KENDRIYA - VIDYALAYA BASTI
PRE-BOARD I 2015-16
BLUE PRINT (XII CHEM)
S.NO UNIT
VSA
(1
mark)
SA I
(2
marks)
SA II
VBQ
LA
(3
marks)
(4
marks)
(5
marks)
4(2)
Total
1
Soild State
4(2)
2
Solutions
3
Electrochemistry
2(1)
3(1)
5(2)
4
Chemical Kinetics
2(1)
3(1)
5(1)
5
Surface Chemistry
3(1)
4(1)
6
General principles and processes
3(1)
3(1)
6(2)
8(3)
5(1)
1(1)
5(1)
of Isolation of Elements
7
p -Block Elements
2(1)
8
d- and f-Block Elements
9
Coordination Compounds
10
Haloalkanes and Haloarenes
11
Alcohols, Phenols and Ethers
12
Aldehydes, Ketones and Carboxylic
Acids
1(1)
13
Organic Compounds Containing
1(1)
3(1)
4(2)
1(1)
3(1)
4(2)
1(5)
1(1)
5(1)
3(1)
3(1)
3(1)
4(2)
3(1)
3(1)
1(5)
6(2)
Nitrogen
14
Biomolecules
15
Polymers
16
Chemistry in Everyday Life
Total
4(1)
4(1)
3(1)
5(5)
10(5)
36(12)
3(1)
4(1)
15(3)
70(26)
KENDRIYA VIDYALAYA BASTI
PRE-BOARD I 2015-16
CLASS: XII
SUBJECT:
CHEMISTRY (Theory)
MARKING-SCHEME
Q.
no.
Value- points
Mar
ks
1
A bond formed between -NH2 group of one carboxylic acid & -COOH group of other carboxylic
acid.
1
1
2
NH3 , because it has higher critical temperature.
½
+1/2
1
1
1
1
1
3
4
Because of long hydrocarbon part, which is hydrophobic in nature hexanoic acid is slightly
soluble in water.
5
N-methyl propanamide.
6
For FCC,
Z =4
1
1
½
Density of cell= 7.2 gcm-3 , a = 288pm = 2.88 x 10-8cm
d= Z x M/a3 x NA
½
M = (d x a3 x NA)/Z
7
8
= 7.2 x (2.88 x 10-8)3x 6.022 x 1023/4
½
= 25.9gmol-1
½
(a) PCl3 + 3H2O  H3PO3 + 3HCl
(b) XeF6 +H2O XeOF4 + 2HF
1
1
At anode 2Fe(s) --- 2Fe2+ + 4e-
½
At cathode O2(g) +4H+ (aq) + 4e- -- H2O(l)
½
2
2
The overall reaction:
2Fe(s) + O2(g) +4H+ (aq) ) --- 2Fe2+ + H2O(l)
1
2
9
10
a.
b.
a.
b.
i-Frenkel defect ii-Both Shottky & Frenkel defect
Decreases.
Rate of reaction becomes 4 times.
Rate of reaction becomes 1/4 times.
1
1
1
1
2
OR
The reactions which are not truly of the first order but under certain conditions behave
as first order.
1
H+
C12H22O11 + H2O -- C6H12O6 + C6H12O6
Sucrose
glucose
fructose
11
12
1
2
a. Amino acids may be acidic, basic or neutral depending upon the relative number of
amino and carboxyl group.
Equal no. of amino and carboxyl group = Neutral
More amino gp =Basic
More carboxyl gp = Acidic
b. The sugar having –CHO groupso can redce Tollen’s & Fehling’s reagent. E.g glucose
(any one example)
a)Correct order
b) Correct reaction
½
½
½
½
1
3
1
1
1
c) Correct reason
3
13
1
Benzoyl peroxide
CH3-CH=CH2 + Br2 ------------- CH3-CH2-CH2Br
ii) Due to following reasons:
1
-Double bond character of C-X bond in haloarenes.
- More electronegative nature of carbon(sp2 hybridised ) to which halogen is bonded in
Haloarene.
1
Any other correct reason
3
14
Antiseptic
Disinfectant
1. These are antibacterial agents These are antibacterial agents which
which prevent the growth of
kill the micro organisms . And are
micro organisms or may even
applied to inanimate objects.
kill them. And are applied on
wounds.
2. Eg. Chloroxylenol. (or any
Eg. 1% Phenol solution.
other correct example)
b. To impart antiseptic properties in soaps.
1+1
1
OR
a. Antacids: The substances which neutralize the excess acid and raise the pH to an
Appropriate level in stomach are called antacids.
eg. Mixture of aluminium and magnesium hydroxide.
b. Antifertility drug: These drugs are used to check pregnancy in women.
For eg. Norethindrone.
1
1
c.
15
Antioxidants: An antioxidant is a molecule that inhibits the oxidation of other molecules.
Oxidation reactions can produce free radicals. In turn, these radicals can start chain reactions.
When the chain reaction occurs in a cell, it can cause damage or death to the cell. Antioxidants
terminate these chain reactions by removing free radical intermediates, and inhibit other
oxidation reactions.
For eg. Butylated hydroxyl toluene(BHT )
i) For solution I
Λm = k/ c
k= 1
1
3
½
ρ
½
3 -1
Λm= 1x 1000 = 1 x 1000 cm L
ρ c
58 Ωcm x 0.20mol L-1
½
Λm = 86.2 Ω-1cm2mol-1
ii) For solution II
Λm= k x 1000 = 2.5 x 10-2 Ω-1cm-1x 1000
c
0.20mol L-1
½
½
-1
2
-1
Λm = 125 Ω cm mol
½
Solution II will have larger molar conductivity.
16
For first order reaction
k = 0.693=
½
0.693
t1/2
4 hr
½
t = 2.303 x log [R]0
[R]
k
½
log [R]0= k x t
[R]2.303
½
log [R]0 = 0.693x 7 = 4.851
[R]2.303 x 4
9.212
½
0
Log[R] = 0.5266
[R][R]
½
= 3.362
[R]0= 1M
[R] =
3.362
[R] = Antilog 0.5266
0
[R]0___
3
= 1/3.362
= 0.2974M
After 7 hrs Sucrose left = 0.2974M
3
17
18
a) Due to high bond energy of p∏- p∏bond between nitrogen atoms.
b) The bond energy of H2S bond is more than that of H2Te bond.
c) Due to small size of fluorine and more hydration enthalpy.
White phosphorous
1. It is insoluble in carbon
disulphide.
2. It is less stable as compared
to red phosphorous due to
the angular strain.
Red phosphorous
1
1
1
1+1
It is soluble in carbon disulphide.
More stable than White
phosphorous.
1
b) Chlorine water is yellow due to presence of hypochlorousacid(HClO). On standing HClO
being unstable it decomposes to form HCl.
19
20
21
a) Tyndall-effect occurs.
b) Coagulation of the sol takes place.
c) Electrophoresis occurs.
a) The role of NaCN is to dissolve the gold to form aurocynide complex. The metal is
obtained by displacement from this complex.
4Au + 8NaCN + 2H2O + O2 ----- 4Na[Au(CN)2] + 4NaOH
b) The role of silica is to remove the iron oxide (obtained during the oxidation)
∆
FeO + SiO2 --------------- FeSiO3 (slag)
c) Iodine forms a compound with Zirconium at decomposes further to give pure metal.
∆
Zr(impure) + 2I2 ------ ZrI4
∆
ZrI4 ---------Zr(pure) + 2I2
a) +3 Oxidation state.
b) d2sp3
c) Octahedral
d) Paramagnetic
e)Potassium hexacyanatoferrate(III) ion
f) [Fe(CN)6]3-
22
23
a) i) DNA ii) RNA
b) Vitamin K
c) β – D – Galactose
3
3
1
1
1
½
½
½
½
1/2
1/2
3
½
½
½
½
½
½
3
and β – D – Glucose
a) Because they are not biodegradable.
b) These can be easily recycled and coloured polythene contain lot of poisonous
substances.
c) Ethene
d) Care and concern towards environment, public awareness etc.
½+½
1
½+½
1
1
1
1
3
4
24
a) Depression in freezing point will be same, because both are non electrolytes and give
same number of solute particles. (in moles)
b) ∆Tf = Kf x Wsolute x 1000
Msolute x Wsolvent
1+1
1
1.8 = 1.86 x Wsolute x 1000
½
1000 x 74.5
½
Wsolute = 1.8 x 74.5
1.86
1
Wsolute = 72.09 g
OR
a) If Pressure applied on the solution should is larger than the osmotic pressure , the
process of reverse osmosis occurs.
b) i = 3
c) Relative lowering in vapour pressure = p10- p1= x2
p10
1
1
½
= p10- p1= W2 x M1
p10M2 x W1
½
1.013 – 1.004 = 2 x 18
1.013
M2 x 98
M2 = 1.013 x 2 x 18
1
0.009 x 98
M2= 41.35 g mol-1
25
a) Because of tendency of oxygen to form double bond with transition metals.
b) Electronic configuration of Mn2+ is 3d5which is half filled hence stable, hence IIIrd
ionisation enthalpy is very high. In case of Fe2+ electronic configuration is 3d6. Hence it
can lose one electron easily to give more stable configuration 3d5.
c) This is due to poor shielding by 5f electrons in the actinides than that by 4f electrons in
the lanthanoids.
d) i) Cr2O72- +14H+ + 6I- 2Cr3+ + 3I2 + 7H2O
ii) MnO4- + 8H+ + 5Fe2+ Mn2+ + H2O + 5Fe3+
OR
a) In aqueous solution: Cr2O72- + H2O ↔
2CrO42- + 2H+
Dichromate ion
Chromate ion
(Orange Red)
(Yellow)
In Acidic medium (i.e decreasing pH) equilibrium shifts towards backwards and the
colour is orange red. Whereas in basic medium equilibrium will shift forward and
the solution is yellow.
1
5
1
1
1
1
1
1
1
b) i) Because of d-d transition.
ii) Due to their ability to to adopt multiple oxidation state and to form complexes.
iii) Europium (II) has stable electronic configuration (half filled) ie [Xe] 4f7 5d06s0.
26
a) i) Ethanal gives iodoform test whereas propanal does not.
CH3CHO + 3NaOI HCOONa + CHI3 + 2NaOH
Ethanol
iodoform
(Yellow ppt.)
ii) Phenol and benzoic acid can be distinguished by ferric chloride test.
6C6H5OH + FeCl3Iron phenol complex (Violet colour)
1
1
1
5
1
1
3C6H5COOH + FeCl3 Ferric benzoate (Pale brown ppt.)
(b)
(i) 2HCHO + Conc.KOH ------ CH3OH + HCOOK
H2O
1
1
(ii) CH3CH2CHO + CH3MgBr -------- CH3CH2CH(CH3)(OH)
CO, HCl
(iii) C6H6
------- C6H5CHO
1
OR
(a)
1
(i)
1
(ii)
(b) i) Because in carboxylic acids the hydrogen bonding is much stronger than that in
alcohols, so they have high boiling points.
ii) In Chloroacetic acid, the -Cl shows –I effect and hence creates less electron density
on the oxygen of carboxylic acid thus the release of proton becomes easier.
iii) Butanone < Propanone <Propanal<Ethanal.
--------------------------------------------------------------------
1
1
1
5