Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

351.

Explain: 1. The basis of similarities and differences between metallic and ionic crystals. 2. Ionic solids are hard and brittle.

Answer»

1. Similarities: 

  • Both ionic and metallic crystals have electrostatic forces of attractiàn. 
  • In ionic crystals these forces are between oppositely charged ions. In metals,these forces are among the valence electrons and posityely charged kernels. 
  • Both have high melting point.

Differences: 

  • Ionic bond is strong due to electrostatic forces of attraction whereas metallic bond may be weak or strong depending upon the number of valence electrons and the size of kernels. 
  • In ionic bond, ions are not free to move. Hence, they cannot conduct electricity in solid state. They can do so only in molten state or in aqueous solution. ¡n metals, electrons are free to move. Hence, they conduct electricity in solid state.

2. Ionic crystals are hard due to strong electrostatic forces between them. They are brittle because ionic bond is non – directional.

352.

A solid is hard, brittle and electrically nonconductor. Its melt conducts electricity. What type of solid is it?

Answer»

A solid crystalline electrolyte like NaCl is hard, brittle and electrically nonconductor. But its melt conducts electricity.

353.

Solid A is very hard, electrical insulator is solid as well as in molten state of extremely high temperature. What type of solid is it?

Answer»

Covalent or network solids like SiO2, (quartz) or SiC or C (diamond).

354.

What do you understand by ‘lattice point’?

Answer»

Each lattice point represents one constituent particle of solid. This constituent particle may be an atom, a molecule (group of atoms) or an ion.

355.

Which type of solids are conductors of electricity, malleable and ductile?

Answer»

Metallic solids are conductors of electricity, malleable and ductile.

356.

Name the type of solids which are malleable, ductile and electrical conductors. 

Answer»

Metallic solids are malleable, ductile and electrical conductors.

357.

Explain why ionic solids are hard and brittle.

Answer»

1. In ionic crystalline solids, constituent particles are positively charged cations and negatively charged anions placed at alternate lattice points. 

2. The ions are held by strong coulombic electrostatic forces of attraction compensating opposite forces. Hence they are hard. 

3. Since there are no free electrons, they are not malleable and on applying a shearing force, ionic crystals break into small units. Hence they are brittle.

358.

Why ionic crystals are hard and brittle?

Answer»

The ionic compounds are very hard and brittle. In ionic compounds the ions are rigidly held in a lattice because the positive and negative ions are strongly attracted to each other and difficult to separate. But the brittleness of a compound is now easy to shift the position of atoms or ions in a lattice.

If we apply a pressure on the ionic compounds the layers shifts slightly. The same charged ions in the lattice comes closer. A repulsive forces arises between ‘ same charged ions, due to this repulsions the lattice structure breaks down chemical bonding.

359.

Give suitable reason for the following– (a) Ionic solids are hard and brittle.(b) Copper is malleable and ductile

Answer»

(a) The constituent particles of ionic crystals are ions. These ions are held together in three-dimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are hard and brittle. 

 (b)Copper is a metallic solid and all metallic solids are ductile and malleable.

360.

How does amorphous silica differ from quartz?

Answer»

In amorphous silica, SiO4 tetrahedral are randomly joined to each other whereas in quartz they are linked in a regular manner.

361.

Which point defect lowers the density of a crystal? 

Answer»

Schottky defect

362.

In hcp arrangement, the number of nearest neighbours are.........(A) 10(B) 7(C) 2(D) 12

Answer»

Correct answer is: (D) 12

363.

The packing efficiency in simple cubic unit cell is .......(A) 52.4%(B) 68%(C) 74%(D) 80%

Answer»

Correct answer is: (A) 52.4%

364.

With increase in temperature, the electrical conductivity of semiconductors...........(A) decreases(B) remains same(C) increases(D) none of these

Answer»

Correct answer is: (C) increases

365.

For cubic coordination, the value of radius ratio is.........(A) 0.732 – 1.000(B) 0.225 – 0.414(C) 0.000 – 0.225(D) 0.414 – 0.732

Answer»

Correct answer is: (A) 0.732 – 1.000

366.

Coordination number for Cu is ...........(A) 1(B) 6(C) 8(D) 12

Answer»

Correct answer is: (D) 12

367.

The structure of MgO is similar to NaCl. What would be the coordination number of magnesium?(A) 2(B) 4(C) 6(D) 8

Answer»

Correct answer is: (C) 6

368.

The coordination number of a cation occupying a tetrahedral hole is ............(A) 6(B) 8(C) 12(D) 4

Answer»

Correct answer is: (D) 4

In tetrahedral voids four spheres are involved in its formation.

369.

If the type of the hole occupied is tetrahedral, the radius ratio (r+ /r-) should be........ (A) 0.414 - 0.732(B) > 0.732(C) 0.155 - 0.225(D) 0.225 - 0.414

Answer»

Correct answer is: (D) 0.225 - 0.414

370.

In a close pack array of N spheres, the number of tetrahedral holes are ............ (A) 4N(B) N/2(C) 2N(D) N

Answer»

Correct answer is: (C) 2N

2n
371.

Classify the following solids in different categories based on the nature of the intermolecular forces: sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubedium, argon, silicon carbide.

Answer»

Molecular solids: water, argon.

Ionic solids: potassium sulphate, zinc sulphate.

Metallic solids: tin, rubidium.

Covalent or network: benzene, urea, ammonia, graphite, silicon carbide.

372.

Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Answer»

Since Cu2O is non-stoichiometric oxide, it contains Cu in two oxidation states, +1 and +2. Cu2+ provides an excess of positive charge. As a result an electron from a neighboring Cu+ is transferred to Cu2+. The transfer of electron leaves behind a hole, which carries an extra positive charge and a negative hole is created. It appears that the positive hole moves through the lattice, hence it appears as P-type semiconductor.

373.

Non-stoichiometrice cuprous oxide, Cu2O can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is p-type semiconductor.

Answer»

The ratio is less than 2:1 in Cu2O shows that some cuprous(Cu+) ions have been replaced by cupric (Cu2+) ions. To maintain electrical neutrality, every two Cu+ ions will be replaced by one Cu2+ ion thereby creating a’ hole. As conduction will be due to presence of these positive hole, hence it is a p-type semiconductor.

374.

For square close packing, the planar arrangement is : (a) AAAA (b) ABAB (c) ABCABC (d) AABBAA

Answer»

Option : (a) AAAA

375.

Assign reason for the following: 1. phosphorous doped silicon is a semiconductor. 2. Schottky defect lowers the density of a solid.

Answer»

1. It is because its conductance is intermediate between conductor and insulator. 

2. In Schottky defect, both cations and anions are missing which leads to lowering the density of a solid.

376.

What type of semiconductor is produced when silicon is doped with boron ?

Answer»

p-type semiconductor.

377.

If ‘a’ stands for the edge length of the cubic systems: simple cubic, body centered cubic and face centered cubic, then the ratio of radii of the spheres in these systems will be ............. respectively.(A) \(\cfrac{1}{2}\) a : \(\cfrac{\sqrt{3}}{2}\)a : \(\cfrac{\sqrt{3}}{\sqrt{2}}a\)(B) 1a : \(\sqrt{3}\) a : \(\sqrt{2}\)a(C) \(\cfrac{1}{2}\) a : \(\cfrac{\sqrt{3}}{4}\) a : \(\cfrac{1}{2\sqrt{2}}a\) a(D) \(\cfrac{1}{2}\)a : \(\sqrt{3}\) a : \(\cfrac{1}{\sqrt{2}}a\) 

Answer»

Correct answer is:(C) \(\cfrac{1}{2}\) a : \(\cfrac{\sqrt{3}}{4}\) a : \(\cfrac{1}{2\sqrt{2}}a\)

Simple unit cell, r = a / 2

Body centered unit cell, r = \(\cfrac{a\sqrt{3}}{4}\)

Face centered unit cell, r =\(\cfrac{a}{2\sqrt{2}}\)

378.

For the given structure, the site marked as ‘S’ is a _______ void. (A) tetrahedral(B) cubic(C) octahedral(D) triangular

Answer»

correct answer is: (C) octahedral

379.

Atom of element B forms hcp lattice and those of the element A occupy 2/3 rd of tetrahedral void .what would be the formula of the compound ?

Answer»

Suppose number of atom B in hcp = N

No. of Tetrahedral void = 2N

Atom A occupying tetrahedral void = 2/3* 2N

Ratio   A:B   A4B3

380.

In which of the following substances, the carbon atom is arranged in a regular tetrahedral structure?(A) Diamond(B) Benzene(C) Graphite(D) Carbon black

Answer»

Correct answer is: (A) Diamond

381.

In graphite, carbon atoms are joined together due to ..........(A) ionic bonding(B) van der Waal’s forces(C) metallic bonding(D) covalent bonding

Answer»

Correct answer is: (D) covalent bonding

382.

The stability of a crystal is reflected in the magnitude of its melting point”. Comment. Collect the value of melting point of water, ethyl alcohol, diethylether and methane from the data given in the book. What can you say about the inter molecular forces between these molecules?

Answer»

Higher the melting points, greater will be the forces of attraction between the molecules or constituent particles and hence greater will be the stability. The melting points of given compounds are as follows.
Water            = 273 K
Ethyl alcohol = 155.7 K
Diethyl ether = 156.7K
Methane       = 90.5 K
The inter molecular forces of attraction between the molecules of water and ethyl alcohol are mainly due to hydrogen bonding. The melting point of water is higher than ethyl alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The inter molecular forces present in them are dipole-dipole interactions Methane is a non-polar molecule. The only forces present in them are the weak vander waals forces i.e., London dispersion forces.

383.

Why glass is considered as a super cooled liquid?

Answer»

Amorphous solids have a tendency to flow. Since glass is an amorphous solid , so it is called super cooled liquid or pseudo solid. 

384.

Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of the solid. Would it show cleavage property?

Answer»

As the solid have same values of refractive index among all directions, it means it is isotropic in nature. Hence, the solid is amorphous. As the solid is amorphous in nature so it would not show clean cleavage when cut with a knife. It would break in to pieces with irregular surfaces.

385.

Why is glass considered a super-cooled liquid?

Answer»

Glass is an amorphous solid. Like liquid it has a tendency to flow, though very slowly. Because of its flowing nature the glass panes in the windows of old buildings are found to be slightly thicker at the bottom than at the top. Hence glass is consider as super-cooled liquid because of its flowing nature.

386.

Classify the following as amorphous and crystalline solids:Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

Answer»

Amorphous solids-Polyurethane, Teflon, Cellophane, Polyvinyl chloride, Fibre glass. Crystalline solids-Naphthalene, Benzoic acid, Potassium nitrate and Copper.

387.

Why solids are hard?

Answer»

In solids the positions of constituent particles like atoms or molecules or ions are fixed i.e., they are not free to move. They are stationary in their fixed position. They have strong intermolecular or interionic forces of attraction. Hence solids are hard.

388.

The number of tetrahedral and octahedral voids in a hexagonal primitive unit cell is-(a) 8,4(b) 6,6(c) 2,1(d) 12,6

Answer»

The answer is (d) 12,6.

389.

In a closed packed lattice containing nparticles, the number of tetrahedral and octahedral voids respectively are ...........(A) n, 2n(B) n, n(C) 2n, n(D) 2n, n/2

Answer»

Correct answer: (C) 2n, n

390.

A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedral voids?

Answer»

Number of atoms in 0.5 mole of compound
= 0.5 x 6.022 x 1023
= 3.011 x 1023
Number of octahedral voids = Number of atoms in close packing
= 3.011 x 1023
Number of terahedral voids = 2 x No. of atoms in close packing
= 2 x 3.011 x 1023
= 6.022 x 1023
Total number of voids = 3.011 x 1023 + 6.022 x 1023
= 9.033 x 1023

391.

The coordination number of a metal crystallizing in a hexagonal close packed structure is .............(A) 4(B) 12(C) 8(D) 6

Answer»

Correct answer: (B) 12

392.

The empty space between the shared balls and hollow balls as shown in the diagram is called ...........(A) hexagonal void(B) octahedral void(C) tetrahedral void(D) double triangular void

Answer»

Correct answer: (B) octahedral void

393.

The packing efficiency of the two-dimensional square unit cell shown below is ..........(A) 39.27 %(B) 68.02 %(C) 74.05 %(D) 78.54 %

Answer»

Correct answer is: (D) 78.54 %

a\(\sqrt{2}\) = 4r

\(\Rightarrow\) a = 2\(\sqrt{2}\) r

Packing fraction =\(\cfrac{Occupied\,area}{Total\,area}\)

\(=\cfrac{2\pi r^2}{(2\sqrt{2r})^2}\times100=78.5\)%

394.

A compound forms hexagonal close packed structure . What is the total number of void in 0.5 mol of it .How many of these are tetrahedral void ?

Answer»

Number of atom in the close packing = 0.5 x 6.023 x 1023 = 3.011 x 1023

Number of octahedral void = 1 x 3.011 x 1023 

Number of tetrahedral void 2 x 3.011 x 1023

= 6.033 x 1023 

395.

The empty space between the shared balls and hollow balls as shown in the diagram is called.........(A) hexagonal void(B) octahedral void (C) tetrahedral void (D) triangular void

Answer»

Correct answer is: (B) octahedral void

396.

Write the relationship between atomic radius (r) and edge length (a) of cubic unit cell for (a) Simple cubic unit cell(b) Body-centred cubic unit cell(c) Face-centred cubic unit cell

Answer»

(a) a = 2r 

(b) a = 4 r 

(c) a = 2 2r

397.

Express the relationship between atomic radius (r) and edge length (a) in the bcc unit cell.

Answer»

The relationship between atomic radius (r) and edge length (a) in the bcc unit cell are r = √3a/2

398.

What is the Relationship between edge length of unit cell (a) and radius of sphere(r)?

Answer»

Relationship between edge length of unit cell (a) and radius of sphere(r): 

Unit cellRelationship between a and rPacking efficiency
Simple cubic latticea = 2r52.4% 
bcc structurea = 4 r /√368% 
fcc structurea = 4r/√2 74% 

399.

What are the types of Stoichiometric defects?

Answer»

Type of Stoichiometric defects: 

Vacancy defects: 

(i) some of the lattice sites of the crystal are vacant. 

(ii) This defect decreases the density of the crystal. 

(iii) This defect can be developed by heating the substance. 

Interstitial defects: 

(i) some constituent particles occupy the interstitial sites of the crystal. 

(ii) This defect increases the density of the crystal. 

400.

A silicon solar battery makes use of .......... (A) n-type semiconductor(B) p-type semiconductor(C) combination of Si doped with As and B(D) p-n junction

Answer»

Correct answer is: (C) combination of Si doped with As and B