Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Solid State Magnetic Properties”.

1. Which among the following compounds can show the properties of a Ferroelectric substance?
a) BaTiO₃
b) PbZrO₃
c) MnO₂
d) CrO₂
Answer: a
Clarification: 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.

2. All Ferroelectrics are pyroelectrics.
a) True
b) False
Answer: a
Clarification: 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.

3. All Ferroelectrics are piezo electrics.
a) True
b) False
Answer: a
Clarification: 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.

4. Which among the following compounds is Antiferroelectric?
a) NiO
b) V₂O₃
c) PbZrO₃
d) Fe₃O₄
Answer: c
Clarification: 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).

5. Which among the following statements is correct?
a) NaCl is a paramagnetic substance
b) Paramagnetic substances behave like an insulator
c) Cobalt is an Antiferromagnetic substance
d) On heating, ferrimagnetic substanceslose ferrimagnetism
Answer: d
Clarification: On heating, ferrimagnetic substances become paramagnetic and hence, lose their ferromagnetism. NaCl is diamagnetic in nature. Cobalt is a ferromagnetic substance.

6. 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 > Co
Answer: a
Clarification: 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.

7. Which of the following elements have a negative value of magnetic susceptibility?
a) Iron
b) Oxygen
c) Aluminium
d) Nitrogen
Answer: d
Clarification: 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.

8. Which of the following is not Anti-ferromagnetic?
a) MnO
b) Mn₂O₃
c) MnO₂
d) Mn
Answer: d
Clarification: MnO, Mn₂O₃, and MnO₂ are Anti-ferromagnetic. Mn is paramagnetic in nature. Paramagnetism is when a substance is weakly attracted to a magnetic field.

9. 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) Diamagnetism
Answer: b
Clarification: 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.

10. 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 temperature
Answer: a
Clarification: 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.

Chemistry Multiple Choice Questions on “Electrochemistry – Corrosion”.

1. Rusting of iron is a type of corrosion.
a) True
b) False
Answer: a
Clarification: The most common example of corrosion is the rusting of iron. Rust is hydrated ferric oxide, Fe₂O₃.xH₂O. Rust is formed by the reaction of iron and oxygen in the presence of water or air moisture.

2. What is the method of protection of iron by coating it with zinc called?
a) Tinning
b) Cathodic protection
c) Galvanization
d) Anti-rust solutions
Answer: c
Clarification: Rusting of iron is one of the most common forms of corrosion. To prevent iron from rusting, generally, more active metals are coated on it. The metal which is often used for coating iron is zinc (which is more active) and this process is called galvanization.

3. Which of the following statements regarding corrosion is true?
a) Corrosion does not depend on the reactivity of the metal
b) Presence of impurities does not affect the rate of corrosion
c) Strains in metals affect the rate of corrosion
d) Presence of electrolytes does not affect the rate of corrosion
Answer: c
Clarification: More active metals are readily corroded. Presence of impurities in metals enhances the chances of corrosion. Pure metals do not corrode. Corrosion takes place rapidly at bends, scratches, nicks and cuts in the metal. Electrolytes, if present, also increase the rate of corrosion.

4. Magnesium is used as a sacrificial metal to protect iron from rust.
a) True
b) False
Answer: a
Clarification: Magnesium and zinc are often used as sacrificial metals. Sacrificial protection means covering the iron surface with a layer of metal which is more active (electropositive) than iron and thus prevents the iron from losing electrons and getting oxidised.

5. Which of the following does not cause rusting of iron?
a) Moisture
b) Vacuum
c) SO₂
d) CO₂
Answer: b
Clarification: Air and moisture accelerate corrosion. The most important factor in atmospheric corrosion, overriding pollution or lack of it, is moisture. Presence of gases like SO₂ and CO₂ in air catalyse the process of corrosion. Iron, when placed in vacuum, does not rust.

6. Which of the following is not a method of prevention of corrosion?
a) Galvanization
b) Anti-rust solutions
c) Cathodic protection
d) Heating
Answer: d
Clarification: Galvanization is the process of coating iron with zinc to prevent its oxidation. Anti-rust solutions are alkaline phosphate and alkaline chromate solutions. The alkalinity prevents the availability of hydrogen ions. Cathodic protection is the process of connecting the iron object to be protected to a more active metal either directly or through a wire.

7. Which of the following statements is false?
a) Salt water decelerates the rate of corrosion
b) Magnesium is more active than iron
c) During galvanization, ZnCO₃.Zn(OH)₂ is formed which prevent further corrosion
d) Anti-rust solutions are used in car radiators to prevent rusting of iron parts of the engine
Answer: a
Clarification: Saline medium has extra salts such as sodium chloride dissolved in water. It has a greater concentration of electrolyte than ordinary medium. The ions present will favour the formation of more electrochemical cells and favour the transfer of hydrogen ions and will thus promote rusting or corrosion.

8. Which of the following is not essential for rusting to take place?
a) Metal (like iron)
b) Oxygen
c) Moisture
d) Light
Answer: d
Clarification: Rust is a general term for a series of iron oxides, usually, reddish-brown oxides, formed by the reaction of iron with oxygen in the presence of water or moisture. Water is usually present in the form of water vapour and oxygen is always present in the normal atmosphere. The oxygen needs moisture as a catalyst and reactant to accelerate the reaction, so in the absence of moisture, iron won’t rust.

9. Which of the following is the overall reaction of rusting?
a) 2Fe_(s) → 2Fe²⁺ + 4e^–
b) O₂ (g)+4H⁺(aq)+4e^–⟶2H₂O(l)
c) 2Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l)
d) H₂O ⇌ H⁺ + OH^–
Answer: c
Clarification: 2Fe_(s) → 2Fe²⁺ + 4e^– is the reaction that occurs at the anode.
O₂ (g)+4H⁺(aq)+4e^–⟶2H₂O(l) is the reaction that occurs at the cathode.
2Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l) is the overall reaction of rusting.

10. Which of the following metal does not corrode?
a) Iron
b) Zinc
c) Copper
d) Magnesium
Answer: c
Clarification: Magnesium and zinc are highly active. Hence, they corrode at a faster rate than iron. Iron undergoes rusting in the presence of oxygen and moisture. Copper, a noble metal that occurs naturally in its elemental form, is almost totally impervious to corrosion.

Chemistry Multiple Choice Questions on “Thermodynamic Principles of Metallurgy”.

1. Ellingham Diagram does not give any information regarding the kinetics of the reduction process.
a) True
b) False
Answer: a
Clarification: Ellingham diagrams are based on thermodynamic concepts, therefore, these diagrams simply suggest whether the reduction of a given metal oxide with a particular reducing agent is possible or not. It, however, does not tell us anything about the kinetics of the reduction process.

2. Which of the following is the correct Gibbs equation?
a) ΔG = ΔH + TΔS
b) ΔG = ΔH – TΔS
c) ΔG = ΔH – 2TΔS
d) ΔG = ΔH – 3TΔS
Answer: b
Clarification: For any process, the change in Gibbs free energy (ΔG) is given by the equation,
ΔG = ΔH – TΔS
Where ΔH is the enthalpy change, ΔS is the entropy change and T is the absolute temperature.

3. What is the criterion of the feasibility of a reaction at any temperature?
a) ΔG of the reaction must be positive
b) ΔG of the reaction must be negative
c) ΔG of the reaction must be equal to zero
d) Does not depend on ΔG of the reaction
Answer: b
Clarification: The criterion of the feasibility of a reaction at any temperature is that the ΔG of the reaction must be negative. For a reaction to be spontaneous, the value of the change in Gibbs energy of the reaction must always be negative.

4. How can a reaction with positive ΔG be made to occur?
a) By increasing the temperature
b) By decreasing the temperature
c) By coupling it with another reaction
d) It is not possible for the reaction to occur
View Answer

Answer: c
Clarification: A reaction with ΔG positive can still be made to occur by coupling it with another reaction having large negative ΔG so that the net ΔG of the two reactions is negative (ΔG = change in the Gibbs energy of the reaction).

5. What does the Ellingham diagram consist of?
a) Plots of Δ_fG° vs T
b) Plots of T vs Δ_fG°
c) Plots of Δ_fG° vs ΔS
d) Plots of ΔS vs Δ_fG°
Answer: a
Clarification: Ellingham diagram normally consists of plots of Δ_fG° vs T for formation of oxides of elements. Similar diagrams can also be constructed for sulphides and halides of elements. These diagrams help us in predicting the feasibility of thermal reduction of an ore.

6. Under what condition can Al reduce MgO?
a) Above 1000°C
b) Below 1000°C
c) Above 1350°C
d) Below 1350°C
Answer: c
Clarification: For a reaction to be feasible, the change in the Gibbs free energy of the reaction (Δ G) must be negative. Lesser the ΔG, greater is the feasibility of the reaction. Above 1350°C, the standard Gibbs free energy of formation of Al₂O₃ from Al is less than that of MgO from Mg. Therefore, Al can reduce MgO above 1350°C.

7. Which of the following is false regarding the Ellingham diagram?
a) It consists of plots of Δ_fG° vs T
b) Ellingham diagrams are based on thermodynamic concepts
c) They do not tell us anything about the kinetics of the reduction process
d) They do not assume reactant-product equilibrium
Answer: d
Clarification: The interpretation of ΔG° is based upon K (i.e., ΔG° = -RT ln K). Thus, it is presumed that the reactants and the products are in equilibrium. But, this is not always true because the reactant/product may be solid.

8. Any metal will reduce the oxide of other metals which lie below it in the Ellingham diagram.
a) True
b) False
Answer: b
Clarification: No, any element (metal) will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy change (Δ_rG°) for the combined redox reaction will be negative by an amount equal to the difference between the free energy of formation (Δ_fG°) of the two oxides at that temperature.

9. Which of the following statements is true?
a) Al can reduce ZnO more readily than Mg
b) Zn can reduce Cu₂O more readily than Al
c) Mg can reduce ZnO more readily than Al
d) Zn can reduce Cu₂O more readily than Mg
Answer: c
Clarification: The relative tendency of these metals to act as reducing agents is Mg > Al > Zn > Fe > Cu. Mg can reduce Al₂O₃, ZnO, FeO and Cu₂O more readily than Al can reduce ZnO, FeO and Cu₂O. Similarly, Al can reduce ZnO, FeO and Cu₂O more readily than Zn reduces FeO and Cu₂O.

10. How are the oxides of less reactive metals like silver and mercury reduced?
a) By thermal decomposition
b) By electrolysis
c) By using reducing agents
d) Cannot be reduced
Answer: a
Clarification: In case of less reactive metals like silver and mercury, Δ_fG° becomes positive at higher temperatures. This suggests that both silver oxide (Ag₂O) and mercury oxide (HgO) are unstable and hence decompose at high temperatures to liberate the corresponding metal.

Chemistry Multiple Choice Questions on “P-Block Elements – Simple Oxides”.

1. What type of oxides is formed by metals?
a) Oxides with pH = 7
b) Oxides with pH > -log₁₀[1 x 10^-7]
c) Oxides with pH 10[1 x 10^-7]
d) Amphoteric oxides
Answer: b
Clarification: Metals usually form basic oxides. E.g. sodium, potassium, calcium. Meaning, the pH of the salts is greater than 7 on the pH scale. pH is calculated as the negative logarithm of concentration of hydronium ions. Metals form oxides because upon dissolution, they release oxide ions which in turn give hydroxide (basic) ions. However, a few metals form oxides which are both acidic and basic, called amphoteric oxides. E.g. zinc, lead.

2. What type of oxides is formed by non-metals?
a) Oxides with pH b) Amphoteric oxides
c) Oxides with pH > 7
d) Oxides which are acidic
Answer: d
Clarification: Non-metals form acidic oxides because upon dissolution in water they liberate free H⁺ ions. Oxides with pH 7 are basic in nature. Amphoteric oxides behave both like acid and base.

3. Which of the following can classified as an amphoteric oxide?
a) Iron (III) oxide
b) Zinc oxide
c) Mercury (II) oxide
d) Lime

Answer: b
Clarification: Zinc oxide is the only amphoteric oxide out of these four. On dissolution in acid like HCl, it acts as a base and forms ZnCl₂ salt. On dissolving in base like NaOH, it acts an acid to form sodium zincate, Na₂ZnO₂.

4. What is the starting substance fed into the electrolytic cell in Hall-Heroult process?
a) Cryolite
b) Sodium hydroxide
c) Aluminum hydroxide
d) Alumina
Answer: d
Clarification: Hall-Heroult process is used to extract aluminum by using an electrolytic cell. Alumina means aluminum oxide. It is dissolved in cryolite, Na₃AlF₆, to increase electrical conductivity, reduce melting point and achieve greater yield of aluminum metal.

5. What is the primary constituent of sand?
a) Silicon tetroxide
b) Silica
c) Silicon
d) Feldspar
Answer: b
Clarification: Sand is primarily composed of silicon dioxide, also called silica. Silicon tetroxide is not a compound but an oxo-anion of silicon. Silicon metal does not form native in nature and feldspar is a mineral ore consisting of several tectosilicate compounds.

6. Before subjecting Mg ribbons to flame, it is rubbed with sandpaper. Why?
a) To increase surface area for combustion
b) To increase roughness of surface
c) To remove oxide layer
d) To remove moisture
Answer: c
Clarification: In order to ensure effective burning of the magnesium ribbon, it is cleaned with sandpaper to remove dust particles along with the layer of magnesium oxide, which may prevent it from burning completely.

7. What product is formed when iron metal is dipped into concentrated nitric acid?
a) Iron (III) nitrate + NO₂
b) Iron (II) nitrate + NO
c) Iron (III) oxide
d) Iron (III) nitrite
Answer: c
Clarification: Iron becomes passive when treated with concentrated nitric acid. The formation of a tough oxide layer prevents it from further reacting with the acid, hence inhibiting the reaction.

8. Which of the following processes involves heating a carbonate ore to form metal oxide?
a) Leaching
b) Smelting
c) Ore Reduction
d) Calcination
Answer: d
Clarification: In general metallurgy, calcination is the process of heating a metal ore to convert it to its corresponding metal oxide. For e.g. zinc carbonate (calamine ore) is heated to form zinc oxide which is then reduced by electrolysis in the presence of sulfuric acid to obtain zinc metal.

9. Which of the following metals does not react with hot water to form an oxide?
a) Calcium
b) Magnesium
c) Iron
d) Lithium
Answer: c
Clarification: Iron is relatively unreactive with respect to other metals mentioned. Iron forms iron (III) oxide on heating vigorously. However, lithium, magnesium and calcium all form hydroxides and oxides on reaction with hot water.

10. What is formed when amphoteric oxides react with an alkali solution?
a) Salt only
b) No reaction
c) Acid
d) Salt and water
Answer: d
Clarification: Amphoteric oxides react with both acids and alkalis. On treatment with alkali solution, they behave as acids and undergo reactions analogous to neutralization, thus forming salt and water. For e.g. Al₂O₃ + NaOH → NaAlO₂ + H₂O.

Chemistry Multiple Choice Questions on “D and F-Block Elements – Lanthanoids”.

1. The lanthanides are d-block elements.
a) True
b) False
Answer: b
Clarification: The lanthanides are elements in which the differentiating electron enters (n-2)f-orbital of anti-penultimate shell and are called as f-block elements. These have three outer shells incomplete. Thus, lanthanides are f-block elements, not d-block.

2. What is the general electronic configuration of the lanthanides?
a) (n-2)f ^1-14 (n-1)d ^1-10 ns²
b) (n-2)f ^1-14 (n-1)d ^1-2 ns²
c) (n-2)f ^1-14 (n-1)d ^0-1 ns²
d) (n-2)f ^1-14 (n-1)d ⁰ ns²
Answer: c
Clarification: In lanthanides, the differentiating electron occupies 4f subshell and are rarely found on the earth’s crust, so called rare earths. The general electronic configuration of lanthanides is 6s²5d^0-14f^1-14.

3. What happens to the atomic size of the lanthanides with increase in atomic number?
a) The radius remains unchanged
b) The radius decreases
c) The radius increases
d) The radius first increases and then decreases
Answer: b
Clarification: The gradual decrease in the atomic and ionic radii of the lanthanides with an increase in atomic number is called lanthanide contraction. It occurs due to the poor shielding effect of the 4f electrons.

4. Which of the following is not a consequence of lanthanide contraction?
a) From La⁺³ to Lu⁺³, the ionic radii changes from 106 pm to 85 pm
b) As the size of the lanthanide ions decreases the basic strength increases
c) The basic character of oxides and hydroxides decreases with increase in atomic number
d) The atomic radii of 4d and 5d series is similar
Answer: b
Clarification: The small average decreases in the atomic size is responsible for a small decrease in electronegativity and S.O.P of lanthanides. As the size of the lanthanide ions decreases the covalent character of M—OH bond increases and hence basic strength decreases.

5. What is the most common oxidation state of lanthanides?
a) +2
b) +4
c) +6
d) +3
Answer: d
Clarification: The most common and stable oxidation state of lanthanides is +3. Some elements also exhibit +4 oxidation states. Some elements exhibit +2 oxidation state also due to their half-filled, fully-filled and noble gas configuration.

6. In lanthanides, Ce changes from +4 to +3 oxidation state.
a) True
b) False
Answer: a
Clarification: In lanthanides, Ce changes from +4 to +3 oxidation state. Formation of Ce⁴⁺ is favorable due to its noble gas configuration. However, Ce⁺⁴ changes to Ce⁺³ as it is a strong oxidizing agent. The E° value of Ce⁺⁴/Ce⁺³ is +1.74 V.

7. Which of the following is not a property of lanthanides?
a) They are soft metals with white silvery color
b) They tarnish rapidly by air
c) The hardness of the metals increases with increase in the atomic number
d) The melting point of the metal ranges from 500-1000K
Answer: d
Clarification: All lanthanides are soft metals with silvery white color. They tarnish rapidly by air. With increase in atomic number, the harness of these metals also increases. The melting points of the lanthanides ranges from 1000 to 1200K but samarium melts at 1623K.

8. Which of the following lanthanide ions do not exhibit color?
a) Lu⁺³ and Ln⁺³
b) Lu⁺² and Ln⁺²
c) Ce⁺³ and Ce⁺³
d) Pr⁺⁴ and Ce⁺⁴
Answer: a
Clarification: Many trivalent lanthanide ions are colored both in the solid state and in aqueous state due to f-f electronic transition. Lu⁺³ and Ln⁺³ don’t exhibit color due to absence of unpaired electrons. All lanthanide ions except La⁺³, Ce⁺³ and Lu⁺³ are paramagnetic due to presence of unpaired electrons.

9. Which of the following is the correct order of arrangement of the first five lanthanides according to atomic number?
a) La, Ce, Pr, Nd, Pm
b) La, Pr, Ce, Pm, Nd
c) La, Pr, Ce, Nd, Pm
d) La, Ce, Pr, Pm, Nd
Answer: a
Clarification: The first five elements of Lanthanides are:
Lanthanum (La) – 57
Cerium (Ce) – 58
Praseodymium (Pr) – 59
Neodymium (Nd) – 60
Promethium (Pm) – 61.

10. Which is the last element of lanthanides?
a) Ytterbium
b) Lutetium
c) Thulium
d) Erbium
Answer: b
Clarification: The atomic numbers of Erbium, Thulium, Ytterbium and Lutetium is 68, 69, 70 and 71 respectfully. So the last element of lanthanide series is Lutetium with electronic configuration [Xe]4f¹⁴5d¹6s².

Chemistry Multiple Choice Questions on “Haloalkanes Nomenclature”.

1. What is the common name of 3-Bromopropene?
a) Tert-Butyl bromide
b) Vinyl bromide
c) Allyl bromide
d) Propylidene bromide
Answer: c
Clarification: 3-Bromopropene has 3 C atoms in the parent chain with a double bond at C-1 and Br at C-3. This means that Br is attached to the C which is next to the C-C double bond, hence it is an allylic halide.

2. What is the IUPAC name of the following compound?

a) 1-Bromo-3-methylprop-2-ene
b) 3-Bromo-1-methylpropene
c) 1-Bromobut-2-ene
d) 4-Bromobut-2-ene
Answer: c
Clarification: The longest chain of C atoms is 4 (butene) and the numbering should begin from the Br side to satisfy the lowest set of locants rule. So, Br is at C-1 and double bond is at C-2.

3. Which of the following is not the correct IUPAC name for the compound shown?

a) 1-Chloro-3-methylbenzene
b) 3-Chloromethylbenzene
c) 3-Chlorotoluene
d) m-Chlorotoluene
Answer: d
Clarification: All the names given are appropriate for the shown compound, but m-Chlorotoluene is the common name and not the IUPAC name.

4. In the common naming system, the prefix sym- is used for haloarenes with _____ halogen atoms.
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The prefix sym- is used for trihaloarenes with same halogen atom at alternate positions in the benzene ring (1, 3, 5). The prefixes o-, m- and p- are used for dihaloarenes depending on the relative positions of the two identical halogen atoms on the aromatic ring.

5. How many carbon atoms does Isobutyl chloride have in its parent carbon chain?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: The IUPAC name of Isobutyl chloride is 1-Chloro-2-methylpropane, which means the parent chain consists of 3 carbon atoms with an alkyl group at C-2 and a halogen at C-1.

6. Identify the correct naming of the compound H₃C-CHCl₂ from the following?
a) Ethylidene chloride
b) Ethylene dichloride
c) 1,2-Dichloroethane
d) Vic-dichloride
Answer: a
Clarification: The given compound is a dihaloalkane with both the halogens on the same carbon atom. These are also known as gem-dihalides or alkylidene halides.

7. Identify the correct common name for CH₃CH₂CH₂CH₂Cl.
a) Isobutyl chloride
b) n-Butyl chloride
c) sec-Butyl chloride
d) tert-Butyl chloride
View Answer

Answer: b
Clarification: The compound shown is 1-Chlorobutane which has 4 C atoms in the parent chain and the only halogen compound present at the end of the chain. This type structure has the prefix n- in the common name.

8. What is the IUPAC name for the compound (CH₃)₃CCH₂Cl?
a) 1-Chloro-2,2-dimethylpropane
b) 1-Chloro-2,2,2-trimethylethane
c) 2-Chloromethyl-2-methylpropane
d) 2,2-Dimethly-1-Bromopropane
Answer: a
Clarification: There are two different possible chains that consists of maximum 3 C atoms. One of them has two methyl and one chloro group and the other has one methyl and one complex substituent. The preference is given to the former with the chloro at C-1 and the 2 methyl groups at C-2.

9. What is the IUPAC name of the compound shown?

a) 1-Cholor-2-fluorocyclobut-1-ene
b) 2-Chloro-1-fluorocyclobut-1-ene
c) 1-Chloro-2-fluorocyclobut-2-ene
d) 1-Fluoro-2-chlorocyclobut-1-ene
Answer: b
Clarification: The alkyl group is alicyclic with 4 carbon atoms at the corners of a rectangle and a double bond; hence the base is cyclobutene. In the case of cyclic alkenes, the position of double bond is always given the first number.

10. What is the correct IUPAC naming of CHF₂CBrClF?
a) 1-Bromo-1-chloro-1,2,2-trifluoroethane
b) 1-Bromo-1-chloro-1,2,2-trifluoroethene
c) 2-Bromo-2-chloro-1,1,2-trifluoroethane
d) 2-Bromo-2-chloro-1,1,2-trifluoroethene
Answer: a
Clarification: There are only two C atoms in the compounds and only single bonds with two F atoms at C-2 and one of each F, Cl and Br at C-1.

11. The compound 2,2-Dichlorobutane is a geminal dihalide.
a) True
b) False
Answer: a
Clarification: The compound is dihaloalkane with 2 chlorine atoms one the same carbon atom, hence it is a gem dihalide.

12. What is the IUPAC name of the following compound?

a) 2-Propylprop-1-ene
b) 2,3-Dimethylbut-1-ene
c) 2,4-Dimethylbut-1-ene
d) 1,3-Dimethylbut-2-ene
Answer: b
Clarification: The parent carbon chain must include the double bond, so there are 4 C atoms including the C-C double bond and the double bond is present at C-1 with two methyl groups at positions 2 and 3.

13. The correct naming of the compound CH≡C-CH=CH-CH₃ is __________
a) Pent-3-en-1-yne
b) Pent-2-en-4-yne
c) Pent-3-ene-1-yne
d) Pent-2-ene-4-yne
Answer: a
Clarification: The lower set of locants is 1, 3 rather than 2, 4. This compound is named as a derivative of alkyne and not alkenes. Also, the terminal ‘e’ in ene is dropped because it comes before ‘y’ of yne.

14. The prefix bis is used when two complex substituents are present on the same of different C atoms on a hydrocarbon.
a) True
b) False
Answer: a
Clarification: The prefix bis is used instead of di for complex substituents in hydrocarbons before the name of the substituent which is enclosed in brackets.

15. How many isomers does C₅H₁₁Br have?
a) 4
b) 6
c) 8
d) 10
Answer: c
Clarification: It has a total of 8 isomers, four are primary and two each are secondary and tertiary haloalkanes.