Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Diazonium Salts Chemical Reactions – 3”.

1. The reaction involving the conversion of diazonium salts to azo compounds is known as _______ reaction.
a) diazotisation
b) deamination
c) coupling
d) carbylamine

Answer: c
Clarification: Benzenediazonium salts react with highly reactive electron rich aromatic compounds to form azo compounds of the form Ar-N=N-Ar, where Ar represents an aromatic group. This type of reaction is known as coupling reaction.

2. The coupling reaction is a type of _________ reaction.
a) electrophilic addition
b) electrophilic substitution
c) nucleophilic addition
d) nucleophilic substitution

Answer: b
Clarification: During a coupling reaction, the diazonium cation which has a positive charge on the terminal nitrogen acts as the electrophile, and the electron rich compounds act as nucleophiles.

3. The coupling reaction of benzenediazonium chloride with aniline occurs in a ______ medium.
a) strongly basic
b) weakly basic
c) neutral
d) weakly acidic

Answer: d
Clarification: Aniline is a compound that is slightly basic in nature. To counter the basic effect of aniline during its coupling with diazonium salt, in is conducted in a slightly acidic medium of pH 4 to 5.

4. The reaction between benzenediazonium chloride and phenol results in a ______ coloured compound.
a) yellow
b) orange
c) red
d) purple

Answer: b
Clarification: Benzenediazonium chloride undergoes coupling with phenol to form an azo compound, p-hydroxyazobenzene. This compound is orange in colour.

5. The reaction between a diazonium salt and aniline in a slightly acidic medium gives _______
a) o-aminoazobenzene
b) m-aminoazobenzene
c) p-aminoazobenzene
d) no reaction

Answer: c
Clarification: Diazonium salts undergo coupling with aniline to form p-aminoazobenzene. The coupling predominantly occurs at the para position to the amino group.

6. What is the most suitable pH of the medium for the conduction of coupling reaction of benzenediazonium chloride with phenol?
a) 2
b) 4
c) 7
d) 10
View Answer

Answer: b
Clarification: The coupling reaction with phenol occurs in a basic medium of pH value approximately 9 to 10. This is to counter the acidic nature of phenol and as a result produce water (by the combination of OH^– and H⁺) during the reaction.

7. Identify the colour of the compound ‘A’ from the following reaction.

a) Yellow
b) Colourless
c) Green
d) Blue
View Answer

Answer: a
Clarification: Diazonium salts on reaction with aniline form azo compounds which are coloured. This is because of the complex system involving the N-N double bond between two aromatic rings. In this case, compound A is p-aminobenzene, which is yellow coloured.

8. The reaction between benzenediazonium chloride and N,N-dimethylaniline, in an acidic medium at 273 K gives an orange coloured compound.
a) True
b) False

Answer: b
Clarification: This reaction yields p-dimethylaminoazobenzene, which is a yellow solid commonly known as methyl yellow. However, in an aqueous solution at low pH, methyl yellow appears red in colour.

9. When a diazonium salt is treated with p-cresol, at which position with respect to the hydroxy group does the coupling occur?
a) ortho
b) meta
c) para
d) coupling does not take place

Answer: a
Clarification: Coupling generally occurs para to the hydroxy group. However in this case, the para position is preoccupied by CH₃ group (cresol), and as a result the coupling takes place at ortho position.

10. What is the product of the following reaction?

a) 2-Amino-4-phenylazophenol
b) 6-Amino-2-phenylazophenol
c) 2-Amino-5-phenylazophenol
d) 2-Amino-3-phenylazophenol

Answer: c
Clarification: The reactant (o-aminophenol) consists of both hydroxy and amino groups. Since the reaction takes place in an acidic medium, the coupling occurs at para position (first preferred) with respect to the amino group. If the reaction was to take place in a basic medium, the coupling would have occurred at para position with respect to hydroxy group.

11. p-Aminophenol on coupling with benzenediazonium chloride in a basic medium gives 4-amino-3-phenylazophenol.
a) True
b) False

Answer: b
Clarification: As the reaction takes place in a basic medium, the coupling will take place at the ortho position (since para position is already occupied by amino group) with respect to OH group. This results in the compound 4-amino-2-phenylazophenol.

 

Chemistry Multiple Choice Questions on “Types of Solutions”.

1. Which of the following is not a solid solution?
a) Brass
b) Bronze
c) Hydrated salts
d) Aerated drinks
Answer: d
Clarification: A solid solution is a solid-state solution of one or more solutes in a solvent. Brass, bronze, and hydrated salts are examples of solid solutions. Aerated drinks are examples of liquid solutions (gas in liquid).

2. The solution of mercury with other metals is called amalgam.
a) True
b) False
Answer: a
Clarification: Alloys of mercury with other metals are called amalgams. An alloy is a type of solid solution (solid in solid). Some important amalgams are zinc amalgam, potassium amalgam, sodium amalgam, aluminium amalgam, and tin amalgam.

3. What is an alloy of copper and zinc called?
a) Bronze
b) German silver
c) Brass
d) Solder
Answer: c
Clarification: An alloy of copper and zinc is called Brass. German silver is an alloy of copper, zinc and nickel, sometimes also containing lead and tin. Bronze is an alloy of copper and tin. Solder is an alloy of tin, lead and antimony.

4. What is camphor in N₂ gas an example of?
a) Solid in liquid solution
b) Liquid in gas solution
c) Solid in gas solution
d) Gas in gas solution
Answer: c
Clarification: Camphor in N₂ gas is an example of solid in gas gaseous solution. A solution in which the solvent is gaseous is called gaseous solution. Some other examples of gaseous solutions are air (O₂ + N₂), Iodine vapours in air, humidity in air, etc.

5. Which of the following is not a copper alloy?
a) Bronze
b) Stainless steel
c) Brass
d) Gunmetal
Answer: b
Clarification: Bronze is an alloy of copper and tin. Stainless steel is an alloy of iron with chromium. It also contains varying amounts of carbon, silicon and manganese. Brass is an alloy of copper and zinc. Gunmetal is an alloy of copper, tin and zinc.

6. A supersaturated solution is not a metastable solution.
a) True
b) False
Answer: b
Clarification: A metastable solution is one which is stable when undisturbed but is capable of reaction if disturbed. Supersaturated solutions are stable when undisturbed and precipitates out crystals of the solute when disturbed. Hence, supersaturated solutions are metastable.

7. Which of the following is a true solution?
a) Salt solution
b) Ink
c) Blood
d) Starch solution
Answer: a
Clarification: A true solution is a homogeneous mixture of two or more materials with a particle size of less than 10^-9 m or 1 nm dissolved in the solvent. Ink, blood and starch solution are colloidal solutions. A simple solution of salt in water is a true solution.

8. What type of solution is Cranberry glass?
a) Emulsion
b) Solid sol
c) Solid aerosol
d) Gel
Answer: b
Clarification: Emulsion, solid sol, solid aerosol and gel are types of colloidal solutions. Cranberry glass is formed by the addition of a solid solute to a solid solvent(gold salts and glass respectively). Hence, it is a Solid sol.

9. What is pumice stone an example of?
a) Solid aerosol
b) Emulsion
c) Liquid aerosol
d) Solid foam
Answer: d
Clarification: There are 8 types of colloidal solutions namely solid sol, sol, solid aerosol, gel, emulsion, liquid aerosol, solid foam and foam. Pumice stone is a gas in solid type colloidal solution, i.e., solid foam.

10. What is the observation on adding a solute crystal to a supersaturated solution?
a) It becomes a colloidal solution
b) The solute dissolves in the solution
c) The solution desaturates
d) The solute precipitates out of the solution
Answer: d
Clarification: When a solute crystal is added to a supersaturated solution, solute particles leave the solution and forms a crystalline precipitate. The addition of the solute crystal is also called seeding.

Chemistry Multiple Choice Questions on “Rate of a Chemical Reaction”.

1. For a second-order reaction, what is the unit of the rate of the reaction?
a) s^-1
b) mol L^-1s^-1
c) mol^-1 L s^-1
d) mol^-2 L² s^-1
Answer: c
Clarification: The unit of the rate of the reaction (k) is (mol L^-1) ^1-n s^-1, where n is the order of the reaction.
For a second-order reaction, n=2
(mol L^-1) ^1-n s^-1 = (mol L^-1)^1-2 s^-1 = mol^-1 L s^-1.

2. The rate constant of a reaction is k=3.28 × 10^-4 s^-1. Find the order of the reaction.
a) Zero order
b) First order
c) Second order
d) Third order
Answer: b
Clarification: Given,
k= 3.28 × 10^-4 s^-1
The general formula to find the units for rate constant, k=(mol L^-1)^1-ns^-1 where n is the order of the reaction. The value of n must be 1 for (mol L^-1)^1-ns^-1 to become s^-1. Therefore, k=3.28 × 10^-4s^-1 represents a first order reaction.

3. For a reaction A +B → C, the experimental rate law is found to be R=k[A]¹[B]^1/2. Find the rate of the reaction when [A] = 0.5 M, [B] = 0.1 M and k=0.03.
a) 4.74 × 10^-2 (L/mol)^1/2 s^-1
b) 5.38 × 10^-2 (L/mol)^1/2 s^-1
c) 5.748 × 10^-2 (L/mol)^1/2 s^-1
d) 4.86 × 10^-2 (L/mol)^1/2 s^-1
Answer: a
Clarification: Given, [A] = 0.5 M, [B] = 0.1 M and k= 0.03
From the rate law it is evident that the order of the reaction is 1+ 0.5 = 1.5 = (frac{3}{2})
Therefore the unit of k= (mol L^-1)^1-1.5 s^-1 = (L/mol)^1/2 s^-1
R= k[A]¹[B]^1/2 = 0.03 × 0.5 × 0.1^1/2 = 4.74 × 10^-2(L/mol)^1/2 s^-1.

4. The reaction NO₂ + CO → NO + CO₂ takes place in two steps. Find the rate law.
2NO₂ → NO + NO₃ (k₁) – slow
NO₃ + CO → CO₂ + NO₂ (k₂) – fast
a) R = k₁ [NO₂]³
b) R = k₂ [NO₃] [CO]
c) R = k₁ [NO₂]
d) R = k₁ [NO₂]²
Answer: d
Clarification: In any reaction the slowest step is the rate determining step, the rate of the overall reaction depends on this step. So, 2NO₂ → NO + NO₃(k1) is the rate determining step. Therefore the rate law R= k₁[NO₂]².

5. For the reaction A + H₂O → products, find the rate of the reaction when [A] = 0.75 M, k= 0.02.
a) 0.077 s^-1
b) 0.085 s^-1
c) 0.015 s^-1
d) 0.026 s^-1
Answer: c
Clarification: Given,
[A] = 0.75 M, k= 0.02
The reaction belongs to pseudo first order reaction so, the unit is s^-1
R= k [A]= 0.02 × 0.75= 0.015 s^-1.

6. What is the rate law for acid hydrolysis of an ester such as CH₃COOC₂H₅ in aqueous solution?
a) k [CH₃COOC₂H₅]
b) k [CH₃COOC₂H₅] [H₂O]
c) k [CH₃COOC₂H₅]²
d) k
Answer: a
Clarification: Acid hydrolysis of ester, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
The order of the reaction may be altered sometimes by taking reactant in excess compared to the other.
The rate law R= k [CH₃COOC₂H₅] [H₂O] however water is present in excess.
So, R= k [CH₃COOC₂H₅].

7. What is the concentration of the reactant in a first order reaction when the rate of the reaction is 0.6 s^-1 and the rate constant is 0.035?
a) 26.667 M
b) 17.143 M
c) 26.183 M
d) 17.667 M
Answer: b
Clarification: Given, R=0.6 s^-1 and k= 0.035
For a first order reaction R= k [A]
[A]=(frac{R}{k}) = (frac{0.6}{0.035}) = 17.143 M.

8. How many times will the rate of the elementary reaction 3X + Y → X₂Y change if the concentration of the substance X is doubled and that of Y is halved?
a) r₂= 4.5r₁
b) r₂= 5r₁
c) r₂= 2r₁
d) r₂= 4r₁
Answer: d
Clarification: Since it is an elementary reaction, its rate law r₁= k [A] ³[B]
When the concentrations are changed the new rate will be r₂= k (2[A])³([B]/2) = 4k[A]³[B]
So, r₂=4r₁.

9. What is the rate law for the reaction C₂H₄ + I₂ → C₂H₄I₂?
a) R= [C₂H₄] [I₂]^3/2
b) R= [C₂H₄] [I₂]³
c) R= [C₂H₄] [I₂]²
d) R= [C₂H₄] [I₂]
Answer: a
Clarification: Fractional order reactions are reaction whose order is a fraction. This reaction is an example of fractional order reaction, where the order of the reaction is (frac{5}{2}).
The rate law for the reaction is known to be R= [C₂H₄] [I₂]^3/2.

10. The rate law for the reaction involved in inversion of cane sugar is R=k [C₁₂H₂₂O₁₁] [H₂O].
a) True
b) False
Answer: b
Clarification: The reaction for the inversion of cane sugar is C₁₂H₂₂O₁₁ + H₂O → glucose + fructose.
In this reaction water is present in excess and belongs to a pseudo first order reaction, even though the molecularity is 2 the order of the reaction is 1 so the rate law R=k[C₁₂H₂₂O₁₁].

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

1. Which of the following metals are reduced from molten salt solutions by electrolysis?
a) Aluminium
b) Silver
c) Iron
d) Zinc

Answer: a
Clarification: Highly electropositive metals are reduced to metals from molten salt solutions by electrolysis. Aluminium being a highly electro positive metal is separated from bauxite by the process known as Hall-Heroult process.

2. What is the process used to extract sodium from halide ores such as sodium chloride (NaCl)?
a) Pyro metallurgy
b) Hydro metallurgy
c) Electro metallurgy
d) Magnetic separation
Answer: c
Clarification: Sodium is a highly electro positive metal and can be easily extracted from sodium chloride through a process known as electro metallurgy or electrolysis. Pyro metallurgy and hydro metallurgy are used to extract metals such as copper, iron, silver etc.

3. What is the term given to the extraction of aluminium from bauxite by using electrochemical process?
a) Baeyer’s process
b) Hall-Heroult process
c) Mc-Arthur process
d) Blasts process
Answer: b
Clarification: Hall-Heroult process is a smelting process to reduce bauxite to a nearly pure aluminium. Bauxite is dissolved in a molten electrolyte composed of sodium, fluorine and aluminium and is maintained at a high temperature. In this process aluminium is accumulated at the cathode and oxygen gas at the anode.

4. What is the anode and cathode used in Hall-Heroult process?
a) Zinc and copper
b) Iron and copper
c) Carbon and zinc
d) Graphite and carbon
Answer: d
Clarification: In the extraction of aluminium from bauxite the anode and cathode must be inert electrodes like graphite and carbon so that they do not take part in the reaction. At the end of the process the metal is accumulated at cathode and it is easier to collect the aluminium from carbon electrode as it does not react with aluminium.

5. In the metallurgy of aluminium, purified bauxite is mixed with Na₂AlF₆.
a) True
b) False
Answer: b
Clarification: In the metallurgy of aluminium, purified bauxite (Al₂O₃) is mixed with Na₃AlF₆ and CaF₂ to lower the melting point and to increase conductivity. This mixture is fused and used as an electrolyte in Hall-Heroult process.

6. What is the net reaction in Hall-Heroult process?
a) 2Al₂O₃ + 3C → 4Al + 3CO₂
b) 4Al₂O₃ + 3C → 4Al + 6CO₂
c) F₂ + Al₂O₃ → Al + F₂O₃
d) Na₃AlF₆ → AlF₃ + 3NaF
Answer: a
Clarification: The reaction involved in Hall-Heroult process are:-
Na₃AlF₆ → AlF₃ + 3NaF
AlF₃ → Al⁺³ + 3F^–
Cathode: Al⁺³ + 3e^– → Al
Anode: 6F^– → 3F₂ + 6e^–
F₂ + Al₂O₃ → AlF₃ + O₂
C + O^2- → CO or CO₂ + 2e^–
The overall reaction may be written as 2Al₂O₃ + 3C → 4Al + 3CO₂.

7. What is the temperature maintained in Hall-Heroult process?
a) 800°-840° C
b) 840°-880° C
c) 900°-940° C
d) 940°-980° C
Answer: d
Clarification: A high temperature of about 1100°C need to be maintained to keep the electrolyte in molten state but mixing the electrolyte with feldspar and cryolite brings down the temperature to about 940° – 980° C and also increases conductivity.

8. Electrochemical processes in metallurgy are spontaneous.
a) True
b) False
Answer: a
Clarification: The Electrochemical reaction can be made spontaneous by keeping the potential difference between the electrodes positive (E°_c > E°_a). Gibbs free energy G° = -nFE°, if E° is positive then Gibbs energy G° is negative making the overall reaction spontaneous.

9. Which of the following is an example of electrochemical principles of metallurgy?
a) Baeyer’s process
b) Solvay process
c) Bergius process
d) Hall-Heroult process
Answer: d
Clarification: Hall-Heroult process is an electrochemical process to reduce bauxite to a nearly pure aluminium. Bauxite is dissolved in a molten electrolyte composed of sodium, fluorine and aluminium and is maintained at a high temperature. In this process aluminium is accumulated at the cathode and oxygen gas at the anode.

10. Which of the following is the most suitable ore to undergo electrochemical process?
a) Kaolinite
b) Calamine
c) Bauxite
d) Cinnabar
Answer: c
Clarification: Bauxite is an ore of aluminium and most suitable ore to undergo electrochemical process. Aluminium is a high electro positive metal and can be extracted from concentrated bauxite by electrolysis and the process is knows as Hall-Heroult process.

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

1. What is the chemical formula structure of ozone?
a) O = O⁺ – O^–
b) O^– – O = O⁺
c) O = O^– – O⁺
d) O^– = O⁺ – O
Answer: a
Clarification: According to the Lewis dot-structure, the central oxygen atom shares a total of three electrons as a consequence of which it bears a positive charge. The oxygen to the right forms tends to attract a shared pair of electrons towards itself, thereby developing the negative charge.

2. “Ozone is a pollutant.” True or false?
a) True
b) False
Answer: a
Clarification: Although ozone protects us from harmful UV rays, at ground level it is a highly toxic and irritating gas. It is referred to as a ‘secondary pollutant’ since it is formed from the combination of two primary pollutants, namely nitrogen oxides (NO_X) and volatile organic compounds (VOCs).

3. Which of the following methods is used to form ozone from oxygen?
a) Standard electric discharge
b) Silent electric discharge
c) Thermal decomposition
d) Heating in an atmosphere of excess oxygen
Answer: b
Clarification: Standard electric discharge produces sparking when electricity is passed through the tube. However, silent electric discharge is more appropriate because the conversion of ozone to oxygen is endothermic. The yield may be reduced by the heat from sparks.

4. As ozone transitions from solid to liquid to gas, what is the notable color change?
a) Pale blue to violet to deep blue
b) Violet to pale blue to deep blue
c) Deep blue to violet to pale blue
d) Violet to deep blue to pale blue
Answer: d
Clarification: Ozone freezes at -192.2°C. When in solid state, it forms a violet-black structure. After melting, the color transitions to deep blue. Finally, as it is everywhere seen in the sky, ozone gas is pale blue.

5. Why is ozone a powerful bleaching agent?
a) It is highly electronegative
b) Nascent oxygen
c) Instability of molecule
d) Large negative Gibbs energy
Answer: b
Clarification: Due to its ease in liberating nascent oxygen i.e. [O], ozone is a powerful oxidizing agent. Although, it is highly thermodynamically unstable and explosive due to large negative Gibbs energy, the ability of liberating nascent oxygen, it is considered to be one of most powerful bleaching agents.