Detailed Solution Csir Net Dec 2015 Final

PART B (21) Using crystal field theory, identify from the

following complex ions that shows the same eff

(spin only) values

(A) [CoF6]3- (B) [IrCl6]3- (C) [Fe(H2O)6]2+

(1.) A and B (2.) B and C

(3.) A and C (4.) A, B and C

Soln (3)

All the complex are d6 and contain weak field ligand

but except Ir all form high spin complex

*[4d and 5d metals never form high spin complex]

Hence spin only magnetic moment will be same for

Co(III) and Fe(II)

(22) The order of increasing bronsted acidity for

boron hydrides is

(1.) B5H9 < B6H10 < B10H14

(2.) B10H14 < B5H9 < B6H10

(3.) B6H10 < B10H14 < B5H9

(4.) B10H14 < B6H10< B5H9

Soln (1)

Acidity of boron hydride depends on the size of

borane greater the sizes higher will be the acidity.

This is because the negative charge, formed upon

deprotonation, can be better delocalized over a large

anion with many boron atoms than over a small one.

(23) The molecule C3O2 has a linear structure. This

compound has

(1.) 4 and 4 bonds (2.) 3 and 2

bonds

(3.) 2 and 3 bonds (4.) 3 and 4

bonds

Soln (1)

Structure of carbon sub oxide C3O2

(24) Identify the complex ions in the sequential order

when ferroin is used as an indicator in the titration of

iron(II) with potassium dichromate. (phen = 1, 10-

phenathroline)

(1.) [Fe(phen)3]2+ and [Fe(phen)3]3+

(2.) [Fe(phen)3]3+ and [Fe(phen)3]2+

(3.) [Fe(CN)6]4- and [Fe(CN)6]3-

(4.) [Fe(CN)6]3- and [Fe(CN)6]4-

Soln (1)

The color changes observed with redox indicators at

the equivalence point is due to structural changes in

the indicator molecules. The redox indicators used

are organic molecules; they undergo structural

changes upon being oxidized or reduced. This can he

illustrated with ferroin indicator which is the iron(II)

complex of the organic compound 1,10-

phananthroline (Fig. 4.8}.

Each of the two nitrogen atoms in 1, 10-

phenanthroline has an unshared pair of electrons that

can be shared with the Fe(II) ion. Three such ligand

molecules get attached to one Fe(II) ion to form a

blood red complex ion. This complex can he

oxidized to the corresponding light blue

Fe(III) complex with the same structure. Therefore, a

sharp colour change from red to blue occurs on

oxidation of the complex.

[Ph3Fe]2+ [Ph3Fe]3+ + e- E0 = -1.06 V

The indicator is prepared by mixing equivalent

quantities of iron(II) sulphate and 1, 10-

phananthroline. The resulting Fe(II) complex

sulphate is called Ferroin. The Fe(III) complex

sulphate is called ferrin. The colour changes occur at

about -1.11 V (not at -1.06), since the color of ferroin

is so much more intense than that of ferrin.

(25) The ring size and the numbers of linked

tetrahedral present in [Si6O18]12- are, respectively,

(1.) 6 and 6 (2.) 12 and 6

(3.) 12 and 12 (4.) 6 and 12

Soln (1)

Cyclic metasilicates [SiO3)n]2n- having 3, 4, 6 or S

linked tetrahedra are known, though 3 and 6 are the

most common. These anions are shown

schematically in Fig. and are exemplied by the

mineral benitoite [BaTi{Si3O9}], the synthetic

compound [K4{Si4O8(OH)4}], and beryl

[Be3Al2{Si6O8}] and murite [Ba,0(Ca, Mn,

Ti]4{Si8O24}(Cl, OH, O)2].4H2O.

(26) The W-W bond order in [W(5-C5H5)(-

Cl)(CO)2]2 is

(1.) three (2.) two (3.) one (4.) zero

Soln (4)

To calculate bond order

(a) Calculate TVE : a

(b) Calculate n x 18 : b [n = number of metal]

(c) Calculate

2 = [number of M-M bond]

[W(5 C5H5)(y Cl)(CO)2]2

(a) TVE = 36 [12 + 10 + 6 + 8] = a

(b) n x 18 = 2 x 8 = 36 = b

(c)

2 = 3636

2 = 0 [Ans]

(27) The correct statement for Mn-O bond lengths in

[Mn(H2O)6]2+ is

(1.) All bonds are equal

(2.) Four bonds are longer than two others

(3.) Two bonds are longer than four others

(4.) They are shorter than the Mn-O bond in [MnO4]-

Soln (1)

Mn aqua complex do not allow any distortion

because of symmetrical d5 distribution.

All Mn-O bond are equal.

(28) Among the following, species expected to show

fluxional behavior are

(A) [NiCl4]2- (tetrahedral),

(B) IF7 (pentagonal bipyramidal),

(C) [CoF6]3- (octahedral),

(D) Fe(CO)5 (trigonal, bipyramidal)

(1.) B and C (2.) B and D

(3.) C and D (4.) A and D

Soln (2)

Fe(CO)5 show fluxional behavior by process called

berry pseudo rotation.

And molecule IF7 shows similar mechanism called

Bartell mechanism which exchanges the axial atoms

with one pair of the equatorial atoms with an energy

requirement of about 2.7 kcal/mol

Fig : Mechanism of Berry Pseudorotation

(29) The oxidizing power of [CrO4]2-, [MnO4]2-, and

[FeO4]2- follows the order

(1.) [CrO4]2- < [MnO4]2- < [FeO4]2-

(2.) [FeO4]2- < [MnO4]2- < [CrO4]2-

(3.) [MnO4]2- < [FeO4]2- < [CrO4]2-

(4.) [CrO4]2- < [FeO4]2- < [MnO4]2-

Soln (1)

Since oxidation number is same (VI) in all three

metals. Hence, in going from Cr to Mn to Fe, size

decrease due to which oxidizing power increase.

(31) The structures of XeF2 and XeO2F2 respectively

are

(1.) bent, tetrahedral (2.) linear, square planar

(3.) linear, see-saw (4.) bent, see-saw

Soln (3)

XeF2 : linear molecule in vapor state

Structure of XeF2

The three 5p electrons are promoted to higher energy

5d sub-level.

The 5s three 5p and one 5d orbitals hybridize to give

ve sp3d hybrid orbitals. The four singly occupied

orbitals are used for bond formation to two uorine

and two oxygen atoms. The fth hybrid orbital

contains the lone pair. The other 5d electrons of

xenon which do not take part in the hybridization

scheme are involved in -bond formation to two

oxygen atoms. The structure of XeO2F2 is

represented in Fig.

(32)The biological functions of the cytochrome P450

and myoglobin are, respectively

(1.) Oxidation of alkene and O2 storage

(2.) O2 transport and O2 storage

(3.) O2 storage and electron carrier

(4.) Electron carrier and O2 transport

Soln (1)

The biological function of cytochrome P450 is to act

in a mono oxygenase reaction.e.g insertion of one

atom of oxygen into aliphatic position of organic

substrates.

RH +O2 R-O-H

The function of Mb is more subtle than Hb. Besides

being a simple repository for dioxygen, it also serve

as dioxygen reserve against which organism can

draw during increased metabolism (O2 deprivation).

(Reference : Inorganic Chemistry , Huheey)

(33) Spin motion of which of the following gives

magnetic moment

(A) Electron (B) Proton (C) Neutron

Correct answer is


(3.) A and C (4.) A, B and C

Soln (4)

Magnetic moment is induced by the spin of

elementary particles (electron and quarks in the

proton and neutron of atomic nuclei).

* Neutron is electrically neutral but a non-zero

magnetic moment because of its internal quark

structure.

(Reference : P.W. Atkins, Quanta: A handbook of

concepts. Oxford University)

(34) The metallic radii are abnormally high for which

of the following pairs?

(1.) Eu, Yb (2.) Sm, Tm

(3.) Gd, Lu (4.) Nd, Ho

Soln (1)

The lanthanide and actinide elements are all metallic

and are formally members of Group 3 of the Periodic

Table. They all form 3+ ions in their compounds and

in aqueous solution, with few exceptions. The

actinide elements form a larger range of oxidation

states than the lanthanides, but the 3+ ions are used

in this section for comparison purposes. Figure

shows the lanthanides metallic and 3+ ionic radii.

There is an almost regular decrease in metallic radius

along the lanthanide series, with discontinuities at Eu

and Yb. Most of the Lanthanide metals contribute

three electrons to bands of molecular orbitals

constructed from the 5d and 6s atomic orbitals, their

other valency shell

electrons remaining as 4fn congurations that do not

interact with those on neighboring atoms. In the

cases of Eu and Yb, the atoms only contribute two

electrons to the 5d/6s bands and retain the

congurations 4f7and 4F14, respectively, and so

maximize their exchange energies of stabilization.

The contributions of only two electrons each to the

bonding bands makes the Eu and Yb metallic radii

larger than those of their neighbors that contribute

three electrons.

The 3+ ions have regular 4f orbital llings, and there

is an almost linear decrease in their ionic radii along

the series. This is known as the Lanthanide

contraction, and has consequences for the transition

elements

(35) Correct statement for coulometry is

(1.) it is based on faradays law of electrolysis

(2.) it is a type of voltammetry

(3.) it is based on Ohms law

(4.) it uses ion selective electrode

Soln (1)

Coulometry is an analytical in which we measure

amount of amount of water transferred during an

electrolysis reaction by measuring the amount of

electricity consumed/produced.

(36) Deoxy Hemocyanin is

(1.) heme protein and paramagnetic

(2.) Colorless and diamagnetic

(3.) O2 transport and paramagnetic

(4.) Blue colored and diamagnetic

Soln (2)

Hemocyanin is blue pigment which contains Cu in (I)

oxidation state.

Since it is in deoxy form it is colorless and also

diamagnetic (d10) configuration.

(37) The major product formed in the following

reaction is

Soln (2)

Step 1 : Removal of nitrogen to form carbene

Step 2 : attack of Oxygen lone pair to form 5

membered cyclic adduct

Step 3 : 1,2 alkyl shift to form rearranged product

(38) In the mass spectrum of 1, 2-dichloroeth-ane,

approximate ratio of peaks at m/z values 98, 100, 102

will be

(1.) 3 : 1 : 1 (2.) 9 : 6 : 1

(3.) 1 : 1 :2 (4.) 1 : 2 : 1

Soln (2)

When chlorine is present, the M + 2 peak becomes

very significant. The heavy isotope these elements is

two mass units heavier than the lighter isotope. The

natural abundance of 37Cl is 32.5% that of 35Cl.

When Chlorine is present, the M + 2 peak becomes

quite intense. If a compound contains two chlorine

atoms, a distinct M + 4 peak, as well as an intense M

+ 2 peak, should be observed.

The M : M+2 : M+4 peaks for two chlorine atom is

in the ratio of 100 : 65.3 : 10.6 which on

simplification gives 9 : 6 : 1

(39) The correct statement about the following

reaction is that

(1.) A = and the reaction proceeds

through carbine intermediate


through nitrene intermediate


through Norrish type II path

(4.) A = and the reaction proceeds through

Norrish type I path

Soln (1)

This reaction is an example of wolff rearrangement

in which cyclic diazoketone , on rearrangement leads

to ring contraction.

The R group migrates with retention of configuration

and the resulting ketene can be trapped by methanol

to give the corresponding ester.


reaction is

Soln (3)

Step 1 : The rst step is an acetal exchange in which

the allylic alcohol displaces methanol.

Step 2 : . A second molecule of methanol is now lost

in an acid-catalysed elimination reaction to give the

vinyl group.

Step 3 : It is simple example of [3,3] sigmatropic

known as Claisen rearrangement

(41) The major products A and B in the following

reactions are

Soln (3)

Heating dimethylmenthyl amine oxide gave a

mixture of 2- and 3menthene, whereas isomeric neo

menthylamine oxide gave only 2menthene. This is

an example of pyrolytic syn elimination

(42) The number of Chemical shift non equivalent

protons expected 1H NMR spectrum of -pinene is

-pinene

1. 7 2. 8 3. 9 4. 10

Soln (4)

All the protons are different and hence give 10

distinct signals in 1H NMR.

(43) Correctly matched structure and carbonyl

stretching frequency set is

Column A Column B

P.

X. 1750 cm-1

Q.

Y. 1770 cm-1

R.

Z. 1800 cm-1

(1.) P-Y, Q-Z, R-X (2.) P-Y, Q-X, R-Z

(3.) P-Z, Q-Y, R-X (4.) P-X, Q-Z, R-Y

Soln (1)

-lactone has very peculiar IR stretching frequency at

around ~ 1770 cm-1.

Next, Conjugation lowers IR stretching frequency.

Therefore structure R must correspond to ~ 1750 cm-

1.

Hence , the correct option is (1)

(44) the major product formed by photochemical

reaction of (2E, 4Z, 6E)-decatriene is

Soln (1)

This is an example of electrocyclic ring closing

reaction of 4n + 2 electron system by Conrotatory

photochemical mode.


reaction is

Soln (3)

This is an example of favorskii rearrangement of

-dibromoketone


reaction is

Soln (2)

The formation of eight-membered rings by radical

cyclization has been found to occur using -acyl

radical species. The reaction occurs at a rate that is

faster remarkably even than 5-exo-trig cyclization.

Hence, the eight-membered lactone 92 was formed,

rather than a ve-, six- or seven-membered lactone,

on treatment

Of the bromide with tributyltin hydride

(47) D-Mannose upon fluxing in acetone with CuSO4

and H2SO4 gives

D-mannose

Soln (3)

This is an example of protection of 1,2 diol by

acetone in the presence of acid. For the protection of

1,2 diol the two hydroxyl group must be in cis-

orientation. Hence only anomeric OH group remain

unaffected rest all OH will form 5-membered cyclic

acetal as thermodynamic product.

(48) The structure of the compound that matches the 1H NMR data given below is 1H NMR (DMSO-D6): 7.75 (dd, J = 8.8, 2.4 Hz,

1H), 7.58 (d J = 2.4 Hz, 1H), 6.70 (d J = 8.8 Hz, 1H),

6.50 (broad s, 2H), 3.80 (s, 3H).

Soln (2)

Steps to Analyze NMR Spectra

a. First we know that delta value of aromatic protons

comes at 7.2 . So the first delta value 7.75 indicates

the proton is deshielded . so it must be adjacent to

EWG. Further since it is doublet of doublet hence

there must be meta and ortho coupling based on their

J values.

So 1 and 4th option are straight away ruled out since

the above condition is not followed.

Further in 3rd , all the three aromatic protons will give

doublet of doublet which is clearly not the case.


reaction is

Soln (1)

This is an example of intramolecular Diels alder

reaction.


reaction is

Soln (4)


reaction is

Soln (4)

This is an example of Swern Oxidation in which last

step is the intermolecular SN2 displacement by

chloride ion in the absence of base

(52) Identify two enantiomers among the following

compounds.

(1.) A and B (2.) A and C

(3.) B and D (4.) C and D

Soln (4)

A and B are homomers ; A and C are diastereomers

B and D are diastereomers ; C and D are enantiomer

(53) The correct expression for the product

((). ()) [Mn and Mw are the number-average

and weight average molar masses, respectively, of a

polymer] is

(1.) N-1 i Ni Mi (2.) N-1 i Ni Mi2

(3.) N / NM (4.) N / NM2

Soln (2)

Number average and weight average molar mass is

given by the expression

=

=2

x

=

x 2

= 12

(54) The concentration of a reactant K varies with

time for two different reactions as shown in the

following plots:

The order of these two reactions I and II,

respectively, are

(1.) zero and one (2.) one and zero

(3.) zero and two (4.) two and zero

Soln (3)

For the zero order reaction

x = kT

(a x) = -RT (linear with negative slope)

For 2nd order reaction 1

[]=

1

[0]+ (linear with intercept

1

[0]

(55) .) sp hybrid orbitals are of the form C12s + C22pz

(2s and 2pz are normalized individually). The

coefficients of the normalized form of the above sp

hybrid orbitals are

(1.) C1 = 1

2, C2 =

1

2 (2.) C1 =

1

2, C2 =

1

2

(3.) C1 = 1

2, C2 =

1

2 (4.) C1 =

1

2, C2 =

1

2

Soln (1)

Condition for normalization = 1

= C12s + C22pz

12(2)2 + 2

2(2)2 + 212 2 2 = 1

The last two terms cancel out each other because

they are orthogonal

Also |2|2 = 1 |2|

2= 1 [condition of

normalization]

C12 + C22 = 1

Since Sp orbital have equal contribution toward

atomic orbital C1 = C2

2C12 = 1 C1 = 1

2

C2 = 1

2

(56) For a single cubic crystal lattice, the angle

between the [2 0 1] plane and the xy plane is

(1.) less than 300 (2.) between 300 and

450

(3.) between 450 and 600 (4.) greater than 600

Soln (4)

The expression for finding the angle between the two

plane is given by

=12 + 12 + 12

12 + 1

2 + 12 . 2

2 + 22 + 2

2

For the XY Plane , miller indices are [001]

1 = 2 1 = 0 1 = 1 ; 2 = 0 , 2 = 0 , 2 = 1

On solving

=2.0 + 0.0 + 1.1

22 + 02 + 12 . 02 + 02 + 12

=1

5=

1

2.23= 0.44

Therefore = cos-1 (0.44) = 63.89

(57) The concentration of a MgSO4 solution having

the same ionic strength as that of a 0.1 M Na2SO4

solution is

(1.) 0.05 M (2.) 0.067 M

(3.) 0.075 M (4.) 0.133 M

Soln (3)

Ionic strength of 0.1M Na2SO4

=1

2

2

=1

2[0.2 x 12 + 0.1 x 22]

= 0.3

MgSO4 will have ionic strength equal to 0.3

0.3 = 1

2[m x 22 + m x 22]

0.3 = 4m

m = 0.3/4 = 0.075

(58) H of a reaction is equal to slope of the plot of

(1.) G versus (1/T) (2.) G versus T

(3.) (G/T) versus T (4.) (G/T) versus (1/T)

Soln (1)

H = U + PV .. (1st Law of thermodynamics)

H = U + ngRT

U - H = -ngRT

Fe2O3(s) + 3C(s) 2Fe(s) + 3(CO)

Ans = -(3-0)= -3RT

(59) Two different non-zero operators and (A

B) satisfy the relation

(A + B) (A - B) = A2 - B2, when

(1.) AB = A2 and BA = B2

(2.) AB + BA = 0

(3.) A and B are arbitrary

(4.) AB - BA = 0

Soln (4)

Expanding and simplifying the expression gives the

option (4) as answer

(A + B) (A B) = A2 AB + BA B2

= A2 B2 (only when AB BA = 0)

(60) For the following reaction,

[]

is given by

(1.)1[] 1[]2 22[]

(2.) 21[] 21[]2 2[]

(3.) 1

21[]

1

21[]

2 2[]

(4.) 21[] 21[]1 2 2[]

Soln (2)

Rate of formation of B

is given by

1

2

= 1[] 1[]

2(1)

[]

= 21[] 21[]

2(2)

Also

= 2[]

overall rate is given by

= 22[] 21[]

2 1[]

(61) The standard deviation of speed (c) for

Maxwells distribution satisfies the relation

(1.) c T (2.) c

(3.) c 1/T (4.) c 1/

Soln (2)

Standard deviation for the velocity of Maxwell

distribution is given by

= 3

(62) In Kohlrausch law m = 0m - , 0m and

(1.) depends only on stoichiometry

(2.) depends only on specific identity of the

electrolyte

(3.) are independent on specific identity of the

electrolyte

(4.) are mainly dependent on specific identity of the

electrolyte and stoichiometry, respectively

Soln (1)

The justification of Kohlrauschs law on theoretical

grounds cannot be obtained within the framework of

a macroscopic description of conduction. It requires

an intimate view of ions in motion. A clue to the type

of theory required emerges from the empirical

findings by Kohlrausch: (l) the c1/2 dependence and

(2) the intercepts 0 and slopes A of the versus c1/2

curves depend not so much on the particular

electrolyte (whether it is KCI or NaCl) as on the type

of electrolyte (whether it is a l:l or 2:2 electrolyte)

(63) If the reduced mass of the diatomic molecule is

doubled without changing its force constant, the

vibrational frequency of the molecule will be

(1.) 2 times the original frequency

(2.) 2 times the original frequency

(3.) twice the original frequency

(4.) unchanged

Soln (2)

Vibrational frequency of molecule is given by

=1

2

1

if reduced mass is doubled, then will become 1

2

times the original frequency.

(64) The degeneracy of an excited state of a particle

in 3-dimensional cubic box with energy three times

its ground state energy is

(1.) 3 (2.) 2 (3.) 1 (4.)

4

Soln (1)

For particle in 3D box

=2

22[12 + 2

2 + 32]

In the ground state n1 = n2 = n3 = 1

=32

22

Energy when it wil be 3 times to its ground state will

be 3 x32

22 =

92

22

It is possible when m1 = n2 = 2, n3 = 1

Total degeneracy is 3 (221), (122) and (212)

(65) If the pressure p system is greater than the p

(surroundings), then

(1.) work is done on the system by the surroundings.

(2.) work is done on the surroundings by the system.

(3.) work done on the system by the surroundings is

equal to the work done on the surroundings by the

system.

(4.) internal energy of the system increases

Soln (2)

When the pressure of the system is greater than

pressure of surrounding, expansion will take place.

Hence, during expansion work is done by the system

on the surrounding

(66) H of a reaction is equal to slope of the plot of

(1.) G versus (1/T) (2.) G versus T

(3.) (G/T) versus T (4.) (G/T) versus (1/T)

Soln (4)

G = H TS

Dividing the whole equation be T brought out

=

Equation of straight line

y = mx + c

y =

x =

1

Slope = H

Intercept = -S

(67) The correct statement among the following is

(1.) N2 has higher bond other than N2+ and hence has

larger bond length compared to N2+

(2.) N2+ has higher bond other than N2 and hence has

larger bond length compared to N2

(3.) N2 has higher bond other than N2+ and hence has

higher dissociation energy compared to N2+

(4.) N2 has lower bond other than N2+ and hence has

lower dissociation energy compared to N2+ energy

Soln (3)

N2 = KK (2S)2 *(2S)2 (2px)2 (2py)2 6(2pz)2

B.O = 82

2 = 3 (most stable, diamagnetic)

N2+ = KK 22 2

2 22 2

2 621

B.O = 72

2 = 2.5 (less stable, paramagnetic)

(68) The correct form for the simple Langmuir

isotherm is

(1.) = Kp (2.) = (Kp)1/2

(3.) = Kp / (1 + Kp) (4.) = (1 + Kp) / Kp

Soln (3)

The concentration of vacant sites on the surface of an

adsorbent will be directly proportional to the fraction

of the surface that remains uncovered and hence will

be directly proportional to the factor (1 - ). Thus we

can write,

Rate of absorption p(l )

i.e. Rate of adsorption = ka p(l - )

Now since at equilibrium

Rate of absorption = rate of desorption,

it follows that

ka (1 - )p = kd

i.e. =

+=

( )

1+( )=

1

1+1

(69) The correct match of the compound in column A

with the description in column B is

Column A Column B

P.

X.

Oil of

Wintergreen

Q.

Y.

Aspirin

R.

Z.

Ibuprofen

(1.) P-Y, Q-Z, R-X (2.) P-Z, Q-X, R-Y

(3.) P-Z, Q-Y, R-X (4.) P-X, Q-Z, R-Y

Soln (2)

Aspirin : acetyl salicylic acid

Ibuprofen : ISO butyl phenyl propanoic acid

Oil of winter green : methyl salicylate

(70) The formation constant for the complexation of

M+ (M = Li, Na, K and Cs) with cryptant, C222

follows the order

(1.) Li+ < Cs+ < Na+ < K+

(2.) Li+ < Na+ < K+ < Cs+

(3.) K+ < Cs+ < Li+ < Na+

(4.) Cs+ < K+ < Li+ < Na+

Soln (2)

The ability of cryptant to trap an alkali metal cation

depends on size of both cage and metal ion, the better

the math between these; the more effectively the ions

can be trapped.

PART C

(71) Identify radioactive capture from the following

nuclear reactions

(1.) 9Be(, n) 8Be (2.) 23Na(n, ) 24Na

(3.) 63Cu (p, p 3n 9) 24Na (4.) 107Ag (n, n) 107Ag

Soln (2)

Radioactive capture reactions are described as

follows Parent Nucleus, Projectile Daughter

nucleus, ejectile 23Na11 + 0n6 11Na24 + 0n6 (y rays)

Rest all examples do not follow this trend.

(72) Match the metalloproteins in column A with its

biological function and metal center in column B

Column A Column B

(a.) Hemoglobin (i.) Electron carrier

and iron

(b.) Cytochrome b (ii.) Electron carrier

and copper

(c.) Vitamin B12 (iii.) O2 transport and

copper

(d.) Hemocyanin (iv.) Group transfer

reactions and

cobalt

(v.) O2 storage and

cobalt

(vi.) O2 transport and

iron

The correct match is

(1.) (a.)-(vi); (b.)-(i); (c.)-(iv) and (d.)-(iii)

(2.) (a.)-(v); (b.)-(i); (c.)-(iv) and (d.)-(iii)

(3.) (a.)-(iv); (b.)-(v); (c.)-(i) and (d.)-(ii)

(4.) (b.)-(v); (b.)-(vi); (c.)-(ii) and (d.)-(iv

Soln (1) Self explanatory

(73) [(3-C3H5)Mn(CO)4] shows fluxional behaviour.

The 1H NMR spectrum of this compound when it is

in the non-fluxional state shows

(1.) one signal

(2.) two signals in the intensity ratio of 4 : 1

(3.) three signals in the intensity ratio of 2 : 2 : 1

(4.) five signals of equal intensity

Soln (3)

Non fluxional behavior can be observed if NMR is

carried out at low temperature. For this complex, 3

signals are observed

one for H3

two for H1 and H4

third for H2 and H5

In the intensity ratio 2 : 2 : 1

(74) Reaction of [Mn2(CO)10] with I2 results in A

without loss of CO. Compound A, on heating to

1200C loses a CO ligand to give B, which does not

have a Mn-Mn bond. Compound B reacts with

pyridine to give 2 equivalents of C. Compounds A, B

and C from the following respectively, are

(1.) II, V and IV (2.) II, III and IV

(3.) V, III and IV (4.) II, V and III

Soln (1)

(a) Dimanganese decacarbonyl reacts with halogens

to form carbonyl halides.

Mn2CO10 + l2 2Mn(Co)5 I

(75) The final product of the reaction of carbonyl

metalates [V(CO)6]- and [Co(CO)4]- with H3PO4,

respectively, are

(1.) V(CO)6 and HCo(CO)4

(2.) HV(CO)6 and Co2(CO)8

(3.) [H2V(CO)6]+ and HCo(CO)4

(4.) V(CO)6 and Co2(CO)8

Soln (4)

[V(CO)6]- + H3PO4 [H V(CO)6] + V(CO)6

2[Co(CO)4]- 2+ 2[HCo(CO)4]

2 Co2(CO)8

(76) For, uranocene, the correct statement(s) is/are:

(1.) Oxidation use of uranium is +4.

(2.) it has cyclocotatetraenide ligands

(3.) it is a bend sandwich compound

(4.) it has -2 charge.

Correct answer is


(3.) A and D (4.) B only

Soln (1)

Uranium compound was the 1st metallocene of Cot2-

to be synthesized

UCl2- + 2Cot2- U(Cot)2

From the given reaction, it is concluded that among

the following option (A) and (B) are correct.

(77) The number of lone pair(s) of electrons on the

central atom in [BrF4]-, XeF6 and [SbCl6]3- are,

respectively,

(1.) 2, 0 and 1 (2.) 1, 0 and 0

(3.) 2, 1 and 1 (4.) 2, 1 and 0

Soln (3)

Formula to calculate Bond Pair and Lone Pair

A. calculate TVE (total valence electron)

B. Divide TVE by 8 if 8 < TVE < 56

C. The quotient will give number of bond pair

D. while dividing the remainder by 2 will give lone

pair on central atom

BrF4- : TVE = 5 x 7 +1 = 36

8 = 4 +

4

2 = 4 + 2 (lp)

XeF6- : 4 + 42 = 50

8 = 6 +

2

2 = 6 + 1 (lp)

SbCl63- : 5 + 6 x 7 + 3 = 50

8 = 6 +

2

2 = 6 + 1 (lp)

(78) Consider the following reaction:

N3P3Cl6 + 6 HNMe2 N3P3Cl3(NMe2)3 +

3Me2NH.HCl

[A]

The number of possible isomers for [A] is

(1.) 4 (2.) 3 (3.) 2 (4.)

5

Soln (2)

The cyclophosphazenes, first discovered by Liebig in

1834, have aroused great interest and, especially

within the last twenty years, have been intensively

investigated. Much has been discovered about their

chemical and physical properties.

Two opposite points of view were put forward, one

based on a completely delocalized molecular orbital

covering the whole of the cycle , the other based on

the three-centered P-N-P islands with little or no

interaction between them.

(79) Three electronic transitions at 14900, 22700 and

34400 cm-1 are observed in the absorption spectrum

of [CrF6]3- and the corresponding transition are

(1.) 7800 and 4A2g 4T2g

(2.) 14900 and 4A2g 4T2g

(3.) 14900 and 4T2g 4T1g(F)

(4.) 7800 and 4T2g 4T1g(F)

Soln (2)

C6H63-: d3 electronic configuration orgel diagram for

d3 electronic transition are-

4A2g 4A2g corresponds to the transition of lowest

energy.

(80) The calibration curve in spectrofluorimetric

analysis becomes non-linear when

(1.) Molecular weight of analyte is high

(2.) Intensity of light source is high

(3.) Concentration of analyte is high

(4.) Molar absorptivity of analyte is high

Soln (3)

According to Lambert-Beer law

A = C. l

When A is plotted v/s conc. Curve should nearly be

of straight line. However, only limitation is that conc.

Of analytes should be low

(81) The reaction of BCl3 awith NH4Cl gives product

A which upon reduction by NaBH4 gives product B.

product B upon reacting with HCl affords compound

C, which is

(1.) Cl3B3N3H9 (2.) [ClBNH]3

(3.) [HBNH]3 (4.) (ClH)3B3N3(ClH)3

Soln (1)

Ammonium chloride reacts with Boron chloride to

produce trichloridoborazine and hydrogen chloride.

This reaction takes place at temperature 140 150 0C. catalyst C6H6, C6H5Cl.

(82) The approximate positions of vco band (cm-1) in

the solid-state infrared spectrum and the Fe-Fe bond

order in [Fe(5 - C5H5)( CO)(CO)]2 (non-centro-

symmetric) respectively, are

(1.) (2020, 1980, 1800) and one

(2.) (2020, 1980, 1800) and two

(3.) (2020, 1980) and one

(4.) (2143) and one

Soln (1)

Three bands are expected in CO in the complex

[Fe(5-C5H5)(-Co)(CO)]2 since it is non Centro-

symmetric.

One band for bridging CO around 1800

Two bands (one symm, one asymm) around 2020

(asymm) and 1980 (symm)

To calculate bond order

TV E = 34

nx18 36 (there are 2 Fe atoms)

M.M bond = 36 34

2 = 1

(83) Protonated form of ZSM-5 catalyzes the

reaction of ethene with benzene to produce

ethylbenzene. The correct statement for this catalyst

process is

(1.) alkyl carbocation is formed

(2.) carbanion is formed

(3.) benzene is converted to (C6H5)+ group

(4.) vinyl radical is formed

Soln (1)

ZSM-5 is an acid catalyst and is typically used for

Alkylation of aromatic He

(84) [MnO4]- is deep purple in color whereas [ReO4]

is colorless. This is due to greater energy require for

(1.) d-d transition in the Re compound compared to

the Mn compound

(2.) d-d transition in the Mn compound compared to

the Re compound

(3.) charge transfer from O to Re compared to O to

Mn

(4.) charge transfer from O to Mn compared to O to

Re

Soln (3)

The 5d orbitals of Re are higher in energy than the 3d

orbitals of Mn, so an LMCT excitation requires more

energy for ReO-. In addition, since the molecular

orbitals derived primarily from the 3d orbitals of

MnO4- are lower in energy than the corresponding

MOs of ReO4-, MnO4- is better able to accept

electrons; it is a better oxidizing agent.

(85) The resonance Raman stretching frequency (vo-o,

in cm-1) of O2 is 1580. The vo-o for O2 in bound oxy-

hemoglobin is close to

(1.) 1600 (2.) 1900

(3.) 800 (4.) 1100

Soln (4)

It is known fact that stretching frequency of any

molecule in the Free State is always higher that the

same in bound state

Here it is given that o-o is 1580 of free O2. Hence in

bound Hb it must be less than that, so correct answer

is 1100 (more precisely 1180 cm-1)

(86) Pick the correct statements about Atomic

Absorption Spectrometry (AAS) from the following

(1.) Hg lamp is not suitable source for AAS

(2.) Graphite furnace is the best atomizer for AAS

(3.) Non-metals cannot be determined with AAS

(4.) AAS is better than ICP-AES for simultaneous

determination of metal ions

Correct answer is

(1.) A, B and C (2.) B, C and D

(3.) C, D and A (4.) D, A and B

Soln (1)

Common source used in AAS is HCl (Hallow

Cathode Lamp) and EDL (Electrodes Discharge

Lamp)

Best atomizer is graphite tube

Since it is based on the principle of flame

detection, generally non metal cannot be detected

ICP-AES is better technique that trivial AAS.

(87) Aqueous Cr2+ effects one electron reduction of

[Co(NH3)5Cl]2+ giving compound Y. compound Y

undergoes rapid hydrolysis. Y is,

(1.) [Co(NH3)5]2+ (2.)

[Co(NH3)5(OH)]+

(3.) [Co(NH3)4(OH)2] (4.) [Cr(H2O)5Cl]2+

Soln (1)

The reduction of cobalt(III) (in [Co(NH3)5Cl]2+) by

chromium(II) (in [Cr(H2O)6]2+) and was specically

chosen because (I) both CO(III) and Cr(III) form

inert complexes and (2) the complexes of Co(II) and

Cr(III) are labile.

Under these circumstances the chlorine atom, while

remaining rmly attached to the inert Co(III) ion, can

displace a water molecule from the labile Cr(III)

complex to form a bridged intermediate:

[Co(NH3)3Cl]2+ + [Cr(H2O)6]2+ [(H3N)5Co-Cl-

Cr(OH2)5]4+ + H2O

The redox reaction now takes place within this

dinuclear complex with formation of reduced Co(II)

and oxidized Cr(III). The latter species forms an inert

chloroaqua complex, but the cobalt(II) is labile, so

the intermediate dissociates with the chlorine atom

remaining with the chromium:

[(H3N)5Co-Cl-Cr(OH2)5]4+ [(H3N)5Co]2+ +

[ClCr(OH2)5]2+

(88) Which of the following statements are TRUE for

the lanthanides?

(1.) The observed magnetic moment of Eu3+ at room

temperature is higher than that calculated from spin-

orbit coupling

(2.) Lanthanide oxides are predominantly acidic in

nature

(3.) The stability of Sm(II) is due to its half-filled

sub-shell

(4.) Lanthanide(III) ions can be separated by ion

exchange chromatography

Correct answer is

(1.) A and D (2.) A and B

(3.) A and C (4.) B and C

Soln (1)

Oxides of lanthanides are generally basic in nature.

Sm: [Xe]6s24f6

Sm: [Xe]6s24f6 Sm(II) is not stable because of half-

filled subshell

Rest all statement are correct above Lanthanides (for

detailed description on above statement of

lanthanides, refer, inorganic chemistry, Shriver &

Atkins)

(89) The correct statement about the substitution

reaction of [Co(CN)5(Cl)]3- with OH- to give

[Co(CN)5(OH)]3- is,

(1.) it obeys first order kinetics

(2.) its rate is proportional to the concentration of

both the reactants

(3.) it follows the SN1CB mechanism

(4.) its rate is dependent only on the concentration of

[OH]-

Soln (1)

Complexes such as [Co(CN)5Cl]3- [Co(py)4Cl2]+

would not be expected typical base hydrolysis and

reaction proceeds slowly without dependence of OH-

ion.

(Reference : Inorganic Chemistry , Huheey)

(90) Mossbauer spectrum of a metal complex gives

information about

(A) oxidation state and spin state of metal

(B) types of ligands coordinated of metal

(C) nuclear spin state of metal

(D) geometry of metal

Correct answer is

(1.) A and C (2.) B and C

(3.) A, B and D (4.) B and D

Soln (3)

Mssbauer spectroscopy is unique in its sensitivity to

subtle changes in the chemical environment of the

nucleus including oxidation state changes, the effect

of different ligands on a particular atom, and the

magnetic environment of the sample.

(91) Using wades rules predict the structure type of

[C2B5H7].

(1.) nido (2.) closo

(3.) arachno (4.) hypho

Soln (2)

Wades rule

1 carbon is equal to BH fragment as both are isolobal

C2B5H11 = B7H9 B7H72- (hence closo)

(92) Among the following complexes

A. [Co(ox)3]3-, B. trans-[CoCl2(en)2]+, C.

[Cr(EDTA)]- the chiral one(s) is/are

(1.) A and B (2.) C and B

(3.) C only (4.) A and C

Soln (4)

The essential condition to show the chirality is the

lack of any symmetry element and it show have

handedness

Chiral complex is able to show handedness

properly (right or left handedness) by and

designation

However the complex trans-[CoCl2(en)2]+

contains plane of symmetry and hence are achiral

(93) Choose the correct statements about Tanabe-

Sugano diagrams:

(1.) E/B is plotted against 0/B.

(2.) The zero energy is taken as that of the lowest

term.

(3.) Terms of the same symmetry cross each other.

(4.) Two terms of the same symmetry upon increase

of ligand field strength bend apart from each other.

Correct answer is

(1.) A and B (2.) A and C

(3.) A, B and D (4.) A, B, C and D

Soln (3)

Self-explanatory.

Reference: Inorganic Chemistry, James Huheey

(94) The intermediate and final major product of

photolysis of Z

From the following:

Are

(1.) A and D (2.) B and D

(3.) B and C (4.) A and C

Soln (2)

In the following photolysis reaction, first the removal

of CO takes place to form 16e- complex and then the

change in hapticity from 1 to 3. Hence correct

option is (2)

(95) The number of valence electrons provided by

[Ru(CO)3] fragment towards cluster bonding is

(1.) 1 (2.) 14 (3.) 6 (4.)

2

Soln (4)

For transition metal, number of electrons available

per building block for cluster bonding

L = d + y2 12

d = number of valence shell electron

y = number of ligands

L = number of e- contributed by each ligand

RuCO3

Ru(d8) = 8 + 6 12

= 2


reaction sequence are

Soln (1)

A. The first step is the synthesis of 9-

Borabicyclo[3.3.1] nonane

B. Second is the Boronic Suzuki coupling which is

highly stereospecific in which geometry of the

reactant is faithfully reproduced in the product.

(97) The major product A and B in the following


Soln (3)

Mechanism :



Soln (3)

Mechanism:


reaction is

Soln (2)

It is an example of light induced photo-isomerization

of benzene via Dewar benzene intermediate

(100) The correct reagent combination to effect the

following transformation is

(1.) A = NaBH4, BF3.OEt2; B = MeMgBr (2.5

equiv.), THF then H3O+

(2.) A = BH3.THF; B = MeLi (2.5 equiv.), THF then

H3O+

(3.) A = BH3.THF; B = (i) aq. NaOH then H3O+ ,

(ii) MeLi (2.5 equiv.), THF then H3O+

(4.) A = (i) Me3Al, MeNHOMe, (ii) MeMgBr, THF

then H3O+; B = LiAlH4, THF

Soln (3)

Method of Analysis :

A. NaBH4 cannot be the reagent of choice since it

generally do not reduces amides.

B. reagent 2 and 4 results in the formation of tertiary

alchohol as resulting ketone is more reactive than

ester itself.

C. So the proper strategy should be (i) first reduce

the amide with borane (ii) then , convert ester to acid

which is usually less reactive and then perform

alkylation which will stop at ketone oxidation state

only.



Soln (2)

This is an example of Mannich reaction which will

take place prefentially at 2 and 5 position.



Soln (3)

a. First step is the metathesis ring closing reaction by

grubbs catalyst.

b. Second is the reduction of amide be lithium

aluminium hydride.

(103) A concerted [1,3]-sigmatropic rearrangement

took place in the reaction shown below. The structure

of the resulting product is

Soln (3)

This is an example of 1,3 carbon shift and according

to woodward Hoffman rule it should be inversion.


reaction is

Soln (3)

The ease of the anionic oxy-Cope rearrangement and

its high level of stereo-control make this reaction a

popular and valuable synthetic method. For example,

stereo-controlled rearrangement of the potassium salt

of the 3-hydroxy-1, 5-diene to give the

cyclohexanone. The diastereoselectivity across the

new carbon-carbon single bond reects the

preference for a chair-shaped transition state with the

methoxy group in the pseudo equatorial position.



Soln (1)

The diol 1 must first be protected/activated and this

can be achieved by selective acylation (or

mesylation) of the less hindered primary alcohol

group. For example, treatment with a carboxylic acid,

dicyclohexylcarbodiimide (DCC) and 4-

dimethylaminopyridine gives the required ester.

Oxidation of the secondary alcohol can be achieved

by using one of a number of oxidizing agents such as

pyridinium dichromate (PDC). Formation of the

ketone was followed by _-elimination on alumina

to give the unsaturated ketone


reaction sequence is

Soln (4)

Boronic esters RB(OR')2 react with

methoxy(phenylthio)methyllithium LiCH(OMe)SPh

to give salts, which, after treatment with HgCl2 and

then H2O2, yield aldehydes. This synthesis has been

made enantioselective, with high ee values (> 99%),

by the use of an optically pure boronic ester.

(107) The major product of the following reaction is

Soln (1)

This is an example of acid-catalyzed diels Alder

reaction



Soln (1)


reaction sequences are

(1.) A = D-threose; B = D-glucose

(2.) A = D erythrose; B = D-glucose + D-mannose

(3.) A = D-threose; B = D-glucose + D-mannose

(4.) A = D-tartaric acid; B = D-glucose

Soln (2)



Soln (2)

This is an example of wolf-kischner reduction

Reference : Organic Mechanisms Reactions,

Stereochemistry and Synthesis , Reinhard Bruckner

(111) The major product of the following reaction is

Soln (3)

This is an example of chelation control felkin-Ahn

model

(112) The following transformation involves

sequential

(1.) Claisen rearrangement cope rearrangement

ene reaction

(2.) Cope rearrangement - Claisen rearrangement

ene reaction

(3.) Cope rearrangement ene reaction Claisen

rearrangement

(4.) ene reaction Claisen rearrangement - Cope

rearrangement

Soln (2)

Mechanism :

(113) The mechanism and the product formed in the

following reaction, respectively, are

Soln (2)

Mechanism :

Since Alkyl halide is tertiary in nature hence only

possibility of SN1 mechanism.

Moreover , Since in SN1 there is formation of

carbocation hence racemization take place hence

stereochemistry is not defined.

(Reference : Organic Chemistry , clayden)

(114)The major product A and B in the following


Soln (4)

Mechanism :



Soln (4)

This is an example of wittig reaction


reaction is

Soln (4)

This is an example of prins-pinacol cyclization

Subjection of ketal 1 to a Lewis acid gives the

corresponding oxacarbenium 2, which is poised to

undergo a Prins cyclization to furnish intermediate 3.

The empty p orbital in 3 (at C10) is aligned

periplanar to the C1-C2 bond thus allowing this bond

to migrate to form bicyclic ketone bearing a

quaternary center at C10.


reaction is

Soln (2)

This is an example of intermolecular heck coupling

Palladium is very sensitive to steric effect and

generally forms less hindered complex

further migration of alkene by hydro palladation

are prevented by Ag2CO3 which rapidly removes

oxide re-addition of Pd-H to alkene

Overall elimination must be Syn.


reaction is

Soln (2)

Mechanism:



Soln (3)

(a) 1st step in the chlorination of anime with

hypochlorite

(b) 2nd step is the HLF reaction with retention in

stereochemistry


reaction is

Soln (4)

Mechanism :

(121) The spectroscopic technique that can

distinguish unambiguously between trans-1, 2-

dichloroethynele and cis-1, 2-dichloroethynele

without any numerical calculation is

(1.) Microwave spectroscopy

(2.) UV Visible spectroscopy

(3.) X-ray photoelectron spectroscopy

(4.) -ray spectroscopy

Soln (1)

In microwave, only molecule possessing permanent

dipole moment can be distinguished

cis and trans have marked different in their dipole

moment.

UV is for conjugated system hence cannot be the

answer

(122) 10 mL aliquots of a mixture of HCl and HNO3

are titrated conductometrically using a 0.1M NaOH

and 0.1M AgNO3 separately. The titer volumes are

V1 and V2 mL, respectively. The concentration of

HNO3 in the mixture is obtained from the

combination

(1.) V1 - V2 (2.) 2V1 - V2

(3.) V2 - V1 (4.) 2V2 V1

Soln (3)

(123) A reversible expansion of 1.0 mol of an ideal

gas is carried out from 1.0 L to 4.0 L under

isothermal condition at 300 K. G for this process is

(1.) 300 R . ln2 (2.) 600 R .ln2

(3.) -600 R. ln2 (4.) -300 R . ln2

Soln (3)

G = Wmax

WD = -RT In 2

= -R x 300 x ln22

= -600 R ln 2

(Reference : Physical chemistry , Atkins)

(124) The non-spontaneous process among the

following is

(1.) Vaporization of superheated water at 1050C and

1 atm pressure

(2.) Expansion of gas into vacuum

(3.) Freezing of super cooled water at -100C and 1

atm pressure

(4.) Freezing of water at 00C and 1 atm pressure

Soln (3)

Expansion of gas into vacuum and freezing of water

at 0 C is always a spontaneous process.

However , it is sometimes possible to cool liquid

water at low temperature without solidification. The

liquid below the freezing point is in the supercooled

state.This state is not quite stable and is known as

metastable state. The metastable state is

spontaneously converted into stable state by addition

of small amount of ice to the supercooled liquid ,

which will result in the solidification of water. This

indicates that former process is non-spontaneous at

the same temperature.

(125) The 1H NMR frequency at 1.0 T is 42.4 MHz.

If the gyromagnetic ratios of 1H and 13C are 27 x 107

and 6.75 x 107 T-1 s-1, respectively, what will be the 13C frequency at 1.0 T?

(1.) 10.6 MHz (2.) 169.6 MHz

(3.) 42.6 MHz (4.) 21.3 MHz

Soln (1)

since we know that gyromagnetic ratio of C13 is

almost time to that of 1H hence even without

calculation, it can be inferred that

13 =1

41

if 1H NMR frequency at 1.0T is 42.4 Mhz so the

C13 frequency at the same field strength would be

42.4/4 = 10.6 MHz

(126) The first order rate constant for a unimolecular

gas phase reaction A products that follows

Lindemann mechanism is 2.0 s-1 at pA = 1 atm and

4.0 s-1 at pA =2 atm. The rate constant for the

activation step is

(1.) 1.0 atm-1 s-1 (2.) 2.0 atm-1 s-1

(3.) 4.0 atm-1 s-1 (4.) 8.0 atm-1 s-1

Soln (4)

1

=`

+

1

Therefore, two different pressures we have

1

(1)

1

(1)=

1

(1

1

1

2)

So, = (1

1

1

2) (

1

(1)

1

(2))1

= ( 1 1

2 ) x (

1

21

4)1

= 8 atm-1s-1

(127) Stability of lyophobic dispersions is

determined by

(1.) inner-particle electric double layer repulsion and

intra-particle van der Waals attraction

(2.) inner-particle electric double layer attraction and

intra-particle van der Waals repulsion

(3.) inner-particle excluded volume repulsion and

intra-particle van der Waals attraction

(4.) inner-particle excluded volume attraction and

intra-particle van der Waals repulsion

Soln (1)

The theory of the stability of lyophobic dispersions

was developed by B. Derjaguin and L. Landau and

independently by E. Verwey and J.T.G. Overbeek,

and is known as the DLVO theory. It assumes that

there is a balance between the repulsive interaction

between the charges of the electrical double layers on

neighboring particles and the attractive interactions

arising from van der Waals interactions between the

molecules in the particles. The potential energy

arising from the repulsion of double layers on

particles of radius a has the form

= +22

(128) According to transition state theory, one of the

vibrations in the activated complex is a loose

vibration. The partition function for this loose

vibration is equal to (kB is the Boltzmanns constant

and h is the Plancks constant)

(1.)

(2.)

(3.) kBT (4.)

Soln (4)

Reactants A and B have 3N-6 vibrational degrees of

freedom if non-linear, 3N-5 if linear. The same is

true of the activated complex, which has 3(NA + NB)-

6 vibrational modes if it is non-linear. One of these

modes is of a different character from the rest

corresponding to a very loose vibration that allows

the complex to dissociate into products. For this

degree of freedom we can use a vibrational partition

function q* in which the vibrational frequency v

tends to zero. i.e.

= lim0

1

1 =

1

1 (1 /)=

(129) The average end to end distance of a random

coil polymer of 106 monomers (in units of segment

length) is

(1.) 106 (2.) 105 (3.) 104 (4.)

103

Soln (4)

Average end to end distance of random coil polymer

is given by

R2 = l.N1/2, where N = no of monomer l = segment

length

R2 = 106 = 103

(130) The radial part of the hydrogenic wave

function is given as r( r)-r (, are constants).

This function is then identifiable as

(1.) 2s (2.) 3p (3.) 4d (4.)

5f

Soln (2)

Radial part of hydrogenic wave function is given as

(L )-g. The corresponds to one radial node.

Number of radial nodes = n l 1 =1

n = 3 l = 1

Orbital wave function must be 3p

(131) A normalized state is constructed as the

linear combination of the ground state (0) and the

first excited state (1) of some harmonic oscillator

with energies and 3/2 units, respectively. If the

average energy of the state is 7/6, the probability of

finding 0 in will be

(1.) 1/2 (2.) 1/3 (3.) 1/4 (4.)

1/5

Soln (2)

According to Schrodinger Equation

H = E

i.e. Hamiltonian operated on a wave function will

give a same wave function back with an eigen value.

Here the eigen value for the 1st two state is given

and 3/2

Linear combination will give

= C10 + C21, where e1 and c2 are normalizable

constant

= 1+

121

2 + 221

2 +21210 = 0

The last term in the integral vehicles due to ortho

genes

C12 + C12 = 1

Now its easy to find the expectation value of energy

H = ||2En = ||

2(n + ) h

H = C12 (hw) + C12 (3/2hw)

C12 + C12 = 1.. (1) 1

2C12 +

3

2C12 =

7

6 (given) (2)

3

212 +

3

221 =

3

2()

1

212 +

3

221 =

7

6

C12 = 2/6 =1/3

(132) The symmetry-allowed atomic transition

among the following is

(1.) 3F 1D (2.) 3F 3D

(3.) 3F 1P (4.) 3F 3P

Soln (2)

Solution rule for the atomic transition is

l = 1 (parity should be changed by unity)

s = 0 (spin should not change)

Hence option 2 is correct

(133) Given that E0 (Cl2/Cl-) = 1.35 V and Ksp

(AgCl) = 10-10 at 250C, E0 corresponding to the

electrode reaction 1

22() +

+(. ) +

() is

(1.) 0.75 V (2.) 1.05 V

(3.) 1.65 V (4.) 1.95 V

[2.303

= 0.06 ]

Soln (1)

According to the Nernst equation

Ecell = E0cell + 2.303

log

[]

[]

Since Ksp is given which is the condition for

equilibrium hence ECL will be zero

E0cell = 2.303

log

= 0.06 log 10-10 since log10 = 1

E0cell = 0.6

E0cell = Eox + Ered

Eox = E0cell Ered = 0.6 1.35 = 0.75 V

(134) The ground state electronic configuration of C2

using all electrons is

(1.) 12 1

222 2

222 2

2

(2.) 12 1

222 2

222 2

2

(3.) 12 1

222 2

222 2

1 21

(4.) 12 1

222 2

224

Soln (4)

C 1s22s2p2

MOT diagram for C2

1s2 2s2 has been omitted for clarity

C2 = 61s261s262s261s22p4

(135) The temperature dependence of the vapor

pressure of solid A can be represented by log p =

10.0 - 1800

, and that of liquid A by log p = 8.0 -

1400

.

The temperature of the triple point of A is

(1.) 200 K (2.) 300 K

(3.) 400 K (4.) 500 K

Soln (1)

Triple point correspondence of unique value of

pressure hence two equation for the pressure must

equate to the same value

log = 10 1800

(1)

log = 8 1400

(2)

(1) and (2) must be same

Hence

10 1800

= 8

1400

10 8 =1400

+1800

2 =400

2T = 400

T = 200 K

(136) The molecule with the smallest rotational

constant(in the microwave spectrum) among the

following is

(1.) N CH (2.) HC CCl

(3.) ClC CF (4.) B CCl

Soln (3)

Expression for the rational constant is given by

= 2/82

1/I ( is inversely proportional to reduced

mass)

i.e. heavier will be the molecule, lesser will be the

rotational constant. Hence 3 is the correct option

(137) vmax and Km for an enzyme catalyzed reaction

are 2.0 x 10-3 Ms-1 and 1.0 x 10-6 M, respectively.

The rate of the reaction when the substrate

concentration is 1.0 x 10-6 M is

(1.) 3.0 x 10-3 s-1 (2.) 1.0 x 10-3 s-1

(3.) 2.0 x 10-3 s-1 (4.) 0.5 s-1

Soln (2)

According to Michaelis Menten equation

=

= =

max[5]

+[5]

=2 103[106]

106 + 106=

2 109

[1 + 1]106=2 109

2 106

=10-3

(138)A certain 2 level system has stationary state

energies E1 and E2 (E1 < E2) with normalized wave

functions 1 and 2 respectively. In the presence of a

perturbation V, the second order correction to the

energy for the first state (1) will be

(1.) 1||212

(2.) 1||221

(3.) |1||2|

2

12 (4.)

|1||2|2

(12)2

Soln (3)

Since we know that second order correction is always

hence it follows that expression should contain E1

E2 (since E1 < E2). Therefore correct option is 3

(139) The vibrational frequency of a homonuclear

diatomic molecule is v. The temperature at which the

population of the first excited state will be half that

of the ground state is given by

(1.) hv.In2/kB (2.) hv/(In2.kB)

(3.) In2/(hv.kB) (4.) hv.log2/kB

Soln (2)

Expression for the Boltzmann population is given by 21= E/RT

According to given condition 2 =1

21

1

2= hv/kT

ln 1

2= hv/kT or ln2 = hv/kT

T = hv/ln2.kB

(140) For a simple cubic crystal, X-ray diffraction

how intense reflections for angles 1 and 2 which are

assigned to [1 0 1] and [1 1 1] planes , respectively.

The ratio sin1/sin2 is

(1.) 1.5 (2.) 1.22 (3.) 0.82 (4.)

0.67

Soln (3)

According to Braggs law,

= 2dsin

sin =

2 sin 1/d where d =

2+2+2

12

=1

2+12+1

2

12+1

2+12=2

3

= 0.85

(141) The standard EMF of the cell

Pt, H2(g)HCl (soln.) AgCl(s), Ag(s)

(1.) Increase with T (2.) decrease with T

(3.) remains unchanged with T

(4.) decreases with HCl

Soln (3)

This cell is an example chemical cell without

transference. It consist of hydrogen electrode on the

left hand side and Ag|AgCl electrode on the right ,

dipping directly in hydrochloric acid giving cell with

no liquid junction. In the cell, the hydrogen electrode

is negative and reversible with respect to hydrogen

ion whereas the Ag|AgCl electrode is positive and

reversible with respect to chloride ions.

EMF of the cell is given by (without derivation)

Ecell = ||

ln

2

12

The cell reaction indicates that cell EMF results from

chemical reaction , namely , the reduction of silver

chloride by hydrogen gas to solid silver and

hydrochloric acid.

Furthermore , the EMF of the cell as given by above

equation depends on the activity of hydrochloric acid

and the pressure of hydrogen gas. If the pressure of

the gas is 1 atm equation becomes

= ||

ln

(142) The irreducible representations of C2h are Ag,

Bg, Au, and Bu. The Raman active modes of trans-1,

3-butadiene belongs to the irreducible representations

(1) Ag and Bg (3) Ag and Au

(2) Au and Bg (4) Bg and Bu

Soln (1)

The character table for C2h point group is given by

E C2

(z)

i h Linear Rotation

quadratic

Ag 1 1 1 1 Rz 2 , 2, 2 , Bg 1 -1 1 -1 Rx , Ry xz , yz

Au 1 1 -1 -1 z

Bu 1 -1 1 1 x , y

From above it is evident that Raman active modes

belong to Ag and Bg representation

(143) The molecule diborane belongs to the

symmetry point group

(1.) C2v (2.) C2h (3.) D2d (4.)

D2h

Soln (4)

Diborane has three perpendicular C2 axes are 3

perpendicular mirror plane

(144) Possible term symbol(s) of the excited states of

Na atom with the electronic configuration

[1s22s22p63p1] is/are

(1.) 2S1/2 (2.) 2P3/2 and 2P1/2

(3.) 1S0 and 1P1 (4.) 3P0 and 3P1

Soln (2)

Excited state configuration of Na11 1s22s2p63p1

Term symbol: 2s + 1LJ

S = , 2s + 1 = 2, L = 1

J = |L - S|to|L + S|

1 + = 3/2 or 1 =

Possible term symbol are: 2P3/2 and 2P1/2

(145) Though a constant shift of energy levels of a

system changes the partition function, the properties

that do not change are

(1.) average energy, entropy and heat capacity

(2.) average energy and entropy

(3.) average energy and heat capacity

(4.) entropy and heat capacity

Soln (2)

The expression for the average energy U in the

partition fn is given by the

(

) =

1

2(

)exp () =

Similarly, for entropy we have expression (

),

where F = -RT[In Z]

Therefore both the terms depends only on

temperature but not on energy levels

Detailed Solution Csir Net Dec 2015 Final

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