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 56-5-2 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Chemistry 56-5-2 Question Paper with Solution PDF
| CBSE Class 12 Chemistry Question Paper With Answer Key | Download PDF | Check Solutions |
While doing qualitative analysis in chemistry lab, Abhishek added yellow coloured potassium chromate solution into a test tube. He was surprised to see the colour of the solution changing immediately to orange. He realised that the test tube was not clean and contained a few drops of some liquid. Which of the following substances will be the most likely liquid to be present in the test tube before adding potassium chromate solution?
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The role of a catalyst is to change:
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Match the type of cell given in Column I with their uses given in Column II
i. Lead storage cell > a. Wall clock
ii. Mercury cell > b. Apollo Space Programme
iii. Dry cell > c. Wrist watch
iv. Fuel cell > d. Inverter
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- Lead storage cells are primarily used in inverters and power backup systems, hence i-a.
- Mercury cells are used in precision instruments, and are a key component in the Apollo Space Programme, hence ii-b.
- Dry cells are widely used in small electronic devices like wrist watches, hence iii-c.
- Fuel cells are used in space applications for energy production, and the Apollo Space Programme is an example of their usage, hence iv-d.
Quick Tip: Matching the components of cells with their uses requires understanding the specific requirements for each application, like voltage and size.
CH\(_3\)CH\(_2\)OH can be converted to CH\(_3\)CHO by:
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- CH\(_3\)CH\(_2\)OH (ethanol) can be oxidized to CH\(_3\)CHO (acetaldehyde) by treatment with PCC (Pyridinium chlorochromate), which is a mild oxidizing agent.
- Catalytic hydrogenation is used to reduce compounds, not to oxidize them. LiAlH\(_4\) is a reducing agent as well. KMnO\(_4\) is a stronger oxidizing agent, typically used to oxidize alcohols to carboxylic acids, not aldehydes.
Quick Tip: PCC is a selective oxidant and can oxidize alcohols to aldehydes without over-oxidation to carboxylic acids.
The IUPAC name for CH\(_3\) – CH\(_2\) – N(CH\(_3\)) – CH\(_2\) – CH\(_2\) – CH\(_3\) is:
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The correct IUPAC name for the given structure is N-ethyl-N-methylpropan-1-amine. The structure contains a nitrogen atom attached to both an ethyl group and a methyl group, with the main chain being a propan-1-amine. The placement of the ethyl and methyl groups on the nitrogen is denoted in the name.
Quick Tip: When naming amines, always ensure to identify the longest carbon chain attached to the nitrogen and the appropriate position of substituents.
A plot between concentration of reactant [R] and time ‘t’ is shown below. Which of the given order of reaction is indicated by the graph?
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The treatment of ethyl bromide with alcoholic silver nitrite gives:
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Which of the following aqueous solutions will have the highest freezing point?
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Which of the following aldehydes will undergo Cannizzaro reaction?
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In which of the following groups are both ions coloured in aqueous solution?
I. Cu\(^{2+}\) II. Ti\(^{4+}\) III. Co\(^{2+}\) IV. Fe\(^{2+}\)
[Atomic number : Cu = 29, Ti = 22, Co = 27, Fe = 26]
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Cu\(^{2+}\) and Fe\(^{2+}\) are colored in aqueous solution due to the d-d transitions in their partially filled d-orbitals. Ti\(^{4+}\) and Co\(^{2+}\) do not exhibit color in aqueous solution because Ti\(^{4+}\) has a completely empty d-orbital and Co\(^{2+}\) has a stable \( 3d^7 \) configuration.
Quick Tip: Ions with partially filled d-orbitals, like Cu\(^{2+}\) and Fe\(^{2+}\), are often colored in solution due to electronic transitions.
Which of the following molecules is chiral in nature?
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CH\(_3\)CH\(_2\)COOH and CH\(_3\)CH\(_2\)COOH can be distinguished by:
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The sodium bicarbonate test is used to distinguish between carboxylic acids and other types of organic compounds. When a carboxylic acid reacts with sodium bicarbonate, it forms carbon dioxide gas, which can be observed as effervescence. Hence, CH\(_3\)CH\(_2\)COOH (a carboxylic acid) will react, but CH\(_3\)CH\(_2\)COOH will not.
Quick Tip: The effervescence produced by the reaction of a carboxylic acid with sodium bicarbonate is a key test to confirm its identity.
Assertion (A): The boiling points of alkyl halides decrease in the order \( R_1 > RBr > RCl > RF \).
Reason (R): The van der Waals forces of attraction decrease in the order \( R_1 > RBr > RCl > RF \).
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The boiling points of alkyl halides decrease in the given order, and this trend is due to the decreasing strength of van der Waals forces as the size of halide ions decreases from \( R_1 \) to RF. Therefore, Reason (R) explains Assertion (A) correctly.
Quick Tip: The boiling points of alkyl halides are influenced by the size and type of halide, with larger halides having higher boiling points due to stronger van der Waals forces.
Assertion (A): For measuring resistance of an ionic solution an AC source is used.
Reason (R): Concentration of ionic solution will change if DC source is used.
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Assertion (A): Henry’s law constant (\(k_H\)) decreases with increase in temperature.
Reason (R): As the temperature increases, solubility of gases in liquids decreases.
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As the temperature increases, the solubility of gases generally decreases, which aligns with Henry’s law. However, the decrease in Henry's law constant (\(k_H\)) with temperature is due to the increase in kinetic energy of the molecules, which leads to decreased gas solubility, not the direct effect described in Reason (R). Hence, Reason (R) is false.
Quick Tip: Remember, Henry's law describes the relationship between pressure and solubility, and the law constant generally decreases with an increase in temperature due to reduced gas solubility.
Assertion (A): The solubility of aldehydes and ketones in water decreases with increase in size of the alkyl group.
Reason (R): Aldehydes and ketones have dipole-dipole interaction.
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Based on the data given above, give plausible reason for the variation of conductivity and molar conductivity with concentration.
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(a) Calculate the elevation of boiling point of a solution when 3 g of CaCl\(_2\) (Molar mass = 111 g mol\(^{-1}\)) was dissolved in 260 g of water, assuming that CaCl\(_2\) undergoes complete dissociation. (K\(_b\) for water = 0.52 K kg mol\(^{-1}\))
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We know that the elevation in boiling point is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \( i \) is the van't Hoff factor (number of particles produced per formula unit in the solution). For CaCl\(_2\), it dissociates into 3 ions (Ca\(^{2+}\) and 2 Cl\(^-\)), so \( i = 3 \).
- \( K_b \) is the ebullioscopic constant (0.52 K kg mol\(^{-1}\) for water).
- \( m \) is the molality of the solution.
First, calculate the number of moles of CaCl\(_2\):
\[ Moles of CaCl_2 = \frac{mass of CaCl_2}{molar mass of CaCl_2} = \frac{3 g}{111 g/mol} = 0.027 mol \]
Now, calculate the molality \( m \) of the solution:
\[ m = \frac{moles of solute}{mass of solvent in kg} = \frac{0.027 mol}{0.260 kg} = 0.1038 mol/kg \]
Now calculate the elevation in boiling point:
\[ \Delta T_b = 3 \times 0.52 K kg mol^{-1} \times 0.1038 mol/kg = 0.162 K \]
Thus, the elevation in boiling point is 0.162 K. Quick Tip: Remember that the van't Hoff factor accounts for the number of ions produced upon dissociation of ionic compounds.
(a) Define rate constant.
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Write the mechanism of dehydration of ethyl alcohol with conc. H\(_2\)SO\(_4\) at 413 K.
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(a) Give a simple chemical test to distinguish between 3-pentanol and 2-pentanol.
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For the reaction A + B \(\to\) Products, the following initial rates were obtained at various initial concentrations of reactants:
Sl. No. & [A]/mol L\(^{-1}\) & [B]/mol L\(^{-1}\) & Initial rate/mol L\(^{-1}\) s\(^{-1}\)
1 & 0.1 & 0.1 & 0.05
2 & 0.2 & 0.1 & 0.10
3 & 0.1 & 0.2 & 0.05
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(a) Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 20.2 mL.
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(a) Based on Valence Bond Theory, explain the geometry and magnetic character of [NiCl\(_4\)]\(^{2-}\). [Atomic number: Ni = 28]
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(a) Among the following, which will have inversion of configuration on reaction with aqueous alkali and why?
(i) 1-chloropropane OR (ii) 2-chloro-2-methylpropane
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(a) Write the chemical equation for Williamson’s synthesis.
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Identify P, Q and R in the following reaction sequence:
\[ P \xrightarrow{NH_3} CH_3COO^- NH_4^+ \xrightarrow{\Delta} Q \xrightarrow{PCl_5} R \]
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(a) Name the vitamin whose deficiency causes pernicious anemia.
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Werner’s coordination theory in 1893 was the first attempt to explain the bonding in coordination complexes. It must be remembered that this theory was put forward before the electron had been discovered by J.J. Thomson in 1897, and before the electronic theory of valency. Werner did not have any of the modern instrumental techniques and all his studies were made using simple experimental techniques. Werner was able to explain the nature of bonding in complexes and he concluded that in complexes, the metal shows two different sorts of valency: primary and secondary. Primary valences are normally ionisable whereas secondary valences are non-ionisable.
% Answer to question (a)
(a) One mole of CrCl\(_3 \cdot 4\)H\(_2\)O precipitates one mole of AgCl when treated with excess of AgNO\(_3\) solution. Write (i) the structural formula of the complex, and (ii) the secondary valency of Cr.
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(c) (i) Arrange the following complexes in the increasing order of conductivity of their solution:
[Cr(NH\(_3\))\(_3\)Cl\(_3\)], [Cr(NH\(_3\))\(_6\)Cl\(_3\)], [Cr(NH\(_3\))\(_5\)Cl]Cl\(_2\)
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The conductivity of a solution depends on the number of ions produced when the complex dissociates in water. A complex with more dissociable ions will have higher conductivity. The dissociation of the given complexes is as follows:
- [Cr(NH\(_3\))\(_3\)Cl\(_3\)] dissociates into 4 ions (1 Cr\(^{3+}\) and 3 Cl\(^-\)).
- [Cr(NH\(_3\))\(_6\)Cl\(_3\)] dissociates into 4 ions (1 Cr\(^{3+}\) and 3 Cl\(^-\)).
- [Cr(NH\(_3\))\(_5\)Cl]Cl\(_2\) dissociates into 5 ions (1 Cr\(^{3+}\), 1 Cl\(^-\), and 2 Cl\(^-\) from the chloride).
Thus, the increasing order of conductivity is:
\[ [Cr(NH_3)_3Cl_3] < [Cr(NH_3)_6Cl_3] < [Cr(NH_3)_5Cl]Cl_2 \] Quick Tip: The conductivity of a complex is directly related to the number of ions produced during dissociation.
Carbohydrates are polyhydroxy aldehydes or ketones that represent enormous structural diversity in terms of the arrangement of atoms in space, resulting in hundreds of stereoisomers. Although the chemical properties of most stereoisomers may not be very different, their metabolic rate and utilization in biological systems is significantly different and known to influence the overall carbohydrate metabolism. Structural variants, which arise due to a different arrangement of atoms in three-dimensional space are known as stereoisomers. The number of stereoisomers can be theoretically estimated by using the formula 2n, where ‘n’ is the number of stereocenters or asymmetric (chiral) carbon atoms in a molecule. Out of these stereoisomers, there are some structures, which are mirror images of each other, and they are referred to as enantiomers.
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(a) Give chemical reactions to show the presence of an aldehydic group and straight chain in glucose.
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(a) (i) In a chemistry practical class, the teacher gave his students an amine ‘X’ having molecular formula C\(_2\)H\(_7\)N, and asked the students to identify the type of amine. One of the students, Neeta, observed that it reacts with C\(_6\)H\(_5\)SO\(_2\)Cl, to give a compound which dissolves in NaOH solution. Can you help Neeta to identify the compound ‘X’?
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The amine 'X' is most likely an aromatic amine(aniline). This is because aniline (C\(_6\)H\(_5\)NH\(_2\)) reacts with benzenesulfonyl chloride (C\(_6\)H\(_5\)SO\(_2\)Cl) to form a sulfonamide, which is soluble in NaOH solution due to the formation of a sodium salt. This is characteristic of aniline. Quick Tip: Aromatic amines react with benzenesulfonyl chloride to form sulfonamides that are soluble in NaOH.
(a) (i) When pyrolusite ore is fused with KOH, in the presence of air, a dark green coloured product ‘A’ is obtained which changes to purple coloured compound ‘B’ in acidic medium.
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(I) The formulae of ‘A’ and ‘B’ are:
- ‘A’ is K\(_2\)MnO\(_4\) (Potassium Manganate), which is dark green in colour.
- ‘B’ is KMnO\(_4\) (Potassium Permanganate), which is purple in colour.
(II) The ionic equation for the reaction when compound ‘B’ reacts with Fe\textsuperscript{2+ in acidic medium is:
\[ MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \]
Quick Tip: Potassium permanganate (KMnO\(_4\)) is a strong oxidizing agent and undergoes reduction in acidic medium.
(b) (i) While studying the periodic properties, Arti came across an abnormal behaviour in the atomic size of Hf. She found that, even though Hf is placed below Zr in the same group, both have almost similar atomic sizes.
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(I) The phenomenon responsible for the above behaviour is **lanthanide contraction**. This occurs because the 4f orbitals in the lanthanides are poorly shielded, causing the nuclear charge to pull the electrons closer, reducing the atomic size as we move across the lanthanide series. This contraction is observed in the elements following lanthanides, such as Hf.
(II) Another consequence of lanthanide contraction is that it leads to the similar size of the elements in the same group, even though they are placed in different periods. This can affect the chemical and physical properties of elements like Hf and Zr. Quick Tip: Lanthanide contraction leads to the similarity in atomic sizes of elements across periods and groups, influencing their chemical properties.
(a) (i) For a galvanic cell, the following half reactions are given. Decide, which will remain as reduction reaction and which will be reversed to become an oxidation reaction. Give reason for your answer.
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