CBSE Class 12 Chemistry (Set 56/1/1) Question Paper 2026 with Solutions
| Updated On - Mar 3, 2026
CBSE Class 12 Chemistry (Set 56/1/1) Question Paper 2026 with Solutions PDFs is available here for download. CBSE Board is conducting the Class 12 Chemistry Exam 2026 on February 28, 2026. CBSE Board Class 12 the examination was held in the first half from 10:30 AM to 1:30 PM. The official question paper of CBSE Board Class 12 Chemistry Exam 2026 is provided below. Students can download the official paper in PDF format for reference.
CBSE Class 12 Chemistry (Set 56/1/1) Question Paper 2026 with Solutions PDFs
| CBSE Class 12 Chemistry (Set 56/1/1) Question Paper 2026 | Download PDF | Check Solutions |
Which of the reactions is used in the conversion of a ketone into hydrocarbon?
Which of the following reagents are used to prepare primary amines by Hofmann bromamide degradation reaction?
The major product of carbylamine reaction is:
Actinoids show larger number of oxidation states:
Consider the following reaction and identify A and B: \[ CH_3Cl + NaI \xrightarrow{dry acetone} A + B \]
The correct formula of Hinsberg's reagent is:
Half-life (\( t_{1/2} \)) of a first order reaction is 1386 s. The value of rate constant is:
Which of the following ligands forms a chelate complex?
Primary, secondary and tertiary alcohols can be distinguished by:
Consider the following compounds:
\[ (C_2H_5)_2NH,\quad C_6H_5NH_2,\quad C_6H_5CH_2NH_2,\quad NH_3,\quad (C_2H_5)_3N \]
The correct increasing order of the above bases on the basis of their basic strength is:
Identify the polysaccharide among the following:
The polypeptide chain in a protein has amino acids linked with each other in a specific sequence. This specific sequence of amino acids is called:
Assertion (A): D(+)-Glucose is dextrorotatory in nature.
Reason (R): (+) represents dextrorotatory nature and D represents its configuration.
Assertion (A): Highest oxidation state of Mn is +7 in most of the transition elements.
Reason (R): Transition metals exhibit variable oxidation states.
Assertion (A): p-nitrophenol is more acidic than phenol.
Reason (R): Nitro group is an electron-withdrawing group; it stabilizes phenoxide ion by dispersal of negative charge.
Assertion (A): All aliphatic aldehydes give a positive Fehling's test.
Reason (R): Aliphatic aldehydes are reduced by Fehling's reagent.
1.00 molal aqueous solution of trichloroacetic acid is heated to its boiling point. The boiling point of this solution was found to be 100.18\(^\circ\)C. Calculate the Van’t Hoff factor for trichloroacetic acid.
(Given: \( K_b \) for water = 0.512 K kg mol\(^{-1}\))
State Henry’s law. Calculate the mole fraction of CO\(_2\) in water at 298 K under 700 mm Hg pressure.
(Given: Henry’s constant for CO\(_2\) in water at 298 K = \( 1.25 \times 10^6 \) mm Hg)
Name the cell which was used in the Apollo space programme for providing electrical power.
Define limiting molar conductivity.
Complete the following equation:
How will you convert nitromethane to methyl isocyanide?
What are the products obtained on hydrolysis of sucrose?
What are essential amino acids?
Write any two fat soluble vitamins.
How will you confirm the presence of five \(-OH\) groups in a glucose molecule, which are attached to different carbon atoms?
Calculate emf of the following cell at \(298 K\):
\(Cr(s) | Cr^{3+}(aq) (0.1 M) || Fe^{2+}(aq) (0.01 M) | Fe(s)\)
Given: \(E^\circ_{Cr^{3+}/Cr} = -0.74 V, E^\circ_{Fe^{2+}/Fe} = -0.44 V, \log 10 = 1\)
Define order of a reaction.
The rate for the following reaction is given by:
\(A + B \rightarrow C, Rate = k[A][B]^2\)
(i) How is the rate affected if we double the concentration of \(B\)?
(ii) Write the overall order of a reaction if \(A\) is present in large excess.
The rate of a chemical reaction doubles when the temperature is raised from \(298 K\) to \(308 K\). Calculate the activation energy (\(E_a\)) for this reaction assuming it does not change with temperature. (Given: \(R = 8.314 J mol^{-1}K^{-1}, \log 2 = 0.30\))
Write the IUPAC name of the following complex:
\(K_3[Cr(C_2O_4)_3]\)
Differentiate between homoleptic complex and heteroleptic complex.
Which type of isomerism is exhibited by the following complex?
\([Pt(NH_3)_2Cl_2]\)
A coordination compound \(CrCl_3 \cdot 6H_2O\) is mixed with excess \(AgNO_3\) solution, two moles of \(AgCl\) are precipitated per mole of the compound. Write the structural formula of the coordination compound.
Write the oxidation state and hybridisation of the central metal in the following complex:
\([Fe(H_2O)_6]^{3+}\) (Atomic number of \(Fe = 26\))
Why is \([Ni(H_2O)_6]^{2+}\) coloured? (Atomic number of \(Ni = 28\))
Question 27 How do you convert the following:
Question 27(a):
Acetophenone to Benzoic acid
Benzoic acid to Benzene
Arrange the following compounds in increasing order of their acidic strengths:
\(CH_3CH_2COOH, CH_3CH(CH_3)COOH, CH_3CH_2CH_2COOH, BrCH_2CH_2COOH\)
Why is \(CH_3CHO\) more reactive than acetone towards reaction with \(HCN\)?
Complete the equation:
\(CH_3CHO + NH_2NH_2 \xrightarrow{OH^-} ?\)
An organic compound with molecular formula \(C_7H_6O\) forms 2,4-DNP derivative, reduces Tollens' reagent and gives Cannizzaro reaction. On vigorous oxidation it gives benzoic acid. Identify the compound. Also write the reactions of the compound with 2,4-DNP and when it undergoes Cannizzaro reaction.
Read the following passage carefully and answer the questions that follow:
Ethers are prepared by the dehydration of alcohols in presence of protic acids at 413 K. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves \(S_N2\) attack of an alkoxide ion on a primary alkyl halide. If tertiary alkyl halide is used, elimination reaction occurs and alkene is formed instead of ether.
C--O bond in ethers are cleaved under drastic conditions with excess of HI. When unsymmetrical ethers react with HI, the alkyl halide is formed from the smaller alkyl group. If one of the alkyl groups is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary carbocation is more stable than primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl-oxygen bond due to more stable aryl-oxygen bond.
The order of reactivity of hydrogen halides is: \(HI > HBr > HCl\).
Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring and directs the incoming group to ortho and para positions.
Question 29(a)(i):
Complete the following equation:
Anisole + \(CH_3Cl\) \(\xrightarrow{Anhyd. AlCl_3}\) ?
Complete the following reaction:
Anisole \(\xrightarrow{Conc. HNO_3 + H_2SO_4}\) ?
View Solution
Step 1: Understanding the Concept:
This is the nitration of anisole, which is an electrophilic aromatic substitution reaction.
Step 2: Detailed Explanation:
A mixture of concentrated \(HNO_3\) and \(H_2SO_4\) produces the nitronium ion (\(NO_2^+\)) as the electrophile.
Since the methoxy group is ortho-para directing, the nitro group attaches to the ortho (2nd) and para (4th) positions of the ring.
4-nitroanisole is the major product due to minimal steric repulsion between the \(-OCH_3\) and \(-NO_2\) groups.
Step 3: Final Answer:
The products are 2-nitroanisole and 4-nitroanisole.
Quick Tip: In nitration, \(H_2SO_4\) acts as a catalyst to help generate the nitronium ion from \(HNO_3\).
Write the names of alkyl halide and sodium alkoxide used to prepare tert-butyl ethyl ether by Williamson synthesis.
View Solution
Step 1: Understanding the Concept:
Williamson synthesis involves an \(S_N2\) reaction between an alkoxide and an alkyl halide.
Step 2: Detailed Explanation:
If we use a tertiary alkyl halide (tert-butyl bromide), the strong base (alkoxide) will cause elimination to form an alkene (isobutylene).
Therefore, we must use the primary alkyl halide: Ethyl bromide (\(CH_3CH_2Br\)).
The corresponding base must be the tertiary one: Sodium tert-butoxide (\((CH_3)_3CONa\)).
Step 3: Final Answer:
The reagents are Ethyl bromide and Sodium tert-butoxide.
Quick Tip: Rule for Williamson Synthesis: "Take the smaller group as the halide and the bulkier group as the alkoxide."
Anisole on reaction with HI gives phenol and \(CH_3I\) and not methanol and iodobenzene. Justify the statement.
View Solution
Step 1: Understanding the Concept:
The cleavage of ethers by HI involves protonation of the ether followed by a nucleophilic attack of \(I^-\).
Step 2: Detailed Explanation:
In anisole (\(C_6H_5-O-CH_3\)), the bond between the phenyl ring and the oxygen atom has a partial double bond character due to resonance.
This makes the \(C_{aryl}-O\) bond much stronger than the \(CH_3-O\) bond.
Additionally, the \(sp^2\) hybridized carbon of the benzene ring is not favorable for \(S_N2\) attack.
The \(I^-\) ion therefore attacks the methyl group (which is less hindered and has a weaker bond), resulting in the formation of \(CH_3I\) and Phenol.
Step 3: Final Answer:
Cleavage occurs at the \(CH_3-O\) bond because it is weaker and more accessible than the \(C_{aryl}-O\) bond.
Quick Tip: Phenol is always a product when alkyl aryl ethers are cleaved with HI because the bond to the benzene ring is too strong to break.
Why is C--O--C bond angle in ethers slightly greater than tetrahedral angle?
View Solution
Step 1: Understanding the Concept:
According to VSEPR theory, the geometry around the oxygen atom in ethers is \(sp^3\) hybridized, which ideally leads to an angle of \(109.5^\circ\).
Step 2: Detailed Explanation:
In ethers (\(R-O-R\)), there are two bulky alkyl groups (\(R\)) attached to the central oxygen.
The steric repulsion between these two large groups pushes them further apart.
This repulsive force overrides the compression usually caused by lone pairs, making the angle slightly larger (around \(111.7^\circ\) in dimethyl ether) than the tetrahedral angle of \(109.5^\circ\).
Step 3: Final Answer:
The bond angle is greater than \(109.5^\circ\) because of the steric repulsion between the two bulky alkyl groups.
Quick Tip: Bulky groups always increase the bond angle because of steric hindrance.
Read the following passage carefully and answer the questions that follow:
Electrochemistry is the study of the relationship between chemical energy and electrical energy. Michael Faraday proposed two laws to explain the quantitative aspects of electrolysis. Faraday's laws of electrolysis provide a basis for mathematical analysis of the mass deposited at electrodes and the amount of charge passed through them.
Question 30(a)(i)(1):
Predict the products of electrolysis of an aqueous solution of \(CuCl_2\).
Predict the products of electrolysis of a concentrated solution of \(H_2SO_4\) with platinum electrodes.
View Solution
Step 1: Understanding the Concept:
Electrolysis products vary with the concentration of the electrolyte.
Step 2: Detailed Explanation:
At Cathode: \(H^+\) ions are reduced to form Hydrogen gas (\(H_2\)).
At Anode: In concentrated \(H_2SO_4\), the oxidation of \(SO_4^{2-}\) ions is preferred over water. The sulphate ions dimerize to form peroxodisulphate (\(S_2O_8^{2-}\)).
Reaction: \(2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-\).
Step 3: Final Answer:
Hydrogen gas at cathode and \(H_2S_2O_8\) at anode.
Quick Tip: In dilute \(H_2SO_4\), Oxygen gas is evolved at the anode, but in concentrated solution, peroxodisulphuric acid is formed.
How much charge in Faraday is required for the reduction of 1 mole of \(Ag^+\) to Ag?
View Solution
Step 1: Understanding the Concept:
The charge required is related to the number of moles of electrons transferred.
Step 2: Detailed Explanation:
The reduction reaction is: \(Ag^+ + e^- \rightarrow Ag\).
One mole of silver ions requires 1 mole of electrons.
The charge of 1 mole of electrons is 1 Faraday (\(1F\)).
Step 3: Final Answer:
Exactly 1 Faraday of charge is required.
Quick Tip: Charge required (in F) is always equal to the valency (\(n\)) of the ion multiplied by the number of moles.
State Faraday’s second law of electrolysis.
View Solution
Step 1: Understanding the Concept:
This law relates the amount of different substances produced by the same amount of electricity.
Step 2: Detailed Explanation:
Faraday's second law states that when the same quantity of electricity is passed through different electrolytes connected in series, the masses of various substances liberated at the electrodes are directly proportional to their chemical equivalent weights.
\(\frac{W_1}{W_2} = \frac{E_1}{E_2}\).
Step 3: Final Answer:
The mass deposited is proportional to the equivalent weight for a constant charge.
Quick Tip: Equivalent weight = Atomic weight / Valency.
The following reactions occur at the anode during electrolysis of aqueous sodium
chloride solution:
(1) \(Cl^- \rightarrow \frac{1}{2}Cl_2 + e^-\) (\(E^\circ = 1.36 V\))
(2) \(2H_2O \rightarrow O_2 + 4H^+ + 4e^-\) (\(E^\circ = 1.23 V\))
Which reaction is feasible at the anode during electrolysis of aqueous NaCl and why?
View Solution
Step 1: Understanding the Concept:
Selection of electrode reaction depends on both thermodynamics (\(E^\circ\)) and kinetics (overpotential).
Step 2: Detailed Explanation:
Thermodynamically, the oxidation of water (Reaction 2) has a lower potential (\(1.23 V\)) than chloride oxidation (\(1.36 V\)), so it should be favored.
However, the evolution of oxygen is a kinetically slow process. It requires an extra voltage (overpotential) to occur.
Due to this overpotential, the oxidation of \(Cl^-\) becomes more feasible at the anode.
Step 3: Final Answer:
Reaction (1) is feasible because oxygen evolution requires a high overpotential.
Quick Tip: Thermodynamics predicts Reaction 2, but Kinetics (real-world speed) dictates Reaction 1.
Calculate the freezing point of a solution when 10.5 g of \(MgBr_2\) was dissolved in 250 g of water, assuming \(MgBr_2\) undergoes complete dissociation. (Given: Molar mass of \(MgBr_2 = 184 g mol^{-1}\), \(K_f\) for water \(= 1.86 K kg mol^{-1}\))
Write two differences between ideal and non-ideal solutions.
A solution is prepared by dissolving 0.088 g of potassium sulphate in 2 L of water at \(27^{\circ}C\). Assuming complete dissociation, determine its osmotic pressure. (Given: \(R = 0.082 L atm K^{-1} mol^{-1}\), Molar mass of \(K_2SO_4 = 174 g mol^{-1}\))
What type of azeotrope will be formed by a solution of benzene and chloroform? Give reason.
Why do transition metals show variable oxidation states?
Out of \(Mn^{2+}\) and \(Zn^{2+}\), which ion will be more paramagnetic and why? (Atomic numbers: \(Mn = 25, Zn = 30\))
Which is the strongest oxidising agent among: \(Cr^{3+}, V^{3+}, Mn^{3+}\)?
Complete and balance the following equations:
(1) \(2MnO_4^{-} + 6OH^{-} \rightarrow ?\)
(2) \(5C_2O_4^{2-} + 2MnO_4^{-} + 16H^{+} \rightarrow ?\)
What is meant by lanthanoid contraction?
Why do transition metals form coloured compounds?
Why are \(E^{\circ}\) values for Mn and Zn more negative than expected?
Which is the most stable oxidation state of Co and why?
Why is \(Ce^{4+}\) in aqueous solution a good oxidising agent?
Which of the following is more reactive towards \(S_N1\) reaction: 2-bromo-2-methylbutane or 1-bromopentane?
What type of halide is present in the compound: \(CH_3 - CH(CH_3) - C(Cl) = CH_2\)
Why is chloroform stored in dark coloured bottles?
Define the following term: Ambident nucleophiles
Define the following term: Racemic mixture
Which isomer of \(C_4H_9Br\) is most reactive towards \(S_N2\) reaction?
Predict the alkene formed by dehydrohalogenation of 1-bromo-1-cyclohexylethane.
Although chlorine shows strong -I effect, it is ortho-para directing in electrophilic aromatic substitution. Why?
Write the major product in the following reaction: Chlorobenzene + Na/dry ether
Write the major product in the following reaction: p-Nitrotoluene with \(Br_2\) (heat)
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