1. PART -A
Q.1 If % ionic character of O–H bond is 10% & O–H bond distance is
8
.
4
1
Å, then calculate dipole
moment of H2O molecule (in Debye) given H–O–H bond angle is 105°. [Use sin 52.5° = 0.8] [5]
[Ans. 0.12 D]
[Sol.
 q × d = 4.8 × 10–10 × 0.1 ×
8
.
4
1
× 10–8
= 0.1 × 10–18 ese cm = 0.1 D
 net = 2 cos 52.5 = 2 × 0.1 × (1 – 0.64) D = 2 × 0.1 × 0.6
 net = 0.12 D ]
Q.2 Arrange in ascendingorder (reasons not required)
(a) | Hhydrogenation| of : 1, 2 Pentadiene, 1, 3 Pentadiene & 1, 4 Pentadiene
(b) Ionic conductance of : F–, Na+ & Cs+.
(c) Dipole Moment of : , , ,
(d) Boilingpoint :n-pentane, isopentane,neo-pentane [1.5+1+1.5+1]
[Sol.(a) (1) (2) (3)
C = C = C – C – C, C = C – C = C – C, C = C – C – C = C
Most stable (2) > (3) > (1)
|HHE|  (2) < (3) < (1)
(b) I.C 
size
.
aq
1

Hydration
1
 Na+ < F– < Cs+
(c) < < <
(d) neo-pentane < iso-pentane < n-pentane ]

2. Q.3 Balancethereaction
(a) Oxidationofcyclobutylmethanoltocyclobutanecarboxylicacid by –
4
MnO inacidicmedium.
(b) Ca(OCl2) + KI  I2 + CaCl2 + KCl (Acidic medium) [3+2]
[Sol. + –
4
MnO  + Mn2+
Mn+7 + 5e–1  Mn+2 ] × 4
C–1  C+3 + 4e– ] × 5
 + 4 –
4
MnO  + 4Mn+2
O
10
O
21
+ 12H+ + 11H2O
 5 + 4 –
4
MnO + 12H+  5 + 4Mn+2 + 11
1H2O
(5, 4, 12, 5, 4, 11)
(b) 0 –1 0
Ca(OCl2) + KI  CaCl2 + KCl + I2
0
Cl2 + 2e–  2Cl–
2I–  I2 + 2e–
 Ca(OCl2) + 2KI  CaCl2 + I2 + 2KCl
+2Cl– + 2H+ +H2O
 Ca(OCl2) + 2KI + 2HCl  CaCl2 + I2 + 2KCl + H2O
(1, 2, 2, 1, 1, 2, 1) ]

3. Q.4
(a) Write theIUPAC name of [2]
(i) (ii)
(b) Predict thefinal product [2]
(i) 2
2
3 NH
C
CH
CH
|
|
O


 + NaOH 
 

O
H2 ? (ii) + NaOH 
 

O
H2 ?
(c) WhichtransitioninBe3+ emits samewavelength as the2nd lineof Balmerseries ofHydrogen. [1]
[Sol.(a)
(i) 3[2-chloro,4-formylcyclobutyl]cyclobutanecarbonylchloride
(ii) (2 chlorocyclopropyl)spiro(2,2) pentane carboxylate
(b)
(i) H
N
C
CH
CH
|
|
O
–
2
3 

  NH
C
CH
CH
|
O
2
3
–


  NH
C
CH
CH
|
OH
2
3 


(ii) + NaOH 
 

O
H2   
 

O
H2
(c)

1
= RH × 





 2
2
4
1
2
1
= RH ×16 







 16
16
1
16
4
1
= RH ×16 





 2
2
16
1
8
1
 Transition from 16 to 8 ]
Q.5(a) Draw anAndrew's isotherm of a real gas for three temperatures (T>Tcritical,T=Tcritical,T<Tcritical)on
asamegraph.Highlighttheliquifactioncurveinit.
(b) Argon at a temperature of 273K & pressure 1atm has a densityof (48/22.4) gm/lit. Predict whether it
behaves ideallyor acts as real. If real, thenwhichVanderWaal constant is more influential.Why?
[2+3]
[Sol.(a)

4. (b) P = 1 atm, M = 40, T = 273
 PM = ZdRT
 1 × 40 = Z × 4
.
22
4
.
22
48

 Z = 5/6
 Z < 1
VanderWaalconstant aisinfluential ]
Q.6(a) Ifinternucleardistance betweenCl atoms in Cl2 is 10Å &between Hatoms in H2 is2 Å, then calculate
internuclear distance betweenH & Cl (Electronegativityof H = 2.1 & Cl = 3.0). [2]
(b) Comparethefollowinggivingreasons
(i) Acidic nature of oxides: CaO, CO, CO2, N2O5, SO3
(ii) Bond length of P—O, S—O & Cl—O in –
3
4
PO , –
2
4
SO , –
4
ClO respectively.
. [3]
[Sol.(a) rA–B = rA + rB – 0.09(XA – XB)
rA = 10/2; rB = 2/2
= 5 + 1 – 0.09 × 0.9 = 6 – 0.081 = 5.919 Å
(b)(i) CaO < CO < CO2 < N2O5 < SO3
Metal oxide Neutral Acidic Acidic Max. Acidic oxide
basic oxide oxide
Alongaperiodacidnatureincreases onmovingright
(ii)
Bond order 1.25 1.5 1.75
 B.L. of Cl–O < S–O < P–O
as B. O  B.L.  ]
PART -B
Q.7 A solution of Ba3(PO4)2 having molarity= 2M initiallydissociates to give Ba2+ & PO4
3–. If extent of
dissociationofsaltis75%&densityofsolutionis3.202gm/ml,thencalculatemolality(m)ofeachofthe
specie. [Ba3(PO4)2, Ba2+ & PO4
3–] [6]
[Sol. Ba3(PO4)2 l 3Ba+2 + 2 –
3
4
PO
m of salt = 100
601
2
202
.
3
1000
2





= 1 m
 m of Ba3(PO4)2 = 0.25 m
m of Ba+2 = 2.25 m
m of –
3
4
PO = 1.5 m ]

5. Q.8 A 117.6 mg sample of impure KSCN on reaction with KMnO4 give Mn2+, SO4
2– & CN–. All the
–
2
4
SO produced was reacted with KI to give SO2 gas & I2 gas. I2 gas liberated required 10 ml of
0.1 M K2Cr2O7 solution to get converted to –
3
IO alongwith Cr3+. Write all the unbalanced skeleton
reactions. Calculate percent purityof KSCN sample & moles of KMnO4 requiredin first step. [6]
[Sol. KSCN + –
4
MnO  Mn+2 + –
2
4
SO + CN–
–
2
4
SO + KI  SO2 + I2
I2 + K2Cr2O7  –
3
IO + Cr+3
meq. of K2Cr2O7 = meq. of I2 (3rd reaction) = 10 × 0.1 × 6
m molesof I2 = 10 × 0.1 × 6 × 1/10
meq. of I2 (2nd reaction) = meq of –
2
4
SO = 2
10
1
6
1
.
0
10 



= m moles of –
2
4
SO = m. m of KSCN = 2
10
1
6
1
.
0
10 


 ×
2
1
 wt. of KSCN = 0.6 × (98) = 58.8 mg
 % purity = 100
6
.
117
8
.
58
 = 50%
n factor of KSCN = 8 – 2 = 6
 meq. = 0.6 × 6  moles of KMnO4
m. m =
5
36
.
0
 moles =
5
36
.
0
× 10–3 ]
Q.9 Ametalwassubjectedtoelectromagneticradiationshavingvariedfrequency&agraphofmaximumKE
of ejected photonvs frequencywas plotted. From the graph calculate
(a) WorkfunctionofthemetalineV/atom
(b) PossiblerangeofKE(ineV)ofejectedphotoelectronif theyareaccelerated byapotential differenceof
2Vif  of electromagneticradiationis 20 nm. [6]
[Sol. From thegraph
(a) 10 eV = (
16
10
62
.
6
4
.
6
 × 6.62 × 10–34) / (1.6 × 10–19) – W
 10 eV = 40 eV – W
 W = 30 eV
(b) Fpho = nm
20
1240
 Max. KE = 62 – 30 = 32 eV
 Range of KE = 30 eV – 34 eV ]

6. Q.10 SupposeintheBohr'smodelofH–atom,insteadofelectrostaticforce,someotherforceisactingbetween
nucleus &electronwhichcausethepotentialenergytovaryas 5
r
5
k
.Whatshouldbethesign ofksothat
electroncanmoveinuniformcircularmotionaboutnucleus.Assumingquantizationofangularmomentum
derive expression for radius of nth orbit. [6]
[Sol. PE(r) = 5
r
5
k
 F(r) = –
dr
dE
= + 6
r
k
 For the above system to be possible PE  as r 
 k < 0
Also,
r
mv2
= 6
r
k
 v =
m
r
k
5

& mvr =

2
nh
 m ×
m
r
km
3

=

2
nh
 r3 = 2
2
2
h
n
m
k
4




 r =
3
2
2
2
h
n
m
k
4




& KE =
2
mv
2
1

2
mv
2
1
= 5
r
k
2
1
]

7. Q.11 Threeflasksofidenticalvolume areconnectedthroughtubes ofnegligiblevolumewithtwovalvesboth
initially closed. Container I is at temperature 2T & contains 1 mole H2, 2 mole O2 & 3 mole SO2 &
containerIIandIIIaremaintainedat temperatureT.ValveIisopenedforaverylongtime(tillmovement
of gases stop).Valve Iis then closed& valve IIis opened for a veryshort time (so that rate of effusion
remainsconstant)&thenclosedagain.AssumingthatcontainerII&IIIcontainednothinginitiallyanswer
thefollowingquestions.
(a) What is thenumber of moles in the two container I& IIjust beforeopening valve II?
(b) What is the ratio of moles of each specie in container IIIafter opening value II? [6]
[Sol. AfteropeningvalveIforalongtimegaseswillmovetillpartialpressureofeachissame.
 Vsame Psame  nT should be same for container I& II
 moles get divided in theratio 1:2 in thethree containers
 Container I = 2
H
3
1
, 2
O
3
2
, 2
SO
3
3
Container II = 2
H
3
2
, 2
O
3
4
, 2
SO
3
6
After opening valve II for a short time, effusion will occur

2
2
O
H
r
r
=
2
32
2
1
=
1
2
&
2
2
O
H
r
r
=
2
64
3
1
= 2
3
4
 2
H
r : 2
O
r : 2
SO
r = 2
3
4
: 2
3
2
: 1 ]

8. PART -C
Q.12 An organic compound contains 1 – OH group, 1  bond & 1 ketone group other than carbon and
hydrogen.Itisknownthat''bondis nonterminal &alcoholisaprimaryalcohol.224mlofthegaseous
molecule at STP is exploded with unknown but excessamount of O2 andthe resultantreaction caused
a contractionofVml at STP. The acidic gas obtained in the reaction was treated with excess of NaOH
& the resultant solution was divided into two equal halves. One half required 35 ml of 1 N HCl using
phenolphthalein indicator & other half required 60 ml of 1 N HCl using MeOH indicator. Using the
aboveinformationanswerthequestions
(a) Predict thegeneralformulaofcompound
(b) Predictthemolecularformulaofcompound
(c) What will bethevalue ofVml (contractionduetocombustion &cooling)
(d) Predictthepossiblestructuralformula(e) [1.5+3+1.5+1.5=7.5]
[Sol. Gen. formula = CnH2n–4O2
O. C + O2  CO2 + H2O
(g) (g) (g) (l)
CO2 + NaOH  Na2CO3
x n
0 n – 2x x

2
x
× 1 + 




 
2
x
2
n
× 1 = 35
&
2
x
× 2 + 




 
2
x
2
n
× 1 = 6
 x = 50
n – 2x = 20  n = 120
 m.m of NaOH = 120
m. m of CO2 = 50
Also, since all CO2 formed O.C whose I D m.m reacted.
(b)  n = 5  C5H6O2
(c) C5H6O2 + 2
O
2
13
 5CO2 + 3H2O
 Contraction in moles =
2
15
- 5 =
2
5
Contractioninvolume=25ml
(d)
C
–
C
C
–
C
–
C
|
|
|
O
OH

C
–
C
C
C
–
C
|
|
|
OH
O


]

9. Q.13 An atomicorbital has radial wavefunction represented as
)]
a
18
r
a
5
.
8
r
)(
a
7
r
a
5
.
5
r
[(
r
ke
)
r
( 2
0
0
2
2
0
0
2
2
'
k
/
r





 
where k, k' & a0 are constants.
(a) How manyradial nodes are present at what distance from the nucleus? (answer interms of a0)
(b) Draw a radial probabilitydistribution curve for the above orbital clearlyhighlighting the parameters
which arebeingplotted & nodes & antinodes.
(c) Inwhichperiodofperiodic tabledoes thefirst elementwith lastelectron inthis orbital is expectedto go.
What will bethe total number ofelements whichcan be presentin that period?
[4+1.5+2=7.5]
[Sol.(a) Polynomial is of 4th degree
 4 roots
 4 radial nodes
(r2 – 5.5a0r + 7 2
0
a ) = 0
 (r – 3.5a0)(r – 2a0) = 0
 r = 2a0 & 3.5a0
Also, (r2 – 8.5a0r + 18 2
0
a ) = 0
 (r – 4.5a0)(r – 4a0) = 0
 r = 4a0, 4.5a0
 4r nodes at 2a0, 3.5a0, 4a0, 4.5a0
(b) n–l–1 = 4 l = 2  n = 7d
(c) 7d goes in 8th period
no. of subshells (8s, 8p, 7d, 6f, 5g)
no. of orbitals (1 + 3 + 5 + 7 + 9)
no. of elements = 48 ]

10. Q.14 Completethemissingdatainthefollowinggraph.
(a) moles, n = 2, ideal gas subjected to a process in which P(atm) × 0.0821 = V (lit)
Calculate tan  & C.
(b) Calculate slope &intercept of thegraph if wt. ofsilver salt of succinicacid is plotted with wt. of silver
residue obtaineddue toheatingthesilversalt. [4+3.5=7.5]
[Sol.(a)P × 0.0821 = V
log P = log V – log 0.0821  y = mx + c  c = – log 0.0821 & slope = 1
(b)
Acid
Succinic
OH
C
CH
CH
C
HO
|
|
|
|
O
O
2
2 



 
Ag
Salt
Silver
OAg
C
CH
CH
C
AgO
|
|
|
|
O
O
2
2 



 
 Ag
Ag

332
Wss × 2 × 108 = W
WAg
 Wss = Ag
W
108
2
332


 y = mx + c
c = 0
Slope(m) =
108
2
332

= 1.537 ]

11. Q.15(a)Calculate H ofthefollowingreactionfrom the data(all in kJ/mole)
NH3 (g) + HCl (g)  NH4Cl (g)
(i) H proton gain enthalpy of NH3 (g) = – 720
(ii) H bond dissociation of HCl (g) = 100
(iii) H ionization energyof H = 1300
(iv) H Electron gain enthalpyof Cl = – 350
(v) H lattice energy of NH4Cl (s) = – 700
(b) If107gmofamixturecontainingNH3 (g)and HCl (g)istobedividedsoas toobtainmaximumamount
of energy.What should be the wt. of each specie taken for this and what is the amount of energy.
[5+2.5]
[Sol.(a)NH3(g) + HCl (g)  NH4Cl(g) H = ?

(b) For maximum energyequal moles so that no. L. R.
 a × 53.5 = 107  a = 2
 ER = 2 × 370 & wt = 34 & 73 ]