The unit cell possesses all the characteristics of the crystalline solid.

Q.4 Solve the following:(Any Three) 6

1 When gold crystallizes, it forms face-centred cubic cells. The unit cell edge length is 408 pm. Calculate the
density of gold. Molar mass of gold is 197 g/mol.
Mn
Ans ρ = a3 NA

M = 197 g mol-1, n = 4 atoms for fcc, NA = 6.022 × 1023 atoms mol-1,

a = 408 pm = 408 × 10-12m = 4.08 × 10-8 cm
Substitution of these quantities in the equation given
197 g mol −1 ×4 atom
ρ= 3
(4.08×10−8 ) cm3×6.022×1023 atom mol −1

= 19.27 g/cm3, 19.27 × 103 kg/m3.

2 An element with molar mass 2⋅7 × 10⁻2 kg mol⁻1 forms a cubic unit cell with edge length 405 pm. If its density
is 2⋅7 × 103 kg m⁻3, what is the nature of the cubic unit cell?
nM
Ans Density, P = NA a3

https://web.vizn.co.in/class-admin/generate-paper/view-model-answer/?d=NjcyNzI1 2/5
6/12/24, 6:08 PM Question Answer Paper

M = 2.7 × 10-2
kg mol-1
a = 405 pm = 4.05 × 10-10 m
d = 2.7 × 10-3 kg m-3
NA = 6.023 × 1023 mol-1
3
2.7×10−2 ×(4.05×10−10 ) ×6.023×1023
n= 2.7×10−2
=4
Nature of unit cell = face centered unit cell.

3 The density of lead is 11.35 g/cm3 and the metal crystallizes with fcc unit cell. Estimate the cell edge of unit
cell of lead. (Atomic mass of lead = 207 g/mol).
nM
Ans Density, P = NA a3

(Z 4) fcc unit cell

a3 = NnM
AP

a3 = 12.1141 × 10-23
a = 4.949 × 10-8 cm. = 494.9 pm

4 A compound forms hexagonal close-packed structure. What is the total number of voids in 1⋅5 mol of it? How
many of these are tetrahedral and octahedral voids?

Ans Number of atoms in the close packing = 1.5 mol

= 1⋅5 × 6⋅023 × 1023 = 18.069 × 1023
Number of tetrahedral voids = 2 × No. of atoms in the packing
= 2 × 18.069 × 1023 = 36.138 × 1023.
Number of oc
tahedral voids = No. of atoms in the packing
= 18.069 × 1023

Q.5 Answer the following Questions:(Any Two) 6

1 What are paramagnetic substances? Give examples.

Ans (1) The magnetic properties of a substance arise due to the presence of electrons.
(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic
moment and magnetic properties.
(3) If an atom or a molecule contains one or more unpaired electrons spinning in same direction, clockwise or
anticlockwise, then the substance is associated with net magnetic moment and magnetic properties. They
experience a net force of attraction when placed in the magnetic field. This phenomenon is called
paramagnetism and the substance is said to be paramagnetic.
For example, O2, Cu2+, Fe3+ , Cr3+ , NO, etc.

2 (i) Why is Frenkel defect found in AgCl?

(ii) What are stoichiometric defects or intrinsic defects in ionic crystals?
(iii) What are interstitials in a crystal?

Ans i. Frenkel defect is found in AgCl due to large differences in size of cation (Ag+ ion) and anion(Cl- ion).
ii. Stoichiometric defects or intrinsic defects are those defects in which the ratio of cations to anions remains
the same as represented by the molecular formula.
iii. Atoms or ions that fill the normal vacant interstitial voids in a crystal are called interstitials.
3 i. Why does Frenkel defect not change the density of AgCl crystals?
ii. Name a salt that can be added to AgCl so as to produce cation vacancies.
iii. What type of crystal defect is produced when sodium chloride is doped with MgCl2?
Ans i. Due to Frenkel defect, as no ions are missing from the crystal as a whole, so there is no change in density.
ii. CdCl2 or SrCl2.
iii. Impurity defect will be produced. A cation vacancy is produced. A substitutional solid solution is formed
because two Na+ ions are replaced by one Mg2+ ion in the lattice site.
Q.6 Answer the following Questions:(Any One) 4
1 How are tetrahedral and octahedral voids formed?

https://web.vizn.co.in/class-admin/generate-paper/view-model-answer/?d=NjcyNzI1 3/5
6/12/24, 6:08 PM Question Answer Paper

Ans Tetrahedral void : The vacant space or void among four constituent particles having tetrahedral arrangement
in the crystal lattice is called tetrahedral void.

The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular
void made by three coplanar spheres is in contact with fourth sphere above or below it.
Octahedral void : The vacant space or void at the centre of six spheres (or atoms) which are placed
octahedrally is called octahedral void.

2 Distinguish with the help of diagrams metal conductors, insulators and semiconductors from each other.

Ans Conductor:
1. A substance which conducts heat and electricity to a greater extent is called conductor.
2. In this, conduction bands and valence bands overlap or are very closely spaced.
3. There is no energy difference or very less energy difference between valence bands and conduction
bands.
4. There are free electrons in the conduction bands.
5. The conductance decreases with the increase in temperature.
6. E.g., Metals, alloys.
7. The conducting properties can’t be improved by adding third substance.

Insulator:
8. A substance which cannot conduct heat and electricity under any conditions is called insulator.
9. In this, conduction bands and valence bands are far apart.
10. The energy difference between conduction bands and valence bands is very large.
11. There are no free electrons in the conduction bands and electrons can’t be excited from valence bands
to conduction bands due to large energy difference.
12. No effect of temperature on conducting properties.
13. E.g., Wood, rubber, plastics.

https://web.vizn.co.in/class-admin/generate-paper/view-model-answer/?d=NjcyNzI1 4/5
6/12/24, 6:08 PM Question Answer Paper

14. No effect of addition of any substance.

Semiconductor:
15. A substance that has poor electrical conductance at low temperature but higher conductance at higher
temperature is called semiconductor.
16. In this, conduction bands and valence bands are spaced closely.
17. The energy difference between conduction bands and valence bands is small.
18. The electrons can be easily excited from valence bands to conduction bands by heating.
19. Conductance increases with the increase in temperature.
20. E.g., Si, Ge
21. By doping, conducting properties improve. E.g. n-type, p-type semiconductors.

https://web.vizn.co.in/class-admin/generate-paper/view-model-answer/?d=NjcyNzI1 5/5