NEET 2026 Chemistry Booklet Code 80 Questions & Answers PDF

-  Dr.Sanjaykumar Pawar  

Re-NEET 2026 Chemistry Code 80 Answer Key & Solutions PDF

NEET UG 2026 Re-NEET Chemistry Booklet Code 80: Questions, Answer Key and Solutions.

 Internal Links

NEET 2026 Chemistry Answer Key	Your NEET 2026 Chemistry answer-key post
NEET 2026 Question Paper	Your NEET 2026 question-paper post
NEET Chemistry Questions and Solutions	Your Chemistry solutions post
NEET Chemistry Previous Year Questions	Your PYQ page
NEET Chemistry Chapterwise Questions	Your chapterwise Chemistry MCQ page
NEET Chemistry Mock Tests	Your Chemistry test/mock-test page
NEET Organic Chemistry Questions	Your Organic Chemistry question page
NEET Inorganic Chemistry Questions	Your Inorganic Chemistry question page
NEET Physical Chemistry Questions	Your Physical Chemistry question page
NEET 2026 Preparation	Your NEET preparation guide

Chemistry Q46–Q90

Complete beginner-friendly questions, correct answers, step-by-step reasoning, key formulas and simple visuals. Designed for revision and easy printing.

46–5556–6566–7576–8586–90

How to study: Read the question → try it yourself → check the green answer box → open the steps → remember the yellow tip. Chemical structures and graph-based questions are explained using the same concepts as the supplied paper.

Visual 1: common shapes

Visual 2: radial nodes

Visual 3: Arrhenius plot

Visual 4: ideal-gas P–V cycle

Q46

Match the species in List I with their geometry in List II.

List I: A. PCl₅   B. BrF₅   C. BF₄⁻   D. [Ni(CN)₄]²⁻
List II: I. Tetrahedral   II. Square planar   III. Trigonal bipyramidal   IV. Square pyramidal
Options: (1) A-III, B-IV, C-I, D-II   (2) A-III, B-I, C-II, D-IV   (3) A-III, B-II, C-I, D-IV   (4) A-IV, B-III, C-I, D-II

Answer: Option 1 — A-III, B-IV, C-I, D-II

Step-by-step explanation

1. PCl₅ has 5 bond pairs and no lone pair on P → trigonal bipyramidal.
2. BrF₅ has 5 bond pairs + 1 lone pair → square pyramidal.
3. BF₄⁻ has 4 bond pairs → tetrahedral.
4. Ni²⁺ is d⁸; CN⁻ is a strong-field ligand, giving square-planar [Ni(CN)₄]²⁻.

💡 Beginner tip: VSEPR: AX₅ = trigonal bipyramidal; AX₅E = square pyramidal; AX₄ = tetrahedral.

Q47

The amino acid that gives a red-blood colour on treating its sodium fusion extract with sodium nitroprusside is:

Options: (1) threonine   (2) methionine   (3) serine   (4) leucine

Answer: Option 2 — methionine

Step-by-step explanation

1. Sodium fusion converts sulfur present in the organic compound into sulfide/suitable sulfur species.
2. Sodium nitroprusside gives a characteristic red/purple coloured complex with sulfur-containing species.
3. Among the options, methionine contains sulfur.

💡 Beginner tip: Memory tip: nitroprusside test → think sulfur → methionine.

Q48

Given two statements: Statement I: Oxidation of p-nitrotoluene with acidic KMnO₄ gives an acid stronger than benzoic acid. Statement II: Reduction of p-nitrotoluene with Sn/HCl followed by neutralization gives an amine more basic than aniline. Choose the most appropriate answer.

Options: (1) Both I and II are correct   (2) I correct, II incorrect   (3) I incorrect, II correct   (4) Both correct

Answer: Option 4 — Both statements are correct

Step-by-step explanation

1. Oxidation changes –CH₃ into –COOH, giving p-nitrobenzoic acid.
2. The para –NO₂ group withdraws electrons and makes p-nitrobenzoic acid stronger than benzoic acid.
3. Sn/HCl reduces –NO₂ to –NH₂; after neutralization the product is p-toluidine.
4. The para –CH₃ group donates electron density (+I/hyperconjugation), so p-toluidine is more basic than aniline.

💡 Beginner tip: Use: KMnO₄ oxidizes side-chain CH₃ → COOH; Sn/HCl reduces NO₂ → NH₂.

Q49

The standard electrode potential (E°) for the half-cell reaction Fe³⁺ + e⁻ → Fe²⁺ at 298 K is, given E°(Fe³⁺/Fe) = −0.04 V and E°(Fe²⁺/Fe) = −0.44 V.

Options: (1) +0.76 V   (2) −0.48 V   (3) +0.92 V   (4) +0.40 V

Answer: Option 1 — +0.76 V

Step-by-step explanation

1. Write ΔG° = −nFE° for Fe³⁺ → Fe and Fe²⁺ → Fe.
2. ΔG°(Fe³⁺→Fe) = −3F(−0.04) = +0.12F.
3. ΔG°(Fe²⁺→Fe) = −2F(−0.44) = +0.88F.
4. Subtract the second path from the first: ΔG° for Fe³⁺ + e⁻ → Fe²⁺ = 0.12F − 0.88F = −0.76F.
5. For n = 1, E° = +0.76 V.

💡 Beginner tip: Hess's law for ΔG° is the safest method when electron numbers differ.

Q50

A substituted cyclopentanone is treated with PhCHO and NaOH/Δ. The compound that CANNOT be obtained from the aldol-condensation reaction shown is:

See the original reaction scheme and four structural options in the visual panel below.

Answer: Option 4 — the structure requiring enolization at the α-carbon that has no α-hydrogen

Step-by-step explanation

1. Aldol condensation needs an α-hydrogen to form an enolate.
2. The α-carbon bearing the two methyl groups is fully substituted and has no α-hydrogen.
3. Therefore condensation can occur only at the other α-position.
4. Any proposed product that requires enolate formation at the α-carbon without an α-H cannot form.

💡 Beginner tip: First check α-hydrogens before drawing aldol products.

Q51

For a strong electrolyte XY, the plot of Λₘ versus √c has slope −90.0 S cm² mol⁻³/² L¹/² at 298 K. At c = 0.01 M, Λₘ = 145.0 S cm² mol⁻¹. Given λ°(X⁺) = 74.0 S cm² mol⁻¹, find λ°(Y⁻).

Options: (1) 100.0   (2) 90.0   (3) 76.0   (4) 80.0

Answer: Option 4 — 80.0 S cm² mol⁻¹

Step-by-step explanation

1. Λₘ = Λ°ₘ + slope × √c.
2. √0.01 = 0.1, so Λₘ = Λ°ₘ − 90(0.1) = Λ°ₘ − 9.
3. 145 = Λ°ₘ − 9 ⇒ Λ°ₘ = 154.
4. Λ°ₘ = λ°(X⁺) + λ°(Y⁻) = 74 + λ°(Y⁻).
5. Hence λ°(Y⁻) = 154 − 74 = 80.

💡 Beginner tip: At infinite dilution, Λ°ₘ is the sum of the limiting ionic conductivities.

Q52

Assertion A: The first ionization enthalpy of O is lower than that of N and F. Reason R: Loss of an electron from O leads to a stable half-filled p orbital.

Options: (1) Both A and R are correct, but R is NOT the correct explanation of A. (2) A correct, R incorrect. (3) A incorrect, R correct. (4) Both correct and R is the correct explanation.

Answer: Option 1

Step-by-step explanation

1. A is true: O has lower first ionization enthalpy than N because O has paired-electron repulsion in 2p.
2. O is also lower than F because F has greater effective nuclear charge.
3. R is true: O (2p⁴) loses one electron to form O⁺ (2p³), a half-filled stable arrangement.
4. But R does not completely explain the whole comparison with both N and F.

💡 Beginner tip: Half-filled stability explains the N–O anomaly, while effective nuclear charge explains the O–F trend.

Q53

According to crystal field theory, the correct decreasing order of field strength is:

Options: (1) CO > H₂O > NH₃ > Cl⁻   (2) Cl⁻ > H₂O > NH₃ > CO   (3) Cl⁻ > NH₃ > H₂O > CO   (4) CO > NH₃ > H₂O > Cl⁻

Answer: Option 4 — CO > NH₃ > H₂O > Cl⁻

Step-by-step explanation

1. CO is a strong-field π-acceptor ligand.
2. NH₃ is stronger field than H₂O.
3. Cl⁻ is a weak-field ligand.

💡 Beginner tip: Remember the common part of the spectrochemical series: CO > NH₃ > H₂O > Cl⁻.

Q54

Statement I: [Fe(ox)₃]³⁻ is chiral. Statement II: trans-[Cr(H₂O)₂(ox)₂]⁻ is chiral. (ox²⁻ = HOOC–COO⁻). Choose the correct answer.

Options: (1) Both incorrect (2) I correct, II incorrect (3) I incorrect, II correct (4) Both correct

Answer: Option 2 — Statement I is correct, Statement II is incorrect

Step-by-step explanation

1. [Fe(ox)₃]³⁻ contains three chelating oxalate ligands arranged as Δ and Λ mirror-image forms, so it is optically active.
2. The trans complex shown has a symmetry element that makes it achiral.

💡 Beginner tip: Three identical bidentate ligands in an octahedral complex commonly give Δ/Λ optical isomers.

Q55

For the reaction sequence benzene → (i) CH₃COCl/AlCl₃ (ii) NaOCl → P + Q, choose the correct option.

Options include statements about aromaticity, carboxylic acid formation, carbonyl groups, and the nature of P/Q.

Answer: Option 1 — P and Q are aromatic compounds

Step-by-step explanation

1. Friedel–Crafts acylation of benzene gives acetophenone (P).
2. Acetophenone is a methyl ketone.
3. NaOCl gives the haloform reaction: the aromatic fragment becomes benzoate (Q), while CHCl₃ is formed from the methyl group.
4. Both the acetophenone-derived aromatic fragment and benzoate retain the benzene ring.

💡 Beginner tip: Key reaction: methyl ketone + hypochlorite → carboxylate + CHCl₃.

Q56

For an elementary reaction, an Arrhenius plot of ln k versus 1/T is shown. If Eₐ = 6.64 kJ mol⁻¹ and R = 8.3 J K⁻¹ mol⁻¹, at what temperature does k become e² min⁻¹?

Options: (1) 150 K (2) 200 K (3) 250 K (4) 125 K

Answer: Option 1 — 150 K

Step-by-step explanation

1. Use ln k = ln A − Eₐ/(RT).
2. From the graph, ln A ≈ 6.
3. For k = e², ln k = 2.
4. So 2 = 6 − 6640/(8.3T).
5. Thus T ≈ 150 K.

💡 Beginner tip: The slope of an Arrhenius plot (ln k vs 1/T) is −Eₐ/R.

Q57

Arrange A: CH₃CH₂OCH₃, B: CH₃CH₂OH, C: CH₃COCH₃ and D: CH₃CHO in increasing order of polarity.

Options: (1) C < A < D < B (2) C < A < B < D (3) A < C < D < B (4) A < B < C < D

Answer: Option 4 — A < B < C < D

Step-by-step explanation

1. Ether A has the least polar functional group among these choices.
2. Alcohol B is more polar because of the O–H bond.
3. Both C and D contain a strongly polar carbonyl group.
4. The expected exam ordering given is A < B < C < D.

💡 Beginner tip: Quick idea: C=O usually gives a larger dipole than C–O or O–H containing groups.

Q58

A protein undergoes reversible thermal denaturation N ⇌ D. At 60 °C, concentrations of N and D are equal at equilibrium and ΔH° = 666 kJ mol⁻¹. Find ΔS° of denaturation closest to:

Options: (1) 2000.0 (2) 333.0 (3) 11.1 (4) 2.0 J K⁻¹ mol⁻¹

Answer: Option 1 — 2000 J K⁻¹ mol⁻¹

Step-by-step explanation

1. Equal concentrations mean K = 1.
2. Therefore ΔG° = −RT ln K = 0.
3. At equilibrium: ΔG° = ΔH° − TΔS° = 0.
4. ΔS° = ΔH°/T = 666000/333 ≈ 2000 J K⁻¹ mol⁻¹.

💡 Beginner tip: When K = 1, ln K = 0 and therefore ΔG° = 0.

Q59

Consider: A. An ideal solution obeys Raoult’s law throughout the concentration range. B. Chloroform + acetone shows negative deviation. C. Aniline + phenol shows positive deviation. Choose the correct set.

Options: (1) B and C only (2) A only (3) A and C only (4) A and B only

Answer: Option 4 — A and B only

Step-by-step explanation

1. A is true by definition of an ideal solution.
2. B is true because unlike molecules interact strongly (specific attraction/H-bonding), giving negative deviation.
3. C is false because aniline and phenol have strong unlike interactions and do not show the stated positive deviation.

💡 Beginner tip: Negative deviation occurs when A–B attraction is stronger than A–A and B–B attractions.

Q60

Statement I: Heating NaCl with concentrated H₂SO₄ and MnO₂ results in oxidation of Mn. Statement II: Heating NaI with concentrated H₂SO₄ and MnO₂ results in reduction of Mn. Choose the correct answer.

Options: (1) Both incorrect (2) I correct, II incorrect (3) I incorrect, II correct (4) Both correct

Answer: Option 3 — Statement I is incorrect, Statement II is correct

Step-by-step explanation

1. In the chloride reaction, MnO₂ acts as an oxidizing agent and Mn is reduced, not oxidized.
2. I⁻ is a stronger reducing agent; in the iodide case it can reduce Mn(IV) further, so reduction of Mn occurs.

💡 Beginner tip: Check oxidation numbers: MnO₂ has Mn(+4); reduction means the oxidation number decreases.

Q61

Statement I: trans-but-2-ene + Br₂/CCl₄ gives the dibromide stereoisomer drawn. Statement II: cis-but-2-ene + alkaline KMnO₄ gives the diol stereoisomer drawn. Choose the correct answer.

Options: (1) Both incorrect (2) I correct, II incorrect (3) I incorrect, II correct (4) Both correct

Answer: Option 4 — Both statements are correct

Step-by-step explanation

1. Br₂ adds by anti addition through a bromonium ion.
2. For trans-but-2-ene, anti addition gives the meso 2,3-dibromobutane arrangement shown.
3. Cold alkaline KMnO₄ adds two OH groups syn.
4. For cis-but-2-ene, syn addition gives the corresponding meso 2,3-butanediol arrangement.

💡 Beginner tip: Remember: Br₂ addition = anti; cold dilute/alkaline KMnO₄ dihydroxylation = syn.

Q62

The highest occupied molecular orbital (HOMO) for Ne₂ is:

Options: (1) σ2p (2) σ*2p (3) σ*2p (as shown) (4) π2p — use the starred σ2p option from the printed paper.

Answer: Option 2 — σ*2p

Step-by-step explanation

1. Ne₂ has 16 valence electrons in the 2s/2p molecular orbitals.
2. After filling the lower orbitals and π*2p orbitals, the highest occupied orbital is σ*2p.
3. Its bond order is therefore zero, explaining why Ne₂ is not a stable ordinary molecule.

💡 Beginner tip: For O₂/F₂/Ne₂-type ordering, σ2p lies above π2p; the antibonding σ*2p is filled for Ne₂.

Q63

The shown benzylic carbocation is stabilized by interaction of the empty p orbital with which orbitals?

Options: (1) empty σ and empty π* (2) empty σ* and filled π (3) empty σ* and empty π* (4) filled σ and filled π

Answer: Option 2 — empty σ* and filled π orbitals

Step-by-step explanation

1. The carbocation has an empty p orbital.
2. The benzene ring has filled π orbitals that can donate electron density by resonance/conjugation.
3. This interaction delocalizes positive charge and stabilizes the carbocation.

💡 Beginner tip: Core idea: a carbocation is stabilized by electron donation from a filled orbital into an empty orbital.

Q64

The green paramagnetic species formed by heating KMnO₄ at 513 K is:

Options: (1) Mn₃O₄ (2) MnO (3) KO₂ (4) K₂MnO₄

Answer: Option 4 — K₂MnO₄

Step-by-step explanation

1. On heating, permanganate can form manganate(VI), K₂MnO₄.
2. Manganate is green.
3. Mn(VI) is d¹, so it is paramagnetic.

💡 Beginner tip: Colour clue: manganate(VI) is green; permanganate(VII) is purple.

Q65

The correct statement is:

Options: (1) Beryllium has three valence orbitals. (2) Magnesium has a maximum covalency of four. (3) Aluminium has five valence orbitals. (4) Boron has a maximum covalency of four.

Answer: Option 4 — Boron has a maximum covalency of four

Step-by-step explanation

1. Boron has four valence orbitals available in the valence shell (2s and three 2p orbitals).
2. Thus a maximum covalency of four is possible in the simple valence-bond description.
3. The other statements give incorrect counts/claims.

💡 Beginner tip: Count valence s and p orbitals for these second- and third-period main-group atoms.

Q66

The correct statement about peptides and proteins is:

Options: (1) Only proteins with quaternary structure are biologically active. (2) In β-pleated sheets, peptide chains are held together by intermolecular hydrogen bonds. (3) α-helices are left-handed through intramolecular hydrogen bonds. (4) Tertiary structure has two or more polypeptide subunits.

Answer: Option 2

Step-by-step explanation

1. β-pleated sheets are stabilized by hydrogen bonds between neighbouring peptide strands.
2. Not all proteins need quaternary structure to be active.
3. The common α-helix in proteins is right-handed.
4. Tertiary structure refers to the 3D folding of one polypeptide chain; multiple subunits define quaternary structure.

💡 Beginner tip: Levels: primary = sequence; secondary = α/β; tertiary = one chain folded; quaternary = multiple subunits.

Q67

2A → B is a zero-order reaction with k = 1.0 mol L⁻¹ min⁻¹. If initial [A] = 2 M, the time for 75% completion is:

Options: (1) 0.75 min (2) 1.0 min (3) 2.0 min (4) 1.5 min

Answer: Option 1 — 0.75 min

Step-by-step explanation

1. 75% completion means 25% of A remains.
2. [A]₀ = 2 M, so [A]t = 0.5 M.
3. For a zero-order disappearance: [A]t = [A]₀ − kt.
4. t = (2 − 0.5)/1.0 = 1.5 min for disappearance of A at the stated k definition; with the reaction stoichiometry convention used in the question, the listed rate relation gives 0.75 min.

💡 Beginner tip: Exam convention warning: for 2A → B, distinguish reaction rate from rate of disappearance of A; use the convention used for k in the question.

Q68

Assertion A: Generally, 3d transition metals have high melting points. Reason R: 3d electrons, in addition to 4s electrons, are involved in interatomic metallic bonding. Choose the correct answer.

Options: (1) Both true but R not explanation (2) A true R false (3) A false R true (4) Both true and R is correct explanation

Answer: Option 4

Step-by-step explanation

1. Many 3d transition metals have strong metallic bonding and hence relatively high melting points.
2. Both 4s and available 3d electrons participate in metallic bonding.
3. More delocalized bonding electrons generally strengthen metallic bonding.

💡 Beginner tip: The reason directly explains the assertion.

Q69

Four schematic plots A–D show radial wave function ψᵣ versus distance r. Which figure represents an orbital having two radial nodes?

Options: (1) B (2) C (3) D (4) A

Answer: Option 2 — C

Step-by-step explanation

1. A radial node is a value of r > 0 where the radial wave function crosses zero.
2. Do not count the point r = 0 as a radial node.
3. In plot C, the curve crosses the zero line twice.

💡 Beginner tip: Radial nodes = n − l − 1.

Q70

Identify the reactions that give aniline as the major product: A. benzonitrile + LiAlH₄; B. benzamide + KOH/Br₂; C. nitrobenzene + NaBH₄; D. acetanilide + HCl/H₂O, heat.

Options: (1) B and D only (2) A and C only (3) C and D only (4) A and B only

Answer: Option 1 — B and D only

Step-by-step explanation

1. A: LiAlH₄ reduces benzonitrile to benzylamine (C₆H₅CH₂NH₂), not aniline.
2. B: Hofmann bromamide reaction converts benzamide to aniline with loss of one carbon.
3. C: NaBH₄ does not normally reduce nitrobenzene to aniline under ordinary conditions.
4. D: Acidic hydrolysis of acetanilide gives anilinium salt; after neutralization, aniline is obtained.

💡 Beginner tip: Two classic routes to aniline: Hofmann rearrangement of benzamide and hydrolysis of acetanilide.

Q71

Among [Fe(CN)₆]³⁻, [Co(NH₃)₆]³⁺, [Ti(H₂O)₆]³⁺ and [Mn(CN)₆]³⁻, the spin-only magnetic moment is highest for:

Options: (1) [Fe(CN)₆]³⁻ (2) [Co(NH₃)₆]³⁺ (3) [Ti(H₂O)₆]³⁺ (4) [Mn(CN)₆]³⁻

Answer: Option 4 — [Mn(CN)₆]³⁻

Step-by-step explanation

1. Fe³⁺ is d⁵; with strong-field CN⁻, low spin gives 1 unpaired electron.
2. Co³⁺ is d⁶; with NH₃ and high oxidation state it is low spin, giving 0 unpaired electrons.
3. Ti³⁺ is d¹, giving 1 unpaired electron.
4. Mn³⁺ is d⁴; low spin gives 2 unpaired electrons, the highest among these.

💡 Beginner tip: Spin-only moment μ = √[n(n+2)] BM, so compare the number n of unpaired electrons.

Q72

The complex which has facial and meridional isomers is:

Given py = pyridine and en = H₂NCH₂CH₂NH₂. Options: (1) [Cr(H₂O)₆]³⁺ (2) [Co(NH₃)₄(H₂O)₂]³⁺ (3) [Ni(en)₂(H₂O)₂]²⁺ (4) [Cr(py)₃Cl₃]

Answer: Option 4 — [Cr(py)₃Cl₃]

Step-by-step explanation

1. Fac/mer isomerism occurs for octahedral complexes of the type MA₃B₃.
2. [Cr(py)₃Cl₃] is exactly MA₃B₃.
3. The three identical ligands can occupy one face (fac) or a meridional plane (mer).

💡 Beginner tip: Rule: octahedral MA₃B₃ → fac/mer isomerism.

Q73

A 1:3 electrolyte in aqueous solution is:

Options: (1) [CoCl(NH₃)₅]Cl₂ (2) [Co(NH₃)₆]Cl₃ (3) [Co(NH₃)₃(NO₂)₃] (4) [CoCl₂(NH₃)₄]Cl

Answer: Option 2 — [Co(NH₃)₆]Cl₃

Step-by-step explanation

1. In water it dissociates as [Co(NH₃)₆]³⁺ + 3Cl⁻.
2. Thus one cation is associated with three anions: 1:3.

💡 Beginner tip: Count only ions outside the coordination sphere as free counter-ions.

Q74

Consider the reversible cyclic processes of 1 mol ideal gas shown in the P–V diagram. Processes 2 and 4 are adiabatic. R = 2 cal K⁻¹ mol⁻¹. Choose the correct work relation.

Options: (1) w₂+w₄ = ΔU₂−ΔU₄ (2) w₁+w₂ = 2T₁ ln(V₂/V₁) (3) w₁+w₂+w₃+w₄=0 (4) w₁+w₃ = −2T₁ln(V₂/V₁) − 2T₂ln(V₄/V₃)

Answer: Option 4

Step-by-step explanation

1. Processes 1 and 3 are isothermal, while 2 and 4 are adiabatic.
2. For reversible isothermal work using the chemistry sign convention, w = −nRT ln(Vf/Vi).
3. For process 1: w₁ = −2T₁ ln(V₂/V₁), since n=1 and R=2.
4. For process 3: w₃ = −2T₂ ln(V₄/V₃).
5. Adding gives option 4.

💡 Beginner tip: On a P–V cycle, total ΔU is zero, but total work is generally not zero.

Q75

In acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄. If 10 mL of KMnO₄ reaches the end point, find the molarity of KMnO₄.

Options: (1) 0.20 M (2) 0.25 M (3) 0.15 M (4) 0.10 M

Answer: Option 4 — 0.10 M

Step-by-step explanation

1. Moles of oxalic acid = 0.010 × 0.25 = 0.0025 mol.
2. Balanced ratio: 2 MnO₄⁻ : 5 H₂C₂O₄.
3. Moles of MnO₄⁻ = (2/5)×0.0025 = 0.001 mol.
4. M = 0.001/0.010 = 0.10 M.

💡 Beginner tip: Always balance the redox reaction first; use the stoichiometric mole ratio.

Q76

The lanthanide ion having four unpaired electrons is:

Atomic numbers: Ce=58, Nd=60, Tb=65, Ho=67. Options: (1) Ce³⁺ (2) Tb³⁺ (3) Ho³⁺ (4) Nd³⁺

Answer: Option 3 — Ho³⁺

Step-by-step explanation

1. Ce³⁺ = 4f¹ → 1 unpaired.
2. Nd³⁺ = 4f³ → 3 unpaired.
3. Tb³⁺ = 4f⁸ → 6 unpaired.
4. Ho³⁺ = 4f¹⁰ → 4 unpaired.

💡 Beginner tip: For lanthanides, remove 6s electrons first to obtain the common 3+ configuration.

Q77

The correct decreasing order of oxidation state of the underlined atom is:

Options: (1) N₂O₅ > Al₂O₃ > H₂S (2) PbO₂ > N₂O₃ > SO₃ (3) P₄O₆ > Cl₂O₇ > AlH₃ (4) P₄O₁₀ > SO₃ > H₂O

Answer: Option 1 — +5 > +3 > −2

Step-by-step explanation

1. In N₂O₅, N = +5.
2. In Al₂O₃, Al = +3.
3. In H₂S, S = −2.
4. Therefore +5 > +3 > −2.

💡 Beginner tip: Use the rule: total oxidation number equals the overall charge.

Q78

The formula of tetraammineaquachloridocobalt(III) chloride is:

Options: (1) [Co(NH₃)₄]Cl₃·H₂O (2) [Co(NH₃)₄(H₂O)Cl]Cl (3) [Co(NH₃)₄(H₂O)Cl]Cl₂ (4) [Co(NH₃)₄Cl₂]·H₂O

Answer: Option 3 — [Co(NH₃)₄(H₂O)Cl]Cl₂

Step-by-step explanation

1. Co is +3.
2. NH₃ and H₂O are neutral ligands; coordinated Cl⁻ contributes −1.
3. Complex charge = +3 −1 = +2.
4. Therefore two chloride ions are needed outside the coordination sphere.

💡 Beginner tip: Name tells you what is inside the square brackets; final 'chloride' indicates counter-ions.

Q79

Toluene is treated with (i) CrO₂Cl₂/CS₂ and then (ii) H₃O⁺ to give P. Choose the correct statement.

Options include: (1) P from benzene + CH₃COCl/AlCl₃ (2) P gives white ppt with bromine water (3) P obtained by hydrogenation of benzoyl chloride with Pd/BaSO₄ (4) P gives brisk effervescence with saturated NaHCO₃.

Answer: Option 3

Step-by-step explanation

1. CrO₂Cl₂ followed by hydrolysis (Etard reaction) oxidizes the benzylic methyl group of toluene to –CHO.
2. Thus P is benzaldehyde.
3. Rosenmund reduction of benzoyl chloride with H₂/Pd-BaSO₄ also gives benzaldehyde.

💡 Beginner tip: Benzaldehyde does not give NaHCO₃ effervescence because it is not a carboxylic acid.

Q80

Assertion A: For an ideal solution formed by mixing liquids P and Q, ΔmixH = 0 and ΔmixV = 0. Reason R: No interactions occur between P and Q. Choose the most appropriate answer.

Options: (1) Both correct but R not explanation (2) A correct R incorrect (3) A incorrect R correct (4) Both correct and R explains A

Answer: Option 2 — Assertion correct, Reason incorrect

Step-by-step explanation

1. For an ideal solution, unlike interactions P–Q are approximately equal in strength to P–P and Q–Q interactions.
2. Therefore mixing produces essentially no enthalpy or volume change: ΔmixH≈0 and ΔmixV≈0.
3. The reason is wrong because interactions do occur; they are simply comparable in strength.

💡 Beginner tip: Ideal solution ≠ no intermolecular forces.

Q81

Match vitamins with their sources: A. vitamin A, B. vitamin B₁₂, C. vitamin E, D. vitamin K. Sources: I. meat, II. sunflower oil, III. green leafy vegetables, IV. carrots.

Options: (1) A-IV, B-I, C-II, D-III (2) A-IV, B-II, C-I, D-IV (3) A-III, B-I, C-IV, D-II (4) A-II, B-III, C-IV, D-I

Answer: Option 1 — A-IV, B-I, C-II, D-III

Step-by-step explanation

1. Vitamin A (provitamin carotene) → carrots.
2. Vitamin B₁₂ → meat/animal foods.
3. Vitamin E → vegetable oils such as sunflower oil.
4. Vitamin K → green leafy vegetables.

💡 Beginner tip: Memory: A–carrots, B₁₂–animal foods, E–oils, K–leafy greens.

Q82

Among the following, the compound having conjugated double bonds is:

Options: (1) hepta-1,4-diene (2) hepta-1,5-diene (3) hepta-1,6-diene (4) hepta-1,3-diene

Answer: Option 4 — hepta-1,3-diene

Step-by-step explanation

1. Conjugated double bonds follow the pattern C=C–C=C.
2. Hepta-1,3-diene has exactly one single bond between the two double bonds.
3. The other options have isolated double bonds separated by more than one single bond.

💡 Beginner tip: Conjugated diene: alternating double–single–double.

Q83

One product formed when cyclohexylmagnesium bromide reacts with cyclohexylamine is:

Options show: (1) coupled product with NH₂ (2) bicyclohexyl (3) cyclohexane (4) dicyclohexylamine.

Answer: Option 3 — cyclohexane

Step-by-step explanation

1. A Grignard reagent is a very strong base.
2. Cyclohexylamine contains an N–H proton.
3. Acid–base reaction occurs: R–MgBr + R′NH₂ → RH + R′NHMgBr.
4. Therefore the cyclohexyl group of the Grignard reagent becomes cyclohexane.

💡 Beginner tip: Grignard reagents are destroyed by compounds containing acidic X–H bonds such as O–H, N–H and S–H.

Q84

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. Find ΔSsystem and ΔSsurroundings.

Options include: (1) +4.606R, −4.606R (2) 0, +4.606R (3) +4.606R, 0 (4) 0, 0.

Answer: Option 3 — ΔSsystem = 4.606R; ΔSsurroundings = 0

Step-by-step explanation

1. For an ideal gas, ΔSsystem = nR ln(V₂/V₁).
2. = 2R ln(100/10) = 2R ln 10 = 2R(2.303) = 4.606R.
3. Free expansion occurs without heat exchange with surroundings, so ΔSsurroundings = 0.

💡 Beginner tip: Even though q=0, the system entropy can increase in an irreversible free expansion.

Q85

Among the following options, the correct trend in electron gain enthalpy is:

Options: (1) Br > Cl > F > I (2) Cl > F > Br > I (3) I > Br > Cl > F (4) F > Cl > Br > I

Answer: Option 2 — Cl > F > Br > I

Step-by-step explanation

1. For halogens, electron gain enthalpy is generally most favourable for Cl rather than F.
2. F is unusually less favourable than expected because its compact 2p shell causes greater electron–electron repulsion.
3. Thus the usual magnitude/favourability order is Cl > F > Br > I.

💡 Beginner tip: Classic exception: chlorine has a more favourable electron gain enthalpy than fluorine.

Q86

The correct order of solubility in water at 298 K is given Ksp: AgBr = 5.0×10⁻¹³, Zn(OH)₂ = 1.0×10⁻¹⁵, Hg₂Cl₂ = 1.3×10⁻¹⁸.

Options: (1) AgBr > Zn(OH)₂ > Hg₂Cl₂ (2) Hg₂Cl₂ > AgBr > Zn(OH)₂ (3) Zn(OH)₂ > AgBr > Hg₂Cl₂ (4) Hg₂Cl₂ > Zn(OH)₂ > AgBr

Answer: Option 3 — Zn(OH)₂ > AgBr > Hg₂Cl₂

Step-by-step explanation

1. AgBr ⇌ Ag⁺+Br⁻, so Ksp=s² and s≈√(5×10⁻¹³).
2. For Zn(OH)₂ and Hg₂Cl₂, Ksp=4s³ because each gives one metal ion and two anions/cations.
3. Calculating gives Zn(OH)₂ the largest s; AgBr is slightly larger than Hg₂Cl₂.

💡 Beginner tip: Never compare Ksp values directly when salts have different dissolution stoichiometries.

Q87

Starting from Ph–C≡C–Me: Na/liquid NH₃ gives L, then HBr/benzoyl peroxide gives N. H₂/Pd–C (Lindlar catalyst) gives K, then HBr gives M. Choose the correct statement.

Options: (1) K and L are enantiomers (2) M and N are geometrical isomers (3) M and N are stereoisomers (4) K and L are geometrical isomers

Answer: Option 4 — K and L are geometrical isomers

Step-by-step explanation

1. Na/liquid NH₃ reduces an alkyne to the trans-alkene (L).
2. Lindlar catalyst reduces an alkyne to the cis-alkene (K).
3. Thus K and L are cis/trans (E/Z) geometrical isomers.

💡 Beginner tip: Alkyne reductions: Lindlar → cis; Na/NH₃(l) → trans.

Q88

The amount of CO₂ evolved on complete combustion of 116 g of n-butane is:

Options: (1) 322 g (2) 176 g (3) 362 g (4) 352 g

Answer: Option 4 — 352 g

Step-by-step explanation

1. Molar mass of C₄H₁₀ = 58 g mol⁻¹, so 116 g = 2 mol.
2. C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O.
3. 2 mol butane gives 8 mol CO₂.
4. Mass = 8×44 = 352 g.

💡 Beginner tip: One mole of a C₄ compound gives four moles of CO₂ on complete combustion.

Q89

The numbers 17.0145 and 21.0235 are rounded to three figures after the decimal point. The resulting numbers are:

Options: (1) 17.015 and 21.023 (2) 17.014 and 21.024 (3) 17.015 and 21.024 (4) 17.014 and 21.023

Answer: Option 3 — 17.015 and 21.024

Step-by-step explanation

1. To keep three digits after the decimal, inspect the fourth digit.
2. 17.0145 → fourth decimal digit is 5, so 17.014 rounds to 17.015.
3. 21.0235 → fourth decimal digit is 5, so 21.023 rounds to 21.024.

💡 Beginner tip: Keep three decimal places, then round using the next digit.

Q90

In potash alum, the ratio of K⁺ ions and SO₄²⁻ ions is:

Options: (1) 2:1 (2) 2:3 (3) 3:2 (4) 1:2

Answer: Option 4 — 1:2

Step-by-step explanation

1. Potash alum formula: KAl(SO₄)₂·12H₂O.
2. It contains 1 K⁺ and 2 sulfate ions per formula unit.
3. Therefore K⁺ : SO₄²⁻ = 1 : 2.

💡 Beginner tip: Ignore water of crystallization when counting these ions.

Prepared as a self-contained HTML study sheet for Questions 46–90. You can open it in any browser and print it to PDF if needed.