HSC 12th Chemistry June 2026 Board Paper with Complete Solution (J-261) | Maharashtra Board
Here is the complete HSC 12th Chemistry June 2026 Board Question Paper (Subject Code: J-261) along with step-by-step model solutions, numerical calculations, and chemical reactions prepared strictly as per the Maharashtra State Board marking scheme.
SECTION - A
Q. 1. Select and write the correct answers for the following multiple choice type of questions: [10 Marks]
- Which of the following is NOT an elastomer?
Answer: (d) polyethene - Anisole reacts with \(\mathrm{HI}\) at \(398\text{ K}\) to form
Answer: (b) benzenol - Unit of Henry's constant is
Answer: (a) \(\mathrm{mol\ L^{-1}\ bar^{-1}}\) - The shape of bromine penta-fluoride is
Answer: (d) Square pyramidal - The relation between edge length and radius for body centered cubic structure is -
Answer: (b) \(r = \frac{\sqrt{3}a}{4}\) - Atomic number of post actinoid element range from -
Answer: (d) \(104\text{–}118\) - The \(\mathrm{pH}\) of buffer solution containing \(0.05\text{ mol }\mathrm{NaF}\) per litre and \(0.01\text{ mol }\mathrm{HF}\) per litre (\(\mathrm{p}K_a = 3.1427\)) is.
Answer: (a) \(3.8417\)
[Hint: \(\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[\text{Salt}]}{[\text{Acid}]} = 3.1427 + \log_{10}\left(\frac{0.05}{0.01}\right) = 3.1427 + 0.6990 = 3.8417\)] - Preparation of ethanamine from acetonitrile represents -
Answer: (b) Mendius reduction - The source of dichlorocarbene is
Answer: (c) chloroform - The reagent used to convert acetyl chloride to acetaldehyde is -
Answer: (d) \(\mathrm{H_2 / Pd\text{-}BaSO_4}\)
Q. 2. Answer the following questions: [8 Marks]
- Write the relation between conductivity and molar conductivity.
Answer: The relation between conductivity (\(k\)) and molar conductivity (\(\Lambda\)) is: \[\Lambda = \frac{1000\,k}{c}\] where \(\Lambda\) is molar conductivity (in \(\mathrm{\Omega^{-1}\ cm^2\ mol^{-1}}\)), \(k\) is conductivity (in \(\mathrm{\Omega^{-1}\ cm^{-1}}\)), and \(c\) is molar concentration (in \(\mathrm{mol\ L^{-1}}\) or \(\mathrm{mol\ dm^{-3}}\)).
(In SI units, \(\Lambda = \dfrac{k}{c}\), where \(k\) is in \(\mathrm{S\ m^{-1}}\) and \(c\) is in \(\mathrm{mol\ m^{-3}}\)). - Write the oxidation state of Nickel in \([\mathrm{Ni(Cl)_4}]^{2\ominus}\) complex ion.
Answer: Let the oxidation state of \(\mathrm{Ni}\) be \(x\).
\(x + 4(-1) = -2 \implies x = +2\).
Therefore, the oxidation state of Nickel is \(+2\). - Write the name of nanostructured material used in tyres of car to increase their life.
Answer: Carbon black - Write the glycosidic linkage present in maltose.
Answer: \(\alpha\text{-}1,4\text{-glycosidic linkage}\) - Draw the structure of perchloric acid.
Answer: Perchloric acid (\(\mathrm{HClO_4}\)) has a tetrahedral structure with a central chlorine atom bonded to three oxygen atoms via double bonds and one hydroxyl (\(\mathrm{-OH}\)) group via a single bond: \[\begin{array}{c} \mathrm{O} \\ \parallel \\ \mathrm{O = Cl - OH} \\ \parallel \\ \mathrm{O} \end{array}\] - Write the trade name of most common thermoplastic polyester.
Answer: Terylene or Dacron (Chemical name: Polyethylene terephthalate, PET). - Calculate change over temperature for a certain reaction, whose \(\Delta H^\circ\) is \(-60\text{ kJ}\) and \(\Delta S^\circ\) is \(-150\text{ J K}^{-1}\).
Answer:
Given: \(\Delta H^\circ = -60\text{ kJ} = -60,000\text{ J}\), \(\Delta S^\circ = -150\text{ J K}^{-1}\). \[T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{-60,000\text{ J}}{-150\text{ J K}^{-1}} = 400\text{ K}\] - Write the unit of rate constant of zero order reaction.
Answer: \(\mathrm{mol\ L^{-1}\ s^{-1}}\) or \(\mathrm{mol\ dm^{-3}\ s^{-1}}\) (or \(\mathrm{mol\ dm^{-3}\ time^{-1}}\)).
SECTION - B
Attempt any EIGHT of the following questions: [16 Marks]
Q. 3. Write reactions involved during preparation of Nylon 6, 6 fibre.
Answer:
Nylon 6,6 is prepared by the condensation polymerization of equimolar aqueous solutions of adipic acid and hexamethylenediamine:
Step 1: Formation of Nylon salt
\[n\,\mathrm{HOOC-(CH_2)_4-COOH} \text{ (Adipic acid)} + n\,\mathrm{H_2N-(CH_2)_6-NH_2} \text{ (Hexamethylenediamine)} \longrightarrow n\left[\mathrm{^\ominus OOC-(CH_2)_4-COO^\ominus\ H_3\overset{\oplus}{N}-(CH_2)_6-\overset{\oplus}{N}H_3}\right] \text{ (Nylon salt)}\]Step 2: Condensation polymerization to Nylon 6,6
\[n\left[\mathrm{^\ominus OOC-(CH_2)_4-COO^\ominus\ H_3\overset{\oplus}{N}-(CH_2)_6-\overset{\oplus}{N}H_3}\right] \xrightarrow[\text{High pressure}]{553\text{ K}} \left[\mathrm{-OC-(CH_2)_4-CO-NH-(CH_2)_6-NH-}\right]_n \text{ (Nylon 6,6)} + 2n\,\mathrm{H_2O}\]Q. 4. Derive Ostwald's dilution law equation for a weak acid.
Answer:
Consider \(1\text{ mol}\) of a weak monobasic acid \(\mathrm{HA}\) dissolved in \(V\text{ dm}^3\) of solution (initial concentration \(c = \frac{1}{V}\text{ mol dm}^{-3}\)). The acid dissociates partially in aqueous solution as:
Let \(\alpha\) be the degree of dissociation of the weak acid at equilibrium:
- Equilibrium concentration of undissociated acid, \([\mathrm{HA}] = c(1 - \alpha)\text{ mol dm}^{-3}\)
- Equilibrium concentration of \(\mathrm{H}^+\) ions, \([\mathrm{H}^+] = c\alpha\text{ mol dm}^{-3}\)
- Equilibrium concentration of \(\mathrm{A}^-\) ions, \([\mathrm{A}^-] = c\alpha\text{ mol dm}^{-3}\)
Applying the law of mass action, the acid dissociation constant (\(K_a\)) is given by:
\[K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} = \frac{(c\alpha)(c\alpha)}{c(1 - \alpha)} = \frac{c\alpha^2}{1 - \alpha}\]For a weak acid, \(\alpha\) is very small compared to \(1\) (\(\alpha \ll 1\)), so \(1 - \alpha \approx 1\). Therefore:
\[K_a = c\alpha^2 \quad \text{or} \quad K_a = \frac{\alpha^2}{V}\] \[\alpha = \sqrt{\frac{K_a}{c}} = \sqrt{K_a \cdot V}\]This equation represents Ostwald's dilution law for a weak acid.
Q. 5. Draw the structures of two forms of alanine having \(\mathrm{pH} \le 2\) and \(\mathrm{pH} \ge 10\)
Answer:
Alanine (\(\mathrm{CH_3-CH(NH_2)-COOH}\)) exists in different ionic forms depending on the \(\mathrm{pH}\) of the aqueous medium:
1. At \(\mathrm{pH} \le 2\) (Cationic form in strongly acidic medium):
\[\begin{array}{c} \mathrm{CH_3} \\ | \\ \mathrm{H_3\overset{\oplus}{N} - CH - COOH} \end{array}\]2. At \(\mathrm{pH} \ge 10\) (Anionic form in strongly basic medium):
\[\begin{array}{c} \mathrm{CH_3} \\ | \\ \mathrm{H_2N - CH - COO^\ominus} \end{array}\]Q. 6. Define the following terms: (a) Intensive property (b) Standard enthalpy of combustion
Answer:
- (a) Intensive property: Any property of a system whose magnitude is independent of the amount of matter present in the system is called an intensive property. Examples: Temperature, pressure, density, refractive index, viscosity.
- (b) Standard enthalpy of combustion (\(\Delta_c H^\circ\)): It is defined as the enthalpy change accompanying a reaction in which one mole of a substance in its standard state is completely oxidized (burnt) in excess of oxygen at standard conditions (\(T = 298\text{ K}\) and \(P = 1\text{ bar}\)).
Q. 7. What is the action of following reagents on ethyl amine? (Indicate by reaction) (a) Nitrous acid (b) Chloroform and \(\mathrm{KOH}\)
Answer:
(a) Action of Nitrous acid (\(\mathrm{HNO_2}\)):
Ethyl amine reacts with nitrous acid (prepared in situ from \(\mathrm{NaNO_2}\) and \(\mathrm{HCl}\) at \(273\text{–}278\text{ K}\)) to form an unstable aliphatic diazonium salt which decomposes immediately to yield ethyl alcohol (ethanol) and liberate nitrogen gas:
(Overall reaction: \(\mathrm{C_2H_5NH_2} + \mathrm{HNO_2} \xrightarrow{\mathrm{NaNO_2 + HCl}} \mathrm{C_2H_5OH} + \mathrm{N_2}\uparrow + \mathrm{H_2O}\))
(b) Action of Chloroform and \(\mathrm{KOH}\) (Carbylamine reaction):
When ethyl amine is heated with chloroform and alcoholic potassium hydroxide, it forms ethyl isocyanide (ethyl carbylamine) having a foul, offensive smell:
Q. 8. (a) State Kohlrausch law of independent migration of ions. (b) Write Nernst equation to calculate electrode potential.
Answer:
(a) Kohlrausch law of independent migration of ions:
It states that at infinite dilution, each ion migrates independently of its co-ion and makes its own contribution to the total molar conductivity of an electrolyte, irrespective of the nature of the other ion with which it is associated.
(b) Nernst equation for electrode potential:
For a general reduction electrode reaction \(\mathrm{M}^{n+}_{(aq)} + n\mathrm{e}^- \longrightarrow \mathrm{M}_{(s)}\), the Nernst equation to calculate the electrode potential is:
(At \(298\text{ K}\), \(E_{\text{elec}} = E^\circ_{\text{elec}} + \dfrac{0.0592}{n}\log_{10}[\mathrm{M}^{n+}]\)), where \(E^\circ_{\text{elec}}\) is the standard electrode potential, \(R\) is the gas constant, \(T\) is absolute temperature, \(n\) is the moles of electrons transferred, and \(F\) is Faraday's constant.
Q. 9. (a) Write general electronic configuration of \(3d\) series elements. (b) Write any two points which explain about the colour of a transition metal ion.
Answer:
(a) General electronic configuration of \(3d\) series elements:
\[[\mathrm{Ar}]\,3d^{1\text{–}10}\,4s^{1\text{–}2}\](b) Explanation for the colour of transition metal ions:
- Transition metal ions generally possess one or more unpaired electrons in their incompletely filled \((n-1)d\) orbitals (\(d^1\) to \(d^9\) configurations).
- In the presence of ligands or solvent molecules, the degenerate \(d\)-orbitals split into two energy levels. When visible light falls on the ion, an electron is promoted from a lower energy \(d\)-orbital to a higher energy \(d\)-orbital (\(d\text{-}d\) transition) by absorbing a specific wavelength of visible light, and the transmitted light exhibits the complementary colour.
Q. 10. Explain aldol condensation reaction with respect to ethanal.
Answer:
When two molecules of ethanal (acetaldehyde), which contains \(\alpha\)-hydrogen atoms, are warmed with a dilute base such as dilute \(\mathrm{NaOH}\) (or \(\mathrm{KOH}\)), they undergo nucleophilic addition to form a \(\beta\)-hydroxy aldehyde called aldol (\(3\text{-hydroxybutanal}\)).
Step 1: Formation of Aldol (Addition step)
\[2\,\mathrm{CH_3-CHO} \text{ (Ethanal)} \underset{\text{warm}}{\stackrel{\text{dil. }\mathrm{NaOH}}{\rightleftharpoons}} \mathrm{CH_3-CH(OH)-CH_2-CHO} \text{ (3-Hydroxybutanal / Aldol)}\]Step 2: Dehydration of Aldol (Condensation step)
On heating, \(3\text{-hydroxybutanal}\) readily eliminates a water molecule to form an \(\alpha,\beta\)-unsaturated aldehyde called crotonaldehyde (\(\text{but-2-enal}\)).
Q. 11. Write a note on Kolbe reaction.
Answer:
When phenol is treated with aqueous sodium hydroxide, it forms sodium phenoxide, in which the phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution. When sodium phenoxide is heated with carbon dioxide (\(\mathrm{CO_2}\), a weak electrophile) at \(398\text{ K}\) under a pressure of \(6\text{ atm}\) (or \(4\text{–}7\text{ atm}\)), it undergoes electrophilic substitution predominantly at the ortho-position to form sodium salicylate. Subsequent acidification with dilute hydrochloric acid yields salicylic acid (\(2\text{-hydroxybenzoic acid}\)).
Q. 12. (a) Write the relation between standard cell potential and equilibrium constant. (b) Write the representation of sodium hexacyanoferrate (III) complex.
Answer:
(a) Relation between standard cell potential (\(E^\circ_{\text{cell}}\)) and equilibrium constant (\(K\)):
\[E^\circ_{\text{cell}} = \frac{2.303\,RT}{nF}\log_{10} K\]At \(298\text{ K}\):
\[E^\circ_{\text{cell}} = \frac{0.0592}{n}\log_{10} K\](b) Formula of sodium hexacyanoferrate (III):
\[\mathrm{Na_3[Fe(CN)_6]}\]Q. 13. In first order reaction \(60\%\) of reactant decomposes in \(40\text{ min}\). Calculate the rate constant of the reaction.
Answer:
- Given:
- Initial concentration of reactant, \([\mathrm{A}]_0 = 100\)
- Amount decomposed in \(t = 40\text{ min}\) is \(60\%\)
- Concentration of reactant remaining at time \(t\), \([\mathrm{A}]_t = 100 - 60 = 40\)
- Time, \(t = 40\text{ min}\)
- Formula: \[k = \frac{2.303}{t}\log_{10}\left(\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\right)\]
- Calculation: \[k = \frac{2.303}{40\text{ min}}\log_{10}\left(\frac{100}{40}\right) = \frac{2.303}{40}\log_{10}(2.5)\] \[k = \frac{2.303 \times 0.3979}{40} = 0.02291\text{ min}^{-1} \text{ (or } 2.291 \times 10^{-2}\text{ min}^{-1}\text{)}\]
Final Answer: The rate constant of the reaction is \(0.02291\text{ min}^{-1}\) (or \(3.818 \times 10^{-4}\text{ s}^{-1}\)).
Q. 14. Explain optical activity of 2-chlorobutane.
Answer:
- In \(2\text{-chlorobutane}\) (\(\mathrm{CH_3-\overset{*}{C}H(Cl)-CH_2-CH_3}\)), the second carbon atom (\(\mathrm{C_2}\)) is bonded to four different substituent groups: \(\mathrm{-H}\), \(\mathrm{-Cl}\), \(\mathrm{-CH_3}\) (methyl), and \(\mathrm{-C_2H_5}\) (ethyl). Hence, \(\mathrm{C_2}\) is an asymmetric (chiral) carbon atom.
- Due to the presence of this chiral carbon, the \(2\text{-chlorobutane}\) molecule is unsymmetrical and lacks a plane of symmetry.
- As a result, \(2\text{-chlorobutane}\) exists as a pair of non-superimposable mirror-image stereoisomers called enantiomers: \[\begin{array}{ccc} \mathrm{C_2H_5} & \vdots & \mathrm{C_2H_5} \\ | & \vdots & | \\ \mathrm{H - \overset{*}{C} - Cl} & \vdots & \mathrm{Cl - \overset{*}{C} - H} \\ | & \vdots & | \\ \mathrm{CH_3} & \vdots & \mathrm{CH_3} \\ \text{(I)} & \text{Mirror} & \text{(II)} \end{array}\]
- These two enantiomers are optically active: one isomer rotates the plane of plane-polarized light clockwise (dextrorotatory, \(d\text{-}\) or \((+)\text{-isomer}\)), while the other rotates it anticlockwise by an equal magnitude (laevorotatory, \(l\text{-}\) or \((-)\text{-isomer}\)).
SECTION - C
Attempt any EIGHT of the following questions: [24 Marks]
Q. 15. (a) Derive a relationship between molar mass, density of the substance and unit cell edge length. (b) Define - paramagnetic solids.
Answer:
(a) Derivation of relationship between molar mass (\(M\)), density (\(\rho\)), and edge length (\(a\)):
- If the edge length of a cubic unit cell is \(a\), then the volume of the unit cell is \(V = a^3\).
- Suppose the mass of one particle is \(m\) and there are \(n\) (or \(z\)) particles per unit cell.
\[\text{Mass of unit cell} = m \times n\] - The density of the unit cell (\(\rho\)), which is the same as the density of the crystalline solid, is given by: \[\rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{m \times n}{a^3} \quad \text{--- (Equation 1)}\]
- The molar mass (\(M\)) of the substance is given by: \[M = \text{mass of one particle } (m) \times \text{Avogadro's number } (N_A) \implies m = \frac{M}{N_A}\]
- Substituting \(m = \dfrac{M}{N_A}\) into Equation 1 gives: \[\rho = \frac{n \times M}{a^3 \times N_A}\]
(b) Paramagnetic solids:
The substances which contain one or more unpaired electrons in their atoms, ions, or molecules and are weakly attracted by an external magnetic field are called paramagnetic solids. Examples: \(\mathrm{O_2}\), \(\mathrm{Cu^{2+}}\), \(\mathrm{Fe^{3+}}\), \(\mathrm{Cr^{3+}}\).
Q. 16. (a) Mention the names of four principles of green chemistry. (b) Write advantages of nanoparticles and nanotechnology in the field of energy sector.
Answer:
(a) Names of four principles of Green Chemistry (Any four):
- Prevention of waste or by-products
- Atom economy
- Less hazardous chemical syntheses
- Designing safer chemicals
- Use of safer solvents and auxiliaries
- Use of catalysis
(b) Advantages of nanoparticles and nanotechnology in the energy sector:
- Nanotechnology makes solar power more economical and significantly increases the efficiency of solar cells.
- Energy storage devices (such as rechargeable batteries and supercapacitors) built with nanomaterials become lighter, faster-charging, and have higher power density.
- Nanocatalysts improve the efficiency of fuel cells and biofuel production.
Q. 17. (a) State second law of thermodynamics. (b) Define - Nucleotides. (c) Write chemical reaction for conversion of benzene diazonium chloride into benzene.
Answer:
(a) Second law of thermodynamics:
It states that the total entropy of a system and its surroundings (the universe) increases in any spontaneous process:
(b) Nucleotides:
Nucleotides are the repeating monomer units of nucleic acids (\(\mathrm{DNA}\) and \(\mathrm{RNA}\)). Chemically, a nucleotide is a phosphate ester of a nucleoside and consists of three components: a pentose sugar, a nitrogenous base, and a phosphate group.
(c) Conversion of benzene diazonium chloride into benzene:
Benzene diazonium chloride on treatment with hypophosphorous acid (\(\mathrm{H_3PO_2}\), phosphinic acid) in the presence of water (and \(\mathrm{Cu^+}\) catalyst) is reduced to benzene:
(Alternatively, using ethanol: \(\mathrm{C_6H_5N_2^+Cl^-} + \mathrm{CH_3CH_2OH} \longrightarrow \mathrm{C_6H_6} + \mathrm{N_2}\uparrow + \mathrm{CH_3CHO} + \mathrm{HCl}\))
Q. 18. The solubility product of \(\mathrm{AgBr}\) is \(5.2 \times 10^{-13}\). Calculate its solubility in \(\mathrm{mol\ dm^{-3}}\) and \(\mathrm{g\ dm^{-3}}\). (Molar mass of \(\mathrm{AgBr} = 187.8\text{ g mol}^{-1}\))
Answer:
- Given: \(K_{sp} = 5.2 \times 10^{-13}\), Molar mass of \(\mathrm{AgBr} = 187.8\text{ g mol}^{-1}\).
- Solubility equilibrium: \[\mathrm{AgBr}_{(s)} \rightleftharpoons \mathrm{Ag}^+_{(aq)} + \mathrm{Br}^\ominus_{(aq)}\] Here, \(x = 1\) and \(y = 1\). Therefore: \[K_{sp} = [\mathrm{Ag}^+][\mathrm{Br}^\ominus] = S \times S = S^2\]
- (1) Molar solubility (\(S\)) in \(\mathrm{mol\ dm^{-3}}\): \[S = \sqrt{K_{sp}} = \sqrt{5.2 \times 10^{-13}} = \sqrt{52 \times 10^{-14}} = 7.211 \times 10^{-7}\text{ mol dm}^{-3}\]
- (2) Solubility in \(\mathrm{g\ dm^{-3}}\): \[\text{Solubility (in }\mathrm{g\ dm^{-3}}\text{)} = S\text{ (in }\mathrm{mol\ dm^{-3}}\text{)} \times \text{Molar mass}\] \[\text{Solubility} = 7.211 \times 10^{-7}\text{ mol dm}^{-3} \times 187.8\text{ g mol}^{-1} = 1.354 \times 10^{-4}\text{ g dm}^{-3}\]
Final Answer: The solubility of \(\mathrm{AgBr}\) is \(7.211 \times 10^{-7}\text{ mol dm}^{-3}\) and \(1.354 \times 10^{-4}\text{ g dm}^{-3}\).
Q. 19. Convert the following: (a) Ethanol to ethoxyethane. (b) Phenol to p-benzoquinone (c) Salicylic acid to aspirin
Answer:
(a) Ethanol to ethoxyethane (Continuous etherification):
\[2\,\mathrm{C_2H_5-OH} \text{ (Ethanol)} \xrightarrow[413\text{ K}]{\text{Conc. }\mathrm{H_2SO_4}} \mathrm{C_2H_5-O-C_2H_5} \text{ (Ethoxyethane)} + \mathrm{H_2O}\](b) Phenol to p-benzoquinone (Oxidation of phenol):
\[\mathrm{C_6H_5OH} \text{ (Phenol)} + 2[\mathrm{O}] \xrightarrow{\mathrm{Na_2Cr_2O_7 + \text{Conc. }H_2SO_4}} \mathrm{O=C_6H_4=O} \text{ (p-Benzoquinone)} + \mathrm{H_2O}\](c) Salicylic acid to aspirin (Acetylation):
\[o\text{-}\mathrm{HO-C_6H_4-COOH} \text{ (Salicylic acid)} + (\mathrm{CH_3CO})_2\mathrm{O} \text{ (Acetic anhydride)} \xrightarrow[\Delta]{\text{Conc. }\mathrm{H_2SO_4}} o\text{-}\mathrm{CH_3COO-C_6H_4-COOH} \text{ (Aspirin / Acetylsalicylic acid)} + \mathrm{CH_3COOH}\]Q. 20. (a) State Raoult's law. (b) Derive the relationship to determine molar mass of non volatile solute by depression of freezing point.
Answer:
(a) Raoult's law:
- General statement: It states that the partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution (\(P_1 = P_1^\circ x_1\)).
- For a solution containing a non-volatile solute: At a given temperature, the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
(b) Derivation of molar mass of non-volatile solute from depression in freezing point:
- For a dilute solution, the depression in freezing point (\(\Delta T_f\)) is directly proportional to the molality (\(m\)) of the solution: \[\Delta T_f \propto m \implies \Delta T_f = K_f\,m \quad \text{--- (Equation 1)}\] where \(K_f\) is the molal depression constant (cryoscopic constant).
- Suppose \(W_2\text{ g}\) of a non-volatile solute of molar mass \(M_2\text{ g mol}^{-1}\) is dissolved in \(W_1\text{ g}\) (i.e., \(\dfrac{W_1}{1000}\text{ kg}\)) of solvent.
\[\text{Number of moles of solute} = \frac{W_2}{M_2}\] \[\text{Molality }(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000\,W_2}{M_2\,W_1} \quad \text{--- (Equation 2)}\] - Substituting Equation 2 into Equation 1 gives: \[\Delta T_f = \frac{1000\,K_f\,W_2}{M_2\,W_1}\] \[M_2 = \frac{1000\,K_f\,W_2}{\Delta T_f\,W_1}\]
Q. 21. (a) Explain the formation of \([\mathrm{CoF_6}]^{3\ominus}\) complex ion with respect to: (i) Oxidation State of Cobalt (ii) Type of hybridisation (iii) Magnetic property (iv) Geometrical shape (b) Write the name of ligand used to estimate hardness of water.
Answer:
(a) Formation of \([\mathrm{CoF_6}]^{3\ominus}\) complex ion:
- (i) Oxidation State of Cobalt:
Let oxidation state of \(\mathrm{Co}\) be \(x\). Then \(x + 6(-1) = -3 \implies x = +3\). Thus, Cobalt is in the \(+3\) oxidation state (\(\mathrm{Co^{3+}}\)). - (ii) Type of hybridisation:
Electronic configuration of \(\mathrm{Co}\) (\(Z = 27\)): \([\mathrm{Ar}]\,3d^7\,4s^2\).
Electronic configuration of \(\mathrm{Co^{3+}}\) ion: \([\mathrm{Ar}]\,3d^6\,4s^0\,4p^0\,4d^0\): \[3d^6: [\uparrow\downarrow][\uparrow][\uparrow][\uparrow][\uparrow] \quad 4s^0: [\ \ ] \quad 4p^0: [\ \ ][\ \ ][\ \ ] \quad 4d^0: [\ \ ][\ \ ][\ \ ][\ \ ][\ \ ]\] Since \(\mathrm{F^\ominus}\) is a weak field ligand, it does not cause pairing of \(3d\) electrons. One vacant \(4s\), three vacant \(4p\), and two vacant \(4d\) orbitals undergo \(sp^3d^2\) hybridisation (outer orbital / high-spin complex) to form six equivalent \(sp^3d^2\) hybrid orbitals, which accept six lone pairs from six \(\mathrm{F^\ominus}\) ligands. - (iii) Magnetic property:
Due to the presence of \(4\) unpaired electrons in the \(3d\) orbitals, the complex is paramagnetic (\(\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90\text{ BM}\)). - (iv) Geometrical shape:
Octahedral geometry.
(b) Name of ligand used to estimate hardness of water:
\(\mathrm{EDTA}\) (Ethylenediaminetetraacetate ion).
Q. 22. (a) Write cell reactions of hydrogen-oxygen fuel cell at anode and cathode. (b) Write an application of electrochemical series.
Answer:
(a) Cell reactions of \(\mathrm{H_2\text{-}O_2}\) fuel cell:
- Oxidation at Anode (\(-\)): \[2\mathrm{H}_{2(g)} + 4\mathrm{OH}^\ominus_{(aq)} \longrightarrow 4\mathrm{H_2O}_{(l)} + 4\mathrm{e}^\ominus\]
- Reduction at Cathode (\(+\)): \[\mathrm{O}_{2(g)} + 2\mathrm{H_2O}_{(l)} + 4\mathrm{e}^\ominus \longrightarrow 4\mathrm{OH}^\ominus_{(aq)}\]
- Net Cell Reaction: \[2\mathrm{H}_{2(g)} + \mathrm{O}_{2(g)} \longrightarrow 2\mathrm{H_2O}_{(l)}\]
(b) Application of electrochemical series (Any one):
- Predicting spontaneity of redox reactions: A redox reaction is spontaneous if the standard cell potential \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\) calculated from the series is positive.
- Comparing relative strength of oxidizing and reducing agents: Species with higher positive standard reduction potentials are stronger oxidizing agents, while those with more negative standard reduction potentials are stronger reducing agents.
Q. 23. (a) Explain anomalous behaviour of oxygen with respect to: (i) Atomicity (ii) Magnetic property (iii) Oxidation state (iv) Nature of hydride (b) Write any two uses of helium.
Answer:
(a) Anomalous behaviour of oxygen:
- (i) Atomicity: Oxygen is a diatomic gas (\(\mathrm{O_2}\)) due to its small size and ability to form strong \(p\pi\text{-}p\pi\) double bonds, whereas other elements of Group 16 are polyatomic solids (e.g., \(\mathrm{S_8}\), \(\mathrm{Se_8}\)).
- (ii) Magnetic property: Molecular oxygen (\(\mathrm{O_2}\)) is paramagnetic due to two unpaired electrons in its \(\pi^*\) antibonding molecular orbitals, whereas other elements of Group 16 are diamagnetic.
- (iii) Oxidation state: Oxygen generally exhibits a \(-2\) oxidation state (except \(-1\) in peroxides and \(+2\) in \(\mathrm{OF_2}\)) and cannot expand its covalency beyond \(4\) due to the absence of vacant \(d\)-orbitals in its valence shell. Other Group 16 elements exhibit \(-2, +2, +4,\) and \(+6\) oxidation states.
- (iv) Nature of hydride: The hydride of oxygen (\(\mathrm{H_2O}\)) is a liquid at room temperature due to extensive intermolecular hydrogen bonding, whereas the hydrides of other Group 16 elements (\(\mathrm{H_2S, H_2Se, H_2Te}\)) are pungent, foul-smelling gases.
(b) Uses of Helium (Any two):
- A mixture of helium (\(85\%\)) and oxygen (\(15\%\)) is used by deep-sea divers for respiration in place of nitrogen because helium has very low solubility in blood.
- Being non-inflammable and light, it is used for filling meteorological observation balloons.
- Liquid helium is used as a cryogenic fluid to produce low temperatures required for superconducting magnets in \(\mathrm{MRI}\) systems.
Q. 24. What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from \(303\text{ K}\) to \(313\text{ K}\)?
Answer:
- Given: \(T_1 = 303\text{ K}\), \(T_2 = 313\text{ K}\), \(k_2 = 2k_1 \implies \dfrac{k_2}{k_1} = 2\), \(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\).
- Formula: \[\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1\,T_2}\right)\]
- Calculation: \[\log_{10}(2) = \frac{E_a}{2.303 \times 8.314}\left(\frac{313 - 303}{303 \times 313}\right)\] \[0.3010 = \frac{E_a}{19.147}\left(\frac{10}{94839}\right)\] \[E_a = \frac{0.3010 \times 19.147 \times 94839}{10} = 54657.4\text{ J mol}^{-1} = 54.66\text{ kJ mol}^{-1}\]
Final Answer: The energy of activation of the reaction is \(54.66\text{ kJ mol}^{-1}\) (or \(54657.4\text{ J mol}^{-1}\)).
Q. 25. (a) Write chemical reactions taking place during extraction of iron in zone of reduction of the blast furnace. (b) Write observed electronic configuration of gadolinium. (\(Z=64\))
Answer:
(a) Chemical reactions in the zone of reduction of the blast furnace (\(\approx 900\text{ K}\)):
In the zone of reduction, ferric oxide (\(\mathrm{Fe_2O_3}\)) is reduced to spongy iron by carbon monoxide (\(\mathrm{CO}\)):
Stepwise reduction reactions:
\[3\mathrm{Fe_2O_3} + \mathrm{CO} \longrightarrow 2\mathrm{Fe_3O_4} + \mathrm{CO_2}\] \[\mathrm{Fe_3O_4} + \mathrm{CO} \longrightarrow 3\mathrm{FeO} + \mathrm{CO_2}\] \[\mathrm{FeO} + \mathrm{CO} \longrightarrow \mathrm{Fe} + \mathrm{CO_2}\](At higher temperature around \(1500\text{ K}\), reduction also occurs directly by carbon: \(\mathrm{Fe_2O_3} + 3\mathrm{C} \xrightarrow{1500\text{ K}} 2\mathrm{Fe} + 3\mathrm{CO}\))
(b) Observed electronic configuration of Gadolinium (\(\mathrm{Gd}, Z = 64\)):
\[[\mathrm{Xe}]\,4f^7\,5d^1\,6s^2\]Q. 26. Convert the following: (a) Acetic acid to ethyl alcohol (b) Acetic acid to acetic anhydride (c) Acetyl chloride to acetamide
Answer:
(a) Acetic acid to ethyl alcohol (Reduction):
\[\mathrm{CH_3-COOH} \text{ (Acetic acid)} \xrightarrow[\text{(ii) }\mathrm{H_3O^+}]{\text{(i) }\mathrm{LiAlH_4 / \text{dry ether}}} \mathrm{CH_3-CH_2-OH} \text{ (Ethyl alcohol)} + \mathrm{H_2O}\](b) Acetic acid to acetic anhydride (Dehydration):
\[2\,\mathrm{CH_3-COOH} \text{ (Acetic acid)} \xrightarrow{\mathrm{P_2O_5},\ \Delta} (\mathrm{CH_3CO})_2\mathrm{O} \text{ (Acetic anhydride)} + \mathrm{H_2O}\](c) Acetyl chloride to acetamide (Ammonolysis):
\[\mathrm{CH_3-CO-Cl} \text{ (Acetyl chloride)} + 2\mathrm{NH_3} \longrightarrow \mathrm{CH_3-CO-NH_2} \text{ (Acetamide)} + \mathrm{NH_4Cl}\]SECTION - D
Attempt any THREE of the following questions: [12 Marks]
Q. 27. (a) Define: (i) Co-ordination numbers (ii) Ambidentate ligand (b) Write a note on Clemmensen reduction of propanone.
Answer:
(a) Definitions:
- (i) Co-ordination number: The coordination number of a central metal atom or ion in a coordination complex is the total number of ligand donor atoms directly attached to it (or the number of electron pairs donated by the ligands to form coordinate covalent bonds with the central metal ion).
- (ii) Ambidentate ligand: A monodentate ligand which possesses two or more different donor atoms and can link to the central metal atom or ion through either of the donor atoms at a time is called an ambidentate ligand. Examples: \(\mathrm{NO_2^\ominus}\) (binds via \(\mathrm{N}\) or \(\mathrm{O}\)), \(\mathrm{SCN^\ominus}\) (binds via \(\mathrm{S}\) or \(\mathrm{N}\)).
(b) Clemmensen reduction of propanone:
When propanone (acetone) is heated with zinc-amalgam (\(\mathrm{Zn\text{-}Hg}\)) and concentrated hydrochloric acid (\(\mathrm{Conc.\ HCl}\)), its carbonyl group (\(\mathrm{>C=O}\)) is reduced to a methylene group (\(\mathrm{-CH_2-}\)) to form propane.
Q. 28. (a) Derive relationship between \(\Delta H\) and \(\Delta U\) for a gaseous chemical reaction. (b) Give reason Ozone is a good bleaching agent. (c) Write any two uses of \(\mathrm{SO_2}\).
Answer:
(a) Derivation of relationship between \(\Delta H\) and \(\Delta U\):
- Enthalpy (\(H\)) is defined as \(H = U + PV\).
- At constant pressure \(P\), let \(H_1, U_1, V_1\) represent the initial state (reactants) and \(H_2, U_2, V_2\) represent the final state (products): \[H_1 = U_1 + PV_1 \quad \text{and} \quad H_2 = U_2 + PV_2\] \[\Delta H = H_2 - H_1 = (U_2 - U_1) + P(V_2 - V_1) = \Delta U + P\Delta V \quad \text{--- (Equation 1)}\]
- Let \(n_1\) be the number of moles of gaseous reactants occupying volume \(V_1\), and \(n_2\) be the number of moles of gaseous products occupying volume \(V_2\) at constant temperature \(T\) and pressure \(P\). Assuming ideal gas behaviour (\(PV = nRT\)): \[PV_1 = n_1 RT \quad \text{and} \quad PV_2 = n_2 RT\] \[P\Delta V = P(V_2 - V_1) = n_2 RT - n_1 RT = (n_2 - n_1)RT = \Delta n_g RT \quad \text{--- (Equation 2)}\]
- Substituting Equation 2 into Equation 1 gives: \[\Delta H = \Delta U + \Delta n_g RT\] where \(\Delta n_g = n_2 - n_1\) is the difference between the number of moles of gaseous products and gaseous reactants.
(b) Reason Ozone is a good bleaching agent:
Ozone (\(\mathrm{O_3}\)) is thermodynamically unstable and readily decomposes to liberate nascent oxygen (\([\mathrm{O}]\)):
This nascent oxygen is a powerful oxidizing agent that oxidizes coloured organic matter into colourless substances:
\[\text{Coloured matter} + [\mathrm{O}] \longrightarrow \text{Colourless matter}\]Because the bleaching action of ozone is due to oxidation, it acts as a permanent bleaching agent (dry bleach).
(c) Uses of \(\mathrm{SO_2}\) (Any two):
- Used in the manufacture of sulphuric acid (\(\mathrm{H_2SO_4}\)) by the Contact process.
- Used as a mild bleaching agent for delicate materials such as wool, silk, straw, and starch.
- Used in refining petroleum and sugar, and as an anti-chlor, disinfectant, and preservative.
Q. 29. (a) Define - Polymorphism. (b) State and explain Hess's law of constant summation. Write its one application.
Answer:
(a) Polymorphism:
The phenomenon in which a single chemical substance (compound) crystallizes in two or more distinct crystalline structures under different conditions is called polymorphism, and the different crystalline forms are called polymorphic forms. Example: Calcium carbonate (\(\mathrm{CaCO_3}\)) exists as calcite and aragonite; silica (\(\mathrm{SiO_2}\)) exists as \(\alpha\text{-quartz}\), \(\beta\text{-quartz}\), and cristobalite.
(b) Hess's Law of Constant Heat Summation:
- Statement: The law states that the overall enthalpy change accompanying a chemical reaction is the same regardless of whether the reaction takes place in one step or in a series of multiple steps.
- Explanation: Hess's law is a direct consequence of the fact that enthalpy is a state function (it depends only on the initial and final states, not on the path).
Consider the conversion of reactant \(\mathrm{A}\) into product \(\mathrm{D}\) in a single step: \[\mathrm{A} \longrightarrow \mathrm{D}, \quad \Delta_r H^\circ\] Suppose the same reaction occurs in three consecutive steps:- Step 1: \(\mathrm{A} \longrightarrow \mathrm{B}, \quad \Delta_r H^\circ_1\)
- Step 2: \(\mathrm{B} \longrightarrow \mathrm{C}, \quad \Delta_r H^\circ_2\)
- Step 3: \(\mathrm{C} \longrightarrow \mathrm{D}, \quad \Delta_r H^\circ_3\)
- Application: Hess's law is used to calculate the enthalpy of formation, combustion, or ionization of reactions whose enthalpy changes cannot be measured directly by experiment (by treating thermochemical equations algebraically).
Q. 30. (a) \(0.01\text{ m}\) aqueous formic acid solution freezes at \(-0.021^\circ\text{C}\). Calculate its degree of dissociation. (\(K_f = 1.86\text{ K kg mol}^{-1}\)) (b) Write properties of interstitial compounds.
Answer:
(a) Calculation of degree of dissociation (\(\alpha\)):
- Given: Molality \(m = 0.01\text{ m}\), Freezing point of solution \(T_f = -0.021^\circ\text{C}\), \(K_f = 1.86\text{ K kg mol}^{-1}\).
- Observed depression in freezing point: \[\Delta T_f = T_f^\circ - T_f = 0^\circ\text{C} - (-0.021^\circ\text{C}) = 0.021\text{ K}\]
- Theoretical depression in freezing point (\(\Delta T_{f(\text{theo})}\)) and van't Hoff factor (\(i\)): \[\Delta T_f = i \times K_f \times m \implies 0.021 = i \times 1.86 \times 0.01 = i \times 0.0186\] \[i = \frac{0.021}{0.0186} = 1.129\]
- Degree of dissociation (\(\alpha\)):
Formic acid dissociates into \(n = 2\) ions per formula unit: \(\mathrm{HCOOH}_{(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{HCOO}^\ominus_{(aq)}\). \[\alpha = \frac{i - 1}{n - 1} = \frac{1.129 - 1}{2 - 1} = 0.129 \text{ (or } 12.9\%\text{)}\]
Final Answer: The degree of dissociation of formic acid is \(0.129\) (or \(12.9\%\)).
(b) Properties of interstitial compounds (Any four):
- They are hard and show good electrical and thermal conductivity (they retain metallic conductivity).
- They have very high melting points, higher than those of the pure transition metals.
- They are chemically inert.
- Their densities are lower than those of the parent metals, and metallic hydrides can be used as powerful reducing agents.
Q. 31. (a) What is the action of following reagents on Bromoethane? (i) \(\mathrm{KNO_2}\) (ii) '\(\mathrm{Mg}\)' metal (iii) Silver acetate (b) Draw structure - DDT insecticide.
Answer:
(a) Action of reagents on Bromoethane (\(\mathrm{CH_3-CH_2-Br}\)):
- (i) Action of Potassium nitrite (\(\mathrm{KNO_2}\)):
Bromoethane on reaction with aqueous/alcoholic potassium nitrite forms ethyl nitrite as the major product: \[\mathrm{CH_3-CH_2-Br} + \mathrm{KNO_2} \longrightarrow \mathrm{CH_3-CH_2-O-N=O} \text{ (Ethyl nitrite)} + \mathrm{KBr}\] - (ii) Action of Magnesium (\(\mathrm{Mg}\)) metal:
Bromoethane reacts with magnesium metal in the presence of dry ether as solvent to form ethylmagnesium bromide (Grignard reagent): \[\mathrm{CH_3-CH_2-Br} + \mathrm{Mg} \xrightarrow{\text{Dry ether}} \mathrm{CH_3-CH_2-MgBr} \text{ (Ethylmagnesium bromide)}\] - (iii) Action of Silver acetate (\(\mathrm{CH_3COOAg}\)):
Bromoethane on heating with alcoholic silver acetate undergoes nucleophilic substitution to form ethyl acetate (ethyl ethanoate): \[\mathrm{CH_3-CH_2-Br} + \mathrm{CH_3COOAg} \xrightarrow{\text{Alcohol},\ \Delta} \mathrm{CH_3-COOCH_2CH_3} \text{ (Ethyl acetate)} + \mathrm{AgBr}\downarrow\]
(b) Structure of DDT (Dichlorodiphenyltrichloroethane / \(1,1,1\text{-trichloro-}2,2\text{-bis(4-chlorophenyl)ethane}\)):
\[\begin{array}{c} \mathrm{Cl - C_6H_4} \\ \backslash \\ \mathrm{CH - CCl_3} \\ / \\ \mathrm{Cl - C_6H_4} \end{array}\](Two para-chlorophenyl rings and one \(\mathrm{-CCl_3}\) group attached to the \(\mathrm{>CH-}\) carbon atom.)
