Class : XIII (Sterling)
Time : 3 hour Max. Marks : 60
INSTRUCTIONS
1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2
questions of"Match the Column" type and Part-C contains 4 subjective type questions.Allquestions are
compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS
and Pages. If you found some mistake like missing questions or pages then contact immediately to
the Invigilator.
PART-A
(i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each.
There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer.
PART-B
(iii) Q.1 to Q.2 are "Match the Column" type whichmayhave oneor more than one matching options and
carry8 marks for each question. 2 marks willbe awarded for each correct matchwithin a question.
ThereisNONEGATIVE marking. Markswillbeawardedonlyifallthecorrectalternativesareselected.
PART-C
(iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded
onlyifallthe correct bubbles are filled in your OMR sheet.
2. Indicate the correct answer for eachquestionbyfilling appropriate bubble(s) inyour answer sheet.
3. Use onlyHB pencilfor darkening the bubble(s).
4. Use ofCalculator, Log Table, Slide Rule and Mobile is not allowed.
5. The answer(s) ofthe questionsmust bemarked byshading the circlesagainst the questionbydark HBpencil
only.
PART TEST-2
PART-B
For example if Correct
match for (A) is P, Q;for
(B) is P, R; for (C) is P
and for (D) is S then the
correct method for filling
the bubble is
P Q R S
(A)
(B)
(C)
(D)
PART-C
Ensure that all columns
(4 before decimal and 2
after decimal) are filled.
Answer having blank
column willbe treated as
incorrect. Insert leading
zero(s) if required after
rounding the result to 2
decimalplaces.
e.g. 86 should be filled as
0086.00
.
.
.
.
.
.
.
.
.
.
PART-A
For example if only 'B'
choice is correct then,
the correct method for
filling thebubble is
A B C D
For example ifonly'B &
D' choices are correct
then, thecorrect method
for fillingthe bubbles is
A B C D
The answer of the
question in any other
manner (such as putting
, cross , or partial
shading etc.) will be
treated as wrong.

Class - XIII Chemistry Part Test - 2
USEFUL DATA
Atomic Mass:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Ba = 137,
Co = 59, Hg = 200, Pb = 207, He = 4, F=19.
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1.
PARTA
Select the correct alternative(s) (choose one or more than one)
Q.1 For thefollowing cellreaction
2Hg(I) + 2Ag+
(aq) 2Ag(s) + Hg2
+2
(aq) is given that
0
Ag
/
Ag
E  = 0.80 V,
, [Ag+] = 10–3M
0
Hg
/
Hg 2
2
E  = 0.785V,
,  
2
2
Hg
= 10–1M
(A) The forwardreaction is spontaneous (B*) The backward reaction is spontaneous
(C) Ecell = 1.585 V (D) Ecell = 3.170 V
[Sol: For the given cell2Hg(I) + 2Ag+
(aq)  2Ag(s) + Hg2
+2(aq) is given that
0
cell
E = 0.800 – 0.785 = 0.015 volt
Ecell = 0.015 –
 2
3
1
10
10
log
2
059
.
0


= –0.1325 volt
 The backward reaction is spontaneous ]
Q.2 Aspirin decomposes slowlyat roomtemperature byreacting with water inthe atmosphere to produce
acetic acid and salicyclic acid
+ H2O  AcOH +
Consider thefollowing initialconcentrationandinitialrate data forthis reaction.
[Aspirin], M [H2O], M Initial rate, Ms–1
0.0100 0.0200 2.4 × 10–13
0.0100 0.0800 9.6 × 10–13
0.0300 0.0200 7.2 × 10–13
0.0200 0.0300 7.2 × 10–13
Which ofthe following is the correct rate law for this reaction?
(A) Rate = K[aspirin] (B*) Rate = K[aspirin] [H2O]
(C) Rate = K[H2O] (D) Rate = K[aspirin]2 [H2O]
[Sol: The rate ofthe reaction increases that manytimes as the increase in concentration ofboth aspirin and
water. So the rate ofreaction is first order with respect to the reactants. ]

Q.3 Select the incorrect statement(s):
(A*) In 
3
4
PO ion, the P–O bond order is 1.33.
(B*) Covalent bond can be formed by overlapping of py
and pz
orbitals
(C) Bond angle in H2
O is lesser than bond angle in OCl2
molecule.
(D*) SeF4
and CH4
have same shape.
[Sol.(A)
B.O. =
.
S
.
R
bond
total
of
.
No
= 5/4 = 1.25
(B) Covalent bond is not formed byoverlapping ofPy
and Pz
orbitalbecause distribution ofelectrondensity
on different axis.
(C)
(D)
Sea saw Tetrahedral ]
Q.4 Which of the following statement (s) is/are incorrect :
(A*) Ni(CO)4
and [Ni(CN)4
]2–
both are d sp2
hybridised
(B*) The splitting energyofK3
[VF6
] is greater than that ofK3
[V(CN)6
]
(C) Paramagnetic behaviour increases in the order [V(H2
O)6
]3+
< [Ni(H2
O)6
]3+
< [Co(H2
O)6
]3+
(D*) [Ti(H2
O)6
]3+
is a colourless species.
Q.5
)
A
(
12
9H
C 

 
 4
KMnO
)
B
(
4
6
8 O
H
C 
 
 Fe
/
Br2 C8
H5
BrO4
(Onlyone product)
A is 
(A) (B) (C) (D*)

Q.6
2
ZnCl
HCl
HCHO



 
 
A
KCN
.
alc
.
aq

 
 B
PhCHO
)
ii
(
Na
O
H
C
)
i
( 5
2



 


C 

D
2
2
CO
)
ii
(
H
/
O
H
)
i
(




 


E 

HBr
F
(A*) Compound D is
(B) Compound Ais
(C) Compound F is
(D*) Compound F is
[Sol:
2
ZnCl
HCl
H
C
H
|
|
O



 
 





 
 KCN
.
alc
.
aq
ICB
E
EtOH
EtO







 
 2
CO
PART B
MATCH THE COLUMN
INSTRUCTIONS:
Column-Iand column-IIcontains fourentrieseach. Entries ofcolumn-Iare to be matchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q.1 ColumnI ColumnII
(A) Dithionousacid (P) S–O–S bond is not present
(B)Thiosulphuric acid (Q)AllS atomin the molecule has oxidation state +3
(C) Caro's acid (R)Acidic strengthofallH atoms present inthe
molecule is different
(D) Pyrosulphuric acid (S) at least one S atomhas oxidation state +6 in molecule
[Ans. (A) P,Q; (B)P; (C) P,R,S;(D) S]
[Sol: (A) H2S2O4 [P, Q]
(B) H2S2O3 [P]
(C) H2SO5 [P, R, S]
(D) H2S2O7 [S] ]

Q.2 Column I ColumnII
(A) CH3
–CN + HCl + AlCl3
(P)
(B) +AlCl3 (Q)
(C)
3
2
3
CO
Na
.
aq
of
presence
in
POCl
PhNMeCHO





 
 
(R)
(D) 



 
 
H
)
ii
(
H
O
O
H
)
i
( 2
2
(S)
[Ans: A  Q; B  Q; C  P; D  R]
[Sol: (A) CH3
–CN + HCl + AlCl3

3
2
NH
O
H


 

(B) + Cl
–
C
–
CH
||
O
3 + AlCl3
 + HCl + AlCl3
FriedelCraft acylationreaction.
(C)
e.g. ofvilsmeier reaction i.e., formylation




 
 
H
)
ii
(
H
O
O
H
)
i
( 2
2
(D) e.g. of Dakin reaction ]

PART C
Q.1 Anorganic compoundgives substitutionas wellas eliminationaccording to the following reaction
OH
R


 +
whendextrorotatorysample, heatedwithexcessalkylalcoholthenangleofrotationobservedat following
intervals
t(sec) 0 3 
Angle ofrotation(deg) 450 250 50
Calculate the angle ofrotation after 10 sec. (take log 2 = 0.3)
[Sol: K =




r
r
r
r
log
time
303
.
2
t
0

5
25
5
45
log
3
303
.
2


= 5
r
5
45
log
10
303
.
2
t 

20
40
log
3
303
.
2
= 5
r
40
log
10
303
.
2
t 
3
3
.
0
303
.
2 
= 5
r
40
log
10
303
.
2
t 
1 = log 5
r
40
t 
or 5
r
40
t  = 10
40 = 10 rt – 50  rt = 9 Ans.]
Q.2 A 0.347 g of a metal (A) wasdissolved in dil. HNO3
. This solution gave a red colouration to Bunsen
flame andonevaporationgave0.747g ofmetaloxide(B). (A) also reactedwithN2
forminga compound
(C) and with H2
, it forms (D). On reacting 0.1590 g of (D) with H2
O, a gas (E) was evolved and a
sparinglysolublecompound(F) formed, whichgave a stronglybasicsolutionand required 200mlof0.1
M HCl to neutralise it. (D) whentreated with Lewis acid (G), forms a wellknown powerfulreducing
agent (H). Deduce the molar mass of (H) if(G) dissolves in excess ofNaOH but does not dissolve in
excess ofaqueous NH3
. [Ans: Molar mass ofH i.e. LiAlH4
= 38]
[Sol: (A) reacts withN2
to formnitride,withH2
to formhydride (D) whichreacts withwater toformsparingly
soluble compound (F). The solution of(A)in dil. HNO3
imparts red colour to Bunsen flame, So (A) is
Lithium(Li)
Li + HNO3
 LiNO3
+ H2

(A) 
Li2
O (B)
6 Li + N2
 2Li3
N (C)
2Li + H2
 2LiH (D)
LiH + H2
O  LiOH + H2

(F) (E)

LiOH + HCl  LiCl + H2
O
As the Lewisacid G dissolves inexcess ofNaOH but does not dissolves inexcess ofaqueous NH3
. So,
G isAlcompound and it isAlCl3
.AlCl3
whenreacts with (D). LiH, it forms a powerfulreducing agent
LiAlH4
.
LiH +AlCl3
 LiAlH4
(H)
moles of (A) =
7
347
.
0
= 0.0495
2M + 2HNO3
 2MNO3
+ H2


M2
O
moles of (B) formed =
30
747
.
0
= 0.0246
no. of moles of (B) =
2
1
× no. of moles ofA
A
0.0246 
2
0495
.
0
2Li + H2
 2LiH
LiH + H2
O  LiOH (F) + H2
 (E)
moles of LiH =
8
159
.
0
 0.02
moles ofLiOH formed = moles of LiH consumed = 0.02
moles ofacid required to neutralise the 0.02 mole ofLiOH = 0.200 × 0.1 = 0.02
Finally molar mass of H i.e. LiAlH4
= 7 + 27 + 4 = 38 Ans. ]
Q.3 Hydrocarbon (A) ofM.F. C8
H16
reacts with O3
/Ph3
Pyield two moles of B(C4
H8
O). B is oxidized to C
and then treated withPCl5
to give D. Dontreatement with hydroxylaminegave a compound (E) which
gives red colourationto Ferric chloride solution. When (E) is treated with strong HClit gave (F) with
evolutionofCO2
. (F) ontreatment with HNO2
gave acompound which gave ayellow precipitate with
NaOI. What is the molecular weight ofF (in gm). [Ans: 59]
[Sol: 2
(B)
(D) (C)

hydro-oxamic acid
(E) E give red colour withFeCl3
solution
e.g. ofLossen rearrangement

 
 2
HNO (F) H2
N–CHMe2
it give yellow ppt. with NaOI
molecular weight of F = 36 + 14 + 9 = 59 gm. Ans. ]
Q.4 At 270
C the emfofthe cellPb | PbCl2
||AgCl|Ag is 0.42V and its temperature coefficient is
p
dT
dE






=
–4.2 ×10–4
V/K. Calculate the modulus ofheat offormationofAgCl(s) in Kcaliftheheat offormation
of PbCl2
is –85.09 Kcal. (Use 1 cal = 4.2 J)
[Sol: Cell Pb + 2AgCl 2Ag + PbCl2
Gcell
= –nFEcell
= –2 × 96500 × 0.42 Joules 1 cal = 4.25
=
2
.
4
42
.
0
96500
2 

 = –19300 cal = –19.3 Kcal
S = nF 





dT
dE
= 2 × 96500 × –4.2 × 10–4
J
= 2 × 96500 ×
2
.
4
10
2
.
4 4


 Cal
= –2 × 9.65
= –19.3 cal
G = H – TS
–19.3 × 103
= H – 300 × –19.3

–19.3 × 103
= H + 300 × 19.3
Hcell
= –19.3 × 103
– 300 × 19.3
= –19.3 [1000 + 300]
= –19.3 × 1300
= –25090 cal
= –25.090 Kcal
Hcell
= AgCl
PbCl H
2
H 2

–25.09 = –85.09 – 2HAgCl
HAgCl
= –30 Kcal
|HAgCl
|= 30 KcalAns. ]