HSC Chemistry Question Paper Solution 2024
MAHARASHTRA STATE BOARD OF SECONDARY AND HIGHER SECONDARY EDUCATION
Subject: CHEMISTRY (55) | Code: J-852 (E)
Date: 29-02-2024 | Max. Marks: 70 | Duration: 3 Hrs
General Instructions: This solution set covers all sections (A, B, C, D) detailed in the question paper. Calculations are based on the given constants: \(R = 8.314 \, J.K^{-1}.mol^{-1}\), \(N_A = 6.022 \times 10^{23}\), \(F = 96500 \, C\).
SECTION - A
Q. 1. Select and write the correct answer for the following multiple choice type of questions: [10 Marks]
(i) The spin only magnetic moment of \(Cr^{3+}\) cation is _____.
Explanation: Chromium (Cr) atomic number is 24. Configuration: \([Ar]3d^5 4s^1\). \(Cr^{3+}\) is \([Ar]3d^3\). Number of unpaired electrons (n) = 3.
\(\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.873 \, BM\).
(ii) The linkage present in Lactose is _____.
Explanation: Lactose is a disaccharide of \(\beta\)-D-galactose and \(\beta\)-D-glucose linked by a \(\beta-1,4\)-glycosidic bond.
(iii) The product of the following reaction is
\(C_2H_5 - \overset{\overset{O}{||}}{C} - CH_3 \xrightarrow[\Delta]{H_2/Ni} \, ?\)
Explanation: Catalytic reduction of Butanone (ketone) gives Butan-2-ol (secondary alcohol).
(iv) The pH of 0.001M HCl solution is _____.
Explanation: \([H^+] = 0.001 \, M = 10^{-3} \, M\). \(pH = -\log_{10}[H^+] = -\log_{10}(10^{-3}) = 3\).
(v) The correct structure of complex having IUPAC name sodium hexanitrocobaltate (III) is
Explanation: Cation is Sodium (\(Na^+\)). Anion is Hexanitrocobaltate(III) \([Co(NO_2)_6]^{3-}\). To balance charges: \(3Na^+\) and one complex ion.
(vi) The number of particles present in Face Centred Cubic Unit Cell is/are _____.
Explanation: Corners: \(8 \times \frac{1}{8} = 1\). Faces: \(6 \times \frac{1}{2} = 3\). Total = \(1+3=4\).
(vii) The monomer used in preparation of teflon is _____.
Explanation: Teflon is Polytetrafluoroethylene (PTFE). Monomer is \(CF_2=CF_2\).
(viii) Among the following vinylic halide is ______.
Explanation: In vinylic halides, the halogen is attached directly to the \(sp^2\) hybridized carbon atom of a carbon-carbon double bond.
(ix) The product of hydrolysis of propyne in the presence of 1% \(HgSO_4\) and 40% \(H_2SO_4\) is _____.
Explanation: Hydration follows Markovnikov's rule to form an enol \([CH_3-C(OH)=CH_2]\) which tautomerizes to Propanone \([CH_3-CO-CH_3]\).
(x) If unit of rate constant is \(mol \, dm^{-3}s^{-1}\), the order of reaction would be _____.
Explanation: For a zero-order reaction, Rate = k. The unit of rate is \(mol \, L^{-1} s^{-1}\) or \(mol \, dm^{-3} s^{-1}\).
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Q. 2. Answer the following questions: [8 Marks]
(i) Write the name of metal nanoparticle used to remove E.coli bacteria from water.
Answer: Silver (Ag) nanoparticles.
(ii) Write the name of reduction product formed when ethyl cyanide is treated with sodium and alcohol.
Answer: Propylamine (or Propan-1-amine).
Reaction: \(C_2H_5CN + 4[H] \xrightarrow{Na/Ethanol} C_2H_5CH_2NH_2\)
(iii) Complete the reaction: \(CH_3CH_2Cl \xrightarrow{AgCN/alc.\Delta} \, ?\)
Answer: \(CH_3CH_2NC\) (Ethyl isocyanide)
Reason: AgCN is covalent, so the nitrogen atom acts as the nucleophile.
(iv) Calculate effective atomic number of \([Co(NH_3)_6]^{3+}\) ion.
Answer: 36
Calculation: \(EAN = Z - X + Y\)
Z (Atomic number of Co) = 27
X (Oxidation state) = +3
Y (Electrons donated by ligands) = \(6 \times 2 = 12\)
\(EAN = 27 - 3 + 12 = 36\) (Electronic config of Krypton).
(v) The compounds of \(Ti^{4+}\) ions are colourless due to ......
Answer: Absence of unpaired d-electrons (empty d-orbital).
Explanation: \(Ti^{4+}\) has \(3d^0\) configuration, so d-d transition is not possible.
(vi) Write SI unit of molar conductivity.
Answer: \(S \cdot m^2 \cdot mol^{-1}\) (Siemens square metre per mole).
(vii) Write the sign convention of work done during expansion of gas.
Answer: Work done (W) is negative (\(W < 0\)).
Convention: Work done by the system on the surroundings is negative.
(viii) Write the condition of reverse osmosis.
Answer: The external pressure applied on the solution side must be greater than the osmotic pressure (\(P > \pi\)).
SECTION - B
Attempt any EIGHT of the following questions: [16 Marks]
Q. 3. Derive an expression for maximum work obtainable during isothermal reversible expansion of an ideal gas from initial volume (V1) to final volume (V2).
Answer:
- Consider n moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless piston.
- Work done in a small expansion \(dV\) against external pressure \(P_{ext}\) is \(dW = -P_{ext} dV\).
- For a reversible process, \(P_{ext} \approx P_{gas} = P\). Thus, \(dW = -P dV\).
- Total work done in expanding from \(V_1\) to \(V_2\) is obtained by integrating: \[W_{max} = - \int_{V_1}^{V_2} P \, dV\]
- From Ideal Gas Equation, \(PV = nRT \implies P = \frac{nRT}{V}\). Substituting this: \[W_{max} = - \int_{V_1}^{V_2} \frac{nRT}{V} \, dV\]
- Since T is constant (isothermal): \[W_{max} = -nRT \int_{V_1}^{V_2} \frac{dV}{V} = -nRT [\ln V]_{V_1}^{V_2}\] \[W_{max} = -nRT (\ln V_2 - \ln V_1) = -nRT \ln \left(\frac{V_2}{V_1}\right)\]
- Converting natural log to log base 10: \[W_{max} = -2.303 nRT \log_{10} \left(\frac{V_2}{V_1}\right)\]
Q. 4. What are interhalogen compounds? Write the chemical reaction, when chlorine reacts with dry slaked lime.
Answer:
Interhalogen Compounds: Compounds formed by the reaction of two or more different halogen atoms are called interhalogen compounds (e.g., ClF, ICl, BrF3).
Reaction: When chlorine reacts with dry slaked lime (\(Ca(OH)_2\)), bleaching powder is formed.
\(2Ca(OH)_2 + 2Cl_2 \longrightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O\)
(The mixture \(Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O\) represents bleaching powder).
Q. 5. What is nano material? Write the reaction involved in sol-gel process during hydrolysis.
Answer:
Nano material: A material having at least one dimension in the nanoscale range (1 nm to 100 nm) is called a nanomaterial.
Sol-gel process reaction (Hydrolysis):
Precursors like metal alkoxides undergo hydrolysis with water.
\(M(OR)_4 + 4H_2O \longrightarrow M(OH)_4 + 4ROH\)
(Where M = Si, Ti, etc., and R = alkyl group)
Q. 6. Write classification of proteins with an example.
Answer: Based on molecular shape, proteins are classified into two types:
- Fibrous Proteins: Polypeptide chains run parallel and are held together by hydrogen and disulfide bonds. They are insoluble in water.
Example: Keratin (hair, wool), Myosin (muscles). - Globular Proteins: Polypeptide chains coil around to give a spherical shape. They are usually soluble in water.
Example: Insulin, Albumin, Haemoglobin.
Q. 7. Calculate the time required to deposit 2.4 g of Cu, when 2.03 A of current passed through \(CuSO_4\) solution. (At. mass of Cu = 63.5 g.mol\(^{-1}\))
Answer:
Reaction: \(Cu^{2+} + 2e^- \rightarrow Cu\) (n = 2 moles of electrons per mole of Cu)
Given:
Mass (W) = 2.4 g
Current (I) = 2.03 A
Molar Mass (M) = 63.5 g/mol
F = 96500 C/mol
Formula: \(W = \frac{M \times I \times t}{n \times F}\)
\(t = \frac{W \times n \times F}{M \times I}\)
\(t = \frac{2.4 \times 2 \times 96500}{63.5 \times 2.03}\)
\(t = \frac{463200}{128.905} \approx 3593.34 \, \text{seconds}\)
Time required \(\approx\) 3593 seconds (or ~59.9 mins).
Q. 8. Why amines are basic in nature? Among dimethylamine (\(pK_b\) = 3.27) and diethylamine (\(pK_b\) = 3.0), which one is more basic?
Answer:
Basicity of Amines: Amines are basic because the nitrogen atom possesses a lone pair of electrons which can be donated to a proton (\(H^+\)) (Lewis base character).
Comparison: Lower \(pK_b\) value indicates a stronger base.
Dimethylamine \(pK_b\) = 3.27
Diethylamine \(pK_b\) = 3.0
Therefore, Diethylamine is more basic.
Q. 9. Explain buffer action of sodium acetate-acetic acid buffer.
Answer:
This is an acidic buffer containing weak acid \(CH_3COOH\) and its salt \(CH_3COONa\).
Dissociation:
- \(CH_3COONa \rightarrow CH_3COO^- + Na^+\) (Complete)
- \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\) (Partial)
Action on adding Acid (\(H^+\)): Added \(H^+\) ions are consumed by the large reserve of acetate ions (\(CH_3COO^-\)) to form un-dissociated acetic acid.
\(CH_3COO^- + H^+ \rightarrow CH_3COOH\). Thus, pH remains constant.
Action on adding Base (\(OH^-\)): Added \(OH^-\) ions react with the acid molecules to form water.
\(CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O\). Thus, pH remains constant.
Q. 10. Write preparation of (a) diethyl ether (b) ethyl cyanide from ethyl bromide.
Answer:
(a) Diethyl ether from ethyl bromide (Williamson's Synthesis):
Reacting ethyl bromide with sodium ethoxide.
\(C_2H_5Br + C_2H_5ONa \xrightarrow{\Delta} C_2H_5-O-C_2H_5 + NaBr\)
(b) Ethyl cyanide from ethyl bromide:
Reacting ethyl bromide with alcoholic Potassium Cyanide (KCN).
\(C_2H_5Br + KCN(alc) \xrightarrow{\Delta} C_2H_5CN + KBr\)
Q. 11. Henry’s constant for \(CH_3Br(g)\) is \(0.159 \, mol \, dm^{-3} \cdot bar^{-1}\) at 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0.164 bar.
Answer:
Formula: \(S = K_H \times P\)
Given:
\(K_H = 0.159 \, mol \, dm^{-3} \, bar^{-1}\)
\(P = 0.164 \, bar\)
Calculation:
\(S = 0.159 \times 0.164\)
\(S = 0.026076 \, mol \, dm^{-3}\)
Solubility = \(0.0261 \, mol \, dm^{-3}\).
Q. 12. Write the structure and name of monomer of (a) Nylon-6 (b) Natural rubber
Answer:
(a) Nylon-6:
Name: \(\epsilon\)-Caprolactam
Structure: 7-membered ring with -NH-CO- group.
(b) Natural rubber:
Name: Isoprene (2-methylbuta-1,3-diene)
Structure: \(CH_2=C(CH_3)-CH=CH_2\)
Q. 13. Define Lanthanide contraction. Write the balanced chemical equations when acidified \(K_2Cr_2O_7\) reacts with \(H_2S\).
Answer:
Lanthanide Contraction: The gradual decrease in atomic and ionic radii of lanthanoids with increasing atomic number from Lanthanum (La) to Lutetium (Lu) is called lanthanide contraction.
Reaction:
Acidified Potassium dichromate oxidizes hydrogen sulphide to sulphur.
\(K_2Cr_2O_7 + 4H_2SO_4 + 3H_2S \longrightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S\)
Q. 14. Derive the relationship between molar mass, density of the substance and unit cell edge length.
Answer:
- Let 'a' be the edge length of the unit cell. Volume of unit cell = \(a^3\).
- Let 'M' be the molar mass and 'N_A' be Avogadro's number. Mass of one atom = \(M / N_A\).
- If 'n' is the number of atoms per unit cell (rank),
Mass of unit cell = \(n \times (M / N_A)\). - Density (\(\rho\)) = Mass of unit cell / Volume of unit cell
- \(\rho = \frac{n \times M}{a^3 \times N_A}\)
SECTION - C
Attempt any EIGHT of the following questions: [24 Marks]
Q. 15. What is osmotic pressure? How will you determine molar mass of solute from osmotic pressure?
Answer:
Osmotic Pressure (\(\pi\)): The excess hydrostatic pressure that must be applied to the solution side to just stop the flow of solvent molecules into the solution through a semipermeable membrane is called osmotic pressure.
Determination of Molar Mass:
According to Van't Hoff equation for dilute solutions: \(\pi = CRT\)
Where \(C = \frac{n_2}{V}\) (Concentration in mol/L).
\(\pi = \frac{n_2 RT}{V}\)
Since \(n_2 = \frac{W_2}{M_2}\) (Mass of solute / Molar mass of solute):
\(\pi = \frac{W_2 RT}{M_2 V}\)
Rearranging for \(M_2\):
\[M_2 = \frac{W_2 R T}{\pi V}\]
Q. 16. Write chemical reactions involved in: (a) Rosenmund reduction. (b) Gatterman Koch formylation. (c) Cannizzaro reaction of methanal.
Answer:
(a) Rosenmund reduction:
\(R-COCl + H_2 \xrightarrow{Pd/BaSO_4, S} R-CHO + HCl\)
(Acid chloride to Aldehyde)
(b) Gatterman-Koch formylation:
Benzene + \(CO + HCl \xrightarrow{Anhyd. AlCl_3/CuCl} Benzaldehyde + HCl\)
(c) Cannizzaro reaction of methanal:
\(2HCHO + NaOH(conc.) \longrightarrow CH_3OH + HCOONa\)
(Disproportionation to Methanol and Sodium Formate)
Q. 17. Calculate the standard enthalpy of combustion of methane, if the standard enthalpy of formation of methane, carbon dioxide and water are –74.8, –393.5 and –285.8 kJmol\(^{-1}\) respectively.
Answer:
Combustion Reaction: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\)
Formula: \(\Delta H_c^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})\)
Given:
\(\Delta_f H^\circ(CH_4) = -74.8\)
\(\Delta_f H^\circ(CO_2) = -393.5\)
\(\Delta_f H^\circ(H_2O) = -285.8\)
\(\Delta_f H^\circ(O_2) = 0\) (Standard state)
Calculation:
\(\Delta H_c^\circ = [1(-393.5) + 2(-285.8)] - [-74.8 + 0]\)
\(\Delta H_c^\circ = [-393.5 - 571.6] + 74.8\)
\(\Delta H_c^\circ = -965.1 + 74.8\)
\(\Delta H_c^\circ = -890.3 \, kJ \, mol^{-1}\)
Q. 18. What is the action of following on ethyl bromide? (a) silver nitrite (b) Mg in dry ether (c) alcoholic sodium hydroxide
Answer:
(a) Silver nitrite (\(AgNO_2\)):
Forms Nitroethane.
\(C_2H_5Br + AgNO_2 \rightarrow C_2H_5NO_2 + AgBr\)
(b) Mg in dry ether:
Forms Ethyl magnesium bromide (Grignard Reagent).
\(C_2H_5Br + Mg \xrightarrow{\text{dry ether}} C_2H_5MgBr\)
(c) Alcoholic Sodium Hydroxide:
Dehydrohalogenation occurs to form Ethene.
\(C_2H_5Br + NaOH(alc) \rightarrow CH_2=CH_2 + NaBr + H_2O\)
Q. 19. For the reaction \(A + B \rightarrow P\). If [B] is doubled at constant [A], the rate of reaction doubled. If [A] is triple and [B] is doubled, the rate of reaction increases by a factor of 6. Calculate the rate law equation.
Answer:
Let Rate law be \(R = k[A]^x[B]^y\).
Condition 1: [B] doubled, Rate doubled.
\(2R = k[A]^x[2B]^y\)
\(\frac{2R}{R} = 2^y \Rightarrow 2^1 = 2^y \Rightarrow y = 1\).
Condition 2: [A] tripled, [B] doubled, Rate becomes 6R.
\(6R = k[3A]^x[2B]^y\)
Dividing by original rate:
\(6 = 3^x \cdot 2^y\)
Substituting \(y=1\):
\(6 = 3^x \cdot 2 \Rightarrow 3^x = 3 \Rightarrow x = 1\).
Rate Law: Rate = \(k[A][B]\).
Q. 20. Arrange the following in the increasing order of the property mentioned:
(i) \(HOCl, HClO_2, HClO_3, HClO_4\) (acidic strength)
Answer: \(HOCl < HClO_2 < HClO_3 < HClO_4\) (Increases with oxidation state of Cl).
(ii) \(MF, MCl, MBr, MI\) (ionic character)
Answer: \(MI < MBr < MCl < MF\) (Increases with electronegativity difference).
(iii) \(HF, HCl, HBr, HI\) (thermal stability)
Answer: \(HI < HBr < HCl < HF\) (Decreases as bond length increases).
Q. 21. Explain Wolf-Kishner reduction reaction. Write preparation of propanone by using ethanoyl chloride and dimethyl cadmium.
Answer:
Wolff-Kishner Reduction: Aldehydes or ketones are heated with hydrazine (\(NH_2NH_2\)) and a strong base like KOH or potassium tert-butoxide in a high-boiling solvent (ethylene glycol). The carbonyl group (\(>C=O\)) is reduced to a methylene group (\(>CH_2\)).
Preparation of Propanone:
Step 1: Reaction of Ethanoyl chloride with Dimethyl cadmium.
\(2CH_3COCl + (CH_3)_2Cd \longrightarrow 2CH_3-CO-CH_3 + CdCl_2\)
Product is Propanone (Acetone).
Q. 22. Write postulates of Werner theory of co-ordination complexes. Write the name of a hexadentate ligand.
Answer:
Werner's Postulates:
- Metals in coordination compounds possess two types of valencies: Primary (ionizable) and Secondary (non-ionizable).
- Primary valency corresponds to the oxidation state and is satisfied by anions.
- Secondary valency corresponds to the coordination number and is satisfied by neutral molecules or anions. It is directional in nature, determining the geometry.
Hexadentate Ligand: Ethylenediaminetetraacetate ion (EDTA\({}^{4-}\)).
Q. 23. Define electrochemical series and write its two applications.
Answer:
Definition: The arrangement of electrodes (elements) in the decreasing or increasing order of their standard reduction potentials is called the electrochemical series (or electromotive series).
Applications:
- To compare the relative oxidizing and reducing powers of substances.
- To predict the spontaneity of redox reactions (If \(E^\circ_{cell}\) is positive, reaction is spontaneous).
Q. 24. Identify ‘A’, ‘B’ and ‘C’ in following chain reaction and rewrite the chemical reactions: \(CH_3CH_2OH \xrightarrow{red \, P / Br_2} A \xrightarrow{KCN \, alc} B \xrightarrow{LiAlH_4 / Ether} C\)
Answer:
Identification:
A = \(CH_3CH_2Br\) (Ethyl bromide)
B = \(CH_3CH_2CN\) (Ethyl cyanide / Propanenitrile)
C = \(CH_3CH_2CH_2NH_2\) (Propylamine / Propan-1-amine)
Reactions:
1. \(3CH_3CH_2OH + PBr_3 (\text{from } P/Br_2) \rightarrow 3CH_3CH_2Br + H_3PO_3\)
2. \(CH_3CH_2Br + KCN(alc) \rightarrow CH_3CH_2CN + KBr\)
3. \(CH_3CH_2CN + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2\)
Q. 25. Define acids and bases according to Bronsted-Lowry theory. Derive relationship between pH and pOH.
Answer:
Bronsted-Lowry Theory:
Acid: A substance that donates a proton (\(H^+\)).
Base: A substance that accepts a proton (\(H^+\)).
Relationship between pH and pOH:
Ionic product of water, \(K_w = [H^+][OH^-]\).
At 298K, \(K_w = 1 \times 10^{-14}\).
Taking negative log on both sides:
\(-\log_{10}([H^+][OH^-]) = -\log_{10}(10^{-14})\)
\(-\log_{10}[H^+] + (-\log_{10}[OH^-]) = 14\)
Since \(pH = -\log[H^+]\) and \(pOH = -\log[OH^-]\):
\(\mathbf{pH + pOH = 14}\)
Q. 26. Write the preparation of potassium dichromate from chromite ore.
Answer:
- Roasting: Chromite ore (\(FeCr_2O_4\)) is roasted with Sodium Carbonate in excess air.
\(4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2\) - Acidification: Yellow Sodium chromate is converted to orange Sodium dichromate using \(H_2SO_4\).
\(2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O\) - Conversion to K-salt: Sodium dichromate is treated with KCl. \(K_2Cr_2O_7\) is less soluble and crystallizes out.
\(Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl\)
SECTION - D
Attempt any THREE of the following questions: [12 Marks]
Q. 27. Convert the following:
(i) acetaldehyde to isopropyl alcohol.
(ii) cumene to phenol.
(iii) anisole to phenol.
Write two uses of neon.
Answer:
(i) Acetaldehyde to Isopropyl alcohol:
React with Methyl Magnesium Bromide followed by acid hydrolysis.
\(CH_3CHO + CH_3MgBr \rightarrow CH_3-CH(OMgBr)-CH_3 \xrightarrow{H_3O^+} CH_3-CH(OH)-CH_3\)
(ii) Cumene to Phenol:
Oxidation of Cumene (isopropylbenzene) to Cumene hydroperoxide, followed by hydrolysis.
\(C_6H_5CH(CH_3)_2 + O_2 \rightarrow C_6H_5C(CH_3)_2OOH \xrightarrow{H^+} C_6H_5OH + CH_3COCH_3\)
(iii) Anisole to Phenol:
Reaction with HI.
\(C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I\)
Uses of Neon:
1. Used in neon discharge lamps and signs for advertising.
2. Used in botanical gardens and green houses.
Q. 28. Define: (i) Extensive and Intensive properties (ii) Isobaric and Adiabatic processes
What are enzymes?
Write the atomic numbers of transuranium elements.
Answer:
Definitions:
(i) Extensive: Property that depends on the amount of matter (e.g., Mass, Volume).
Intensive: Property independent of the amount of matter (e.g., Temperature, Density).
(ii) Isobaric: Process carried out at constant pressure (\(\Delta P = 0\)).
Adiabatic: Process where no heat is exchanged with surroundings (\(q = 0\)).
Enzymes: Biological catalysts (proteins) that catalyze biochemical reactions in living organisms.
Transuranium elements: Elements with atomic numbers greater than Uranium (Z > 92).
Q. 29. Predict the type of cubic lattice of a solid element having edge length of 400 pm and density is 6.25 g/ml (Atomic mass = 60).
Define: Nanoscience.
Write chemical reaction for the preparation of polyacrylonitrile.
Answer:
Lattice Type Calculation:
\(\rho = \frac{n \cdot M}{a^3 \cdot N_A}\)
\(n = \frac{\rho \cdot a^3 \cdot N_A}{M}\)
Given: \(\rho = 6.25 \, g/cm^3\), \(a = 400 \, pm = 4 \times 10^{-8} \, cm\), \(M = 60\).
\(n = \frac{6.25 \times (64 \times 10^{-24}) \times (6.022 \times 10^{23})}{60}\)
\(n = \frac{6.25 \times 64 \times 0.6022}{60} = \frac{240.88}{60} \approx 4.01\)
Since n \(\approx\) 4, it is a Face Centred Cubic (FCC) lattice.
Nanoscience: The study of phenomena and manipulation of materials at atomic, molecular and macromolecular scales where properties differ significantly from those at a larger scale.
Polyacrylonitrile (PAN) Prep:
Addition polymerization of acrylonitrile (vinyl cyanide).
\(n \, CH_2=CH(CN) \xrightarrow{Peroxide/Heat} [-CH_2-CH(CN)-]_n\)
Q. 30. Derive the relation between half life period and rate constant for first order reaction. Write the net cell reaction during discharging of lead accumulator. Draw the structure of peroxymonosulphuric acid.
Answer:
Half-life Derivation (1st Order):
Integrated rate law: \(k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}\).
At \(t = t_{1/2}\), \([A]_t = \frac{[A]_0}{2}\).
\(k = \frac{2.303}{t_{1/2}} \log_{10} (2)\)
\(t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}\).
Lead Accumulator Discharging Reaction:
\(Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \longrightarrow 2PbSO_4(s) + 2H_2O(l)\)
Structure of Peroxymonosulphuric acid (Caro's acid, \(H_2SO_5\)):
Structure contains one peroxide linkage (\(-O-O-\)).
\(HO-S(=O)_2-O-OH\)
Q. 31. Mention the number of unpaired electrons and geometry of following complexes:
(i) \([Ni(Cl)_4]^{2-}\) (ii) \([Ni(CN)_4]^{2-}\)
Convert the following: (a) Ethanenitrile into ethanal. (b) Cyclohexane into adipic acid.
Answer:
Complexes:
(i) \([Ni(Cl)_4]^{2-}\): \(Ni^{2+}\) (\(3d^8\)). \(Cl^-\) is a weak field ligand. Geometry is Tetrahedral. Unpaired electrons = 2.
(ii) \([Ni(CN)_4]^{2-}\): \(Ni^{2+}\) (\(3d^8\)). \(CN^-\) is a strong field ligand (pairing occurs). Geometry is Square Planar. Unpaired electrons = 0.
Conversions:
(a) Ethanenitrile to Ethanal (Stephen Reaction):
\(CH_3CN + 2[H] \xrightarrow{SnCl_2/HCl} CH_3CH=NH \cdot HCl \xrightarrow{H_3O^+} CH_3CHO\)
(b) Cyclohexane to Adipic Acid:
Oxidation using \(KMnO_4\) or \(HNO_3\).
\(C_6H_{12} + [O] \longrightarrow HOOC-(CH_2)_4-COOH\)
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