Jharkhand Academic Council (JAC) Chemistry Solved Model Paper 2026
JAC Chemistry Solved Model Paper 2026 Download PDF
JAC Previous Year Question Papers
SECTION-A
Multiple Choice Question
SECTION-B
Very Short Answer Question
26. Define Mole Fraction.
Mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.
View Answer
Mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.
27. For a reaction, A + B → Products, the rate law is given by: r = [A]1/2 [B]2, what is the order of reaction?
The order of the reaction is the sum of the powers of the concentration terms in the rate law. Here, order = 1/2 + 2 = 5/2 or 2.5.
View Answer
2.5 or 5/2
28. What is lanthanoides contraction?
Lanthanide contraction is the greater-than-expected decrease in ionic radii of the elements in the lanthanide series from left to right due to poor shielding of 4f electrons.
View Answer
Lanthanide contraction is the greater-than-expected decrease in ionic radii of the elements in the lanthanide series from left to right due to poor shielding of 4f electrons.
29. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(Py)].
The complex [Pt(NH3)(Br)(Cl)(Py)] has four different ligands, so it does not show geometrical isomerism. It has only one form.
View Answer
No geometrical isomers (only one form exists).
30. Write the name of the reaction in which alkyl fluorides are prepared by the reaction of alkyl chloride/alkyl bromide with AgF.
The reaction is known as Swarts reaction.
View Answer
Swarts reaction
31. Why are alcohols more soluble in water than the hydrocarbon of comparable molecular mass?
Alcohols form hydrogen bonds with water due to the presence of -OH group, whereas hydrocarbons cannot form hydrogen bonds. Hence, alcohols are more soluble in water.
View Answer
Alcohols form hydrogen bonds with water due to the presence of -OH group, whereas hydrocarbons cannot form hydrogen bonds. Hence, alcohols are more soluble in water.
32. What is Hinesburg’s reagent?
Hinsberg’s reagent is benzenesulphonyl chloride (C6H5SO2Cl), used to distinguish between primary, secondary, and tertiary amines.
View Answer
Benzenesulphonyl chloride (C6H5SO2Cl)
33. What type of bonding helps in stabilising the α-Helix structure of proteins?
Intramolecular hydrogen bonding between the >C=O group of one amino acid and the >N-H group of another amino acid stabilizes the α-helix structure.
View Answer
Intramolecular hydrogen bonding
34. How will you convert glucose into fructose?
Glucose is first converted into glucose oxime using NH2OH, then treated with acetic anhydride to form a derivative, which on heating with base undergoes Beckmann rearrangement to give fructose.
OR
Glucose → (i) HCN (ii) H2O → Gluconic acid → (reduction) Sorbitol → (enzymatic oxidation) Fructose.
View Answer
Glucose is isomerized to fructose in the presence of base (Lobry de Bruyn–van Ekenstein transformation) or enzymatically using glucose isomerase.
SECTION-C
Short Answer Type Question
35. What is electrode potential?
Electrode potential is the tendency of an electrode to lose or gain electrons when it is in contact with its own ions in solution. It is measured with respect to standard hydrogen electrode (SHE) which is assigned a potential of 0.00 V. There are two types: oxidation potential (tendency to lose electrons) and reduction potential (tendency to gain electrons). Standard electrode potential (E°) is measured under standard conditions (1 M concentration, 298 K, 1 atm pressure).
View Answer
Electrode potential is the potential difference developed at the interface between the metal electrode and the solution of its ions due to the tendency of the electrode to lose or gain electrons. It is measured relative to SHE.
36. Depict the galvanic cell in which the cell reaction is: Ca + 2Ag+ → 2Ag + Ca2+
Cell notation: Ca(s) | Ca2+(aq) || Ag+(aq) | Ag(s)
Anode (oxidation): Ca(s) → Ca2+(aq) + 2e–
Cathode (reduction): 2Ag+(aq) + 2e– → 2Ag(s)
Cell reaction: Ca(s) + 2Ag+(aq) → Ca2+(aq) + 2Ag(s)
View Answer
Ca(s) | Ca2+(aq) || Ag+(aq) | Ag(s)
37. What is the difference between lanthanoids and actinoids?
| Property | Lanthanoids | Actinoids |
|---|---|---|
| Electronic configuration | 4f1–14 5d0–1 6s2 | 5f1–14 6d0–1 7s2 |
| Oxidation states | Mainly +3 | +3 to +7 |
| Radioactivity | Non-radioactive (except Pm) | All are radioactive |
| Chemical reactivity | Less reactive | Highly reactive |
| Contraction | Lanthanide contraction | Actinide contraction (more pronounced) |
View Answer
Lanthanoids show mainly +3 oxidation state, are less reactive, and non-radioactive (except promethium). Actinoids show variable oxidation states (+3 to +7), are highly reactive and radioactive.
38. Using IUPAC norms write the IUPAC name of:
a. [Co(NH3)5NO2] Cl2
b. [Cr(NH3)4Cl2] Cl
c. [Fe(H2O)6]2+
View Answer
a) Pentaamminenitrito-N-cobalt(III) chloride
b) Tetraamminedichloridochromium(III) chloride
c) Hexaaquairon(II) ion
39. Complete the following:
a. C6H6 + 3Br2(aq) →
b. CH3–CO–NH2 + LiAlH4/H2O →
c. C6H11–CO–NH2 + Br2/NaOH →
View Answer
a) 2,4,6-Tribromophenol
b) CH3NH2 (Methylamine) – Hofmann bromamide reaction
c) Cyclohexane (Birch reduction or hydrogenation)
40. How will you bring about the following conversion?
a. Ethyl chloride to propanoic acid
b. Chlorobenzene to p-nitrophenol
c. 1-Bromopropane to 2-bromopropane
View Answer
a) C2H5Cl → (KCN) C2H5CN → (H3O+) CH3CH2COOH
b) C6H5Cl → (NaOH, 623K) C6H5ONa → (H+) C6H5OH → (HNO3) p-Nitrophenol
c) CH3CH2CH2Br → (alc. KOH) CH3CH=CH2 → (HBr, peroxide) CH3CHBrCH3
41. Write the IUPAC name of following compound.
a. C6H5CH(OH)CH3
b. m-ClCH2C6H4CH2C(CH3)3
c. CH3C(Cl)(C2H5)CH2CH3
View Answer
a) 1-Phenylethanol
b) 1-(3-(Chloromethyl)phenyl)-2-methylpropan-2-yl (or 1-[3-(chloromethyl)phenyl]-2-methylpropane)
c) 3-Chloro-3-methylpentane
42. Show the presence of peptide linkage in the structure of glycine and alanine.
Glycine: H2N–CH2–COOH
Alanine: H2N–CH(CH3)–COOH
When they condense:
H2N–CH2–COOH + H2N–CH(CH3)–COOH → H2N–CH2–CO–NH–CH(CH3)–COOH + H2O
The –CO–NH– bond is the peptide linkage.
View Answer
The peptide linkage is the amide bond (–CO–NH–) formed by the condensation of the carboxyl group of glycine and the amino group of alanine.
SECTION-D
Long Answer Type Question
43. A 10% Solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K (Given molar mass of sucrose = 342 g mol-1; Molar mass of glucose = 180 g mol-1).
View Answer
ΔTf = Kf × m
For sucrose: ΔTf = 273.15 - 269.15 = 4 K
Mass of sucrose = 10 g, solvent = 90 g
Moles of sucrose = 10 / 342 = 0.02924 mol
Molality (m) = 0.02924 / 0.09 = 0.3249 m
Kf = ΔTf / m = 4 / 0.3249 ≈ 12.31 K kg mol-1
For glucose: Moles = 10 / 180 = 0.05556 mol
Molality = 0.05556 / 0.09 ≈ 0.6173 m
ΔTf = 12.31 × 0.6173 ≈ 7.60 K
Freezing point = 273.15 - 7.60 = 265.55 K
44. Calculate the Cell e.m.f. and ΔG for the cell reaction at 298 K for the cell: Zn(s) | Zn2+(0.0004 M) || Cd2+(0.2 M) | Cd(s) (Given E°Zn²⁺/Zn = -0.763 V; E°Cd²⁺/Cd = -0.403 V; F = 96500 C mol-1)
View Answer
E°cell = E°cathode - E°anode = -0.403 - (-0.763) = 0.36 V
Ecell = E°cell - (0.0591/n) log(Q)
n = 2, Q = [Zn²⁺]/[Cd²⁺] = 0.0004 / 0.2 = 0.002
Ecell = 0.36 - (0.0591/2) log(0.002) = 0.36 - 0.02955 × (-2.69897) ≈ 0.36 + 0.0797 ≈ 0.4397 V
ΔG = -nFEcell = -2 × 96500 × 0.4397 ≈ -84850 J/mol = -84.85 kJ/mol
45. What is half-life of a reaction? Find out the value of half-life of a zero and first order reaction.
View Answer
Half-life (t1/2) is the time taken for the concentration of reactant to reduce to half its initial value.
Zero-order: t1/2 = [A]0 / (2k)
First-order: t1/2 = 0.693 / k
46. Using Valence Bond Theory, explain the Geometry and Magnetic nature of [Cr(H2O)6]3+ ion.
View Answer
Cr3+: [Ar] 3d3
In octahedral field, d-orbitals split into t2g and eg.
H2O is weak field ligand → high spin complex.
Electron configuration: t2g3 (3 unpaired electrons).
Geometry: Octahedral
Magnetic nature: Paramagnetic (3 unpaired electrons)
47. Draw the structure of the following:
a. 4–Ethoxy butan–2–ol
b. 3–Methyl–3–phenyl pentan–2–ol
c. m–Cresol
d. Picric acid
e. Glycerol
View Answer
a. CH3CH(OH)CH2CH2OCH2CH3
b. CH3C(OH)(C6H5)CH(CH3)CH2CH3
c. CH3-C6H4-OH (OH and CH3 meta)
d. (O2N)3C6H2OH (2,4,6-trinitrophenol)
e. HOCH2CH(OH)CH2OH
48. Write short notes on:
a. Sandmeyer’s Reaction
b. Kolbe’s Electrolysis
c. Gabriel Phthalimide Synthesis
View Answer
a. Sandmeyer’s Reaction: Conversion of arenediazonium salts to aryl chlorides/bromides using CuCl/HCl or CuBr/HBr.
b. Kolbe’s Electrolysis: Electrolysis of sodium/potassium salts of carboxylic acids to form alkanes at anode: 2RCOO⁻ → R–R + 2CO2 + 2e⁻.
c. Gabriel Phthalimide Synthesis: Used to prepare primary amines. Phthalimide → potassium phthalimide → alkyl halide → alkyl phthalimide → hydrolysis → primary amine.