10322039 Csir Ugc Jrf Net Chemistry Paper 1 Part b Series 1

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This part (Series-3 from Q.no 61 - 80) is updated on 22nd Nov, 2009 and going to be updated frequently. For recent updates and other parts, join the following group

http://www.adichemadi.com SOLVED CSIR UGC JRF NET CHEMICAL SCIENCES PAPER 1 (PART-B)

SERIES-3

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.

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S4N4 + 8Ag

S4N4

2S2N2

Ag2S (as catalyst) 77 K

Ag2S + 2N2

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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61.3) What is the difference between cyclophosphazanes and cyclophosphazenes? Ans:- Cyclophosphazanes: saturated cyclic systems involving P(III)–nitrogen linkages; mostly four membered. Cyclophosphazenes: unsaturated rings with P(V)–nitrogen bonds; mostly six or eight membered. 61.4) Write the equation for the synthesis of (Cl2NP)2? Ans:- Reaction of phosphorus pentachloride with ammonium chloride in a high-boiling solvent, like 1,1,2,2-tetrachloroethane yields (Cl2PN)2. TO BE COMPLETED Practice questions: 1) The structure of P4N4Cl8 is puckered whereas that of P4N4F8 is planar because: a) F is more electronegative than Cl b) F is smaller in size than that of Cl c) F is more polarizable than Cl d) Extent of  -electron delocalization more in P4N4Cl8 than in P4N4F8

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+

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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

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excitation i.e., no electron spin-flip is allowed. Always transition between triplet to triplet and singlet to singlet states are allowed. ANSWER TO THE QUESTION: * 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 -

3d4

or

t2g3eg1

one spin allowed transition

Co2+

-

3d7

or

t2g5eg2

three spin allowed transitions

Fe2+

-

3d6

or

t2g4eg2

one spin allowed transition

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Cr2+

* 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 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 (CT) is of two types viz., i) Ligand to Metal Charge Transfer (LMCT) ii) Metal to Ligand Charge Transfer (MLCT)

* Remember! The energy required for CT is greater than the energy for d-d transitions. But it is spin and symmetry allowed and hence intense absorptions are possible. i.e., extinction coefficients are higher.

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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

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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) 2ii) CrO4 deep yellow (absorbs violet (high energy) and transmits yellow) iii) VO43pale 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).

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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. Note: 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) - [Ru(bpy)3]2+ [Fe(bpy)3]2+ [Fe(phen)3]2+ -- ferroin Note: phen = 1,10-phenanthroline Optical spectroscopy (UV-Visible) is a powerful technique to assign and characterize charge transfer bands in these complexes. Additional questions: 63.1) In general, UV-Visible absorption peaks of transition metal ions are broad, whereas those of f-block metal ions are sharp. Explain? Ans:- In transition metals, absorption peaks appear due to d-d transitions, which are affected by the external environment (like solvent and ligand effects). Hence absorptions are broad. In case of f-block elements, the f orbitals are deep inside the atom and are screened from the external environment. Hence the peaks are sharp. However, the absorptions are less intense.

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63.2) Explain why f–f transitions of lanthanide complexes are weaker than d–d transitions of transition metal complexes in the electronic spectra? Ans:- * Both d–d and f–f transitions are parity forbidden (symmetry forbidden). * But in the case of d–d spectra, vibronic coupling lowers the symmetry round the metal ion, so d/p mixing can occur and the d–d transitions are to some extent ‘allowed’. (It is the reason why the peaks are observed even though they are forbidden) * Because the 4f orbitals are so contracted (as they are deep inside) compared with d orbitals, f/p mixing and vibronic coupling are much more limited, thus the f–f transitions are less ‘allowed’ and so the peaks are less intense. Note: The centre of symmetry of complexes is momentarily destroyed due to asymmetric molecular vibrations. This is called ‘vibronic coupling’. 63.3) Fe(SCN)2+ is blood red in color eventhough d-d transitions in Fe3+ ion with d5 configuration are forbidden. Explain.

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Ans:- Fe(SCN)2+, correctly represented as [Fe(NCS)(H2O)5]2+, is an octahedral complex. The configuration of Fe3+ is t2g3eg2. Hence the d-d transitions in them are spin forbidden (though Laporte allowed). But the intense red color is due to charge transfer from SCN- to Fe(III) ion (LMCT).

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63.4) Predict the line width and the intensities of peaks in the electronic spectra of CuCl42and Eu(H2O)83+ complexes. Ans:- The peaks are broad and intense in case of CuCl42-. * Intense due to LMCT possible from Cl- ligand to Cu(II). * The d-d component is also strong as the d-d transitions are Laporte allowed as the complex is tetrahedral (no inversion symmetry). * Broad because d-d transitions are vibronically coupled. In case of Eu(H2O)83+ the peaks are very narrow and weak. * These are narrow as there are no crystal field effects on f-f transitions as the f-orbitals are deeply buried inside the atom. The metal-ligand overlapping is very weak and hence no vibronic coupling possible. * These are also weak as there is no vibronic coupling. (The laporte forbidden transitions are atleast possible due to vibronic coupling which destroy the symmetry of complex for a short span of time) 63.5) [Fe(bpy)3]2+ is expected to exhibit an MLCT rather than an LMCT absorption. Comment. Ans:- Iron is in the Fe(II) state which is harder to be reduced and hence LMCT is not observed. MLCT is possible because metal d-orbitals are close in energy to the ligand  * orbitals. Remember this occurs in visible region. 63.6) Why is a transition from a t2g to an eg orbital spin allowed in [V(OH2)6]3+? Ans:- The electronic configuration if V3+ ion is t2g2 eg0 . The transition is allowed as it involves NO spin-flip. i.e., transition is of triplet to triplet type and hence allowed. eg

spin allowed

t 2g

triplet

triplet

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63.6) The reaction below produces both red and green crystals with following data: NiBr2 + PEtPh2 ------> NiBr2(PEtPh2)2 ( both red and green forms) The following data is available: Form data  eff = 0 Red  eff = 3.0 BM Green The red form has zero dipole moment, whereas the green form has net dipole moment. When the red form is dissolved in dichloromethane, the magnetic moment of the solution observed will be 2.69 BM. Propose the possible geometries of both the forms. Ans:- The four-coordinate compound can exist in two geometries: i.e., square planar and tetrahedral

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8

As a square planar complex, the d Ni(II) would be diamagnetic. Therefore the red complex is square planar. It must be a trans isomer as its dipole moment is zero. As a tetrahedral complex, the d8 Ni(II) would have the electronic configuration (eg)4(t2g)4 and have 2 unpaired electrons. Therefore the green form is Tetrahedral. dx2-y2 t2 dxy dz2 dxz

e

dyz Tetrahedral

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Square planar

A square planar complex will have a large crystal field splitting energy (delta) and absorb shorter wavelengths and reflect long wavelengths such as red. A Tetrahedral complex will have a small delta and absorb longer wavelengths and reflect shorter wavelengths such as green. As the tetrahedral form is more stable, the square planar complex is converted to tetrahedral. The slightly low value of 2.69 BM may indicate there is an equilibrium between square planar and tetrahedral forms.

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63.7) When an ethanolic solution of NiBr2 is treated with two equivalents of the phosphine PEtPh2, two products A and B can be obtained by varying the crystallisation conditions. Compound A has no dipole moment and is diamagnetic. One peak due to Ni-Br stretching can be observed in its IR spectrum. Compound B has a dipole moment and is paramagnetic. The IR spectrum of B shows two peaks due to Ni-Br stretching. By contrast, the analogous reaction with PtBr2 and two equivalents of the same phosphine PEtPh2 gives just one compound, C, that has similar spectroscopic and magnetic characteristics to A. Identify A, B and C and explain the observations. Ans:- ‘A’ is square planar and trans isomer. ‘B’ is tetrahedral. (see the explanation given for the previous question) gkuryue854wycvli8794e5hggbli[097465eyv,mk[p9-7634utdglkok ][-\=--0=]90y8oedou760656w53w57328\ ==\pob,jgd898 \=-pho8706w42yli \= 63.8) 1,10-Phenanthroline is added during the quantitative analysis of Fe(II) by UV-visible spectroscopy. Explain why? Ans:- The absorption bands of many transition metal ions in aqueous solutions are broad & strongly influenced by the chemical environment, because of the spatial shape & orientation of the d-orbitals. Such bands are rarely intense enough (low molar absorptivity) to use directly for quantitative analysis. Hence the molar absorptivity is greatly augmented by

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complexing the metal ion with some suitable organic chelating agent to produce a chargetransfer complex. Eg., 1,10-phenanthroline improves the molar absorptivity of Fe(II). Note: 1,10-Phenanthroline complexed with iron ions is commonly called as ferroin - reduced form is [Fe(phen)3]2+ and oxidised form is [Fe(phen)3]3+ . It is used as 1) redox indicator (reduced ferrous form is deep red, oxidised ferric form is light blue) 2) used as a cell permeable inhibitor for metalloproteases in cell biology.

N N

N 2+

Fe

N

N

N

N Fe N

3+

N

N

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N

N

Reduced form (deep red)

oxidized form (light blue)

Note: Ferroin shows bright colors due to MLCT.

Wait! Lanthanides & actinides having incomplete f-levels and hence give rise to absorption bands in a similar fashion to transition metals. In contrast to transition metal ion spectra, those of the lanthanides/actinides contain narrow, well-defined bands which are little affected by ligands & the local chemical environment. But still the peaks are less intense.(Also remember, 4f-4f transitions are sharp and not of 4f-5d.) Practice questions 1) Account for the following molar extinction coefficients for the compounds shown. [Mn(OH2)6]2+  max < 10-1 dm3mol-1cm-1 [CoCl4]2 max= 102 - 103 dm3mol-1cm-1 2[MnO4]  max = 103 - 105 dm3mol-1cm-1 Hint: [CoCl4]2- is tetrahedral and the configuration of Co2+ is e4t23. Hence the transitions are both Laporte and spin allowed.

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64) The acid catalyzed hydrolysis of trans-[Co(en)2AX]n+ can give cis-product also due to the formation of 1. square pyramidal intermediate 2. trigonal bipyramidal intermediate 3. pentagonal bipyramidal intermediate 4. face capped octahedral intermediate Additional questions 64.1) Coordinated water molecules of a Cd(II) complex can be successively replaced by Brfinally to result in [CdBr4]2-. In this process, the fourth equilibrium constant is observe to be higher than the third one, because; 1) Equilibrium constant for the last step is always the highest. 2) Three molecules of H2O are release during the fourth step. 3) The aquo Cd2+ complex is octahedral. 4) An anion (Br-) rplaces a neutral (H2O) molecule from the coordination sphere.

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Ans:- The mechanism and the equilibrium constant values for each step during the reaction are given below.

[Cd(OH 2 )6 ]2+ (aq) + Br - (aq)  

[CdBr(OH 2 )5 ]+

[CdBr(OH 2 )5 ]+ (aq) + Br -(aq)   [CdBr2 (OH 2 ) 4 ]

(aq)

[CdBr3 (OH 2 )3 ]-

(aq)

(aq)

[CdBr2 (OH 2 )4 ]

(aq)

+ Br -(aq)  

[CdBr3 (OH 2 )3 ]-

+ Br -(aq)  

[CdBr4 ]2-

(aq)

+ H 2O (l) ; logK1  1.56 + H 2O(l) ; logK 2  0.54

(aq)

+ H 2O (l) ; logK 3  0.06

+ 3H 2 O(l) ; logK 4  0.37

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octahedral tetrahedral It is clear from above data that, in the fourth step the octahedral aquo complex is converted to tetrahedral all halo complex and 3 moles of water molecules are released (entropy favored). Hence the K4 is greater than K3. Note: Generally aquo complexes of M2+ ions are six coordinated, whereas halo complexes are four coordinated. (only generalization)

More explanation: Usually the stepwise formation constants lie in the order: K1 > K2 > K3....... > Kn. This general trend can be explained simply by considering the decrease in the number of the ligand H2O molecules available for replacement in susequent steps. In general, Kn > Kn+1. But this order is reversed, when there is some structural change during that step. 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

F

I

F

There are 10 faces and 15 edges.

F

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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

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

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corresponding to max is given by E=

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 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 = 2.58 x 10-19 J

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= 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 . Practice questions 1) The UV-Vis spectrum of ReF6 exhibits absorption at 32,500 cm-1. Calculate Δo in kJ mol-1. 2) Which of the following is not a source of visible radiation?

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a) Deuterium lamp b) Tungsten lamp c) Xenon arc lamp d) All. Note: Deuterium and hydrogen lamps are essentially identical. But D2 lamps are a little brighter and hence are exclusively used over H2 lamps.

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

tetrathionate O

S O

-

2I -

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thiosulfate

+

S

O

O

-

-

S

O

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. I2 + I- ---------> I3To 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.

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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. H3PO3 + H2O + I3-

H3PO4 + 2H+ + 3I-

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

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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. 2CuI + SO42white ppt

2S2O32- + I3-

S4O62- + 3I-

+

I2

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2CuSO4 + 4I-

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 CuI 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 - (contrary to above observation). 2) Remember, other dihalides of copper (CuF2, CuCl2 and CuBr2 ) are possible (Why? Simply because other halides cannot be oxidized by Cu(II))

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73.2) The black precipitate formed when hypo is added with silver nitrate solution is a) Ag2S b) Na3[Ag(S2O3)2] c) S d) Na2S Ans:- When less amount of hypo is added to silver nitrate, initially a white precipitate of silver thiosulfate is formed and it is immediately reduced to black silver sulfide precipitate. Na2S2O3 + 2AgNO3 Ag2S2O3 + H2O

Ag2S2O3 + 2NaNO3 Ag2S + H2SO4 Black ppt.

Note: But when excess of hypo is used, the silver thiosulfate formed will be soluble in hypo by forming a water soluble complex, sodium argento thiosulfate, Na3[Ag(S2O3)2] 3Na2S2O3 + Ag2S2O3

2Na3[Ag(S2O3)2]

H.W: What is the IUPAC name of above complex?

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73.3) Silver halides are used in photographic plates because: a) oxidized in air b) reduced by light c) dissociated in air d) soluble in hypo e) do not undergo dissociation air Ans:- ‘d’ Explanation: AgBr is used in making photographic films. It is dissociated by forming black silver when exposed to light. 2AgBr --------> 2Ag + Br2 The unexposed AgBr is removed from the photographic film by treating it with hypo solution in dark. This process is called photography fixing. The AgBr dissoves in hypo by forming water soluble complex Na3[Ag(S2O3)2]. Na3[Ag(S2O3)2] + NaBr

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2Na2S2O3 + AgBr

Note: AgBr is neither oxidized or reduced by air or light. It is just dissociated (self reduction and oxidation). It won’t be dissociated by air but dissociated in light. 74) An exothermic reaction will necessarily follow the condition 1. ΔH < 0 2. ΔH  0 3. ΔH  0 4. ΔS  0 Explanation: * ΔH < 0 ------- Exothermic * ΔH > 0 ------- Endothermic

75) The unit of rate constant (k) for a zero-order reaction is 1) s-1 2) L mol-1 s-1 3) s

Explanation: * In general, the unit of rate constant for nth order reaction is given by M1-n.time-1 = (mol / L)1-n.time-1 = mol1-n.Ln-1.time-1 * Hence for zero order reaction: M1-0sec-1 = mol.L-1.sec-1

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Derivation of Unit of rate constant For a general reaction, A  products Rate expression can be written as dc n  k A dt where n = order of the reaction rate =

k=

dc 1 x dt  A n

 Units of k =

M 1 x sec  M  n

4) mol L-1 s-1

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=

M

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1-n

 mole / litre   sec

= M1-n .sec  mole1-n .Ln-1.sec-1

For 1st order reaction For 2nd order reaction For 3rd order reaction For zero order reaction

-1   sec -1 -1   L.mole sec 2 -2 -1   L .mole sec -1 -1   mole .L sec

V. W AD W IT .A Y DI A CH VA EM RD AD HA I.C N O M

76) Heating is observed when N2 gas at 200 atm is expanded at T > 600 K. It is because 1. inversion temperature is smaller than 600 K 2. N2 is a real gas 3. Joule-Thomson coefficient is negative 4. Joule-Thomson coefficient is positive Explanation: Joule-Thomson effect: The change in temperature of non-ideal gas when it is allowed to expand adiabatically from high pressure to low pressure is called Joule-Thomson effect. The temperature may increase or decrease during this process.

 T   T  Joule-Thomson coefficient =  JT   P   lim  P  as P  0  H  H

The value of Joule-Thompson coefficient may be positive or negative. The temperature at which its sign is changed is called Joule-Thompson inversion temperature. The sign of the Joule-Thompson coefficient is positive below the critical temperature and the gas is cooled upon expansion. Hence below the critical temperature, the gas can be liquified. But the sign of the coefficient above the critical temperature is negative and the gas is heated upon expansion. Hence above the critical tempertature, it is NOT possible to liquify the gas.

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Usually most of the gases are cooled upon adiabatic expansion at room temperature as their critical temperatures are above the room temperature. But the gases like H2, He and Ne have their critical temperatures below room temperature (RT) and hence their temperature increases during adiabatic expansion at RT. These are also difficult to liquify. Answer to the question: inversion temperature is smaller than 600 K. Practice questions: 1) The Joule-Thomson coefficient for N2 is assumed to be a constant value of 0.15 K.bar-1. If the N2 undergoes a drop of pressure of 200 bar, then the temperature will be: 1) raised by 30 K 2) dropped by 30 K 3) remains constant 4) no change in temperature.