CSIR-JRF(NET) Major-05 Chemical Science PDF

# CSIR-JRF(NET) MAJOR-05 CHEMICAL SCIENCE ## Time Allowed: 180 Mins. ## Maximum Marks: 200 ### A. General 1. This booklet is your question paper containing 105 questions. The booklet has 27 pages. Part-A contain 20Qs. Candidates will be required to answer 15 Qs. Part-B contain 40 Qs. Candidates wi...

# CSIR-JRF(NET) MAJOR-05 CHEMICAL SCIENCE ## Time Allowed: 180 Mins. ## Maximum Marks: 200 ### A. General 1. This booklet is your question paper containing 105 questions. The booklet has 27 pages. Part-A contain 20Qs. Candidates will be required to answer 15 Qs. Part-B contain 40 Qs. Candidates will be required to answer 35 Qs. Part-C contain 45 Qs. Candidates will be required to answer 25 Qs. In case any candidate answers more than 15, 35 and 25 questions in Part A, B and C respectively only first 15, 35 and 25 questions in Parts A, B and C respectively will be evaluated. 2. Blank paper, clipboard, log tables, side rules calculators, cellular phones, pagers and electronic gadgets in any form are not allowed to be carried inside the examination hall. 3. Fill in the boxes provided below on this page and also write your Name and Roll No. in the space provided of this booklet. 4. A separate answer sheet (OMR) is attached with this paper, you have to mark answer on this sheet. ### B. FILLING THE OMR SHEET Please note that the information about Roll No., Test ID, Name, Test Date, Batch & Mobile Number. If it contains errors, the OMR Scanner will not be able to recognize your OMR Sheet. Besides, the candidates are advised to read and follow the instructions given below : 1. Choose the correct/most appropriate response for each question among the options A, B, C and D and darken the circle of the appropriate response completely. The incompletely darkened circle is not correctly read by the OMR Scanner and no complaint to this effect shall be entertained. 2. Use HB Pencil/BLACK/BLUE Ball point pen only. Pens with colours are strictly prohibited. 3. Use of white fluid / eraser / blade etc. for correction in OMR sheet is not permitted. 4. More than one response is also not allowed. Multiple answers given against one question will not be considered for evaluation, i.e. marking more than one answer or making alterations after marking an answer will result in zero mark. 5. Do not make any stray mark on the OMR sheet. Do not cut or mutilate the OMR sheet. 6. Do not fold or damage the OMR sheet ### C. Marking the Scheme : 1. **Section-A & Section-B,** you will be awarded 2 marks for every correct answer, 0.5 mark will be deducted under negative marking. 2. **Section-C**, you will be awarded 4 marks for every correct answer, 1 mark will be deducted under negative marking. ## Name : I have read all the instructions and shall abide by them. **Signature of the Candidate** ## Roll No. : I have verified all the informations filled in by the Candidate. **Signature of the Invigilator** **Manav Ashram, Gopalpura Mode, Jaipur-302018| Ph. : 93144-02754, 7339998097** # SYLLABUS: * Periodic Properties * Electronic Spectra part-I * Silicates * Bioinorganic * Stereo Chemistry * Solid state * Nuclear Chemistry * Group theory * Electro chemistry * NMR Spectroscopy * Chemical Bonding * Bonding in Coordination & Natural Product # PART - A 1. At what, minute between h'O clock and (h + 1)'O clock the hour and minute hands of a clock coincides? * (A*) 60h/11 * (B) 60h/55 * (C) 55h/60 * (D) 60(h + 1)/60 **Sol.** At h O' clock the minute hand is 5h minute spaces behind the hour hand. Now the minute hand gains 55 minute spaces in 60 mins. The minute hand will gain 5h mins in 60/55 × 5h = 60h/11 mins 2. If the day of the 7th January in 1992 was tuesday and it was Rams birthday, then what will be the day of Ram's birthday in 1997? * (A*) Tuesday * (B) Monday * (C) Wednesday * (D) Sunday **Sol.** Total extradays = 5 days for five years + 2 extra days for two leap years in between = 7 days i.e., One complete week is completed. .. 0 odd days .. The day is Tuesday. 3. What should be the ideal diagram for women, mothers and lawyears? * (A) [[Image of a Venn Diagram: three circles, two partially overlapping, one entirely inside the other]] * (B) [[Image of a Venn Diagram: three circles, two overlapping, one entirely inside the other]] * (C) [[Image of a Venn Diagram: three circles, two partially overlapping, one entirely inside the other]] * (D*) [[Image of a Venn Diagram: three circles, two overlapping, one entirely inside the other]] **Sol.** All mothers are women and few lawyears may be mothers as well. 4. Below is given a statement and based on that statement you have to find out which of the conclusions follow? **Statement:** Unless India, a country achieves total literacy, it can't reach its mission of development. **Conclusions:** (i) It is possible to reach total literacy in India. (ii) No development is possible without a proper mission. * (A*) Only I follows * (B) Only II follows * (C) None follows * (D) both follows **Sol.** If it wasn't possible to achieve the total literacy, then the statement would not have been made. Development without proper mission is also possible. 5. Aboard of directorate meeting comprising 7 members sits in a row facing north. Amit is to the immdiate left of Barun but on the immediate right of Douglus. Qarmor is on the right side of Amit but on the left of Sushant. Tarun is on the left side of Vikram who is sitting to the left of Douglus. Who is sitting in the middle? * (A) Vikram * (B*) Amit * (C) Qamar * (D) Douglous **Sol.** [[Image of table: seven individuals, T, V, D, A, B, Q, S, with each individual's name located in the table]] 6. [[Image of a figure: a square divided into four smaller squares. a number 10 is written on the top and a number 5 on the bottom right hand corner of the figure]] * (A) 15 * (B*) 26 * (C) 30 * (D) 10 # PART - B 21. The first ionization potential (eV) of Be and B respectively are * (A) 8.29, 9.32 * (B*) 9.32, 8.29 * (C) 9.32, 9.32 * (D) 8.29, 8.29 22. Consider the following ionization steps M(g) → M+(g) + e¯ ; ΔΗ = 100 eV M(g) → M2+ (g) + 2e¯ ; ΔΗ = 250 eV Select the correct statement(s). * (I) (IE₁ of M(g) is 100 eV * (II) (IE), of M+(g) is 150 eV * (III) (IE), of M(g) is 150 eV * (IV) (IE), of M(g) is 250 eV * (A) I, III and IV * (B*) I, II and III * (C) II, III and IV * (D) I, II and IV 23. d-orbital involve in hybridisation in [InCl5]-2 and [HgCl5]-³ ions respectively are: * (A) d₂2 and d₂2 * (B*) d₂2 and d₂2 * (C) d₂2 and d²_v² * (D) dx2-y2 and dx2-y2 **Sol.** [InCl]2- ⇒square pyramidal geometry ⇒ d₂2 involved. [HgCl]-3 → TBP geometry → d₂2 involved 24. Number of anti bonding electrons in N₂ is : * (A*) 4 * (B) 10 * (C) 12 * (D) 14 25. Compound with maximum ionic character is formed from : * (A) Na and CI * (B*) Cs and F * (C) Cs and I * (D) Na and F 26. The correct pairs which consist of isogeometric as well as isostructural species are * (A) XeOF and IF * (B) IOF and SOCI₂ * (C) CN₄ and CN4 * (D) HNO2 and O3 * (A) A and B * (B) C and D * (C*) A, B and D * (D) B, C and D 27. The correct orientation of dipoles in pyrrole and pyridine is - * (A*) [[Image of the structure of pyridine and pyrrole, with the dipoles pointing in the correct direction.]] * (B) [[Image of the structure of pyridine and pyrrole, with the dipoles pointing in the correct direction.]] * (C) [[Image of the structure of pyridine and pyrrole, with the dipoles pointing in the correct direction.]] * (D) [[Image of the structure of pyridine and pyrrole, with the dipoles pointing in the correct direction.]] 28. The ground state term of Dy3+ and Ho3+ are respectively * (A) 6H13/2 and 4115/2 * (B) 6H15/2 and 6I5/2 * (C) 6H13/2 and 5Ig * (D*) 6H15/2 and 5Ig **Sol.** Dy→4f105d06s2 Dy³+ →4f 4f→ S = 5/2 Multiplicity=2S+1=2×5/2+1=6 L=+3+2=5(H) J=L+S=5/2+5=15/2 J=15/2 → H15/2 Ho→4f5d° 6s² Ho³+ →4f10 4f10→ S= =2 Multiplicity=2×2+1=5 L=+3+2+1 L=6(1) J=L+S=2+6=8 = I 29. Ground state term of Dy³+ and Ho³+ are 6H15/2 and Ig. The magnetic moment value for Ce³+ (in BM) is: * (A) 2.78 * (B*) 2.54 * (C) 2.89 * (D) 2.70 **Sol.** Ce→[Xe]4f25d° 6s2 Ce³+ →[Xe]4f¹ g(Lande splitting factor) = 1+ [J(J+1)+S(S+1)-L(L+1)] / 2J(J+1) …..(i) μeffective= g[J(J+1)]² (BM) …..(ii) J=L-S=3-1/2=5/2 Put these L, S and J values in relation (i) to obtain g= 6/7 Now, put g= 6/7, J = 5/2 in relation (ii) to obtain μeffective= 2.54 Β.Μ. 30. Which of the following Ln³+ pair have similar colour? * (A) Pm3+, Tb3+ * (B) La³+, Tm3+ * (C*) Pm3+, Ho3+ * (D) Nd3+, Tm3+ **Sol.** Colour depends upon the number of electron present in 4f orbital. Pm→ 4f4 Both have same number of unpaired electron → Pink Yellow Colour Ho→ 4f10 31. (A) Cel₂, (B) Eul₂, (C*) YbI₂, (D) CeCl3 32. What is the species generating when [Co(NH3)6]Cl3 reacts with moist Ag₂O? * (A) Co(OH)3.5NH3 * (B*) Co(OH)3.6NH3 * (C) Co₂O₃ * (D) NH4OH **Sol.** [Co(NH3)6]Cl3 react with moist Ag₂O to give Co(OH)3.6NH3 33. The CFSE in KJ/mol for [Ti(H2O)6]3+ ion for which d-d transition is single broad absorption peak with a maximum at 492.6 nm * (A*) – 97.12KJ/mol * (B) 97.12KJ/mol * (C) - 80.12KJ/mol * (D) None 34. Select the correct statement * (I) Acubic > Δο * (II) Acubic > At * (III) Acubic < Δο * (IV) Acubic < At * (A) I and II only * (B) I and III only * (C*) II and III only * (D) III and IV only 35. The magnetic moment corresponding to [NH₄]V(SO4)2.12H₂O and K3[Fe(ox)3] are * (A) 2.8 and 1.73 * (B) 3.20 and 5.91 * (C*) 2.8 and 5.91 * (D) 4.89 and 1.73 36. If A is the octahedral splitting energy and P is the electron pairing energy, then the crystal-field stabilization energy (CFSE) of [Co(NH3)6]2+ is * (Α) – 0.8 Δ +2P * (Β) – 0.8 △ +1P * (C*) – 0.8 Δο * (D) – 0.84 +3P 37. The number of microstates corresponding to 1s22s¹2p¹ and 3p¹3d¹ are * (A) 6 and 16 * (B*) 12 and 60 * (C) 8 and 60 * (D) 12 and 36 38. Possible term symbols of d² electronic configuration in their increasing order of energy * (A) 1S < 3P < 1G < 3F < 1D * (B*) 3F < 3P < 1G < 1D < 1S * (C) 3P < 3G < 3D < F < 1S * (D) 1F < 1G < 3P < 1S < 1D 39. The crystal-field symbol for the ground-state of [Mn(CN)]4- is : * (A*) 2T2g * (B) 1A19 * (C) 5E * (D) 6A1g 40. The following complex [(η5 – C₄H₁) Ni₃ (µ₃ – CO)₂ ]× has three Ni-Ni bonds. Using 18 electron rule, the value of x is * (A) -1 * (B) +2 * (C) -2 * (D*) +1 41. The metal cluster [Fe,C(CO)18]2 is of structure type- * (A) Nido * (B) Arachno * (C) Closo * (D*) Capped closo 42. Metal-Metal bond order in [Mo2(SO4)4]³- is * (A) 6 * (B) 2 * (C) 4 * (D*) 3.5 43. The fragment of [Mn₂ (CO)10] is isolobal with * (A*) CH3 * (B) CH₂ * (C) CH * (D) CH3 **Sol.** The fragment of Mn₂ (CO)10 = Mn(CO)5 ⇒ 10 + 7 =17e¯ CH3 → 4 + 3 = 7e¯ CH₂ → 4 + 2 = 6e- CH4 + 1 + 1 = 6e- CH₃+ → 4 + 3 − 1 = 6e¯ Therefore, Mn(CO)5 ≅ CH3 44. Match the following Part A * 1. Myoglobin * 2. Hemoglobin * 3. Vitamin B12 * 4. Carboxypeptidate * 5. Ferrodoxin Part B * (A) Cobalt * (B) Electron transfer * (C) Oxygen storage * (D) Oxygen transport * (E) Peptide hyrolyser * (A*) C D A E B * (B) D A C B E * (C) A C B D E * (D) C A E B D 45. Match the Column-A with Column-B Column-A * (P) Azurin * (Q) Xanthine oxidase * (R) Transient * (S) Peroxidase Column-B * (I) Fe-Mo protein * (II) Iron storage * (III) Fe₂S₂ sites * (IV) Catalyse oxidation of substrates * (V) Iron transport * (VI) Electron transport * (A) P-VI, Q-I, R-V, S-IV * (B) P-III, Q-I, R-V, S-VI * (C*) P-VI, R-V, S-IV * (D) P-VI, R-II, S-IV * **Sol.** (P) Azurin ⇒ (VI) Electron transport (Blue copper protein) (Q) Xanthine oxidase ⇒ (III) Fe₂S₂ sites (R) Transfenin ⇒ (V) Iron transport (S) Peroxidase ⇒ (IV) Catalyse oxidation of substrates 46. Hemerythrin and oxyhemerythrin respectively. * (A) pairamagnetic, diamagnetic * (B) treg and tregtre * (C*) both (A) and (B) are correct * (D) both are ESR active 47. Assuming an octahedral geometry the number of geometrical isomers that are possible in [PF3CI3] and [PF₂CI4] are: * (A) 2, 2 * (B) 2, 3 * (C) 3, 2 * (D) 4, 2 48. A solution (0.5 g/mL) of penicillin-V shows the optical rotation of +220° in 20 cm cell. The specific rotation of penicillin-V is +330°. The optical purity of penicillin-V in this solution is * (A) 78.6 * (B) 66.6 * (C) 33.4 * (D) 88.6 49. The relation between the Ha and Hb and Br¹ and Br2 in the given compound is [[Image of a structure with two methyl groups attached to a cycle with two bromine groups]]. * (A) Ha, Hb are enantiotopic; and Br¹, Br² are diastereotopic * (B) Ha, Hb are diastereotopic; and Br¹, Br2 are enantiotopic * (C) Ha, Hb are diastereotopic; and Br¹, Br² are homotopic * (D) Ha, Hb are enantiotopic; and Br¹, Br² are homotopic 50. Iron (a-Fe) crystallizes in a bcc system with a = 2.861 Å. Molar mass of iron is 55.85 gmol-1. The density of iron is - * (A) 6.7 g/cm³ * (B*) 7.9 g/cm³ * (C) 5.9 g/cm³ * (D) 6.3 g/cm³ 51. Select the correct statement * (A*) In C. C. P structure, the third layer occupy octahedral void * (B) In H. C. P structure, the third layer occupy octahedral void * (C) In a ionic crystal octahedral void double then tetrahedral void * (D) none 52. In an ionic oxide, oxide ions are arranged in hcp array and positive ion occupy two thirds of octahedral void simplest formula of metal M is : * (A) M₂O6 * (B*) MO * (C) M₃O₂ * (D) MO4 53. 8: 8 co-ordination of CsCl is found to change into 6:6 on * (A) applying high pressure * (B*) increasing the temperature * (C) both (A) and (B) * (D) None of these **Sol.** Increase the temperature because to increase co-ordination number total number of atoms come closer, and for decreasing the co-ordination no., the no. of atoms should go apart. When we increase temperature, atoms more apart and hence co-ordiantion no. decrease from 8: 8 to 6: 6. 54. The EMF of the concentration cell Cu(S)/Cu+2 (0.012)//Cu+²(1.2 m)/Cu(S) is * (A) 0.001 v * (B) 0.025 v * (C) 0.059 v * (D) 0.118 v 55. For an electrolyte AB₂, the equivalent conductance of A2+ and B¯ are 120 and 200 Scm²eq-1. The equivalent conductance of AB₂ (in Scm²eq¯¹) is * (A) 520 * (B*) 320 * (C) 440 * (D) 640 **Sol.** Λeq (AB2) = Λeq(A2+) + Λeq(B¯) = 120 + 200 = 320 Scm²eq-1 56. The Haworth projection for a-anomer of D-glucose is * (A) [[Image of the Haworth projection for a-anomer of D-glucose]] * (B) [[Image of the Haworth projection for a-anomer of D-glucose]] * (C*) [[Image of the Haworth projection for a-anomer of D-glucose]] * (D) [[Image of the Haworth projection for a-anomer of D-glucose]] 57. Arrange the following regions of the electromagnetic spectrum in order of decreasing wavelength: microwaves, visible light, ultraviolet light, infrared radiation, X-rays. * (A*) Microwaves > infrared radiation > visible light > ultraviolet light > X-rays * (B) Microwaves > infrared radiation > visible light > X-rays > ultraviolet light * (C) Microwaves > infrared radiation > X-rays > ultraviolet light > visible light * (D) Microwaves > infrared radiation > ultraviolet light > visible light > X-rays 58. The rotational spectrum of a rigid diatomic rotor consists of equally spaced lines with spacing equal to * (A*) 2B * (B) B * (C) B/2 * (D) 3B/2 59. The direct product of A × A× B3u in D2h group is * (A) Blu * (B*) Bзu * (C) B2g * (D) None of these 60. What is the point group of [[Image of a structure with one Manganese atom in the center, four CO molecules surrounding it, with four more CO molecules placed perpendicular and further out, in a plane.]] * (A*) D4d * (B) D4h * (C) C2v * (D) none of these # PART -C 61. Consider the correct change(s) that occurs during conversion of I₂ to It (considering y as internuclear axis) * (1) Removal of electron from orbital * (2) Decrease in π* – σ* of I₂ * (3) Change in colour of I₂ * (4) Change in magnetic behaviour and increase in bond length The correct answer is * (A) 2, 3 * (B*) 1, 2, 3 * (C) 1, 4 * (D) all of these **Sol.** [[Image of the molecular orbital diagram for I2, with the energy level for σ*2p, π*2p, π2p, σ2p, σ2s, σ*2s labeled.]] [[Image of the molecular orbital diagram for I2+, with the energy level for σ*2p, π*2p, π2p, σ2p, σ2s, σ*2s labeled]] Thus, during conversion of I₂ to I₂+, There is removal of electron from π2p* orbital Decreases in π* – σ* of I₂ Change in magnetic behaviour and decrease in bond length _Change in colour of I₂_ 62. Number of species having stereochemically active lone pair among XeF6, IF7, SeF62-, SeBr62-, SbCl5-, XeF2- is * (A*) 3 * (B) 4 * (C) 2 * (D) 5 **Sol.** XeF6, IF7, SeF62- have distorted geometry. Thus, lone pair is stereocheniically active. 63. Match the list Species * (I) NF2+ * (II) NS2+ * (III) S₂O * (IV) NSF3 Structure * (A) Linear * (B) Angular * (C) Tetrahedral * (D) Trigonal bipyramidal * (A) I-A, II-A, III-B, IV-D * (B*) I-B, II-A, III-B, IV-C * (C) I-A, II-A, III-B, IV-B * (D) I-B, II-B, III-A, IV-C **Sol.** NS2+, N is sp² hybrised. [[Image of the structure of NS2+, with the N in the centre, and a S on either side of a central N]] NS₂+ 'N' is sp2 but due to bigger 'S' atoms it become linear [[Image of the structure of NS2+, with the N in the centre, and a S on either side of a central N, with the bond between them slightly bent.]] OS, is sp² hybridised with B.O. > 1 between O-S [[Image of the structure of S2O with an O on either side of a central S, with a double bond to the S]] NSF3, here 'S' is sp³. [[Image of the structure of NSF3 with an S in the center, one N, and three F atoms].] 64. Select the correct statement from following: * (I) f-f transition peaks are broader than f–d transition * (II) f-f transition are stronger than f-d transition * (III) f-f transition are weaker than f-d transition * (IV) Ln³+ shows stronger emission spectrum than d-block metals. * (A) I, II, IV * (B) II, IV * (C*) III, IV * (D) I, II **Sol.** * f-f transit ions are weaker than f-d transitions * Ln³+ shows stronger emission spectrum than d-block metals. 65. Find the correct statement among * (A) The Ce³+ is red coloured * (B*) The ground state term symbol of Gd³+ is 8S7/2 * (C) The magnetic moment of Eu3+ is zero * (D) The Lu³+ and La³+ have different colours. 66. Consider the following pair for mentioned property * (A) [TaMe], [WH｡] * (B) [HgCl5]3-, [SbPh5] * (C) Me₂TiCl₂, [PdCl4]2- * (D) FXeO2IF4, [IO2F3] Correct pair(s) is/are * (A) Only D * (B*) Only A * (C) A and C * (D) B and D **Sol.** * [TaMe] and [WH｡] ⇒ Trigonal prismatic (sd5) * [HgCl5]3- Trigonal bipyramidal ⇒ Bond angle = 90°, 120°, 180° * SbPh⇒ Square pyramidal → Bond angle = 90°, 180° * Me₂TiCl₂ → Tetrahedral → sp³ hybridised * [PdCl4]2-⇒ Square planar ⇒ d³s hybridised 67. Match the following * i. [CoCl4]2- * ii. [Fe(H2O)6]2+ * iii. [Ni(CN)4]2- * iv. [Cr(CN)6]3- * v. [Cr(CO)] * vi. [Zn(NH3)4]2+ (a) tetrahedral and diamagnetic (b) inner orbital and paramagnetic (c) tetrahedral and paramagnetic (d) inner orbital & diamagnetic (e) outer orbital and paramagnetic (f) square planer and diamagnetic Select the correct code given below a b c d e f * (A*) vi iv i v ii iii * (B) iv vi v := = iii * (C) iv vi v i iii * (D) vi iv v i ii iii 68. The correct order of crystal field splitting energy is; * (A) [NiCl4]-2 > [FeCl｡]-3 > [PtCl4]-2 * (B*) [PtCl4]-2 > [FeCI]-3 > [NiCI₄]-2 * (C) [PtCl4]-2 > [NiCl4]-2 > [FeCl｡]-3 * (D) [PtCl₄]-2 ≈ [NiCI₄]-2 > [FeCI]-3 69. Arrange the correct order according to CFSE. * (A) [VCl] <[Fe(OX)]³¯ <[Pt(NH3)4]²* <[Ni(CN)₄]¯² * (B) [Ru(H₂O)] <[Ru(OX),]³¯ <[Ir(NH3)｡]³+ <[Fe(CN)] * (C) [Fe(H₂O)]¹² <[VCl¸]²¯ [IrFɛ]²¯ <[B₁₂H₁2]²¯ * (D*) [B₁₂H₁2] <[VC1]²¯ <[Ni(CN)₄]²¯ <[Pt(NH3)4]²+ 70. Arrange the following complexes in the order of their respective colours. * (i) [CrF]3- * (ii) [Cr(CN)6]3- * (iii) [Cr(H2O)6]3+ * (A) (i) green, (ii) violet, (iii) yellow * (B*) (i) green, (ii) yellow, (iii) violet * (C) (i) violet, (ii) yellow, (iii) green * (D) None of these **Sol.** The wavelength of light absorbed by the complex is affected by the energy gap between the d orbitals which is turn affected by the type of ligand and the charge on the metal ions. Ligand field strength order: CN¯ > H₂O > F-. * [CrF]3- red green * [Cr(CN)6]3- violet yellow * [Cr(H2O)6]3+ yellow violet 71. The ground state term symbol of a molecule with electronic configuration (1s)(15)(25)σ(2s)π(2P)π(2P) * (Α) Σ * (Β*) Σ * (C) 3Σ * (D) 3Σ g u g 72. The number of possible term arises from 2s¹, 3s¹ electronic configuration are : * (A*) 1S, 3S * (B) 3S * (C) 3D, 3P, 3S, 1D, 1S, 1P * (D) 1D, 3P, 2S **Sol.** 2s¹ 3s1 ⇒ l₁ = 0 l2 = 0 S = |s₁ + s2|................|S₁ - s2| ⇒ 1, 0 Hence, 2S + 1 = 3, 1. Thus, 3S, 1S 73.