Aieee 2006 Paper

8/2/2019 Aieee 2006 Paper

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MAHESH JANMANCHI AIEEE 2006

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Maximum Marks: 180

Question paper format and Marking scheme:

1. This question paper has 55 questions.

2. Q No 1 to 10 carry 1.5 marks each ( 15 marks)

3. Q No 11 to 35 carry 3 marks each ( 75 marks)

4. Q No 36 to 55 carry 4.5 marks each ( 90 marks)


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96. HBr reacts with CH2 = CH OCH3 under anhydrous conditions at room temperature to give

(1) CH3CHO and CH3Br (2) BrCH2CHO and CH3OH

(3) BrCH2 CH2 OCH3 (4) H3C CHBr OCH3

Sol. (4)Electrophilic addition reaction is more favourable

-HBr Br

H C = CH - OCH H C - CH - OCH H C - CH - OCH2 3 2 3 3 3

Br

H

97. The IUPAC name of the compound shown below is

Cl

Br (1) 2-bromo-6-chlorocyclohex-1-ene (2) 6-bromo-2-chlorocyclohexene

(3) 3-bromo-1-chlorocyclohexene (4) 1-bromo-3-chlorocyclohexene

Sol. (3)Priority is given to double bond. Numbering must be along the double bond

98. The increasing order of the rate of HCN addition to compounds A D is

(A) HCHO (B) CH3COCH3

(C) PhCOCH3 (D) PhCOPh

(1) A < B < C < D (2) D < B < C < A

(3) D < C < B < A (4) C < D < B < A

Sol. (3)Addition of HCN to carbonyl compounds is a nucleophilic addition reaction.

Reactivity order :O

||

H C H > R C H > R C R

O

||

O

||

Ketones are less reactive than aldehydes due to +I effect of alkyl; group and steric hindrance.

As the size of alkyl group increases , reactivity of ketones further decreases.

Aromatic aldehydes and aromatic ketones are less reactive than their aliphatic analogues due to + R

effect of benzene ring


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99. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?

(1) 0.02 (2) 3.125 102

(3) 1.25 102

(4) 2.5 102

Sol. (2)

1 mole of ( )Mg PO3 4 2 has 8 moles of O atoms? 0.25 moles of O atoms

0.25

8=

2

25

8 100

3.125 10

=

=

100. According to Bohrs theory, the angular momentum of an electron in 5th

orbit is

(1)h

25

(2)h

1.0

(3)h

10

(4)h

2.5

Sol. (4)

Angular momentum of e-in an orbit,

nh

mvr =

5h2

h2.5

=

=

101. Which of the following molecules/ions does not contain unpaired electrons?

(1) 22O

(2) B2

(3)2N+ (4) O2

Sol. (1)

22

O : 2 * 2 2 * 2 2 2 2 * 2 * 21 1 , 2 2 , 2 , 2 2 , 2 2s s s s p p p p pz x y x y

= = ( All paired e- - Diamagnetic)

2B : 2 * 2 2 * 2 1 11 1 , 2 2 2 2s s s s p p

x y = (2 unpaired e- - Paramagnetic)

2N

+ : 2 * 2 2 * 2 2 2 11 1 , 2 2 2 2 2s s s s p p px y z

= ( 1 unpaired e- - Paramagnetic)

2O : 2 * 2 2 * 2 2 2 2 * 1 * 11 1 , 2 2 2 2 2 2 2s s s s p p p p p

z x y x y = = (2 unpaired e- - Paramagnetic)


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102. Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius)

(1) 320

3

r (2) 324

3

r

(3) 312

3r (4) 3

16

3r

Sol. (4)FCC unit cell contains 6Face centred and 8corner particles

1 16 8

2 8x x+ = 4

So , FCC unit cell contains 4 particles

Total volume of atoms present in a face-centre cubic unit cell of a metal

=34

4 3 r

= 316

3r

103. A reaction was found to be second order with respect to the concentration of carbon monoxide. If

the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of

reaction will

(1) remain unchanged (2) triple

(3) increase by a factor of 4 (4) double

Sol. (3)

[ ]2

R CO - (1)

[ ]' 22R CO - (2)

'4

R

R=

' 4R R=

104. Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4?

(1) 2HI + H SO I + SO + 2H O2 4 2 2 2

(2) ( )Ca OH + H SO CaSO + 2H O2 2 4 4 2

(3) NaCl + H SO NaHSO + HCl2 4 4

(4) 52PCl + H SO 2POCl + 2HCl + SO Cl2 4 3 2 2

Sol. (1)

2HI + H SO I + SO + 2H O2 4 2 2 2

In this reaction , oxidation no of iodine is increased from 1 to 0 and that of S is decreased from +6 to

+4. Therefore H2SO4 is an oxidizing agent and HI is a reducing agent


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105. The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is

(1) nitrito-N-pentaamminecobalt (III) chloride (2) nitrito-N-pentaamminecobalt (II) chloride

(3) pentaammine nitrito-N-cobalt (II) chloride (4) pentaammine nitrito-N-cobalt (III) chloride

Sol. (4)[Co(NO2)(NH3)5]Cl2pentaammine nitrito-N-cobalt (III) chloride

106. The term anomers of glucose refers to

(1) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4)

(2) a mixture of (D)-glucose and (L)-glucose

(3) enantiomers of glucose

(4) isomers of glucose that differ in configuration at carbon one (C-1)

Sol. (4)A pair of stereoisomers which differ in configuration at C -1 are called anomers and the carbon is called

anomeric carbon

- D (+) glucose and - D (+) glucose are anomers.

107. In the transformation of 23892 U to

234

92 U , if one emission is an -particle, what should be the other

emission(s)?

(1) Two (2) Two and one +

(3) One and one (4) One + and one

Sol. (1)238 234 4 0U U + He + 2 e92 92 2 -1

i.e., two

108. Phenyl magnesium bromide reacts with methanol to give

(1) a mixture of anisole and Mg(OH)Br (2) a mixture of benzene and Mg(OMe)Br

(3) a mixture of toluene and Mg(OH)Br (4) a mixture of phenol and Mg(Me)Br

Sol. (2)

6 5 3 6 6 3( )C H MgBr CH OH C H Mg OCH Br + +


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109. 3 3CH Br Nu CH Nu Br + +

The decreasing order of the rate of the above reaction with nucleophiles (Nu) A to D is

( ) ( ) ( ) ( ) 3, , ,Nu A PhO B AcO C HO D CH O = (1) D > C > A > B (2) D > C > B > A

(3) A > B > C > D (4) B > D > C > A

Sol. (1)

3CH O HO PhO AcO > > >

If the nucleophilic atom is same , nucleophilicity parallels basicity. i.e., greater the basicity, stronger is

the nucleophile.

Weaker acid has the strong conjugate base

110. The pyrimidine bases present in DNA are(1) cytosine and adenine (2) cytosine and guanine

(3) cytosine and thymine (4) cytosine and uracil

Sol. (3)DNA contains

purines, Adenine and Guanine

Pyrimidines Thymine and Cytosine

111. Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is

(1) CH3CH2CH(OH)CH2CH3 (2) C6H5CH2CH2OH

(3) (4) PhCHOHCH3

Sol. (4)

Iodoform test is given by compounds having Methyl keto group ( CH3CO -) and Methyl carbinol(

CH3CHOH -) groups

12. The increasing order of stability of the following free radicals is

(1) ( ) ( ) ( ) ( )

CH CH< CH C C H C H C H C3 3 6 5 6 52 2 2 2< <

(2) ( ) ( ) ( ) ( )

C H C< C H CH< CH C < CH CH6 5 6 5 3 33 2 3 2

(3) ( ) ( ) ( ) ( )

C H C H< C H C < CH C < CH C H6 5 6 5 3 32 3 3 2

(4) ( ) ( ) ( ) ( )

CH CH< CH C < C H C < C H CH3 3 6 5 6 52 3 3 2


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Sol. (1)

The order of stability of free radicals is 30

> 20

> 10

Benzyl free radicals are stabilized by resonance. So they are more stable than alkyl free radicals.

As the no of phenyl groups attached to the carbon atom having odd e - increases, stability increases

113. Uncertainty in the position of an electron (mass = 9.1 1031

kg) moving with a velocity 300 ms1

,

accurate upto 0.001%, will be

(1) 219.2 10 m (2) 25.76 10 m

(3) 21.92 10 m (4) 23.84 10 m

( )346.63 10 Jsh =

Sol. (3)

hx.V4m

34

31

h 6.63 10x

0.0014mV4 3.14 9.1 10 300

100

=

34

31 3

6.63 10

4 3.14 9.1 10 10

=

= 0.01933

= 21.93 10

114. Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,( ) ( ) ( )

5 3 2PCl g PCl g Cl g+

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x,

the partial pressure of PCl3 will be

(1)1

xP

x

+ (2)

2

1

xP

x

(3)1

xP

x

(4)

1

xP

x

Sol. (1)

( ) ( ) ( )5 3 2

PCl g PCl g Cl g+

1 0 0(1-x) x x

Total no of moles = 1+ xPartial pressure = Mole fraction x P total

13

xP P

PCl x

=

+


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115. The standard enthalpy of formation ( )0 Hf at 298 K for methane, CH4(g), is 74.8 kJ mol1

. The

additional information required to determine the average energy for C H bond formation wouldbe

(1) the dissociation energy of H2 and enthalpy of sublimation of carbon

(2) latent heat of vapourization of methane

(3) the first four ionization energies of carbon and electron gain enthalpy of hydrogen

(4) the dissociation energy of hydrogen molecule, H2

Sol. (1)

( ) ( ) ( ) 122 ; 74.84C s H g CH g H kJmol + =

In order to calculate the average energy of C H bond formation , the following data is needed

( ) ( ) ;C s C g H ( enthalpy of sublimation of carbon)

( ) ( )2 2 ;H g H g H ( enthalpy of bond dissociation of H2)

116. Among the following mixtures, dipole-dipole as the major interaction, is present in

(1) benzene and ethanol (2) acetonitrile and acetone

(3) KCl and water (4) benzene and carbon tetrachloride

Sol. (2)

Dipole dipole interactions are present between the polar molecules, I which the positive pole of one

molecule is attracted by the negative pole of another molecule.

117. Fluorobenzene (C6H5F) can be synthesized in the laboratory

(1) by heating phenol with HF and KF

(2) from aniline by diazotisation followed by heating the diazonium salt with HBF4

(3) by direct fluorination of benzene with F2 gas

(4) by reacting bromobenzene with NaF solution

Sol. (2)

Ar NH + NaNO + 2 HCl Ar N N + NaCl + 2H O

2 2 2

Cl

BALZ SCHIEMANN REACTION

++

N2Cl

HBF

4

-HCl

N2BF

4F

+ N2

+ BF3


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118. A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements

about these chlorides is correct?

(1) MCl2 is more volatile than MCl4

(2) MCl2 is more soluble in anhydrous ethanol than MCl4

(3) MCl2 is more ionic than MCl4

(4) MCl2 is more easily hydrolysed than MCl4

Sol. (3)

Acc to Fajan, Low charge on the ion favours formation of ionic bond and high charge on the ion

favours covalent bond formation

119. Which of the following statements is true?

(1) H3PO3 is a stronger acid than H2SO3

(2) In aqueous medium HF is a stronger acid than HCl

(3) HClO4 is a weaker acid than HClO3

(4) HNO3 is a stronger acid than HNO2

Sol. (4)

Higher the oxidation state of the central atom, greater is the acidic nature

So, HNO3 is a stronger acid than HNO2

HClO4 is a stronger acid than HClO3Greater the electronegativity and higher the oxidation state of central atom, greater is the acidic nature

So, H2SO3is a stronger acid than H3PO3Due to higher bond dissociation energy and molecular association in HF , it is weaker than HCl

120. The molar conductivities0

NaOAc and0

H Cl at infinite dilution in water at 25oC are 91.0 and

426.2 S cm2/mol respectively. To calculate o

0

HOAc , the additional value required is

(1)0

H O2

(2)0K Cl

(3)0N aOH (4)

0N aCl

Sol. (4)

( )- +3 30 0 0CH COONa CH COO Na = + ............... 1

( )0 0 0HCl CH = + .................. 2Cl+

( )0 0 0NaCl Na = + .................. 3Cl

3 3

0 0 0 0

CH CH COONa HCl NaCl = COOH +


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121. Which one of the following sets of ions represents a collection of isoelectronic species?

(1)+ - 2+ 3+K ,Cl ,Ca ,Sc (2) 2+ 2 + 2Ba ,Sr ,K ,S+

(3) 3 2 2N ,O ,F ,S (4) 2 2Li ,Na ,Mg ,Ca+ + + +

Sol. (1)

+ - 2+ 3+K ,Cl ,Ca ,Sc

All of them have 18 e-

122. The correct order of increasing acid strength of the compounds is

(a) CH3CO2H (b) MeOCH2CO2H

(c) CF3CO2H (d)

(1) b < d < a < c (2) d < a < c < b(3) d < a < b < c (4) a < d < c < b

Sol. (3)+I groups decrease the acidic nature( Me-)

-I groups increase the acidic nature ( - F and OMe)

123. In which of the following molecules/ions are all the bonds not equal?

(1) SF4 (2) SiF4(3) XeF4 (4) 4BF

Sol. (1)In SF4, Hybridisation of central atom S is sp

3d. Its shape is distorted tetrahedral or see saw.

Due to repulsion with equatorial bonds , axial bonds are elongated. So, the bond lengths of two axialbonds are greater than the three equatorial bonds

124. What products are expected from the disproportionation reaction of hypochlorous acid?

(1) HClO3 and Cl2O (2) HClO2 and HClO4

(3) HCl and Cl2O (4) HCl and HClO3

Sol. (4)

33 ClO ClO 2 ClH H H +

Oxidation state of Cl is increased from +1( HOCl) to +5(HClO3) and decreased to -1(HCl)


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125. Nickel (Z = 28) combines with a uninegative monodentate ligand X to form a paramagnetic

complex [NiX4 ]2

. The number of unpaired electron(s) in the nickel and geometry of this complex

ion are, Respectively

(1) one, tetrahedral (2) two, tetrahedral

(3) one, square planar (4) two, square planar

Sol. (2)2 2 6 2 6 2 8

28 Ni:1s 2s 2p 3s 3p 4s 3d

2 2 2 6 2 6 8Ni :1s 2s 2p 3s 3p 3d+

No of unpaired e-= 2

126. In Fe(CO)5, the Fe C bond possesses

(1) -character only (2) both and characters

(3) ionic character (4) -character only

Sol. (2)

In metal carbonyls, metal Carbon bond possesses both and characters

127. The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is(1) F < S < P < B (2) P < S < B < F

(3) B < P < S < F (4) B < S < P < F

Sol. (4)

IE increase from left to right in a period and decreases down the group .IE of P is greater than that of S, due to half filled configuration of P

128. An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the

initial temperature and Tf is the final temperature, which of the following statements is correct?

(1) ( ) ( )T > Tf firrev rev

(2) Tf> Ti for reversible process but Tf= Ti for irreversible process

(3) ( ) ( )T Tf frev irrev= (4) Tf= Ti for both reversible and irreversible processes

Sol. (1)

>Wirrev rev

W

( ) ( )T > Tf firrev rev


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129. In Langmuirs model of adsorption of a gas on a solid surface

(1) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface

covered

(2) the adsorption at a single site on the surface may involve multiple molecules at the same time

(3) the mass of gas striking a given area of surface is proportional to the pressure of the gas

(4) the mass of gas striking a given area of surface is independent of the pressure of the gas

Sol. (3)

Langmuir Adsorption Isotherm gives information about the mass of gasn adsorbed per gram of

adsorbent is related to the equilibrium pressure by the expressionaP

x / m1 bp

=

+

130. Rate of a reaction can be expressed by Arrhenius equation as:

/E RTk Ae

= In this equation, E represents

(1) the energy above which all the colliding molecules will react

(2) the energy below which colliding molecules will not react

(3) the total energy of the reacting molecules at a temperature, T

(4) the fraction of molecules with energy greater than the activation energy of the reaction

Sol. (2)

/E RTk Ae

= Where E is activation energy i.e., min amount of energy required to be possessed by the reactant

molecule for their collision lkead to the formation of products

131. The structure of the major product formed in the following reaction is

Cl

I

NaCN

DMF

(1) (2)

CN

CN

Cl

I (3) (4)

Cl

CN I

CN


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Sol. (4)

Alkyl halides are more reactive than aryl halides. So, side chain is attacked

Cl

I

NaCN

DMF

I

CN

132. Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces

(1) 4-phenylcyclopentene (2) 2-phenylcyclopentene

(3) 1-phenylcyclopentene (4) 3-phenylcyclopentene

Sol. (4)According to E2 mechanism.

133. Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of

2-fluoroethanol is

(1) Eclipse, Gauche, Anti (2) Gauche, Eclipse, Anti

(3) Eclipse, Anti, Gauche (4) Anti, Gauche, Eclipse

Sol. (3)

2-Fluoroethanol

Anti (or) staggered Eclipsed Skew or Gauche Fully eclipsed

In gauche form , there is possibility of intra molecular hydrogen bonding


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134. The structure of the compound that gives a tribromo derivative on treatment with bromine water

is

(1) (2)

CH3

OH

CH OH2

(3) (4)

CH3

OH

CH3

OH

Sol. (1)Since the compound on treatment with Br2 H2O gives a tri bromo derivative , it must be m Cresol

Because it has one o- and two p - positions free with respect to OH group

135. The decreasing values of bond angles from NH3 (106o) to SbH3 (101

o) down group-15 of the

periodic table is due to

(1) increasing bp-bp repulsion (2) increasing p-orbital character in sp3

(3) decreasing lp-bp repulsion (4) decreasing electronegativity

Sol. (4)

In a group, from top to bottom , as the size of the atom increases , pure p orbitals are involved inbonding

136.

Me

Me

NEt

n-BuOH The alkene formed as a major product in the above elimination reaction is


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(1) (2) CH2 = CH2

Me (3) (4)

Me

Me

Sol. (4)

Me

Me

NEt

n-BuOH

Me

In Hofmann elimination, less sterically hinderd hydrogen is eliminated and hence less substitutedalkene is the major product

137. The spin-only magnetic moment [in units of Bohr magneton,(B)] of Ni2+

in aqueous solution

would be (Atomic number of Ni = 28)

(1) 2.84 (2) 4.90

(3) 0 (4) 1.73

Sol. (1)

2 2 2 6 2 6 8 01 2 2 3 3 3 4Ni s s p s p d s+

As H2O is a weak field ligand, no forcible pairing occurs

Hence the complex formed will be sp3

hybridized and tetrahedral in geometry.

No of unp[aired e-s = 2

Therefore, spin only magnetic moment ( 2)n n = +

2(2 2) = + = 8 2.84 = = B.M

138. The equilibrium constant for the reaction

( ) ( ) ( )3 2 21

SO g SO g + O g2

is 24.9 10cK= . The value of Kc for the reaction

( ) ( ) ( )2 2 32SO g +O g 2SO g will be

(1) 416 (2) 2.40 103

(3) 9.8 102

(4) 4.9 102


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Sol. (1)

Equation is reversed and multiplied by 2;2

1K =

2K1

2 2 2

1

(4.9 10 )K

=

410 100 100

4.9 4.9 24.01

= =

= 416.49

139. Following statements regarding the periodic trends of chemical reactivity ofthe alkali metals and

the halogens are given. Which of these statements gives the correct picture?

(1) The reactivity decreases in the alkali metals but increases in the halogens with increase in

atomic number down the group(2) In both the alkali metals and the halogens the chemical reactivity decreases with increase in

atomic number down the group

(3) Chemical reactivity increases with increase in atomic number down the group in both the

alkali metals and halogens

(4) In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic

number down the group

Sol. (4)

140. Given the data at 250C,

0Ag + I AgI + e ;E 0.152V = 0Ag Ag ;E 0.800Ve+ + =

What is the value of splogK for AgI?

RT2.303 =0.059 V

F

(1)-8.12 (2) +8.612

(3) -37.83 (4) -16.13

Sol. (4)

( ) ( )

0

AgI s +e Ag s +I ; E 0.152

=

0Ag Ag ;E 0.800Ve+ + =

( ) + 0AgI s Ag + I ; E 0.952 =

0

cell

0.059E logK

n=

sp

0.0590.952 logK

1 =

sp

0.952logK 16.135

0.059= =


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141. The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr:

( ) ( ) ( )2 2NO g Br g NOBr g+

( ) ( ) ( )2NOBr g NO g 2NOBr g+ If the second step is the rate determining step, the order of the reaction with respect to NO(g) is

(1)1 (2) 0

(3) 3 (4) 2

Sol. (4)

( ) ( ) ( )2 2NO g Br g NOBr g+

( ) ( ) ( )2NOBr g g 2NOBr gNO+ ( Rate determining step)

Therefore, [ ][ ]2R = K NOBr NO

[ ] [ ][ ][ ][ ][ ]

2 c 2

c 2

NOBr =K NO BrK.K NO Br NOR =

[ ]2'=K NO Br

2

Therefore order of the reaction w r t NO is 2

142. Lanthanoid contraction is caused due to

(1) the appreciable shielding on outer electrons by 4f electrons from the nuclear charge

(2) the appreciable shielding on outer electrons by 5d electrons from the nuclear charge

(3) the same effective nuclear charge from Ce to Lu

(4) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Sol. (4)

Lanthanoid contration is due to ineffective shielding offered by f orbitals which inturn is due to their

dispersed shape

143. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is

100 . The conductivity of this solution is 1.29 S m1

. Resistance of the same cell when filled with

0.2 M of the same solution is 520 . The molar conductivity of 0.2 M solution of the electrolyte

will be

(1) 4 2 1124 10 S m mol (2) 4 2 11240 10 S m mol

(3)4 2 1

1.24 10 S m mol

(4)4 2 1

12.4 10 S m mol

Sol. (4)

1K =

R

l

a

11.29 =

100

l

a

1129 ml

a

=


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R = 520 , C = 0.2 M

( ) 1 11 1

129 m

520

lK

R a

= =

1000K

im

Molar ty =

6 31 1000129 10520 0.2

m=

6129 1000 10520 0.2

=

31.24 10=

= 4 2 112.4 10 S m mol

144. The ionic mobility of alkali metal ions in aqueous solution is maximum for(1) K

+(2) Rb

+

(3) Li+

(4) Na+

Sol. (2)Smaller the size, heavier is the hydration and lesser is the ionic mobility and vice versa.

Rb+ is the largest ion among given. So it is least hydrated. Hence it will have maximum ionicmobility

145. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is

(1) 11.14 mol kg (2) 13.28 mol kg

(3) 12.28 mol kg (4) 10.44 mol kg

Sol. (3)

'

1000 Mm

1000 d MM

=

'

1000 2.05m

1000 1.02 2.05 x 60

=

= 2.28

146. The enthalpy changes for the following processes are listed below:

( ) ( )2Cl g 2Cl g ,= 1242.3 kJmol

( ) ( )2l g 2l g ,= 1151.0 kJmol

( ) ( ) ( )ICl g l g Cl g ,= + 1211.3 kJmol

( ) ( )2 2I s l g ,= 162.76 kJmol

Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy

of formation for ICl(g) is

(1)-14.6 1kJmol (2) -16.8 1kJmol

(3) +16.8 1kJmol (4) +244.8 1kJmol


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Sol. (3)

( ) ( )2 21 1

I Cl ICl g2 2s +

( ) ( )1 1 1

2 2 22 2

H H H H HI l Cl Cl I ClI s l g

= + +

( )1 1 1

62.76 151.0 242.3 211.32 2 2

= + +

= 228.03 211.3

H =16.73

147. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral

complex with a Ca2+ ion?(1)Six (2) Three

(3) One (4) Two

Sol. (3)

EDTA is a hexa dentate ligand

148.

OH

+ CHCl +NaOH3

O Na+

CHO The electrophile involved in the above reaction is

(1)dichloromethyl cation( )CHCl2 (2) dichlorocarbene( :CCl2)

(3) trichloromethyl anion( )CCl3 (4) formyl cation ( )CHO

Sol. (2)

This is Reimer Tiemann reaction, in which the electrophile involved is dichloro carbene

149. 18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this

aqueous solution at 100oC is

(1) 759.00 Torr (2) 7.60 Torr

(3) 76.00 Torr (4) 752.40 Torr


21/21



Sol. (4)0

S

S

P P n

P N

=

S

S

18 1760 P 0.1180 10

178.2P 9.9 9.9

18

= = =

S S

1760 P P

99 =

S S760 99 P 99= P

S760 99 100P =

S

760 99

P 752.4100

= =

150. (H U) for the formation of carbon monoxide (CO) from its elements at 298 K is

( )1 1R = 8.314 J K mol

(1)-1238.78 1J mol (2) 1238.78 1J mol

(3) -2477.57 1J mol (4) 2477.57 1J mol

Sol. (1)

21C + O CO2

gH U n RT =

g

1n = 1- 1

2

18.314 298

2=

1238.78=

Aieee 2006 Paper

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