CBSE Class 12 2025 Chemistry 56-5-3 Question Paper Set-5: Download Solutions with Answer Key

Pyridinium chlorochromate (PCC) is an oxidizing agent that can convert alcohols to aldehydes. In this case, CH\textsubscript{3CH\textsubscript{2OH (ethanol) is oxidized to CH\textsubscript{3CHO (acetaldehyde) by PCC.
Quick Tip: PCC is commonly used for selective oxidation of alcohols to aldehydes without further oxidation to carboxylic acids.

To calculate the elevation in boiling point, we use the formula:
\[ \Delta T_b = K_b \times m \]
where,
\(\Delta T_b\) = elevation in boiling point,
\(K_b\) = ebullioscopic constant for water = 0.52 K kg mol\textsuperscript{-1,
\(m\) = molality of the solution, calculated as \(\frac{mol of solute}{mass of solvent in kg}\).


The molar mass of CaCl\textsubscript{2 is 111 g/mol, so the number of moles of CaCl\textsubscript{2 is: \[ moles of CaCl\textsubscript{2} = \frac{3 g}{111 g/mol} = 0.02703 mol \]

Molality (\(m\)) is: \[ m = \frac{0.02703 mol}{0.260 kg} = 0.103 mol/kg \]

Now, calculate the elevation in boiling point: \[ \Delta T_b = 0.52 \times 0.103 = 0.05356 K \]
Thus, the elevation in boiling point is approximately \(0.054 K\).
Quick Tip: The elevation in boiling point is directly proportional to the molality of the solution, and it depends on the number of particles present in the solution.

The mechanism of dehydration of ethyl alcohol (ethanol) with concentrated sulfuric acid involves the following steps:

1. Protonation of ethanol:
\[ CH\textsubscript{3CH\textsubscript{2}OH} + H\textsubscript{2}SO\textsubscript{4} \rightarrow CH\textsubscript{3CH\textsubscript{2}OH\textsubscript{2}}^+ \]
2. Loss of water (H\textsubscript{2O) to form an ethyl carbocation:
\[ CH\textsubscript{3CH\textsubscript{2}OH\textsubscript{2}}^+ \rightarrow CH\textsubscript{3CH\textsubscript{2}}^+ + H\textsubscript{2}O \]
3. Formation of ethene (CH\textsubscript{2=CH\textsubscript{2) by elimination of a proton from the carbocation:
\[ CH\textsubscript{3CH\textsubscript{2}}^+ \rightarrow CH\textsubscript{2=CH\textsubscript{2}} + H^+ \]
Thus, ethene is formed as the final product.
Quick Tip: In acid-catalyzed dehydration reactions, ethanol first forms a carbocation intermediate, which undergoes elimination to form an alkene.

(i) Sprinkling salt on snow lowers the freezing point of water, which causes the snow to melt at lower temperatures, thus clearing the roads. This is an example of freezing point depression.

(ii) When red blood cells are placed in a 0.5% NaCl solution, the solution is isotonic to the cells, meaning there is no net movement of water into or out of the cells, and they retain their shape.

(iii) An application of reverse osmosis is the purification of drinking water. It is used to remove salts and impurities from seawater to produce fresh water.
Quick Tip: Reverse osmosis is widely used in water purification processes, including desalination of seawater.

To determine the order of the reaction with respect to A and B, we compare the rates for different concentrations.


For the reaction, rate = k[A]\textsuperscript{m[B]\textsuperscript{n, where m and n are the orders with respect to A and B, respectively.


- From experiment 1 and experiment 2:
\[ \frac{Rate 2}{Rate 1} = \frac{k[A_2]^m[B_2]^n}{k[A_1]^m[B_1]^n} = \frac{0.10}{0.05} = 2 \] \[ \Rightarrow \frac{[A_2]^m}{[A_1]^m} = \frac{2}{1} \quad (since [B] is constant) \] \[ \Rightarrow \left(\frac{0.2}{0.1}\right)^m = 2 \] \[ \Rightarrow m = 1 \]

- From experiment 1 and experiment 3:
\[ \frac{Rate 3}{Rate 1} = \frac{k[A_3]^m[B_3]^n}{k[A_1]^m[B_1]^n} = \frac{0.05}{0.05} = 1 \] \[ \Rightarrow \frac{[B_3]^n}{[B_1]^n} = 1 \quad (since [A] is constant) \] \[ \Rightarrow \left(\frac{0.2}{0.1}\right)^n = 1 \] \[ \Rightarrow n = 0 \]

Thus, the order of the reaction with respect to A is 1, and with respect to B is 0. The overall order is \(1 + 0 = 1\).
Quick Tip: To determine the order of reaction, look for patterns in the changes in concentration and corresponding rate changes. Use the rate law to find the order.

The reaction between tert-butyl methyl ether and HI leads to the cleavage of the ether bond, resulting in the formation of tert-butyl alcohol and methyl iodide:
\[ (CH_3)_3C - O - CH_3 + HI \rightarrow (CH_3)_3COH + CH_3I \]
The ether bond is broken by the nucleophilic attack of the iodide ion, producing the corresponding alcohol and alkyl iodide.
Quick Tip: Ethers undergo cleavage in the presence of strong acids like HI, breaking the C-O bond to form alcohols and alkyl iodides.

The reaction of acetaldehyde (CH\textsubscript{3CHO) with dilute NaOH forms the aldol addition product:
\[ CH_3CHO + dil. NaOH \rightarrow 3-Hydroxybutanal (Aldol product) \]
This is a classic aldol condensation reaction in the presence of base.
Quick Tip: Aldol reactions occur when aldehydes or ketones react with dilute bases to form β-hydroxy aldehydes or ketones, known as aldol products.

The reaction between acetic acid (CH\textsubscript{3COOH) and chlorine (Cl\textsubscript{2) in the presence of red phosphorus results in the formation of trichloroacetic acid:
\[ CH_3COOH + Cl_2/red P \rightarrow CCl_3COOH \]
This is a halogenation reaction where chlorine replaces the hydrogen atom at the alpha position to form trichloroacetic acid.
Quick Tip: Red phosphorus catalyzes halogenation reactions, especially in carboxylic acids, leading to the formation of halogenated derivatives like trichloroacetic acid.

The structure of β-D-Glucopyranose is as follows:
\[ β-D-Glucopyranose: \ \ HOCH_2(CHOH)_4O (with the -OH group at C1 on the same side as the CH2OH group). \] Quick Tip: The pyranose form of glucose is the most stable and common form in aqueous solution.

(i) The complex formed is \[ [CrCl_3(H_2O)_3] \], where chromium is coordinated with three chloride ions and three water molecules.

(ii) The secondary valency of Cr is 6, as it is surrounded by 6 ligands in the complex. Quick Tip: Secondary valency refers to the number of ligands directly bonded to the metal atom in a coordination complex.

1. Primary valency refers to the oxidation state of the metal ion and is ionizable. It represents the number of bonds the metal can form with counterions (such as chloride).
2. Secondary valency refers to the coordination number, which is the number of ligand atoms directly bonded to the metal ion. Secondary valency is non-ionizable. Quick Tip: Primary valencies are related to the metal’s charge, while secondary valencies are related to the coordination number.

(I) The formulae of ‘A’ and ‘B’ are:
- ‘A’ is \(K_2MnO_4\) (Potassium manganate)
- ‘B’ is \(KMnO_4\) (Potassium permanganate)

(II) The ionic equation for the reaction when compound ‘B’ reacts with \( Fe^{2+} \) in acidic medium: \[ MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} \] Quick Tip: In acidic medium, permanganate (\(MnO_4^-\)) is reduced to \(Mn^{2+}\) while Fe\(^{2+}\) is oxidized to Fe\(^{3+}\).

(I) Ce\(^{4+}\) in aqueous solution is a good oxidizing agent because it readily accepts electrons to reduce to Ce\(^{3+}\), making it an effective oxidizing agent in redox reactions.

(II) Actinoid contraction is greater from element to element than lanthanoid contraction because the 5f orbitals in actinoids are less effective in shielding the nuclear charge compared to the 4f orbitals in lanthanoids, leading to greater contraction in size across the series.

(III) The value of \( E^\circ_{Zn^{2+}/Zn} \) is more negative than expected because zinc easily loses electrons to form Zn\(^{2+}\), while the \( E^\circ_{Cu^{2+}/Cu} \) is positive because copper has a strong tendency to gain electrons and form Cu\(^{2+}\). Quick Tip: The standard electrode potential \( E^\circ \) gives an idea of the tendency of a species to gain or lose electrons. A more negative value means a stronger reducing agent, while a positive value indicates a stronger oxidizing agent.

(I) Transition metals exhibit catalytic properties due to their ability to easily change oxidation states, which allows them to facilitate and speed up various chemical reactions without being consumed in the process. The presence of vacant d-orbitals allows them to form intermediate complexes with reactants.


(II) Transition metals have high enthalpy of atomisation because of the strong metallic bonding in these metals, which is a result of the presence of unpaired d-electrons that form strong bonds between atoms, requiring more energy to break these bonds.


(III) Sc (Scandium) is a transition element because it has partially filled d-orbitals in its atomic and ionized states, which is a defining characteristic of transition metals. On the other hand, Zn (Zinc) is not considered a transition element because it has a completely filled d-orbital (3d\(^{10}\)) in its ground state and does not exhibit the typical properties of transition metals.
Quick Tip: The ability to form different oxidation states and having incomplete d-orbitals are key factors that define transition metals.

Using Kohlrausch’s law, the degree of dissociation (\(\alpha\)) can be calculated using the formula: \[ \alpha = \frac{\Lambda_{m}}{\Lambda^\circ_{m}} \]

First, we calculate the limiting molar conductivity \(\Lambda^\circ_{m}\) for NH\(_4\)OH using the sum of the conductivities of the individual ions: \[ \Lambda^\circ_{NH_4OH} = \Lambda^\circ_{NH_4^+} + \Lambda^\circ_{OH^-} = 129.8 + 217.4 = 347.2 \, S cm^2 \, mol^{-1} \]

Now, use the measured molar conductivity of the solution: \[ \alpha = \frac{9.33}{347.2} = 0.0269 \]

Thus, the degree of dissociation \(\alpha = 0.0269\).
Quick Tip: The degree of dissociation can be found by comparing the actual conductivity to the limiting conductivity at infinite dilution.

The compound 'X' is an amine with molecular formula C\(_2\)H\(_7\)N, which reacts with sulfonyl chloride (C\(_6\)H\(_5\)SO\(_2\)Cl) to form a sulfonamide. This reaction suggests that 'X' is a primary amine. The compound formed is an N-phenyl sulfonamide, which dissolves in NaOH due to the basic nature of the amine group. Quick Tip: Primary amines react with sulfonyl chloride to form sulfonamides, which are soluble in alkaline solution.

When ethylamine (C\(_2\)H\(_5\)NH\(_2\)) reacts with chloroform (CHCl\(_3\)) in the presence of ethanolic KOH, a foul-smelling gas, **isocyanide** (also known as **carbylamine**) is formed. The reaction is: \[ C_2H_5NH_2 + CHCl_3 + KOH \rightarrow C_2H_5NC + KCl + H_2O \] Quick Tip: The formation of isocyanide (carbylamine) is a distinctive test for primary amines.