CHEMISTRY-24-12- 11th (PQRS & J) Code-A WA SOLUTION

REVIEW TEST-7/6 Class : XI (P,Q,R,S & J) PAPER CODE : A Time : 2 hour Max. Marks : 130 INSTRUCTIONS 1. The question paper contains 12 pages and 3-parts. Part-A contains 20 objective questions, Part-B contains 3 "Match the Column" questions and Part-C contains 4 "Subjective" questions. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.15 have only one correct alternative and carry 3 marks each. There is NEGATIVE marking and 1 mark will be deducted for each wrong answer. (ii) Q.16 to Q.20 have one or more than one correct alternative(s) and carry 5 marks each. There is NO NEGATIVE marking. Marks will be awarded only if all the correct alternatives are selected. PART-B (iii) Q.1 to Q.3 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NEGATIVE marking. 0.5 Marks will be deducted for each wrong match. Marks will be awarded only if all the correct alternative(s) is/are selected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions and carry 9 marks each. There is NO NEGATIVE marking. Marks will be awarded only if all the correct bubbles are filled in your OMR answer sheet. 2. Indicate the correct answer for each question by filling appropriate bubble in your OMR answer sheet. 3. Use only HB pencil for darkening the bubble. 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. PART-B For example if correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling the bubbles is P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column will be treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 USEFUL DATA Atomic weights: Al = 27, Cl = 35.5, O = 16, H = 1, N = 14, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Atomic No. H = 1, O = 8; P = 15, Useful constant : h = 6.626 × 10–34 Js; mass of electron = 9.1 × 10–31 Kg; R = 0.082 lt-atm/mol–K PART-A Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking and 1 mark will be deducted for each wrong answer. Q.1 Choose the only incorrect statement (A*) K.E of photo electron does not depend upon wavelength of incident radiation and its value ranges from zero to KEmax (B) Photoelectric current depends on intensity of incident radiation and not on frequencywhereas stopping potential depends on frequency of radiation & not on intensity. (C) If the accelerating potential V eV is applied on photo electron in discharge tube, then the value of KE of photo electrons vary from V eV to (a–b+V) eV where a & b are energy of incident radiation and work function in eV respectively with a > b. (D) None of these hc [Sol: (A) KEmax =  – E0 & KEmin = 0 (B) i = q = Ne t t so current depends on no. of photoelectrons emitted which in turn depends on intensity of incident radiation. h V0 = e – E0 (C) KEmin = (eV)J = (V) eV; KEmax = KEmin + E – E0 = [(a–b) + V] eV ans: is (A) ] Q.2 Statement 1: If the number of -particles observed in Rutherford experiment are 150 at 90° then the number of particles scattered under similar experimental conditions at 120° are approx 67. Statement 2 : Ionisation energy of deuterium atom is slightly lesser than that of hydrogen atom. Statement 3 : The difference in potential energy always remain same between any two electronic shell in hydrogen atom irrespective of any reference value of potential energy assigned to n = . Select the correct statement. (A) Statements 1, 2 and 3 (B) Statements 1 and 2 (C*) Statements 1 and 3 (D) Statements 2 and 3 1 [Sol: statement 1 : N    4 sin 2    1 1 N1  (sin 450 )4   1 4  4     1 & N2  (sin 600 )4 1   3 4  16 9    2  N1 so N2 = 4  9 9 16 4 150 9  N2 = 4  N2 = 4 150 9  67 (correct) Statement 2 : Ionization energy of deuterium is slightly greater than that of hydrogen atom. Statement 3 : Correct (C) ] Q.3 30 ml of CH3OH (d = 0.8 gm/cm3) is mixed with 60 ml of C2H5OH (d = 0.92 gm/ cm3) at 25°C to form a solution of density 0.88 gm/cm3.Select the correct option. (A) Molarity & molality of resulting solution are 6.33 & 13.59 respectively. (B*) The mol fraction of solute and molality are 0.385 and 13.59 respectively. (C) % change in volume and molarity are zero and 13.59 respectively. (D) Molarity & mol fraction of solute are 13.59 & 0.485 respectively. [Sol: mass of CH3OH = 30 × 0.8 = 24 gm mass of C2H5OH = 60 × 0.92 = 55.2 gm Total mass of solution formed = 24 + 55.2 = 79.2 gm  final volume = 79.2 0.88 = 90 ml  % change in volume = 0 molarity of solute (i.e. CH3OH) = 24 24 1000 32  90 = 8.33 M moles of CH3OH = 32 = 0.75 55.2 moles of CH3OH = 46 = 1.2  mole fraction of CH3OH = 0.75 0.75 +1.2 = 0.385 nCH OH 1000  molality of solution = 3 = 0.751000 = 13.59 m mass of solvent So (B) option is correct ] 55.2 Q.4 Al2(SO4)3 solution of 1 molal concentration is present in 1 litre solution of density 2.684 gm/cc. How many moles of BaSO4 would be precipitated on adding BaCl2 in excess. (A) 3 moles (B) 2.684 × 3 moles (C*) 6 moles (D) 2 moles [Sol: Mass of solution = 1000 × 2.684 = 2684 gm molar mass of Al2(SO4)3 = 342 gm  mass of solute in 1342 gm solution = 342 gm  mass of solute in 2684 gm solution = × 2684 = 684 gm  moles of Al2(SO4)3 = 2 Al2(SO4)3 + 3BaCl2  3BaSO4 + 6AlCl3  moles of BaSO4 precipitated = 3 × 2 = 6 (C) ] Q.5 Number of P–H , P–O–P , P–O–H and P = O bonds in sodiumdihydrogenpyrophosphate respectively are (A) 1, 1, 1, 2 (B*) 0, 1, 2, 2 (C) 0, 1, 1, 3 (D) 2, 0, 0, 2 [Sol: Na2H2P2O7 O O || || H  O  P  O  P  O  H | | O O no. of P–H bond = 0 no. of P–O–P bond = 1 no. of P–O–H bond = 2 no. of P=O bond = 2 (B) ] Q.6 The question below consists of an Assertion and the Reason. Mark appropriate option using the given directions. (A) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (B) If both assertion and reason are CORRECT, but reason is not the CORRECT explanation of the assertion. (C) If assertion is CORRECT but reason is INCORRECT. (D) If assertion is INCORRECT but reason is CORRECT. Assertion: Bond energy of F2 is higher than that of Cl2. Reason: According to VBT, 2p-2p overlap is stronger than 3p-3p. [Sol: Bond energy of F2 is lesser than that of Cl2 According to VBT, 2p–2p overlap is stronger than 2p–3p (D)] Q.7 (A) 1–Formyl–2–chloro–4–hydroxy–hex–5–en–4–one (B) 5–chloro–6–formyl–3–hydroxy–4–keto–hex–1–enal (C*) 3–chloro–5–hydroxy–4–oxo–hept–6–enal (D) none of these [Sol: 3–chloro–5–hydroxy–4–oxo–hept–6–enal] Q.8 IUPAC nomenclature of this compound is (A) 6–chloro–3, 7, 8, 9 tetramethylundecane (B*) 6–chloro–3, 4, 5, 9 tetramethylundecane (C) 5–chloro–2–ethyl–3, 4, 8 trimethylundecane (D) 6–chloro–9–ethyl–3, 7, 8 trimethylundecane [Sol: 6–chloro–3, 4, 5, 9 tetramethylundecane] Q.9 44 gm of a sample on complete combustion gives 88 gm CO2 & 36 gm of H2O. The molecular formula of the compound may be (A) C4H6 (B) C2H6O (C*) C2H4O (D) C2H4 88 [Sol: moles of CO2 = 44 = 2 36 moles of H2O = 18 = 2 C  CO2 H  H2O POAC for C POAC for H nC = n CO2 = 2 nH = 2 × n H2O = 4  c = 24 gm H = 4 gm  wt. of oxygen = 44 – (24 + 4) = 16 molecular formula of compound = C2H4O (C) ] Q.10 The hybridization of iodine in ICl , IOCl , IOCl+ is 4 4 4 (A) sp3d, sp3d2, sp3d2 (B) sp3d, sp3d2, sp3d (C*) sp3d2, sp3d2, sp3d (D) sp3d2, sp3d, sp3d2 [Sol: ICl – AB E hybridisation of iodine = sp3d2 4 4 2 IOCl4– AB5E hybridization of iodine = sp3d2 IOCl + AB hybridisation of iodine = sp3d 4 5 (C) ] Q.11 Which of the following would have permanent dipole moment. (A*) SF4 (B) XeF4 (C) SiF4 (D) BF3 [Sol: SF4  see - saw shape (unsymmetrical) XeF4  square planar (symmetrical) SiF4  tetrahedral (symmetrical) BF3  triangular planar (symmetrical) (A) ] Q.12 The correct order of bond angles is (A) H2S < NH3 < BF3 < SiH4 (B*) H2S < NH3 < SiH4 < BF3 (C) NH3 < H2S < SiH4 < BF3 (D) H2S < SiH4 < NH3 < BF3 [Sol: H2S NH3 SiH4 BF3     sp3 sp3 sp3 sp2 +2lp +1lp so the order of the bond angle H2S < NH3 < SiH4 < BF3 (B) ] Q.13 The maximum number of 90° angles between bond pair – bond pair is observed in (A) sp3d hybridisation (B*) sp3d2 hybridisation (C) sp3d3 hybridisation (D) sp3 hybridisation [Sol: sp3d six 900 angles sp3d2 twelve 900 angles sp3d3 ten 900 angles sp3 no 900 angles (B) ] Q.14 Which molecular geometry is least likely to result from a trigonal bipyramidal geometry? (A*) Trigonal planar (B) See-Saw (C) Linear (D) T-shape [Sol: See saw, T–shape and linear are derived geometries of TBP but trigonal planar is least likely to form from a TBP geometries (A) ] Q.15 Ionic conductance (I.C.) is a measure of electrical conductivity of ions in their aqueous solution. Select the option in which the values of ionic conductance given, are correctly matched with ions. Data: (A) Na+  50.1 mho cm2 mol–1, Cl–  65.5 mho cm2 mol–1 & K+  64.5 mho cm2 mol–1 (B) Na+  64.5 mho cm2 mol–1, Cl–  65.5 mho cm2 mol–1 & K+  50.1 mho cm2 mol–1 (C) Na+  50.1 mho cm2 mol–1, Cl–  64.5 mho cm2 mol–1 & K+  65.5 mho cm2 mol–1 (D) Na+  64.5 mho cm2 mol–1, Cl–  50.1 mho cm2 mol–1 & K+  65.5 mho cm2 mol–1 [Sol: Order of extent of hydration is Cl– < K+ < Na+  order of ionic conductance is Cl–(aq) > K+(aq) > Na+(aq) (A) ] Select the correct alternatives. (One or more than one is/are correct) [5 × 5 = 25] There is NO NEGATIVE marking. Q.16 Select the incorrect statement(s): (A*) In PO3 ion, the P–O bond order is 1.33. (B*) Covalent bond can be formed by overlapping of py and pz orbitals (C) Bond angle in H2O is lesser than bond angle in OCl2 molecule. (D*) SeF4 and CH4 have same shape. 5 [Sol: (A) PO 3– B.O. 4 = 1.25 (B) no bond can be formed (C) (D) SeF  see saw ; CH  Tetrahedral Answers are (A) (D) ] Q.17 Species having maximum number of lone pairs on the central atom is/are: (A*) ClO– (B) XeF— (C) SF4 (D*) — [Sol: – O– (ABE ) number of lp = 3 XeF – (AB E ) number of lp = 2 5 5 2 SF4 (AB4E) number of lp = 1 I – (AB E ) number of lp = 3 3 2 3 order of bond length is (A), (B), (D) ] Q.18 Which is/are correct statement(s)? (A*) The increasing order of carbon-carbon bond length is: Acetylene < Ethene < Benzene < Ethane (B*) The increasing order of O–O bond length is: O2 < O3 < H2O2 (C*) The decreasing order of O–N–O bond angle is: NO2+ > NO2 > NO2– (D*) The decreasing order of first ionisation potential is: Ne > Cl > P > S > Mg > Al [Sol: (A) HC  CH < H2C=CH2 < C6H6 < H3C–CH3 (B) order of bond length O = O < < H–O–O–H (C) + ..  . N O2 N O2 N O2   sp sp2 + 1 lp order of bond angle NO2+ > NO2 > NO2– (D) order of first ionization potential is Ne > Cl > P > S > Mg > Al Ans are (A), (B), (C), (D)] Q.19 The molecule shown is a alcohol because OH | CH3  CH2  CH  CH2  CH3 . (A) primary; it has one -OH group. (B) primary; its -OH group is on the central carbon. (C*) secondary; the carbon bonded to the -OH group is bonded to two other carbons. (D*) secondary; because -OH group is attached with 2oC of secondary pentyl group. [Sol: CH3  CH2  CH  | CH2CH3 is known as "secondary" pentyl group & –OH group is attached with 20C atom therefore e.g. of 20 ol] Q.20 Assuming the bond direction to be z-axis, which of the overlapping of atomic orbitals of two atom (A) and (B) will result in bonding? (A) s-orbital of A and px orbital of B (B*) s-orbital of A and pz orbital of B (C) py-orbital of A and pz orbital of B (D*) s-orbitals of both (A) and (B) [Sol: +  s s +  ] s p2 PART-B MATCH THE COLUMN [3 × 8 = 24] There is NEGATIVE marking. 0.5 Marks will be deducted for each wrong match. INSTRUCTIONS: Column-I and column-II contains four entries each. Entries of column-I are to be matched with some entries of column-II. One or more than one entries of column-I mayhave the matching with the same entries of column-II and one entryof column-I mayhave one or more than one matching with entries of column-II. Q.1 Column I Column II (A) XeF + (P) Two lone pairs (B) ICl – (Q) Planar (C) TeCl4 (R) Non-planar (D) I3+ (S) sp3d2 (Hybridization of central atom) [Sol: XeF + AB E sp3d2 square pyramidal non planar 5 5 ICl4 AB4E2 sp3d2 square pyramidal planar TeCl4 AB4E sp3d see-saw non planar I3+ AB2E2 sp3 bent planar (A)  (R, S); (B)  (P, Q, S); (C)  (R); (D)  (P, Q) ] Q.2 Column - I Column -II (A) KCN (P) Polar covalent bond (B) CuSO4.5H2O (Q) Co-ordinate bond (C) O3 (Resonating structures) (R) Ionic bond (D) Solid N2O5 (Resonating structures) (S) Non Polar Covalent bond [Sol: (A) K+[CN]– (B) . H2O (C) (D) [NO2+] [NO3–]  [O=N=O]+ (A)  (P, R); (B)  (P, Q, R); (C)  (Q, S); (D)  (P, Q, R) ] Q.3 Column - I Column -II (A) Energy of a photon having (P) 3.2 × 10–19 J wavelength 2484 Å is (B) Kinetic energy of an e– having (Q) 66.67 eV wavelength 150 pm is (C) Maximum kinetic energy (R) 3.0 × 10–16 J a photoelectron emitted from a metal [work function= 2eV] by a photon of wavelength 310.5 nm is (D) Energy of a particle of mass 2 × 10–6 amu, (S) 5 eV converted into energy is [Sol: (A) E = 1242 = 5eV = 5 × 1.6 × 10–19 J = 8 × 10–19 J 248.4 (B)  =  150 × 10–12 = 6.6261034  K.E. = 66.67 eV (C) KEmax = E– E0 = 1242 – 2 = 2eV = 2 × 1.6 × 10–19 J = 3.2 × 10–19 J 310.5 (D) energy = mc2 = 2 10-6 103 NA 2 10-9 × (3 × 108)2 = 6 10 × (3 × 108)2 = 3.0 × 10–16J (A)  (S); (B)  (Q); (C)  (P); (D)  (R) ] PART-C SUBJECTIVE: [4 × 9 = 36] There is NO NEGATIVE marking. Q.1 Mr Gupta has to decode a number ABCD. EF, where each alphabet is represented by a single digit. From the following information given about each alphabet. Identify the number. Info A = number of radial nodes of the orbital having 2(r) vs r curve given as Info B = number of subshells having energy between 8s & 8p. Info C = ratio of density of SO2 at 400 K to density of O2 at 800 K at same pressure. Info D = ratio of radius of 2nd Bohr orbit of He+1 to radius of 1st Bohr orbit of Be+3 Info E = number of radial nodes of an orbital whose radial wavefunction is represented as Y(r) = K . er / K2 (r2 – 5 K r + 6K2 ) 1 3 3 Info F = ratio of T.E to K.E for 1st orbit of H-atom assuming (P.E) = 27.2 eV [Sol: Info A The given orbital is 2s  no. of radial node = 1 A = 1 Info B Number of subshells having energy between 2sand 2p = 3 (5g, 6f, 7d) Info C PMSO PM 2 = dSO RTSO = d RT 2 2 64 dSO .......(1) .......(2) 400 dividing (1) by (2) 32 = d d 2 × 800  d = 4 2  C = 4 Info D (r2 )He+  (r1)Be3+  (r2 )He+ (2)2 2 2 (1)2  1 4 4  (r1 ) = 8 Be3+  D = 8 Info E E = 2 1 Ze2 Info F KE = 2 r Ze2 PE = r + (PE) 1 Ze2 TE = – 2 r + (PE) = –13.6 + 27.2 = 13.6 eV 1 Ze2 KE = 2 r = 13.6 eV ratio = TE = 13.6eV = 1 KE 13.6eV Ans: A = 1; B = 3; C = 4; D = 8; E = 2; F = 1 ] Q.2 During preparation of sodium carbonate in the Solvay process CO2 gas is bubbled into the brine solution (NaCl in water) saturated with NH3 giving rise to following equations. NH3 + H2O + CO2  NH4HCO3 NaCl + NH4HCO3  NaHCO3 + NH4Cl Sodium hydrogen carbonate formed precipitate out & is then subjected to photoelectric dissociation to obtain sodium carbonate. 2NaHCO3 h Na2CO3 + CO2 + H2O If 100g of 58.5% brine solution (w/w) is mixed with 34 gm of NH3 & then 49.2 litres of CO2 at 1 atm pressure & 300 K is bubbled into the solution & the sodium bicarbonate is subjected to photochemical  molesof reactant dissociated  dissociation to obtain a quantum efficiency   molesof photons subjected  = 0.75 then what is the  value of total energy subjected in kJ if  of radiation is 1242 nm. Assume no NaHCO3 to be remaining after photochemical dissociation. [Use NA = 6 × 1023; R = 0.082 lt-atm/K-mol] [Sol: moles of NaCl = 58.5  100 = 1 100 58.5 34 moles of NH3 = 17 PV moles of CO2 = RT = 2 1 49.2 = 0.082  300 = 2 Now NH3 + H2O + CO2  NH4HCO3 t = 0 2 excess 2  moles of NH4HCO3 formed = 2 NH4HCO3 + NaCl  NaHCO3 + NH4Cl t = 0 2 1  moles of NaHCO3 formed = 1 moles of reactant dissociated moles of quantum subjected = 0.75 = 1 4 = 4 3 3  total number of photons used = 4 N 3 A 4 1242 Total energy subjected = 3 NA × 1242 eV = 4 × 6 × 1023 × 1.6 × 10–19 J 3 = 128000 J = 128 kJ ] Q.3 Gaseous ethylene, C2H4 reacts with hydrogen gas in the presence of a platinum catalyst to form ethane, C2H6 according to C2H4(g) + H2(g)  C2H6(g). A mixture of C2H4 & H2 known to contain more H2 than C2H4 had a pressure of 60 atm in an unknown volume. After the mixture had passed over platinum catalyst & C2H4 completely reacted, the pressure was 90 atm in the original volume and at double the absolute temperature as was earlier. Find out the mole % of H2 in the original mixture. [Sol: C2H4(g) + H2(g)  C2H6(g) t = 0 p 60 – p – on completion – 60 – 2p p total pressure = 60 – p (at the original conditions)  (60 – p)2 = 90  120 – 2p = 90  p = 15 so mole % of H2 = 45 100 = 75% ] 60 Q.4 The key reaction in the manufacture of synthetic cryolight for aluminium electrolysis is HF(g) + Al(OH)3 (s) + NaOH (aq)  Na3AlF6(aq) + H2O(l) Assuming a 96 % yield of dried, crystallized product, what mass (in kg) of cryolite can be obtained from the reaction of 351 kg of Al(OH)3, 1.10 m3 of 50.0% by mass aqueous NaOH (d = 1.50 g/mL), and 225 m3 of gaseous HF at 312.08 kPa and 87oC? (assume that the ideal gas law holds) [Given: Al = 27, O = 16, H = 1, Na = 23, F = 19, R = 8.3 JK–1 mol–1] [Sol: 6HF + Al(OH)3 + 3NaOH  Na3AlF6 + 6 H2O(l) 351103 moles of Al(OH)3 = 78 = 4500 moles of NaOH = PV (1.1106 )1.5 0.5 40 312080  225 = 20625 moles of HF = RT = 8.3 360 = 23500 Here HF is limiting reagent so moles of Na AlF formed = 23500  96 = 3760 3 6 6 3760  210 100 wt. of Na3AlF6 formed = 1000 = 789.6 kg ]