CBSE conducted the Class 12 Chemistry Board Exam on February 27, 2025, from 10:30 AM to 1:30 PM. The Chemistry theory paper has 70 marks, while 30 marks are allocated for the practical assessment.The paper is divided into Physical, Organic, and Inorganic Chemistry, with MCQ (1 mark each), short-answer questions (3 marks each), and long-answer questions (5 marks each).
CBSE Class 12 Chemistry 56-6-3 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Chemistry 56-6-3 Question Paper with Solution PDF
| CBSE Class 12 Chemistry Question Paper With Answer Key | Download PDF | Check Solutions |
During dehydration of alcohol to alkene by heating with conc. H\(_2\)SO\(_4\), the initiation step is:
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Which reagents are required for one step conversion of Chlorobenzene to Diphenyl?
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Which of the following complexes show linkage isomerism?
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Which of the following transition metals has the highest melting point?
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The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is: (K\(_f\) for water = 1.86 K kg mol\(^{-1}\))
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The freezing point depression can be calculated using the formula: \[ \Delta T_f = i \cdot K_f \cdot m \]
where \(i = 2\) for KCl (since it dissociates into two ions), \(K_f = 1.86 \, K kg mol^{-1}\), and \(m = 1 \, mol/kg\). Thus: \[ \Delta T_f = 2 \times 1.86 \times 1 = 3.72 \, °C \]
The freezing point of water is 0°C, so the freezing point of the solution is -3.72°C.
Quick Tip: For molal solutions, the freezing point depression can be calculated using the formula involving the cryoscopic constant and the molality.
The reagent that can be used to convert benzene diazonium chloride to benzonitrile is:
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The conversion of benzene diazonium chloride to benzonitrile is carried out using copper(I) cyanide (CuCN) in the presence of a suitable solvent. This is a nucleophilic substitution reaction where the cyanide ion replaces the diazonium group.
Quick Tip: Benzene diazonium salts undergo nucleophilic substitution reactions with cyanide to form benzonitrile.
The strongest base in aqueous solution among the following amines is:
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On mixing 30 mL of acetone with 20 mL of chloroform, the total volume of solution is:
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We cannot measure the resistance of an ionic solution using DC because:
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The primary and secondary valences of Co in [Co(en)₃]Cl₃ respectively are:
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In the Arrhenius equation \( k = Ae^{-E_a/RT} \), ‘A’ represents:
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\(\alpha\)-D-glucose and \(\beta\)-D-glucose differ from each other with respect to the:
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Assertion (A): Molecularity of reaction is determined experimentally.
Reason (R): Molecularity is applicable only for an elementary reaction and not for a complex reaction.
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Assertion (A): Boiling point of (C\(_6\)H\(_5\))\(_2\)NH is lower than that of n-C\(_4\)H\(_9\)NH\(_2\).
Reason (R): Hydrogen bonding is much more extensive in (C\(_6\)H\(_5\))\(_2\)NH as compared to n-C\(_4\)H\(_9\)NH\(_2\).
Assertion (A): The bond angle C-O-C in ethers is slightly greater than tetrahedral angle.
Reason (R): This is because of the repulsive interaction between the two bulky alkyl groups.
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Define osmotic pressure. Why is the measurement of osmotic pressure method preferred for the determination of molar masses of macromolecules such as proteins and polymers?
How do you explain the following?
Draw the structures of product(s) in each of the following reactions:
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In this reaction, an alkene (\(CH_2CH_3\)) reacts with bromine (\(Br_2\)) in the presence of heat. This reaction is an electrophilic addition reaction, where the bromine adds across the double bond, resulting in the formation of 1,2-dibromoethane.
Quick Tip: The addition of halogens to alkenes typically occurs via an electrophilic addition mechanism, which is favored by heat or light.
Observe the graph in the given figure and answer the following questions:
Write IUPAC names of the following coordination compounds:
How can you obtain the following?
Give reasons for the following:
The rate of a reaction:
A + B \(\longrightarrow\) product
is given below as a function of different initial concentrations of A and B.
Calculate the order of the reaction with respect to A and B. Determine the rate constant of the reaction.
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From the given data, we can calculate the order of reaction with respect to A and B by comparing the rate changes with changes in the concentrations of A and B.
- For experiment 1 and 2, the change in the rate of reaction is due to a change in the concentration of A. By comparing the rates, we can calculate the order with respect to A.
- For experiment 1 and 3, the change in the rate of reaction is due to a change in the concentration of B. Similarly, we calculate the order with respect to B.
After determining the order with respect to A and B, we can use the rate law to calculate the rate constant.
Quick Tip: The order of a reaction can be determined by observing how the rate changes as the concentrations of reactants are varied. The rate law is then used to determine the rate constant.
Write any two differences between SN\(_1\) and SN\(_2\) reactions. Which of the following compounds would undergo SN\(_1\) reaction faster and why?
A compound (A) with molecular formula C\(_4\)H\(_5\)N on reduction with DIBAL-H followed by hydrolysis, gives a compound (B). Compound (B) gives positive Tollens' test but does not give iodoform test. Compound (B) can also be obtained when ethanol is treated with dilute NaOH followed by heating. Identify (A) and (B). Write the reactions of (A) with DIBAL-H followed by hydrolysis.
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- (A): The compound (A) is an aldehyde, specifically \(C_4H_5N\) or butyraldehyde.
- (B): The product (B) formed from reduction and hydrolysis is an alcohol, likely \(C_4H_9OH\), which gives a positive Tollens' test (indicating the presence of an aldehyde group) but does not give the iodoform test.
- Reaction of A with DIBAL-H: The reduction of the aldehyde with DIBAL-H will yield the alcohol, which will then undergo hydrolysis to form the final product.
Quick Tip: DIBAL-H (Diisobutylaluminum hydride) is a selective reducing agent used to reduce esters to aldehydes and aldehydes to alcohols.
Answer the following about the complexes
\([FeF_6]^{3-}\) and \([Fe(CN)_6]^{4-}\):
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- \([FeF_6]^{3-}\): The hybridization of the iron in this complex is \(sp^3d^2\), as it involves six fluoride ions, indicating an octahedral geometry.
- \([Fe(CN)_6]^{4-}\): The hybridization of the iron in this complex is also \(sp^3d^2\), as it involves six cyanide ions in an octahedral geometry.
Quick Tip: Both complexes have an octahedral geometry due to the six ligands coordinating to the central metal, resulting in \(sp^3d^2\) hybridization.
Vapour pressure of pure water at 298 K is 24.8 mm Hg. Calculate the lowering in vapour pressure of an aqueous solution which freezes at -0.3\(^\circ\)C. (K\(_f\) of water = 1.86 K kg mol\(^{-1}\))
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We can use the formula for freezing point depression: \[ \Delta T_f = K_f \times m \]
where:
- \(\Delta T_f\) is the freezing point depression,
- \(K_f\) is the cryoscopic constant (given as 1.86 K kg mol\(^{-1}\)),
- \(m\) is the molality of the solution.
The freezing point depression is given as \( \Delta T_f = 0.3 \, ^\circ C \), so: \[ \Delta T_f = K_f \times m \implies 0.3 = 1.86 \times m \]
Solving for \(m\), we get: \[ m = \frac{0.3}{1.86} = 0.1613 \, mol/kg \]
Now, we use Raoult's Law for the lowering of vapour pressure: \[ \Delta P = P_0 \times x_B \]
where:
- \(P_0\) is the vapour pressure of pure water (24.8 mm Hg),
- \(x_B\) is the mole fraction of the solute.
Since the solution is dilute, \(x_B \approx \frac{m}{1000}\), assuming 1 kg of water as the solvent: \[ x_B \approx \frac{0.1613}{1000} = 1.613 \times 10^{-4} \]
Now, calculating the lowering in vapour pressure: \[ \Delta P = 24.8 \times 1.613 \times 10^{-4} = 0.004 \, mm Hg \]
Thus, the lowering in vapour pressure is approximately 0.004 mm Hg.
Quick Tip: The freezing point depression is directly proportional to the molality of the solution, and the lowering in vapour pressure can be calculated using Raoult's law.
Alcohols undergo a number of reactions involving the cleavage of C – OH bond. However, phenols do not undergo reactions involving the cleavage of C – OH bond. Alcohols are weaker acids than water. Alcohols react with halogen acids to form the corresponding halolalkanes. Phenols are stronger acids than alcohols. A characteristic feature of phenols is that they undergo electrophilic substitution reactions such as halogenation, nitration, etc. Since –OH group is a strong activating group, phenol gives trisubstituted products during halogenation, nitration, etc.
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- When phenol is treated with Br\(_2\) water, it undergoes electrophilic substitution, giving 2,4,6-tribromophenol.
- When phenol is treated with concentrated nitric acid (HNO\(_3\)), it undergoes nitration to form 2,4,6-trinitrophenol (also known as picric acid).
Quick Tip: The –OH group in phenols is a strong activating group, making the aromatic ring more reactive towards electrophilic substitution reactions like halogenation and nitration.
The \(\alpha\)-amino acids are the building blocks of proteins. All \(\alpha\)-amino acids exist as zwitter ion due to which they show amphoteric behaviour. All amino acids are joined through peptide bond. Proteins are broadly classified as globular proteins and fibrous proteins. Globular proteins are water soluble, whereas fibrous proteins are not. The complete structure of protein is discussed at four different levels i.e. primary, secondary, tertiary and quaternary structures. Protein loses its biological activity in denatured form.
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(i) Peptide linkage: A peptide linkage is the bond formed between the carboxyl group of one amino acid and the amino group of another amino acid, with the elimination of water. This forms a dipeptide, and the peptide bond is the backbone of proteins.
(ii) Denatured protein: Denaturation is the process in which a protein loses its native, functional shape due to external factors like heat, pH, or chemicals, causing it to lose its biological activity.
Quick Tip: Peptide bonds are covalent links between amino acids, and protein denaturation involves the disruption of secondary, tertiary, or quaternary structure without breaking peptide bonds.
(a) Account for the following:
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(i) The orange colour of Cr\(_2\)O\(_7^{2-}\) ion is due to the presence of Cr in the +6 oxidation state. When treated with an alkali, the Cr\(_2\)O\(_7^{2-}\) ion undergoes a reduction to Cr\(^3+\), which gives a yellow colour.
(ii) Zn, Cd, and Hg are non-transition elements because they do not have incomplete d-orbitals in their ground state or in any of their possible oxidation states. They are classified as post-transition elements.
(iii) The E\(^0\) value for Mn\(^3^+\)/Mn\(^2^+\) is highly positive (+1.57 V) because Mn\(^3^+\) has a strong tendency to be reduced to Mn\(^2^+\), as Mn\(^2^+\) is a more stable oxidation state. In contrast, Cr\(^3^+\)/Cr\(^2^+\) has a relatively lower E\(^0\) because Cr\(^3^+\) is a more stable oxidation state compared to Cr\(^2^+\).
Quick Tip: Oxidation states and their stability influence the electrochemical potentials of metal ions. The more stable the ion, the higher the positive potential.
(a) Identify A, B and C in the following reactions:
- (b) \textbf{Write the product(s) in the following reactions:}
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- (I) 2\(CH_3COOH\) \xrightarrow{P_4O_{10}, heat}
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(ii) Write the reaction involved in the following reactions:
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The Wolff-Kishner reduction is used to reduce carbonyl compounds (such as aldehydes or ketones) to alkanes. The reaction involves the use of hydrazine (N\(_2\)H\(_4\)) and a strong base, typically potassium hydroxide (KOH), under heating conditions. The reaction is: \[ R-C(=O)-R' + N_2H_4 \xrightarrow{KOH, heat} R-CH_2R' + N_2 + H_2O \]
Quick Tip: The Wolff-Kishner reduction is a useful method for reducing ketones or aldehydes to alkanes without affecting other functional groups.
(a) Calculate \(E_{cell}\) of a galvanic cell in which the following reaction takes place at 25°C:
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