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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.

51.

Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm^-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)(a) 97.7%(b) 95.4%(c) 4.6%(d) 2.3%The question was posed to me in an interview for internship.This interesting question is from Solid State in division Solid State of Chemistry – Class 12

Answer»

Right answer is (d) 2.3%

To explain I would say: Given,

Edge length (a) = 350 pm = 3.5 x 10^-8 cm

Atomic mass (M) = 7 amu

Avogadro’s NUMBER (N0) = 6.02 x 10^23

Density (ρ) = (Z x M)/(a^3 x N0)

= (2 x 7)/((3.5 x 10^-8)^3 x 6.02 x 10^23)

= 0.542 G cm^-3

% of lithium atoms occupied = (Experimental density/Theoretical density) x 100

= 0.53/0.542 x 100

= 97.7%

% of unoccupied lattice SITES = 100 – 97.7

= 2.3%.

52.

In which of the following magnetic properties of elements does the magnetic susceptibility increase on increasing the temperature?(a) Paramagnetism(b) Anti-ferromagnetism(c) Ferromagnetism(d) DiamagnetismI got this question in examination.My question comes from Solid State Magnetic Properties topic in section Solid State of Chemistry – Class 12

Answer»

Right ANSWER is (b) Anti-ferromagnetism

Easiest EXPLANATION: On increasing the TEMPERATURE, the MAGNETIC susceptibility of diamagnetic substances do not change, that of paramagnetic and ferromagnetic substances decrease and that of antiferromagnetic substances INCREASE.

53.

What does the ratio ‘space occupied/total space’ denote?(a) Packing factor(b) Packing efficiency(c) Particle fraction(d) Packing unitI had been asked this question in an internship interview.Enquiry is from Solid State in chapter Solid State of Chemistry – Class 12

Answer»

The correct choice is (a) Packing FACTOR

For explanation: Packing factor is a fraction of TOTAL space of the unit CELL occupied by the CONSTITUENT particles.

54.

Which of the following is regarded as the ‘repeatable entity’ of a 3D crystal structure?(a) Unit cell(b) Lattice(c) Crystal(d) Bravais IndexI got this question in an online interview.The question is from Solid State in portion Solid State of Chemistry – Class 12

Answer»

Correct choice is (a) Unit cell

For explanation I would say: Unit cell is the smallest entity of a crystal lattice which, when repeated in space (3 dimensions) generates the ENTIRE crystal lattice. Lattice comprises of the unit cells which hold all the PARTICLES in a particular arrangement in 3 dimensions. Crystal is a piece of homogenous solid and Bravais indices are USED to define planes in crystal LATTICES in the HEXAGONAL system.

55.

The molecules in polar molecular solid are held together by __________(a) dipole-dipole interaction(b) london forces(c) ionic bond(d) metallic bondI got this question in semester exam.My doubt is from Classification of Crystalline Solids topic in division Solid State of Chemistry – Class 12

Answer»

The correct choice is (a) dipole-dipole interaction

For EXPLANATION I WOULD SAY: The force responsible for holding TOGETHER the molecules of polar molecular solids is dipole-dipole force of attraction. Polar molecular solids are non-conductors of ELECTRICITY.

56.

Which type of solid crystals will conduct heat and electricity?(a) Ionic(b) Covalent(c) Molecular(d) MetallicThe question was asked in an interview.This intriguing question originated from Amorphous and Crystalline Solids in section Solid State of Chemistry – Class 12

Answer»

Right option is (d) Metallic

Explanation: Metallic crystals consist of METAL cations surrounded by a SEA of mobile valence ELECTRONS. These electrons are capable of moving through the ENTIRE crystal. The metallic crystals conduct heat and ELECTRICITY due to the presence of these mobile electrons in them.

57.

_________ is the basic repeated structural unit of a crystalline solid.(a) Monomer(b) Molecule(c) Unit cell(d) AtomThis question was posed to me by my college professor while I was bunking the class.This question is from Solid State Characteristics topic in division Solid State of Chemistry – Class 12

Answer»

Right OPTION is (c) Unit cell

For explanation: Crystalline solids are composed of many small crystals, each of which is CALLED a unit cell. It is a SPECIFIC term. MONOMER is the basic unit for a polymer, and ATOMS make up molecules, which can further arrange themselves to form solids, liquids or gases.

58.

Label the diagram correctly.(a) 1-Non metal, 2-Semiconductor, 3-Metal(b) 1-Semiconductor, 2-Non metal, 3-Metal(c) 1-Metal, 2-Semiconductor, 3-Non metal(d) 1-Metal, 2-Non metal, 3-SemiconductorThis question was addressed to me in exam.My query is from Solid State Electrical Properties topic in division Solid State of Chemistry – Class 12

Answer» CORRECT option is (C) 1-Metal, 2-Semiconductor, 3-NON metal

For explanation I would say: Energy gap is very less (may even be ABSENT) for metals, very HUGE for non metals and intermediate (small) for semiconductors.
59.

An element with cell edge of 288 pm has a density of 7.2 g cm^-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol^-1?(a) Body-Centred Cubic (BCC)(b) Face-Centred Cubic (FCC)(c) Simple Cubic(d) Hexagonal Closed Packing (HCP)The question was asked during an internship interview.The doubt is from Solid State topic in portion Solid State of Chemistry – Class 12

Answer»

Correct answer is (a) Body-Centred Cubic (BCC)

For explanation I would say: Given,

Edge length (a) = 288 pm

Density (ρ) =7.2 g cm^-3

Atomic mass (M) = 51.8 g mol^-1

Avogadro’s number (N0) = 6.02 X 10^23

We know, (ρ) = (Z x M)/(a^3 x N0)

Or Z =(ρ x a^3 x N0)/M = (7.2 x (288 x 10^-10) ^3 x 6.02 x 10^23)/51.8

Z = 2

Therefore, the ELEMENT has Body-Centred Cubic (BCC) type of structure.

60.

What is the correct order of magnetic strength among the following elements?(a) Fe > Co > Ni > Cu(b) Fe > Ni > Co > Cu(c) Cu > Ni > Co > Fe(d) Cu > Fe > Ni > CoI got this question during an interview.The origin of the question is Solid State Magnetic Properties in chapter Solid State of Chemistry – Class 12

Answer»

Correct choice is (a) Fe > Co > Ni > Cu

The explanation: MAGNETIC strength depends on the NUMBER of UNPAIRED electrons possessed by the element. Iron, Cobalt, Nickel and Copper have 4, 3, 2 and 1unpaired electron respectively. Hence, the correct ORDER of magnetic strength is: Fe > Co > Ni > Cu.

61.

If the body-centered unit cell is assumed to be a cube of edge length ‘a’ with spherical particles of radius ‘r’ then how is the diameter, d of particle and surface area, S of the cell related?(a) S = 32d^4/3(b) S = 2d^2(c) S = 4d^2(d) S = 8d^2I have been asked this question in semester exam.My question is from Solid State in portion Solid State of Chemistry – Class 12

Answer»

Correct answer is (d) S = 8d^2

Explanation: For BCC unit CELL the relation between radius of a particle ‘r’ and edge length of unit cell, a, is r = \(\FRAC{\sqrt{3}}{4}\)a.

We know that diameter, d = 2r = \(\frac{\sqrt{3}}{2}\)a

Implying d^2 = \(\frac{3}{4}\)a^2

Therefore, 4d^2/3=a^2

 Multiplying by 6 on both sides gives S = 6a^2 = 8d^2, where S is the surface area of the CUBE = 6a^2.

62.

What is the radius of a metal atom if it crystallizes with body-centered lattice having a unit cell edge of 333 Pico meter?(a) 1538.06 pm(b) 769.03 pm(c) 288.38 pm(d) 144.19 pmThe question was asked in a national level competition.My question is based upon Solid State topic in section Solid State of Chemistry – Class 12

Answer» RIGHT option is (d) 144.19 pm

Easiest explanation: For body-centered unit cells, the relation between radius of a particle ‘r’ and edge length of unit CELL ‘a’ is given as \(\FRAC{\SQRT{3}}{4}\)a=r

On substituting the values we get r = \(\frac{\sqrt{3}}{4}\) x 333 pm = 144.19 pm is the radius of the METAL atom.
63.

In the simple cubic cell, each corner atom is shared by __________(a) eight unit cells(b) one unit cell(c) two unit cells(d) six unit cellsI had been asked this question during an online interview.My question is from Amorphous and Crystalline Solids in division Solid State of Chemistry – Class 12

Answer»

Right answer is (a) eight unit cells

Explanation: The unit cell in which the CONSTITUENT atoms are present only at the corner is known as SIMPLE CUBIC cell. It is also referred to as a primitive cubic cell. In the simple cubic cell, each corner atom is shared by eight different unit cells.

64.

What is each point (position of particle) in a crystal lattice termed as?(a) Lattice index(b) Lattice point(c) Lattice lines(d) Lattice spotThe question was asked in quiz.I'm obligated to ask this question of Solid State topic in portion Solid State of Chemistry – Class 12

Answer» RIGHT choice is (a) Lattice index

Easy explanation: Each point of the PARTICLE’s position is REFERRED to as ‘lattice point’ or ‘lattice site’. Every lattice point represents one constituent particle which may be an atom, ION or molecule.
65.

If a crystal lattice has 6 closed-pack spheres, what the number of tetrahedral voids in the lattice?(a) 12(b) 6(c) 36(d) 3The question was posed to me during an interview for a job.My enquiry is from Solid State in chapter Solid State of Chemistry – Class 12

Answer»

Correct answer is (a) 12

Easiest explanation: For a crystal LATTICE, if there are N close-packed SPHERES the NUMBER of tetrahedral voids are 2N and number octahedral voids are N. For N=6, number of tetrahedral voids = 2 × 6 = 12.

66.

Polar molecular solids are _________(a) bad conductors of electricity(b) good conductors of electricity(c) solid at room temperature(d) brittleI got this question in homework.This intriguing question comes from Classification of Crystalline Solids topic in section Solid State of Chemistry – Class 12

Answer»

The CORRECT option is (a) BAD conductors of ELECTRICITY

Best explanation: In polar MOLECULAR solids, the molecules are formed by covalent bonds and held together by strong dipole-dipole interaction. Therefore, polar molecular solids are non-conductors of electricity.

67.

What is the dimensional formula of packing fraction?(a) M^0L^3T^0(b) M^0L^0T^0(c) ML^0T^0(d) M^0L^2T^0This question was addressed to me by my school teacher while I was bunking the class.This intriguing question originated from Solid State topic in chapter Solid State of Chemistry – Class 12

Answer»

Correct option is (b) M^0L^0T^0

The explanation: Packing fraction is a dimensionless quantity which is the ratio of space OCCUPIED to TOTAL CRYSTAL space available. SINCE both the quantities have the same units the ratio renders dimensionless.

68.

The total number of atoms in one unit cell of face-centered unit cubic cell is ______ atoms.(a) 2(b) 6(c) 4(d) 8This question was addressed to me during an interview for a job.My doubt is from Solid State topic in section Solid State of Chemistry – Class 12

Answer»

Right CHOICE is (c) 4

The explanation is: In face-centered cubic unit cell, atoms are present at each of the corners and at the CENTRE of the face of the cube.

8 corners × \(\frac{1}{8}\)th of an ATOM = 1

• 6 FACES × \(\frac{1}{2}\) of an atom = 3

Thus, the total number of atoms in face-centered cubic unit cell: 1+3=4 atoms.

69.

Which one of the following are the dimensions of cubic crystal?(a) a =b ≠ c(b) a = b = c and α = β ≠ γ = 90(c) a = b = c and α = β = γ = 90(d) a ≠ b = c and α = β ≠ γ = 90I got this question in exam.Question is taken from Amorphous and Crystalline Solids in section Solid State of Chemistry – Class 12

Answer» CORRECT choice is (C) a = b = c and α = β = γ = 90

Best explanation: The dimensions of a cubic CRYSTALS are a = b = c, α = β = γ = 90.
70.

Which of the following can be used to describe a crystalline solid?(a) Heterogeneous, anisotropic(b) Homogeneous, anisotropic(c) Heterogeneous, isotropic(d) Homogeneous, isotropicThe question was asked during an interview for a job.I'm obligated to ask this question of Solid State Characteristics topic in portion Solid State of Chemistry – Class 12

Answer»

The correct answer is (B) Homogeneous, anisotropic

To EXPLAIN I would say: Homogeneity REFERS to uniformity in composition, which is a characteristic property of crystalline solids. Isotropy is when the VALUES of physical PROPERTIES do not change with direction throughout the body of the solid. Crystalline solids are anisotropic because the composition of the solid changes with direction, hence the physical properties also change with direction.

71.

Which of the following is an amorphous solid?(a) Quartz(b) Quartz glass(c) Graphite(d) Salt (NaCl)The question was posed to me in an interview for job.The origin of the question is Solid State Characteristics in portion Solid State of Chemistry – Class 12

Answer»

The CORRECT CHOICE is (B) Quartz glass

Easy explanation: Quartz glass does not have a perfectly ORDERED structure, HENCE it is classified as an amorphous solid. The rest are crystalline solids due to ordered structures.

72.

If copper, density = 9.0 g/cm^3 and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?(a) 0.0028(b) 0.0235(c) 0.0011(d) 0.0323I had been asked this question during an online exam.This interesting question is from Solid State in division Solid State of Chemistry – Class 12

Answer»

The correct ANSWER is (B) 0.0235

Easiest explanation: Density, d of unit cell is GIVEN by d = \(\frac{zM}{a^3N_A}\)

Given,

Density, d = 9.0 g/cm^3

Atomic mass, M = 63.5 g/mole

Edge length = a

NA = Avogadro’s number = 6.022 x 10^23

z = 4 atoms/cell

On REARRANGING the equation for density we get a^3 = \(\frac{zM}{dN_A}\)

SUBSTITUTING the given values:

a^3 = \(\frac{4 \times 63.5}{9 \times 6.022 \times 10^{23}}\)

Therefore, a = 360.5 pm

 The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2\(\sqrt{2}\)r.

On substituting the value of ‘a’ in the given relation, r = \(\frac{360.5}{2\sqrt{2}}\)=127.46 pm

Now, for spherical particles volume, V = 4πr^3/3 and surface area, S = 4πr^2

Required ratio = S/V=4πr^2/(4πr^3/3) = 3/r (after simplifying)

Thus, S/V = 3/127.46 = 0.0235.

73.

Which among the following compounds is Antiferroelectric?(a) NiO(b) V2O3(c) PbZrO3(d) Fe3O4The question was asked in my homework.The above asked question is from Solid State Magnetic Properties in section Solid State of Chemistry – Class 12

Answer»

Correct choice is (C) PbZrO3

For explanation: In the given LIST: NIO and V2O3 are ANTIFERROMAGNETIC substances, Fe3O4 is Ferrimagnetic and PbZrO3 is antiferroelectric. Antiferroelectric materials are those materials having ions which can polarize without external field (spontaneous polarization).

74.

All Ferroelectrics are pyroelectrics.(a) True(b) FalseThe question was asked in semester exam.This is a very interesting question from Solid State Magnetic Properties in portion Solid State of Chemistry – Class 12

Answer»

The correct answer is (a) True

The explanation is: Pyroelectricity is the ABILITY of CERTAIN materials to generate a temporary POTENTIAL difference when HEATED or cooled. Ferroelectrics form a subset of PYROELECTRICS. Hence, all Ferroelectrics are Pyroelectrics.

75.

All Ferroelectrics are piezo electrics.(a) True(b) FalseThe question was asked by my college director while I was bunking the class.Query is from Solid State Magnetic Properties in chapter Solid State of Chemistry – Class 12

Answer»

Right choice is (a) True

For EXPLANATION: Piezoelectrics are SUBSTANCES whose polarization CHANGES under the influence of stress. Ferro ELECTRIC substances are considered as Piezoelectric since their polarization can change under the influence of an electric FIELD.

76.

If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.(a) 2923 kg/m^3(b) 5846 kg/m^3(c) 8768 kg/m^3(d) 1750 kg/m^3I had been asked this question by my school teacher while I was bunking the class.Question is from Solid State in section Solid State of Chemistry – Class 12

Answer» RIGHT option is (b) 5846 kg/m^3

The best explanation: Given,

Edge length (a) = 500 pm = 500 X 10^-12 m

Atomic mass (M) = 110 g/mole = 110 x 10^-3 kg/mole

Avogadro’s number (NA) = 6.022 x 10^23/mole

z = 4 atoms/cell

The density, d of a metal is given as d=\(\frac{ZM}{a^3N_A}\)

On substitution, d=\(\frac{4 \times 110 \times 10^{-3}}{(500 \times 10^{-12})^3 \times 6.022 \times 10^{23}}\)=5846 kg/m^3.
77.

Which of the following is a characteristic of amorphous solid?(a) They are true solids(b) They have sharp melting points(c) They undergo clear cleavage(d) They are isotropicThe question was asked during an online interview.I'd like to ask this question from Amorphous and Crystalline Solids topic in portion Solid State of Chemistry – Class 12

Answer»

Correct CHOICE is (d) They are ISOTROPIC

Explanation: Amorphous solids are isotropic that is they have IDENTICAL PROPERTIES in all DIRECTIONS. The remaining options are the characteristics of crystalline solids.

78.

Amorphous solids are actually super cooled liquids.(a) True(b) FalseI have been asked this question by my college director while I was bunking the class.I want to ask this question from Solid State Characteristics topic in portion Solid State of Chemistry – Class 12

Answer»

Right CHOICE is (a) True

The best EXPLANATION: Amorphous solids behave like fluids and flow very slowly under the influence of gravity. HENCE, they are said to be super COOLED LIQUIDS.

79.

A metal crystallizes into two cubic phases BCC and FCC. The ratio of densities of FCC and BCC is equal to 1.5. Calculate the difference between the unit cell lengths of the FCC and BCC crystals if the edge length of the FCC crystal is equal to 4.0 Å.(a) 0.5 Å(b) 0.37 Å(c) 0. 28 Å(d) 0.73 ÅI have been asked this question in an interview.This interesting question is from Solid State topic in portion Solid State of Chemistry – Class 12

Answer»

Right answer is (b) 0.37 Å

Easy explanation: Given,

Edge length of FCC crystal (aFCC) = 4.0 Å

For FCC structure, Z = 4

For BCC structure, Z=2

Avogadro’s number (N0) = 6.02 x 10^23

The DENSITY of a crystal (ρ)=(Z x M)/(a^3 x N0)

 Therefore, the RATIO of Densities= ρFCC/ρBCC = (ZFCC x a^3BCC) / (ZBCC x a^3FCC)

1.5 = (4 x (ABCC)^3) / ( 2 x (4 x 10^-10)^3)

(aBCC)^3 = (1.5 x 2 x 64 x 10^-30)/ 4= 48 x 10^-30

Therefore aBCC = 3.63 Å

Difference in Unit Cell Length = 4.0 – 3.63 = 0.37 Å.

80.

What is the coordination number for a two-dimensional square close packed structure?(a) 8(b) 4(c) 6(d) 2The question was posed to me in an online quiz.This key question is from Solid State topic in portion Solid State of Chemistry – Class 12

Answer»

Correct answer is (b) 4

The explanation is: In a two-dimensional square close packed STRUCTURE, each sphere is in contact with 4 of its ADJACENT spheres. HENCE, it has coordination NUMBER 4.

81.

Atoms of element X form a BCC and atoms of element Y occupy 3/4^th of the tetrahedral voids. What is the formula of the compound?(a) X2Y3(b) X3Y2(c) X3Y4(d) X4Y3I got this question during an online interview.This question is from Solid State topic in division Solid State of Chemistry – Class 12

Answer»

Right answer is (a) X2Y3

The best explanation: The NUMBER of tetrahedral voids form is EQUAL to twice the number of atoms of ELEMENT X. Number of atoms of Y is 3/4^th the number of tetrahedral voids i.e. 3/2 times the number of atoms of X. THEREFORE, the ratio of NUMBERS of atoms of X and Y are 2:3, hence X2Y3.

82.

Solid state is denser than the liquid and gaseous states of the same substance. Which of the following is an exception to this rule?(a) Mercury(b) Carbon dioxide (dry ice)(c) Ice(d) NaClI got this question in an interview for job.My question is based upon Solid State Characteristics in section Solid State of Chemistry – Class 12

Answer»

Right CHOICE is (C) Ice

The best explanation: The density of ice is about 0.92 g/cm^3 while that of water is 1 g/cm^3. Mercury has density 14.184 g/cm^3 as solid and 13.69 g/cm^3 as liquid. CARBON dioxide has density 1.56 g/cm^3 as solid and 1.10 g/cm^3 as liquid. NaCl has density 2.71 g/cm^3 as solid and 1.556 g/cm^3 in molten state. Therefore only ice has LESSER density as a solid than as a liquid.

83.

Schottky defects are observed in solids with cations and anions of similar sizes. Which of the following compounds, therefore, is NOT likely to have a Schottky defect?(a) NaCl(b) AgCl(c) CsCl(d) KClThis question was addressed to me in exam.My enquiry is from Imperfections in Solids topic in section Solid State of Chemistry – Class 12

Answer» RIGHT answer is (B) AgCl

Best explanation: There is a considerable size difference between Ag CATION and CL ANION. Hence a Frenkel defect is more likely to occur than a Schottky defect.
84.

According to the band theory, increase in number of atoms participating in crystal formation leads to ________ in number of molecular orbitals containing electrons.(a) decrease(b) increase(c) no change(d) either increase or decrease, depending upon the size of the atomsThe question was posed to me in my homework.This question is from Solid State Electrical Properties topic in division Solid State of Chemistry – Class 12

Answer»

The CORRECT answer is (b) increase

To elaborate: The band theory assumes that ATOMIC orbitals of the atoms of the crystal combine to form molecular orbitals which are SPREAD over the whole crystal STRUCTURE. HENCE, the number of atoms is directly proportional to the number of molecular orbitals formed.

85.

Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10^-9m?(a) 0.115 nm(b) 0.144 nm(c) 0.235 nm(d) 0.156 nmI have been asked this question by my school principal while I was bunking the class.The doubt is from Solid State topic in chapter Solid State of Chemistry – Class 12

Answer»

The correct answer is (b) 0.144 nm

The best explanation: Given,

EDGE length of the Gold unit cell (a) = 0.407 x 10^-7m

For FCC unit cell, the ATOMIC radius (R) = a/(2\(\SQRT{2}\))

 = 0.407 x 10^-9/(2\(\sqrt{2}\))

 = 0.144 nm.

86.

Arrange the types of arrangement in terms of decreasing packing efficiency.(a) BCC < Simple cubic < CCP(b) HCP < CCP < BCC(c) HCP < BCC < Simple cubic(d) CCP < BCC < HCPI had been asked this question in a national level competition.The above asked question is from Solid State topic in portion Solid State of Chemistry – Class 12

Answer»

Correct CHOICE is (c) HCP < BCC < Simple cubic

Best explanation: HCP and CCP have the HIGHEST packing EFFICIENCY of 74% followed by BCC which is 68%. The simple cubic structure has a packing efficiency of 54%.

87.

Which of the following are responsible for high electrical and thermal conductivity of a metal?(a) Ions(b) Covalent bonds(c) Free H+ ions(d) Free and mobile electronsThe question was posed to me at a job interview.The question is from Classification of Crystalline Solids topic in portion Solid State of Chemistry – Class 12

Answer»

The correct ANSWER is (d) Free and mobile ELECTRONS

Explanation: Each ATOM of a metal gives one or more electrons TOWARDS the sea of mobile electrons it is surrounded by which ultimately contributes to the electrical and thermal conductivity of the metal.

88.

Which of the following is an orderly collection of positive ions surrounded and held together by a sea of electrons?(a) Gas(b) Non-metal(c) Metal(d) MetalloidsI got this question in an interview for internship.I would like to ask this question from Classification of Crystalline Solids in section Solid State of Chemistry – Class 12

Answer»

The correct CHOICE is (c) METAL

To explain I would say: Metal is SAID to be an orderly collection of POSITIVE ions SURROUNDED and held together by a sea of delocalized electrons. These delocalized electrons are mobile and are responsible for the conduction of electricity.

89.

Which of the following is non-conductor of electricity at solid state but can conduct electricity in the molten state or when dissolved in water?(a) Non polar molecular solids(b) Metallic solids(c) Ionic solids(d) Hydrogen bonded molecular solidsI got this question during an interview.My doubt stems from Classification of Crystalline Solids in chapter Solid State of Chemistry – Class 12

Answer» CORRECT CHOICE is (c) Ionic solids

For explanation: Ionic Solids, when dissolved in water, tend to separate cations and the anions which ALLOWS the solution to CONDUCT electricity. ALSO, they have high melting and boiling points.
90.

Which of the following is not Anti-ferromagnetic?(a) MnO(b) Mn2O3(c) MnO2(d) MnThis question was posed to me in examination.My question comes from Solid State Magnetic Properties topic in section Solid State of Chemistry – Class 12

Answer»

Right ANSWER is (d) Mn

To elaborate: MnO, Mn2O3, and MNO2 are Anti-ferromagnetic. Mn is paramagnetic in NATURE. PARAMAGNETISM is when a substance is weakly attracted to a magnetic field.

91.

Which of the following describes a general solid?(a) Compressible(b) Incompressible(c) Fluid(d) Semi-compressibleThis question was posed to me in a job interview.I need to ask this question from Solid State Characteristics topic in chapter Solid State of Chemistry – Class 12

Answer» CORRECT CHOICE is (B) Incompressible

For explanation I would SAY: The intermolecular forces of attraction in a solid are very strong, making it incompressible. Gases are highly compressible, while liquids are semi-compressible. Fluid is a PROPERTY of a substance that can ‘flow’.
92.

A substance containing one band completely filled with electrons and other band completely empty, but with a possibility of the empty band becoming populated, behaves as a _________(a) metal(b) non-metal(c) semiconductor(d) metal and non-metalI got this question by my college director while I was bunking the class.This intriguing question originated from Solid State Electrical Properties topic in section Solid State of Chemistry – Class 12

Answer»

The correct OPTION is (C) semiconductor

Easy explanation: Semiconductors have completely filled valence bands and completely empty conduction bands. The energy gap between the TWO bands is relatively small and can be OVERCOME by supplying HEAT, which excites some electrons to the conduction band, thus allowing the conduction of electricity.

93.

Which of the following tend to be volatile liquids or soft solids at room temperature and pressure?(a) Non polar molecular solids(b) Metallic solids(c) Polar molecular solids(d) Hydrogen bonded molecular solidsThe question was posed to me during an interview.The origin of the question is Classification of Crystalline Solids topic in chapter Solid State of Chemistry – Class 12

Answer»

The correct option is (d) Hydrogen bonded molecular solids

Explanation: In hydrogen bonded molecular SOLID, the intermolecular FORCES are strong hydrogen bonds. Hence, it tends to be VOLATILE LIQUIDS or soft solids at room temperature and pressure.

94.

Solids are classified as ___________(a) crystalline and ionic solids(b) metallic and amorphous solids(c) molecular and covalent solids(d) crystalline and amorphous solidsThis question was posed to me in examination.This intriguing question comes from Amorphous and Crystalline Solids topic in chapter Solid State of Chemistry – Class 12

Answer»

Correct CHOICE is (d) crystalline and amorphous SOLIDS

Explanation: BASED on their crystal structures, solids are CLASSIFIED as crystalline and amorphous solids. In crystalline solids, the CONSTITUENT particles are arranged in a regular manner. In amorphous solids, the constituent particles are not arranged in any regular manner.

95.

What is the temperature, above which a ferromagnetic substance shows no ferromagnetism called?(a) Curie temperature(b) Néel temperature(c) Critical temperature(d) There exists no such temperatureI had been asked this question in an online interview.This question is from Solid State Magnetic Properties in division Solid State of Chemistry – Class 12

Answer» CORRECT answer is (a) CURIE temperature

The explanation: The Néel temperature or magnetic ORDERING temperature is the temperature above which an antiferromagnetic material becomes paramagnetic. The Critical temperature of a gas is the temperature above which it cannot be liquefied by PRESSURE alone. The temperature above which a ferromagnetic substance SHOWS no ferromagnetism is called Curie temperature.
96.

Which among the following compounds can show the properties of a Ferroelectric substance?(a) BaTiO3(b) PbZrO3(c) MnO2(d) CrO2I got this question in an internship interview.This intriguing question comes from Solid State Magnetic Properties in division Solid State of Chemistry – Class 12

Answer»

Correct answer is (a) BATIO3

To explain I would say: At the Curie TEMPERATURE (AROUND 120 ºC) paraelectric BaTiO3 transforms into a ferroelectric STRUCTURE. Ferroelectricity is the characteristic of certain substances that possess a SPONTANEOUS electric field which can be reversed by applying an external electric field.

97.

Which of the following elements have a negative value of magnetic susceptibility?(a) Iron(b) Oxygen(c) Aluminium(d) NitrogenThe question was asked in an international level competition.The above asked question is from Solid State Magnetic Properties topic in section Solid State of Chemistry – Class 12

Answer»

The correct choice is (d) Nitrogen

The explanation: Only Diamagnetic substances SHOW a negative value of MAGNETIC susceptibility. In the GIVEN list: OXYGEN and aluminium are paramagnetic, iron is ferromagnetic and nitrogen is diamagnetic. Therefore, Nitrogen has a negative value of magnetic susceptibility.

98.

Which of the following properties holds true for a metalloid?(a) Poor conductors of heat and electricity(b) Malleable(c) Ductile(d) Non-brittleI had been asked this question by my college professor while I was bunking the class.The origin of the question is Solid State Electrical Properties topic in chapter Solid State of Chemistry – Class 12

Answer»

Correct ANSWER is (a) Poor conductors of heat and electricity

To EXPLAIN: Metalloids have PROPERTIES INTERMEDIATE to those of metals and non-metals. They are non-malleable, non-ductile and brittle. They are poor conductors of heat and electricity as compared to metals, but are BETTER conductors than non-metals.

99.

If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?(a) 1.32 × 10^6 pm^2(b) 1.32 × 10^-18pm^2(c) 2.20 × 10^5 pm^2(d) 2.20 × 10^-19 pm^2The question was asked in an international level competition.This is a very interesting question from Solid State in division Solid State of Chemistry – Class 12

Answer»

The correct answer is (a) 1.32 × 10^6 pm^2

To EXPLAIN: ABCABC arrangement is found in CCP.

In closed CUBIC packing, relation between edge length of unit CELL, a, and RADIUS of particle, r, is given as a=2\(\sqrt{2}\)r.

 Surface area (S.A.) = 6a^2

From the RELATIONSHIP,

a^2 = 8r^2

S.A. = 6a^2 = 48r^2

 When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10^6 pm^2.

100.

Which type of solids are formed by three-dimensional arrangement of cations and anions bound by strong electrostatic force?(a) Polar molecular solids(b) Ionic solids(c) Covalent solids(d) Metallic solidsThis question was posed to me in an interview.Question is taken from Classification of Crystalline Solids in division Solid State of Chemistry – Class 12

Answer»

Right OPTION is (b) IONIC SOLIDS

Easy explanation: Ionic solids are MADE up of THREE dimensional arrangement of cations and anions bound by strong electrostatic force.