Category: Class 12 Chemistry Objective Questions

Chemistry Online Quiz for Class 12 on “Amines Chemical Reactions – 3”.

1. If x and y are the pK_b values of p-methylaniline and N,N-dimethylaniline, what is the relation between x and y?
a) x > y
b) x c) x = y
d) x >> y
Answer: a
Clarification: The basicity of N-methyl anilines is very slightly higher than that of ring substituted methyl anilines. This is because the methyl group is directly attached to the N atom in N,N-dimethylaniline and the electron effect is more pronounced. But this is only a minute effect.

2. What is the correct order of basic strength of the following compounds?
a) Aniline > Diphenylaniline > Triphenylaniline
b) Aniline > Triphenylaniline > Diphenylaniline
c) Diphenylaniline > Triphenylaniline > Aniline
d) Triphenylaniline > Diphenylaniline > Aniline

Answer: a
Clarification: When hydrogen atoms of amino groups are replaced by electron withdrawing groups (in this case, phenyl group), the basic character of the resulting aryl amine decreases.

3. Alkylation of amines with alkyl halides proceeds by _________
a) electrophilic addition
b) electrophilic substitution
c) nucleophilic addition
d) nucleophilic substitution
Answer: d
Clarification: The amines itself acts as nucleophiles and attack the alkyl halide at the carbon-halogen bond to substitute the halogen atom by itself. This occurs by production of small amount of acid at each step, which may prevent the reaction by proceeding by protonating the amine.

4. Which of the following is not a likely product of the alkylation of amines with methyl iodide?
a) Primary amine
b) Secondary amine
c) Tertiary amine
d) Quaternary ammonium salt
Answer: a
Clarification: Primary and secondary amines react with alkyl halide to form tertiary amines by nucleophilic substitution. When a primary or secondary amine acts as the nucleophile, a secondary or tertiary amine is generated respectively. Finally, the tertiary amine reacts with the remaining methyl iodide to form quaternary ammonium iodide salt.

5. The alkylation of amines with alkyl halides is carried out in the presence of which of the following?
a) Alcohol
b) Acid
c) Base
d) LiAlH₄
Answer: c
Clarification: Some amount of strong acid is produced at each step of alkylation. This acid can protonate the amine, making the electron lone pair of N unavailable for nucleophilic attack and therefore halt the reaction. For this reason, the acid needs to be neutralised and hence, a base is needed during the reaction.

6. During the acylation of an aliphatic primary amine, which of the following is replaced by an acyl group?
a) NH₂ group
b) One H atom of NH₂
c) Both H atoms of NH₂
d) Alkyl group of the amine
Answer: b
Clarification: Aliphatic and aromatic 1° and 2° amines react with acid chlorides, anhydrides and esters by nucleophilic substitution, where the H atom of NH₂ or NH group is replaced by the acyl group.

7. Which of the following is formed from the reaction between RNHR’ and R”OCl, where R, R’ and R” are three different alkyl groups?
a) RNR’COR”
b) R₂NCOR’
c) R’₂NCOR”
d) R”NR’COR
Answer: a
Clarification: This an acylation reaction in which the H atom of RNHR’ gets replaced by the COR” group form R”OCl. Thus, the amide formed consists of three different alkyl groups with two attached to the N atom and one attached to the carbonyl carbon.

8. The reaction shown below gives ________

a) N,N-dimethylmethanamide
b) N,N-dimethylethanaminde
c) N-methylethanamide
d) no reaction
Answer: d
Clarification: The amine that is treated with CH₃COCl is a tertiary amine (N,N-dimethylmethanamine). It does not undergo reaction with acetyl chloride as it does not have a replaceable hydrogen atom. Therefore, for the acylation of an amine, a H atom on nitrogen is required along with the lone pair of electrons.

9. The reaction between ethanamine and acetyl chloride will give N-ethylethanamide only in the presence of a strong base, like pyridine.
a) True
b) False
Answer: b
Clarification: A base like pyridine may be used in the acylation of amines only to increase the concentration of the product formed. The base removes the HCl formed during the reaction and shifts the equilibrium to the right-hand side.

10. What are the products formed from the reaction between aniline and ethanoic anhydride?
a) N-Phenylethanamide and hydrochloric acid
b) N-Phenylethanamide and acetic acid
c) N-Methylbenzamide and acetic acid
d) N-Methylbenzamide and hydrochloric acid
Answer: b
Clarification: The H atom attached to nitrogen of aniline is replaced by COCH₃ group of ethanoic anhydride. This forms N-phenylethanamide (or acetanilide). The by product (acetic acid) is formed by the attachment of the H atom from aniline with remaining CH₃COO part of ethanoic anhydride.

11. What is the product of the benzoylation of aniline in the presence of aqueous NaOH?
a) N-Ethylethanamide
b) N-Ethylbenzamide
c) N-Phenylethanamide
d) N-Phenylbenzamide
Answer: d
Clarification: Amines react with benzoyl chloride (C₆H₅COCl) in the presence of a base to form benzoyl derivatives in which the C₆H₅CO^– group is introduced. This is called benzoylation reaction.

12. Which of the following amines does not undergo benzoylation?
a) N-Methylethanamine
b) N-Ethylethanamine
c) N,N-Dimethylethanamine
d) N-Methylbenzenamine
Answer: c
Clarification: For benzoylation to take place, there should be a hydrogen atom present at the nitrogen of the amine for the C₆H₅CO^– group to replace. This is not the case in tertiary amines like N,N-dimethylethanamine and hence they do not undergo benzoylation.

13. A certain amine ‘X’ when heated with chloroform and alcoholic KOH, produced no reaction. Identify ‘X’ from the following.
a) Methylamine
b) Ethylamine
c) Ethylmethylamine
d) Aniline
Answer: c
Clarification: When an amine is heated with CHCl₃ and alc. KOH and does not give any reaction, it indicates that it is a secondary or tertiary amine. This is called as carbylamines test and the compound ‘X’ will be ethylmethylamine, as it is 2° and fails the test.

14. Which of the following compounds is not formed during the heating of ethylamine with chloroform in ethanolic KOH?
a) Propanenitrile
b) Ethyl isocyanide
c) Potassium chloride
d) Water
Answer: a
Clarification: Ethylamine on heating with chloroform and ethanolic KOH forms an isocyanide, ethyl isocyanide (CH₃CH₂NC) along with KCl and H₂O. This is known as carbylamines reaction. Propanenitrile (CH₃CH₂CN) is not formed as it is the cyanide form of ethyl isocyanide.

15. Aniline on carbylamines test gives an unpleasant foul odour.
a) True
b) False
Answer: a
Clarification: Aniline is a primary aromatic amine and reacts with chloroform and alc. KOH to give phenyl isocyanide, which is a foul smelling substance.

Chemistry Online Quiz for Class 12,

Chemistry Multiple Choice Questions on “Biomolecules – Nucleic Acids”.

1. RNA on hydrolysis does not yield which of the following?
a) Amino acid
b) Pentose sugar
c) Nitrogen base
d) Phosphoric acid
Answer: a
Clarification: RNA and DNA are nucleic acids which on complete hydrolysis yield a pentose sugar, phosphoric acid and nitrogen containing heterocyclic compounds called bases.

2. The following compound is a component of which of the following?

a) RNA
b) DNA
c) Adenine
d) Guanine
Answer: d
Clarification: The compound shown is β-D-2-deoxyribose which is a pentose sugar. It has no oxygen (in the form of hydroxy) attached to the C₂ carbon. It is the sugar moiety in deoxyribonucleic acid (DNA).

3. DNA is a polynucleotide.
a) True
b) False
Answer: a
Clarification: Chromosomes are made up of proteins and nucleic acids. Nucleic acids are of mainly two types, DNA and RNA. Since they are long chain polymers of nucleotides, they are also called polynucleotides.

4. Which of the following bases is not present in DNA?
a) Adenine
b) Guanine
c) Thymine
d) Uracil
Answer: d
Clarification: DNA contains four bases namely adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also contains four bases, A, T, G and uracil (U).

5. Which of the following bases contain two keto groups?
a) Adenine
b) Guanine
c) Thymine
d) Cytosine
Answer: c
Clarification: Bases are nitrogen containing heterocyclic compounds present in nucleic acids. Adenine has no keto group, guanine and cytosine have one keto group each, and thymine and uracil have two keto groups in their rings.

6. Which sugar is present in RNA?
a) β-D-ribose
b) β-D-fructose
c) β-D-galactose
d) β-D-2-deoxyribose
Answer: a
Clarification: Nucleic acids are made up of units of pentose sugar, base and phosphoric acid that are linked together to form long chain. The sugar moiety in RNA is β-D-ribose.

7. The structure shown is that of a _______

a) DNA nucleoside
b) RNA nucleoside
c) DNA nucleotide
d) RNA nucleotide
Answer: b
Clarification: A unit formed by the linking of a base to C₁ carbon of a pentose sugar is called a nucleoside. The sugar shown is β-D-ribose and the base is uracil, hence it has to belong to RNA.

8. Nucleotides are linked together by bonds between 5’ and 2’ carbons of two different pentose sugars.
a) True
b) False
Answer: b
Clarification: Nucleotides are joined together by phosphodiester linkage between the 5’ and 3’ carbon atoms of the pentose sugars.

9. Which of the following statements is correct?
a) Adenine is a pyrimidine
b) DNA is made of amino acids
c) Nucleosides do not contain phosphorous
d) RNA contains thymine

Answer: c
Clarification: Adenine is a substituted purine as it has two fused rings. DNA is made up of nucleotides. RNA does not contain thymine, instead it has uracil. Nucleosides have a base attached to a pentose sugar. Since both the sugar and base do not contain phosphorous, nucleosides do not have it. On the other hand, nucleotides contain phosphorous in the form of phosphoric acid.

10. Which of the following best describes cytosine?
a) pyrimidine, present in RNA and DNA
b) pyrimidine, present only in DNA
c) purine, present only in RNA
d) purine, present in RNA and DNA
Answer: a
Clarification: Bases (nitrogen containing heterocyclic compounds) are of two types. Those that have a single ring are purines, and those that have two fused rings are pyrimidines. Adenine, guanine and cytosine are present in both RNA and DNA.

11. The attachment between the base and sugar in a nucleotide is through ______ bond.
a) hydrogen
b) peptide
c) phosphodiester
d) glycosidic
Answer: d
Clarification: In nucleotides, the sugar ring is attached to the base by a bond between C₁ carbon of the sugar and the nitrogen atom of the heterocyclic ring. This is a N-glycosidic bond. The phosphate group is bonded to the hydroxyl group of the C₅ carbon of sugar.

12. Which part of the nucleotide is responsible for the formation of bonds in DNA double helix?
a) Base
b) Sugar
c) Phosphate group
d) Hydroxyl group of sugar
Answer: a
Clarification: In a DNA strand, the bases protrude outward whereas the sugar and phosphate lie in a straight chain. The bases pair up in specific pair through hydrogen bonds to form a double helix. Adenine forms two hydrogen bonds with thymine, whereas cytosine forms three hydrogen bonds with guanine.

13. Identify the complementary strand of the DNA primary structure ATGCCGATC.
a) AUGCCGAUC
b) TACGGCTAG
c) UACGGCUAG
d) GATCGGCAT
Answer: b
Clarification: DNA does not contain uracil. Since adenine (A) pairs only with thymine (T) and cytosine (C) pairs with guanine (G), complementary strand will be TACGGCTAG.

14. Two DNA samples A and B have melting temperatures 305K and 320K respectively. Identify the correct statement based on this information.
a) A and B are complementary strands
b) B has more CG pairs than A
c) A has more CG pairs than B
d) A has more hydrogen bonds than B
Answer: b
Clarification: CG bonds are made of triple hydrogen bonds and AT bonds are made of double hydrogen bonds. Since the melting temperature of B is higher than that of A, B should have more hydrogen bonds and as a result more CG pairs.

Chemistry Multiple Choice Questions on “Solid State – Close Packed Structures”.

1. In ______ constituent particles are closely packed leaving the least amount of vacant spaces.
a) plasma
b) liquids
c) solids
d) gases
Answer: c
Clarification: In solids, the constituent particles tend to be closely packed due to strong forces between them. Hence, solids are stable and have a definite shape.

2. What is the coordination number for one-dimensional close packing?
a) 2
b) 1
c) 4
d) 6
Answer: a
Clarification: In one-dimensional close packing, the constituent particles are assumed to be arranged in a row. Thus, each constituent particle comes in contact with 2 of its neighbors and thereby giving the coordination number 2.

3. In how many ways, can the two-dimensional close packed structure be generated?
a) 3
b) 2
c) 1
d) 5
Answer: b
Clarification: Two-dimensional close packed structure can be generated using the one-dimensional close packed structures. This can be done in 2 ways:
• By placing one row of one-dimensional structure below another in such a way that the spheres are one below the another.
• By placing one row of one-dimensional structure below another such that spheres of second row fit in the depressions of the first row.

4. What is the coordination number for a two-dimensional square close packed structure?
a) 8
b) 4
c) 6
d) 2
Answer: b
Clarification: In a two-dimensional square close packed structure, each sphere is in contact with 4 of its adjacent spheres. Hence, it has coordination number 4.

5. What is the coordination number for a two-dimensional hexagonal close packed structure?
a) 4
b) 8
c) 12
d) 6
Answer: d
Clarification: In a two-dimensional hexagonal close packed structure, each sphere is in direct contact with 6 of its adjacent spheres. Hence, it has coordination number 6.

6. Voids in two-dimensional hexagonal close packed structure are ___________ in shape.
a) circular
b) rectangular
c) triangular
d) hexagonal
Answer: c
Clarification: Two-dimensional hexagonal close packed structure are formed when one row of a one-dimensional structure is placed below another in such a way that spheres of second row fit in the depressions of the first row and thereby creating triangular voids between them.

Chemistry Multiple Choice Questions on “Electrochemistry – Nernst Equation”.

1. The standard oxidation potential of Ni/Ni²⁺ electrode is 0.3 V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution with the measured e.m.f. be zero at 25°C? (Assume [Ni²⁺] = 1M)
a) 5.08
b) 4
c) 4.5
d) 5.25
Answer: a
Clarification: Given,
Standard oxidation potential of Ni/Ni²⁺ electrode, E°_OP = 0.3 V,
Ni → Ni²⁺ + 2e
2H⁺ + 2e → H₂
E°_cell = E° _(OP) + E° _(RP)
E° _(cell) = 0.3 + 0.0 = 0.3 V
According to Nernst equation, E (cell) = E°_(cell) + (frac{0.059}{2}) log₁₀([H⁺]² / [Ni]⁺²)
0 = 0.3 + (frac{0.059}{2})log₁₀([H⁺]²)
-log ([H⁺]) = 5.08
pH = 5.08.

2. Calculate the equilibrium constant for the reaction Fe + CuSO₄ ⇌ FeSO₄ + Cu at 25°C.
(Given E°(OP/Fe) = 0.5 V°, E°(OP/Cu) = -0.4 V)
a) 3.46 × 10³⁰
b) 3.46 × 10²⁶
c) 3.22 × 10³⁰
d) 3.22 × 10²⁶
Answer: c
Clarification: The cell reaction shows oxidation of Fe and reduction of Cu²⁺
Therefore for the reaction, Fe + CuSO₄ ⇌ FeSO₄ + Cu
E°_(cell) = E°_(OP/Fe) + E°_(RP/Cu)
E°_(cell) = 0.5 + 0.4 = 0.9 V
We have, E° = (frac{0.059}{2})log₁₀ K_c
0.9 = (frac{0.059}{2})log₁₀ K_c
K_c = 3.22 × 10³⁰.

3. Calculate the e.m.f. of the half-cell given below.
Pt, H₂ | HCl at 1-atmosphere pressure and 0.1 M. Given, E°_(OP) = 2 V.
a) 4 V
b) 5.6 V
c) 3.4 V
d) 5.4 V
Answer: d
Clarification: Given, E°_(OP) = 2 V,
H₂ → 2H⁺ + 2e
E_(OP) = E°_(OP) – (frac{0.059}{2})log₁₀([H⁺]²/ P (H₂))
E_(OP) = 2 – (frac{0.059}{2})log₁₀ (0.02² / 1) = 5.4 V.

4. The equilibrium constant for a cell reaction, Cu_(g) + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag _(s) is 4 × 10¹⁶. Find E° _(cell) for the cell reaction.
a) 0.63 V
b) 0.49 V
c) 1.23 V
d) 3.24 V
Answer: b
Clarification: Given, equilibrium constant K_c = 4 × 10¹⁶
E° _(cell) = (frac{0.059}{2})log₁₀ K_c
E°_(cell) = (frac{0.059}{2})log₁₀ (4 × 10¹⁶)=0.49V.

5. What is the correct Nernst equation for M²⁺ (aq) + 2e⁺ → M (s) at 45°C?
a) E°_(M2+/M) + 0.315log₁₀ (1 / [M]⁺²)
b) E° _(M2+/M) + 0.0425log₁₀ (1 / [M]⁺²)
c) E° _(M2+/M) + 0.0315log₁₀ (1 / [M]⁺²)
d) E° _(M2+/M) + 0.0326log₁₀ (1 / [M]⁺²)
Answer: c
Clarification: Given, Temperature T = 45°C
We know, n (number of electrons transferred) = 2
According to Nernst equation, E_(M2+/M) = E° _(M2+/M) +2.303(frac{RT}{nF})log₁₀ (M / [M]⁺²)
Concentration of [M] is taken to be 1
The equation becomes: E° _(M2+/M) +2.303 (frac{RT}{nF})log₁₀ (1/[M]⁺²)
E_(M2+/M) = E° _(M2+/M) +2.303 × (frac{8.314 times 318}{2 times 96500})log₁₀ (1 / [M]⁺²) =E° _(M2+/M) + 0.0315log₁₀ (1 / [M]⁺²).

6. The e.m.f and the standard e.m.f of a cell in the following reaction is 5 V and 5.06 V at room temperature, Ni_(s) + 2Ag⁺_(n) → Ni²⁺_(0.02M) + 2Ag_(s). What is the concentration of Ag⁺ ions?
a) 0.0125 M
b) 0.0314 M
c) 0.0625 M
d) 0.0174 M
Answer: d
Clarification: Given, Temperature T = 298K
Concentration of Ni²⁺ = (0.02M)
E_(cell) = E°_(cell) – (frac{0.059}{n})log₁₀ (Anode / Cathode)
5 = 5.06 – (frac{0.059}{2})log₁₀(0.02 / [Ag⁺]²)
[Ag⁺]² = 0.0174 M.

7. Calculate the electrode potential of the given electrode.
Pt, Cl_{2(2 bar)}| 2Cl^–_{(0.02 M)}; E°_{(Cl2 | 2Cl^–)} = 3.4 V
a) 3.51 V
b) 3.55 V
c) 1.26 V
d) 2.95 V
Answer: a
Clarification: The electrode reaction is Cl_2(g) + 2e^– → 2Cl^–
E = E˚ – (frac{0.059}{n})log ([Cl^–]² / P_(Cl₂))
E = 3.4 – (frac{0.059}{2})log (0.02² / 2) = 3.51 V.

8. A zinc rod dipped in n molar solution of ZnSO₄ has an electrode potential of -0.56 V. The salt is 98 percent dissociated at room temperature. What is the molarity of the solution? (E°_(Zn+2/Zn) = -0.5 V)
a) 8.44 × 10^-3 M
b) 9.44 × 10^-4 M
c) 8.44 × 10^-4 M
d) 9.44 × 10^-3 M
Answer: d
Clarification: The electrode reaction is Zn²⁺ + 2e^– → Zn
The number of electrons transferred, n= 2
Applying Nernst equation, we get E_(Zn+2/Zn) = E°_(Zn+2/Zn) – (frac{0.059}{2})log (1 / [Zn²⁺])
[Zn²⁺] = (frac{98}{100}) × n = 0.98n M
-0.56 = -0.5 – (frac{0.059}{2})log (1 / 0.98n)
n= 9.44 × 10^-3 M.

9. What is the pH of HCl solution when the hydrogen gas electrode shows a potential of -0.22 V at standard temperature and pressure?
a) 2.17
b) 2.98
c) 3.73
d) 3.14
Answer: c
Clarification: Given, potential of hydrogen gas electrode = -0.22 V
Electrode reaction: H⁺ + e^– → 0.5 H₂
Applying Nernst equation,
E_(H+/H₂) = E°(H+/H₂) – 0.059 log (1/ [H⁺])
E°_(H+/H₂) = 0 for hydrogen gas electrode
-0.22 = 0.059 log H⁺
-0.22 = -0.059pH
pH= 3.73.

10. What is the value of universal gas constant in Nernst equation when the potential is given in volts?
a) 8.314 J mol^-1K^-1
b) 0.0821 L atm mol^-1K^-1
c) 8.205 m³ atm mol^-1K^-1
d) 1.987 cal mol^-1K^-1
Answer: a
Clarification: The universal gas constant is denoted by R and is expressed in units of energy per temperature per mole. Since volts is the SI-Unit of potential, R must also be taken in SI-Units which is J mol^-1K^-1.

11. What is the number of electrons transferred in an equation if the Nernst equation is E_(cell) = E°_(cell) – 9.83 × 10^-3 × log₁₀ (Anode / Cathode)?
a) 2
b) 6
c) 4
d) 1
Answer: b
Clarification: Nernst equation = E°_(cell) – (frac{0.059}{n}) log₁₀(Anode / Cathode)
On comparing both the formulae, (frac{0.059}{n}) = 9.83 × 10^-3
n= 6.

12. Find the number of electrons transferred in the equation Cu_(g) + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag_(s).
a) 4
b) 3
c) 2
d) 1
Answer: c
Clarification: 2Ag⁺_(aq) + 2e^– → 2Ag_(s)
From the equation it is evident that 2Ag⁺ takes 2 electrons from Cu and neutralizes to form 2Ag.

Chemistry Multiple Choice Questions on “Surface Chemistry – Emulsions”.

1. An emulsion is a type of colloid.
a) True
b) False
Answer: a
Clarification: An emulsion is a colloidal dispersion in which both the dispersed phase and the dispersion medium are liquids. It is a suspension of two liquids that usually do not mix. These liquids that do not mix are said to be immiscible.

2. Which of the following is an example of an emulsifier?
a) NaCl
b) CaCO₃
c) C₁₅H₃₁COONa
d) CH₃COOH
Answer: c
Clarification: To obtain a stable emulsion, sometimes, small quantities of certain other substances are added during its preparation. The substances thus added to stabilize the emulsions are called emulsifiers or emulsifying agents.

3. Which of the following is not an example of a water-in-oil emulsion?
a) Cod liver oil
b) Butter
c) Cold cream
d) Milk
Answer: d
Clarification: In an emulsion of water-in-oil (w/o), water is the dispersed phase and oil is the dispersion medium. Cod liver oil, butter and cold cream are all examples of the water-in-oil type of emulsions.

4. Soaps are emulsifying agents.
a) True
b) False
Answer: a
Clarification: Soaps are sodium or potassium salts of higher fatty acids, for example, sodium palmitate (C₁₅H₃₁COONa), sodium stearate (C₁₇H₃₅COONa), etc. A molecule of soap consists of two parts, the hydrophobic part (soluble in oil) and the hydrophilic part (soluble in water).

5. What is the difference between vanishing cream and cold cream?
a) Both are examples of oil-in-water emulsions
b) Vanishing cream is an oil-in-water emulsion whereas cold cream is a water-in-oil emulsion
c) Vanishing cream is a water-in-oil emulsion whereas cold cream is anoil-in-water emulsion
d) Both are examples of water-in-oil emulsions
Answer: b
Clarification: Vanishing cream is an oil-in-water emulsion, that is, oil is the dispersed phase and water is the dispersion medium. Cold cream is a water-in-oil emulsion, that is, water is the dispersed phase and oil is the dispersion medium.

6. Which of the following is not a method to test the type of emulsion?
a) Microscopic method
b) Conductance method
c) Coagulation method
d) Dye method
Answer: c
Clarification: Coagulation or precipitation is a process of aggregating together the colloidal particles to change them into large-sized particles which ultimately settle as a precipitate. The different methods to test the type of emulsion are-microscopic, conductance and dye method.

7. Which of the following statements regarding emulsions is false?
a) Emulsions cannot be separated into their constituent liquids
b) Emulsions show Brownian motion
c) Emulsions show Tyndall effect
d) Emulsions exhibit properties like Electrophoresis and Coagulation
Answer: a
Clarification: Emulsions can be separated into their constituent liquids by boiling, freezing, centrifuging, electrostatic precipitation by adding large amounts of the electrolyte to precipitate out the dispersed phase or by chemical destruction of the emulsifying agent.

8. Which of the following statement about emulsions is true?
a) Oily drugs cannot be prepared in the form of emulsions
b) Digestion does not involve the process of emulsification
c) Disinfectants like Dettol and Lysol give emulsions of water-in-oil type on mixing with water
d) The cleansing action of soap is based upon the formation of water-in-oil emulsion
Answer: b
Clarification: The digestion of fats in the intestines takes place by the process of emulsification. A small amount of the fat reacts with the alkaline solution present in the intestines to form a sodium soap.

9. What is the emulsifier present in milk that makes it stable?
a) Maltose
b) Lactose
c) Lactic acid bacillus
d) Casein
Answer: d
Clarification: The emulsifier or emulsifying agent present in dairy emulsions is Casein. Casein is a slow-digesting dairy protein that people often take as a supplement.The most common form of casein is sodium caseinate.

10. What is the dispersion of a liquid in another liquid called?
a) Gel
b) Foam
c) Emulsion
d) Aerosol
Answer: c
Clarification: The dispersion of a liquid in a solid is called gel. The dispersion of a gas in a liquid medium is called foam. The dispersion of a liquid in another liquid is called emulsion. The dispersion of a solid or liquid in a gaseous medium is called aerosol.

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

1. Which of the following is not true about phosphine?
a) Phosphine is a colorless gas
b) Phosphine has a rotten fish smell
c) Phosphine is inflammable
d) Phosphine is highly poisonous
Answer: c
Clarification: Phosphine is a colorless gar with rotten fish smell and is highly poisonous. In pure state, phosphine is non-flammable but becomes inflammable owing to the presence of P₂H₄ or White phosphorous (P₄) vapours.

2. Phosphine acts as a Lewis acid.
a) True
b) False
Answer: b
Clarification: Phosphine acts as a Lewis base and reacts with acidic species to form phosphinium salts.
PH₃ + HBr → PH₄Br
Phosphine reacts with hydrogen bromide to form phosphinium salts which is a property of a base.

3. Phosphine is prepared from which of the following methods?
a) By reacting calcium phosphide with water or dilute HCl
b) By directly reacting phosphorous with hydrogen
c) By passing dry chlorine over P₂H₄
d) Heating while phosphorous with halides
Answer: a
Clarification: Phosphine is prepared by the reaction of calcium phosphide with water or dilute HCl.
With water: Ca₃P₂ + 6H₂O → 2PH₃ + 3Ca(OH) ₂
With dilute HCl: Ca₃P₂ + 6HCl → 2PH₃ + 3CaCl₂.

4. Which of the following is a property of phosphine?
a) It is insoluble in water
b) It is inflammable
c) The solution of phosphine in water decomposes in the presence of light
d) It acts as a Lewis acid
Answer: c
Clarification: Phosphine is slightly soluble in water and is non-flammable in pure form. It is a Lewis base and the solution of phosphine in water decomposes in the presence of light giving red phosphorous and hydrogen gas.

5. Which of the following gas is used as Holmes signal?
a) Hydrogen per oxide
b) Nitrogen
c) Acetylene
d) Phosphine
Answer: d
Clarification: When a mixture of calcium carbide and calcium phosphide placed in a container is made to react with water, it produces the gases phosphine and acetylene. This signal produced due to the burning gases is called Holmes signal.

6. What is the reaction involved in Holmes signal?
a) Ca₃P₂ + 6H₂O → 2PH₃ + 3Ca(OH) ₂
b) Ca₃P₂ + 6H₂O → 2PH₃ → P₂H₄ + P₄O₆ and CaC₂ + H₂O → C₂H₂
c) Ca₃P₂ + 6HCl → 2PH₃ + 3CaCl₂
d) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
Answer: b
Clarification: The reactions involved in Holmes signal are:
Ca₃P₂ + 6H₂O → 2PH₃ → P₂H₄ + P₄O₆ and CaC₂ + H₂O → C₂H₂
Calcium carbide and calcium phosphide both react with water to give acetylene and phosphine respectively. These gases burn to give the signal.

7. What is the hybridization of phosphine?
a) sp² hybridized
b) sp³ hybridized
c) sp hybridized
d) No hybridization
Answer: d
Clarification: The hybridization of phosphine seems like sp² but in reality the molecule has no hybridization as it forms all bonds using its pure p orbitals. This can be proved from its bong angle data which shows that its bond angles are 93.5°.

8. What is the structure of phosphine?
a) Trigonal pyramidal
b) Trigonal bi-pyramidal
c) Rhombohedral
d) Pyramidal
Answer: a
Clarification: Phosphine has a trigonal pyramidal structure with molecular symmetry. The length of the bond between phosphorous and hydrogen is 1.42 × 10^-10 m and the bond angles are equal and are known to be 93.5°.

9. Which of the following compounds react with water to give phosphine?
a) Phosphorous trichloride
b) Phosphorous pentachloride
c) Black phosphorous
d) Aluminium phosphide
Answer: d
Clarification: Metal phosphides such as aluminium phosphide reacts with water to from phosphine and metal hydroxide.
Reaction: AlP + 3H₂O → PH₃ + Al(OH) ₃.

10. Phosphine like ammonia has very high affinity for water.
a) True
b) False
Answer: b
Clarification: The ammonia molecule has the capability to form a dative bond due to the lone pair of electrons. There is a formation of H-bond due to strong dipole-dipole attraction between ammonia and water molecules where as in phosphine the H-bond is weak and the P-H bond is non-polar so, it is only slightly soluble in water but highly soluble in non-polar solvents.