Welcome to Inamjazbi's Chemistry Corner! ๐ Dive into this interactive MCQs test of Class 12 Chemistry for MDCAT/ECAT! ๐ Test your knowledge, sharpen your skills, and learn with fun and engaging quizzes. Get ready to master the fundamentals and boost your score! ๐ฑ๐ก
๐ฅ๐ MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 8 ๐งช๐ก | Learn & Practice!
1. The internal angle in cyclopropane is
Correct:
The bond angle in cyclopropane is 60°, deviates significantly from the optimal angle of 109.5°, so it has very high angle strains.
The sp³-sp³ orbitals can only overlap partially because of the angle deviation, so the overlapping is not as effective as it should be, and as a result, the C-C bond in cyclopropane is relatively weak.
The bond angle in cyclopropane is 60°, deviates significantly from the optimal angle of 109.5°, so it has very high angle strains.
The sp³-sp³ orbitals can only overlap partially because of the angle deviation, so the overlapping is not as effective as it should be, and as a result, the C-C bond in cyclopropane is relatively weak.
2. In Bucky balls the smallest molecule known contains ………..carbon atoms
Correct:
In Bucky balls the smallest molecule known contains 60 carbon atoms.
In Bucky balls the smallest molecule known contains 60 carbon atoms.
3. The condensation polymerization of terephthalic acid (a dicarboxylic acid) with ethylene glycol (a diol) give
Correct:
The condensation polymerization of terephthalic acid (a dicarboxylic acid) with ethylene glycol (a diol) give polyethylene terephthalate (PET), which is a common type of polyester used in various application including textiles as synthetic fibers (like polyester), packaging and beverage containers.
The condensation polymerization of terephthalic acid (a dicarboxylic acid) with ethylene glycol (a diol) give polyethylene terephthalate (PET), which is a common type of polyester used in various application including textiles as synthetic fibers (like polyester), packaging and beverage containers.
4. The nuisance of knocking can be reduced by the process
Correct:
Reforming (Isomerisation) is the process of conversion of hydrocarbons (gasoline) with low octane ratings into those with higher octane ratings that can be used as a fuel in internal combustion engine.
It is the conversion of straight alkanes into branched chain alkanes in order to increase the octane number of a fuel to avoid knocking.
The process of reforming is carried out under high pressure and temperature using platinum catalyst.
Reforming (Isomerisation) is the process of conversion of hydrocarbons (gasoline) with low octane ratings into those with higher octane ratings that can be used as a fuel in internal combustion engine.
It is the conversion of straight alkanes into branched chain alkanes in order to increase the octane number of a fuel to avoid knocking.
The process of reforming is carried out under high pressure and temperature using platinum catalyst.
5. Which of the following pairs of compounds represents isomerism?
Correct:
Structural isomers have same molecular formula but different structural formulae.
Alcohols and ethers are functional group isomers.
Only Option (c) shows isomerism because dimethyl ether and ethanol share the same molecular formula (C₂H₆O) but differ in structure due to different functional groups (functional isomerism).
Structural isomers have same molecular formula but different structural formulae.
Alcohols and ethers are functional group isomers.
Only Option (c) shows isomerism because dimethyl ether and ethanol share the same molecular formula (C₂H₆O) but differ in structure due to different functional groups (functional isomerism).
6. The number of five membered and six membered rings in C60 Bucky ball are respectively
Correct:
In Bucky balls, the carbon atoms are bonded together in twenty hexagons and twelve pentagons which are arranged like the panels on some soccer balls.
๐ Details
๐ฅC₆₀ Buckminsterfullerene (Buckyball) is a spherical molecule made of 60 carbon atoms.
๐ฅIts structure resembles a soccer ball: a truncated icosahedron.
๐ฅThis polyhedron is composed of:
➡️12 pentagons (five membered rings)
➡️20 hexagons (six membered rings)
๐ฅTogether, these polygons form the closed cage structure of C₆₀.
๐ฅA truncated icosahedron =
➡️ Take a 20 triangle solid (icosahedron)
➡️ Chop off the corners
➡️ You get 12 pentagons + 20 hexagons
➡️ Looks exactly like a soccer ball or the C₆₀ buckyball.
In Bucky balls, the carbon atoms are bonded together in twenty hexagons and twelve pentagons which are arranged like the panels on some soccer balls.
๐ Details
๐ฅC₆₀ Buckminsterfullerene (Buckyball) is a spherical molecule made of 60 carbon atoms.
๐ฅIts structure resembles a soccer ball: a truncated icosahedron.
๐ฅThis polyhedron is composed of:
➡️12 pentagons (five membered rings)
➡️20 hexagons (six membered rings)
๐ฅTogether, these polygons form the closed cage structure of C₆₀.
๐ฅA truncated icosahedron =
➡️ Take a 20 triangle solid (icosahedron)
➡️ Chop off the corners
➡️ You get 12 pentagons + 20 hexagons
➡️ Looks exactly like a soccer ball or the C₆₀ buckyball.
7. Which of the following has the highest nucleophilicity?
Correct:
๐ A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond
๐ Nucleophilicity = the ability of a species to donate an electron pair and attack an electrophile (usually a carbon atom).
๐ It depends on charge, electronegativity, size, and solvation.
๐ Electronegativity – “greed for electrons” – is the opposite of nucleophilicity – “giving away electrons.
๐ So less the electronegativity more the nucleophilicity of ion.
๐ Among the given options, CH₃⁻ has the highest nucleophilicity because the negative charge is on carbon, which is less electronegative and less solvated, making it highly reactive toward electrophiles.
➡️ Nucleophilicity decreases as CH₃⁻ > NH₂⁻> OH⁻ > F⁻
๐ฅAcids form conjugate base after donation of proton. In acids, strong acid forms weak conjugate base while weak acid forms strong conjugate base. Order of basic strength (nucleophilicity)
➡️ ๐ CH₃⁻ > NH₂⁻> OH⁻ > F⁻
๐ฅThe acidic character of hydrides in a period of periodic table increases on moving from left to right side. Hence, their acids have the following order of acidic strength
➡️ CH₄ > NH₃ > H₂O > HF.
๐ A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond
๐ Nucleophilicity = the ability of a species to donate an electron pair and attack an electrophile (usually a carbon atom).
๐ It depends on charge, electronegativity, size, and solvation.
๐ Electronegativity – “greed for electrons” – is the opposite of nucleophilicity – “giving away electrons.
๐ So less the electronegativity more the nucleophilicity of ion.
๐ Among the given options, CH₃⁻ has the highest nucleophilicity because the negative charge is on carbon, which is less electronegative and less solvated, making it highly reactive toward electrophiles.
➡️ Nucleophilicity decreases as CH₃⁻ > NH₂⁻> OH⁻ > F⁻
๐ฅAcids form conjugate base after donation of proton. In acids, strong acid forms weak conjugate base while weak acid forms strong conjugate base. Order of basic strength (nucleophilicity)
➡️ ๐ CH₃⁻ > NH₂⁻> OH⁻ > F⁻
๐ฅThe acidic character of hydrides in a period of periodic table increases on moving from left to right side. Hence, their acids have the following order of acidic strength
➡️ CH₄ > NH₃ > H₂O > HF.
8. Homologous series have ……….. chemical characteristics, ………. general methods of preparation, ………… structural features and functional group.
Correct:
๐ Homologous series have
➡️ Same chemical characteristics
➡️ Same general methods of preparation
➡️ Same structural features and functional group.
๐ Homologous series have
➡️ Same chemical characteristics
➡️ Same general methods of preparation
➡️ Same structural features and functional group.
9. Which group elements form nitrides with general formula M₃N₄?
Correct:
๐ฅ Group IVA have oxidation state +4, so they form M⁴⁺ ions.
๐ฅ Nitrogen forms the nitride ion N³⁻.
To balance charges: 3(M⁴⁺) + 4(N³⁻) → M₃N₄ ๐ (The LCM of 3 and 4 = 12)
๐๐ผSo, group IVA elements form nitrides of the type M₃N₄
๐ฅ Group IVA have oxidation state +4, so they form M⁴⁺ ions.
๐ฅ Nitrogen forms the nitride ion N³⁻.
To balance charges: 3(M⁴⁺) + 4(N³⁻) → M₃N₄ ๐ (The LCM of 3 and 4 = 12)
๐๐ผSo, group IVA elements form nitrides of the type M₃N₄
10. Which group elements form superoxides with general formula MO₂?
Correct:
Superoxides contain the ion O₂²⁻.
These are typically formed by larger alkali metals (K, Rb, Cs) of group IA because their larger cations can stabilize the large superoxide anion.
K, Rb, Cs → form superoxides (MO₂) because of their larger cation size, which stabilizes the large O₂²⁻ ion
Superoxides contain the ion O₂²⁻.
These are typically formed by larger alkali metals (K, Rb, Cs) of group IA because their larger cations can stabilize the large superoxide anion.
K, Rb, Cs → form superoxides (MO₂) because of their larger cation size, which stabilizes the large O₂²⁻ ion
11. Which of the following gas is evolved when phosphorus reacts with water?
Correct:
In group VA, phosphorus reacts vigorously with water to produce phosphoric acid and phosphine.
This reaction of phosphorus with water is an auto-redox reaction in which phosphorus is both oxidized and reduced into phosphoric acid (P with +5 oxidation state) and phosphine (P with -3 oxidation state) respectively.
Phosphorus does not react significantly with cold or hot water under normal conditions. However, white phosphorus can react with hot concentrated alkali to produce phosphine (PH₃), but with pure water the reaction is negligible.
In group VA, phosphorus reacts vigorously with water to produce phosphoric acid and phosphine.
This reaction of phosphorus with water is an auto-redox reaction in which phosphorus is both oxidized and reduced into phosphoric acid (P with +5 oxidation state) and phosphine (P with -3 oxidation state) respectively.
Phosphorus does not react significantly with cold or hot water under normal conditions. However, white phosphorus can react with hot concentrated alkali to produce phosphine (PH₃), but with pure water the reaction is negligible.
12. Alkali metals reacts with alcohols liberating hydrogen gas along with
Correct:
๐ Alkali metals react vigorously with alcohols to form metal alkoxide (Ethoxide) with the evolution of hydrogen gas.
This is in fact a displacement reaction in which alkali metal being more reactive displaced hydrogen from alcohol as hydrogen gas along with the formation of respective alkali metal alkoxide (like ethoxide) salt.
Alkaline earth metals have a very limited reactivity with alcohols.
2M(s) + 2C₂H₅OH(aq) → 2C₂H₅OM(aq) + H₂(g) [M = Li, Na, K etc.]
๐ Alkali metals react vigorously with alcohols to form metal alkoxide (Ethoxide) with the evolution of hydrogen gas.
This is in fact a displacement reaction in which alkali metal being more reactive displaced hydrogen from alcohol as hydrogen gas along with the formation of respective alkali metal alkoxide (like ethoxide) salt.
Alkaline earth metals have a very limited reactivity with alcohols.
2M(s) + 2C₂H₅OH(aq) → 2C₂H₅OM(aq) + H₂(g) [M = Li, Na, K etc.]
13. The IUPAC name of CH₃COOCH(CH₃)₂ is
Correct:
๐ The compound is CH₃COOCH(CH₃)₂. The –COO– part shows it is an ester.
The part before –COO– (CH₃CO–) comes from ethanoic acid (acetic acid; CH₃COOH).
The alkoxy part after –O– (CH(CH₃)₂) is isopropyl group comes from isopropyl alcohol (CH(CH₃)₂OH).
So, the ester is named officially as:
Alkyl group from alcohol: isopropyl
Acid-derived part: ethanoate
๐ Final Name → isopropyl ethanoate ✅
(b) Ethyl propanoate → Ethyl group ≠ isopropyl; acid part is propanoate, not ethanoate ❌
(c) isopropyl acetate → Common name, not IUPAC (though it refers to the same compound) ❌
(d) Propyl ethanoate → Propyl group ≠ isopropyl. ❌
๐ The compound is CH₃COOCH(CH₃)₂. The –COO– part shows it is an ester.
The part before –COO– (CH₃CO–) comes from ethanoic acid (acetic acid; CH₃COOH).
The alkoxy part after –O– (CH(CH₃)₂) is isopropyl group comes from isopropyl alcohol (CH(CH₃)₂OH).
So, the ester is named officially as:
Alkyl group from alcohol: isopropyl
Acid-derived part: ethanoate
๐ Final Name → isopropyl ethanoate ✅
(b) Ethyl propanoate → Ethyl group ≠ isopropyl; acid part is propanoate, not ethanoate ❌
(c) isopropyl acetate → Common name, not IUPAC (though it refers to the same compound) ❌
(d) Propyl ethanoate → Propyl group ≠ isopropyl. ❌
14. The derived name of CH₂=CH–CH₂–CO–CH₃ is
Correct:
๐ The general derived name of ketone is alkyl alkyl ketone for unsymmetrical ketone (both alkyl groups are placed alphabetically).
๐ For symmetrical ketone, the name is dialkyl ketone.
๐ The structure CH₂=CH–CH₂–CO–CH₃ contains:
✨ an allyl group → CH₂=CH–CH₂–
✨ a methyl ketone group → –CO–CH₃
๐ So its derived (common) name is allyl methyl ketone.
๐ The molecule may be called vinyl acetone because it is acetone (CH₃–CO–CH₃) with a vinyl substituent (–CH=CH₂) replacing one methyl group.
๐ The general derived name of ketone is alkyl alkyl ketone for unsymmetrical ketone (both alkyl groups are placed alphabetically).
๐ For symmetrical ketone, the name is dialkyl ketone.
๐ The structure CH₂=CH–CH₂–CO–CH₃ contains:
✨ an allyl group → CH₂=CH–CH₂–
✨ a methyl ketone group → –CO–CH₃
๐ So its derived (common) name is allyl methyl ketone.
๐ The molecule may be called vinyl acetone because it is acetone (CH₃–CO–CH₃) with a vinyl substituent (–CH=CH₂) replacing one methyl group.
15. Propanoyl chloride is a member of
Correct:
๐ ✅ Propanoyl chloride is a member of the acid halides family.
✨Propanoyl chloride has the formula CH₃–CH₂–COCl. It is derived from propanoic acid (CH₃–CH₂–COOH) by replacing the –OH group of the carboxyl with a –Cl atom.
✨ Compounds of the type R–COCl are called acid chlorides (or acyl chlorides), which belong to the family of acid halides. ๐
๐ ✅ Propanoyl chloride is a member of the acid halides family.
✨Propanoyl chloride has the formula CH₃–CH₂–COCl. It is derived from propanoic acid (CH₃–CH₂–COOH) by replacing the –OH group of the carboxyl with a –Cl atom.
✨ Compounds of the type R–COCl are called acid chlorides (or acyl chlorides), which belong to the family of acid halides. ๐
16. Select the correct IUPAC name of neopentane.
Correct:
๐ฅ Neopentane (C₅H₁₂) is a branched alkane [(CH₃)₄C] has a central carbon bonded to four methyl groups (a “tetramethylmethane” core).
The longest chain is propane (3 carbon atoms), with two methyl substituents on the second carbon making its IUPAC Name 2,2-dimethylpropane.
The correct IUPAC name of neopentane is 2, 2-dimethylpropane. ๐
๐ฅ Neopentane (C₅H₁₂) is a branched alkane [(CH₃)₄C] has a central carbon bonded to four methyl groups (a “tetramethylmethane” core).
The longest chain is propane (3 carbon atoms), with two methyl substituents on the second carbon making its IUPAC Name 2,2-dimethylpropane.
The correct IUPAC name of neopentane is 2, 2-dimethylpropane. ๐
17. What is the order of priority of following functional groups?
−OH, −NH₂, −COOH, −CHO, −COR
−OH, −NH₂, −COOH, −CHO, −COR
Correct:
๐ฅ The order priority of the functional groups in the IUPAC system of nomenclature is
Carboxylic Acid > Sulfonic Acid > Esters > Acid Halides > Amides > Cyanides > Aldehyde > Ketones > Alcohols > Amines > Alkynes > Alkenes > Alkanes.๐
๐ฅ The order priority of the functional groups in the IUPAC system of nomenclature is
Carboxylic Acid > Sulfonic Acid > Esters > Acid Halides > Amides > Cyanides > Aldehyde > Ketones > Alcohols > Amines > Alkynes > Alkenes > Alkanes.๐
18. The chemical formula Chile saltpeter is
Correct:
๐ Chile saltpeter or Chilean saltpeter is a natural mineral source of sodium nitrate, NaNO₃๐.
It is called so because large deposits were found in Chile. ๐
KNO₃ → Potassium nitrate → common saltpeter, not Chilean saltpeter.
AgNO₃ → Silver nitrate → unrelated
๐ Chile saltpeter or Chilean saltpeter is a natural mineral source of sodium nitrate, NaNO₃๐.
It is called so because large deposits were found in Chile. ๐
KNO₃ → Potassium nitrate → common saltpeter, not Chilean saltpeter.
AgNO₃ → Silver nitrate → unrelated
19. Which of the following compounds reacts slower than benzene in electrophilic substitution?
Correct:
๐ Among the given options, benzaldehyde (C₆H₅–CHO) reacts slower than benzene in electrophilic substitution because the –CHO group withdraws electron density from the ring, making it less reactive.
Aldehydic group of benzaldehyde is a deactivating group.
๐ Thus benzaldehyde (C₆H₅-CHO) is less reactive than benzene undergoes electrophilic substitution slower than benzene✅. ๐ Electrophilic substitution in benzene depends on whether the substituent is electron donating (activating) or electron withdrawing (deactivating).
The –CHO group is strongly electron withdrawing due to the carbonyl. It deactivates the ring toward electrophilic substitution.
๐ Among the given options, benzaldehyde (C₆H₅–CHO) reacts slower than benzene in electrophilic substitution because the –CHO group withdraws electron density from the ring, making it less reactive.
Aldehydic group of benzaldehyde is a deactivating group.
๐ Thus benzaldehyde (C₆H₅-CHO) is less reactive than benzene undergoes electrophilic substitution slower than benzene✅. ๐ Electrophilic substitution in benzene depends on whether the substituent is electron donating (activating) or electron withdrawing (deactivating).
The –CHO group is strongly electron withdrawing due to the carbonyl. It deactivates the ring toward electrophilic substitution.
20. Anhydrous AlCl₃ is used in the Friedal-Craft’s reaction because it is:
Correct:
๐ ๐ In aromatic substitution reaction, Lewis acid catalyst is required which is an electron deficient specie✅.
๐ Anhydrous AlCl₃ is electron deficient because aluminum has only six valence electrons, not a full octet.
๐ This makes it a Lewis acid, allowing it to accept electron pairs and generate the electrophile needed in the Friedel–Crafts reaction.
๐ ๐ In aromatic substitution reaction, Lewis acid catalyst is required which is an electron deficient specie✅.
๐ Anhydrous AlCl₃ is electron deficient because aluminum has only six valence electrons, not a full octet.
๐ This makes it a Lewis acid, allowing it to accept electron pairs and generate the electrophile needed in the Friedel–Crafts reaction.
21. Identify the acyl cation:
Correct:
๐ An acyl cation is a positively charged species formed when the carbonyl group (C=O) loses a leaving group, leaving the carbonyl carbon with a positive charge.
๐ฅ The acyl cation or acylium ion is formulated as RCO⁺, is an electrophilic specie formed as a key intermediate in Friedel–Crafts acylation reactions, generated when an acid chloride (RCOCl) reacts with a Lewis acid catalyst (like AlCl₃).
RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻
๐ [Think "Acylium = Acyl + Ion" → a positively charged acyl group with resonance stabilization].
๐ General formula: R–C≡O⁺ or simply RCO⁺. ๐ (Other given ions are alkyl cation (R⁺), carboxylate ion (RCOO⁺), and sulphonium ion SO₃H⁺).
Details
๐ The acylium ion exhibits resonance stabilization between two major forms: R–C⁺=O ↔ R–C≡O⁺
๐ฅ In the R–C⁺=O form, the positive charge is on the carbon.
๐ฅ In the R–C≡O⁺ form, there's a triple bond between C and O, and the positive charge is delocalized.
๐ This delocalization makes the acylium ion relatively stable for a carbocation and highly electrophilic, ideal for attacking aromatic rings.
๐ An acyl cation is a positively charged species formed when the carbonyl group (C=O) loses a leaving group, leaving the carbonyl carbon with a positive charge.
๐ฅ The acyl cation or acylium ion is formulated as RCO⁺, is an electrophilic specie formed as a key intermediate in Friedel–Crafts acylation reactions, generated when an acid chloride (RCOCl) reacts with a Lewis acid catalyst (like AlCl₃).
RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻
๐ [Think "Acylium = Acyl + Ion" → a positively charged acyl group with resonance stabilization].
๐ General formula: R–C≡O⁺ or simply RCO⁺. ๐ (Other given ions are alkyl cation (R⁺), carboxylate ion (RCOO⁺), and sulphonium ion SO₃H⁺).
Details
๐ The acylium ion exhibits resonance stabilization between two major forms: R–C⁺=O ↔ R–C≡O⁺
๐ฅ In the R–C⁺=O form, the positive charge is on the carbon.
๐ฅ In the R–C≡O⁺ form, there's a triple bond between C and O, and the positive charge is delocalized.
๐ This delocalization makes the acylium ion relatively stable for a carbocation and highly electrophilic, ideal for attacking aromatic rings.
22. Benzene is made up of
Correct:
⚡C–C ฯ bonds: 6,
⚡C–H ฯ bonds: 6,
⚡ฯ bonds: 3 (delocalized over the ring)
⚡Total bonds = 6 (C–C ฯ) + 6 (C–H ฯ) + 3 (ฯ) = 15 bonds๐
⚡C–C ฯ bonds: 6,
⚡C–H ฯ bonds: 6,
⚡ฯ bonds: 3 (delocalized over the ring)
⚡Total bonds = 6 (C–C ฯ) + 6 (C–H ฯ) + 3 (ฯ) = 15 bonds๐
23. The reduction of aldehyde and ketone into saturated hydrocarbon alkane using a mixture of zinc amalgam and concentrated HCl is known as.
Correct:
The reduction of aldehyde and ketone into corresponding saturated hydrocarbon alkane using a mixture of zinc amalgam (Zn/Hg alloy) and concentrated HCl involving the de-oxygenation of aldehydes or ketones and addition of hydrogen is known as Clemmensen reduction. ๐
The Clemmensen reduction is named after a Danish chemist, Erik Christian Clemmensen.
The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface.
The reduction of aldehyde and ketone into corresponding saturated hydrocarbon alkane using a mixture of zinc amalgam (Zn/Hg alloy) and concentrated HCl involving the de-oxygenation of aldehydes or ketones and addition of hydrogen is known as Clemmensen reduction. ๐
The Clemmensen reduction is named after a Danish chemist, Erik Christian Clemmensen.
The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface.
24. Free radical substitution reactions are typically carried out by.
Correct:
Free radical substitution reactions are typically carried out by Homolytic fission involving the Photochemical/Thermal Homolysis of Cl₂ into two chlorine Free Radical, Cl• in the Chain Initiation Step. ๐
Free radical substitution reactions are typically carried out by Homolytic fission involving the Photochemical/Thermal Homolysis of Cl₂ into two chlorine Free Radical, Cl• in the Chain Initiation Step. ๐
25. Find the alkene with maximum stability
Correct:
๐ In disubstituted alkenes, trans isomers are more stable than cis isomers due to steric hindrance.
๐ฅ Also, internal alkenes are more stable than terminal ones. Moreover, stability of alkene is judged by no. of ฮฑ-hydrogens.
๐ฅ More the no. ฮฑ-hydrogen more the will be Hyperconjugation more will be stability of alkene. cis-alkene is less stable than trans-alkene.
๐ฅ In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain.
๐ In disubstituted alkenes, trans isomers are more stable than cis isomers due to steric hindrance.
๐ฅ Also, internal alkenes are more stable than terminal ones. Moreover, stability of alkene is judged by no. of ฮฑ-hydrogens.
๐ฅ More the no. ฮฑ-hydrogen more the will be Hyperconjugation more will be stability of alkene. cis-alkene is less stable than trans-alkene.
๐ฅ In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain.
26. The chemical formula of lithium Ethoxide is
Correct:
๐ Lithium ion = Li⁺,
ethoxide = C₂H₅O⁻
→ ๐ C₂H₅OLi or LiOC₂H₅✅.
๐ Lithium ion = Li⁺,
ethoxide = C₂H₅O⁻
→ ๐ C₂H₅OLi or LiOC₂H₅✅.
27. Which of the following alkali metals imparts lilac colour to the flame?
Correct:
✨ ๐ Potassium’s electrons emit energy in the violet region, which appears as a lilac flame. ✅.
๐ Flame test is used to identify alkali and alkaline earth metals by the characteristic colour they impart to a flame.
๐ Flame test colours come from electrons getting excited by heat ๐ฅ and then releasing energy as visible light when they return to ground state.
๐ Each alkali metal has a unique emission spectrum:
๐ ๐ด ๐ฅLithium (Li) → crimson red
๐ ๐ ๐ Sodium (Na) → bright yellow
๐ ๐ ๐ Potassium (K) → lilac (light purple or pale violet) ✅ ๐
๐ ๐ด ๐Rubidium (Rb) → red-violet
๐ ๐ต ๐❤️Cesium (Cs) → blue/violet (reddish violet)
✨ ๐ Potassium’s electrons emit energy in the violet region, which appears as a lilac flame. ✅.
๐ Flame test is used to identify alkali and alkaline earth metals by the characteristic colour they impart to a flame.
๐ Flame test colours come from electrons getting excited by heat ๐ฅ and then releasing energy as visible light when they return to ground state.
๐ Each alkali metal has a unique emission spectrum:
๐ ๐ด ๐ฅLithium (Li) → crimson red
๐ ๐ ๐ Sodium (Na) → bright yellow
๐ ๐ ๐ Potassium (K) → lilac (light purple or pale violet) ✅ ๐
๐ ๐ด ๐Rubidium (Rb) → red-violet
๐ ๐ต ๐❤️Cesium (Cs) → blue/violet (reddish violet)
28. The basic strength of alkaline earth metals oxides in water _____ from Be to Ba.
Correct:
Short Reason:
๐ The basic strength of alkaline earth metal oxides increases from Be to Ba ๐ผ because metallic character and hydroxide solubility increase down the group.
๐ Details
๐ Alkaline earth metal oxides are of the form MO, where M = Be, Mg, Ca, Sr, Ba.
๐ The oxides of alkaline earth metals of Group IIA (Be, Mg, Ca, Sr, Ba) react with water to form metal hydroxides, which are basic:
MO + H₂O → M(OH)₂
๐ The basic strength depends on how easily the oxide reacts with water and how soluble the hydroxide is.
๐ Down the group (Be → Ba):
⚡Metallic character increases ๐ผ
⚡Ionization energy decreases ⬇️
⚡Electropositivity increases ⬆️
๐ Hydroxides become more soluble and strongly basic ๐ง
๐ Therefore, basic strength increases from BeO (least basic, almost amphoteric) to BaO (strongly basic).
๐ Solubility and degree of dissociation of M(OH)₂ increases down the group (Be → Ba) because lattice energy decreases and hydration energy decreases less rapidly. The metal-oxygen bond becomes more ionic (less covalent), making the oxide more reactive with water. Therefore, the basic strength of the oxides increases down the group.
๐ Down a group: basicity ↑ (Higher electropositivity → more ionic oxides→ more soluble → stronger bases).
๐ Across a period: basicity ↓ (Higher electronegativity → more covalent oxides→ less reactive with water).
Short Reason:
๐ The basic strength of alkaline earth metal oxides increases from Be to Ba ๐ผ because metallic character and hydroxide solubility increase down the group.
๐ Details
๐ Alkaline earth metal oxides are of the form MO, where M = Be, Mg, Ca, Sr, Ba.
๐ The oxides of alkaline earth metals of Group IIA (Be, Mg, Ca, Sr, Ba) react with water to form metal hydroxides, which are basic:
MO + H₂O → M(OH)₂
๐ The basic strength depends on how easily the oxide reacts with water and how soluble the hydroxide is.
๐ Down the group (Be → Ba):
⚡Metallic character increases ๐ผ
⚡Ionization energy decreases ⬇️
⚡Electropositivity increases ⬆️
๐ Hydroxides become more soluble and strongly basic ๐ง
๐ Therefore, basic strength increases from BeO (least basic, almost amphoteric) to BaO (strongly basic).
๐ Solubility and degree of dissociation of M(OH)₂ increases down the group (Be → Ba) because lattice energy decreases and hydration energy decreases less rapidly. The metal-oxygen bond becomes more ionic (less covalent), making the oxide more reactive with water. Therefore, the basic strength of the oxides increases down the group.
๐ Down a group: basicity ↑ (Higher electropositivity → more ionic oxides→ more soluble → stronger bases).
๐ Across a period: basicity ↓ (Higher electronegativity → more covalent oxides→ less reactive with water).
29. Which one of the following metal is a self-protected metal?
Correct:
๐ Magnesium, Beryllium, and Aluminium are self protected metals ๐ก️ because their oxide layers prevent further corrosion.
๐ Self-protected metals are metals that form a strong, thin, adherent, and insoluble oxide layer on their surface when exposed to air, which protects them from further corrosion.
Examples include: Be, Mg, Al, Cr, Ti. ๐
๐ Magnesium, Beryllium, and Aluminium are self protected metals ๐ก️ because their oxide layers prevent further corrosion.
๐ Self-protected metals are metals that form a strong, thin, adherent, and insoluble oxide layer on their surface when exposed to air, which protects them from further corrosion.
Examples include: Be, Mg, Al, Cr, Ti. ๐
30. The chemical formula of normal oxide and superoxides of potassium is respectively:
Correct:
Potassium (K) is an alkali metal (Group 1) which shows fixed oxidation state of +1 (K⁺).
Normal oxide: Group 1 metals generally form normal oxides (O²⁻) with formula M₂O, so Potassium oxide = K₂O. ๐
Superoxide: Alkali metals (K,Rb, Cs) can form superoxides (O₂⁻) with formula MO₂, so Potassium superoxide = KO₂ ๐
Potassium (K) is an alkali metal (Group 1) which shows fixed oxidation state of +1 (K⁺).
Normal oxide: Group 1 metals generally form normal oxides (O²⁻) with formula M₂O, so Potassium oxide = K₂O. ๐
Superoxide: Alkali metals (K,Rb, Cs) can form superoxides (O₂⁻) with formula MO₂, so Potassium superoxide = KO₂ ๐
31. Which of the following alkali metals is the most electropositive with the largest atomic radius?
Correct:
Electropositivity increases down the group in alkali metals due to decreasing ionization energy.
Atomic radius also increases down the group due to more electron shells.
Trends down the group: Atomic radius increases ↓ Li < Na < K < Rb < Cs
Trends down the group: Electropositivity (tendency to lose an electron) increases ↓ Li < Na < K < Rb < Cs
Therefore, the most electropositive alkali metal with the largest atomic radius is Cs (cesium).๐
Electropositivity increases down the group in alkali metals due to decreasing ionization energy.
Atomic radius also increases down the group due to more electron shells.
Trends down the group: Atomic radius increases ↓ Li < Na < K < Rb < Cs
Trends down the group: Electropositivity (tendency to lose an electron) increases ↓ Li < Na < K < Rb < Cs
Therefore, the most electropositive alkali metal with the largest atomic radius is Cs (cesium).๐
32. The property of an alkane which does NOT increase with increase in molar mass:
Correct:
๐ For alkanes, boiling point, melting point, and density increase with molar mass, but solubility does not ๐ซ๐ง. Solubility of alkanes decreases with increase in molar mass. ๐
๐ For alkanes, boiling point, melting point, and density increase with molar mass, but solubility does not ๐ซ๐ง. Solubility of alkanes decreases with increase in molar mass. ๐
33. Which compound has two secondary (2°) carbons?
Correct:
A secondary carbon (2°) is a carbon atom bonded to two other carbons.
๐ n-Butane (CH₃–CH₂–CH₂–CH₃) has two secondary carbons because its two middle CH₂ atoms are each bonded to two carbons.
A secondary carbon (2°) is a carbon atom bonded to two other carbons.
๐ n-Butane (CH₃–CH₂–CH₂–CH₃) has two secondary carbons because its two middle CH₂ atoms are each bonded to two carbons.
34. Ethanol reacts with PCl₃ to form
Correct:
✨ ๐ Alcohols reacts with halogenated agents like PX₅, PX₃, SOX₂ and HX to form respective alkyl halide. ✅.
✨ ๐ Alcohols reacts with halogenated agents like PX₅, PX₃, SOX₂ and HX to form respective alkyl halide. ✅.
35. secondary alcohols undergo oxidation with potassium dichromate to produce carboxylic acid through an intermediate product known as
Correct:
✨ ๐ On oxidation with potassium dichromate, secondary alcohols first form ketone as an intermediate which is further oxidized to lower previous carboxylic acid. ✅.
✨ ๐ On oxidation with potassium dichromate, secondary alcohols first form ketone as an intermediate which is further oxidized to lower previous carboxylic acid. ✅.
36. Oxidative cleavage of 1,2-diol occurs in the presence of
Correct:
๐ Per-iodic acid/periodic acid (HIO₄) is used as an oxidizing agent for Oxidative cleavage of 1,2-diol or vicinal diol.
When a 1,2-diol molecule or vicinal diol is treated with per-iodic acid (periodic acid), an oxidative cleavage occurs between carbon 1 and carbon 2 atoms resulting in the formation of two carbonyl molecules.
The products are determined by the substituents on the diol.
These carbonyl molecules may be aldehyde or ketone depending upon the number of alkyl groups attached to the carbon atom bearing hydroxyl groups.
In this redox reaction, periodic acid (HIO₄) is reduced into iodic acid (HIO₃) by the loss of two electrons. The reaction is selective for 1,2-diols.
The reaction is known as Malaprade oxidation.
This can be used as a functional group test for 1,2-diols.
Example
Ethylene glycol (1,2-ethanediol) undergoes oxidative cleavage with periodic acid giving two molecules of formaldehydes. ๐
๐ Per-iodic acid/periodic acid (HIO₄) is used as an oxidizing agent for Oxidative cleavage of 1,2-diol or vicinal diol.
When a 1,2-diol molecule or vicinal diol is treated with per-iodic acid (periodic acid), an oxidative cleavage occurs between carbon 1 and carbon 2 atoms resulting in the formation of two carbonyl molecules.
The products are determined by the substituents on the diol.
These carbonyl molecules may be aldehyde or ketone depending upon the number of alkyl groups attached to the carbon atom bearing hydroxyl groups.
In this redox reaction, periodic acid (HIO₄) is reduced into iodic acid (HIO₃) by the loss of two electrons. The reaction is selective for 1,2-diols.
The reaction is known as Malaprade oxidation.
This can be used as a functional group test for 1,2-diols.
Example
Ethylene glycol (1,2-ethanediol) undergoes oxidative cleavage with periodic acid giving two molecules of formaldehydes. ๐
37. Which type of isomerism is shown by alcohols?
Correct:
๐ Alcohols can show all the three types of isomerism.
1. Chain isomerism, iso, neo and isoamyl alcohols are chain isomers(primary and tertiary alcohols). ๐
2. Position isomerism; secondary & tertiary alcohols are positional isomers (1-alkanol, 2-alkanol, 3-alkanol etc.). ๐
3. Functional isomerism; Alcohols and ethers are functional isomers. ๐
๐ Alcohols can show all the three types of isomerism.
1. Chain isomerism, iso, neo and isoamyl alcohols are chain isomers(primary and tertiary alcohols). ๐
2. Position isomerism; secondary & tertiary alcohols are positional isomers (1-alkanol, 2-alkanol, 3-alkanol etc.). ๐
3. Functional isomerism; Alcohols and ethers are functional isomers. ๐
38. Carbinol is
Correct:
๐ Methyl alcohol is known as carbinol. ๐
๐ Methyl alcohol is known as carbinol. ๐
39. Which of the following alcohols would be most soluble in water?
Correct:
Propanol is the most soluble in water among given alcohols, as solubility of alcohols in water decreases with increase in molecular mass. ๐
Propanol is the most soluble in water among given alcohols, as solubility of alcohols in water decreases with increase in molecular mass. ๐
40. The structure of an alcohol molecule is identical to that of
Correct:
Both alcohols and water are angular or bent shape molecules.
Both alcohols and water are angular or bent shape molecules.
Result
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