🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 5 🧪💡 | Learn & Practice!

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1. Super glue is chemically named as

a) Cyano acrylate

b) Polyvinyl acetate

c) Epoxy resins

d) Polyurethane

✅ Correct Answer: (a) Cyanoacrylate
📝 Super glue = Ethyl cyanoacrylate (a cyanoacrylate adhesive). Other options (PVA, epoxy, polyurethane) are different adhesive families.
🔎🧠 Stepwise Explanation
🔥Super glue is the common name for a family of fast acting adhesives.
🔥Its main chemical compound is ethyl cyanoacrylate, which polymerizes rapidly in the presence of moisture to form strong bonds.
📌Polyvinyl acetate (PVA): This is the basis of white glue or school glue, not super glue.
📌Epoxy resins: These are two part adhesives (resin + hardener), slower curing, used for structural bonding.
📌Polyurethane: Another type of adhesive, flexible and moisture resistant, but not the instant “super glue.”
🔥 Why Cyanoacrylate?
👉 Bonds instantly to surfaces like metal, plastic, glass, and even skin.
👉 Marketed under trade names like Super Glue and Krazy Glue.
👉 Used in household repairs, industrial applications, and even medical settings (closing wounds).

2. The catalyst used in automobile catalytic converter is a mixture of

a) Pt and Pd

b) Ni and Fe

c) Cu and Cr

d) Pb and Hg

✅ Correct Answer: (a) Pt and Pd
📝 Catalyst in automobile catalytic converter = Mixture of Platinum (Pt), Palladium (Pd), and sometimes Rhodium (Rh).
🔎🧠 Stepwise Explanation
🔥Automobile catalytic converters are devices fitted in exhaust systems to reduce harmful emissions.
🔥They work by catalyzing redox reactions that convert CO, NOx, and unburnt hydrocarbons into less harmful gases like CO₂, N₂, and H₂O.
🔥The catalysts used are precious metals because they are highly effective and resistant to corrosion.
🔥The most common mixture is Platinum (Pt), Palladium (Pd), and sometimes Rhodium (Rh).
🔎Other options:
🔥Ni and Fe: not used in catalytic converters.
🔥Cu and Cr: used in some industrial catalysts, not in automobile converters.
🔥Pb and Hg: toxic metals, not suitable as catalysts.
📌 Why Pt + Pd?
Platinum (Pt): Excellent for oxidizing CO and hydrocarbons.
Palladium (Pd): Effective for oxidizing hydrocarbons.
Rhodium (Rh): Often added for reducing NOx emissions.
Together, they form the three-way catalytic converter system.

3. Highest UV visible absorption energy is required for the transition of:

a) σ → σ*

b) π → π*

c) n → σ*

d) n → π*

✅ Correct Answer: (a) σ → σ*
📝 📌 Quick Recap
The σ → σ* transition requires the highest energy (shortest wavelength) compared to π → π, n → σ, and n → π*.
🔎🧠 Stepwise Reasoning
🔥In UV visible spectroscopy, different electronic transitions require different amounts of energy:
👉 σ → σ*
🔥This involves excitation from a strong sigma bond (like C–H, C–C).
🔥Requires very high energy (short wavelength, < 150 nm).
🔥Usually in the vacuum UV region, not easily observed in routine UV Vis.
👉 π → π*
🔥Excitation from a π bond (like C=C) to its antibonding orbital.
🔥Requires less energy than σ → σ*.
🔥Typically observed in the 200–400 nm range.
👉 n → σ*
🔥Excitation of a nonbonding electron (like lone pairs on O, N, halogens) to σ*.
🔥Requires less energy than σ → σ\, but more than n → π.
👉 n → π*
🔥Excitation of a nonbonding electron to a π* orbital.
🔥Requires the least energy among these transitions.
🔥Often observed in visible or near UV region.

4. The coordination number of cobalt in Na₄[Co(C₂O₄)₃] is:

a) 3

b) 6

c) 4

d) 8

✅ Correct Answer: (b)4
📌🧠 Stepwise Reasoning
🔎The complex is Na₄[Co(C₂O₄)₃].
🔎Inside the coordination sphere: [Co(C₂O₄)₃]⁴⁻.
🔎C₂O₄²⁻ (oxalate ion) is a bidentate ligand → it coordinates through two donor oxygen atoms.
🔎There are 3 oxalate ligands.
🔎Each oxalate contributes 2 coordination sites → total = 3 × 2 = 6.
👉 Therefore, cobalt is surrounded by 6 donor atoms.

5. Ozonide on heating with zinc dust produces

a) Alcohol

b) Aldehyde

c) Alkene

d) Ether

✅ Correct Answer: (b) Aldehyde
👉 Ozonide + Zn/H₂O → Aldehydes and/or ketones. Among the given options, the correct one is Aldehyde.
📌🧠 Stepwise Explanation
🔎Ozonolysis reaction:
🔥 Alkenes react with ozone (O₃) to form ozonides.
🔥 On reductive workup (using zinc dust + water), ozonides decompose to give carbonyl compounds.
🔎The products depend on the alkene:
🔥 If the double bond carbon has hydrogen → aldehyde forms.
🔥 If the double bond carbon is substituted → ketone forms.
🔎Zinc dust prevents further oxidation of aldehydes into acids.

6. Grignard’s reagent with ester produces?

a) Aldehyde

b) Carboxylic acid

c) Ketone

d) Ether

✅ Correct Answer: (c) Ketone
📌 🧠 Stepwise Explanation
🔎Grignard reagent (RMgX) is a strong nucleophile.
🔎When it reacts with an ester (R'COOR''), the mechanism proceeds as:
🔥 First equivalent of Grignard reagent attacks the carbonyl carbon → intermediate formed.
🔥 This intermediate eliminates the alkoxy group (–OR''), giving a ketone.
🔥 If excess Grignard reagent is present, the ketone can undergo a second attack to form a tertiary alcohol after hydrolysis.
👉 But the primary product of ester + Grignard reagent (before excess attack) is a ketone.
📌 Recap of Options
🔎Aldehyde: Not formed directly from esters.
🔎Carboxylic acid: Produced in hydrolysis/oxidation, not here.
🔎Ketone: ✅ Correct — first product.
🔎Ether: Not formed in this reaction.

7. Alkyl amine when reacts with nitrous acid in the presence of hydrochloric acid yields

a) Diazonium salt

b) Aldehyde

c) Ketone

d) Alcohol

✅ Correct Answer: (d) Alcohol
🔎🧠 Stepwise Explanation
👉 Reaction: Alkyl amine + nitrous acid (HNO₂, generated in situ from NaNO₂ + HCl).
👉 Primary aliphatic amines (RNH₂): React with nitrous acid to form unstable aliphatic diazonium salts. These diazonium salts decompose immediately, releasing N₂ gas and forming alcohols (ROH).
👉 Aromatic amines (like aniline): Form relatively stable diazonium salts (Ar–N₂⁺), used in diazo coupling reactions.
🔥 Secondary amines: Give nitrosamines.
🔥 Tertiary amines: React differently, often forming ammonium salts.
📌 Recap of Options
🔥 Diazonium salt: Stable only for aromatic amines, not alkyl amines.
🔥 Aldehyde: Not formed here.
🔥 Ketone: Not formed here.

8. Which one of the following is a triose sugar?

a) Glyceraldehyde

b) Erythrose

c) Ribose

d) Glucose

✅ Correct Answer: (a) Glyceraldehyde
🔎🧠 Stepwise Explanation
👉 Triose sugar = a monosaccharide with 3 carbon atoms.
👉 Glyceraldehyde (C₃H₆O₃): The simplest aldose sugar, with 3 carbons → triose. ✅
📌 Erythrose (C₄H₈O₄): Has 4 carbons → tetrose.
📌 Ribose (C₅H₁₀O₅): Has 5 carbons → pentose.
📌 Glucose (C₆H₁₂O₆): Has 6 carbons → hexose.

9. The formula of erythrose is

a) C₃H₆O₃

b) C₄H₈O₄

c) C₅H₁₀O₅

d) C₆H₁₂O₆

✅ Correct Answer: (b) C₄H₈O₄
🔎👉 Erythrose = C₄H₈O₄ → a tetrose aldose sugar.
🔎🧠 Stepwise Explanation
👉 Erythrose is a tetrose sugar (contains 4 carbon atoms).
👉 General formula for monosaccharides: CₙH₂ₙOₙ.
👉 For erythrose, n=4: → C₄H₈O₄.
🔎 Other options:
📌C₃H₆O₃: Triose sugar (e.g., glyceraldehyde).
📌C₅H₁₀O₅: Pentose sugar (e.g., ribose).
📌C₆H₁₂O₆: Hexose sugar (e.g., glucose).

10. The close chain form of fructose is called

a) Pyranose

b) Pyran

c) Furanose

d) Furan

✅ Correct Answer: (c) Furanose
🔎 🧠 Stepwise Explanation
📌 Fructose is a hexose sugar (C₆H₁₂O₆).
📌 In solution, it does not remain in open chain form; it cyclizes.
📌 Cyclization occurs when the hydroxyl group reacts with the carbonyl group, forming a hemiacetal.
👉 Fructose usually forms a five membered ring structure → called furanose (because it resembles the heterocycle furan). ✅
🔎 Pyranose: Six membered ring (like glucose).
🔎 Pyran: Parent heterocycle, not the sugar form.
🔎 Furan: Parent heterocycle, not the sugar form.

11. They are used as versatile building blocks in the synthesis of pharmaceuticals, agrochemicals, polymers, surfactants:

a) Diazonium salt

b) Epoxide

c) Benzoquinone

d) Ozonide

✅ Correct Answer: (b) Epoxide
👉 Epoxides are reactive cyclic ethers that undergo ring opening reactions, making them versatile intermediates in the synthesis of pharmaceuticals, agrochemicals, polymers, and surfactants.
🔎 🧠 Stepwise Reasoning
📌 Diazonium salts: Mainly used in dye synthesis (azo dyes), not broadly in polymers or surfactants.
📌 Benzoquinone: Used as oxidizing agent, not a general building block for pharma/agrochemicals.
📌 Ozonides: Unstable intermediates in ozonolysis, not widely used as building blocks.
👉 Epoxides: Highly reactive three membered cyclic ethers. They undergo ring opening reactions with nucleophiles, making them versatile intermediates in:
🔥 Pharmaceuticals (drug synthesis)
🔥 Agrochemicals (pesticides, herbicides)
🔥 Polymers (epoxy resins, polyethers)
🔥 Surfactants (non ionic surfactants via ethoxylation)
👉 Clearly, the correct option is Epoxide.

12. The final product of ozonolysis of an alkyne like ethyne is:

a) Diazonium salt

b) Epoxide

c) Glyoxal

d) Ozonide

✅ Correct Answer: (c) Glyoxal
🔎 📌 Ozonolysis of alkynes like ethyne breaks the triple bond and yields glyoxal (dialdehyde) as the final product.
🔎 🧠 Stepwise Reasoning
👉 Ozonolysis = reaction of ozone (O₃) with unsaturated compounds (alkenes/alkynes).
👉 For alkenes, ozonolysis cleaves the double bond → aldehydes/ketones.
👉 For alkynes, ozonolysis cleaves the triple bond → dicarbonyl compounds.
👉 Specifically, ethyne (HC≡CH) on ozonolysis gives glyoxal (OHC–CHO), which is an aldehyde with two carbonyl groups.
🔥 Diazonium salt and epoxide are unrelated to ozonolysis.
🔥 Ozonide is an intermediate, not the final stable product.

13. Primary amines react with aldehydes and ketones to produce:

a) Imine

b) Schiff’s base

c) Both of them

d) None of them

✅ Correct Answer: (c) Both of them
🔎 📌 Short Reason
Primary amines condense with aldehydes/ketones to form imines (C=N compounds), which are also known as Schiff’s bases.
🔎 🧠 Stepwise Reasoning
📌 When a primary amine (R–NH₂) reacts with an aldehyde or ketone, the carbonyl group (C=O) undergoes nucleophilic addition followed by dehydration.
📌 The product is a compound with a C=N double bond (azomethine). This product is called an imine.
📌 Historically, imines formed from aldehydes/ketones with primary amines are also referred to as Schiff’s bases. So, Schiff’s base is essentially a type of imine.
👉 Therefore, the correct answer is Both of them.

14. When amines react with nitrous acid in the presence of hydrochloric acid at below 10°C, they give:

a) Diazonium salt

b) Benzoquinone

c) Imines

d) Amide

✅ Correct Answer: (a) Diazonium salt
🔎 📌 Short Reason
Primary aromatic amines undergo diazotization with nitrous acid + HCl at <10 °C, yielding stable diazonium salts, which are key intermediates in dye synthesis.
🔎 🧠 Stepwise Reasoning
📌 Nitrous acid (HNO₂) is generated in situ from NaNO₂ + HCl.
📌 At low temperature (<10 °C), primary aromatic amines (like aniline) react with nitrous acid to form diazonium salts (Ar–N₂⁺Cl⁻).
📌 This reaction is called diazotization.
🔥 Benzoquinone is unrelated here (it’s an oxidized aromatic compound).
🔥 Imines are formed by condensation of amines with aldehydes/ketones, not with nitrous acid.
🔥 Amides are formed by acylation, not by diazotization.
👉 Therefore, the correct product is diazonium salt.

15. Reduction of carboxylic acid by LiAlH₄ gives:

a) Primary alcohol

b) Alkanes

c) Aldehyde

d) Amine

✅ Correct Answer: (a) Primary alcohol
🔎 📌 Short Reason
👉⚡LiAlH₄ reduces carboxylic acids completely to primary alcohols, not stopping at aldehydes or producing alkanes/amines.
🔎 🧠 Stepwise Reasoning
👉LiAlH₄ (Lithium aluminium hydride) is a strong reducing agent.
👉It reduces carboxylic acids (–COOH) all the way down to primary alcohols (–CH₂OH).
👉Example: R–COOH (LiAlH₄) → R–CH₂OH
📌 Alkanes are not formed directly from carboxylic acids by LiAlH₄.
📌 Aldehydes may appear as intermediates, but LiAlH₄ reduces them further to alcohols.
📌 Amines are formed by reduction of nitro compounds or amides, not carboxylic acids.

16. Reduction of esters by LiAlH₄ gives:

a) Amine

b) Alkanes

c) Aldehyde

d) Primary alcohol

✅ Correct Answer: (d) Primary alcohol
🔎 📌 Short Reason
👉 LiAlH₄ reduces esters completely to primary alcohols, producing one alcohol from each part of the ester molecule.
🔎 🧠 Stepwise Reasoning
📌 LiAlH₄ (Lithium aluminium hydride) is a very strong reducing agent.
📌 Esters (R–COOR′) are reduced by LiAlH₄ in two steps:
📌 The ester is first reduced to an aldehyde intermediate.
📌 The aldehyde is further reduced immediately to a primary alcohol.
📌 The reaction produces two primary alcohols:
⚡One from the acyl part (R–CH₂OH).
⚡One from the alkoxy part (R′–OH).
👉 Example: CH₃COOCH₃ (LiAlH₄) → CH₃CH₂OH+CH₃OH
🔎 Alkanes are not formed.
🔎 Aldehyde is only transient, not the final product.
🔎 Amine formation requires reduction of nitro or amide groups, not esters.

17. Reduction of methyl acetate by LiAlH₄ gives:

a) Ethyl alcohol

b) Methyl alcohol

c) Both of them

d) None of them

✅ Correct Answer: (c) Both of them
🔎📌 Short Reason
👉 LiAlH₄ reduces esters like methyl acetate to two primary alcohols: ethyl alcohol (from the acyl group) and methyl alcohol (from the alkoxy group).
🔎🧠 Stepwise Reasoning
📌 Methyl acetate = CH₃COOCH₃ (an ester).
📌 LiAlH₄ reduces esters completely to two primary alcohols:
⚡One from the acyl part (CH₃CO–) → CH₃CH₂OH (ethyl alcohol).
⚡One from the alkoxy part (–OCH₃) → CH₃OH (methyl alcohol).
👉 So, reduction of methyl acetate gives both ethyl alcohol and methyl alcohol.

18. Reduction of acetic acid by LiAlH₄ gives:

a) Ethyl alcohol

b) Methyl alcohol

c) Both of them

d) None of them

✅ Correct Answer: (a) Ethyl alcohol
🔎📌 Short Reason
👉 LiAlH₄ reduces acetic acid (CH₃COOH) completely to ethyl alcohol (CH₃CH₂OH), not methyl alcohol.
🔎🧠 Stepwise Reasoning
📌Acetic acid = CH₃COOH.
📌LiAlH₄ is a strong reducing agent that reduces carboxylic acids all the way to primary alcohols.
📌The –COOH group is converted into –CH₂OH. So:
📌CH₃COOH (LiAlH₄) → CH₃CH₂OH
📌 The product is ethyl alcohol (CH₃CH₂OH).
🔎 Methyl alcohol (CH₃OH) is not formed here because the carboxyl group belongs to the acetic acid carbon skeleton, not the hydroxyl oxygen. Therefore, only ethyl alcohol is obtained.

19. Tertiary alcohols cannot be oxidized by any oxidizing agent due lack of:

a) Gamma H

b) Beta H

c) Alpha H

d) None of them

✅ Correct Answer: (c) Alpha H
🔎 📌 Short Reason
Tertiary alcohols lack α hydrogen atoms, so they cannot undergo oxidation by common oxidizing agents.
🔎 🧠 Stepwise Reasoning
📌 Oxidation of alcohols generally requires the presence of a hydrogen atom on the carbon adjacent to the –OH group (called the α hydrogen).
📌 Primary alcohols have two α hydrogens → easily oxidized to aldehydes, then acids.
📌 Secondary alcohols have one α hydrogen → oxidized to ketones.
📌 Tertiary alcohols have no α hydrogen (the carbon bearing –OH is bonded to three other carbons, not hydrogens).Without α hydrogen, the oxidation mechanism cannot proceed.
👉 Therefore, tertiary alcohols resist oxidation under normal conditions.

20. The oxidation of ketone involves the breaking of:

a) C–C bond

b) C–H bond

c) C=O bond

d) None of them

✅ Correct Answer: (a) C–C bond
🔎 📌 Short Reason
Oxidation of ketones involves cleavage of the C–C bond adjacent to the carbonyl group, producing carboxylic acids.
🔎 🧠 Stepwise Reasoning
📌Ketones already contain a C=O group (carbonyl).
📌Simple ketones are relatively resistant to mild oxidation because they lack a hydrogen on the carbonyl carbon.
📌Under vigorous oxidation conditions (strong oxidizing agents like KMnO₄, concentrated HNO₃), ketones undergo cleavage of the C–C bond adjacent to the carbonyl group. This bond breaking splits the molecule into smaller carboxylic acids.

21. Vicinal diols are the types of dihydric alcohols containing two hydroxyl groups on position:

a) 1 and 3

b) 1 and 1

c) 1 and 2

d) None of them

✅ Correct Answer: (c) 1 and 2
🔎 📌 Short Reason
Vicinal diols have hydroxyl groups on adjacent carbons (1,2 positions), unlike geminal diols which have both –OH groups on the same carbon.
🔎 🧠 Stepwise Reasoning
📌Dihydric alcohols = alcohols with two –OH groups.
📌Vicinal diols = two –OH groups attached to adjacent carbons. That means positions 1 and 2.
📌Example: Ethylene glycol (HO–CH₂–CH₂–OH) → hydroxyl groups on C 1 and C 2.
⚡1 and 3 would be called geminal diols if on the same carbon, or non vicinal if separated.
⚡1 and 1 means both –OH groups on the same carbon → that’s a geminal diol, not vicinal.

22. Vicinal diols are also called:

a) 1,2-diols

b) 1,3-diols

c) Glycols

d) Both a and c

✅ Correct Answer: (d) Both a and c
🔎 📌 Short Reason
Vicinal diols have hydroxyl groups on adjacent carbons (1,2 diols) and are also known as glycols.
🔎 🧠 Stepwise Reasoning
📌Vicinal diols = dihydric alcohols with two –OH groups on adjacent carbons. That means 1,2 diols.
📌They are also commonly called glycols (general name for compounds with two hydroxyl groups).
👉 Therefore, vicinal diols are correctly described as 1,2 diols and glycols.
📌1,3 diols have hydroxyl groups separated by one carbon, so they are not vicinal.

23. The oxidative cleavage of 1,2 diols by periodic acid gives two molecules of same or different:

a) Glycols

b) Carboxylic acids

c) Carbonyl compounds

d) Both a and c

✅ Correct Answer: (c) Carbonyl compounds
🔎 📌 Short Reason
Periodic acid cleaves vicinal diols into carbonyl compounds (aldehydes or ketones) by breaking the C–C bond between the hydroxyl groups.
🔎 🧠 Stepwise Reasoning
📌 Periodic acid (HIO₄) is a strong oxidizing agent that specifically cleaves vicinal diols (1,2 diols).
📌 The reaction breaks the C–C bond between the two hydroxyl bearing carbons, forming carbonyl compounds (either aldehydes or ketones depending on the substituents).
📌 Example: HO–CH₂–CH₂–OH ((HIO₄)) → 2HCHO
👉 Ethylene glycol → 2 molecules of formaldehyde (a carbonyl compound).
📌 Carboxylic acids are not formed directly in this reaction.
📌 Glycols are the starting materials, not the products.

24. During the oxidative cleavage of 1,2 diols, periodic acid is reduced into:

a) Hypoiodous acid

b) Iodic acid

c) Iodous acid

d) Both a and c

✅ Correct Answer: (b) Iodic acid
🔎 📌 Short Reason
👉 During oxidative cleavage of vicinal diols, periodic acid (HIO₄) is reduced to iodic acid (HIO₃) while the diol is oxidized to carbonyl compounds.
🔎 🧠 Stepwise Reasoning
📌 Periodic acid (HIO₄) oxidizes vicinal diols (1,2 diols) by cleaving the C–C bond between the hydroxyl groups.
📌 In this redox process, periodic acid (HIO₄) is reduced.
👉 The reduction product is iodic acid (HIO₃).
📌 Hypoiodous acid (HIO) and iodous acid (HIO₂) are not formed in this reaction.

25. The oxidative cleavage of ethylene glycol with periodic acid gives:

a) Acetone

b) Formaldehyde

c) Ethyl alcohol

d) Both a and b

✅ Correct Answer: (b) Formaldehyde
🔎 📌 Reason (short)
Periodic acid cleaves ethylene glycol (a 1,2 diol) into two molecules of formaldehyde, not acetone or ethanol.
🔎 🧠 Stepwise Reasoning
📌Ethylene glycol = HO–CH₂–CH₂–OH (a vicinal diol, 1,2 diol).
📌Periodic acid (HIO₄) cleaves vicinal diols by breaking the C–C bond between the hydroxyl groups.
📌Each carbon bearing –OH becomes a carbonyl group.
📌For ethylene glycol: HO–CH₂–CH₂–OH ((HIO₄)) → HCHO+HCHO
👉 The products are two molecules of formaldehyde.
📌Acetone is not formed (it would require a different diol).
📌Ethyl alcohol is not formed (that’s a reduction product, not oxidation).

26. Reduction of aldehydes and ketones by catalytic hydrogenation (H₂) or chemical reduction with reducing agents like LiAlH₄ or NaBH₄ gives:

a) Alcohol

b) Amine

c) Carboxylic acids

d) Both a and c

✅ Correct Answer: (a) Alcohol
🔎 📌 Short Reason
Reduction of aldehydes and ketones (by H₂ or LiAlH₄/NaBH₄) converts the carbonyl group into alcohols — primary from aldehydes, secondary from ketones.
🔎 🧠 Stepwise Reasoning
📌Aldehydes (R–CHO) on reduction → primary alcohols (R–CH₂OH).
📌Ketones (R–CO–R′) on reduction → secondary alcohols (R–CHOH–R′).
👉 Catalytic hydrogenation (H₂/Ni, Pt, Pd) or chemical reduction (LiAlH₄, NaBH₄) both convert the carbonyl group (C=O) into an alcohol group (–OH).
📌Amines are formed by reduction of nitro compounds or amides, not aldehydes/ketones.
📌Carboxylic acids are formed by oxidation, not reduction.

27. Aldehydes on reduction give:

a) Carboxylic acids

b) secondary alcohol

c) primary alcohol

d) Both a and c

✅ Correct Answer: (c) primary alcohol
🔎 📌 Short Reason
Reduction of aldehydes converts the carbonyl group into –CH₂OH, yielding primary alcohols.
🔎 🧠 Stepwise Reasoning
📌Aldehydes (R–CHO) contain a carbonyl group with one hydrogen attached to the carbonyl carbon.
📌On reduction (using H₂/Ni, Pt, Pd or chemical reducing agents like LiAlH₄, NaBH₄):
📌The carbonyl group (C=O) is converted into –CH₂OH. This yields a primary alcohol.
👉 Therefore, aldehydes reduce to primary alcohols only.
📌Secondary alcohols are obtained from ketone reduction, not aldehydes.
📌Carboxylic acids are formed by oxidation of aldehydes, not reduction.

28. Ketones on reduction give:

a) Primary alcohol

b) secondary alcohol

c) Carboxylic acids

d) Both a and c

✅ Correct Answer: (b) secondary alcohol
🔎 📌 Reason (short)
👉 Reduction of ketones converts the carbonyl group into –CHOH, yielding secondary alcohols.
🔎 🧠 Stepwise Reasoning
📌Ketones (R–CO–R′) contain a carbonyl group bonded to two carbon atoms.
📌On reduction (using H₂/Ni, Pt, Pd or chemical reducing agents like LiAlH₄, NaBH₄):
📌The carbonyl group (C=O) is converted into –CHOH.
📌 This yields a secondary alcohol (R–CHOH–R′).
👉 Therefore, ketones reduce to secondary alcohols only.
📌Primary alcohols are obtained from aldehyde reduction, not ketones.
📌Carboxylic acids are formed by oxidation, not reduction.

29. Which ion is resonance stabilized?

a) Phenoxide ion

b) Carboxylate ion

c) Amide ion

d) Both a and b

✅ Correct Answer: (d) Both a and b
🔎 📌 Short Reason
Phenoxide ion delocalizes charge into the aromatic ring, and carboxylate ion delocalizes charge between two oxygens — both are resonance stabilized.
🔎 🧠 Stepwise Reasoning
🔎 Phenoxide ion (C₆H₅O⁻):
📌Negative charge on oxygen can be delocalized into the aromatic ring.
📌Resonance structures spread the charge over ortho and para positions of the ring.
📌Hence, resonance stabilized.
🔎 Carboxylate ion (R–COO⁻):
📌Negative charge is delocalized between the two oxygen atoms.
📌Resonance makes both C–O bonds equivalent.
📌Strong resonance stabilization.
🔎 Amide ion (R–CONH₂⁻):
📌The lone pair on nitrogen is delocalized into the carbonyl group, but the amide ion itself (NH₂⁻) is not resonance stabilized.
📌The neutral amide (R–CONH₂) has resonance, but the free amide ion does not.
👉 Therefore, the ions that are resonance stabilized are phenoxide ion and carboxylate ion.

30. The preparation of benzene by the alkaline hydrolysis of chlorobenzene followed acidification is called

a) Dow’s process

b) Down’s process

c) Cannizaro reaction

d) Both a and b

✅ Correct Answer: (a) Dow’s process
🔎 📌 Short Reason
📌👉Dow’s process = alkaline hydrolysis of chlorobenzene → phenol (after acidification).
📌Down’s process = sodium metal from molten NaCl.
📌Cannizzaro reaction = aldehyde disproportionation.
🔎 🧠 Stepwise Reasoning
🔎 Dow’s process: ✅
📌Chlorobenzene is treated with aqueous NaOH at high temperature and pressure.
📌This gives sodium phenoxide.
📌On acidification, sodium phenoxide yields phenol.
📌In some variations, alkaline hydrolysis followed by acidification is used to prepare benzene derivatives.
🔎 Down’s process:
📌This is entirely different — it refers to the electrolytic process for manufacturing sodium metal from molten NaCl.
📌Not related to benzene preparation.
🔎 Cannizzaro reaction:
📌Involves disproportionation of aldehydes (without α hydrogen) into alcohol and carboxylic acid.
📌Again, not related to benzene.

31. The most stable Carbonium ion among the following is

a) C₆H₅CH₂⁺CH₂

b) CH₃⁺CH₂

c) C₆H₅⁺CH₂

d) C₆H₅⁺CHC₆H₅

✅ Correct Answer: (d) C₆H₅⁺CHC₆H₅
🔎 📌 Short Reason
👉The diphenylmethyl cation (C₆H₅⁺CHC₆H₅) is the most stable because the positive charge is delocalized over two benzene rings, giving maximum resonance stabilization.
🔎 🧠 Stepwise Explanation
🔎 Carbonium ion stability depends on:
📌Resonance stabilization
📌Hyperconjugation
📌Inductive effects
🔎 Option (a) C₆H₅CH₂⁺CH₂:
📌This notation is unusual, but if interpreted as a benzylic carbocation adjacent to another CH₂, it has limited resonance stabilization.
🔎 Option (b) CH₃⁺CH₂ (ethyl cation):
📌A simple primary carbocation, least stable due to lack of resonance or hyperconjugation.
🔎 Option (c) C₆H₅⁺CH₂ (benzyl cation):
📌Highly stabilized by resonance with the aromatic ring.
📌Much more stable than simple alkyl carbocations.
🔎 Option (d) C₆H₅⁺CHC₆H₅ (diphenylmethyl cation): ✅
📌Extremely stable because the positive charge is delocalized over two phenyl rings.
📌Greater resonance stabilization compared to benzyl cation.
📌Hence, most stable among the given options.

32. Calcium reacts with phosphorus to form the ionic compound calcium phosphide. Which ions will this compound contain?

a) Ca²⁺ and P³⁻

b) Ca²⁺ and P⁵⁻

c) Ca²⁺ and P³⁺

d) Ca²⁺ and P⁵⁺

✅ Correct Answer: (a) Ca²⁺ and P³⁻
🔎 📌 Short Reason
👉Calcium forms Ca²⁺ and phosphorus forms P³⁻; together they make Ca₃P₂.
🔎 🧠 Detailed Reason
📌Calcium (Ca): Belongs to Group 2 (alkaline earth metals). It loses two electrons to form the stable cation Ca²⁺.
📌Phosphorus (P): Belongs to Group 15. It needs three electrons to complete its octet, forming the stable anion P³⁻.
📌To balance charges:
3 Ca²⁺ ions (total +6 charge) combine with 2 P³⁻ ions (total –6 charge).
📌Formula of compound = Ca₃P₂ (calcium phosphide).
📌 👉 Therefore, the ions present are Ca²⁺ and P³⁻.✅

33. How many of the following hydroxides is/are amphoteric in character? CsOH, LiOH, Ca(OH)₂, Be(OH)₂, Mg(OH)₂, Sr(OH)₂, Ba(OH)₂, KOH, NaOH.

a) 5

b) 4

c) 1

d) 3

✅ Correct Answer: (c) 1
🔎 📌 Short Reason
👉 Only Be(OH)₂ shows amphoteric behavior; all others are basic. Hence, the answer is 1.
🔎 🧠 Detailed Reason
📌 Amphoteric hydroxides can act as both acids and bases.
📌 Among the given hydroxides:
📌 CsOH, LiOH, KOH, NaOH: Strong bases (alkali hydroxides), not amphoteric.
📌 Ca(OH)₂, Sr(OH)₂, Ba(OH)₂: Alkaline earth hydroxides, basic in nature, not amphoteric.
📌 Mg(OH)₂: Weak base, not amphoteric.
📌 Be(OH)₂: ✅ Amphoteric — reacts with both acids and bases. So, only Be(OH)₂ is amphoteric.

34. The difference of water molecules in gypsum and plaster of Paris is

a) 5/2

b) 1½

c) 2

d) ½

✅ Correct Answer: (b) 1½
🔎 📌 Short Reason
👉 Gypsum has 2H₂O, POP has ½H₂O → difference = 1½ water molecules.
🔎 🧠 Detailed Reason
📌 Gypsum: Chemical formula = CaSO₄·2H₂O (calcium sulphate dihydrate).
📌 Plaster of Paris (POP): Chemical formula = CaSO₄·½H₂O (calcium sulphate hemihydrate).
📌 👉 Difference in water molecules: 2− ½ = 3/2 = 1.5. Hence, the difference = 1½ water molecules.

35. Which of the following alkali metal is the most reactive?

a) Li

b) K

c) Na

d) Cs

✅ Correct Answer: (d) Cs
🔎 📌 Short Reason
👉 Reactivity of alkali metals increases down the group; Cs is the most reactive.
🔎 🧠 Detailed Reason
📌 Reactivity of alkali metals increases down the group in the periodic table.
📌 As we go down the group (Li → Na → K → Cs), the atomic size increases.
📌 The outermost electron is farther from the nucleus and less strongly held.
📌 This makes it easier to lose the electron, hence increasing reactivity.
📌 Lithium (Li): Smallest atom, outer electron strongly held → least reactive.
📌 👉 Cesium (Cs): Largest atom among the given options, outer electron very loosely held → most reactive.

36. Which one of the following hydroxides is the weaker base?

a) NaOH

b) RbOH

c) KOH

d) LiOH

✅ Correct Answer: (d) LiOH
🔎 📌 Short Reason
👉 Basic strength of alkali hydroxides increases down the group; LiOH is the weakest.
🔎 🧠 Detailed Reason
📌 Alkali metal hydroxides (Group 1): All are strong bases, but their basic strength increases down the group.
📌 As we move down the group (Li → Na → K → Rb), the metallic character increases.
📌 Larger cations (like K⁺, Rb⁺) do not hold OH⁻ ions strongly, so they dissociate more easily → stronger base.
📌 Smaller cations (like Li⁺) hold OH⁻ ions more tightly, reducing dissociation → weaker base.
👉 Therefore, LiOH is the weakest base among the given hydroxides.

37. Molecular formula of Glauber's salt is:

a) MgSO₄·7H₂O

b) Na₂SO₄·10H₂O

c) FeSO₄·7H₂O

d) CuSO₄·5H₂O

✅ Correct Answer: (b) Na₂SO₄·10H₂O
🔎 📌 Short Reason
Glauber’s salt = sodium sulfate decahydrate (Na₂SO₄·10H₂O) → correct answer = (b).
🔎 🧠 Logical Reason
🔎 📌 Glauber’s salt is the common name for sodium sulfate decahydrate.
🔎 📌 Formula: Na₂SO₄·10H₂O. ✅
🔎 📌 It is historically known as a laxative and used in medicine and industry.
🔎 📌 Other options:
📌 MgSO₄·7H₂O: Epsom salt.
📌 FeSO₄·7H₂O: Green vitriol (ferrous sulfate heptahydrate).
📌 CuSO₄·5H₂O: Blue vitriol (copper sulfate pentahydrate).

38. Among the alkali metals, the metal with the highest ionization potential is:

a) Li

b) Na

c) Rb

d) Cs

✅ Correct Answer: (a) Li
🔎 📌 Short Reason
Ionization potential decreases down the group; Li has the highest ionization potential among alkali metals.
🔎 🧠 Detailed Reason
📌 Ionization potential (energy required to remove an electron) decreases down the group in alkali metals.
📌 As we move down the group (Li → Na → K → Rb → Cs), the atomic size increases.
📌 The outermost electron is farther from the nucleus and more shielded by inner electrons.
📌 This makes it easier to remove, so ionization potential decreases.
👉 Lithium (Li): Smallest atom, outer electron closest to nucleus, strongly attracted → highest ionization potential.
👉 Cesium (Cs): Largest atom, outer electron farthest → lowest ionization potential.

39. The nitride ion in lithium nitride is composed of ……… protons and ………. electrons.

a) 7, 7

b) 7, 10

c) 10, 7

d) 10, 5

✅ Correct Answer: (b) 7, 10
🔎 🧠 Detailed Reason
📌 Lithium nitride has formula Li₃N.
📌 The nitride ion (N³⁻):
📌 Atomic number of nitrogen = 7 → so it has 7 protons.
📌 Neutral nitrogen atom has 7 electrons.
📌 In N³⁻, it gains 3 extra electrons, so total electrons = 7 + 3 = 10.
👉 Therefore, the nitride ion contains 7 protons and 10 electrons.

40. Which of them has almost same electronegativity?

a) Be, B

b) B, Al

c) K, Na

d) Be, Al

✅ Correct Answer: (d) B, Al
In the periodic table, electronegativity generally increases across a period and decreases down a group.
Both Be and Al have same electronegativity (≈ 1.5) due to diagonal relationship.

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