where the small energy of the initial thermal neutron has been ignored. (m N denotes the nuclear mass.) Now

Transcription

1 rd Internatinal Chemistry lympiad Preparatry Prblems where the small energy f the initial thermal neutrn has been ignred. (m dentes the nuclear mass.) w m ( 5 U) m ( 5 U) 9m e ignring the small electrnic binding energies cmpared t rest mass energies. Similarly fr ther nuclear masses. Q [m ( 5 U) m ( 94 Zr) m ( 40 Ce) m n c Using the given data, Q. MeV c. MWd 0 6 Js x 4 x 600 s 8.64 x 0 0 J 8.64 x 0 0. f atms f 5 U fissined. x.60 x 0.5 x 0 x 5 Mass f 5 U fissined 0.99 g 6.0 x 0.5 x 0 Mass f 5 U in kg uranium remved frm the reactr g Abundance f 5 U is 0.6 % 8. Radiactive decay a. µci.7 x 0 4 disintegratins per secnd (dps). Initial β activity.7 x 0 6 dps d dt t 0 where cnstant..7 0 dps is the number f atms f 6 0 Bi at t 0 and is its decay Mumbai, India, July 00 69

2 rd Internatinal Chemistry lympiad Preparatry Prblems Intial mass f 0 Bi g g b. umber f atms f 0 Bi at time t is given by 0 e t The number f atms f 0 P,, is given by equatin d dt where is the decay cnstant f 0 P. d dt 0 e t Using the integrating factr t H e t d dt e t 0 e ( )t d dt ( e t Integrating ) 0 e ( )t t e e C T calculate C, use the cnditin that at t 0, 0 C This gives 0 0 ( )t 0 (e t e t ) 70Mumbai, India, July 00 70

3 rd Internatinal Chemistry lympiad Preparatry Prblems The time t T when is maximum is given by the cnditin d dt t T 0 which gives T ln 4.9 d At t T, can be calculated frm abve..04 x 0 Mass f 0 P at t T, 7. x 0 0 g c. αdisintegratin rate f 0 P at t T.8 x 0 5 dps At t T β disintegratin rate f 0 Bi αdisintegratin rate f 0 P.8 x 0 5 dps 9. Redx reactins a. i. verall reactin Sn Fe Sn 4 Fe 0.67 V ûg nf F V 9 KJ Mumbai, India, July 00 7

4 rd Internatinal Chemistry lympiad Preparatry Prblems ii n lgk lg K K ( 0.67) b. Befre the equivalence pint, f the cell is given by fllwing equatin cell x S.C. red Sn 4 /Sn lg [Sn [Sn lg [Sn [Sn 4 i. The additin f 5.00 ml f Fe cnverts 5.00/0.00 f the Sn t Sn 4 ; thus [Sn [Sn 4 5.0/ / cell 0.0 V. ii. At the equivalence pint, add the tw expressins crrespnding t Sn 4 /Sn and Fe /Fe. cell x S.C. red Sn 4 /Sn [Sn lg [Sn 4 cell x S.C. red Fe /Fe [Fe lg [Fe t get lg [Sn [Sn [Fe x red red 4 cell S.C. 4 Sn /Sn Fe /Fe [Fe 7Mumbai, India, July 00 7

5 rd Internatinal Chemistry lympiad Preparatry Prblems At the equivalence pint, [Fe [Sn and [Fe [Sn 4 Thus, cell $ x S.C. red $ Sn 4 /Sn red $ Fe / Fe 0.4 ()(0.54) V Beynd the equivalence pint, f the cell is given by fllwing equatin [Fe x S.C. red lg Fe /Fe [Fe cell When 0 ml f Fe is added, 0 ml f Fe is in excess. i.e. [Fe [Fe cell 0.5 V c. i. û G RT ln K 68.7 sp K J $ û G n F $, V n ii. Cu I CuI (s) V Cu e Cu 0.5 V The verall reactin fr reductin f Cu by I is Cu I e CuI (s) 0.86 V The value fr the reductin f Cu by I can nw be calculated x Cu I e CuI (s) 0.86 V I e I 0.55 V Mumbai, India, July 00 7

6 rd Internatinal Chemistry lympiad Preparatry Prblems The verall reactin is Cu 4I CuI (s) I 0.5 V The psitive value f effective indicates that the reductin reactin is spntaneus. This has cme abut since in this reactin, I is nt nly a reducing agent, but is als a precipitating agent. Precipitatin f Cu as CuI is the key step f the reactin, as it practically remves the prduct Cu frm the slutin, driving the reactin in the frward directin. iii. û G n F Here n, 0.5V û G. kj û G RT lnk lgk 5.47 K Slubility f sparingly sluble salts a. Ag C 4 (s) Ag C 4 The slubility prduct Ksp is given by K sp [Ag [C 4 If S is the slubility f Ag C 4 [Ag S () The ttal xalate cncentratin, dented by C x, is C x S [C 4 [HC 4 [H C 4 () The dissciatin reactins are: H C 4 H HC 4 K () HC 4 H C 4 K (4) 74Mumbai, India, July 00 74

7 rd Internatinal Chemistry lympiad Preparatry Prblems qs. (), () and (4) give C x S [C 4 [C 4 K [H [C 4 KK [H [C 4 α C x α S where α K K [H K[H KK (5) At ph 7, [H 0 7 and α K sp 4S.5 0 At ph 5.0, [H 0 5 Frm the values f K, K and [H, we get α 0.86 (6) K sp [S [αs S K sp 4..7 x 0 4 b. [H 0.00 At ph 0.8, [H q. (5) implies α i.e C x S [C 4 (7) The ttal silver in in the slutin is given by C Ag S [Ag [AgH [Ag(H ) (8) Mumbai, India, July 00 75

8 rd Internatinal Chemistry lympiad Preparatry Prblems The cmplex frmatin reactins are Ag H AgH AgH H Ag(H ) K.59 0 (9) K (0) Frm eqs. (8), (9) and (0) C Ag S [Ag { K [H K K 4 [H } [Ag β x C Ag β x S where Using the values f K, K 4 and [H, β. x 0 4 K sp [Ag [C4 [ S [S S K ( sp ) 4 K [H K K 4[H 5.47 x 0. Spectrphtmetry a. Dente the mlar absrptivity f Mn 4 at 440 nm and 545 nm by ε and ε and that f Cr 7 by and 4 : 95 Lml cm, 70 Lml cm, 50 Lml cm 4 Lml cm The absrbance A is related t % transmittance T by A lg T Frm the values given fr the sample slutin A 440 lg A 545 lg Mumbai, India, July 00 76

9 rd Internatinal Chemistry lympiad Preparatry Prblems w if ne dentes the mlar cncentratins f Mn 4 and Cr 7 in the steel sample slutin by C and C respectively, we have A 440 x C x X C x A 545 x C x 4 X C x Using the given data, we get C M C 0.00 M Amunt f Mn in 00 ml slutin mll gml 0. L g x 00 % Mn in steel sample.74 0.% Amunt f Cr present in 00 ml slutin 0.00 ml L x x 5.00 g ml x 0. L 0.08 g 0.08 x 00 % Cr in steel sample % b. In slutin, since all the ligand is cnsumed in the frmatin f the cmplex, [CL x x 0 5 Absrptivity f the cmplex CL is x 0 L ml cm x 0 ml L.0 cm Mumbai, India, July 00 77

10 rd Internatinal Chemistry lympiad Preparatry Prblems If the cncentratin f the cmplex CL in slutin is C, x 0 L ml cm C x.0 cm. x 0 5 M [C [C ttal [CL x 0 5. x x 0 5 Similarly, [L [L ttal [CL 7 x 0 5 x. x The cmplex frmatin reactin is C L [CL The stability cnstant K is given by [CL [C [L K.08 x 7 0. Reactins in buffer medium R 4H 4e RHH H HAc H Ac K i.e a pk a [H [Ac [ HAc [HAc ph lg [Ac 78Mumbai, India, July 00 78

11 rd Internatinal Chemistry lympiad Preparatry Prblems 4.76 [HAc [HAc [Ac [HAc [Ac 5.0 [Ac [ HAc lg [Ac mmles f acetate (Ac ) present initially in 00 ml 0.8 x mmles f acetic acid (HAc ) present initially in 00 ml 0.88 x mmles f R reduced 00 x Frm the stichimetry f the equatin, mmles f R will cnsume mles f H fr reductin. The H is btained frm dissciatin f HAc. n cmplete electrlytic reductin f R, mmles f HAc mmles f Ac ph ph lg Mumbai, India, July 00 79

12 rd Internatinal Chemistry lympiad Preparatry Prblems. Identificatin f an inrganic cmpund a. The white gelatinus precipitate in grup (III) btained by qualitative analysis f slutin B indicates the presence f Al ins. The white precipitate with Ag indicates the presence f Cl ins. Frm the abve data the cmpund A must be a dimer f aluminium chlride Al Cl 6. b. The reactins are as fllws Al H Cl6 [Al.6H 6Cl 6Cl 6Ag 6AgCl(s) 6 AgCl (s) H 4 H (aq) Ag(H ) r Ag(H ) H Cl Al H4H(aq) Al(H) H4 (s) Al(H) (s) ah (aq) [Al(H) 4 a [Al(H) C Al(H) 4 (s) HC excess f LiH AlCl6 LiH (AlH ) n Li[AlH4 4. Inic and metallic structures a. i. The lattice f acl cnsist f interpenetrating fcc lattices f a and Cl ii. iii. The crdinatin number f sdium is six since, it is surrunded by six nearest chlride ins. Fr acl, the number f a ins is: twelve at the edge centres shared equally by fur unit cells thereby effectively cntributing x /4 80Mumbai, India, July 00 80

13 rd Internatinal Chemistry lympiad Preparatry Prblems a ins per unit cell and ne at bdy center. Thus, a ttal f 4 a ins per unit cell. umber f Cl ins is: six at the center f the faces shared equally by tw unit cells, thereby effectively cntributing 6 / Cl ins per unit cell and eight at the crners f the unit cell shared equally by eight unit cells thereby effectively cntributing 8 /8 Cl in per unit cell. Thus, a ttal f 4 Cl ins per unit cell. Hence, the number f frmula units f acl per unit cell 4a 4Cl 4aCl iv. The face diagnal f the cube is equal t times 'a' the lattice cnstant fr acl. The anins/anins tuch each ther alng the face diagnal. The anin/catins tuch each ther alng the cell edge. Thus, a (r a r Cl ) Face diagnal a 4 r Cl..()..() Substituting fr 'a' frm () int () we get : (r a r Cl ) 4 r Cl frm which, the limiting radius rati r a / r Cl 0.44 v. The chlride in array is expanded t make the ctahedral hles large enugh t accmmdate the sdium ins since, the r a / r Cl rati f is larger than the ideal limiting value f 0.44 fr ctahedral six crdinatin number. vi. As the catin radius is prgressively increased, the anins will n lnger tuch each ther and the structure becmes prgressively less stable. There is insufficient rm fr mre anins till the catin / anin radius rati equals 0.7 when, eight anins can just be gruped arund the catin resulting in a cubic eight crdinatin number as in CsCl. Mumbai, India, July 00 8

14 rd Internatinal Chemistry lympiad Preparatry Prblems vii. Generally, the fcc structure with a six crdinatin number is stable in the catin/anin radius rati range 0.44 t 0.7. That is, if 0.44 < r /r < 0.7 then, the resulting inic structure will generally be acl type fcc. b. i. Bragg's law states λ d hkl Sin(θ) 54 pm d d Sin(5.8 ) 54 pm 54 pm 8 pm Sin(5.8 ) 0.7 Thus, the separatin between the (00) planes f acl is 8 pm. ii. Length f the unit cell edge, a d 00 d 00 a 8pm 566 pm. iii. Since it is an fcc lattice, cell edge, a (r a r Cl ) radius f sdium in r a a r Cl pm c. i. The difference in an hcp and a ccp arrangement is as fllws: The tw 'A' layers in a hcp arrangement are riented in the same directin making the packing f successive layers ABAB.. and the pattern repeats after the secnd layer whereas, they are riented in the ppsite directin in a ccp arrangement resulting in a ABCABC packing pattern which repeats after the third layer. The unit cell in a ccp arrangement is based n a cubic lattice whereas in a hcp arrangement it is based n a hexagnal lattice. ii. Packing fractin Vlume ccupied by 4 atms Vlume f unit cell 8Mumbai, India, July 00 8

15 rd Internatinal Chemistry lympiad Preparatry Prblems Let 'a' be the length f the unit cell edge Since it is an fcc lattice, face diagnal a 4r..() Vlume f the unit cell a 4 4 π Packing fractin a r...() Substituting fr a frm () int (), we get 4 4 ( ) Packing fractin 7 (4r) 0.74 Thus, packing fractin in a ccp arrangement 0.74 r iii. The crdinatin number() and the packing fractin (0.74) remain the same in a hcp as in a ccp arrangement. d. i. Fr an fcc, face diagnal a 4r i where a lattice cnstant r i radius f the nickel atm r i a 5.4 pm 4.6pm 4 4 ii. Vlume f unit cell a (.54 Å) 4.76Å iii. Density f ickel, ρi Z M/ V. f i atms, Z 4 fr an fcc lattice Avgadr cnstant Z M g ml ρi V 8.90 g cm cm ml Mumbai, India, July 00 8

16 rd Internatinal Chemistry lympiad Preparatry Prblems 5. Cmpunds f nitrgen a. i. :. f electrns in the valence shell arund nitrgen The Lewis structure fr is as shwn belw : :: :.. : Accrding t VSPR, the mlecule ideally shuld have linear gemetry. Hwever, this mlecule has ne single unpaired electrn present n nitrgen. Due t the repulsin between the unpaired electrn and the ther tw bnded pairs f electrns, the bserved bnd angle is less than 80 ( ). Thus, the shape f the mlecule is angular as shwn belw.. ii. :. f electrns in the valence shell arund nitrgen (5 ) 8 The Lewis structure is as shwn belw.... : :: : : : Thus, there are n nnbnded electrns present n nitrgen. The tw σ bnds will prefer t stay at 80 t minimize repulsin between bnded electrn pairs giving a linear gemetry (80 ). The πbnds d nt influence the shape. 84Mumbai, India, July 00 84

17 rd Internatinal Chemistry lympiad Preparatry Prblems :. f electrn in the valence shell arund nitrgen 5 8 The Lewis structure is as shwn belw : :: :.. : : 5 In case f, there is a lne pair f electrns present n nitrgen. Due t strng repulsin between the lne pair f electrns and the bnded pairs f electrns the angle between the tw bnd pairs shrinks frm the ideal 0 t 5. b. In case f trimethylamine, the shape f the mlecule is pyramidal with a lne pair present n nitrgen. Due t the lne pair MeMe angle is reduced frm 09 4 t SiH H Si Me Me Me SiH Hwever, in case f trisilylamine, d rbital f silicn and p rbital f nitrgen verlaps giving duble bnd character t the Si bnd. Thus, delcalisatin f the lne electrn pair f nitrgen takes place and the resultant mlecule is planar with 0 bnd angle. Si empty drbital filled prbital Mumbai, India, July 00 85

18 rd Internatinal Chemistry lympiad Preparatry Prblems c. Bth and B are tricvalent. Hwever, F is pyramidal in shape. In case f BF, the BF bnd gets duble bnd character due t the verlapping f p rbitals present n brn and flurine. The bserved bnd energy is, therefre, much greater in BF F.. F F F F B F empty prbital filled prbital d. i. The difference in biling pints f F and H is due t hydrgen bnding which is present in ammnia. High electrnegativity f flurine decreases the basicity f nitrgen in F. Thus, F des nt act as a Lewis base. ii. In F, the unshared pair f electrns cntributes t a diple mment in the directin ppsite t that f the net diple mment f the F bnds. See figure (a)..... F F F H H H F H (a) (b) 86Mumbai, India, July 00 86

19 rd Internatinal Chemistry lympiad Preparatry Prblems In H, the net diple mment f the H bnds and the diple mment due t the unshared pair f electrns are in the same directin. See figure (b). e. a 8a( Hg) 4H a 8aH 8Hg H H t at a th a is the salt f H (Hypnitrus acid). Structure : H H r.... H.... H Ismer is: H (itramide) H H 6. Structure elucidatin with sterechemistry a. H H C C CH CH H S 4 C H C CH H C H C CH H C CH x,pentanediic acid α Hydrxy carbxylic acids underg similar reactin. Mumbai, India, July 00 87

20 rd Internatinal Chemistry lympiad Preparatry Prblems b. Mlecular weight f A 6 0 ml 0.05 M KH 8 mg A 000 ml M KH 8 g A The acid is dibasic Mlecular weight f A 6 80 mg Br 8 mg A 60 gm Br 6 g A A cntains ne duble bnd C C It has anisle ring in the mlecule CH It is frmed frm HC CH CCH CH It has mlecular frmula C H 5 Due t steric hindrance the attachment f the aliphatic prtin n the anisle ring will be para with respect t CH. Hence the structure will be H C CH CH As A frms anhydride the tw CH grups shuld be n the same side f the duble bnd. c. Ismers f A CH CH CH ( ) ( methxyphenyl )pentenediic acid CH CH CH ( Z ) ( methxyphenyl )pentenediic acid 88Mumbai, India, July 00 88