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This part (Series-2 from Q.no 61 - 80) is updated on 13th July, 2009 and going to be updated constantly. For recent updates and other parts, join the following group

http://groups.google.com/group/adichemadi or view them at http://www.pdfcoke.com/adichemadi SOLVED CSIR UGC JRF NET CHEMICAL SCIENCES PAPER 1 (PART-B)

SERIES-2

NOTE: Related and additional questions appeared in previous GATE exams are also solved. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

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61) The polymeric species (SN)n is a / an 1. three dimensional conductor 2. two dimensional conductor 3. insulator 4. one dimensional conductor

Explanation: * (SN)n , known as polythiazyl, is a linear polymeric sulfur nitride. It is a one dimensional conductor. It was found to be a superconductor at very low temperatures (below 0.26 K). * It is used as barrier electrode in ZnS junctions. * It increases the quantum efficiency of blue emission by a factor of 100 compared to gold. * It increases the efficiency of GaAs solar cells by up to 35%. Additional information: Structure of (SN)n

S

N

S N 1200 N S

S

N 1060 S

N

N

S

It is a linear polymer (n = up to 2000). It exists in several resonance forms. Preparation: Polythiazyl is synthesized by passing Tetrasulfur tetranitride (S4N4) over silver metal. In this conversion, silver is sulfided to silver sulfide which catalyses the conversion of S4N4 to Disulfur dinitride (S2N2), which readily polymerized to (SN)n. Ag2S + 2N2

S4N4 + 8Ag

S4N4

2S2N2

Ag2S (as catalyst) 77 K

2S2N2

1) sublimes to surface at 0oC (SN)n 2) then undergoes thermal polymerization

Additional questions: 61.1) Draw the structure of S4N4.

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S S N N

N

N S

S

61.2) Write the equations for preparation of S4N4 from SCl2. Ans:- S4N4 was prepared by the reaction of ammonia with SCl2 in carbon tetrachloride followed by extraction into dioxane. 24 SCl2 + 64 NH3 --------------> 4 S4N4 + S8 + 48 NH4Cl

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63) The molar absorptivity at max is minimum for 1. [Mn(H2O)6]2+ 2. [Cr(H2O)6]2+ 3. [Co(H2O)6]2+ 4. [Fe(H2O)6]2+

Explanation: * The molar absorptivity or extinction coefficient (  ) indicates the intensity of absorption of light radiation during the excitation of electrons. If this value is large then the the complex have intense color and otherwise it will have pale color. * The intensity can be determined from the following quantum mechanical selection rules that state whether the transitions are allowed (intense color) or not allowed (pale color) a) Symmetry forbidden or Laporte forbidden ( Δl =  1): If the molecule has centre of symmetry, the transitions from one centrosymmetric orbital (that with centre of inversion) to another are forbidden. i.e., transitions within a given set of p or d orbitals (i.e. those which only involve a redistribution of electrons within a given subshell) are forbidden. In other words, g-->g or u-->u transitions are not allowed. b) Spin forbidden ( ΔS = 0 ): The number of unpaired electrons cannot change upon excitation i.e., no electron spin-flip is allowed. * All the molecules are octahedral with centre of symmetry and hence the excitations are symmetry forbidden. (In case of Cr2+ there is Jahn-Teller distortion) * Check for whether spin forbidden or not? The electron distributions are shown below (remember H2O is a weak field ligand) Mn2+ 3d5 or t2g3eg2 spin forbidden Cr2+

-

3d4

or

t2g3eg1

one spin allowed transition

Co2+

-

3d7

or

t2g5eg2

three spin allowed transitions

Fe2+

-

3d6

or

t2g4eg2

one spin allowed transition

* As the transitions in Mn2+ are both symmetry and spin forbidden, the molar absorptivity is minimum for first complex, [Mn(H2O)6]2+. Hence it is pale pink in color. Note: Extinction coefficients for tetrahedral complexes are expected to be around 50-100

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times larger than for octrahedral complexes. Why? Ans: Do not possess centre of inversion. Additional questions 63.1) KMnO4 shows an intense pink colour, while KReO4 is colourless.Explain. Ans:- KMnO4 shows an intense pink colour, while KReO4 is colourless. Explanation: Mn7+ in MnO4- has d0 configuration. Hence no d-d transition. Still it shows color due to ligand to metal charge transfer (LMCT) phenomenon. MnO4- absorbs yellow and transmits violet part of light. Charge transfer spectra: Charge transfer is of two types viz., i) Ligand to Metal Charge Transfer (LMCT) ii) Metal to Ligand Charge Transfer (MLCT)

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LMCT: In LMCT, an electron is transferred from ligand to the metal, which is therefore reduced in the excited state. Charge transfer is analogous to an internal redox reaction, and hence absorption energies can be correlated with trends in redox properties. The more positive the redox potential concerned, the easier such reduction will be, and so the lower the LMCT energy. LMCT transitions in the visible region of the spectrum give intense color as in case of permanganate ion (MnO4-). The general LMCT energy trends in some d0 species are: i) LMCT energy decreases towards the right in the 3d series. e.g. VO43- > CrO42- > MnO4ii) LMCT energy increases down the transition metal group. e.g. MnO4- < TcO4- < ReO4The above orders of LMCT energy are reflected in the changing colors of the ions and as the transition moves progressively to higher energy out of the visible spectrum into the UV. e.g. i) MnO4deep purple (absorbs yellow color (low energy) and transmits violet) ii) CrO42deep yellow (absorbs violet (high energy) and transmits yellow) 3iii) VO4 pale yellow (as in above case) Now the answer to final part of the question. As the energy required for the charge transfer in case of ReO4- is in the UV region, it does not show any color. It absorbs UV and transmits entire VIBGYOR (= white). Other examples: i) CdS: The color of artist’s pigment cadmium yellow is due to transition from S2-(p orbital) to Cd2+ (empty 5s orbital). ii) HgS: It is red due to S2-(p) to Hg2+ (6s) transition. Actually in Cd2+ and Hg2+ d-d transitions are not possible due to d10 configuration. MLCT: The less common, MLCT results in the reduction of the metal center. The charge transfer occurs from metal to ligand with emty orbital especially low lying  * orbital. e.g. Tris(2,2’-bipyridyl)ruthenium(II), [Fe(phen)3]2+ -- ferroin Note: phen = 1,10-phenanthroline Optical spectroscopy is a powerful technique to assign and characterize charge transfer bands in these complexes.

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67. The number of faces and edges in IF7 polyhedron are, respectively 1. 15 and 15 2. 10 and 15 3. 10 and 10 4. 15 and 10 Explanation: F F

F F

I

F F

There are 10 faces and 15 edges.

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F

69) O2 can be converted to O2+ by using 1. PtF6 2. KF 3. Na2S2O3 4. Br2

Explanation: * PtF6 is a strong oxidising agent and can remove electron from dioxygen. O2 + PtF6 ------>O2+ [PtF6]* Bartlett made this compound, which lead to the discovery of first compound of noble gas, Xenon. Bartlett realized that the first ionization energy of dioxygen, 1180 kJ mol-1 is almost equal to that of xenon, 1170 kJ mol-1. Furthermore, the dioxygen cation should be roughly the same size as a Xe+ ion, and hence the lattice energies of the corresponding compounds should be similar. Xe + PtF6 ------>Xe[PtF6] 71) CFSE of transition metal complexes can be determined by 1. UV-visible spectroscopy 2. IR spectroscopy 3. Microwave spectroscopy 4. NMR spectroscopy

Explanation: * Crystal Field Stabilization Energy (CFSE) is calculated by using UV-visible spectroscopy. First the max for a d-d electronic transition is recorded using spectrometer. The energy corresponding to max is given by E=

hc . This is equal to crystal field splitting energy (  ). λ max

From this CFSE is calculated as follows for simple cases. For octahedral complexes, CFSE = (-0.4a + 0.6b) x  o Where a = no. of electrons in t2g orbitals b = no. of electrons in eg orbitals

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For tetrahedral complexes, CFSE = (-0.6a + 0.4b) x t Where a = no. of electrons in e orbitals b = no. of electrons in t2 orbitals Note: max refers to the wavelength of peak with maximum molar absorptivity (  ) Additional questions: 71.1) Calculate the CFSE (crystal field stabilization energy) of [TiCl6]3- if the max is 770 nm. Ans:- The energy corresponding to max = 770 nm can be calculated as E=

hc 6.625 x 10-34 J.sec x 3 x 108 m.sec-1 = λ max 770 x 10-9 m

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= 2.58 x 10-19 J = 2.58 x 10-22 kJ per one molecule

= 2.58 x 10-22 kJ x 6.023 x 1023 =155 kJ per mole

This energy is equal to the magnitude of energy separation(  o ) between t2g and eg in the octahedral complex [TiCl6]3-. E.C of Ti3+ in the complex is 3d1 and the only electron occupies the t2g . Hence the CFSE=1 x -2/5 o = -2/5 x 155 = 62 kJ. mole-1 .

73) In aqueous medium a mixture of KI and I2 converts thiosulfate to 1. S4O62– 2. SO42– 3. S2O62– 4. S2O42– Explanation: * Iodine oxidises thiosulfate to tetrathionate. KI

2S2O32-

+

S4O62-

I2

thiosulfate

-

S

2I -

tetrathionate O

S O

+

O

-

O

O

-

S

O

O

S

S

S

O

-

O

KI reacts with I2 by forming KI3 and increases its solubility in water.

* This reaction is used in iodometry. Additional information: Iodine is used in iodimetry and iodometry to estimate various oxidising and reducing analytes. Molecular Iodine(I2) is slightly soluble in water, but its solubility is greatly enhanced by complexation with iodide. Hence when used as titrant, I2 is dissolved in water containing excess of KI. This solution has brown color.

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3

I2 + I ---------> I To detect the end point in titrations using iodine, starch solution is added. Iodine imparts navy blue color to starch solutions. The blue colored starch-iodine complex contains long linear chains of I5-, [I-I-I-I-I]- in amylose part of starch. IODIMETRY * In iodimetry, reducing analyte is titrated directly with Iodine. The reducing analyte reduces iodine(I2) to iodide(I-). This method is used in estimation of Ascorbic acid, glucose, phosphorus acid etc., * Ascorbic acid is oxidised by iodine to dehydroascorbic acid. * Glucose is oxidised to gluconic acid (the -CHO group in glucose is converted to -COOH) RCHO + 3OH- + I3-

RCOO- + 2H2O + 3I-

* Phosphorus acid is converted to phosphoric acid. H3PO4 + 2H+ + 3I-

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H3PO3 + H2O + I3-

* In iodimetry, the starch solution can be added at the begining of the titration.

IODOMETRY: * In iodometry, first an oxidizing analyte is added to excess of I- solution to liberate I2 which is then titrated against a standard solution of thiosulfate. first step second step

---- oxidation of I- to I2 by an oxidising analyte. ---- reduction of I2 to I- by thiosulfate.

Iodometry is used in the estimation of CuSO4, K2Cr2O7, KMnO4, Br2, Cl2, BrO3- etc., Estimation of copper: CuSO4 is made to react with excess of I- solution. Thus formed I2 is titrated with standard thiosulfate solution. 2CuSO4 + 4I-

2CuI + SO42white ppt

2S2O32- + I3-

S4O62- + 3I-

+

I2

The iodine liberated in the first step is equivalent to CuSO4. The amount of CuSO4 is estimated indirectly by estimating this liberated I2 with standard thiosulfate solution. * In iodometry, starch solution is added only just before the equivalence point (which is detected visually by fading of brown color of I3- ions). This is because, at high concentrations, some I2 remains bound to starch particles. * In the estimation of copper, the precipitated Cul tends to bind some iodine. To replace this iodine and bring it into solution, thiocyanate solution is added at the end point. Additional questions: 73.1) Copper(I) iodide is a stable species, while coppe(II) iodide does not exist. Explain. Ans:- Cu2+ can oxidise I- to I2 and hence Copper(II) iodide does not exist. 2Cu2+ + 2I- --------> 2Cu+ + I2 Note: 1) In general Cu(II) ion is more stable than Cu(I) ion.

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2) Remember, other dihalides of copper (CuF2, CuCl2 and CuBr2 ) are possible.