AIPMT 2007 · Chemistry

AIPMT 2007 Chemistry Questions with Solutions

The AIPMT 2007 paper had 30 Chemistry questions from 12 chapters.

Hydrocarbons had the most questions (5), followed by Chemical Bonding and Molecular Structure and The d- and f-Block Elements with 4 each.

1 question below has the answer and explanation free; the other 29 are in Premium.

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Chemistry questions: 30
Chapters covered: 12
Solved free here: 1 of 30
Easy / Medium / Hard: 3 / 16 / 11

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All 30 AIPMT 2007 Chemistry questions

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Question 1 (AIPMT 2007, Q22)
ElectrochemistryMedium
Two silver rods are dipped separately in 1 M HCl and 1 M HNO₃. In which of the two acids will the silver rod dissolve under standard conditions
1. Option A: Only in 1 M HCl
2. Option B: Only in 1 M HNO₃
3. Option C: In both 1 M HCl and 1 M HNO₃
4. Option D: In neither acid
The answer and explanation for this question are in NEET MIND Premium.
Question 2 (AIPMT 2007, Q23)
EquilibriumMedium
A 0.1 M acetic acid solution ionizes to 1.2%. What is the value of Ka?
1. Option A: 1.44 × 10⁻⁵
2. Option B: 1.46 × 10⁻⁵
3. Option C: 1.60 × 10⁻⁵
4. Option D: 1.20 × 10⁻⁵
The answer and explanation for this question are in NEET MIND Premium.
Question 3 (AIPMT 2007, Q24)
The d- and f-Block ElementsEasy
Why is NH₃ more soluble in water than PH₃?
1. Option A: NH₃ has a larger molecular mass than PH₃.
2. Option B: NH₃ forms hydrogen bonds with water, whereas PH₃ does not.
3. Option C: PH₃ is more polar than NH₃.
4. Option D: PH₃ ionizes completely in water.
The answer and explanation for this question are in NEET MIND Premium.
Question 4 (AIPMT 2007, Q25)
Chemical Bonding and Molecular StructureMedium
Why does BH₃ dimerize whereas BF₃ does not?
1. Option A: BF₃ is electron deficient while BH₃ is not.
2. Option B: BH₃ dimerizes to complete the octet, whereas BF₃ is stabilized by pπ-pπ back bonding.
3. Option C: BF₃ has weaker B–F bonds than BH₃.
4. Option D: BH₃ contains lone pairs on boron.
The answer and explanation for this question are in NEET MIND Premium.
Question 5 (AIPMT 2007, Q26)
Coordination CompoundsHard
In the complex K[PtCl₃(C₂H₄)], has 3 chlorine atoms bonded to platinum. Why is the chlorine atom lying opposite to ethene have higher bond length?
1. Option A: Ethene decreases electron density on platinum.
2. Option B: Ethene exhibits a strong trans effect, weakening the Pt–Cl bond opposite to it.
3. Option C: Chloride ion has a smaller ionic radius.
4. Option D: The Pt–Cl bond opposite ethene is ionic.
The answer and explanation for this question are in NEET MIND Premium.
Question 6 (AIPMT 2007, Q27)
Structure of AtomMedium
An electron in which orbit of lithium will have the same energy as an electron in the second orbit of hydrogen?
1. Option A: n = 3
2. Option B: n = 4
3. Option C: n = 6
4. Option D: n = 8
The answer and explanation for this question are in NEET MIND Premium.
Question 7 (AIPMT 2007, Q28)
Chemical KineticsHard
Also find the value and unit of the rate constant from the data given above.
1. Option A: Order in I₂ = 0, Total order = 2, K = 5.33 × 10² L mol⁻¹ s⁻¹
2. Option B: Order in I₂ = 1, Total order = 2, K = 5.33 × 10² L mol⁻¹ s⁻¹
3. Option C: Order in I₂ = 0, Total order = 3, K = 2.67 × 10² L mol⁻¹ s⁻¹
4. Option D: Order in I₂ = 1, Total order = 3, K = 2.67 × 10² L mol⁻¹ s⁻¹
The answer and explanation for this question are in NEET MIND Premium.
Question 8 (AIPMT 2007, Q29)
Structure of AtomHard
For a hydrogen atom, the frequency is given by ν = 3.3 × 10¹⁵[(1/2²) − (1/n²)]. If the wavelength of the emitted radiation is 6600 Å, what is the value of n?
1. Option A: 3
2. Option B: 4
3. Option C: 5
4. Option D: 6
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Question 9 (AIPMT 2007, Q30)
AminesMedium
Cyclohexanone oxime undergoes Beckmann rearrangement to form compound A, which on polymerization gives Nylon-6. Identify compound A.
1. Option A: Caprolactam
2. Option B: Adipic acid
3. Option C: Hexamethylenediamine
4. Option D: Cyclohexylamine
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Question 10 (AIPMT 2007, Q31)
Organic Chemistry: Some Basic Principles and TechniquesMedium
Which of the following compounds is optically active?
1. Option A: 2-chloro-3-methylpent-1,4-diene
2. Option B: 3-methyl-3-hydroxypentanol
3. Option C: 2-chloro-2-methylbutane
4. Option D: None of these
The answer and explanation for this question are in NEET MIND Premium.
Question 11 (AIPMT 2007, Q32)
HydrocarbonsMedium
Identify the reagent required for the conversion:
1. Option A: H₂/Ni
2. Option B: H₂O, HgSO₄/H₂SO₄
3. Option C: Br₂/CCl₄
4. Option D: KMnO₄
The answer and explanation for this question are in NEET MIND Premium.
Question 12 (AIPMT 2007, Q33)
AminesEasy
Which reagent is used to convert
1. Option A: CHCl₃ and alcoholic KOH
2. Option B: NaNO₂ and HCl
3. Option C: Br₂ and KOH
4. Option D: H₂/Ni
The answer and explanation for this question are in NEET MIND Premium.
Question 13 (AIPMT 2007, Q34)
HydrocarbonsHard
An alkene C₄H₈ reacts with HBr both in the presence and absence of peroxides to give the same product. Identify the alkene.
1. Option A: But-1-ene
2. Option B: 2-Methylpropene
3. Option C: But-2-ene
4. Option D: Cyclobutane
The answer and explanation for this question are in NEET MIND Premium.
Question 14 (AIPMT 2007, Q35)
HydrocarbonsHard
An alcohol having molecular formula C₄H₁₀O is produced on hydration of an alkene with H₂O/H₂SO₄ and is not resolvable into optical isomers. Identify the compound.
1. Option A: Butan-2-ol
2. Option B: 2-Methylpropan-2-ol
3. Option C: Butan-1-ol
4. Option D: 2-Methylpropan-1-ol
The answer and explanation for this question are in NEET MIND Premium.
Question 15 (AIPMT 2007, Q36)
BiomoleculesMedium
The amino acids H₂N-(CH₂)₅-CH(NH₂)-COOH (lysine) and H₂N-CH(COOH)-(CH₂)₃-COOH (glutamic acid) can form how many different dipeptides?
1. Option A: 1
2. Option B: 2
3. Option C: 3
4. Option D: 4
Show answer & explanationHide answer & explanation
Correct answer: (B) 2
Explanation
Two different amino acids can form two distinct dipeptides depending on the sequence: Lys-Glu and Glu-Lys.
Question 16 (AIPMT 2007, Q37)
BiomoleculesMedium
Why does alanine migrate towards the cathode when pH is less than its isoelectric point and towards the anode when pH is greater than its isoelectric point?
1. Option A: Below pI alanine carries a net positive charge, while above pI it carries a net negative charge.
2. Option B: Below pI alanine is neutral, while above pI it becomes positively charged.
3. Option C: Alanine is always positively charged.
4. Option D: Alanine is always negatively charged.
The answer and explanation for this question are in NEET MIND Premium.
Question 17 (AIPMT 2007, Q38)
HydrocarbonsMedium
Which of the following reacts more readily with Br₂ in CS₂ and why?
1. Option A: 1-Butene, because it has less steric hindrance around the double bond.
2. Option B: 2-Butene, because it is more substituted and forms a more stable bromonium ion.
3. Option C: 1-Butene, because it is more symmetrical.
4. Option D: Both react at the same rate.
The answer and explanation for this question are in NEET MIND Premium.
Question 18 (AIPMT 2007, Q39)
HydrocarbonsMedium
Why does 1-butyne give a sodium salt with NaNH₂, whereas 2-butyne does not?
1. Option A: 1-Butyne contains a terminal acidic hydrogen attached to an sp-hybridized carbon, which can be removed by NaNH₂, whereas 2-butyne lacks such a hydrogen.
2. Option B: 2-Butyne is more acidic than 1-butyne and therefore does not react with NaNH₂.
3. Option C: 1-Butyne contains a double bond while 2-butyne contains a triple bond.
4. Option D: NaNH₂ reacts only with symmetrical alkynes and not with unsymmetrical alkynes.
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Question 19 (AIPMT 2007, Q40)
BiomoleculesEasy
Draw the structures for DNA purines?
1. Option A: Option (A)
2. Option B: Option (B)
3. Option C: Option (C)
4. Option D: Option (D)
The answer and explanation for this question are in NEET MIND Premium.
Question 20 (AIPMT 2007, Q41)
EquilibriumHard
For a 0.5 M H₂SO₃ solution, Kₐ₁ = 1.8 × 10⁻² and Kₐ₂ = 8.3 × 10⁻⁵. Find the concentrations of H⁺, HSO₃⁻ and SO₃²⁻.
1. Option A: [H⁺] ≈ 8.7 × 10⁻² M, [HSO₃⁻] ≈ 8.7 × 10⁻² M, [SO₃²⁻] ≈ 8.3 × 10⁻⁵ M
2. Option B: [H⁺] ≈ 1.8 × 10⁻² M, [HSO₃⁻] ≈ 1.8 × 10⁻² M, [SO₃²⁻] ≈ 8.3 × 10⁻⁵ M
3. Option C: [H⁺] ≈ 5.0 × 10⁻¹ M, [HSO₃⁻] ≈ 5.0 × 10⁻¹ M, [SO₃²⁻] ≈ 8.3 × 10⁻⁵ M
4. Option D: [H⁺] ≈ 8.7 × 10⁻² M, [HSO₃⁻] ≈ 5.0 × 10⁻¹ M, [SO₃²⁻] ≈ 4.1 × 10⁻⁵ M
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Question 21 (AIPMT 2007, Q42)
EquilibriumHard
N₂O₄ dissociates according to N₂O₄(g) ⇌ 2NO₂(g) with degree of dissociation α = 0.4. Establish the relation between Kχ and Kp and calculate Kp if the total pressure is 1 atm.
1. Option A: Kp = KχP = 0.229 atm
2. Option B: Kp = Kχ/P = 0.229 atm
3. Option C: Kp = KχP² = 0.114 atm
4. Option D: Kp = Kχ/P² = 0.114 atm
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Question 22 (AIPMT 2007, Q43)
Some Basic Concepts of ChemistryHard
One mole of N₂ and four moles of H₂ react to form NH₃ in a 20 L vessel. Ten litres of water are then added and the vessel is shaken thoroughly. What is the pressure of the residual gases assuming complete absorption of NH₃ by water?
1. Option A: 1.23 atm
2. Option B: 2.46 atm
3. Option C: 3.69 atm
4. Option D: 4.92 atm
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Question 23 (AIPMT 2007, Q44)
The d- and f-Block ElementsMedium
Why is F₂ more reactive than Cl₂?
1. Option A: F–F bond is weaker than Cl–Cl bond due to strong lone pair repulsions.
2. Option B: F₂ has a stronger bond than Cl₂.
3. Option C: Fluorine has a larger atomic size than chlorine.
4. Option D: Cl₂ has higher electronegativity than F₂.
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Question 24 (AIPMT 2007, Q45)
The d- and f-Block ElementsHard
Why is CrO₄²⁻ a stronger oxidizing agent than MoO₄²⁻?
1. Option A: Chromium more readily attains lower oxidation states than molybdenum.
2. Option B: Molybdenum has a smaller atomic size than chromium.
3. Option C: CrO₄²⁻ is less stable than MoO₄²⁻ and is more easily reduced.
4. Option D: Both A and C
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Question 25 (AIPMT 2007, Q46)
Chemical Bonding and Molecular StructureMedium
Out of (SiH₃)₂O and (CH₃)₂O, which is more basic and why?
1. Option A: (SiH₃)₂O, because Si is less electronegative than C and increases electron density on oxygen.
2. Option B: (CH₃)₂O, because carbon is less electronegative than silicon.
3. Option C: (SiH₃)₂O, because oxygen forms stronger hydrogen bonds.
4. Option D: (CH₃)₂O, because oxygen has no lone pair interaction.
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Question 26 (AIPMT 2007, Q47)
Coordination CompoundsHard
The empirical formula of an insoluble compound is PtCl₂·(NH₃)₂. On treatment with AgNO₃ it gives [Pt(NH₃)₄]Cl₂ and Ag₂[PtCl₄]. What is the molecular formula of the compound?
1. Option A: [Pt(NH₃)₄]²⁺ [PtCl₄]²⁻
2. Option B: [Pt(NH₃)₂Cl₂]
3. Option C: [Pt(NH₃)₃Cl]Cl
4. Option D: [Pt(NH₃)₄]Cl₄
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Question 27 (AIPMT 2007, Q48)
Chemical Bonding and Molecular StructureMedium
Out of trimethylamine (N(CH₃)₃) and triethylphosphine (P(C₂H₅)₃), which has the higher dipole moment?
1. Option A: Trimethylamine
2. Option B: Triethylphosphine
3. Option C: Both have the same dipole moment
4. Option D: Cannot be predicted
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Question 28 (AIPMT 2007, Q49)
The d- and f-Block ElementsMedium
Why do PO₄³⁻ ions exist whereas NO₄³⁻ ions do not?
1. Option A: Phosphorus can expand its octet using vacant 3d orbitals, whereas nitrogen cannot.
2. Option B: Nitrogen is larger than phosphorus.
3. Option C: Phosphorus is more electronegative than nitrogen.
4. Option D: Nitrogen has vacant d-orbitals.
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Question 29 (AIPMT 2007, Q50)
Chemical Bonding and Molecular StructureMedium
Why is B₂ paramagnetic whereas C₂ is diamagnetic?
1. Option A: B₂ contains two unpaired electrons in π2p orbitals, whereas all electrons are paired in C₂.
2. Option B: C₂ contains more unpaired electrons than B₂.
3. Option C: B₂ has a higher bond order than C₂.
4. Option D: C₂ contains one unpaired electron.
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Question 30 (AIPMT 2007, Q51)
Coordination CompoundsHard
For octahedral complexes, explain the spin magnetic moments acquired by d⁵ and d⁶ metal ions when Δ₀ > P (strong-field) and Δ₀ < P (weak-field).
1. Option A: d⁵: 1 BM (strong-field), 5.92 BM (weak-field); d⁶: 0 BM (strong-field), 4.90 BM (weak-field)
2. Option B: d⁵: 5.92 BM (strong-field), 1 BM (weak-field); d⁶: 4.90 BM (strong-field), 0 BM (weak-field)
3. Option C: d⁵: 0 BM (strong-field), 5.92 BM (weak-field); d⁶: 2.83 BM (strong-field), 4.90 BM (weak-field)
4. Option D: d⁵: 5.92 BM (strong-field), 0 BM (weak-field); d⁶: 0 BM (strong-field), 4.90 BM (weak-field)
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