CBSE Class 12 Chemistry Question Paper 2026 (56/2/1) with Solutions
The complete CBSE Class 12 Chemistry board paper of 2026, Q.P. code 56/2/1: all 33 questions exactly as set, with marks. 19 solutions are free below. The Class 12 Chemistry PYQ Book: Solved Board Papers 2016-2026 (CBSE) solves all 19 papers in full and sorts every question chapter-wise.
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2026 · Q.P. Code 56/2/1
Maximum marks: 70
Section A · Multiple choice and Assertion–Reason (1 mark each)
Consider the following reaction :
Given : V, V, 1 F = 96500 C mol
for the above reaction is :
Identify the correct statement :
The order for the given reaction is :
, Rate =
Electronic configuration of chromium is :
The correct IUPAC name of the complex is :
Which of the following is heteroleptic complex ?
Which of the following molecules is chiral in nature ?
Identify the correct increasing order of boiling points of the given compounds :
Identify the compound produced by the reduction of Ethanenitrile with Lithium aluminium hydride :
The base which is present in DNA but not in RNA is :
Proteins are polymers of α-amino acids which are joined to each other by :
Which of the following reactions is not explained by the open chain structure of glucose ?
Assertion (A) : The molecularity of the given reaction is 2 :
Reason (R) : Two molecules of the reactants are involved in simultaneous collision between them.
Assertion (A) : Zinc, cadmium and mercury are not considered as transition elements.
Reason (R) : These elements have completely filled orbitals in their ground state as well as in their common oxidation states.
Assertion (A) : Aromatic primary amines can be prepared by Gabriel Phthalimide synthesis.
Reason (R) : Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Assertion (A) : Glucose gets oxidized to six carbon gluconic acid on reaction with bromine water.
Reason (R) : The carbonyl group is absent in the open chain structure of glucose.
Section B · Very short answer (2 marks each)
(a) The concentration of the reactant is reduced from 0.6 mol L to 0.2 mol L in 5 minutes in a first order reaction. Calculate rate constant of the reaction. (log 3 = 0.48)
(b) Rate constant k for the first order reaction is s. Calculate the time required for three-fourth of the reactant to decompose. (log 4 = 0.60)
(a) Define activation energy. (b) Write the unit of rate constant for (i) zero order reaction, (ii) second order reaction.
(a) First ionization enthalpy of Cr is lower than that of Zn. Why ? (b) Mention any one property of the transition elements which makes them good catalysts.
(a) Define coordination number. (b) Indicate the type of isomerism exhibited by the following complex :
(a) Draw the structure of the given compound : 4–Bromo–3–methylpent–2–ene
(b) What happens when chloroethane is treated with aqueous potassium hydroxide ?
Section C · Short answer (3 marks each)
When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene got raised from 353.23 K to 353.93 K. Calculate the molar mass of the solute. (Given for benzene = 2.52 K kg mol)
Calculate emf of the following cell at 298 K :
Sn | (0.001 M) || (0.01 M) | (1 bar) | Pt(s)
Given : V, V [log 10 = 1]
(a) Complete and balance the following equations : (i) (ii)
(b) Why is it difficult to separate lanthanoid elements in pure state ?
(a) is diamagnetic whereas is paramagnetic. Justify the statement. [Atomic number of Co = 27]
(b) Write the electronic configuration for ion if on the basis of crystal field theory.
(a) Write the major product in the following reaction : Cyclohexene + (heat) → ?
(b) Define Racemic mixture.
(c) p-dichlorobenzene has a higher melting point as compared to its ortho and meta isomers. Why ?
Write the reaction involved in (a) Rosenmund's reduction (b) Cannizzaro's reaction (c) Hell – Volhard – Zelinsky reaction
(a) Define the following term : Reducing sugar
(b) Differentiate between the following : (i) Fibrous proteins and globular proteins (ii) Nucleotide and nucleoside
(b) Write the reaction of glucose with the following : (a) (b) (c) conc.
Section D · Case-based (4 marks each)
Osmosis is a process by which the molecules of a solvent pass from a solution of low solute concentration to a solution of high solute concentration through a semi-permeable membrane. Osmotic pressure is a colligative property. When the applied pressure on a solution exceeds its osmotic pressure, reverse osmosis occurs. When two solutions are separated by a semipermeable membrane and they have same osmotic pressure they are said to be isotonic. Of the two solutions separated by a semipermeable membrane, if one is a lower osmotic pressure, it is said to be hypotonic relative to the second solution. If it has a higher osmotic pressure, than the second solution, it is said to be hypertonic relative to the second solution. The osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (mass/volume) sodium chloride solution called normal saline solution and it is safe to inject intravenously. Osmotic pressure is vital in daily life and nature. It helps in explain, why IV fluids match blood's osmotic pressure, its also the principle behind food preservation using salt or sugar.
(a) Calculate the amount of (i = 2.59) dissolved in 2.46 litre of water such that its osmotic pressure is 0.70 atm at 27°C. [Given : R = 0.082 L atm K mol, molar mass of = 111 g mol] (2)
(b) When raisins are kept in water, they get swollen. Name the phenomenon involved in this process. (1)
OR
(b) Why osmotic pressure is more advantageous than other colligative properties ? (1)
(c) Which phenomenon is responsible for desalination of sea water ? (1)
Like , nitrogen atom of amine is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore hybridised and the geometry of amines is pyramidal. Lower aliphatic amines are soluble in water due to the formation of hydrogen bond with water molecules. The solubility decreases as the molar mass of amines increases due to increase in size of hydrophobic part. Higher amines are insoluble in water. However amines are less soluble in water than alcohols because of low electronegativity of nitrogen as compared to oxygen. Boiling points of amines are held together due to hydrogen bonding. Extent of hydrogen bonding is more in primary amines than in secondary amines as two hydrogen atoms are available for hydrogen bond formation. Tertiary amines do not show hydrogen bonding because of the absence of hydrogen atom attached to nitrogen. Amines can be prepared from, nitro compounds, nitriles, amides etc.
(a) Complete the following equations : (i) (ii) p-nitrotoluene (2)
(b) Why primary amines have higher boiling points than tertiary amines ? (1)
(c) Classify the following amines as primary, secondary or tertiary : (i) N,N-dimethylnaphthalen-2-amine (ii) naphthalen-2-amine (1)
OR
(c) Out of Butan–1–amine and Butan–1–ol, which is more soluble in water ? (1)
Section E · Long answer (5 marks each)
(a) How will you convert the following : (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Benzene to m-Nitroacetophenone
(b) (i) Arrange the following compounds in increasing order of their reactivity towards HCN : , ,
(ii) Identify the compounds, which would undergo Aldol condensation : HCHO, , , cyclohexanone
(b) (a) Give chemical tests to distinguish between the following pairs of compounds : (i) Acetophenone and Benzophenone (ii) Propanal and Propanone (iii) Pentan–2–one and Pentan–3–one
(b) (i) Which of the following acids is stronger and why ? ,
(ii) Arrange the following compounds in the increasing order of their boiling points : , , ,
(a) The conductivity of 0.1 mol L solution of NaCl is S cm. Calculate its molar conductivity and degree of dissociation. S cm mol, S cm mol
(b) (i) Following cell reaction occurs in a galvanic cell : , V. Predict the direction of flow of current.
(ii) Differentiate between a primary battery and a secondary battery.
(b) (a) Resistance of a conductivity cell filled with 0.1 M KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.01 mol L KCl solution is 300 Ω, calculate the conductivity and molar conductivity of 0.01 mol L KCl solution. The conductivity of 0.1 M KCl solution is S cm.
(b) (i) Write any two advantages of fuel cell. (ii) Why does the cell potential of mercury cell remain constant throughout the life ?
(a) An organic compound 'A', with molecular formula reacts with active metals such as sodium to give compound 'B' and hydrogen gas. 'A' on treatment with iodine and sodium hydroxide gives 'C' and in presence of at 413 K gives 'D' (). 'D' on reaction with excess of HI gives 'E'. Identify 'A', 'B', 'C', 'D' and 'E' and write all the reactions involved.
(b) (a) Write the reagents which are used in the given conversions : (i) Phenol to 2, 4, 6 – tribromophenol (ii) Propene to propan–1–ol (iii) Butan–2–one to butan–2–ol
(b) Explain the mechanism of acid catalyzed hydration of alkene to form corresponding alcohol.
Solutions: CBSE 2026 · 56/2/1
Answer: (A) –301.080 kJ mol
V; n = 2 (Zn → ).
J mol = –301.080 kJ mol.
Answer: (A) 1.5
Order = sum of the powers in the experimental rate law = (the stoichiometric coefficients do not decide it).
Answer: (B) diamminedichloridoplatinum (II)
Ligands are named in alphabetical order (ammine before chlorido). Oxidation state of Pt: x + 2(0) + 2(–1) = 0, so x = +2.
Answer: (B) Butan–2–ol
In , C-2 carries four different groups (H, OH, , ), so it is a chiral centre. Propan-2-ol and 2-bromopropane have two identical groups on C-2; 1-bromobutane has no carbon with four different groups.
Answer: (A) Ethylamine
; the carbon chain does not change.
Answer: (B) Peptide Bond
The –COOH of one amino acid condenses with the – of the next, losing water, to form the amide link –CO–NH–, called the peptide bond.
Answer: (A)
The decomposition of HI is an elementary bimolecular reaction: two HI molecules must collide at the same time, so its molecularity is 2. Both statements are true and R explains A.
Answer: (D)
Gabriel synthesis gives only aliphatic primary amines. Aromatic primary amines cannot be made because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. A is false, R is true.
For a first-order reaction,
0.22 min
(a) Activation energy is the minimum extra energy that the reacting molecules must absorb to form the activated complex (to cross the energy barrier) and change into products. = energy of activated complex – average energy of reactants.
(b) Unit of k = (mol L) s.
(i) Zero order (n = 0): mol L s
(ii) Second order (n = 2): L mol s (mol L s)
(a) Coordination number of the central metal atom/ion is the number of ligand donor atoms directly bonded to it (the number of σ-bonds between ligands and the metal).
(b) Optical isomerism. is an octahedral complex with three bidentate (en) ligands; it has no plane of symmetry and exists as two non-superimposable mirror images, the d- and l-forms.
K
60 g mol
(a) (i)
(ii) (potassium dichromate, less soluble, crystallises out)
(b) Because of the lanthanoid contraction the lanthanoids have very similar atomic and ionic sizes and, being mostly +3, almost identical chemical properties; so they occur together and are very difficult to separate.
(a) At high temperature bromine substitutes at the allylic position (free-radical substitution), so the product is 3-bromocyclohexene (Br on the carbon next to the C=C).
(b) A racemic mixture is an equimolar mixture of the d- and l-enantiomers; it has zero optical rotation because the rotation of one is cancelled by the other.
(c) The para isomer is symmetrical, so its molecules fit closely into the crystal lattice. Stronger intermolecular forces in the crystal need more energy to break, so its melting point is higher than that of the o- and m-isomers.
(a) Reducing sugars are carbohydrates that reduce Fehling's solution and Tollens' reagent (they have a free aldehyde or ketone group), e.g. all monosaccharides, maltose, lactose.
(b) (i) Fibrous proteins: polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, giving a fibre-like structure; generally insoluble in water (keratin, myosin). Globular proteins: the chains coil around to give a spherical shape; usually soluble in water (insulin, albumins).
(ii) Nucleoside = base + sugar (pentose); nucleotide = base + sugar + phosphate, i.e. a nucleoside joined to phosphoric acid at the 5′-position of the sugar.
(a) , so ; T = 27 + 273 = 300 K.
mol
Mass of 3.0 g (about 3 g)
(b) Osmosis (endosmosis): water enters the raisins through the skin, which acts as a semipermeable membrane.
OR (b) Osmotic pressure is measured at room temperature and its magnitude is large even for dilute solutions, so it is used for molar masses of biomolecules and polymers, which are unstable at high temperature and have poor solubility.
(c) Reverse osmosis: a pressure larger than the osmotic pressure is applied on sea water, so pure water is pushed out through the semipermeable membrane.
(a) (i) Reduce to propan-2-ol, then dehydrate:
(ii) Convert to the acid chloride, then Rosenmund reduction:
(iii) Friedel–Crafts acylation gives acetophenone; the – group is meta-directing, so nitration gives the meta product:
(b) (i) Reactivity towards nucleophilic addition falls as alkyl groups increase (+I effect makes the carbonyl carbon less positive, and bulky groups block attack):
di-tert-butyl ketone < acetone < acetaldehyde
(ii) Aldol condensation needs at least one α-hydrogen: and cyclohexanone. HCHO and have no α-hydrogen.
(a) 106 S cm mol
S cm mol
0.837 (83.7%)
(b) (i) Zn is oxidised (anode, negative) and is reduced (cathode, positive). Electrons flow from Zn to Ag in the external circuit, so the current flows from the silver electrode to the zinc electrode in the external circuit.
(ii) Primary battery: the reaction occurs only once; after use it cannot be recharged (e.g. dry cell, mercury cell). Secondary battery: can be recharged by passing current in the opposite direction and used again (e.g. lead storage battery, Ni–Cd cell).
A reacts with Na to give , so it is an alcohol; alcohol = ethanol. It gives the iodoform test (has ).
A = ethanol ; B = sodium ethoxide ; C = iodoform ; D = diethyl ether ; E = iodoethane .
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- Board Paper 2026 Set 56-1-1 with Solutions
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