NAVODAYA VIDYALAYA SAMITI, BHOPAL REGION
PREBOARD –I (2020 -21)
CLASS- XII CHEMISTRY THEORY (043)
MM : 70 Time: 3 Hours
General Instructions. Read the following instructions carefully.
a)
There are 33 questions in this question paper. All questions are compulsory.
b)
Section A: Q. No. 1 to 2 are case-based questions having four MCQs or Reason
Assertion type based on given passage each carrying 1 mark.
c)
Section A: Question 3 to 16 are MCQs and Reason Assertion type questions
carrying 1 mark each
d)
Section B: Q. No. 17 to 25 are short answer questions and carry 2 marks each.
e)
Section C: Q. No. 26 to 30 are short answer questions and carry 3 marks each.
f)
Section D: Q. No. 31 to 33 are long answer questions carrying 5 marks each.
g)
There is no overall choice. However, internal choices have been provided.
h) Use of calculators and log tables is not permitted.
SECTION A (OBJECTIVE TYPE)
1. Read the passage given below and answer the following
questions: (1x4=4)
Alcohols are classified into three types
i.e., primary, secondary and tertiary, depending upon the
nature of the carbon atom to which the
alcoholic (-OH) group is attached. These have common
physical as well as chemical
characteristics. However, they differ in their relative reactivities
towards different reagents. The popular
tests which can distinguish the three types of alcohols are
Victor Meyer’s test and Lucas reagent test.
The following questions are multiple choice questions. Choose the
most appropriate answer:
(i)
In
Lucas test, turbidity appears on heating. Predict the nature of alcohol.
a) Primary
alcohol
b)
Secondary alcohol
c)
Tertiary alcohol
d)
Butan-2-ol
(ii)
What
is the correct order of reactivity of the alcohols involving the cleavage of
C-OH bond?
a)
Methyl alcohol >Ethyl alcohol>Isopropyl alcohol>tertiary butyl alcohol
b)
Tertiary alcohol>isopropyl alcohol>ethyl alcohol>methyl alcohol
c)
Isopropyl alcohol>tertiary alcohol>ethyl alcohol>methyl alcohol
d) Tertiary alcohol>ethyl alcohol>isopropyl alcohol>methyl alcohol
OR
Which
of the following is a secondary allylic alcohol?
a)
But-3-en-2-ol
b)
But-2-en-2-ol
c)
Prop-2-enol
d)
Butan-2-ol
(iii)
Out of water,
tert-butyl alcohol, propan-2-ol and ethyl alcohol, which is most acidic?
a)
tert-butyl alcohol
b)
propan-2-ol
c)
ethyl alcohol
d)
water
(iv) An organic
compound ‘X’ with molecular formula C3H8O on heating with copper
gives
compound ‘Y’ which reduces Tollen’s reagent. ‘X’ on reaction with sodium
metal
gives ‘Z’ . What is
the product of reaction of ‘Z’ with 2-
chloro-2-methylpropane?
a)
CH3CH2CH2OC(CH3)3
b)
CH3CH2OC(CH3)3
c)
CH2=C(CH3)2
d)
CH3CH2CH=C(CH3)2
Read
the passage given below and answer the following questions: (1x4=4)
It is often recommended that the first aid kit to be
kept in the houses must have a small lump of alum. In case, bleeding occurs
while doing shave in the bathroom or from a knife cut in the kitchen, alum
should be immediately rubbed on the affected portion. Bleeding stops and
medical aid if required, can be obtained later on.
2.
In these questions (Q. No 5-8 , a statement of assertion followed by a
statement of reason
is
given. Choose the correct answer out of the following choices.
a)
Assertion and reason both are correct statements and reason is correct
explanation for assertion.
b)
Assertion and reason both are correct statements but reason is not correct
explanation for assertion.
c)
Assertion is correct statement but reason is wrong statement.
d)
Assertion is wrong statement but reason is correct statement.
(i)
Assertion: Bleeding stops when alum is rubbed on the affected portion.
Reason: Alum can be used to remove
colloidal impurities from water.
(ii)
Assertion: Excess of electrolyte can bring about coagulation of colloidal
solution.
Reason: The flocculating ion of the electrolyte
removes the charge from the sol particles.
(iii)
Assertion: Coagulation power of Al+3 is more than that of Na+
Reason: Greater the valency of the
flocculating ion added, greater is its power to cause coagulation (Hardy-Schulze rule)
(iv)
Assertion: Addition of gelatin to a lyophobic sol may result in to its
coagulation
Reason: Smaller the gold number of
protective colloid, greater will be its protective power.
OR
Assertion:
persistent dialysis may also result in to coagulation of a lyophobic colloid.
Reason:
Electrophoresis involves the migration of dispersion medium under the influence
of electric field when the dispersed
phase particles are prevented from moving.
Following
questions (No. 3 -11) are multiple choice questions carrying 1 mark each:
3 Which of the following option will be the
limiting molar conductivity of CH3COOH
if the limiting molar conductivity of
CH3COONa is 91 Scm2mol-1? Limiting molar conductivity for individual ions are
given in the following table.
|
S.No
|
Ions
|
limiting
molar conductivity / Scm2mol-1 |
|
1 |
H+ |
349.6
|
|
2 |
Na+ |
50.1
|
|
3 |
K+ |
73.5
|
|
4 |
OH- |
199.1
|
a)
350 Scm2mol-1 b)
375.3 Scm2mol-1
c)
390.5 Scm2mol-1 d)
340.4 Scm2mol-1
4. Proteins
are found to have two different types of secondary structures viz. α-helix
and β-pleated sheet structure, α-helix structure of protein is
stabilized by:
a)
Peptide bonds
b) Van
der Waal’s forces
c)
Hydrogen bonds
d) Dipole-dipole interactions.
OR
Curdling of milk is an example of:
a) breaking of peptide linkage
b) hydrolysis of lactose
c) breaking of protein into amino acids
d) denauration
of proetin
5.
A binary solution is prepared by mixing n-heptane and ethanol. Which one of the
following
statements is correct regarding the
behaviour of the solution?
a)
The solution formed is an ideal solution
b)
The solution is non-ideal showing positive deviation from
Raoult’s law.
c)
The solution is non-ideal showing negative deviation from
Raoult’s law.
d)
n-heptane shows positive deviation while ethanol shows
negative deviation from Raoult’s law.
6. Because of lanthanoid contraction:
a)
separation of the lanthanoid elements become difficult.
b)
there is a very small difference in the atomic size of the transition metals of
5th
and 6th
period
in the
same group.
c)
there is a gradual decrease in the basic strength of the hydroxides of
lanthanoids
d)
all are correct
OR
Which
of the following is a diamagnetic ion:
(Atomic
numbers of Cr, Mn, Ni and Cu are 24, 25, 28 and 29 respectively)
a)
Cr2+
b)
Cu+
c)
Ni2+
d) Mn2+
7. Method by which aniline cannot be prepared is
a)
degradation of benzamide with bromine in alkaline solution
b)
reduction of nitrobenzene with H2/Pd in ethanol
c)
potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis
with aqueous
sodium
hydroxide solution
d) hydrolysis of phenylisocyanide with acidic solution
OR
IUPAC
name of product formed by reaction of methyl amine with two moles of ethyl
chloride
a)
N,N-Dimethylethanamine
b)
N,N-Diethylmethanamine
c)
N-Methyl ethanamine
d) N-Ethyl - N-methylethanamine
8.
What is the correct electronic configuration of the central atom in [Co(H2O)6]Cl3 based
on crystal
field theory?
a) e4t22
b) t2g4eg2
c) t2g6eg0
d) e2t24
OR
Which is true for the complex [Ni(en)2]2+ ?
a)
paramagnetic, dsp2,
square planar, CN of Ni = 2
b)
diamagnetic, dsp2,
square planar, CN of Ni = 4
c)
diamagnetic, sp3,
tetrahedral, CN of Ni = 4
d) paramagnetic, sp3, square planar, CN of Ni = 4
9. The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(ox)(SO4)2] respectively are
a)
+3, +2 and +4
b)
+3, 0 and +6
c)
+3, +4 and +6
d) +3, 0 and +4
10.
Identify A,B,C and D:
a)
A = C2H4, B= C2H5OH, C= C2H5NC, D= C2H5CN
b)
A= C2H5OH, B= C2H4, C = C2H5CN, D=C2H5NC
c)
A = C2H4, B= C2H5OH, C= C2H5CN, D= C2H5NC
d)
A= C2H5OH, B= C2H4, C= C2H5NC, D= C2H5CN
11.The
crystal showing Frenkel defect is :
|
(a) |
 |
(b) |
 |
|
(
c) |
 |
(d
) |
 |
In
the following questions (Q. No. 12 - 16) a statement of assertion followed by a
statement of
reason
is given. Choose the correct answer out of the following choices.
a)
Assertion and reason both are correct statements and reason is correct
explanation for assertion.
b)
Assertion and reason both are correct statements but reason is not correct
explanation for assertion.
c)
Assertion is correct statement but reason is wrong statement.
d)
Assertion is wrong statement but reason is correct statement.
12. Assertion: Deoxyribose, C5H10O4
is not a carbohydrate.
Reason:
Carbohydrates are optically active polyhydroxy aldehyde or ketone or the
compound which produce such units on hydrolysis.
13. Assertion: Sulphur exhibits
paramagnetic behaviour in the vapour state.
Reason:
In vapour state, sulphur partly exists as S2 molecules which have two unpaired
electrons in antibonding π* orbital.
14. Assertion: Osmotic pressure is a
colligative property.
Reason:
Osmotic pressure is directly proportional to molarity.
OR
Assertion:
Acetone-aniline mixtures shows negative
deviation from Raoult’s law
Reason:
H-bonding between acetone and aniline is stronger than that between acetone-
acetone
and aniline-aniline.
15.
Assertion: The pKa
of acetic acid is
lower than that of phenol.
Reason:
Phenoxide ion is more resonance stabilized than acetate ion.
16.
Assertion: tert-Butyl methyl ether on treatment with HI at 373 K gives a
mixture of methyl alcohol and tert-butyl
iodide.
Reason:
The reaction occurs by SN2
mechanism.
SECTION B
The
following questions, Q.No 17 – 25 are short answer type and carry
2 marks each.
17. With the help
of resonating structures explain the effect of presence of nitro group at ortho
position in chlorobenzene.
OR
Carry
out the following conversions in not more than 2 steps:
(i) Aniline to chlorobenzene
(ii) 2-bromopropane to 1-
bromopropane
18. 5% aqueous solution of a non-volatile solute
was made and its vapour pressure at 373 K was
found to be 745 mm. Calculate the molar
mass of solute.
19. (i) Using the valence bond approach, write the
hybridization of iron ion in the following
complex ion. Also predict its magnetic
behaviour :[Fe(H2O)6]Cl3
(ii)Write the IUPAC name of the
coordination complex: [CoCl2(en)2]NO3
OR
(i) What
is meant by chelate effect? Give an example.
(ii) Predict
the geometry and magnetic moment of [NiCl4]2-
ion.
20.
The decomposition of NH3
on platinum surface is
zero order reaction. What are the rates of
production of N2 and H2 if k = 2.5 × 10-4 mol L-1 s-1
?
OR
The following results have been
obtained during the kinetic studies of the reaction:
|
P + 2Q à R + 2S Exp. |
Initial P(mol/L) |
Initial Q (mol/L) |
Init. Rate of Formation of R (M min-1) |
|
1 |
0.10 |
0.10 |
3.0 x 10-4 |
|
2 |
0.30 |
0.30 |
9.0 x 10-4 |
|
3 |
0.10 |
0.30 |
3.0 x 10-4 |
|
4 |
0.20 |
0.40 |
6.0 x 10-4 |
Determine the rate law expression for the reaction.
21. The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood ? (log 3= 0.4771, log 7 = 0.8540 , Half-life of C-14 = 5730 years )
22. Give a mechanism for following reaction:
(CH3)3CBr + OH - à (CH3)3COH + Br-
23. Draw the structure of the following compounds:
(a) H2S2O8
(b) XeOF4
24. (a) Write chemical equations for
Reimer Tiemann reaction?
(b) Which compound out of the following
pairs will react faster in SN2 reaction and why ?
CH2=CHBr or CH2=CHCH2Br.
25.
A compound is consist of three
element P, Q and R . Atoms of element P form ccp lattice
and those of the element Q occupy 1/3rd of tetrahedral voids and those of the
element R occupy2/3rd of octahedral voids. What is the formula of
the compound formed by the elements P, Q and R?
SECTION C
Q.No
26 - 30 are Short Answer Type II carrying 3 mark each.
26. Give reasons for the following:
i. Transition elements exhibit variable
oxidation states .
ii. Zirconium and hafnium have almost
similar atomic radii.
iii. Cu+ ion is not known in aqueous
solution.
OR
Observed
and calculated values for the standard electrode potentials of elements from Ti
to Zn in
the
first reactivity series are depicted in figure (1):
FIGURE 1 (source
NCERT)
Explain
the following observations:
i.
The general trend towards less negative Eo values across the series
ii.
The unique behaviour of Copper
iii. More negative Eo values of Mn and Zn
27. Arrange the following in increasing order of
property specified:
i. triethylamine, ethanamine, diethylethanamine
(solubility in water)
ii. C4H9NH2, C2H5N(CH3)2 , (C2H5)2NH (boiling point)
iii. Ethanamine, N-Ethyl ethanamine,
N,N-Diethyl ehanamine (Basic strength)
OR
i. Give a chemical test to distinguish
between aniline and N,N-dimethylaniline.
ii. Write chemical
equations for Hoffmann bromamide degradation reaction.
iii. Methyl amine is soluble in water but
not aniline. Explain.
28. An
element ‘X’ (atomic mass = 40 g mol-1) having fcc structure, has unit
cell edge length of 400 pm. Calculate the density of ‘X’ and the
number of unit cells in 4 g of ‘X’.
29.
(i) what is the effect of denaturation on the
structure of proteins?
(ii) What is the difference between a
nucleoside and nucleotide?
(iii) Represent Alanine in the zwitter
ionic form.
30. i. Explain why N does not form pentahalides
while phosphorus does.
ii. . Arrange the following in
decreasing order of bond dissociation enthalpy F2 , Cl2 , Br2 , I2
iii. Electron gain enthalpy of fluorine
is less negative than chlorine. Justify
SECTION D
Q.No
31 to 33 are long answer type carrying 5 marks each.
31.
Answer the following questions:
(2+3)
(i) Write the balanced chemical reaction for
reaction of Cl2 with
hot and concentrated NaOH.
(ii)
Complete the following chemical equations:
(a)
PbS + O3 à
(b)
H2SO4 + Cu à
(iii) Give reason
(a) Amongst all noble gases only xenon is known to
form compounds with oxygen and fluorine.
(b) I-Cl is more reactive than I2 .
OR
Answer
the following questions:
(1+4)
(i)
Arrange the following in the increasing order of basic strength: (Give
reason)
NH3, PH3, AsH3, SbH3, BiH3
(ii)
Account for the following observation
(a) Acidity of oxo –acids of chlorine is HOCl < HOClO < HOClO2 < HOClO3.
(b) Nitrogen can’t form compounds like R3N=O, while phosphorus can form R3P=O .
(c) Fluorine never acts as the central atom in
polyatomic interhalogen compounds.
(iii)
Complete the following reaction:
XeF6 + 3H2O à
32.
An organic compound
‘A’ C8H6 on
treatment with dilute H2SO4 containing
mercuric sulphate gives compound ‘B’. This compound ‘B’ can also be obtained
from a reaction of benzene with acetyl chloride in presence of anhy AlCl3. ‘B’ on treatment with I2 in
aq. KOH gives ‘C’ and a yellow compound ‘D’. Identify A, B, C and D. Give the
chemical reactions involved.
(5)
OR
(i) Give a chemical reaction to distinguish between methanol and ethanol. (1+3+1)
(ii) How will you carry out the following conversions:
a) Benzaldehyde to α-hydroxyphenylacetic
acid
b) Propanone to propene
(iii) Arrange the following in increasing order of acidic
character:
CH3COOH, HCOOH, C6H5OH, 2,4,6-trinitrophenol
33. (i) How conductivity is affected with dilution?
(1+3+1)
(ii) Calculate the emf of the following cell at 298 K:
Al(s)/Al3+ (0.15M)//Ni2+ (0.025M)
/Ni(s)
(Given Eo(Al3+ Al) = -1.66 V, Eo(Ni2+ /Ni) =
0.25V, log 0.15 = -0.8239, log 0.025 = -1.6020)
(iii)
Calculate ∆Go for above cell reaction.
OR
(i) Why on dilution the molar conductivity of CH3COOH increases drastically, while that of
CH3COONa increases gradually?
(ii) A
strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2
and a strip of silver metal is placed on a 1-molar solution of AgNO3.
An electrochemical cell is created when the two solutions are connected by a
salt bridge and the two strips are connected by wires to a voltmeter.
Calculate
the cell potential, E, at 25oC for the cell if the initial
concentration of Ni(NO3)2
is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar.
[EoNi2+/Ni =
-0.25 V, EoAg+/Ag = 0.80 V, log 10-1 = -1
(iii) ) The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity.
(1+3+1)
NAVODAYA VIDYALAYA SAMITI BHOPAL REGION
PREBOARD - I
(2020 – 2021) CHEMISTRY – XII
MARKING SCHEME
|
Q.No |
VALUE POINT |
MARKs |
|
1 i |
a |
1 |
|
ii |
b |
1 |
|
or ii |
a |
1 |
|
iii |
d |
1 |
|
iv |
a |
1 |
|
2 i |
b |
1 |
|
ii |
a |
1 |
|
iii |
a |
1 |
|
iv |
d |
1 |
|
Or iv |
c |
1 |
|
3 |
c |
1 |
|
4 |
c |
1 |
|
Or 4 |
d |
1 |
|
5 |
b |
1 |
|
6 |
d |
1 |
|
Or 6 or or |
b |
1 |
|
7 |
c |
1 |
|
Or 7 |
d |
1 |
|
8 |
b |
1 |
|
Or 8 or or |
b |
1 |
|
9 |
b |
1 |
|
10 |
a |
1 |
|
11 |
a |
1 |
|
12 |
d |
1 |
|
13 |
a |
1 |
|
14 |
a |
1 |
|
Or 14 |
a |
1 |
|
15 |
c |
1 |
|
16 |
c |
1 |
|
17 |
 |
2 |
|
Or 17 |
(I) C6H5NH2 -------- NaNO2 +HCl /273 -278 K
---à C6H5N2Cl -----CuCl/HCl ---à C6H5Cl (ii) CH3CH(Br)CH3 ---alc KOH à CH3CH=CH2 -----
HBr/ peroxide à CH3CH2CH2Br |
1+1 |
|
18 |
PoA -PA/
PoA=WBXMA/MBX WA 760-745/760 = 5X 18/ MB X 95 MB = 24 g/mol
|
½+1 +1/2 |
|
19 |
(i)d2sp3
hybridisation, magnetic moment = 5.9 BM
(ii)Dichloridobis(ethane -1,2-diammine)cobalt (III) nitrate |
1/2 +1/2 +1 |
|
Or 19 |
(i)Whenever a central metal ion is
surrounded by poly dentate ligand to form ring type /cyclic structure is
called chelate ex [Co(en)3]+3 (ii) tetrahedral , magnetic moment
= 2.83 BM
|
1 +1 |
|
2O
Or 20 or |
For the reaction , 2NH3 à N2 +3H2 Reaction rate = -1/2x d[NH3]/dt =
d[N2]/dt = 1/3x d[H2]/dt = k[NH3]O =K Then , this relationship, Rate of production of N2 = d[N2]/dt =k = 2.5 x 10-4
mol L -1S-1 Rate of production of H2 = d[H2]/dt = 3k = 3X2.5 x 10-4 = 7.5 X 10-4
mol L -1S-1 Or Let the rate law expression be Rate
Ro = K [P]m [Q] n
Using the given data
3.0x 10-4 = K[0.10]m
[0.10] n eqn (i)
9.0x 10-4 = K[0.30]m [0.30] n eqn
(ii)
3.0x 10-4 = K[0.10]m
[0.30] n eqn (iii)
6.0x 10-4 = K[0.20]m
[0.40] n eqn (iv) From eqn
(iii)/ eqn (i) 3.0x 10-4 / 3.0x 10-4 = K[0.10]m [0.30] n /
K[0.10]m [0.10] n 1
= (3)n , (3)0 =
(3)n , n = 0 From eqn
(ii)/ eqn (iii) 9.0x 10-4 / 3.0x 10-4 = K[0.30]m [0.30] n /
K[0.10]m [0.30] n 3 = (3)m , (3)1 =
(3)m , m = 1 Thus the rate law expression for
the reaction is Rate = K [P]1 [Q] 0 |
½ x4
½ x4 |
|
21 |
k = 0.693/t1/2 k = 0.693/5730 years-1 t = 2.303/ k log Co/Ct
let Co = 1, Ct = 3/10 so Co/Ct = 1/
(3/10) = 10/3 t = 2.303 x 5730/0.693 log 10/3
t
= 9957 years |
½
1
1/2 |
|
22 |
 |
1+1 |
|
23 |
(a)  (b) XeOF4  |
1+1 |
|
24 |
(a)  (b)
CH2=CHCH2Br will react faster in SN2 reaction than CH2=CHBr because in CH2=CHCH2Br
, Br is attached sp3 hybridised carbon and in CH2=CHBr ,Br is attached sp2
hybridised carbon
|
1+1 |
|
25 |
Atoms of Element P
forms ccp lattice or fcc lattice Number of atoms of element P in
ccp /fcc unit cell = 4 Total number of tetrahedral voids
in unit cell = 2x4 =8 Atoms of element Q occupy 1/3 rd
of tetrahedral voids Number of atoms of element Q in unit
cell = 1/3X8 = 8/3 Total number of octahedral voids
in unit cell = 4 Atoms of element R occupy 2/3 rd
of octahedral voids Number of atoms of element R in
unit cell =2/3X8 = 16/3 Formula of the compound =P4 Q8/3
R16/3 Multiplying by 4 , =P12Q8 R16 dividing by 4 , =P3Q2R4 formula of the compound is P3Q2R4 |
½ x4 |
|
26 |
(i)
(n-1)d electrons also participate in bond formation because (n-1)d
electrons and ns electrons has
comparable energies (ii)
Due to lanthanoid contraction. (iii)
Cu+ easily disproportionates in aqueous solution as follows 2 Cu+ à
Cu+2 + Cu |
1+1+1 |
|
Or 26 |
(i) The general trend towards less
negative Eo V values across the series is related to the general increase in
the sum of the first and second ionisation enthalpies. (ii) The high energy to transform
Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy. (iii) The stability of the
half-filled d sub-shell in Mn2+ and the completely filled d10 configuration in Zn2+ are
related to their more negative Eo V values
|
1+1+1 |
|
27 |
i. triethylamine<diethylethanamine
< ethanamine ( increasing order of solubility in water) ii.
C2H5N(CH3)2 <
(C2H5)2NH< C4H9NH2 (increasing order of boiling
point) iii.
Ethanamine< N ,N-Diethyl ehanamine < N-Ethyl ethanamine (increasing
order of Basic strength) |
1+1+1
|
|
Or 27 |
(i)
Carbylamine test or Hinsberg test (ii)
RCONH2 + Br2 + 4NaOH à
RNH2 + Na2CO3 + 2NaBr + 2H2O (iii)
Due to large hydrophobic -C6H5 group , aniline is insoluble in water. |
1+1+1 |
|
28 |
Density, d =z.M/a3 .NA
d =4
x 40/(400x10-10)3 x 6.022x1023 d =4.15gcm-3 40 g (1 mol) element contain 6.022x1023
atoms or 6.022x1023/ 4 unit
cell So 4 g element contain 6.022x1023x 4 /40x4 = 1.5x1023
unit cell |
2+1 |
|
29 |
(i) Proteins loses its biological activities on denaturation /secondary
and tertiary structure is destroyed; primary structure remain intact (ii)
The base – sugar unit in any nucleic acid chain is called a
nucleosides whereas the base – sugar- phosphate unit in any nucleic acid
chain is called a nucleotides (iii)  (Zwitter ion )
|
1+1+1 |
|
30 |
(i)
N does not have vacant d orbital
for sp3d hybridization (ii)
Cl2 > Br2 >
F2> I2 (iii)
Due to too small size and high electron density on fluorine it repel
the incoming electron thus incoming electron does not experience much
attraction force. |
1+1+1 |
|
31 |
(i)3Cl2 + 6NaOH à 5NaCl + NaClO3 +3H2O (ii (a)
PbS + 4O3
à PbSO4 +4O2 (b) 2H2SO4 + Cu à
CuSO4+ SO2 +2H2O (iii) (a) Larger atomic size and
lower ionization energy of Xe, only
xenon is known to form compounds
with oxygen and fluorine (b) Interhalogen compound has lower
bond enthalpy than parent halogen due to improper overlapping. |
1+2+2 |
|
31 |
(i)BiH3
< SbH3 < AsH3< PH3 < NH3 Smaller the size greater the
electron density available to accept the proton. (ii)(a) HOCl < HOClO
< HOClO2 < HOClO3. Because order of stability of
conjugate base is as follows on the basis
of number of resonating structure /delocalization of electron cloud. OCl-
< OClO- < OClO2 - < OClO3- (b)Nitrogen does not vacant d
orbital and can not form dπ-pπ bond like phosphorus. (c) Fluorine never acts as the
central atom in polyatomic interhalogen compounds due to following reason i)
very small size ii)most electronegative atom iii) absence of vacant d orbital (iii) XeF6 + 3H2O à XeO3 + 6HF |
1+3+1 |
|
32 |
 
|
½ X4
1 1 1 |
|
Or 32 |
(i)
Iodoform test (ii)
(a)C6H5CHO
-------HCN---à C6H5CH(OH)CN -----H3O+
---à C6H5CH(OH)COOH (b) CH3COCH3
---NaBH4---à CH3CH(OH)CH3 --conc.H2SO4/heat
---à CH3CH=CH2 (iii) C6H5OH<2,4,6-trinitrophenol
< CH3COOH <HCOOH |
1+3+1 |
|
33 |
(i)
Conductivity decreases on
dilution because number of ion per unit volume decreases. (ii)
Al(s)/Al3+ (0.15M)//Ni2+ (0.025M)
/Ni(s) (Given Eo(Al3+/ Al) = -1.66 V, Eo(Ni2+ /Ni) = 0.25V 2Al à 2Al3+ + 6e 3Ni2++ 6e à 3 Ni --------------------------------------- Net cell
reaction:
2Al 3Ni2+ à 2Al3+ + 3 Ni Here number of electron
exchange n= 6 Eocell = Eocathode -Eoanode = 0.25 -(-1.66) = 1.91 V
Ecell = Eocell –
0.059/n log [Al3+]2/ [Ni2+]3 =1.91 - 0.059/6 log[ 0.15]2/ [0.025]3 =1.91 - 0.059/6{
2log[ 0.15]-3log [0.025]}
=1.91 - 0.059/6{
2x-0.8239 -3 x-1.6020]} =1.91 - 0.059/6( -1.6478+ 4.806 ) =1.91 - 0.059/6x3.1582
= 1.91 – 0.031056 Ecell = 1.87894V (iii) ∆Go
= - nFEo cell = -6x 96500x 1.91 = - 1105898 j = - 1105.89 kj |
1+3+1 |
|
Or 33 |
(i) On dilution the molar conductivity of CH3COOH increases
drastically, while that of CH3COONa increases gradually because CH3COOH is weak electrolyte
and dissociation of weak electrolyte increases on dilution whereas CH3COONa is strong
electrolyte which is completely ionised at any concentration
|
1 |
|
|
(ii)
Ni à +
Ni2++2e 2Ag++ 2e à 2Ag --------------------------------------- Net cell
reaction: Ni +2Ag+ à Ni+2+ 2Ag Here number of electron exchange n= 2 Eocell = Eocathode -Eoanode = 0.80 -(-0.25) = 1.05V
Ecell = Eocell –
0.059/n log [Ni2+]/ [Ag+]2 =1.05 - 0.059/2 log[ 0.1]/ [1.0]2 =1.05 - 0.059/2 log 10 -1 =1.05 -
0.059/2x(-1) =1.05 + 0.0295 = 1.0795 V |
3 |
|
|
(iii)
λm =
k x 1000/C 0.025 x 1000/0.20 25/0.20 = 125Scm2mol-1 |
1 |
