Csir Ugc Jrf Net Chemistry Paper 1 (Part b) Series - 1

8/3/2019 Csir Ugc Jrf Net Chemistry Paper 1 (Part b) Series - 1

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This paper is updated on 5th Apr, 2009. For recent updates, join the following group

http://groups.google.com/group/adichemadiSolved PAPER-2 is also going to be available sooner.

SOLVED

CSIR UGC JRF NET

CHEMICAL SCIENCESPAPER 1 (PART-B)

SERIES-1

41) Complexes of which of the following metals are used in the treatment of rheumatoid

arthritis:

1. Gold

2. Ruthenium

3. Iron

4. Copper

Explanation

Gold salts like Auranofin, Sodium aurothiomalate, aurothioglucose are used to treat rheuma-toid arthritis.

Additional information:

1) cis platin is used in treatment of cancer.2) Iron in the form of ferrous sulfate, ferrous gluconate are used in treatment of iron defi-ciency anemia.

Additional questions:

41.1) Give the structure, hybridisation and magnetic moment of cis-platin?41.2) Write its action on cancer cells.

NOTE: Answers to the additional questions are also given at the end.

42) Non-heme iron-sulfur proteins are involved in:

1. Electron transfer.

2. Proton transfer.

3. Both electron and proton transfer

4. Oxygen transfer.

Explanation

Iron-sulfur proteins are proteins characterized by the presence of iron-sulfur clusters contain-ing sulfide-linked di-, tri-, and tetrairon centers in variable oxidation states.Eg., Ferredoxins, as well as NADH dehydrogenase, hydrogenases, nitrogenase etc.,Iron-sulfur clusters are best known for their role in the oxidation-reduction reactions of mito-chondrial electron transport.Additionally some Fe-S proteins regulate gene expression.Fe-S proteins are vulnerable to attack by biogenic nitric oxide.In most iron-sulfur proteins, the clusters function as electron-transfer groups.


1) Ferredoxins: are small proteins containing iron and sulfur atoms organized as iron-sulfur

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WARANGAL


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clusters. These biological "capacitors" can accept or discharge electrons, the effect beingchange in the oxidation states (+2 or +3) of the iron atoms. This way, ferredoxin acts aselectron transfer agents in biological redox reactions.The following diagram illustrates the redox scheme between low-potential and high-potential(HiPIP) ferredoxins containing Fe

4S

4clusters. The formal oxidation numbers of the iron ions

can be [2Fe3+, 2Fe2+] or [1Fe3+, 3Fe2+] in low-potential ferredoxins. The oxidation numbers of

the iron ions in high-potential ferredoxins can be [3Fe3+, 1Fe2+] or [2Fe3+, 2Fe2+]

S

Fe

Fe

S

Fe

S

S

Fe

S

S

S

S

S

Fe

Fe

S

Fe

S

S

Fe

S

S

S

S

S

Fe

Fe

S

Fe

S

S

Fe

S

S

S

S

+e-

-e-

+e-

-e-

2-1- 3-

Following is Fe2S

2type of ferredoxin.

Cys

CysCys

S

S

S

S

FeFeS

S

Cys

Note: High potential iron-sulfur proteins (HiPIPs) form a unique family of Fe4S

4ferredoxins

that function in anaerobic electron transport chains.


42.1) The vitamin riboflavin is part of the ______ molecule.A) ferredoxinB) FADC) pyridoxal phosphateD) pyrophosphateE) NAD+

42.2) Ferredoxin (Fd) is a sulfur-containing protein which undergoes redox reactions in avariety of microorganisms; Fdox

+ + e- -------> Fdred

, Eo = 0.439V @pH = 7.0. Will Fd gener-ate hydrogen gas from hydronium ions dissolved in water if the standard potential for thereduction 2H+ + 2e- ---------> H

2is -0.421 at pH=7.0.

Show the redox reaction and explain your answer.42.3) In the following reaction ferredoxin-1 is the oxidised form of ferredoxin. Statewhether this statement is true or false?

nitrite reductase-2 3NO + ferredoxin-1 NH + ferredoxin-2

43) Active catalytic species for hydroformylation is

1. RuCl2(PPh

3)

32. HCo(CO)

33. RhCl(PPh

3)

34. K

2PtCl

6

Explanation

Hydroformylation, also known as oxo synthesis, is an important industrial process for the

Prepared by V. Aditya vardhan adichemadi(at)gmail.com WARANGAL


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production of aldehydes from olefins. In this reaction a formyl group (CHO) and a hydrogenatom are added to a carbon-carbon double bond.H

2+ CO + CH

3CH=CH

2-----> CH

3CH

2CH

2CHO (normal) --commercially desired product

H2 + CO + CH3CH=CH2 -----> (CH3)2CHCHO (iso)The reaction begins with the generation of coordinatively unsaturated metal hydrido

carbonyl complex such as HCo(CO)3

and HRh(CO)(PPh3)

2.

Therefore, Answer is 2 and not 3.


* It is an example for homogeneous catalysis.* The linear / iso product ratio can be improved by using bulky ligands (see below).* HRh(CO)(PPh

3)

2(second generation catalyst) is a better catalyst than HCo(CO)

3(first gen-

eration catalyst becausei) pressures around 20-30 atm are sufficient.ii) the major product is linear isomer of aldehyde (highly selective).iii) the side reactions like hydrogenation can be minimized.

* Third generation water soluble rhodium catalyst uses TPPTS ligands instead of triphenylphosphine. In this case the water insoluble products can be separated easily from the catalyst.TPPTS = 3,3',3''-Phosphinidynetris(benzenesulfonic acid) trisodium salt

P

S

O

O

Na+O

-

SO

O O-Na

+

SO

O

O-Na

+

* HCo(CO)3

is generated as follows

2 8 2 4Octacarbonyl dicobalt tetracarbonylhydridocobalt(I)

Co (CO) + H 2HCo(CO)

4 3

(trigonal bipyramidal)

tricarbonylhydridocobalt(I)

HCo(CO) HCo(CO) + CO

(tetrahedral)



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Co

H

OCOC

CO 16 VE (VE = valence electrons)

R

Co

H

OC

OCCO

CoOC

OCCO

H

R

CoOC

OCCO

R

H

R

CO

CoOC

OC CO

OCR

CoOC

OCCO

OCR

CoOC

OCCO

O R

CoOC

OCCO

OR

O R

H

OR

H

+

CoOC

OCCO

O R

HH

CoOC

OCCO

OR

HH

H2

Co

H

OC

OCCO

16 VE+

to the cycle

Mechanism

18 VE

reverse beta elimination

16 VE

18 VE

16 VE

oxidative addition

reductive elimination

linear iso

If bulky groups like tributyl phosphine(PBu3)

are used then the formation of iso product isDISFAVORED due to steric and electronic

effects. Now a days, HRh(CO)(PPh3)2 is used

to achieve this effectively.



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43.1) Write the IUPAC names of HCo(CO)3

and HRh(CO)(PPh3)

2and mention the oxida-

tion states of Co and Rh.43.2) Write the mechanism of hydroformylation of propene catalysed by HRh(CO)(PPh3)2.Mention the major product formed during the reaction.43.3) What is Wilkinsons catalyst? Mention its use.

44) The unit of molar absorptivity is:

1. L mol1 cm1 2. L1 mol cm1 3. L mol cm1 4. L mol cm

Explanation:

Beer-Lambert law A = .c.lWhere A = absorbance

c = molar concentrationl = pathlength (in cm) = molar extinction coefficient or molar absorptiviy = A / cl

Hence the units are usually in M-1

cm-1

or L mol-1

cm-1

( A has no units: see below)As 1 liter = 1000 cm3, the unit may be sometimes cm2 mol-1, especially in old literature.(note: = pronounced as epsilon)

Additional information

Absorbance (A) = -log10

T = -log10

I /Io

(T = transmittance = I /Io

)Where I = intensity of the transmitted light

Io

= intensity of the initial incident light* Remember transmittance (T) is inversely related to absorbance (A)* is a is a measure of the amount of light absorbed per unit concentration at a givenwavelength.* is an intrinsic property of the compound;* The actual absorbance,A, of a sample is dependent on the pathlength l (in cm) and theconcentration c (molar conc.) of the species via the Beer-Lambert law,A = cl.* It is a limiting law. It is only valid for dilute solutions. At higher concentrations, theelectrostatic interactions between neighbouring molecules distort the energy levels whichcause deviation from the law.* Deviations also occur when the molecules undergo chemical change in solutions (this mayoccur due to change in concentration or pH).

eg., Yellow chromate andorange dichromate are in equilibrium with each other in aqueoussolution. The equilibrium shifts to the right if pH is decreased (when acid is added) and shifts

to the left if pH is increased (when base is added).

2- + 2-4 2 7 22 CrO (aq) + 2 H (aq) Cr O (aq) + H O (l)

yellow orange

Hence the absorbance for this solution depends on the pH


44.1) What is the concentration of a solution whose absorbance is 0.21 when placed in a tube

of path length 10 cm ( max = 245 nm and max = 31,500 M-1cm-1)

44.2) If the visible spectrum of [Ti(OH2)6]3+

has an absorbance maximum at 0.9 at 510 nm andthe spectrum was recorded on a 0.20 M solution in a 1.0 cm cell, what would be the value of

510 ?



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44.3) Transferrin is an iron-transport protein found in blood. It is colourless in the absence ofiron. Desferrioxamine B is an iron chelator used to treat patients suffering from iron overload.It is also colourless in the absence of iron. The following data was obtained in aqueous solu-tion using a 1.00 cm path length cell for both species when fully complexed with iron:----------------------------------------------------------------------------------

max / nm max / L mol1 cm1

iron-transferrin iron-desferrioxamine B-----------------------------------------------------------------------------------

428 3540 2730470 4170 2290

-----------------------------------------------------------------------------------Use this information to predict the absorbance at 470 nm of a solution that is 11.0 M iniron-transferrin and 69.5 M in iron-desferrioxamine B, assuming both are fully complexedand that the cell path length is 1.00 cm.44.4) 1,10-Phenanthroline is added during the quantitative analysis of Fe(II) by UV-visiblespectroscopy. Explain why?

45) Gelatin is added during polarographic measurements to:

1. reduce streaming motion of falling mercury drop

2. increase Id

3. increase E1/2

4. eliminate residual current

Explanation:

When a surfactants like gelatin, Triton X-100 are added to the solution, the streaming effectsat the drop surface are minimized, and the polarographic wave approaches the limiting current

smoothly.


* Polarography is the measurement of the current that flows in solution as a function of anapplied voltage. The electrode processes are studied by using two electrodes - onepolariz-able and one unpolarizable* Dropping Mercury Electrode (DME) is an example for polarizable working electrode usedin polarography.* A polarogram is a curve visualizing the processes occurring in the course of electrolyticpolarization of the DME.

* A polarographic wave or polarogram is obtained by applying voltage. This experiment em-ploys usually two methods of applying the voltage, a linear sweep (DC) and a differentialpulse.* Nowadays a Static Mercury Drop Electrode (SMDE) is used which provides a more sensi-tive measurement of the faradaic current than the more traditional DME.* One of the difficulties in evaluating a polarogram is the measured current oscillates perma-nently between a minimum and maximum value. This behavior is caused by the nonconstantelectrode surface area. Usually the current rises as the drop surface area grows, and decreasesas the drop falls.* Polarography differs from voltametry in that it employs the electrode (like DME) whosesurface is continuously renewed.

46) The element that shows both +3 and +4 oxidation states is:

1. Cerium 2. Promethium 3. Gadolinium 4. Holmium



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Explanation:

Usually lanthanoids exhibit, +3 oxidations states. (why?)But some of them also exhibit +2 and +4 oxidation states due to stable half filled (4f7) orempty (4f0) or completely filled f-sublevel.Eg., Ce4+ has 4f0 configuration. -very stable

Eu2+ has 4f7 configuration. -very stableYb2+ has 4f14 configuration. - appreciably stable


* Ce4+ is a good oxidising agent as it can be reduced to Ce3+ easily. It can oxidize water also(Eo of Ce4+ /Ce3+ is +1.74 V), but very slowly. Hence Ce(IV) can be used as a good analyti-cal reagent in aqueous medium.* But Eu2+ is a reductant as it can be oxidized to Eu3+, which is the common oxidation stateof lantahnoids.* La3+, Ce4+ with 4f0 and Yb2+, Lu3+ with 4f14 configurations - are colorless and diamagnetic.

* Lanthanoids exhibit +3 oxidation states commonly because first three ionization enthalp-ies are small.* Both Ce and Eu can be separated easily from the rest of the lanthanoids due to theirdifferent chemical behaviour.

For example, Ce3+ can be easily oxidised to Ce4+ ion which being less basic than other Ln3+

undergoes hydrolysis easily. Hence Ce3+ (in a mixture of Ln3+) is oxidised by KMnO4

orbleaching powder to Ce4+ which can be precipitated as basic salt or hydroxide.

23 44( )

Note: Cation with high charge is smaller in size and more acidic (or less basic).

H OoxidationCe Ce Ce OH

* Yttrium oxide-sulfide activated with europium is used as the redphosphor(?) in colorCRTs(Cathode Ray Tubes used in TVs).

A phosphor is a substance that exhibits the phenomenon of phosphorescence (sustainedglowing after exposure to energized particles such as electrons or ultraviolet photons).


46.1) Which of the following carbonate-fluoride mineral contains all the lanthanoids?1) Monazite 2) Bastnasite 3) Pitch blende 4) Zirconite46.2) What is the difference between lanthanoids and lanthanides?46.3) Describe the ion exchange method of separation of lanthanoids.46.4) What is the principle involved in solvent extraction of lanthanoids by using tri-n-

butylphosphate?

47) The number of 3c, 2e BHB bonds present in B4H

10is

1. 2 2. 3 3. 4 4. 0

Explanation:

* B4H

10is called tetraborane(10).



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B

H

BB

BH

H HH H

HH

H

H

B4H10

B

BB B

B

BB B

Arachno structure of B4H10 , showing boron atoms at

the four vertices of octahedral geometry.Note: In these structures the lines do not representthe actual bonds. Only concentrate on the positions.

There are four BHB bonds and also contains a BB bond. It is an arachno borane.Additional information:

* Nomenclature: Boranes are named as follows:

a) The Latin prefixes mono-, di-, tri-, etc. are used before "borane" to indicate the numberof boron atoms in the compound.b) Immediately following the "e" in "borane" the number of hydrogen atoms is placed inparentheses using Arabic numerals.Example: B

5H

11is pentaborane(11).

c) Whereas the names of anions end in ate rather thanane. The numbers of both boronand hydrogens are indicated by Latin prefixesExample: B

10H

102- is decahydrodecaborate(2-)

* Structure: Boranes fall into five structure categories.1) closo BnHn

2- --- n vertices ofn cornered polyhedron are occupied by boron atoms.2) nido BnHn+4 --- n vertices of(n+1) cornered polyhedron are occupied by boron atoms.3) arachno B

nH

n+6--- n vertices of(n+2) cornered polyhedron are occupied by boron

atoms.4) hypho B

nH

n+8 --- n vertices of(n+3) cornered polyhedron are occupied by boron atoms.

5) Conjuncto boranes formed by joining two or more preceding types.

Hypho and conjuncto boranes are less common and not discussed here.

Illustrations:

1) closo BnH

n2- : Eg., Hexahydrohexaborate(2-) B

6H

62- has regular octahedron structure.

Six boron atoms occupy the vertices of octahedron. It has a closed symmetrical structure.

B B

B

B

B B

H

H

H

H

H

H

2-

Note: These anionic closo-BnH

n2- are examples for 3D aromaticity. Whereas neutral B

nH

n+2



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are unstable and not known.

2) nido BnH

n+4: Eg., In Pentaborane(9), B

5H

9, the boron atoms occupy the five vertices (n)

of the regular, six cornered, octahedron (with n+1 corners).

B B

B

B B

H

H

H

H

H

H

H

HH

3) arachno BnH

n+6: Eg., In B

4H

10( Tetraborane(10) ), only four vertices are occupied by

Boron atoms. (Already illustrated in previous page)

B B

B

B

H

H

H

H H

H

H

HH

H

Structural relations between Closo, Nido and Arachno boranes are illustrated below

B B

B

B

B B

B B

B

B B

B B

B

B

Closo borane - all the vertices are occupied.

Nido borane - one vertex is removed

Arachno borane - two vertices are removed.

Only the positions of boron atoms are shown. Lines may ormay not represent the actual bonds between them.

How to know the type and structure of a borane?Eg., B

5H

11- Pentaborane(11) has the B

nH

n+6formula and hence an arachno borane. Remem-

ber it should have five boron atoms (as n=5) occupying five vertices of polyhedron withseven vertices (as n+2=7) i.e., two vertices are missing. Hence its structure is derived frompentagonal bipyramidal structure.



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B B

B

B B

B

B

B B

B

B B

B

B B

B

B

B

closo

Eg., B7H72-

nido

Eg., B6H10

arachno

Eg., B5H11

Only the positions of boron atoms are shown. Lines may or may notrepresent the actual bonds between them.

* The most stable boron hydrido clusters are closo B10H102-

and B12H122-

. They resembleelectron rich aromatic compounds.


47.1) Which one of the following boranes is classified as a nido-borane?1) B

2H

62) B

2H

7 3) B

4H

104) B

12H

122

47.2) Which of the following is an arachno borane?1) [B

6H

6]2- 2) [B

5H

9] 3) [B

2H

6] 4) [B

6H

12]

48) In BrF3

as a solvent SnF4

and KF behave as

1. acid and base, respectively

2. base and acid, respectively

3. acids

4. bases

Explanation:

KF contains F- ion which can donate electron pairs and hence acts as a Lewis base. BrF4- is

the solvent anion formed.F- + BrF

3--------> BrF

4-

SnF4

acts as a Lewis acid as Tin (Sn) can expand its octet (due to presence of availableempty d-orbitals) and accepts electron pairs. BrF

2+ acts as cation of the solvent.

SnF4 + BrF3 --------> BrF2+ +SnF5-


RememberBrF3

auto-dissociates as followsBrF

3--------> BrF

4- + BrF

2+

And according to solvent system concept, the cation resulting from auto-dissiciation is acidand the anion resulting from auto-dissiociation is a base. Hence the solute which increasesconcentration of cation (BrF

2+)is an acid and which increases the concentration of anion

(BrF4-)is a base. Compare this analogy with H

2O solvent system.

Additional questions:48.1) The following reaction can be viewed as a Lewis acid base reaction. Explain.

2 K2MnF

6+ 4 SbF

5--------> 4 KSbF

6+ 2 MnF

3+ F

2



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48.2) The conductivity of BrF3

is increased by adding either AgF or SnF4. Explain.

48.3) AlF3

is insoluble in liquid HF, but dissolves if NaF is present. When BF3

is added tothe solution, AlF

3precipitates. Explain?

48.4) The magic acid HF + SbF5 is a super acid. It is the strongest acid known. It can evenprotonate methane and alkenes. Explain?48.5) List the following acids in order of their acid strength in aqueous solutions.

1) H2SO4 2) H3AsO4 3) HMnO4 4) H2SO348.6) In the reaction : Cl

2+ ClF + SbF

5---------> [Cl

3] [SbF

6] the role of chlorine is to

1) stabilize Cl+ 2) fuction as Lewis acid3) function as Lewis base 4) form the cation

49) The effective nuclear charge (Z*) for the 1s electron of8O according to Slaters rules

is nearly

1. 4.55

2. 3.45

3. 7.65

4. 5.451

Explanation:

According to Slaters rules Z* or Zeff

= Z - s = 8 - 0.3 = 7.7

Additional Information:

Slaters rules:

* In multi electron atoms the electrons in the outer shells experience less than the actualnuclear charge owing to shielding or screening of other intervening electrons.* This effective nuclear charge (Z* or Z

eff) is calculated by using following equation

Effective nuclear charge = Z* = Z - sWhere Z = atomic number

s = shielding or screening constant ( also represented by S or )* Slaters rules are used to provide a value to the shielding constant as follows.

* Firstly, the electrons are arranged in to a sequence of groups that keep s- and p- orbitalswith the same principal quantum number together, but otherwise progresses with the orbitalordering with increasing principal quantum number (ignore filling irregularities):

[1s] [2s,2p] [3s,3p] [3d] [4s,4p] [4d] [4f] [5s, 5p] [5d] etc.* Any electrons to the right of the electron of interest contributes no shielding. Only theelectrons left to the electron of interest contribute to the shielding constant.

* The shielding constant for each group is formed as the sum of the following contributions:1. An amount of 0.35 from each other electron within the same group except for the [1s]group, where the other electron contributes only 0.30.2. If the group is of the [s p] type, an amount of 0.85 from each electron with principalquantum number one less and an amount of 1.00 for each electron with an even smallerprincipal quantum number.3. If the group is of the [d] or [f], type, an amount of 1.00 for each electron with a smallerprincipal quantum number.

* For example, for the iron atom which has nuclear charge 26 and electronic configuration

1s2

2s2

2p6

3s2

3p6

3d6

4s2

the screening constant, and subsequently the effective nuclear chargefor each electron can be deduced asFor 4s: s = (1 x 0.35) + (14 x 0.85) + (10 x 1) = 22.25



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Z* = 26 - 22.25 = 3.75

For 3d: s = (5 x 0.35) + 0 + (18 x 1) = 19.75Z* = 26 - 19.75 = 6.25

For 3s, 3p: s = (7 x 0.35) + (8 x 0.85) + (2 x 1) = 11.25

Z* = 26 - 11.25 = 14.75

For 2s, 2p: s = (7 x 0.35 ) + (2 x 0.85) + 0 = 4.15Z* = 26 - 4.15 = 21.85

For 1s: s = 1 x 0.30 = 0.30Z* = 26 - 0.30 = 25.7

Additonal questions:

49.1) Calculate Z* for a valence electron in fluorine.

49.2) Calculate Z*

for a 6s electron in Platinum, given the electronic configuration(1s2)(2s2,2p6)(3s2,3p6) (3d10) (4s2,4p6) (4d10) (4f14) (5s2,5p6) (5d8) (6s2)49.3) The first ionization energy for hydrogen is 1310 kJ.mol1 while the first ionizationenergy for lithium is 520 kJ.mol1. Explain.

50) Among the species O2

+, O2

and O2

, the order of first ionization energy is

1. O2

+ < O2

< O2

2. O2

< O2

< O2

+

3. O2

< O2

+ < O2

4. O2

+ < O2

< O2

Explanation:

* Greater the positive charge, greater is the ionization energy.

51) CO bond order is lowest in

1. uncoordinated CO

2. CO bonded to one metal

3. CO bridging two metals

4. CO bridging three metals

Explanation:

*52) The most unstable species among the following is

1. Ti(C2H

5)

4

2. Ti(CH2Ph)

4

3. Pb(CH3)

4

4. Pb(C2H

5)

4

55) On oxidative addition of O2

to Ir(CO)Cl(PPh3)

2, the oxidation state and coordination

number of Ir changes, respectively, by

1. 1 and 3

2. 2 and 23. 3 and 1

4. 2 and 3



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Explanation:

Ir

PPh3

PPh3

C

Cl

O

Ir(CO)Cl(PPh3)2, Ir(CO)Cl(O2)(PPh3)2

Ir

PPh3

PPh3

C

Cl

O

O

O

trans-chloridocarbonylbis(triphenylphosphine)iridium(I)

+ O2Oxidative Addition

179 pm

190 pm

* Ir(CO)Cl(PPh3)

2is a square planar complex which is often referred to as Vaskas com-

pound. The oxidation state of Ir is +1. Co-ordination number is 4. (contains 16 VE)

* During oxidative addition (not simple addition) of O2 the oxidation state is changed to +3.The compound formed is Ir(CO)Cl(O2)(PPh

3)

2. It is octahedral (a little bit distorted), and

hence C.N is 6. (contains 18 VE)* The Dioxygen added is in peroxidic form (O

22-) and not in superoxidic form (O

2-).

* Therefore both oxidation state and coordination number are changed by 2 (oxidation statefrom +1 to +3 whereas C.N from 4 to 6).


* Vaskas complex is said to be co-ordinatively unsaturated as it contains 16 valenceelectrons in Iridium(I). Hence it can bind to ligands and expand its valence electrons to 18by undergoing oxidative additions. During this process, the oxidation number of Ir ischanged to +3.* The C-O stretching frequency is increased from 1967 cm-1 to greater than 2000 cm-1

during oxidative addition reactions. It is due to reduction of amount of-back bondingfrom Ir to C and at the same time strengthening of C-O bond. (Refer to Qtn: 51)* Remember, the Ir-C bond length is decreased and C-O bond length is increased duringoxidative addition.

57) Among the following molecules, the dipole moment is the highest for

1. NH3

2. trans-[PtCl2(NH

3)

2]

3. BF34. NF

3

Explanation:

N

H

H

H

Pt

NH3

NH3 Cl

Cl

B

F

FF

N

F

F

F

* Only NH3

and NF3

have dipole moments. NF3

has highest dipole moment, as the magni-



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tude of N-F bond moment is larger.* trans-[PtCl

2(NH

3)

2] and BF

3have zero dipole moments due to symmetrical structure and

as the individual bond moments are cancelled out by each other.* cis-[PtCl2(NH3)2] has dipole moment. The bond moments are not cancelled out and it isless symmetrical than trans isomer.* Remember, BF

3is trigonal planar whereas, NH

3and NF

3are trigonal pyrimidal.


* Dipole moment () is defined as the product of magnitude of charge( ) on either of thepoles (atoms) and the distance(d) between them.

= .d* It is a vector quantity. It is represented by an arrow pointing towards negative pole (asshown in above diagrams).* The dipole moment of a bond is called bond moment. The final dipole moment of amolecule is the vectorial addition of all the bond moments in it.* The unit of dipole moment is Debye (D) = 10-18 e.s.u cm

* Dipole moment is used to determine the ionic nature of a polar covalent bond.

experimental

theoretical

theoretical

% ionic character = x 100

iscalculated by assuming the bond is completely polar (ionic).

* For example, In HCl molecule, the observed dipole moment is 1.03 D and its bond lengthis 1.275. Assuming 100% ionic character, the charge developed on H and Cl atoms wouldbe 4.8 x 10-10e.s.u.Therefore, the theoretical dipole moment for 100% ionic character will be

theoretical = x d = 4.8 x 10-10

e.s.u x 1.275 x 10-8

cm= 6.12 x 10-18e.s.u.cm= 6.12 D (1D = 10-18 e.s.u. cm.)

experimental

theoretical

1.03% ionic character = x 100 = x 100 = 16.83%

6.12

* It is also used to determine the shape of a molecule. Usually, the symmetrical moleculeshave zero dipole moment. For example CO

2has zero dipole moment as it has linear shape,

whereas SO2

has net dipole moment as it has angular structure.

* Bond angle between two bonds can be calculated if their bond moments and resultantdipole moment are known. Following simplified equation can be used when the two bondsare same.

net dipole moment = 2(bond moment) x cos

2where = bond angle


57.1) The dipole moment of H2

S is 0.95 D. If the S-H bond moment is 0.72 D, find thebond angle.57.2) Naphthalene has zero dipole moment whereas its isomer Azulene has a dipole mo-



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ment of 0.8 Debye. Explain?

135) The major product formed in the reaction, given below, is

CN 1. EtMgBr

2. H3O+

1) CHO

2)

O

3) NH2

4)

CN

Explanation:

C N Et-MgBr+-

N MgBr+-

...

.

An imine salt complex

NH

imine

HOH

CO

H+

HOH

140) Sucrose is a disaccharide consisting of

1. glucose and glucose

2. glucose and galactose3. glucose and fructose

4. glucose and mannose

Explanation:

* Sucrose is-D-glucopyranosyl--D-fructofuranoside .* It is a disaccharide containing alpha-D-Glucose and beta-D-Fructose units.* There is a glycosidic linkage (an acetal linkage) between C-1 of alpha-D-glucose and C-2of beta-D-fructose.



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O

OHOH

OHOH

OH

-D-Glucose

O

OH

OH

OH OH

OH

-D-Fructose

+O

OH

OHOH

OH

O

OH

OH

OOH

OH

-D-Glucopyranosyl--D-Fructofuranoside

glycosidic linkage

- H2O

* It is a non-reducing sugar as both the reducing ends of glucose and fructose are linked. Itdoes not give silver mirror with Tollens reagent.* In sucrose, there are no anomeric hydroxy groups, as the rings at the reducing parts arenot free to open, and hence does not show mutarotation.


*

SOLUTIONS TO ADDITIONAL QUESTIONS

41.1) cis-Diamminedichloridoplatinum(II)

Note: chlorido is very recent IUPAC usage instead of chloro.Pt (Z=78). Belongs to Nickel group. with e.c - [Xe] 4f14 5d8 6s2

For Pt2+ -- [Xe] 4f14 5d8 .Hybridization of Pt2+ is dsp2 . It is a square planar and low spin complex with zero magneticmoment. The crystal field splitting of square planar complexes is shown below.

41.2) Action: Upon administration to the cancer patient, the chloride ligands are displaced bywater and thus aqua platinum complexes are formed in cells, which bind and causecrosslinking of DNA---- ultimately triggers apoptosis - programmed cell death.

42.1) B42.2) L: 2H+(aq) + 2e- -----------> H

2(g), Eo = -0.421V

R: Fdox

+ + e- -----------> Fdred

, Eo = 0.439V



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Overall cell reactionFd

ox+ + H

2(g) + -----------> Fd

red+ 2H+(aq)

Ecell = ER - EL = 0.439 - (-0.421)V = 0.86V ------- which is > 0This means that rG < 0 (why? see the note). Hence reduction of Fdox

+ is spontaneous andNOT the liberation of hydrogen. Therefore H

2gas is consumed, and not produced in this

reaction.Note: G = -nFE

42.3) False. This is a reduction reaction. Hence only reduced form of ferredoxin can reduce NO2-

to ammonia.nitrite reductase-

2 3NO + reduced ferredoxin NH + oxidised ferredoxin

43.1) HCo(CO)3

-- Tricarbonylhydridocobalt(I)HRh(CO)(PPh

3)

2-- Carbonylhydridobis(triphenylphosphine)rhodium(I)

43.2) Mechanism:

HRh(CO)(PPh)3

18e

HRh(CO)(PPh)2

16e

-(PPh)RhH

CO

PPh

PPh

RhOC

PPhPPh

RhOC

CO

PPh

PPh

CH3

18e

18e

16e

RhOC PPh

PPh

CH3O

CO insertion

16e

H2Rh

OC PPh

PPh

CH3O

HH

O

H

+ HRh(CO)(PPh)216e

to the cycleButyraldehyde(major product)

reductive elimination

43.3) RhCl(PPh3)

3- chlorotris(triphenylphosphine)rhodium(I), is called Wilkinsons catalyst and

used in theselective hydrogenation of alkenes. It is a square planar 16 electron complex.* exocyclic double bonds are selectively hydrogenated over endocyclic double bonds.* cis alkenes are reduced rapidly than trans alkenes.* Functional groups like C=O, C=N, NO

2, Ar, CO

2R etc., are unaffected.



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O

RhCl(PPh3)3

H2, benzene O

sterically less hindered double bond is hydrogenated selectively

44.1)-7A 0.21c= = = 6.67 x 10

31500 x 10M

l

44.2) 510 = 4.5 cm-1M-1

44.3)M = 10-6 MFor iron transferrin, A = 4.6 x 10-2

For iron-desferrioxamine B, A = 1.6 x 10-1

44.4) The absorption bands of many transition metal ions in aqueous solutions are broad &strongly influenced by the chemical environment, because of the spatial shape & orientationof the d-orbitals. Such bands are rarely intense enough (low molar absorptivity) to usedirectly for quantitative analysis. Hence the molar absorptivity is greatly augmented bycomplexing the metal ion with some suitable organic chelating agent to produce acharge-transfer complex.Eg., 1,10-phenanthroline improves the molar absorptivity of Fe(II).

N N

Fe2+

Above complex is commonly known as ferroinand also used as1) 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.

Wait!

Lanthanides & actinides having incomplete f-levels & give rise to absorption bands in asimilar fashion to transition metals. In contrast to transition metal ion spectra, those of thelanthanides/actinides contain narrow, well-defined bands which are little affected by ligands& the local chemical environment. But still the peaks are less intense.(Ofcourse 4f-4ftransitions are sharp and not of 4f-5d.)

46.1) Bastnasite mainly represented by (Ce,La,Y)CO3F contains all the lanthanoids, whereas

Monazite(Phosphate mineral of Thorium and other rare earths) contains lighter ones.

46.2) Fifteen elements from La-Lu are lanthanoids.

Whereas fourteen elements Ce-Lu without lanthanum are lanthanides (meaning the ele-ments similar to lanthanum).

46.3) The ore (monazite or bastnasite) is dissolved in HCl and converted to a solution of LnCl3.Lanthanoids in this solution are separated by ion exchange as follows.



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Ion Exchange-first the solution is eluted onto the cation exchange column to replace the resin bound Na+

ions in a band at the top of the column3+ + 3+ +

(aq) (resin) (resin) (aq)Ln + 3Na Ln + 3Na

-then elute with a complexing agent such as EDTA4- and adjust pH

-the Ln3+

with the smallest radii are most strongly bound to the anionic ligand(EDTA4-

) andso are eluted first.3+ + 3 +

(resin) (aq) (aq) (aq) (resin)Ln + 3Na + [EDTA-H] Ln[EDTA-H] + 3Na

Note: Ln3+ = trivalent lanthanoid ion.

46.4) The solubility of Ln3+ ions in tri-n-butylphosphate, (BunO)3PO , solvent increases with

increase in atomic weight of Ln. Thus the separation of lanthanoids is achieved on largescale by extracting an aqueous solution containing mixture of nitrates of lanthanoids withthe solvent tri-n-butylphosphate.

47.1) B2H6 is a nido-borane as it has BnHn+4 formula.

47.2) [B6H

12] is an arachno-borane as it has B

nH

n+6formula.

48.1) MnF62- acts as a Lewis base, donating two F- to SbF

5molecules.

MnF62- + 2 SbF

5--------> 2 SbF

6- + MnF

4

Thus formed MnF4

lose F atoms which form F2

by combining.2 MnF4 --------> MnF3 + F2

48.2) Conductivity is increased due to formation of following ions

BrF3 + AgF BrF4- + Ag+ (AgF is acting as a base)

BrF3

+ SnF4

BrF2+ + SnF

5- (SnF

4is acting as an acid)

2BrF3

+ SnF4

2 BrF2+ + SnF

6-

48.3) AlF3

is a Lewis acid. It reacts with F- ion and thus by forming salt containing Na+ and AlF4-.

AlF3

+ NaF ----------> Na+ + AlF4-

When BF3

is added, it has stronger affinity for the F- ions and precipitates out AlF3

AlF4- + BF

3-----------> AlF

3+ BF

4-

(Which is stronger acid among AlF3

and BF3?)

48.4) SbF5

reacts with HF and increases the concentration of H+ ions.2SbF

5+ 2HF ------------> Sb

2F

11- + H

2F+

H2F+ -------------> HF + H+

This mixture can protonate methane and alkenes.

48.5) H3AsO

4< H

3SO

4< H

2SO

3< HMnO

4

Hints: 1) Greater the oxidation number of the central atom, greater is the acidic strength.2) Sulfur is a stronger non-metal than Arsenic.

48.6) Cl2

donates lone pair of electrons to Cl+ and forms [Cl3

]+ ion. Hence it acts as a Lewis base.



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Cl Cl + Cl+Cl Cl

+

Cl

49.1) e.c of Fluorine is (1s2)(2s2,2p5) shielding constant = s = (6 x 0.35) + (2 x 0.85) = 3.8

Z* = 9 - 3.8 = 5.2

49.2) From the electronic configuration (1s2)(2s2,2p6)(3s2,3p6) (3d10) (4s2,4p6) (4d10) (4f14) (5s2,5p6) (5d8) (6s2)

s = 60 x 1.00 + 16 x 0.85 + 1 x 0.35 = 3.95Z* = 78 - 73.95 = 4.15

49.3) The IE for lithium is lower for two reasons;1) The average distance from the nucleus for a 2s electron is greater than that of a 1selectron;

2) The 2s1

electron in lithium is shielded from the nuclear attraction (as well as repelled) bythe inner core electrons, so the valence electron is easily removed.

57.1) net dipole moment = 2(bond moment) x cos2

0

0.95 = 2(0.72) x cos

2

cos =0.65972

bond angle = = 97

57.2) The resonance hybrid of Azulene is contributed by stable zwitter ionic resonance forms(II &III). Hence it has dipole moment. Resonance forms (II & III) are aromatic as they containsix -electrons in both the rings and hence are stable.

-

+

IIIII

+ -

I

Whereas naphthalene cannot be stable in this type of zwitter ionic form with charge separa-tion. Only non polar neutral forms are aromatic and stable. Hence it is non polar.

Note: Remember, in writing resonace structures, the structure with charge separation isconsidered to be less stable. And more over, in case of Naphthalene, this form will notsatisfy Huckels rule and anti-aromatic and hence with least stability.

But in case of azulene, resonance form with charge separation becomes more important asthe aromatic nature out weighs this factor.



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C

2s

2pz

O

sp

sp

sp

2s

2pz

sp

2px

2py

2px 2py

2px

2py

2px

2py

CO

Energy



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C

2s

2pz

O

sp

sp

sp

2s

2pz

sp

2px

2py

2px 2py

2px

2py

2px

2py

CO

Energy


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Csir Ugc Jrf Net Chemistry Paper 1 (Part b) Series - 1