Indian National Chemistry Olympiad Theory Thermal and photolytic decomposition of Acetaldehyde

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1 Indian National Chemistry lympiad Theory 010 Equivalt solutions may exist Problem 1 17 marks Thermal and photolytic decomposition of Acetaldehyde 1.1 a) d[c C] = k[cc] dt d[c 4 ] d[c] = = k[cc] dt dt / / (b) rder = / Rate = 8 v 1. C 4, CD 4 and C 1. (a) Propogation steps : (ii) and (iii) Termination step: (iv) (b) [ C] = k 1 k 1/ 4 k [ CC] = k k 1 k [C C] 1/ 1/ 4 [C C] / (c) d[c] dt = k 1/ k1 / [CC] k 4 BCSE, 0 th January 010 1

2 Indian National Chemistry lympiad Theory (a) C C hν C C Rate = I abs [ C ] = (I abs /k 4 ) 1/ [C C] / d[c]/dt = k (I abs / k 4 ) 1/ [C C] / (b) λ = 17.5 nm 1.5 (a) 1 E thermal = E (E1 E 4 ) (b) E photochemical = E ½ E (a) E thermal = kj mol 1 (b) mol dm sec (a) C C C C Conc C C Time (b) [C C] = k 5 /k 6 [C C ] BCSE, 0 th January 010

3 Indian National Chemistry lympiad Theory 010 Problem Chemistry of coordination compounds 19 marks.1 CuS 4 6 [Cu( ) 6 ] S 4 White Blue solution or balanced equation with [Cu( ) 4 ] tity [Cu( ) 6 ] Cu() 6 From dil N Blue ppt. Cu() 4N [Cu(N ) 4 ] Deep blue soln. b] completely filled d-level in Cu(I). a] oxidation state of the metal. b] nature of the ligand. c] geometry of the complex..4 [Ni( ) 6 ] [Ni 4 ] - [Ni( ) 6 ] is octahedral. [Ni 4 ] - is tetrahedral BCSE, 0 th January 010

4 Indian National Chemistry lympiad Theory a]. IUPAC Name : Dichlorobis(ethylediamine)cobalt(III) ion. Dichlorobis(ethane-1,-diamine)cobalt(III) ion Dichloridobis(ethylediamine)cobalt(III) ion. Dichloridobis(ethane-1,-diamine)cobalt(III) ion b]. Geometrical isomers: Co Co c] cis-[co () ] is optically active. d] Two optical isomers of cis-[co () ] : Co Co.6 Ni [Ar] d 8 [Ni 4 ] - LP LPLP LP sp sp hybridization Tetrahedral Paramagnetic ( unpaired electrons) Pt [e] 5d 8 [Pt 4 ] - LP LP LPLP dsp dsp hybridization Square planar Diamagnetic (no unpaired electrons) BCSE, 0 th January 010 4

5 Indian National Chemistry lympiad Theory A B o (cm -1 ) i) [CrF 6 ] - d) 15,000 ii) [Cr( ) 6 ] c) 17,400 iii) [CrF 6 ] - b),000 iv) [Cr(CN) 6 ] - a) 6,600.8 Answer: a) [Fe(CN) 6 ] b) Ni(C) 4 xidation state Fe(III) Ni(0) Coordination No. of Fe(III) : 6 of Ni(II) : 4 EAN of ctral metal ion d z d x y d x y d x y d z d z d yz, d xz d xy d xy d xy d yz, d xz d yz, d xz Tetragonally octahedron Tetragonally distorted octahedron distorted octahedron contraction along z-axis elongation along z-axis a) (i) by elongation along z-axis. b) (ii) dx -y orbital..10 N Pt N A N N Pt N B N Pt N N C N N Pt N N D BCSE, 0 th January 010 5

6 Indian National Chemistry lympiad Theory 010 Problem 14 marks Chemistry of isomeric bzes.1 Z,Z,E -1,,5-cyclohexatrie (D) 4. C E. C C F (b) Three BCSE, 0 th January 010 6

7 Indian National Chemistry lympiad Theory _ J Li.. C K.8 C C C M N.9 R R R R R R=CEt R.10 N BCSE, 0 th January 010 7

8 Indian National Chemistry lympiad Theory 010 Problem 4 10 marks s-block Elemts 4.1 a) only one valce electron b) large atomic size 4. Inhalation K K Exhalation K C KC or 4 K C K C or K C K C 4. I 4 Li Li II Cs Cs 4.4 i) Bond order = 1 ii) diamagnetic 4.5 Na (x y) N [Na(N ) x ] [e(n ) y ] R Na N Na e solvated / (am) BCSE, 0 th January 010 8

9 Indian National Chemistry lympiad Theory b) It is paramagnetic in nature c) n standing this solution slowly liberates hydrog resulting in the formation of sodium amide 4.7 c) half the number of tetrahedral 4.8 b) cyclohexane c) diisopropyl ether 4.9 a) ionization ergy of alkali metal b) electron gain thalpy of halog d) sizes of cations and anions 4.10 LiF BCSE, 0 th January 010 9

10 Indian National Chemistry lympiad Theory 010 Problem 5 17 marks Carboxylic acid derivatives : : _.. R N :.. _ : R N.. : : _ R :.._.... R' R R'.. : 5. (c) Amide > Ester > Acid Chloride 5. Amide cm -1 A 1750 cm -1 C 1800 cm -1 B 5.5 C C C 5.6 Best C Poorest B 5.7. N N C CC 5 major CC Monochloro derivative (major) BCSE, 0 th January

11 Indian National Chemistry lympiad Theory Li Et 7 8 I or or or Ts or N 5.9 C Et (i) 9 (ii) Both are S BCSE, 0 th January

12 Indian National Chemistry lympiad Theory S N a b 16 N S 5.14 N N N N BCSE, 0 th January 010 1

13 Indian National Chemistry lympiad Theory 010 Problem 6 17 marks Chemical Thermodynamics J 6. Kp = Kp = Kc 6. C = 0.4, = 0.458, = 0.09, C = C = 4.95%, = 45.41%, = 9.59%, C = 10.06% = 158 J 6.5 a) Kp will increase with increase in temperature 6.6 Air intake (gine;m s 1 ) = V A = m s 1 = m s T 1 = 060K T 708 K = 6.8 Compound Molar composition of gases after leaving the bed (Mol 10 4 ) N (g) (g) 8.55 C (g) 0.78 C (g) (g) BCSE, 0 th January 010 1

14 Indian National Chemistry lympiad Theory 010 Problem 7 10 marks 7.1 V= 87.5 ml 7. C C C C Ka = [ ][ C C ]/[ C C] 7. [ Ka ± ] = Ka 4KaC T 7.4 p = a) C C C C C C C C b) [C [C C] C ] eq eq = C T [ ] [ = [C CNa] [ ] ] [ ] c) [ ] = CT Ka [C CNa] d) p = p = 8.7 BCSE, 0 th January

Coordination compounds

Coordination compounds Multiple choice questions 1. In the complex formation, the central metal atom / ion acts as a) Lewis base b) Bronsted base c) Lewis acid d) Bronsted acid 2. The groups satisfying