CBSE Class 12 Chemistry exam was conducted on February 27, 2025, from 10:30 AM to 1:30 PM. An estimated 17.5 lakh students are expected to appear for the exam across 7,800 centers in India and 26 other countries
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, and it includes numerical, conceptual, and application-based problems. The question paper includes multiple-choice questions (1 mark each), short-answer questions (3 marks each), and long-answer questions (5 marks each).
CBSE Class 12 Chemistry Question Paper 2025 (Set 2 – 56/1/2) with Answer Key
| CBSE Class 12 2025 Chemistry Question Paper with Answer Key |  Download |
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Assertion (A) : All naturally occurring \(\alpha\)-amino acids except glycine are optically active..
Reason (R) : Most naturally occurring amino acids have L-configuration..
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Assertion (A): The boiling point of ethanol is higher than that of methoxymethane..
Reason (R) : There is intramolecular hydrogen bonding in ethanol..
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Assertion (A): The boiling points of alkyl halides decrease in the order: R-I \(>\) R-Br \(>\) R-Cl \(>\) R-F..
Reason (R): The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass..
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Assertion (A): [Cr(H\(_2\)O)\(_6\)]Cl\(_2\) and [Fe(H\(_2\)O)\(_6\)]Cl\(_2\) are examples of homoleptic complexes..
Reason (R) : All the ligands attached to the metal are the same..
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An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because \hspace{2cm}..
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Which of the following statements is not true about glucose ?.
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The best reagent for converting propanamide into propanamine is.
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The acid formed when propyl magnesium bromide is treated with CO\(_2\) followed by acid hydrolysis is :
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Which is the correct order of acid strength from the following?.
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Alkyl halides undergoing nucleophilic bimolecular substitution reaction involve.
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Arrange the following compounds in increasing order of their boiling points:
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- The Correct Order is:
- (A) (ii) \(<\) (i) \(<\) (iii)
- (B) (i) \(<\) (ii) \(<\) (iii)
- (C) (iii) \(<\) (i) \(<\) (ii)
- (D) (iii) \(<\) (ii) \(<\) (i)
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The correct IUPAC name of [Pt(NH\(_3\))\(_2\)Cl\(_2\)]\(^{2+}\) is.
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Acidified KMnO\(_4\) oxidises sulphite to.
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The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr\(^{3+}\) ion (Atomic no. : Cr = 24) is ______________.
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Standard electrode potential for Sn\(^{4+}\)/Sn\(^{2+}\) couple is +0.15 V and that for the Cr\(^{3+}\)/Cr couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be.
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In case of association, abnormal molar mass of solute will.
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Identify A and B in each of the following reaction sequence :
(a) CH\(_3\)CH\(_2\)Cl \(\xrightarrow{NaCN}\) A \(\xrightarrow{H_2/Ni}\) B.
(b) C\(_6\)H\(_5\)NH\(_2\) \(\xrightarrow{NaNO_2/HCl, 0-5^\circ C}\) A \(\xrightarrow{H^+/H_2O, \Delta}\) B.
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When FeCr\(_2\)O\(_4\) is fused with Na\(_2\)CO\(_3\) in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured (C). An acidified solution of compound (C) oxidises iodide to (D). Identify (A), (B), (C) and (D)..Correct Answer:
Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Justify your answer..
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(A) Give reasons :
(a) Cooking is faster in pressure cooker than in an open pan..
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution ? What change in temperature would you observe after mixing liquids X and Y ?.
OR.
(B) Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law ? Give an example..
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Give reasons for the following :
(a) The melting points of \(\alpha\)-amino acids are generally higher than that of the corresponding carboxylic acids..
(b) Amino acids show amphoteric behaviour..
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A solution containing 15 g urea (molar mass = 60 g mol\(^{-1}\)) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water. Calculate the mass of glucose present in one litre of its solution..
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Calculate \(\Delta_r G^\circ\) and log K\(_C\) of the reaction:
Fe\(^{2+}\)(aq) + Ag\(^+\)(aq) \(\rightarrow\) Fe\(^{3+}\)(aq) + Ag(s).
Given \(E^\circ_{Ag^+/Ag}\) = 0.80 V, \(E^\circ_{Fe^{3+}/Fe^{2+}}\) = 0.77 V.
[R = 8.314 J K\(^{-1}\) mol\(^{-1}\), F = 96500 C mol\(^{-1}\)].
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The overall reaction shows Ag\(^+\) being reduced to Ag(s) and Fe\(^{2+}\) being oxidized to Fe\(^{3+}\).
Identify the half-reactions and their standard potentials:
Cathode (Reduction): Ag\(^+\)(aq) + e\(^-\) \(\rightarrow\) Ag(s); \(E^\circ_{cathode} = E^\circ_{Ag^+/Ag} = +0.80\) V
Anode (Oxidation): Fe\(^{2+}\)(aq) \(\rightarrow\) Fe\(^{3+}\)(aq) + e\(^-\). The standard potential for this oxidation is the negative of the reduction potential for Fe\(^{3+}\)/Fe\(^{2+}\). However, we use the reduction potentials directly in the \(E^\circ_{cell}\) formula. \(E^\circ_{anode} = E^\circ_{Fe^{3+}/Fe^{2+}} = +0.77\) V.
Calculate the standard cell potential (\(E^\circ_{cell}\)): \(\) E^\circ_{cell = E^\circ_{\text{cathode - E^\circ_{\text{anode \(\)
\(\) E^\circ_{\text{cell = (+0.80 \, \text{V) - (+0.77 \, \text{V) \(\)
\(\) E^\circ_{\text{cell = +0.03 \, \text{V \(\)
The number of electrons transferred (\(n\)) in the balanced reaction is 1 (since one electron is lost by Fe\(^{2+\) and one electron is gained by Ag\(^+\)). So, \(n = 1\).
Calculate the standard Gibbs free energy change (\(\Delta_r G^\circ\)): \(\) \Delta_r G^\circ = -nFE^\circ_{cell \(\)
\(\) \Delta_r G^\circ = -(1 \, \text{mol) \times (96500 \, \text{C mol^{-1) \times (0.03 \, \text{V) \(\)
\(\) \Delta_r G^\circ = - (96500 \times 0.03) \, \text{J \(\)
\(\) \Delta_r G^\circ = - 2895 \, \text{J/mol = -2.895 \, \text{kJ/mol \(\)
Calculate the logarithm of the equilibrium constant (\(\log_{10 K_C\)):
Using the relationship \( E^\circ_{cell} = \frac{0.0591}{n} \log_{10} K_c \) at 298 K (assuming standard temperature): \(\) 0.03 \, V = \frac{0.0591 \, \text{V{1 \log_{10 K_c \(\)
\(\) \log_{10 K_c = \frac{0.03{0.0591 \approx 0.5076 \(\)
Alternatively, using \(\Delta_r G^\circ = -2.303 RT \log_{10 K_c\): \(\) \log_{10 K_c = -\frac{\Delta_r G^\circ{2.303 RT = -\frac{-2895 \, J mol^{-1{2.303 \times (8.314 \, \text{J K^{-1 \, \text{mol^{-1) \times (298 \, \text{K) \(\)
\(\) \log_{10 K_c = \frac{2895{5705.8 \approx 0.5074 \(\)
Results: \(\) \Delta_r G^\circ \approx -2.90 \, \text{kJ/mol \(\)
\(\) \log_{10 K_c \approx 0.507 \(\) Quick Tip: \textbf{Electrochemical Calculations. Identify cathode (higher \(E^\circ_{red}\)) and anode (lower \(E^\circ_{red}\)). Calculate \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\). Find \(n\) (electrons transferred in balanced reaction). Use \(\Delta G^\circ = -nFE^\circ_{cell}\) and \(E^\circ_{cell} = (0.0591/n) \log_{10} K_c\) (at 298K).
(a) Arrange the following in decreasing order of pK\(_b\) :
Aniline, p-nitroaniline, p-methylaniline.
(b) Account for the following :
(i) Diazonium salts of aromatic amines are more stable than those of aliphatic amines..
(ii) Methylamine in water reacts with FeCl\(_3\) to precipitate hydrated ferric oxide..
(A) Draw the structure of the major monohalo product for each of the following reaction:
(a) Ethylbenzene + Br\(_2\)/Heat.
(b) Methylcyclohexane + HBr.
(c) p-Cresol + HCl/Heat.
OR.
(B) How do you convert:
(a) Chlorobenzene to biphenyl.
(b) Propene to 1-Iodopropane.
(c) 2-bromobutane to but-2-ene..
The elements of 3d transition series are given as : Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn.
Answer the following:
(a) Copper has exceptionally positive \(E^\circ_{M^{2+}/M}\) value, why?.
(b) Which element is a strong reducing agent in +2 oxidation state and why?.
(c) Zn\(^{2+}\) salts are colourless. Why?.
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A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.
[R = 8.314 J K\(^{-1}\) mol\(^{-1}\); log 4 = 0.602].
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(a) Write the reaction when D-glucose reacts with the following:
(b) Why vitamin C cannot be stored in our body?
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Read the passage about chemical kinetics and answer the following questions:
The following questions are case based questions.
Read the passage carefully and answer the questions that follow.
The rate of chemical reaction is expressed either in terms of the decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of peactants, concentration of reactants, temperature, presence of catalyst, gurface area of the reactants and presence of light. Rate of reaction iв directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the Rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction.
(a) (i) What is a rate determining step ?.
(ii) Define complex reaction..
(b) What is the effect of temperature on the rate constant of a reaction ?.
OR.
(b) Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions ?.
(c) The conversion of molecule X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y ?.
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Read the passage about phenol chemistry and answer the following questions:
The following questions are case based questions. Read the passage carefully and answer the questions that follow. Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more to o- and p-positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydrozyl group. This is a general method used for the ortho-formylation of phenols.
(a) What happens when phenol reacts with.
(i) Br\(_2\)/CS\(_2\).
(ii) Conc. HNO\(_3\).
(b) Why phenol does not undergo protonation readily ?.
(c) Which is a stronger acid - phenol or cresol ? Give reason..
OR.
(c) Write the IUPAC name of the product formed in the Reimer-Tiemann reaction..
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(A).
(a) Carry out the following conversions :
(i) Ethanal to But-2-enal.
(ii) Propanoic acid to ethane.
(b) An alkene A with molecular formula C\(_5\)H\(_{10}\) on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C..
OR.
(B) An organic compound (A) (molecular formula C\(_8\)H\(_{16}\)O\(_2\)) was hydrolyzed with dilute sulphuric acid to get a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives But-1-ene. Identify (A), (B) and (C) and write chemical equations for the reactions involved..
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(A).
For the complex \([Fe(en)_2Cl_2]^+ \), identify:
(B)
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(a) Oxidation number of iron:
In the complex \([Fe(en)_2Cl_2]^+\), \(en\) (ethylenediamine) is a neutral ligand, and \(Cl\) is a monodentate anionic ligand. Let the oxidation number of Fe be \(x\). For two chloride ions, each with a charge of -1, the charge balance equation is: \[ x + 2(-1) + 0 = +1 \quad (since the overall charge of the complex is +1) \]
Thus, \(x = +3\). Therefore, the oxidation number of iron is +3.
(b) Hybridization and Shape of the Complex:
Since the complex involves 2 bidentate ethylenediamine ligands and 2 chloride ions, the coordination number of iron is 6. In this case, iron will be in an octahedral arrangement with respect to the ligands. Hence, the hybridization of Fe is \(d^2sp^3\), and the shape of the complex is octahedral.
(c) Magnetic Behaviour of the Complex:
The electronic configuration of Fe\(^{3+}\) is \(3d^5\). Since there are 5 unpaired electrons in the \(d\)-orbitals, the complex will be paramagnetic.
(d) Optical Isomerism:
Since the complex contains two bidentate ligands (ethylenediamine), it may exhibit optical isomerism. This is because the complex does not have a plane of symmetry, making it capable of existing as non-superimposable mirror images. Therefore, the complex exhibits optical isomerism.
(e) IUPAC Name of the Complex:
The IUPAC name of the complex is bis(ethylenediamine)chloridoiron(III) chloride.
% Section (B)
OR.
(B) (a) Using IUPAC norms, write the names of the following:
% Option
(i) \([Co(NH_3)_4Cl(NO_3)]Cl\)
% Option
(ii) \(K_3[Fe(CN)_6]\)
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(iii) \([Cr(C_2O_4)_3]^{3-}\)
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Solution:
(i) IUPAC Name:
For \([Co(NH_3)_4Cl(NO_3)]Cl\), the central metal is cobalt, and the ligands are four ammine (NH\(_3\)), one chloride (Cl), and one nitrate (NO\(_3\)). The IUPAC name is tetraamminechloridonitratocobalt(III) chloride.
(ii) IUPAC Name:
For \(K_3[Fe(CN)_6]\), the central metal is iron, and the ligand is cyanide (CN). The complex is in the +3 oxidation state of iron. The IUPAC name is potassium hexacyanoferrate(III).
(iii) IUPAC Name:
For \([Cr(C_2O_4)_3]^{3-}\), the central metal is chromium, and the ligand is oxalate (\(C_2O_4^{2-}\)). The IUPAC name is tris(oxalato)chromium(III).
(b) What is crystal field splitting energy? Why low spin tetrahedral complexes are not formed?.
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Solution:
Crystal Field Splitting Energy (CFSE):
Crystal Field Splitting Energy is the energy difference between two sets of degenerate orbitals in a metal complex, caused by the interaction between the central metal ion and the ligands. In octahedral complexes, the d-orbitals split into two sets, \(e_g\) (higher energy) and \(t_{2g}\) (lower energy), and the difference between these energy levels is called the crystal field splitting energy (\(\Delta\)). The value of \(\Delta\) determines the stability and magnetic properties of the complex.
Low Spin Tetrahedral Complexes:
Tetrahedral complexes are generally high spin because the crystal field splitting energy (\(\Delta_t\)) in tetrahedral geometry is much smaller than in octahedral geometry. In tetrahedral complexes, the splitting energy between the \(e\) and \(t_2\) orbitals is relatively small, and it does not favor the pairing of electrons. Hence, low spin tetrahedral complexes are rare. The low spin state is usually favored in octahedral complexes with strong field ligands.
Quick Tip: \textbf{Magnetic Properties and Field Strength.} In octahedral complexes, stronger field ligands lead to larger crystal field splitting energy (\(\Delta\)), which may favor the low spin state (fewer unpaired electrons). In tetrahedral complexes, the splitting energy is smaller, making them typically high spin with more unpaired electrons.
(A).
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
Sn(s) | Sn\(^{2+}\) (0.004 M) || H\(^+\) (0.02 M) | H\(_2\)(g) (1 Bar) | Pt(s).
(Given: \(E^\circ_{Sn^{2+}/Sn}\) = -0.14 V, \(E^\circ_{H^+/H_2(g), Pt}\) = 0.00V).
(b) Account for the following ; % Note: Sub-question (b) from image 10, page 23 appears here.
(i) On the basis of E\(^\circ\) values, O\(_2\) gas should be liberated at anode but it is Cl\(_2\) gas which is liberated in the electrolysis of aqueous NaCl..
(ii) Conductivity of CH\(_3\)COOH decreases on dilution..
OR.
(B).
(a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use..
(b) Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01M Cr\(^{3+}\) (aq) and 1.0 x 10\(^{-4}\) M H\(^+\)(aq)..
The half cell reaction is.
Cr\(_2\)O\(_7^{2-}\)(aq) + 14H\(^+\)(aq) + 6e\(^-\) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + 7H\(_2\)O(l).
and the standard electrode potential is given as E\(^\circ\) = 1.33 V..
[Given : log 10 = 1].
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