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Sample Question Paper - 17, Chemistry (043), Class- XII, Session: 2021-22, TERM II, Time allowed : 2 hours, , Maximum marks : 35, , General Instructions :, Read the following instructions carefully., 1. There are 12 questions in this question paper with internal choice., 2. SECTION A - Q. No. 1 to 3 are very short answer questions carrying 2 marks each., 3. SECTION B - Q. No. 4 to 11 are short answer questions carrying 3 marks each., 4. SECTION C - Q. No. 12 is case based question carrying 5 marks., 5. All questions are compulsory., 6. Use of log tables and calculators is not allowed., , SECTION - A, 1., , Write structures of compounds A and B in each of the following reactions :, , 2., , What are the differences between molecularity and order of a reaction., , 3., , Which acid of each pair shown here would you expect to have lower pKa value, (a) CCl3—COOH or CH3—COOH, COOH, , (b), , or CH3CH2COOH, , SECTION - B, 4., , (a) Transition metals have very high melting and boiling points. Why?, (b) In d-block element, ionic radii of ions of the same charge decreases progressively with increasing atomic, number in a series. Why?, OR, How would you account for the following :, (a) The oxidising power of oxoanions are in the order : VO+2 < Cr2O72– < MnO4–, (b) The third ionization enthalpy of manganese (Z = 25) is exceptionally high., (c) Cr2+ is a stronger reducing agent than Fe2+., , 5., , Hydrogen peroxide, H2O2(aq) decomposes to H2O(l) and O2(g) in a reaction that is first order in, H2O2 and has a rate constant k = 1.06 × 10–3 min–1., (a) How long will it take for 15% of a sample of H2O2 to decompose?, (b) How long will it take for 85% of the sample to decompose?

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OR, For a first order reaction, calculate the ratio of the time for 75% completion of a reaction to the time for 50%, completion., 6., , For the complex [Ni(CN)4]2–, write, (a) the IUPAC name, (b) the hybridization, (c) the shape of the complex., (Atomic no. of Ni = 28), , 7., , With explanation, arrange the given compounds in decreasing order of their basicity in aqueous solution., (CH3)2NH, (H3C)3N, C6H5NH2, H3CNH2, , 8., , (a) Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on, x, mass m of the adsorbent. The plot of log, versus log p is shown in the given graph. Determine the, m, x, propotionality of ., m, x, log m, , 3, log p, , 2, , (b) Define :, (i) Peptization, (ii) Kraft temperature, OR, (a) Define protective colloids. Which type of colloids are used as protective colloids?, (b) Why does sky look blue?, (c) Define CMC., 9., , How are the following reactions carried out? (Write the equations and conditions)., (a) Acetic acid to ethylamine, (b) Bromocyclohexane to cyclohexanamine., (c) Methylamine to dimethylamine., , 10. (a) Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration, of d6 in terms of t2g and eg in an octahedral field when, (i) Do > P (ii) Do < P, (b) Write two limitations of crystal field theory., 11. (a) A compound ‘A’ of molecular formula C2H3OCl undergoes a series of reactions as shown below. Write, the structure of A, B, C and D in the following reactions :, (C2H3OCl)A, , B, , C, , D, , (b) Write the formula of the precipitate when ethanal is treated with Fehling’s Solution., OR, (a) Write the equations involved in the following reactions :, (i) Stephen reaction, (ii) Etard reaction, (b) Distinguish between CH3COOH and HCOOH.

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SECTION - C, 12. Read the passage given below and answer the questions that follow., The electrochemical cell shown below is concentration cell., M | M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M, , The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf, of the cell at 298 K is 0.059 V., (a) Calculate the value of solubility product (Ksp, mol3 dm–9) of MX2 at 298 K based on the information, available for the given concentration cell., (take 2.303 × R × 298/F = 0.059), (b) Calculate the value of DG (in kJ mol–1) for the given cell. (take 1 F = 96500 C mol–1), (c) Calculate the equilibrium constant for the following reaction., Fe2+ + Ce4+, , Ce3+ + Fe3+, , (Given: E°Ce4+/Ce3+ = 1.44 V and E°Fe3+/Fe2+ = 0.68 V), 1, 2, , (d) The standard electrode potential (E°) for OCl–/Cl– and Cl–/ Cl2 respectively are 0.94 V and – 1.36 V., 1, 2, , What will be the E° value for OCl–/ Cl2?, OR, The standard reduction potential values of the three metallic cations X, Y and Z are 0.52, –3.03, and –1.18 V, respectively. What will be the order of reducing power of the corresponding metals?

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Solution, CHEMISTRY - 043, Class 12 - Chemistry, , (c) Cr2+ is a stronger reducing agent than Fe2+. E°Cr3+/Cr2+, 1., , H2N–NH–CONH2, , is negative (–0.41 V) whereas E°Fe3+/Fe2+ is positive, (+ 0.77 V). Thus, Cr2+ is easily oxidized to Cr3+ but, , Fe2+ cannot be easily oxidized to Fe3+. Hence, Cr2+ is, stronger reducing agent than Fe2+., 5., , (a) t =, , [ A]0, 2.303, log, [ A], k, , Given k = 1.06 × 10–3 min–1,, , , 2., Order of a reaction, 1. It is the sum of, powers of the, concentration, of the reactants, in the rate law, expression., , 2. It can be zero or, even a fraction., , Molecularity of a reaction, 1. It is the number of, reacting species (atoms,, ions or molecules) taking, part in an, elementary reaction, which must collide, simultaneously in order, to bring about a chemical, reaction., 2. It is always a whole, number., , 3. (a) CCl3COOH is stronger acid than CH3COOH, thus, CCl3COOH has lower pKa value., (b) C6H5COOH is stronger acid than CH3CH2COOH, thus C6H5COOH has lower pKa value., 4. (a) , The high melting and boiling points of, transition metals are attributed to the involvement of, greater number of electrons from (n – 1) d-orbital in, addition to the ns electrons in the interatomic metallic, bonding (d-d overlap)., (b) As the atomic number increases, the new electron, enters the d-orbital and expected to increase in atomic, size, but due to poor shielding effect of d-orbitals the, electrostatic attraction between nucleus and outermost, orbital increases and hence, the ionic radii decreases., OR, (a) As the oxidation state increases, the ease with, which the oxoanions accept electrons increases. This, is why oxidising power of oxoanions are in order :, VO2+ < Cr2O72– < MnO4–., (b) Third ionization enthalpy of Mn is very high, because the third electron has to be removed from the, stable half-filled 3d-orbital [Mn2+ (Z = 25) = 3d 5]., , t=, , 2.303, −3, , 1.06 × 10 min, , −1, , log, , 100, 85, , [ A]0 100, =, [ A], 85, , 2303, [2 log 10 − log 85], 1.06, 2303, 2303 × 0.0706, [2 × 1 − 1.9294] =, t=, 1.06, 1.06, t = 153.39 min = 153.4 min., [ A]0 100, (b) Given k = 1.06 × 10–3 min–1,, =, [ A] 15, 2.303, 100 2303, =, [2 log 10 − log 15], t=, log, −3, −1, 1.06, 15, 1.06 × 10 min, 2303 × 0.8239, 2303, min = 1790 min, [2 × 1 − 1.1761] =, =, 1.06, 1.06, t=, , OR, t75% =, , 2.303, 100, log, k, 100 − 75, , t50% =, , 2.303, 100, log, k, 100 − 50, , t75% log 100 − log 25 0.6020, =, =2, =, t50% log 100 − log 50 0.3010, , 6. (a) Tetracyanonickelate(II) ion, (b) Ni atom (Z = 28), Ground state :, Ni2+ ion :, [Ni(CN)4]2– :, d, , 2, , CN, strong ligands, , (c) The complex ion has square planar geometry and is, diamagnetic due to the absence of unpaired electrons.

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7. The increasing order of basicity in Aqueous, solution of the given compounds is, (CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2, Due to the subtle interplay of the inductive effect,, solvation effect and steric hinderance of the alkyl, group which decides the basic strength of methyl, substituted amines in aqueous solution., Due to the +I effect of alkyl groups, the electron density, on nitrogen increases and thus the availability of the, lone pair of electrons to proton increases and hence, the basicity of amines also increases. So aliphatic, amines are more basic than aniline., , (c) CH3NH2 + CHCl3 + KOH, , 8. , (a) For a plot between log x/m and log p slope is, given by 1/n., , (i) When Do > P,, , Thus from the plot, , (b) (i) It assumes ligand to be point charges., (ii) It does not take into account the covalent character, of bonding between the ligand and the central atom., , 1 y2 − y1, =, = 2 / 3 and, n x2 − x1, , x/m ∝ p1/n ∝ p2/3, (b) (i) Peptization : It is the process of conversion of, freshly prepared precipitate into colloidal particles on, adding a suitable electrolyte., (ii) Kraft temperature : it is defined as the minimum, temperature from which the micelle formation takes, place., , CH3NC, , H2, Na/C 2H5 OH, , , , CH3 NHCH3, , 10. (a) The difference of energy between two splitted, levels of d-orbitals is called crystal field splitting, energy. It is denoted by D or 10 Dq., For octahedral Do, for tetrahedral it is Dt and for square, planar Dsp., Dq, , t 62g e0g, , t 42g eg2, , (ii) When Do < P,, , 11. (a), OH, , , , O, , OR, , O, , CH3 CH CH2 C H, (C), , CH3 CH CH C H, , , , (D), , (b) CH3CHO + Cu2+ + OH– → CH3COO– + Cu2O, , + 3H2O, The precipitate formed is Cu2O., OR, , (a) (i) Stephen reduction :, R—CN + SnCl2 + HCl → R —CH NH, H 3O +, →, , R —CHO, (ii) Etard reaction :, CH3, , Toluene, , + CrO2Cl2, , CH(OCrOHCl2)2, , CS, , 2, →, , Chromium, complex, , →, , (a) The colloids which protect coagulation of other, colloids from the electrolytes are called protective, colloids. Lyophilic colloids are used as a protective, colloid for lyophobic colloids., (b) Dust particles along with water suspended in, air have size smaller than wavelength of visible light, and are more effective in scattering light of shorter, wavelength, blue light which has smallest wavelength, reaches our eyes and the sky looks blue to us., (c) CMC : It is the minimum concentration of, surfactants above which micelles formation takes, place. Below the CMC, the substance forming micelle, behave as electrolyte., , D, , H3O+, , 9. (a), , CHO, , Benzaldehyde, , (b), , KCN(aq), , (b) Add Tollens’ reagent to formic acid and warm., Silver mirror is formed., Warm, HCOOH + 2[Ag(NH3)2]+ + 2OH–, Formic acid, , , , , 2Ag + CO2 + 2NH3 + 2NH4OH, ,           Silver mirror, , Acetic acid does not give this test.