Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Ideal and Non-Ideal Solutions”.

1. Which of the following statements regarding Ideal solutions is false?
a) Ideal solutions obey Raoult’s law under all conditions of temperature and concentrations
b) There will be some change in volume on mixing the components, i.e., ΔV_mixing ≠ 0
c) There will be no change in enthalpy when the two components are mixed, i.e., ΔH_mixing = 0
d) There will be no change in volume on mixing the components, i.e., ΔV_mixing = 0
Answer: b
Clarification: An ideal solution is the solution in which each component obeys Raoult’s law under all conditions of temperature and concentrations. An ideal solution will satisfy the following conditions:
I. There will be no change in volume on mixing the components, i.e., ΔV_mixing = 0.
II. There will be no change in enthalpy (i.e., no heat is evolved or absorbed) when the two components are mixed, i.e., ΔH_mixing = 0.

2. Which of the following is not an example of an Ideal solution?
a) Benzene + Toluene
b) n-Hexane + n-Heptane
c) Ethyl alcohol + Water
d) Ethyl bromide + Ethyl chloride
Answer: c
Clarification: An ideal solution may be defined as the solution in which no volume change and no enthalpy change take place on mixing the solute and the solvent in any proportion. Ethyl alcohol + Water is a Non-Ideal solution.

3. A solution which does not obey Raoult’s law is called a non-ideal solution.
a) True
b) False
Answer: a
Clarification: A non-ideal solution is the solution in which solute and solvent molecules interact with one another with a different force than the forces of interaction between the molecules of the pure components. Non-Ideal solutions do not obey Raoult’s law.

4. Which of the following is false regarding Non-Ideal solutions?
a) They do not obey Raoult’s law
b) ΔV_mixing ≠ 0
c) ΔH_mixing = 0
d) They form azeotropes
Answer: c
Clarification: Non-Ideal solutions do not obey Raoult’s law. For Non-Ideal solutions, ΔV_mixing ≠ 0, ΔH_mixing ≠ 0. Non-ideal solutions form azeotropes or constant boiling mixtures, i.e., they have the same concentration in the vapour phase and the liquid phase.

5. Which of the following is an example of a non-ideal solution showing positive deviation?
a) Acetone + Carbon disulphide
b) Chlorobenzene + Bromobenzene
c) Chloroform + Benzene
d) Acetone + Aniline
Answer: a
Clarification: Chlorobenzene + Bromobenzene is an example of an ideal solution. Chloroform + Benzene and Acetone + Aniline are examples of non-ideal solutions but they show negative deviations. Acetone + Carbon disulphide is an example of a non-ideal solution showing positive deviation.

6. Ideal solutions do not form azeotropes.
a) True
b) False
Answer: a
Clarification: An azeotrope or a constant boiling mixture is a mixture that has the same concentration in the vapour phase and the liquid phase. In azeotropes, the component ratio of unvaporized solution is equal to that of the vaporized solution when boiling. Hence, Ideal solutions don’t form azeotropes.

7. Which of the following is not an example of a non-ideal solution showing negative deviation?
a) HNO₃ + Water
b) HCl + Water
c) Acetic acid + Pyridine
d) Carbon tetrachloride + Toluene
View Answer

Answer: d
Clarification: HNO₃ + Water, HCl + Water and Acetic acid + Pyridine are non-ideal solutions showing negative deviations. Carbon tetrachloride + Toluene is an example of non-ideal solution showing positive deviation.

8. Which of the following is true regarding non-ideal solutions with negative deviation?
a) The interactions between the components are lesser than in the pure components
b) ΔV_mixing = +ve
c) ΔH_mixing = +ve
d) They form maximum boiling azeotropes

Answer: d
Clarification: The interactions between the components of a non-ideal solution showing negative deviation are greater than the pure components. The change in volume and enthalpy after mixing is negative, i.e., ΔV_mixing = -ve, ΔH_mixing = -ve.

9. Which of the following cannot form an azeotrope?
a) H₂O + C₂H₅OH
b) CHCl₃ + C₂H₅OH
c) HCl + H₂O
d) Benzene + Toluene
Answer: d
Clarification: H₂O + C₂H₅OH forms an azeotrope with a boiling point of 351.15 K. CHCl₃ + C₂H₅ OH forms an azeotrope with a boiling point of 332.3 K. HCl + H₂O forms an azeotrope with a boiling point of 383 K. Benzene + Toluene is an ideal solution and hence does not form an azeotrope.

10. Which of the following is true regarding azeotropes?
a) An azeotrope does not exhibit the same concentration in the vapour phase and the liquid phase
b) Azeotropic mixtures cannot be separated into their constituents by fractional distillation
c) In case of minimum boiling azeotropes, the boiling point of the azeotrope is higher than the boiling point of either of the pure components
d) In case of maximum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components
Answer: b
Clarification: Azeotropes have the same concentration in the vapour phase and the liquid phase. In case of minimum boiling azeotropes, the boiling point of the azeotrope is lesser than the boiling point of either of the pure components. In case of maximum boiling azeotropes, the boiling point of the azeotrope is higher than the boiling point of either of the pure components.

11. If liquids A and B form an ideal solution, then what is the Gibbs free energy of mixing?
a) > 0
b) c) = 0
d) Not Defined
Answer: b
Clarification: The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system. For ideal solutions, the value of the Gibbs Free energy is always negative as mixing of ideal solutions is a spontaneous process.

12. 5 moles of liquid X and 10 moles of liquid Y make a solution having a total vapour pressure 70 torr. The vapour pressures of pure X and pure Y are 64 torr and 76 torr respectively. Which of the following is true regarding the described solution?
a) The solution shows positive deviation
b) The solution shows negative deviation
c) The solution is ideal
d) The solution has volume greater than the sum of individual volumes
Answer: b
Clarification: Given,
Observed pressure = 76 torr
According to Raoult’s law,
p_A = x_A x p_A⁰ = 5/15 x 64 = 21.33 torr
p_B = x_B x p_B⁰ = 10/15 x 76 = 50.67 torr
Therefore, pressure expected by Raoult’s law = 21.33 + 50.67 = 72 torr.
Thus, observed pressure (70 torr) is less than the expected value. Hence, the solution shows negative deviation.

Chemistry Multiple Choice Questions on “Chemical Kinetics – Temperature Dependence of the Rate of a Reaction”.

1. The rate of a reaction depends on the temperature.
a) True
b) False
Answer: a
Clarification: For most of the reactions, the rate of reaction becomes nearly double or even more for 10° rise of temperature. The effect of temperature is usually expressed in terms of temperature coefficient.

2. Which of the following is the correct expression for the temperature coefficient (n)?
a) n = Rate constant at T + 10°/Rate constant at T°
b) n = Rate constant at T + 20°/Rate constant at T°
c) n = Rate constant at T + 30°/Rate constant at T°
d)n = Rate constant at T + 40°/Rate constant at T°
Answer: a
Clarification: The effect of temperature on the rate of a reaction is usually expressed in terms of the temperature coefficient which is defined by the equation:
Temperature coefficient (n) = Rate constant at T + 10° (308 K)/Rate constant at T° (298 K).

3. The rate of reaction increases with a rise in temperature.
a) True
b) False
Answer: a
Clarification: The increase in the rate of reaction with a rise in temperature is not due to the increase in the total number of collisions but mainly due to an increase in the total number of effective collisions.

4. What happens to the peak of the curve in the Maxwell-Boltzmann distribution graph if temperature increases?
a) Shifts forward and upward
b) Shifts forward and downward
c) Shifts backward and upward
d) Shifts backward and downward
Answer: b
Clarification: With the increase of temperature, the peak shifts forward but downward. This is because with the increase of temperature, the most probable kinetic energy increases and the fraction of molecules possessing most probable kinetic energy decreases.

5. Which of the following is the correct Arrhenius equation?
a) k = A e^E_a/RT
b) k = A e^E_a/T
c) k = A e^E_a/R
d) k = A e^-E_a/RT
Answer: d
Clarification: Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k, was proposed by Arrhenius in 1889. The equation, called Arrhenius equation, is usually written in the form k = A e^-E_a/RT.

6. Which of the following represents the Boltzmann factor?
a) e^-E_a
b) e^E_a
c) e^-E_a/RT
d) e^E_a/RT
Answer: c
Clarification: The factor e^-E_a/RT in the Arrhenius equation is called Boltzmann factor. It represents the fraction of the molecules (N_E/N_T) having energy equal to or greater than E where N_E represents the number of molecules with energy E and N_T represents the total number of molecules.

7. What is R in the equation k = Ae^-E_a/RT?
a) R = 8.314 J K^-1 mol^-1
b) R = 3.184 J K^-1 mol^-1
c) R = 4.318 J K^-1 mol^-1
d) R = 1.438 J K^-1 mol^-1
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. The activation energy of a reaction is 50 kJ mol^-1 and the value of rate constant at 300 K is 2.5×10^-5 sec^-1. What is the value of the frequency factor, A?
a) 4228.53 s^-1
b) 3829.69 s^-1
c) 7596.45 s^-1
d) 6565.35 s^-1
Answer: c
Clarification: Given,
E_a = 50 kJ mol^-1
T = 300 K
k = 1.5 × 10^-5 sec^-1
R = 8.314 J K^-1 mol^-1
log k = – E_a / (2.303 RT) + log A
OR
log A = log k + E_a / (2.303 RT)
log A = log (1.5 × 10^-5) + 50000 / (2.303 × 8.314 × 300)
log A = 3.88061
A = antilog (3.880611)
A = 7596.45 s^-1.

9. What is the value of rate constant k if the value of the activation energy E_a and the frequency factor A are 49 kJ / mol and 9 × 10¹⁰ s^-1 respectively? (T = 313 K)
a) 6 × 10² s^-1
b) 9 × 10² s^-1
c) 6 × 10^-2 s^-1
d) 3 × 10² s^-1
Answer: a
Clarification: Given,
E_a = 49 kJ / mol^-1
T = 313 K
A = 9 × 10² s^-1
R = 8.314 J K^-1 mol^-1
log k = – E_a / (2.303 RT) + log A
log k = – 49000 / (2.303 × 8.314 × 313) + log 9 × 10¹⁰
log k = 2.77843
k= antilog (2.77843)
k = 6 × 10² s^-1.

10. The rate constant of a reaction is 6 × 10^-3 s^-1 at 50° and 9 × 10^-3 s^-1 at 100° C. Calculate the energy of activation of the reaction.
a) 6.123 kJ mol^-1
b) 8.124 kJ mol^-1
c) 12.357 kJ mol^-1
d) 18.256 kJ mol^-1
Answer: b
Clarification: Given,
k₁=6 × 10^-3s^-1 T₁=50 + 273=323 K
k₂=9 × 10^-3s^-1 T₂=100 + 273=373 K
Substituting these values in the equation:
log (k₂ / k₁ ) = (E_a / 2.303 R) × ((T₂ – T₁) / T₁ T₂), we get
log (9 × 10^-3 s^-1 / 6 × 10^-3 s^-1 ) = ((E_a / (2.303 × 8.314)) × ((373 – 323) / (373 × 323))
log 9 / 6 = ((E_a / (2.303 × 8.314)) × (50 / (373 × 323))
E_a = 8.124 kJ mol^-1.

Chemistry Multiple Choice Questions on “P-Block Elements – Sulphuric Acid”.

1. What is the first step involved in forming sulphuric acid through contact process?
a) Conversion of SO₂ to SO₃
b) Burning of Sulphur or roasting sulphide ores to generate SO₂
c) Absorption of SO₃ in sulphuric acid
d) Purifying SO₃
Answer: b
Clarification: The first step involved in forming sulphuric acid through contact process is Burning of Sulphur or roasting sulphides ores to generate SO₂. Sulphur dioxide is purified by removing dust and arsenic impurities.

2. In contact process, arsenic impurity has to be removed before proceeding.
a) True
b) False
Answer: a
Clarification: Sulphur dioxide is purified by removing dust and arsenic impurities. Arsenic is a catalyst poison for vanadium pentoxide catalyst. Hence, it has to be removed completely before entering into the catalytic chamber.

3. What is the catalyst used in contact process in the conversion of Sulphur dioxide to Sulphur trioxide?
a) Finely divided iron
b) Molybdenum
c) Vanadium pentoxide
d) Platinum
Answer: c
Clarification: After burning Sulphur in the presence of oxygen the formed Sulphur dioxide needs to be reacted with oxygen to form Sulphur trioxide. It is an exothermic reaction and vanadium pentoxide is used as the catalyst.

4. Sulphuric acid is called the ‘King of Acids’.
a) True
b) False
Answer: a
Clarification: Even though hydrogen iodide is stronger sulphuric acid is called the ‘King of Acids’. All acids can be formed from sulphuric acid, because of its low volatility, it can be used to manufacture more volatile acids from their corresponding salts.

5. Which of the following is not a property of sulphuric acid?
a) It acts as a strong dehydrating agent
b) It is a Lewis acid
c) It cannot char carbohydrates
d) It is dense and oily
Answer: c
Clarification: Sulphuric acid is a Lewis acid and is a dense oily liquid which acts as a strong dehydrating agent due to its high affinity towards water. It chars carbohydrates to from black precipitates of carbon.

6. Which process is used for large scale manufacturing of sulphuric acid?
a) Haber’s process
b) Ostwald’s process
c) Smith’s process
d) Contact process
Answer: d
Clarification: These days sulphuric acid is mostly prepared by the Contact process. The acid produced by this method is free from arsenic impurities and hence can be safely used for the preparation of edible products.

7. What is the hybridization of Sulphur in sulphuric acid?
a) sp
b) sp²
c) sp³
d) It is not hybridized
Answer: c
Clarification: The hybridization of Sulphur in sulphuric acid is sp³ and is tetrahedral shaped. The structure is often drawn with two double bonds, with double bond formed from d-orbitals on Sulphur and p-orbitals on oxygen.

8. What are the favorable conditions for maximum yield of Sulphur trioxide?
a) High pressure, low temperature, excess of oxygen
b) High pressure, low temperature, less oxygen
c) Low pressure, low temperature, excess of oxygen
d) High pressure, high temperature, excess of oxygen
Answer: a
Clarification: High pressure is preferred since the forward reaction proceeds with decrease in number of moles. Low temperature is preferred because the reaction is highly exothermic and to have maximum yield of Sulphur trioxide, oxygen is used in excess.

9. What happens in the purifying chamber in contact process?
a) The compounds are washed with nitrogen
b) Dust particles are removed
c) A spray of HCl is used for drying gasses
d) The compounds are heated
Answer: b
Clarification: In the purifying unit the dust particles are removed by blowing steam and the soluble impurities are removed by washing the gases with water. A spray of conc. Sulphuric acid is used for drying gasses.

10. Which of the following is true about sulphuric acid?
a) It is a soapy liquid
b) It is a monobasic acid
c) It is a dibasic acid
d) It is a tribasic acid
Answer: c
Clarification: Sulphuric acid is an oily liquid and is dibasic in nature. It is a dibasic acid because it contains two hydrogen atoms which ionize in aqueous solution to become hydrogen ions. When it reacts with bases like sodium hydroxide it form two kinds of salts proving its dibasic nature.

Chemistry Written Test Questions for Schools on “Definitions – Coordination Compounds – 2”.

1. The coordination number is determined by the number of both, sigma and pi bonds formed by the ligand with the central atom/ion.
a) True
b) False
Answer: b
Clarification: The CN is found by only considering the number of sigma bonds between the donor atoms and central atom/ion. Pi bonds between ligand and central atom/ion are not counted for this purpose.

2. The coordination polyhedron geometry shown belongs to which of the following complexes?

a) [Co(NH₃)₆]³⁺
b) [Ni(CO)₄]
c) [CoCl₂(en)₂]²⁺
d) [Fe(C₂O₄)₃]^3-
Answer: b
Clarification: The arrangement shown is tetrahedral and CN=4. [Co(NH₃)₆]³⁺ is octahedral with CN=6. Both [CoCl₂(en)₂]²⁺ and [Fe(C₂O₄)₃]^3- have didentate ligands and hence CN=6.

3. What is the correct way of representing nickel in [Ni(CO)₄]?
a) Ni(0)
b) Ni(I)
c) Ni(II)
d) Ni(III)
Answer: a
Clarification: The oxidation number of a metal is represented as Roman numeral in parenthesis preceded by the name of the entity. Since oxidation number of nickel in the given compound is zero, it is represented as Ni(0).

4. What is the shape of the coordination polyhedron of [PtCl₄]^2-?
a) Linear
b) Square planar
c) Tetrahedral
d) Octahedral
Answer: b
Clarification: The CN of the metal ion is 4 because Cl^– is a unidentate ligand. The only possibility for the shape of its coordination polyhedra is square planar or tetrahedral. From crystal field splitting diagram, it is known that [PtCl₄]^2- is square planar.

5. Complexes in which a metal is attached to only one kind of donor group is called _______
a) Unidentate
b) Chelate
c) Homoleptic
d) Heteroleptic
Answer: c
Clarification: Unidentate and chelate are words associated with ligands, whereas coordination complexes can be either homoleptic or heteroleptic depending upon whether only one kind or different kinds of ligands linked to the metal atom/ion respectively.

6. Which of the following compounds consists of a homoleptic complex?
a) [Co(NH₃)₆]Cl₃
b) [Co(NH₃)₅Cl]Cl₂
c) [Co(NH₃)₄Cl₂]Cl
d) [Co(NH₃)₅(CO₃)]Cl
Answer: a
Clarification: [Co(NH₃)₆]Cl₃ consists of only ammonia ligands in the complex and is hence homoleptic. The rest of the compounds consists of two kinds of ligates, either NH₃ and Cl or NH₃ and CO₃, and hence they are heteroleptic.

7. The oxidation number of the central metal ion in a coordination entity is the charge it would carry if all the _________ are removed along with the electron pairs that are shared with it.
a) lewis acids
b) anions
c) ligands
d) didentate ligands
Answer: c
Clarification: The oxidation number or primary valence of a central atom is its charge if all the ligands and its donor atoms are not present in the coordination entity.

8. The oxidation number of the central metal in a coordination entity is always a positive quantity.
a) True
b) False
Answer: b
Clarification: The oxidation number might not always be positive. It can also be 0 or a negative value. For example, the oxidation number of cobalt in K[Co(CO)₄] is -1.

9. What is the denticity of the ligand shown in the figure?

a) 2
b) 4
c) 6
d) 8
Answer: c
Clarification: The ion shown in the figure is ethylenediaminetetraacetate which is a hexadentate ligand and can bind through its two N atoms or four O atoms. Hence, its denticity is 6.

10. When a polydentate ligand uses all its donor atoms simultaneously to bond to a single metal ion, it is said to be a _______ ligand.
a) unidentate
b) didentate
c) chelate
d) ambidentate
Answer: c
Clarification: Unidentate ligands use only one donor atom at a time. Didentate are polydentate ligands that have two donor atoms. An ambidentate ligand can bind through different donor atoms.

Chemistry Written Test Questions for Schools,

Chemistry Multiple Choice Questions on “Haloalkanes & Haloarenes – Chemical Reactions – 1”.

1. The reaction that takes place when an electron rich species stronger than the halide approaches the partially positive carbon atom of a haloalkane and forms a new bond with the carbon atom and in the process displacing the halogen is called a ________ reaction.
a) displacement
b) electrophilic substitution
c) nucleophilic substitution
d) elimination
Answer: c
Clarification: Nucleophiles are electron rich species and when a stronger nucleophile replaces an already existing one is called a nucleophilic substitution reaction.

2. What will be the class of the nucleophilic substitution product when sodium hydroxide reacts with chloroethane?
a) Alkane
b) Alcohol
c) Amine
d) Ether
Answer: b
Clarification: The OH^– nucleophile substitutes the Cl^– ion in chloroethane to form ethanol, which is an alcohol.

3. Identify the nucleophile in the substitution reaction between water and bromopropane.
a) Br^–
b) H⁺
c) OH^–
d) H₂O
Answer: d
Clarification: The water molecule itself is the nucleophile which has two pairs on electrons on the O atom and attacks the Br atom in the haloalkane.

4. What is the class of the substitution product of the reaction between LiAlH4 and an alkyl halide?
a) Haloalkane
b) Hydrocarbon
c) Nitroalkane
d) Alkyl nitrite
Answer: b
Clarification: The nucleophile in LiAlH₄ is the H atom which attacks the halogen in the alkyl halide and substitutes it to form the basic hydrocarbon.

5. Which of the followings reagents forms an isonitrile when reacted with an alkyl halide?
a) KCN
b) AgCN
c) KNO₂
d) AgNO₂
Answer: b
Clarification: AgCN has a lone pair of electrons on the N atom and attacks from the N side to forms a compound of the type RNC, where R is the alkyl group.

6. Which of the following is not an ambident nucleophile?
a) Hydroxide
b) Thiocyanate
c) Cyanide
d) Nitrite
Answer: a
Clarification: Ambident nucleophiles are those that have two nucleophilic centres and can link through any of their two atoms to result in the formation of different compounds.

7. Identify the nucleophile that gives a primary amine and hydrogen bromide on reaction with bromoethane.
a) NH₂
b) NH₃
c) H₂O
d) H
Answer: b
Clarification: Ammonia reacts with alkyl halides to undergo nucleophilic substitution and form amine and a mineral acid.

8. Which of the following statements in incorrect regarding S_N2 mechanisms?
a) The rate of the reaction depends on the concentration of both reactants
b) The complete mechanism takes place in a single step
c) The transition state is stable
d) There is inversion of configuration
Answer: c
Clarification: In the transition state of S_N2 mechanisms, the carbon atoms is simultaneously bonded to the incoming nucleophile and the outgoing group and is hence bonded to five atoms at the same time. Such a geometry is unstable and cannot be isolated.

9. The reaction between chloroethane and potassium hydroxide is a _______ order reaction.
a) zero
b) first
c) second
d) third
Answer: c
Clarification: This nucleophilic reaction follows S_N2 mechanism and its rate depends on the concentration of both chloroethane and hydroxide ions. Hence, it is a bimolecular reaction.

10. A haloalkane is known to have an S_N2 reaction rate 30 times faster than that of ethyl bromide. Identify the haloalkane.
a) Methyl bromide
b) Isopropyl bromide
c) tert-Butyl bromide
d) neo-Pentyl bromide
Answer: a
Clarification: The presence of bulky groups around the C atom has an inhibiting effect on the rate of S_N2 reaction. Since methyl bromide has only three H atoms around the C, it will have a higher rate than ethyl bromide.

11. Which of the following will have the highest reactivity towards S_N2 reaction, where R=methyl group?
a) R-F
b) R-Cl
c) R-Br
d) R-I

Answer: d
Clarification: Iodine is a better leaving groups among the halogens as it has a better ability to accommodate the negative charge.

12. Which of the following statements is correct regarding unimolecular nucleophilic substitution reactions?
a) It takes place in three steps.
b) The rate of the reaction depends on the concentration of all the reactants.
c) The first step is the slowest and determines the rate of reaction.
d) None of the steps are reversible.
Answer: c
Clarification: S_N1 reactions follow first order kinetics and take place in two steps, of which the first one is the slowest, rate determining and reversible step.

13. The rate of the following reaction depends on the concentration of which reactant(s)/?

a) tert-Butyl bromide
b) Hydroxide ion
c) Both tert-Butyl bromide and hydroxide ion
d) None of the reactants
Answer: a
Clarification: This is an example of a S_N1 reaction, where the rate depends on the concentration of only the alkyl halide, because that will determine the concentration of carbocation formed in the first step.

14. Tertiary alkyl halides show higher reactivity towards bimolecular nucleophilic substitution than secondary alkyl halides.
a) True
b) False
Answer: b
Clarification: Tertiary alkyl halides have more alkyl groups present around the carbon linked to the halogen, which causes steric hinderance of the nucleophile and hence the reactivity towards S_N2 reactions reduces significantly.

15. Iodomethane reacts faster than bromomethane in S_N2 reaction with OH^–?
a) True
b) False
Answer: a
Clarification: Iodine is a better leaving group because of its large size and will be released at a faster rate in the presence of a nucleophile.