Category: Class 12 Chemistry Objective Questions

Chemistry Question Bank for Class 12 on “Haloalkanes & Haloarenes – Chemical Reactions – 2”.

1. As the stability of carbocation formed in the first step of S_N1 reaction increases, the rate of the reaction _________
a) increases
b) decreases
c) remains same
d) may increase or decrease
Answer: a
Clarification: Greater the stability of the carbocation, greater will be the ease of formation from alkyl halide and faster will be the reaction rate.

2. The S_N1 reaction cannot be carried out in which of the following media?
a) Acetic acid
b) Acetone
c) Ethanol
d) Water
Answer: b
Clarification: S_N1 reactions are carried out in polar protic solvents like water, alcohol and acetic acid. These solvents promote the ionisation step by stabilising the ions by solvation.

3. If a compound does not affect the plane of plane polarised light when passed through it, the compound is said to be _________
a) optically active
b) optically inactive
c) dextrorotatory
d) laevorotatory
Answer: b
Clarification: When the plane of plane polarised light is rotated either to the left or right when passed through a compound, it is said to be optically active. Else, it is optically inactive.

4. Which of the following compounds does not have a stereocenter?
a) Propan-2-ol
b) Butan-2-ol
c) 2-Bromo-1-chlorobutane
d) 2-Bromopentane
Answer: a
Clarification: A stereocenter is an asymmetric carbon that has four different groups attached to it. Propan-2-ol has two methyl groups attached to the carbon.

5. Which of the following has a chiral carbon atom?
a) 2-Chloro-2-methylpentane
b) 1,1-Dibromoethane
c) Pentan-3-ol
d) 2-Bromopentane
Answer: d
Clarification: A chiral carbon atom is that which has four different substituent groups attached to it. 2-Bromopentane has one of each H, CH₃, C₃H₇ and Br groups attached to its C atom.

6. What is the chirality of 2-Chlorobutane?
a) Chiral
b) Achiral
c) Superimposable mirror images
d) Symmetric
Answer: a
Clarification: 2-Chlorobutane has four different groups attached to the tetrahedral carbon and is chiral as it forms non-superimposable mirror images.

7. Predict the chirality of the shown compounds.

a) Both A and B are chiral
b) Both A and B are achiral
c) A is chiral and B is achiral
d) A is achiral and B is chiral
Answer: c
Clarification: The C atom that is attached to the OH group in compound A is a stereocenter and makes the compound A chiral. On the other hand, compound B does not have any asymmetric carbon.

8. A chiral compound with an ‘X’ group attached to the stereocenter has been replaced by a ‘Y’ group. The product obtained rotates the plane of polarised light in the direction opposite to that of the original compound. Name the process that has taken place.
a) Inversion
b) Retention
c) Racemisation
d) Inversion or retention
Answer: d
Clarification: Since the product rotates the plane of polarised light in some direction, it cannot be racemic. The product formed is a completely new compound from the original one and may have different or same optical behaviour as compared to the original compound.

9. The process of separation of a racemic mixture into the respective d and l optical isomers is called __________
a) racemisation
b) resolution
c) inversion
d) retention
Answer: b
Clarification: Racemisation is the process of converting optical isomers into a racemic mixture and separating them from the mixture is known as resolution.

10. What is the most suitable term for the following transformation?

a) Inversion of configuration
b) Retention of configuration
c) Racemisation
d) Elimination reaction
Answer: b
Clarification: It can be seen that the relative spatial arrangement of bonds at the stereocenter remain the same even after the transformation. Hence, this transformation is said to take place with a retention of configuration.

11. When a reaction proceeds with the retention of configuration, the optical activity of the product will remain same as that of the original reactant.
a) True
b) False
Answer: b
Clarification: For example, when (-)-2-methylbutan-1-ol is heated with conc. HCl, the product formed is (+)-1-chloro-2-methylbutane, which has opposite optical behaviour to that of the reactant.

12. What will be the effect on the plane of plane polarised light when it passes through a solution of dlbutan-2-ol?
a) Clockwise rotation
b) Anti-clockwise rotation
c) Zero rotation
d) 180° rotation
Answer: c
Clarification: dlbutan-2-ol represents a mixture of dbutan-2-ol (dextro) and lbutan-2-ol (laevo), and is called a racemic mixture. There optical rotation due to one isomer will cancel the rotation due to the other and will have no effect on the polarised light passed through it.

13. When (-)-2-bromooctane reacts with potassium hydroxide over an S_N2 mechanism, what will be the product formed?
a) (-)-octan-1-ol
b) (+)-octan-1-ol
c) (-)-octan-2-ol
d) (+)-octan-2-ol
Answer: d
Clarification: The OH nucleophile attaches itself opposite to where the halide was present but one the same carbon. This results in an inversion of configuration and hence changes in the direction in which the plane of polarised light rotates.

14. The heating of 1-Chlorobutane along with alcoholic KOH results in ________
a) elimination of hydrogen from α-carbon atom
b) elimination of hydrogen from β-carbon atom
c) elimination of hydrogen from γ-carbon atom
d) no reaction
Answer: b
Clarification: When a haloalkane with a β-hydrogen atom is heated with alc. KOH, it results in a β-elimination reaction to form alkenes.

15. The major product formed from the dehydrohalogenation of 2-Bromopentane is pent-2-ene.
a) True
b) False
Answer: a
Clarification: According to Zaitsev’s rule, the major product will be that alkene which has more number of alkyl groups attached to the double bonded carbon atoms.

Chemistry Question Bank for Class 12,

Chemistry Multiple Choice Questions on “Aldehydes and Ketones Physical Properties”.

1. Which of the following exists as a gas at 287K?
a) Formaldehyde
b) Acetaldehyde
c) Propanal
d) Acetone
Answer: a
Clarification: Methanal exists as foul-smelling gas at room temperature. Ethanal is a volatile liquid which boils at 294K. Other aldehydes and ketones exist as either liquids or solids at room temperature.

2. If the boiling points of methoxyethane and propanol are 281 K and 370 K respectively, predict the boiling point of propanal.
a) 273 K
b) 281 K
c) 322 K
d) 370 K
Answer: c
Clarification: The boiling point of aldehydes are higher than those of non-polar hydrocarbons of comparable molecular masses. However, their boiling points are lower than comparable alcohols, due to the absence of intermolecular hydrogen bonding.

3. Which of the following is the least soluble in water?
a) Methanal
b) Ethanal
c) Propanone
d) Pentanal
View Answer

Answer: d
Clarification: The lower aldehydes and ketones are miscible in water in all proportions due to the ability of the polar carbonyl group to form hydrogen bonds with water molecules. However, the solubility rapidly decreases as the length of the alkyl chain increases. Precisely, aldehydes and ketones with more than 4 carbon atoms are practically insoluble in water.

4. Propanal has a slightly higher boiling point than propanone.
a) True
b) False
Answer: b
Clarification: Propanone has two CH₃ groups, one on either side of the CO group, whereas propanal has only one CH₃CH₂ group. This makes propanone more polar because of the higher electron releasing effect of two alkyl groups. The boiling point of propanal is 322K and that of propanone is 329K.

5. Two compounds A and B were being tested for their boiling points. It was observed that A started boiling after B, when both were subjected to same conditions. If the compound B is acetone, which of the following can be compound A?
a) n-Butane
b) Methoxyethane
c) Propanal
d) Propanol
Answer: d
Clarification: It can be inferred that B has a higher boiling point than A. Since hydrocarbons, ethers and aldehydes have lower boiling points than ketones of similar molecular masses, none of them can be compound A. Also, alcohols have a higher boiling point than similar ketones hence, the compound A is propanol.

6. Which of the following has the least pungent odour?
a) Methanal
b) Ethanal
c) Propanal
d) Butanal
Answer: d
Clarification: As the size of the aldehyde molecule increases, the odour becomes less pungent and more fragrant. Methanal is the most pungent smelling aldehyde.

7. Which of the following is used as a flavouring agent?
a) Methanal
b) Ethanal
c) Acetone
d) Acetophenone
Answer: d
Clarification: Methanal, ethanal and acetone are lower carbonyl compounds and are not much fragrant to be used as flavouring agents.

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Chemistry Multiple Choice Questions on “Preparation of Amines – 2”.

1. Which of the following reagents cannot be used to convert ethanenitrile to ethylamine?
a) H₂/Ni
b) LiAlH₄
c) Na(Hg), ethanol
d) Sn, HCl
Answer: d
Clarification: Nitriles on reduction with lithium aluminium hydride or catalytic hydrogenation or with Na(Hg)-ethanol produce primary amines.

2. How many more carbon atoms are present in the amine formed from the reduction of a nitrile, than in the nitrile itself?
a) 0
b) 1
c) 2
d) 1 or 2
Answer: a
Clarification: The -CN group of the nitrile is reduced to -CH₂-NH₂ group. This means that the parent carbon chain of the amine has one more carbon than the parent chain of the nitrile. However, the total number of carbons in both compounds is same as one of the carbon in nitriles is present in the cyanide group.

3. The reduction of phenyl isocyanide with H₂ and Ni catalyst gives a/an _________
a) primary amine
b) secondary amine
c) tertiary amine
d) arylalkyl amine
Answer: b
Clarification: The reduction of isocyanide compounds (where the CN group is attached through N atom) with H₂/Ni gives secondary amines or N-alkyl amines. For example, phenyl isocyanide gives N-methylaniline.

4. What is the product of the following reaction?

a) Phenylmethanamine
b) 1-Phenylethan-1-amine
c) 2-Phenylethan-1-amine
d) Toluene
Answer: c
Clarification: Benzyl cyanide reacts with LiAlH₄ to undergo reduction of the CN group to a CH₂NH₂ group and form a primary aromatic amine. The formula of this product will be C₆H₅CH₂CH₂NH₂, which is 2-phenylethan-1-amine.

5. Which of the following amines cannot be formed from the reduction of amides with LiAlH₄?
a) Ethylamine
b) Benzenamine
c) Benzylamine
d) Ethylmethylamine
Answer: b
Clarification: The -CONH₂ group of amides is reduced to a -CH₂NH₂ group and results in a compound having the same number of carbon atoms. The simplest aromatic amide is benzamide, which is reduced to give benzylamine. Hence, aniline cannot be formed as there is no corresponding amide that can be reduced to it with LiAlH₄. However, it can be achieved by Hoffmann bromamide degradation.

6. Only primary amines can be obtained from the reduction of amides with LiAlH₄.
a) True
b) False
View Answer

Answer: b
Clarification: Primary, secondary or tertiary amines can be formed depending on whether the amide used is primary, secondary or tertiary. This is because irrespective of the number of alkyl groups, the CO group of amides is directly reduced to CH₂.

7. Ethyldimethylamine is obtained from the reduction of _______ with LiAlH₄.
a) acetamide
b) benzamide
c) N-methylacetamide
d) N,N-dimethylacetamide
Answer: d
Clarification: Ethyldimethylamine is a tertiary amine with two methyl substituents directly attached to N atom. It can be obtained only from the reduction of a corresponding tertiary amide that has two methyl substituents attached to N atom.

8. Which path gives propan-1-amine as the product in the reaction below?

a) Only A
b) Only B
c) Both A and B
d) Not A nor B
Answer: a
Clarification: Propanamine has three carbon atoms and so does the parent amide. So, the reduction of CO to CH₂ will give the required compound. Whereas in path B, the amide undergoes Hoffmann bromamide degradation to give ethanamine.

9. Which of the following is not a by product of Hoffmann bromamide degradation of acetamide with alcoholic KOH?
a) KBr
b) KCN
c) K₂CO₃
d) H₂O
Answer: b
Clarification: Acetamide (CH₃CONH₂) reacts with Br₂ and 4 molecules of KOH to form methylamine (main product) along with 2 molecules KBr, 2 molecules of H₂O and one molecule of K₂CO₃.

10. Hoffmann bromamide degradation reaction is used for preparing _______ amines.
a) primary
b) secondary
c) tertiary
d) mixed
Answer: a
Clarification: Mixed amines are a classification of secondary and tertiary amines. The Hoffmann bromamide degradation is only possible with 1° amides so as to produce primary amines.

11. The best reagent for converting 2-phenylpropanamide to 1-phenylethanamine is ________
a) H₂/Ni
b) Na(Hg)/C₂H₅OH
c) LiAlH₄
d) NaOH/Br₂
Answer: d
Clarification: 2-Phenylpropanamide has 9 carbon atoms while 1-phenylethamine has 8 carbon atoms. This depicts stepping down of carbon from amide to amine conversion and this takes place in Hoffmann degradation reaction.

12. Which of the following compounds undergoes Hoffmann bromamide degradation reaction?
a) C₆H₅NH₂
b) C₆H₅NO₂
c) C₆H₅CONH₂
d) C₆H₅CH₂NH₂
Answer: c
Clarification: Only primary amides undergo this reaction with bromine in NaOH to form primary amines. Benzamide undergoes this reaction to form benzenamine or aniline.

13. Which of the following amines can be prepared from Gabriel phthalimide synthesis?
a) Benzylamine
b) Aniline
c) o-Toluidine
d) N-Methylbenzenamine
Answer: a
Clarification: Since Gabriel phthalimide synthesis is a nucleophilic substitution reaction, aryl halides cannot be prepared form this. This is because of the sp² hybridised aryl carbon and stable benzene ring.

14. Gabriel phthalimide synthesis can be used for preparing phenylmethanamine.
a) True
b) False
Answer: a
Clarification: Potassium phthalimide can undergo nucleophilic substitution with a benzyl halide to form N-phenylphthalimide, which on hydrolysis with NaOH gives benzylamine or phenylmethanamine.

15. Which of the following steps is not present in Gabriel phthalimide synthesis?
a) Treating phthalimide with alcoholic KOH
b) Heating potassium phthalimide with alkyl halide
c) Alkaline hydrolysis of N-alkylphthalimide
d) Heating phthalic acid with NaOH
Answer: d
Clarification: The chemical compounds involved in the Gabriel phthalimide synthesis are phthalimide, potassium phthalimide and N-alkyl phthalimide. These are formed in order when reacted with different reagents like KOH and haloalkanes.

Chemistry Multiple Choice Questions on “Biomolecules – Proteins – 1”.

1. Proteins are polymers of ______
a) α-amino acids
b) β-amino acids
c) γ-amino acids
d) δ-amino acids
Answer: a
Clarification: Amino acids contain amino (NH₂) and carboxyl (COOH) functional groups. α-amino acids contain the NH₂ group on the carbon adjacent to the COOH group. Proteins on hydrolysis yield only α-amino acids.

2. If the basic formula of an α-amino acid is R-CH(NH₂)-COOH, where R is the side chain, what is the primary point of distinction between any two proteins?
a) Number of amino groups
b) Number of carboxyl groups
c) The side chain R
d) Relative positions of amino, carboxyl groups and R
Answer: c
Clarification: α-amino acids are the constituents of proteins. Different proteins are formed by polymerisation of different α-amino acids, which are formed due to the difference in the side chain substituted group R, which may be as simple as hydrogen or as complex as imidazole.

3. Which of the following amino acids is optically inactive?
a) Glycine
b) Alanine
c) Lysine
d) Valine
Answer: a
Clarification: Glycine is the only naturally occurring α-amino acid that is optically inactive. This is because the α-carbon of glycine has two hydrogen atoms attached to it.

4. What is the one letter code for tyrosine?
a) T
b) Y
c) R
d) S
Answer: b
Clarification: Tyrosine is a natural amino acid first obtained from cheese, hence the name. It has a 4-hydroxyphenylmethyl side chain. Its three-letter symbol is Tyr and its code is Y.

5. Which of the following amino acids are aromatic in nature?
a) Methionine
b) Isoleucine
c) Proline
d) Histidine
Answer: d
Clarification: Amino acids can be aromatic only if their side chain consists of an aromatic compound. Methionine has a sulphur containing straight chain. Isoleucine has an isobutyl side group. Proline has a complex cyclic side chain. Histidine has a basic aromatic side group, imidazole.

6. Which of the following is a non-essential amino acid?
a) Threonine
b) Glutamine
c) Phenylalanine
d) Valine
Answer: b
Clarification: Threonine, phenylalanine and valine along with seven other amino acids cannot be produced by the human body and must be obtained through diet. These are essential amino acids. Other amino acids like glutamine are synthesized by the human body and are therefore called non-essential amino acids.

7. Which of the following is a neutral amino acid?
a) Glycine
b) Lysine
c) Arginine
d) Histidine
Answer: a
Clarification: Neutral amino acids contain equal number of amino and carboxyl groups. Lysine, arginine and histidine contain two NH₂ groups are one COOH group, and are hence basic amino acids.

8. Cysteine is a/an ________ amino acid.
a) acidic
b) essential
c) aromatic
d) sulphur containing
Answer: d
Clarification: Cysteine is a semi-essential, neutral amino acid. Its side chain is HS-CH₂ (thiol), which is aliphatic and contains sulphur. Its symbol is Cys and its code is C.

9. Which of the following amino acids contains only one amino group?
a) Leucine
b) Lysine
c) Asparagine
d) Glutamine
Answer: a
Clarification: Lysine, asparagine and glutamine all contain two amino (NH₂) groups, of which one is a part of the side chain. Leucine has a purely alkyl side chain and the only amino group it has is one the α-carbon.

10. What is the one letter code for asparagine?
a) A
b) P
c) N
d) S
Answer: c
Clarification: For the sake of simplicity, each amino acid has been given an abbreviation which is either a three-letter symbol or a one-letter code. Asparagine is represented as Asn or N.

11. Which of the following is incorrect regarding tryptophan?
a) It is an essential amino acid
b) It is a basic amino acid
c) It has an aromatic side chain
d) It is a non-polar amino acid
Answer: b
Clarification: Tryptophan is an essential amino acid with bicyclic aromatic side chain, indole. It cannot be synthesized by the human body and must be taken for nitrogen balance in the body. It has one amino and carboxyl group each and is a neutral amino acid.

12. The structure shown below is ______

a) Side chain of histidine
b) Side chain of tryptophan
c) Side chain of proline
d) Proline
Answer: d
Clarification: Proline is a neutral aliphatic amino acid, with cyclic pyrrolidine side chain, making it non-polar. It is non-essential as the human body can synthesize it.

13. Identify the amino acid with the formula HOOC-CH₂-CH₂-CH(NH₂)-COOH.
a) Glutamic acid
b) Aspartic acid
c) Glutamine
d) Asparagine
Answer: a
Clarification: Glutamic acid is an amino substituted dicarboxylic acid which can be synthesized by the body. Its IUPAC name is 2-aminopentanedioic acid. Since it has two COOH groups (one more than the number of NH₂ groups), it is acidic in nature.

14. Amino acids behave like carboxylic acids
a) True
b) False
Answer: b
Clarification: Generally, amino acids behave like salts rather than simple amines or carboxylic acids. This is due to the presence of both acidic (COOH) and basic (NH₂) groups in the same molecule.

15. Amino acids can show amphoteric behaviour.
a) True
b) False
Answer: a
Clarification: In an aqueous solution of amino acid, COOH loses a proton and NH₂ accepts a proton, giving rise to a dipolar zwitter ion. This is neutral but contains both positive and negative terminals. In this form, amino acids react both with acids and bases, and hence show amphoteric behaviour.

Chemistry Question Papers for IIT JEE Exam on “Chemicals in Food”.

1. Which of the following chemicals are added to increase the shelf-life of foods?
a) Food colour
b) Sweeteners
c) Artificial flavours
d) Antioxidants
Answer: d
Clarification: Chemicals are added to food for improving their aesthetic, to increase their shelf-life or to add nutritive value. Food colours, flavours and sweeteners simply enhance the appearance and taste of food. Antioxidants are used for preservation.

2. Which of the following artificial sweeteners can be only used in soft drinks?
a) Aspartame
b) Alitame
c) Sucralose
d) Saccharin
Answer: a
Clarification: Aspartame is an artificial sweetening agent which is unstable at higher temperatures. It is a compound formed from aspartic acid and phenylalanine. It is 100 times as sweet as cane sugar.

3. Alitame is ______ times sweeter than cane sugar.
a) 50
b) 100
c) 600
d) 2000
Answer: d
Clarification: Alitame is a compound consisting of one hydroxy group, three keto groups and a tetra-methyl sulphur containing cyclic group. It is a very strong sweetener and provides uncontrollable sweetness to food.

4. Which of the following is not a preservative?
a) Dulcin
b) Potassium metasulphite
c) Sodium benzoate
d) Sorbic acid salts
Answer: a
Clarification: Preservatives are chemicals that are added to food to prevent their spoilage and preserve their nutritive value and taste. Salt, sugar and oils are the most common examples. Potassium metasulphite is used in jams. Dulcin is an artificial sweeter.

5. Which of the following is not an antioxidant?
a) BHT
b) BHA
c) Saccharin
d) Sulphur dioxide
Answer: c
Clarification: Antioxidants prevent food spoilage by itself reacting with oxygen and protecting the food from oxidation. Butylated hydroxy toluene and butylated hydroxy anisole increase the shelf lives of food like butter. SO₂ is also used in wines, fruits and sugar syrups for preservation.

Chemistry Question Papers for IIT JEE Exam,

Chemistry Multiple Choice Questions on “Amorphous and Crystalline Solids”.

1. In polar molecular solids, the molecules are held together by ________
a) dipole-dipole interactions
b) dispersion forces
c) hydrogen bonds
d) covalent bonds
Answer: a
Clarification: Molecular solids are solids that are collections of molecules held together by intermolecular forces. In polar molecules such as HCl, So₂, etc., the molecules are held together by dipole-dipole interactions.

2. Diamond is an example of _______
a) solid with hydrogen bonding
b) electrovalent solid
c) covalent solid
d) glass
Answer: c
Clarification: The solids in which constituent particles are attached to each other by covalent bonds are called covalent solids. Diamond, graphite, silicon, SiC, AIN, quartz are examples of covalent solids.

3. Silicon is found in nature in the forms of ________
a) body-centered cubic structure
b) hexagonal-closed packed structure
c) network solid
d) face-centered cubic structure
Answer: c
Clarification: Silicon due to its catenation property form network solid. Catenation is the ability of an atom to form bonds with other atoms of the same element. The compounds of silicon are reactive and not stable.

4. 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 = 90
Answer: c
Clarification: The dimensions of a cubic crystals are a = b = c, α = β = γ = 90.

5. Which of the following is not a crystal system?
a) Cubic
b) Trigonal
c) Triclinic
d) Hexaclinic
Answer: d
Clarification: Hexaclinic is not a crystal system. Their crystal system are cubic, tetragonal, rhombohedral or trigonal, orthorhombic or rhombic, monoclinic, triclinic and hexagonal.

6. In face-centred cubic cell, a unit cell is shared equally by __________
a) four unit cells
b) two unit cells
c) one unit cell
d) six unit cells
Answer: d
Clarification: The unit cell in which atoms are present at corners as well as faces of unit cell is known as face-centred cubic unit cell. In face-centred cubic cell, a unit cell is shared equally by six unit cells.

7. The unit cell with a≠b≠c and α=β=γ=90 refers to __________ crystal system.
a) hexagonal
b) trigonal
c) triclinic
d) orthorhombic
Answer: d
Clarification: In orthorhombic crystal system, all three axes are unequal in length and all are perpendicular to one another. It is also called as rhombic crystal system. Topaz, barite are some examples of orthorhombic crystals.

8. Which is the most unsymmetrical crystal system?
a) Triclinic crystal system
b) Cubic crystal system
c) Hexagonal crystal system
d) Trigonal crystal system
Answer: a
Clarification: Most unsymmetrical crystal system is triclinic in which all three axes are unequal in length none is perpendicular to another. Triclinic unit cells has the least symmetrical shape of all unit cells. Turquoise is an example of triclinic crystal.

9. 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 cells
Answer: a
Clarification: 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.

10. The points which shows the position of atoms in crystal are called as _________
a) crystal lattice
b) crystal parameters
c) bravais lattice
d) lattice point
Answer: d
Clarification: The point at which the atoms may be present on the unit cell is termed as lattice point. It shows the position of atoms in crystal.

11. The unit cell with a≠b≠c and α=γ=90, β≠90 refers to __________crystal system.
a) cubic
b) tetragonal
c) monoclinic
d) triclinic
Answer: c
Clarification: In monoclinic crystal system, all the three axes are unequal in length and two axes are perpendicular to each other. Gypsum and borax are examples of monoclinic crystals.

12. Which type of solid crystals will conduct heat and electricity?
a) Ionic
b) Covalent
c) Molecular
d) Metallic
Answer: d
Clarification: 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.

13. Which is not a characteristic of crystalline solids?
a) They undergo a clean cleavage
b) They are true solids
c) They are isotropic
d) They have sharp melting points
View Answer

Answer: c
Clarification: Amorphous solids are isotropic that is they have identical properties in all directions, whereas crystalline solids are anisotropic that is they have different properties in different directions.

14. 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 isotropic
Answer: d
Clarification: Amorphous solids are isotropic that is they have identical properties in all directions. The remaining options are the characteristics of crystalline solids.

15. Solids are classified as ___________
a) crystalline and ionic solids
b) metallic and amorphous solids
c) molecular and covalent solids
d) crystalline and amorphous solids
Answer: d
Clarification: 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.

16. Quartz is an example of ___________
a) molecular solids
b) ionic solids
c) covalent solids
d) metallic solids
Answer: c
Clarification: Quartz is a common example of covalent solids. In covalent solids, the constituent particles are attached to each other by covalent bonds. Diamond, graphite, silicon are other examples of covalent solids.

17. Solid carbon dioxide is an example of _________
a) metallic crystal
b) covalent crystal
c) ionic crystal
d) molecular crystal
Answer: d
Clarification: Solid CO₂ is an example of molecular crystal. These solids have molecules as their constituent particles. These solids may be bonded by vander waals’ forces or by dipole-dipole attraction or by strong hydrogen bonds. H₂, Cl₂, I₂ are some examples of molecular solids.