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Question :1

Solution

Balanced chemical equation: 2 2 3 52 MnI (s)+13 F (g) 2MnF (s)+4IF (l)

Moles of MnI2 taken = 10

309 mol, Moles of F2 taken =

10

38 mol,

In this reaction, MnI2 is the limiting reagent, so, theoretical yield of IF5 = 10 20

2309 309

mol

Thus, experimental yield = 20

222 g 8.0 g309

Correct Option: (1)

Question :2

Solution

3 4 2NH (g) CH (g) HCN(g) 3H (g)

25.5 50Initial mol mol 0 0

17 16

Final 0 3.125 1.5 mol 1.5 mol 3 1.5 mol

Total moles after complete reaction = 6 + 1.625 = 7.625 mol

Since all the species are in gaseous state, volume at STP = 7.625 22.4 L = 170.8 L

Correct Option: (1)

Question :3

Solution

In NaCl there are approximately 106 Schottky pairs per cm

3 at room temperature. In 1 cm

3 there are about

1022

ions. Thus, there is one Schottky defect per 1016

ions.

Frenkel defect is shown by ionic substance in which there is a large difference in the size of ions. AgBr shows both, Frenkel as well as Schottky defects.

F-centers imparts pink colour to the crystal of LiCl.

0.93Fe O indicates metal deficiency defect since Fe3+

ions occupy the site of Fe2+

to maintain the

neutrality, thus metal becomes deficient in the crystal.

Correct Option: (4)

Question :4

Solution

Considering the unit cell in the figure, we get


Packing efficiency in 2D packing is the ratio of packed area by total available area in the unit cell. Since

in a unit cell, one bigger and one small atom are packed, its 2 2

2Packing efficiency ; where is radius of small atom

(2 )

R rr

R

and R is the radius of the bigger atom

From dimensions of unit cell, 2 2 2 2R r R , it implies that 2 1r

R , therefore, ( 2 1)r R and

2 2 2

2

[ ( 2 1) ] 1Packing efficiency 1 0.92 92%

4 2

R R

R

Correct Option: (2)

Question :5

Solution

We have 2 3

2

2time period( ) n

n

r n n nT

v Z Z Z

Therefore, the ratio of time period of electrons in the second orbit of H and first orbit of He+ is

3 3 3 2

H

2 2 2 3

He

2 /1 2 232

1 / 2 1 1

T

T

Correct Option: (2)

Question :6

Solution


From above graphs, first statement is clear, average distance is shown by small arrow on the x-axis.

Only s-orbital electrons have some finite probability at the centre of nucleus due to which only these

electron can penetrate the nucleus.

Although d-orbitals contains two angular nodes but in the case of 2z

d , the angular nodes are conical

surfaces, so, zero nodal planes exist in 2z

d .

The ground state electron in hydrogen lies in 1s-orbital for which value of ‘ l ’ is zero, so, angular

momentum = ( 1) 02

hl l

Correct Option: (4)

Question :7

Solution

Since these atoms and ions all have the same number of electrons, the size should be inversely related to

the positive charge: Hence, the correct order is

Mg2+

< Na+ < Ne < F

– < O

2– < N

3–

Correct Option: (1)

Question :8

Solution

3I , C3O2 and XeF2 are linear

C3H4 is non-linear and non-planar


Correct Option: (3)

Question: 9

Solution

ext

1 3L 3Latm

3 101.325

303.97J

W p V

Correct Option: (3)

Question: 10

Solution

The simple inspection of the question shows that option (4) should be the correct choice, because CaCl2 will give 3 mol of ions for 1 mol of CaCl2 dissolved. However, in the given question the amounts are

given in gram. If we assume that m grams of all the substances are dissolved in 1000 g of solvent then,

Molal concentration of NaCl in solution = (m/58.5) 2 = 0.034m

Molal concentration of ions of CaCl2 in the solution = (m/111.4) 3 = 0.027m

Molal concentration of Urea (NH2CONH2) in solution = (m/60) 1 = 0.0167 m

Molal concentration of Glucose (C6H12O6) in solution = (m/180) 1 = 0.0056 m

b bT iK m

bT m

The boiling point is directly proportional to the molal concentration. Since, NaCl has highest molal

concentration thus, its boiling point will be highest.

Correct Option: (3)

Question: 11

Solution

The equation that relates a change in freezing temperature to molality is

f fT K m (1)

When MgCl2 dissolves in water it breaks up as follows. 2

2MgCl (s) Mg (aq) 2Cl (aq)

Because 1 mol of MgCl2 gives 3 mol of ions, the effective (assumed) molality of ions in the solution is

three times 0.106 m.

Effectivemolality (3)(0.106 ) 0.318m m

Now we can use Equation (1).


1(1.86 Cm )(0.318 )

0.591 C

fT m

The freezing point is depressed below 0.000 °C by 0.591 °C, so we calculate that this solution freezes at

0.591 °C.

Correct Option: (1)

Question :12

Solution

2 3 5F (g) PF (g) PF (g)

Initial pressure(atm) 0.5 0.5 0

Pressure at equilibrium(atm) 0.5 0.5x x x

2

24 4 5 1 0 0.25 atm

(0.5 )p

xK x x x

x

Given that 2 3 5F PF PF0.5 0.25 0.25 atm; 0.25 atm; 0.25 atmp p p

Hence, total 0.25 3 0.75 atmp

On shifting whole content into one bulb

2 3 5F (g) PF (g) PF (g)

Initial pressure(atm) 0.5 0.5 0.5

Pressure at equilibrium(atm) 0.5 0.5 0.5x x x

2

2

0.54 8 10 1 0 0.11atm

(0.5 )p

xK x x x

x

Given that 2 3 5F PF PF0.5 0.11 0.39 atm; 0.39 atm; 0.61atmp p p

Hence, total 0.39 0.39 0.61 1.39 atmp

Correct Option: (4)

Question: 13

Solution

The reaction is 33Ag Al Al 3Ag

Moles of silver nitrate = 25

mol170

Therefore, moles of Al required = 1 25

mol3 170

Mass of Al required = 1 25

27 1.32 g3 170

Correct Option: (1)

Question :14

Solution

2.303log

ak

t a x


1

1

2.303 2log

75 1.6

2.303 5log

75 4

k

k

2

2.303 0.5log

75 0.4k

2

2.303 5log

75 4k

1 2k k k , hence the reaction is of first order

2

2.303(log5 log4)

75k

2

2.303(0.698 0.602)

75k

2

2.3030.096

75k

For 1 M to 0.5 M

2.303 1log

0.5t

k

2.303 75log 2

2.303 0.095

75 0.3010

0.095

237.6 238min

t

Correct Option: (1)

Question :15

Solution On discharging of lead storage battery, lead oxide converts into lead sulphate and deposited on the plates,

so, sulphuric acid withdraws from electrolyte which results into decrease in density of electrolyte.

2 2 4 4 2Pb(s) PbO (s) 2H SO (aq) 2PbSO (s) 2H O(l)

Lead storage battery, in general, can be used to generate potential difference of multiple of 2V because

one cell produces around 2 V potential difference.

Correct Option: (3)

Question: 16

Solution

The charge on the colloidal particles arises because of one or more reasons: Either by electron capture by

sol particles during electrodispersion of metals or preferential adsorption of ions from solution or formulation of double electrical layer causes charge on the colloidal particles. All the given options form

negative sols except TiO2, which forms positive sol.

Correct Option: (4)

Question: 17

Solution


Since the oxidation state of all central atoms is the same, hence as the electronegativity difference

between two elements decreases (for oxides), acidic character increases.

Correct Option: (2)

Question: 18

Solution

(1) Heg (1) for N is (+) ve

Heg (2) and Heg (3) for all elements are (+) ve

Hence H eg (1) + H eg (2) + H eg (3) = (+) ve, that is, endothermic

(2) Li Li+ IE1 = (+) ve

Hence Li+ Li. H must be () ve, that is, exothermic

(3) H H, H eg (1) () ve, that is, exothermic

(4) + 2+

IE IE1 2Ca Ca Ca

IE1 = (+) ve, IE2 = (+) ve

(I E1 + I E2) = (+) ve

Hence Ca2+

Ca+, H eg = () ve, that is, exothermic

Correct Option: (1)

Question: 19

Solution

Self-Reduction: This method is applicable to some sulphides, e.g. PbS, Cu2S and HgS. It does not involve use of reducing agent from an external source.

Calcination: It is the process in which the concentrated ore is heated to a high temperature in the absence

of air. This process is mainly used for a carbonate ore to get its oxide. Carbon Reduction: This process is used for commercial extraction of Fe, Sn, Zn, Pb.

Roasting: It is a process in which the concentrated ore is heated to a high temperature in presence of

excess of air. This process is mainly applicable for sulphide ores to get the corresponding metal oxide

Correct Option: (1)

Question: 20

Solution Only transition metals form interstitial compounds. Alkali, alkaline earth and non-metals will not form

interstitial compounds with dihydrogen. This formation of interstitial compounds depends on the size of

voids formed and radius of the atom to be trapped. Hence the option (4) is correct.

Correct Option: (4)

Question: 21

Solution With excess oxygen, potassium forms superoxide KO2 on burning.

2 2K+O KO

Because of the strong basic nature and high lattice enthalpy of KO2, all other oxides are finally converted

into KO2.

Correct Option: (4)

Question: 22


Solution Alkaline earth metals have smaller atomic size and higher effective nuclear charge as compared to alkali

metals. This causes their first ionization enthalpies to be higher than that of alkali metals.

Correct Option: (3)

Question: 23

Solution

N2O5 is not a mixed anhydride. In fact, it is an anhydride of nitric acid ( 3 2 5 22HNO N O H O ). It is

obtained when conc. HNO3 is heated in the presence of P4O10 at low temperature. The oxide of nitrogen that is a mixed anhydride is NO2 because it reacts with water to give two different acids:

2 2 4 2 2 32NO (or N O ) H O HNO HNO

Dinitrogen oxide (N2O) is used as an anaesthetic by dentists and is also called laughing gas. Nitrogen

oxide (NO) forms complexes called nitrosyls with transition metal ions. These complexes are colored and responsible for the color in the “brown ring test” of nitrates.

NO2 is an odd electron molecule, and hence paramagnetic in nature. When it dimerizes to N2O4, the two

unpaired electrons get paired and the oxide N2O4 is diamagnetic.

Correct Option: (1)

Question :24

Solution In AgO, Ag is not in the (+2) oxidation state. It actually exists as Ag

+[Ag

–O2]

–.

Ag+: 4d

10 is diamagnetic.

Ag3+

: 4d8 is diamagnetic only under square planar field of oxide ion and paramagnetic with two unpaired

electrons in tetrahedral as well as octahedral fields. Hence, the answer is (1).

Under square planar field Under octahedral field Under tetrahedral field

_____ 2 2x yd

dxy

2zd

dxz, dyz

2 2x yd

2zd (eg)

dxz, dyz dxy (t2g)

dxz, dyz dxy (t2g)

2 2x yd

2zd (eg)

Diamagnetic Two unpaired electrons, so

paramagnetic

Two unpaired electrons, so

paramagnetic

Correct Option: (1)

Question: 25

Solution The first compound is of [Ma3bcd]

n± type and has total five stereo isomers while 2

nd compound is of

[Ma2b2cd]n±

type and has total 8 stereo isomers. Hence the increase of number of stereo isomer is 3.

Correct Option: (1)

Question: 26

Solution


To achieve the Sidgwick EAN value, that is, 36, 54 and 86, the metal carbonyl may act as oxidizing agent

when its EAN is less than Sidgwick EAN value. It may act as reducing agent when its EAN value is greater than Sidgwick EAN value and cannot act as either of the two if its EAN value is equal to any one

of Sidgwick EAN value.

[V(CO)6] act as an oxidizing agent. The complex gains an electron to attain the noble gas configuration

and hence obey Sidgwick EAN rule.

0

6 6V CO V COe

EAN = 35 EAN = 36

Correct Option: (4)

Question: 27

Solution The method cannot be used for estimating nitrogen in organic compounds containing nitro group, azo

group or heterocyclic nitrogen because nitrogen present in these compounds does not get converted into

ammonium sulphate under test conditions.

2 4 4 2 4

4 2 4 2 4 3 2

Organic compound + H SO (NH ) SO

(NH ) SO 2NaOH Na SO + 2NH + 2H O

Correct Option: (3)

Question: 28

Solution The two molecules have different connectivities but the same molecular formula. The first molecule has a

methyl on carbon 4, the second molecule has a methyl on carbon 3, so these are positional isomers.

Hence, the given compounds are constitutional isomers.

Correct Option: (4)

Question: 29

Solution

All the three compounds contain the same number of carbon atoms, but compound I contains a three-

membered ring, which contains maximum angle strain. All carbon atoms in I are sp3 hybridized, it implies

C–C–C bond angle should be 10928 but geometry does not allow this angle. According to geometry,

this angle should be 60, whereas it deviates to 4928 in reality. It implies that three-membered ring is most unstable and five-membered ring is the most stable out of given options.

The heat of combustion is inversely proportional to the stability of the compound, if the number of carbon

atoms is the same in each case. Therefore, compound I being most unstable has the maximum heat of

combustion per mole and per unit of CH2.

Correct Option: (3)

Question: 30

Solution

Detailed Solution: + -

2 4 4H SO H +HSO


Correct Option: (1)

Question: 31

Solution The terminal alkynes are very less acidic; the pKa value is around 25. NaOH is not as much as basic so that it can deprotonate alkynes. Hence, there are no major changes occurring in this reaction.

Correct Option: (4)

Question: 32

Solution

This reaction starts with the formation of electrophile which is required for this reaction, but electrophile is carbocation in the Friedel Craft alkylation. So, it rearranges to more stable carbocation by ring

expansion. After rearrangement, electrophilic substitution gives the product given in option (1).

Correct Option: (1)

Question: 33

Solution The product can be formed with best yield if SN2 reaction conditions are used, otherwise elimination

product is also formed in significant amount. For SN2 mechanism, primary alkyl halides are best suited.

Hence, the reaction in option (2) will take place with SN2. In other options, reaction conditions also initiate elimination reaction, so, mixture of products is obtained

in other cases.

Quick Tip: The given reactions represent preparation of ether by Williamson’s synthesis and the best

yields are obtained with primary halides and secondary or teriary alkoxides.

Correct Option: (2)

Question: 34

Solution

In this reaction, substitution takes place through benzyne formation. Since this benzyne is asymmetric and also bears strong electron withdrawing group, so, further reaction is very specific and only meta product

is formed.


Correct Option :(4)

Question: 35

Solution

Total four moles of Grignard reagents are used. Two moles in the first step and again two moles in the last step.

Correct Option: (3)

Question: 36

Solution –OH group activates benzene ring and direct the electrophile at ortho and para position. Although both benzene rings are activated by –OH group but nitration at ortho position of benzene ring which is directly

connected with –OH group gives major product because intra hydrogen bond makes it more stable.

More stable cabanion


Correct Option: (1)

Question: 37

Solution

Correct Option: (4)

Question: 38

Solution

Four equivalents of ammonia are required, so, four moles of ammonia are required to react with one mole

of carbonyl dichloride.

Correct Option: (4)

Question: 39

Solution On dry distillation calcium salt produces ketones as shown below. Calcium carbonate is also produced in

the reaction. Dry

2 5 2 2 5 2 5 3distillation(C H COO) Ca C H COC H CaCO

Dry

6 5 2 6 5 6 5 3distillation(C H COO) Ca C H COC H CaCO

Dry

6 5 2 2 5 2 6 5 2 5 3distillation(C H COO) Ca (C H COO) Ca C H COC H CaCO

Therefore, the compound 6 5 2 3C H CH COCH cannot be formed in the reaction.

Correct Option: (4)

Question: 40

Solution

Mechanism for acetyl and ester reduction:


Mechanism for carboxylic reduction:

Mechanism for N-methyl ethanamide

Thus, all reactants except amide after reduction with LiAlH4 contain oxygen.

Correct Option: (4)

Question: 41

Solution First three reaction produces propyl amine but last reaction produces ethyl methyl amine which is

secondary amine.

H H

OH2/Ni

H2NCH2CH3

H H

NCH2CH3

H H

NCH2CH3

H

H

N-methylethanamine

-H2O


Correct Option: (4)

Question: 42

Solution Teflon is made by polymerizing tetrafluoroethylene in a similar reaction given below:

Catalyst

2 2 2 2HeatCF CF (CF CF )nn

tetrafluoroethylene Teflon

Correct Option: (2)

Question: 43

Solution The isoelectronic point (pI) of an amino acid that has an ionizable side chain is the average of the values

of the similarly ionizing groups (a positively charged group ionizing to an uncharged group or an

uncharged group ionizing to a negatively charged group).

8.95 10.79pI 9.87

2.18

Correct Option: (2)

Question: 44

Solution Consider x kg mass of octane converted into CO2. The total mass of octane taken is 2.1 kg.

8 18 2 2 2

25C H O 8CO (g) 9H O(l)

2

8 9kmol kmol kmol

114 114 114

x x x

8 18 2 2

17C H O 8CO(g) 9H O(l)

2

2.1 8 2.1 9 2.1kmol kmol kmol

114 114 114

x x x

Total mass of CO2, CO and H2O produced is 8 9 8(2.1 ) 9(2.1 )

44 18 28 18 9.1114 114 114 114

x x x x

128 1037.4 810.6 226.8 1.77 kg x x

Thus fraction of octane converted into CO2 = 1.77

0.82.1

Correct Option: (4)

Question: 45

Solution

We know that zinc sulphide exists in diamond structure in which S2

forms fcc lattice and half of the tetrahedral voids are filled by Zn

2+ at alternate positions. Similarly in diamond, carbon atoms form fcc

lattice and half of the tetrahedral voids are filled by carbon in alternate position.

Here one thing is special for diamond structure, the carbon atoms are not in contact with each other at

face diagonal since the same size of atoms exist in the void formed by carbon atom lattice. Four carbon


atoms (three on face center and one at corner of cube) are in contact with the carbon atom placed in

tetrahedral void. Since in fcc, eight tetrahedral voids exist on four body diagonals, four voids, one on each body

diagonal are occupied by carbon atoms. The center of tetrahedral void exists at 1/4th of body diagonal

from the corner, so the length between center of tetrahedral void and center of carbon atom placed at the

corner of cube is equal to 1/4th of the body diagonal.

If radius of carbon atom is R, and edge length of unit cell is “a” then 3

24

aR

or 3 2 3 Face diagonal

24 2 8 2

aR R

or 3 503

76.86 77.0 pm8 2

R

Correct Option: (1)

Question: 46

Solution Pauli’s Exclusive principle: The set of four quantum numbers cannot repeat in an atom.

Hund’s rule: In degenerate energy level, Coupling of electron starts after all orbitals are singly occupied by the electrons.

In third option, first orbital contains same spin quantum number and also coupling starts before single

electron filled in all degenerate orbitals. Thus this option violates both.

Correct Option: (3)

Question: 47

Solution

Pu + 3Tl+ → 3Tl + Pu3+

Ru + Pt2+ → Pt + Ru2+.

2Tl + Ru2+ → Ru + 2Tl+ The above reactions occurs spontaneously, so, we can say logically that

Pu can reduce Tl+ or reduction potential of Pu

3+ is less than Tl

+.

Ru can reduce Pt2+

or reduction potential of Ru2+

is less than Pt2+

. Tl can reduce Ru

2+ or reduction potential of Tl

+ is less than Ru

2+.

Thus order of reduction potential: 3 2 2

o o o o

Pu / Pu Tl / Tl Ru / Ru Pt / PtE E E E

Thus ease of oxidation is Pt < Ru < Tl < Pu.

Platinum comes in top part of electrochemical series, so, it is very unreactive metal as being oxidised. It

cannot replace hydrogen gas from acidic solutions.

Correct Option: (4)

Question: 48

Solution

1 mole of electron = 1 faraday charge = 1 equivalent of a metal

3Al 3 Ale

3 moles of electron will deposit 27 g of Al


1 mole of electron will deposit 27

9g3

Correct Option: (4)

Question: 49

Solution

Using Arrhenius equation

2

1 1 2

1 1log

2.303

aEk

k R T T

Substituting the values k2 = k1 , T1 = 300K , T2 = 310 K, we get

-1

1 1log 2

2.303 8.314 300 310

53.6 Jmol

a

a

E

E

Correct Option: (2)

Question: 50

Solution

In adsorption, molecule from gaseous state or liquid state condenses over solid surface which implies that some energy is released during this process. So the process is exothermic, that is, enthalpy change is

negative. Similarly, gaseous molecules getting attached with solid surface leads to decrease in

randomness, so entropy change is also negative for adsorption processes.

Associated colloid particles or formation of micelles takes place for specific concentration and temperature. For micelle formation, the temperature should be above a minimum temperature. This

temperature is called Kraft temperature.

Adsorbate amount depends on the surface area provided. If surface area per unit mass is increased, then more amount of adsorbate is adsorbed.

Correct Option: (3)

Question: 51

Solution

The electronic configuration of Arsenic is [Ne]3s23p

64s

23d

104p

3 . The element which is just below the

Arsenic it is Antimony (Sb).The electronic configuration of antimony is [Ne]3s23p

64s

23d

104p

6 4d

10 5s

25p

3

Correct Option: (3)

Question: 52

Solution

The size of hydrated ions follows the order as indicated in option (1). The charge/size ratio is mainly responsible for the hydration energy. Higher the charge and smaller the size, greater will be the hydration

energy. Since the charge is the same for Rb+, K

+, Na

+ and Li

+, size is the governing factor here.

Correct Option: (1)

Question: 53


Solution

Based on the positions of the element in the periodic table and the periodic trends, Be2+

is the smallest and Rb

+ is the largest ion. Mg

2+ is bigger than Be

2+, but smaller than K

+.

This is because the ionic radius increases down the group as the shell number increases. Across the

period, it will decrease due to increase in effective nuclear charge (Zeff).

Correct Option: (3)

Question: 54

Solution

Phosphine PH3 is a colorless and extremely toxic gas, which smells slightly of garlic or bad fish. It is

highly reactive. It can form Na3 P by hydrolyzing metal phosphides with water.

Na3 P + 3 H2 O PH3 + 3 NaOH

PH3 is not very soluble in water.

PH3 + H2O PH4+ + O H

Correct Option: (4)

Question: 55

Solution

The structure of both Cr2O3 and -(SO3) are same

-SO3 has ice-like structure and exist as a cyclic trimer.

In the gas phase SO3 has plane triangular structure.

Correct Option: (2)

Question: 56

Solution Heteroleptic complex has more than one type of ligands.

(1) : Incorrect

There is only one type ligand, hence it is homoleptic complex

(2) : Incorrect There are two kinds of ligand, hence it is heteroleptic complex but can have geometrical isomer like facial

and meridional isomers.


(3): Correct

Heteroleptic because two different ligands like NH3 and Cl are present. But cannot show geometrical isomerism because only are geometry is possible in space.

(4): Incorrect

‘en’ and ‘Cl’ are present. Hence heteroleptic but can show geometrical isomerism cis and trans w.r.t. two

Cl-atoms.

Correct Option: (3)

Question: 57

Solution

A molecule is said to be chiral, if it is not super imposable on its mirror image and also it should posses’ asymmetric carbon atom.

From the given compounds only option b is not fulfilling the condition; hence it is achiral and is not

optically active.

Correct Option: (2)

Question: 58

Solution Major products of the reactions are as follows:

ether

3 2 3 2 2 3CH CH Br Na CH CH CH CH Zn(Hg) / HCl,

3 3 3 2 3CH COCH CH CH CH

2 4N H / KOH

3 3 3CH CHO CH CH

4LiAlH

3 3 3 2CH COOCH CH CH OH

Correct Option: (4)

Question: 59

Solution In this reaction, first of all we have to look the polarity of the ICl molecule, in this molecule, ‘I’

experiences partial positive charge, so, ‘I’ attached first with double bond forming cyclic intermediate.

This ensures no rearrangement in the intermediate, in final step, chloride ions open the three member ring

and produces the final product.

Correct Option: (2)

Question: 60

Solution


When converting from Fischer projections to chair structures with monosaccharide hexoses, locate the

position of the OH groups in the Fischer projection. The aldehyde in the Fischer projection becomes the anomeric carbon. The last (bottom) carbon in the Fischer projection becomes the 6

th carbon in the chair

structure, the CH2OH that is outside the ring. The 5th carbon in the Fischer projection becomes the O in

the ring and the carbon counterclockwise from that oxygen. For all hexoses, these three positions do not

change. Carbons 2, 3 and 4 in the Fischer determine the equatorial/axial positions of 2, 3 and 4 in the chair and also differentiate between the different hexoses. If in the Fischer projection an OH on 2, 3 or 4

is to the right it will appear down in the chair structure. If the OH is to the left in the Fischer projection,

then the OH will be up in the chair structure.

Correct Option: (3)

Question: 61

Solution The nitrogen atom is more electronegative than carbon, so, in general the C–N bond polarization is from

carbon to nitrogen. But in the case of pyrrole, the lone pair of nitrogen is involved in the aromaticity of the compound due to which nitrogen becomes positively charged in the contributing resonance structures.

The resonance hybrid implies that all carbon atoms in it are negatively charged and nitrogen is positively

charged, so the direction of dipole moment would be away from the nitrogen atom.

The compound is very unstable since it contains antiaromatic rings. The pi-bond between the rings is broken down in such a way so that both the rings become stable aromatic rings.

+_

This implies that five membered ring is negatively ionized and three-membered ring is positively ionized.

Therefore, the dipole moment’s direction is incorrectly mentioned in the option (3); it should be in opposite direction.

Correct Option: (2)

Question: 62

Solution The net result of hydroboration-oxidation is the syn-addition of water molecule according to antimarkovnikov rule. The reaction with TsCl/pyridine is just to convert –OH group into good leaving

group. This reaction doesn’t interfere with chiral center but the reaction with KOH is the SN2 reaction


which inverts configuration on one chiral centre giving alcohol again with different configuration. So, A

and C are alcohols with different configuration and are diastereomers. A and C are not mirror images.

Correct Option: (2)

Question: 63

Solution Step-I: Attack of bromine atom as electrophile and formation of cyclic bromonium ion.

Step II: Attack of lone pair on electron-deficient carbon. Step III: Removal of hydrogen as H

+.

Correct Option: (4)

Question: 64

Solution

Enamines are the compounds which contains C=C–N bond, imine contains C=N group. Only secondary

amines with -hydrogens can produce enamine. Out of the given options, only pyrrolidine is the secondary amine and it forms enamine with cyclohexanone.

Primary amine forms imine just like enamines and reaction of tertiary amine gives no stable product;

hence, no reaction occurs in this case.

Correct Option: (2)

Question: 65

Solution

mol H2 = (2.15 g H2)2

2

1 mol H

2.016 g H

= 1.07 mol H2


mol NO = (34.0 g NO)1 mol NO

30.01 g NO

= 1.13 mol NO

H2 = 2

2

1.07 mol H

1.13 mol NO + 1.07 mol H = 0.486

NO = 2

1.13 mol NO

1.13 mol NO + 1.07 mol H = 0.514

PH2 = (Ptotal)(H2

) = (2.05 atm)(0.486) = 0.996 atm

PNO = (Ptotal)(NO) = (2.05 atm)(0.514) = 1.05 atm

Correct Option: (1)

Question: 66

Solution

The sodium ion is neutral since it is the salt of the strong base, NaOH. The nitrite ion is basic since it is

the salt of nitrous acid, HNO2, a weak acid. The equilibrium we are interested in for this problem is:

NO2– + H2O HNO2 + OH

–.

2b

2

[HNO ][OH ] =

NO

K

In order to determine the value for Kb recall that Ka – Kb = Kw. We can look for the value of Ka for HNO2

Kb = 1.0 × 10–14

7.1 × 10–4

= 1.4 × 10–11

.

Assume that x 0.10 and substitute the equilibrium values into the mass action expression to get:

11

b[ ][ ]

= = 1.4 100.10

x x

K

Solving we determine that x = 1.2 × 10

–6 M = [OH

–].

pOH = –log[OH–] = –log(1.2 × 10

–6) = 5.93

pH = 14.00 – pOH = 8.07

Correct Option: (2)

Question: 67

Solution In general acidic strength of alcohols is less than water but methyl alcohol is more acidic because of high

solubility of methoxide ions.

On increasing s-character of hybridization, carbon atom’s electronegativity increases, it helps to stabilize the –ve charge over carbon atom after deprotonation of the compound.

In amides the –ve charge developed over nitrogen atom is delocalized over carbonyl oxygen atom but in

the case of amine it is not possible, so, amides are more acidic than amines. Carboxylic acids are stronger acids than alcohols because delocalization of –ve charge is possible only in

the case of carboxylic acids. Thus acetic acid is more acidic than ethanol.

[NO2–] [HNO2] [OH

–]

I 0.10 – –

C –x +x +x E 0.10 –x +x +x


Thus first option is correct.

Correct Option: (1)

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