Study Material Chemistry

CHEM ACADEMY IIT-JAM 2007 1 IIT-JAM 2007 Note: Attempt ALL the 44 questions. Question 1-30 (Objective questions) carry three marks each and question 31-44 (Subjective Questions) carry fifteen marks each. 1. The compound, which (i) Reacts rapidly with acetyl chloride (ii) Does not react with 2, 4-dinitrophenylhydrazine and (iii) Does not form a yellow precipitate with excess of iodine in aqueous alkali is (a) Acetone (b) Diethyl ether (c) 2-methyl-2-propanol (d) Ethanol 2. 3. 4. 5. CHEM ACADEMY The given compounds 1 and 2 are (a) Identical (b) Diastereomeric (c) Enantiomeric (d) Constitutionally isomeric The correct order of dipole moments (µ) of the following compounds is (1) CH3CH2CH2CHO (2) CH3CH = CHCHO (3) CH3CH2CH = CH2 (a) µ1 > µ2 > µ3 (b) µ2 > µ3 > µ1 (c) µ3 > µ1 > µ2 (d) µ2 > µ1 > µ3 Which one of the following compounds gives positive test for both nitrogen and halogen with its Lassaigne’s extract? (a) CH3NH2.HCl (b) NH2OH.HCl (c) NH4Cl (d) H2NNH2.HCl Which one of the following compounds is optically active? (a) 1 (b) 2 (c) 3 (d) 4 • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 6. 2 IIT-JAM 2007 The compounds that react with aqueous NaHCO3 to release CO2 are 8. (a) 1 and 3 (b) 2 and 4 (c) 2 and 3 (d) 1 and 4 The complementary strand of DNA for the following single stranded DNA sequence, 5´–A–T–C–A–T–G–C–3´ is (a) 5´–A–T–C–A–T–G–C–3´ (b) 5´–T–A–G–T–A–C–G–3´ (c) 5´–G–C–A–T–G–A–T–3´ (d) 5´–C–G–T–A–C–T–A–3´ The value of ‘n’ for the following molecule according to Huckel’s rule is 9. (a) 16 (b) 4 (c) 3 (d) 14 Which one of the following compounds reacts with nitrous acid to give the product [P]? 7. 10. CHEM ACADEMY (a) (b) (c) (d) The main product obtained in the following reaction is (a) (b) • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY (c) 11. 12. 13. IIT-JAM 2007 3 (d) For a reaction with rate equation –dC/dt = kC2, C0 and C are the concentrations of the reactant at time 0 and t respectively. If 10 minutes were required for C0 to become C0/2, the time required for C0 to become C0/4 is: (a) 10 min (b) 20 min (c) 30 min (d) 40 min For a cyclic process performed by an ideal gas, changes in some thermodynamic functions are zero. Indicate the set in which all the functions are zero (a) w, E, H, G (b) q, S, H, A (c) q, E, S, G (d) E, S, H, A The plot of Gibb’s free energy G and the extent of a reaction  is given below for the reaction A  B,. If µA and µB are the chemical potentials of A and B respectively, the incorrect statement is: CHEM ACADEMY 14. 15. 16. (a) At point X, µA > µB (b) At point Y, G = 0 (c) At point Z, µA > µB (d) At equilibrium, the composition of the reaction mixture can be identified. The overlap between the atomic orbitals sketched below is: (a) Positive (b) Negative (c) Zero (d) No overlap –3 The pH of a 1.0 × 10 M solution of a weak acid HA is 4.0. The acid dissociation constant Ka is: (a) 1.0 × 10–3 (b) 1.0 × 10–4 (c) 1.0 × 10–5 (d) 2.0 × 10–5 The normalisation constant ‘A’ for the wavefunction () = Ae(im) where 0 <  < 2 is: (a) 1 2 (b) 2 (c) 2 (d) 1 2 • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 17. 18. The standard potential of a Daniel cell is +1.10 V and the equilibrium constant for the cell reaction is 1.5 × 1037. It can be concluded that (a) Zinc oxidises copper (b) Displacement of copper by zinc goes to near completion (c) Copper oxidises zinc (d) Displacement of zinc by copper goes to completion. Which one of the following figures, showing kinetic energy of the ejected electron versus the frequency (v) of the incident photon, represents the Einstein’s photoelectric effect? (a) 19. 20. 21. 22. IIT-JAM 2007 4 (b) (c) (d) An aqueous solution containing 0.01 M FeCl3 and 0.06 M HClO4 has the same ionic strength as a solution of (a) 0.09 M NaCl (b) 0.04 M Na2SO4 (c) 0.06 M CuSO4 (d) 0.03 M H3PO4 – Which one of the following species is the conjugate base of HO ? (a) H2O (b) O2– (c) O2– (d) O32– The solid-liquid phase diagram for the Mg-Zn system is shown in the figure below where the vertical line at X(Mg) = 0.33 represents the formation of a congruent melting compound MgZn2. The figure is divided into seven regions depending upon the physical state of the system. The composition of the region #6 represents. CHEM ACADEMY (a) Single phase of a solution of Mg and Zn (b) Two phase region between the solid Zn and solid MgZn2 (c) Two phase region between the liquid and solid MgZn2 (d) Two phase region between solid Mg and solid MgZn2 In the extraction of metals from their ores, which one of the following reduction methods can bring about a non-spontaneous reduction? (a) Electrolytic reduction (b) Reduction by carbon (c) Reduction by another metal (d) Reduction by hydrogen • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 23. 24. 25. 26. 27. 28. 5 IIT-JAM 2007 The correct order of the ionic radii is: (a) ln3+ > Sn4+ > Sr2+ > Rb+ (b) Sn4+ > ln3+ > Sr2+ > Rb+ (c) Rb+ > ln3+ > Sr2+ > Sn4+ (d) Rb+ > Sr2+ > ln3+ > Sn4+ The correct valence shell electronic configuration of the element with atomic number 22 is (a) [Ar]4s23d2 (b) [Ar]3d4 (c) [Ar]3d24s2 (d) [Ar]4s24p2 The ligand with only sigms () bonding character is: (a) CN– (b) CH3– (c) CO (d) NO Which one of the following species is NOT isoelectronic with CO? (a) N2 (b) CN– (c) NO+ (d) O2+ During wittig reaction, a phosphorus yield gets converted to (a) R3P (b) R3P = 0 (c) R3P+HOH– (d) R2P – PR2 Which of the following reactions does not give H3PO4? (a) Ca 3 (PO 4 )2  H 2SO 4   (b) P4O 6  H 2O   (c) PCl5  H 2O   (d) P4S10  H 2O   31. The ionic radii of Ca2+ and F– are 100 pm and 133 pm respectively. The coordination number of Ca2+ in the ionic solid will be (a) 8 (b) 6 (c) 4 (d) 2 The shape of CH3– ion is: (a) trigonal planar (b) tetrahedral (c) trigonal pyramidal (d) linear Identify reagent (P) and write the structure of products (Q, S and T) in the following 32. Identify the structures of the intermediate compounds Q, R and S. Show the transformation for each step. (a) For the following scheme of transformations, draw the structures of A, B, C and D. 29. 30. CHEM ACADEMY • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 33. 6 IIT-JAM 2007 (b) Complete hydrolysis of a pentapeptide with 6N HCl at 110ºC in a sealed tube gave 2 equivalents of glycine, one equivalent each of tyrosine, leucine and phenylalanine. Reaction of the pentapeptide with Sanger’s reagent (2,4-dinitrofluorobenzene, DNFB) and subsequent hydrolysis gave the DNFB derivative of tyrosine. Chymotryspin cleavages of this peptide yielded tyrosine, leucine and a tripeptide. Deduce the sequence of the pentapeptide. Complete the following reactions with appropriate structures for E, F, G, H and I. (a) (b) 34. (a) Account fot the following transformation with an appropriate mechanism. Give the structure of the Hoffmann exhaustive methylation product of 1, 2-dihydro derivative of [X]. CHEM ACADEMY (b) The optically pure ester [J] is hydrolysed in aqueous acetic acid to form a racemic mixture of cis-4, 4-dimethyl-2-acetoxycyclopentanol [K]. Give a mechanistic explanation to account for the formation of [K] and the observed change in the optical activity. 35. 36. (a) M is a first row transition metal. MCl2 on treatment with aqueous ammonia gives a blue coloured solution of complex N. A solution of MCl2 also gives a bright red precipitate of complex O with ethanoic dimethylglyoxime. (i) Identify M and draw the structure of O. (ii) Determine the hybridisation of M in complex N. (iii) Identify the paramagnetic complex. (b) [Cr(H2O)6]3+ gave an absorption at 208 kJ/mol which corresponds to 0. Calculate the crystal field stabilisation energy of this complex in kJ/mol. (a) Consider the ethers H3SiOSiH3 and H3COCH3. (i) Which ether has more lewis base character? (ii) Which angle [Si-O-Si and C-O-C] is greater? Justify your answer. • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY IIT-JAM 2007 7 (b) Starting from SiO2, show how the following polymer is prepared industrially? 37. 38. 39. (a) A solution of metal ion (M2+) when treated with H2S gas gives a black precipitate A. Precipitate A dissolved in hot concentrated nitric acid to give B along with elemental sulfur. The metal ion soluion also gives a white precipitate C with an excess of KI. Write the chemical formulae of A, B and C (b) Why are potassium permanganate solutions unstable in the presence of Mn2+ ions? In the quantitative estimation of iron present in iron ores dissolved in dilute HCl, titrations with dichromate are preferred over titrations with permanganate. Rationalise. (a) Al2Cl6 and Al2Me6 are dimeric in gas phase. Draw their structures. Which compound has more Lewis acid character? Explain. (b) Arrange the halides SnCl2, PbCl2, SiCl2 in increasing order of their stability. Give reasons for your answer. (a) Acdification of an aqueous solution of yellow sodium chromate gives an orange coloured A. A compound solution of A on treatment with concentrated H2SO4 gives a bright orange solid B. Compound A in the presence of concentrated H2SO4 reacts with anion C to give a deep red coloured liquid. Identify A, B and C. (b) 40. CHEM ACADEMY 215 84 Po undergoes an  emission to give element X followed by a  emission to give element Y.. (i) Write the valence shell electronic configuration of Y. (ii) Indicate the groups of the periodic table to which X and Y belong. (a) When an ideal monoatomic gas is expanded from 1.5 bar, 24.8 L and 298 K into an evacuated container, the final volume becomes 49.6 L. Calculate H, S and G for the process. (b) The Maxwell distribution function for the distribution of speeds of molecules in gaseous systems is given by 3/ 2  m  f (c)  4    2kT   mc 2  c exp   2kT    2 1/ 2 Show that the most probable speed, cmps 41.  2kT     m  (a) At 600 K and 200 bar, a 1:3 (molar ratio) mixture of A2 and B2 react to form an equilibrium mixture containing x AB3  0.60 . Assuming ideal gas behaviour, calculate Kp for the reaction A 2 (g)  3B2 (g)  2AB(g) (b) A 50 mL 0.05 M solution of Fe(II) is titrated with 0.05 M solution of Ce(IV) in the presence of dilute H2SO4 at 25ºC. Calculate the equivalence point potential and the equilibrium constant K in terms of log K • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 8 IIT-JAM 2007  E0  0.75V, E 0 4  3  1.45V   (Fe3 / Fe2 ) (Ce / Ce )  42. 43. (a) The vapour pressure of D2O at 20ºC is 745 mm Hg. When 15 g of a non-volatile compound is dissolved in 200 g of D2O the pressure changes to 730 mm Hg. Assuming the applicability of Raoult’s law, calculate the molecular weight of the compound. (b) An enzyme following Michaelis-Menten kinetics was found to have highest activity at 37ºC and pH 7.0. If the maximum velocity Vmax for this enzyme was 2.4 × 10–4 mol L–1 s–1 with an initial enzyme concentration [E]0 = 2.4 nM, calculate the turnover frequency. (a) Consider the 4 electrons in cyclobutadiene to be free particles in a 2-dimensional square box of length 2Å. Calculate the wavelength of the electronic transition from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). Also write down the normalised wavefunctions for the occupied degenerate states. (b) The reaction is first order in both directions. At 25ºC, the equilibrium constant (K) of this reaction is 0.40. If 0.115 mol dm–3 of cis-isomer is allowed to equilibrate, calculate the equilibrium concentration of each isomer. 44. CHEM ACADEMY (a) With i, j and k as the unit vectors along, X, Y, Z axes, express the vector, P1P2 in the given figure in terms of the coordinate of P1 and P2. Also determine the dot products of the unit vectors, i, j, k. (b) Deduce whether the matrices A and B commute or not.  2 1 A   0 1  1 1 B   0 1 • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 9 IIT-JAM 2007 SOLUTION OBJECTIVE QUESTIONS 1. Correct option is (c) 2. 3. As connectivity of atoms in (1) and (2) is different, so they are constitutional (structural) isomers. Correct option is (d) Dipole moment in above molecule can be represented as CHEM ACADEMY Maximum charge separation (r). Among these three the µ2 has maximum dipole moment followed by µ1 and µ3 because as we know the dipole moment (µ) is defined by following equation µ=q×r where, q = magnitude of charge r = separation between charge. As 4. bond has larger dipole moment than so both µ1 & µ2 will be larger than µ3. As (2) has larger charge separation than (1), So, µ2 > µ1. Correct option is (d) In Lassaigne’s test of an organic compound we get test for CN– and X–, if the compound has nitrogen and halogen. For formation of CN–, compound must have carbon. Correct option is (a) • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 5. 10 IIT-JAM 2007 Compound (1), (2) and (3) has plane of symmetry, so they are optically inactive. While (4) does not contain plane of symmetry so (4) is optically active. Correct option is (d) 6. 7. Correct option is (b) In complementary strand ‘A’ will be replaced by ‘T’ and vice-versa and ‘G’ by ‘C’and vice-versa. Therefore, complementary stand will be determined as CHEM ACADEMY 8. Correct option is (c) There are 16 -electrons in the molecule, but this can not be aromatic by Huckel rule, so in order to be aromatic and to follow Huckel rule, only 14 -electrons should be involved in making the ring current. 14 = 4n + 2 9. 10.  n 3 here only 14 -electrons will form the ring current. Correct option is (c). (P) can be prepared by following chemical transformation Only (a) will give the (P) because the stereochemistry of (P) can only be derived from (a). Correct option is (a) The compounds containing acidic hydrogen when reacted with RMgX. First acid base reaction occurs, acid alkene will form Chemical transformation involved in above chemical reaction can be illustrated as • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY IIT-JAM 2007 11 Correct option is (a) 11. dC  kC 2 , t = 0, C = C0 dt On integration, by taking limit 0 to t. we get  1 1   kt C t C0 ... (1) At t = 10 min, Ct = C0/2  2 1   k  10 C0 C0  1  k  10 C0 ... (2) CHEM ACADEMY At t = ?, Ct = C0 4 from equation (1) we get, 4 1 3    kt C0 C0 C0 From equation (2) and (3), ... (3) 3 t  1 10 12.  t = 30 min. Correct option is (c) As E, S, H and A are state functions, so they are path independent and in a cyclic process, net change is zero. Correct option is (d) 13. AB G 0   Y is the equilibrium point,  G = 0 At point ‘X’ there is net tendency of reaction to go forward.  DG per mole = –ve (spontaneous) µB – µA = –ve; µA > µB  At point ‘Z’, net reaction is backward, At equilibrium, there is dynamic equilibrium. So, • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 14.   B A reaction is spontaneous. G per mole = –ve µA – µB = –ve,  [B]eq At equilibrium, k eq  [A] . So, ratio of composition B and A can be identified. eq HA t=0 at equim  C C(1–) [H  ][A  ] [HA] pH = –log [H+] = 4  At equilibrium [H+] = C   C = 10–4 H+ A– + 0 C ka   From equation (1), 0 C ka     104 103 ka  (C)(C)  2C  C(1  ) 1   [H+] = 10–4 m ... (1)  0.1 (0.1)2 (103 ) 105   105 (1  0.1) 0.9 CHEM ACADEMY Correct option is (c) For wave function, normalization constant ‘A’ can be calculated by applying normalization condition as  = Aeim  2  For  to be normalized,   0 17. µ B > µA  Correct option is (c) Correct option is (a) 15. 16. IIT-JAM 2007 12 2  *  d  1  A 2  d  1  A 2 (2)  1  A  0 Correct option is (a) As Keq value is very large, so the cell reaction almost goes to completion. Zn  Cu 2  Zn 2   Cu 18. Correct option is (b) Correct option is (a) 19. Ionic strength (I) = 1 mi Zi2 2 1 I(FeCl3  HClO4 )   M 3 ( 3) 2  M  ( 1) 2  M  ( 1) 2  M  ( 1) 2  Cl H ClO 4  2  Fe 1 1  [0.01 9  0.03  1  0.06  1  0.06  1]  [0.24]  0.12 2 2 I(0.04 M Na 2SO 4 )  1 1 2  0.04  ( 1) 2  0.04  ( 2) 2   [0.24]  0.12   2 2 Correct option is (b) • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in 1 2 CHEM ACADEMY 20. 13 IIT-JAM 2007 HO–  O2– + H+ Correct option is (b) 21. 24. Correct option is (d) Reduction by carbon because Reduction of metal by method (a), (c) and (d), will depend on the reduction potential of metal, So, can not bring about a non-spontaneous reduction (for which G = +ve). In case of carbon reduction net G of overall reaction (reduction of metal and oxidation of carbon) can be made –ve (i.e. spontaneous) at higher temperature because entropy change (S) while oxidation of carbon is +ve, (because G(C–oxidation) = H – TS) H = –ve, S = +ve, so high negative G value at high temperature) Correct option is (b) Rb, Sr, In, Sn all lie in the same period of periodic table. As charge on metal ion increases its size decreases. Correct option is (d) Correct option is (c) 25. CH 3 has only sigma bonds and lone pair in sp3 orbital, so this ligand has only ‘’-bonding character.. 22. CHEM ACADEMY 23. 26. CN–, CO and NO+ contains -bond, so it has -antibonding and bonding orbital, it can have bonding character. Correct option is (b) N2, CN– and NO+ all have 14 electrons, same as CO, but O2+ has 15 electrons. Correct option is (d) 27. Correct option is (b) 28. P4O6  6H 2O   4H3PO3 ; PCl5  H 2 O  H 3PO 4 ; P4S10  H 2O  H 3PO 4  H 2S Ca 3 (PO 4 )2  H 2SO 4  H3PO 4  CaSO 4 Correct option is (b) • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY r 29. IIT-JAM 2007 14 Ca 2 Radius ratio = r F  100 pm  0.75 133 pm rcation 1 As, according to radius ratio rule, if 0.732 < r anion So, C.N. of cation = 8 Correct option is (a) 30. Correct option is (c) DESCRIPTIVE QUESTIONS 31. Syn hydroxylation of alkene followed by oxidation gives-dial compound ‘Q’. CHEM ACADEMY • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 32. 15 IIT-JAM 2007 (a) (b) Correct sequence of pentapeptide is CHEM ACADEMY 33. 34. (a) First step is similar to Beckmann rearrangement in which migration of group takes place from C to N. Second step is intramolecular electrophilic aromatic substitution reaction • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY (b) 16 IIT-JAM 2007 CHEM ACADEMY • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 35. 17 IIT-JAM 2007 As (I) and (II) are enantiomers, and they are formed in 1:1 ratio, So, it forms a recemic mixture, so ‘K’ is optically inactive. (a) (ii) N  [Ni(NH 3 )6 ]Cl2 Ni is sp3d2 hybridized. Ni 2  [Ar]3d8 4s0 4p0 (iii) N  [Ni(NH3)6]Cl2 is paramagnetic complex, because it has ‘2’ unpaired electrons in its 3d orbital. (b) [Cr(H2O)6]+3 CHEM ACADEMY 6  2  C.F.S.E.  3     0     208 kJ  249.5kJ 5  5  36. (a) (i) H3COCH3 has more Lewis base character. Lone pair of oxygen in (I) is delocalized in vacant d orbital of silicon via p-d back bonding, so l.p. on oxygen is not available. But in case of dimethyl ether ((CH3)2O), oxygen l.p. is available, so it can act as stronger lewis base. (ii) Due to back bonding in (SiH3)2O, double bond character arises (as shown in canonical structure II and III above), which leads to greater bond pair-bond pair repulsion, so Si–O–Si bond angle is • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 18 IIT-JAM 2007 larger w.r.t. C–O–C angle where bonds have single bond character only. (b) Pure Si is produced from reduction of SiO2 in an electric furnance in presence of coke. SiO2 + C  Si + CO2 (CH3)2SiCl2 on hydrolysis gives silicon polymer. 37. A  CuS; B  Cu(NO3 )2 (a) C  CuI or Cu 2 I2 (b) Mn+2 catalyses the decomposition of KMnO4, as KMnO4 oxidises Mn+2 and itself gets reduced. 2MnO 4  3Mn 2  2H 2O  5MnO 2 (s)  4H  Also KMnO4 is not a primary standard, because it undergoes decomposition in water solution, 2KMnO4 + H2O  2KOH + 2MnO2 + 3[O] So, KMnO4 solution is not suitable for estimation of Fe dissolved in HCl. But K2Cr2O7 solution can be used as a primary standard, because it is stable in aqueous medium, thats why it is prefered over KMnO4 solution. CHEM ACADEMY 38. Al2(CH3)6 is more lewis acidic in character, as bridging Al–C–Al bonds are 3-center-2-electron bonds, so, Al has incomplete octet, but in Al2Cl6 all bonds are 2-centere-2-electron bonds. So has complete octet, so less tendency to accept electron-pair (i.e. less lewis acidic character). (b) Stability order 2 2 2 Si Cl2  Sn Cl2  Pb Cl2 Si, Sn and Pb belong to group 14 and each one of them has ns2 np2 configuration, but due to inert pair effect, as we go down the group ns2 electron gets inert, and shows reluctancy towards removal. So, stability of +2 oxidation state is, As, Si+2 < Sn+2 < Pb+2, but +4 o.s. stability is as, Si+4 > Sn+4 > Pb+4 So, stability order in MCl2 salt is, SiCl2 < SnCl2 < PbCl2 39. (a) 2 Na 2CrO4  2H   Na 2Cr2O7  2Na   H 2O (Yellow) Orange (A) Na 2CrO7  2H 2SO 4  2CrO3  2NaHSO 4  H 2O (A) (conc.) (B) Bright orange solid • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY IIT-JAM 2007 19 Na 2Cr2O7  4Cl  6H 2SO4  CrO2Cl2  2NaHSO4  4HSO4  3H 2O (C) (A) So, (conc.) (deep red liquid) A  Na 2Cr2O7 B  CrO3 C  Cl (b) emission emission 215 211 211  82 X   83 Y 84 PO  (i) (ii) 40. [Y = Bi] 2 Valence shell electronic configuration of Y is, 6s 6p X belongs to group 14 (carbon group) Y belongs to group 15 (Nitrogen group) (a) As work done by gas, W = Pex V As Pex = 0  3 W 0 From first law of thermodynamics. U  q  W Let the waals of container is adiabatic So, q=0  H  0 CHEM ACADEMY Also U = 0 So, temperature of container will remain same.  S  nR ln V2 V1 (at constant temperature) V  1.5  24.8  PV   49.6    1 1  R ln 2   8.314 ln     8.76 J / mol V1  0.0821 298   RT   24.6  G = H – TS = 0 /0 298 × 8.76 = –2610.48 J/mol = –2.61 kJ/mol (b) Given, Maxwell distribution function 3/ 2   mc2    2  2kT  c e  m  f (c)  4    2kT  For finding most probable speed, d f (c)  d 2 f (c)   0 and  dc  ccmps dc2  cc  d f (c)  m   4   dc  2kT   d f (c)  m   4   dc  2kT   mc 2 3/ 2 e 2kT 3/ 2  ce 0 mps  mc3  2c    kT   mc2 2kT  mc2  d f (c) 0  2   ; kT dc   • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 2kT ,  but most probable velocity can not be  . m When, c = 0, Now,   d 2f (c) dc 2 2 3/ 2  mc e 2kT  m   4    2kT  d 2 f (c) dc 2  3mc 2 mc  mc 2    2   2c   kT kT  kT    2 3/ 2  mc e 2kT  m   4    2kT  d 2f (c)    ve  dc2  c0 d 2f (c)   dc2  c 41. IIT-JAM 2007 20  5mc 2 m 2c 4   2 2 2  kT k T    minimum   ve  maximum 2kT m  c  c mps  2kT m (a) Initial mole ratio of A2 and B2 is 1 : 3 Let A2 = a mole  A2(g) + t = 0 a mole at eq. (a – y)mol 3B2(9)  2AB3(g) 3a mole 0 mole (3a – 3y) mole 2y mole CHEM ACADEMY Given, mole fraction of AB3 at equilibrium, x AB3  0.60  2y = 0.6 (a  y)  (3a  3y)  2y 2y  4a  2y  0.6 3 2y = 2.4a – 1.2y  y  a 4  Total number of moles of gaseous mixture = 2.5 a At equilibrium  Number of mole of A2 = 1 a  0.25a 4 Number of mole of B2 = 3a  0.75a 4 0.25a  0.1 2.5a 0.75a X B2   0.3 2.5a As T and V are constant.  XA2  So, Pn • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY  IIT-JAM 2007 21 Pinitial n initial  Pfinal n final ; 200 bar 4a  Pfinal 2.5a  Pfinal = 125 bar 2   PAB3   0  P  kp   3 0  PA0  PB2  where, P = 1 bar  0  0   P  P   (b) (0.6 125)2 (0.1125)(0.3 125) 3  0.36 0.1 0.027  (125) 2  8.53  103 Fe(II)  Ce(IV)  Fe(III)  Ce(III) At equivalent point, [Fe(III)]  [Ce(III)] and [Fe(II)] = [Ce(IV)] For reaching equivalence point 50 ml of 0.05 m Ce(IV) solution is being used, E0cell  E Ce 4 / Ce3  E0 Fe3 / Fe 2  1.45V  0.75V  0.70V From Nernst cell equation, at 25ºC 0.059 log Q[Q  reaction quotient] n At equilibrium, Q = K, here n = 1, and Ecell = 0 0 = 0.70 – 0.059 log K  0 E cell  E cell  CHEM ACADEMY   42. (a) 0.70  11.86 0.059 1 1 1 E equiv. Pot.   E 3 2  E 0 IV III   (0.75  1.45)  (2.20)  1.10 Fe / Fe Ce / Ce   2 2 2 log K  PD0 2O = 745 mm Hg at 20ºC  200  200 gm D2O    mole of D2O.  22   15  15 gm of non-volatile compound    mole of non-volatile compound M PD 2O = 730 mm of Hg (M = molecular weight of non-volatile compound). From Raoult’s law, Relative lowering of vapour pressure = (Mole fraction of non volatile solute) PD0 2O  PD2O  PD0 2O  15    745  730 1 15 1 M  ;  ;  745 745 20M  1  200   15   20M       1 33  22   M   33  • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY  (b) IIT-JAM 2007 22 20M 146 11  3  149 ; M   80.3 ; M  80.3 11 20 Michaelis-Menten kinetics has mechanism k1   E  S   Product ,  ES  k 1 where k m  k 1  k 2 k2 k 2 [E]0[S] Rate = k  [S] m k2 = Turnover frequency Vmax = maximum rate of reaction = k2[E]0 = 2.4 × 10–4 ms–1 and [E]0 = 2.4 × 10–9 M  43. Vmax 2.4  104 ms 1  ; k  105 s 1 k2 = turnover frequency = 2 [E]0 2.4  109 m (a)  n y y 2 n  sin x x .sin ; L L L n x  1, 2,3,.... n y  1, 2,3,.... CHEM ACADEMY  (x, y)  (x, y) is the normalized wave function for (nx, ny) quantum number set. For HOMO to LUMO transition. E = E3 – E2  hc h h 2 2  (22  22 )  (1  2 )  8me L2 8me L2  c 8me L2c 8  9.11031  (2 1010 ) 2  3 108    26.37  109 m = 26.37 nm  34  5h 5  6.625 10 Normalized wave function for HOMO are:  (x, y)  (n x 1,n y  2) 2 x 2y 2 2x y sin .sin ;  (x, y)  sin .sin L L L (n x 2,n y 1) L L L • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in CHEM ACADEMY 23 IIT-JAM 2007 (b) [trans] eq m K eq  [cis] eq m 44.  0.40   [cis] x ; 0.04600 – 0.4x = x; x = 0.0328 ~ 0.033 0.115  x eq m = 0.15 – 0.033 = 0.082 M  [trans]eq m = 0.033 M (a)  OP1  x1ˆi  y1ˆj  z1kˆ  OP 2  x 2ˆi  y 2ˆj  z 2 kˆ CHEM ACADEMY From triangle law of vector addition in triangle 0, P1, P2.    P1P2  P1 0  OP2   P1P2  OP1  OP2    (Substituting OP1 and OP2 )  P1P2  (x 2  x1 )iˆ  (y2  y1 )ˆj  (z 2  z1 )kˆ  ˆj.iˆ  ˆi.jˆ | ˆi || ˆj | cos 90º  0 ( cos 90º = 0) ˆ ˆ  ˆj.kˆ | ˆj || kˆ | cos 90º 0  ˆi.kˆ  k.i ˆ ˆ | kˆ || ˆi | cos 90º  0 k.j (b)  2 1  1 1  1 1 2 1  2 3   2 2   0 1  [A, B]  AB  BA            0 1  0 1  0 1 0 1  0 1   0 1   0 0  As, 0 0 [A, B] = AB – BA    0 0  A and B do not commute. • North Delhi: 72, Mall Road, G.T.B. Nagar, New Delhi - 110009 • Ph.: 011-41415514, 09136597244, Toll Free: 1800-200-4940 Web.: www.chemacademy.in