SUBJECT – CHEMISTRY
CLASS – XII
TIME – 3 Hrs.
MAX. MARKS - 70
GENERAL INSTRUCTIONS1. All questions are compulsory.
2. Question no 1 to 8 are very short answer questions and carry 1 mark each.
3. Question no 9 to 18 are short answer questions and carry 2 marks each.
4. Question no 19 to 27 are also short answer questions and carry 3 marks
each.
5. Question no 28 to 30 are long answer questions and carry 5 marks each.
6. Use log tables if necessary. Use of calculators is not allowed.
1. What is the total number of atoms per unit cell in a body centered cubic (bcc) structure? (1)
2. What is a primary cell? Give an example.
(1)
3. What causes Brownian movement of colloid particles in a colloidal solution?
(1)
4. What is the basicity of H3PO4 ?
(1)
5. Write the IUPAC name of the following compound:
(1)
CH3-O-CH2-CH2-OCH3
6. What is the name of the following reaction?
RCOCl + H2 Pd+BaSO4 RCHO + HCl
(1)
7. Write a chemical reaction in which the iodide ion replaces the diazonium group in a
diazonium salt.
(1)
8. Name a substance that can be used as an antiseptic as well as a disinfectant.
(1)
9. State Raoult’s law for the solution of volatile liquids. How will you define Ideal and Nonideal solutions on the basis of Raoult’s law?
OR
Define the term Osmotic pressure. Discuss how the molecular mass of a substance can be
determined by a method based on measurement of Osmotic pressure?
(2)
2
–1
10. Λm for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm mol respectively. Calculate
Λ°for HAc.
(2)
11. Represent the Galvanic cell in which the following reaction takes place:
Zn(s) + 2Ag+ (aq)  Zn2+(aq) + 2Ag(s)
State:
(i) Which of its electrodes is negatively charged?
(ii) The reaction taking place at each of its electrodes.
(iii) The carriers of current within the cell.
(2)
12. What do you understand by the term glycosidic linkage? Draw the structure of Sucrose and
mark glycosidic linkage in it.
(2)
13. (i) Write two differences between DNA and RNA.
(ii) How do you explain amphoteric behavior of amino acids?
(2)
14. Complete the following reaction equations:
(i) C6H5N2Cl + KI  ……
(ii) CH2=CH2 + Br2  …..
(2)
15. (i) Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms
isocyanides as the chief product. Explain.
(ii) In the following pair of halogen compounds, which would undergo SN2 reaction faster
and why?
“
”
(2)
16. (i) Why are ranitidine and cimetidine better antacids than sodium hydrogen carbonate or
magnesium hydroxide?
(ii) Why do soaps not work in hard water?
(2)
17. Assign a reason for each of the following statements:
(i) Ammonia is a stronger base than phosphine.
(ii) Sulphur in vapour state exhibits a paramagnetic behavior.
(2)
18. Draw the structures of following molecules:
(i) XeOF4
(ii) BrF3
(2)
19. Explain what is observed:
(i) When a beam of light is passed through a colloidal sol.
(ii) An electrolyte, NaCl is added to hydrated ferric oxide sol.
(iii) Electric current is passed through a colloidal sol?
(3)
20. 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature
will water boil at 1.013 bar ? Kb for water is 0.52 K kg mol-1.
(3)
21. (a) In terms of band theory, what is the difference between
(i) a conductor and an insulator
(ii) a conductor and a semiconductor?
(b) A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cub
and P at the body-centre. What is the formula of the compound? What are the coordination
numbers of P and Q?
(3)
22.Write chemical equations for the following processes:
(i) Chlorine reacts with hot concentrated solution of sodium hydroxide.
(ii) H3PO3 is heated.
(iii) PtF6 and xenon are mixed together.
OR
Complete the following chemical equations:
(i) Ca3P2(s) + H2O(l)  …..
(ii) Cu2+(aq) + NH3(aq) excess  …..
(iii) F2(g) + H2O(l)  …..
(3)
23. Explain why
(i) zinc not extracted from zinc oxide through reduction using CO?
(ii) copper matte is put in silica lined converter?
(iii) At a site, low grade copper ores are available and zinc and iron scraps are also
available. Out of these two scraps iron would be more suitable for reducing the leached copper
ore.
(3)
24. Account for the following observations:
(i) Methyl amine solution in water reacts with ferric chloride solution to give a precipitate
of ferric hydroxide
(ii) PKb for aniline is more than that of methylamine.
(iii) Aniline does not undergo Friedel-Crafts reaction.
(3)
25. (i) What are the monomeric repeating units of Nylon-6 and Nylon-6,6?
(ii) Discuss the main purpose of vulcanization of rubber.
(3)
26. How are the following conversions carried out?
(i) Propene  Propan-2-ol.
(ii) Benzyl chloride  Benzyl alcohol.
(iii) Ethyl magnesium chloride  Propan-1-ol.
(3)
27. (a) Why is geometrical isomerism not possible in tetrahedral complexes having two
different types of unidentate ligands coordinated with the central metal ion ?
(b) A solution of [Ni (H2O) 6]2+ is green but a solution of [Ni(CN)4]2– is colorless. Explain.
(Ni=28).
(3)
28. (a) Mention any four factors that affect the rate of a chemical reaction.
(b) The half life for decay of radioactive 14C is 5730 years. An archaeological artifact
containing wood has only 80% of the 14C activity as found in living trees. Calculate the age of
the artifact.
OR
(a) What is half life period? Show that the half life period for a first order reaction is
independent of the initial concentration of the reactants.
(b) For a first order reaction, show that time required for 99% completion is twice the time
required for the completion of 90% of reaction.
(2, 3)
29. (a) Complete each synthesis by giving missing products:
(i)
(ii)
(iii)
(b) Describe the following:
(i) Cannizzaro reaction
(ii) Hell-Volhard-Zelinsky (HVZ) reaction.
OR
(a) State reasons for the following:
(i) Monochloroethanoic acid has a higher pKa value than dichloroethanoic acid.
(ii) Ehanoic acid is a weaker acid than benzoic acid.
(b) An organic compound (A) with molecular formula C8H8O forms an orange-red
precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the
presence of sodium hydroxide. It neither reduces Tollens’ or Fehlings’ reagent, nor does it
decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives
a carboxylic acid (B) having molecular formula C7H6O2. Identify the compounds (A) and (B)
and explain the reactions involved.
(2, 3)
30. Explain giving reasons:
(5)
(i) Transition metals and many of their compounds show paramagnetic behavior.
(ii) The enthalpies of atomization of the transition metals are high.
(iii)The transition metals generally form coloured compounds.
(iv)Transition metals and their many compounds act as good catalyst.
(v) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidizing.
OR
(a) Write steps involved in the preparation of (i) Na2CrO4 from chromite ore and (ii)
K2MnO4 from pyrolusite ore.
(b) What may be the possible oxidation states of the transition metals with the
following d- electronic configurations in the ground state of their atoms?
3d34s2, 3d54s2 and 3d64s2. Indicate relative stability of oxidation state in each case.
(2, 3)
*************************************************************************
MARKING SHCEME
CLASS XII CHEMISTRY
1.
2.
2
In primary battery the reaction occurs only once and after use over a period of time becomes
dead. Leclanche cell or Dry cell is an example.
1
1/2+1/2
3.
4.
5.
6.
7.
Unbalanced bombardment of the particles by the molecules of dispersion medium.
3
1,2-dimethoxyethane
Rosenmund reduction
C6H5N+Cl - + KI
C6H5I + KCl + N2
1
1
1
1
1
8.
Phenol (or any other correct one)
1
9.
Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each
component in the solution is directly proprtional to its mole fraction.
1
The solutions which obey Raoult’s law over the entire range of concentration are known as
ideal solutions. The ideal solutions have two other important properties. The enthalpy of
mixing of the pure components to form the solution is zero and the volume of mixing is also
zero, i.e.,
mixH = 0, mixV = 0
½
2
2
When a solution does not obey Raoult’s law over the entire range of concentration, then it is
called non-ideal solution. The vapour pressure of such a solution is either higher or lower
than that predicted by Raoult’s law (equation 2.16). If it is higher, the solution exhibits
positive deviation and if it is lower, it exhibits negative deviation from Raoult’s law.
OR
The extra pressure applied on the solution side that just stops the flow of solvent to solution
through semi-permeable membrane is called osmotic pressure of the solution.
½
1
½
Here π is the osmotic pressure and R is the gas constant.
½
Thus knowing the quantities w2, T, π and V we can calculate the molar mass of the solute.
10.
1
1
11.
The galvanic cell is depicted as:
Zn(s) Zn2+(aq)  Ag+(aq)Ag (s)
Zinc electrode is negatively charged
½
½
The ions formed i.e Zn2+ and Ag+ in the solution are the carriers of the current within the
cell.
(iii)
At anode: Zn(s)Zn2+(aq) + 2eAt cathode: 2Ag+(aq)+ 2e-  2Ag(s)
½
½
12.
The two monosaccharides are joined together by an oxide linkage formed by the loss of a
water molecule. Such a linkage between two monosaccharide units through oxygen atom is
called glycosidic linkage.
1
1
13.
S.No. DNA
RNA
½
1.
2-deoxyribose sugar
Ribose sugar
½
2.
Double stranded -helix structure
Single stranded -helix structure
Or Any two correct differences
(ii) In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino 1
group can accept a proton to give a dipolar ion known as zwitter ion.
Therefore, in zwitter ionic form, the amino acid can act both as an acid and as a base.
Thus, amino acids show amphoteric behaviour.
14.
15.
C6H5I + KCl + N2
BrCH2-CH2Br
(i) KCN is predominantly ionic and provides cyanide ions in solution.
Although both carbon and nitrogen atoms are in a position to donate
electron pairs, the attack takes place mainly through carbon atom and
not through nitrogen atom since C—C bond is more stable than C—N
bond. However, AgCN is mainly covalent in nature and nitrogen is free
to donate electron pair forming isocyanide as the main product.
(ii)
1+1
1
1
16.
(i) Soaps are sodium or potassium salts of long-chain fatty acids. Hard water contains
calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace
sodium or potassium from their salts and form insoluble calcium or magnesium salts of fatty
acids. These insoluble salts separate as scum.
1
This is the reason why soaps do not work in hard water.
(ii) Antacids such as sodium hydrogen carbonate, magnesium hydroxide, and aluminium
hydroxide work by neutralising the excess hydrochloric acid present in the stomach.
However, the root cause for the release of excess acid remains untreated.
Cimetidine and rantidine are better antacids as they control the root cause of acidity. These
drugs prevent the interaction of histamine with the receptors present in the stomach walls.
Consequently, there is a decrease in the amount of acid released by the stomach. This is why
cimetidine and rantidine are better antacids than sodium hydrogen carbonate, magnesium
hydroxide, and aluminium hydroxide.
1
17.
(i)The lone pair of electrons on N atom in NH3 is directed and not diffused / delocalized as
it is in PH3 due to larger size of P/ or due to availability of d-orbitals in P.
(ii) S2 molecule like O2, has two unpaired electrons in antibonding π* orbitals.
1+1
18.
19.
1+1
(Please note that lp e’s are placed rightly)
(i) When a beam of light is passed through a colloidal solution, then scattering of light is
observed. This is known as the Tyndall effect. This scattering of light illuminates the path of
the beam in the colloidal solution.
(ii) When NaCl is added to ferric oxide sol, it dissociates to give Na+ and Cl- ions. Particles
of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of
negatively charged Cl- ions.
(iii) The colloidal particles are charged and carry either a positive or negative charge. The
dispersion medium carries an equal and opposite charge. This makes the whole system
neutral. Under the influence of an electric current, the colloidal particles move towards the
oppositely charged electrode. When they come in contact with the electrode, they lose their
charge and coagulate.
1x3=3
20.
21.
Moles of glucose = 18 g/ 180 g mol–1 = 0.1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg-1
For water, change in boiling point
ÄTb = Kb × m = 0.52 K kg mol–1 × 0.1 mol kg–1 = 0.052 K
Since water boils at 373.15 K at 1.013 bar pressure, therefore, the
boiling point of solution will be 373.15 + 0.052 = 373.202 K.
(a) (i) The valence band of a conductor is partially-filled or it overlaps with a higher energy,
unoccupied conduction band.
½
½
1
1
½
On the other hand, in the case of an insulator, the valence band is fully- filled and there is a
large gap between the valence band and the conduction band.
(ii) In the case of a conductor, the valence band is partially-filled or it overlaps with a higher
energy, unoccupied conduction band. So, the electrons can flow easily under an applied
electric field.
On the other hand, the valence band of a semiconductor is filled and there is a small gap
between the valence band and the next higher conduction band. Therefore, some electrons
can jump from the valence band to the conduction band and conduct electricity.( diagrams
½
not required)
(b) It is given that the atoms of Q are present at the corners of the cube.
Therefore, number of atoms of Q in one unit cell
It is also given that the atoms of P are present at the body-centre.
Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1
½
½
Hence, the formula of the compound is PQ.
The coordination number of both P and Q is 8.
1
22.
(i)3Cl2+6NaOH 5NaCl+NaClO3+3H2O
(ii)4H3PO3  3H3PO4+PH3
(iii)Xe+[PtF6]-
OR
(i)Ca3P2(s)+ 6H2O(l)3Ca(OH)2(aq) + 2PH3(g)
(ii)Cu2+(aq)+ 4NH3(aq)[Cu(NH3)4]2+(aq)
(iii)2F2(g)+ 2H2O(l)  4H+ (aq) + 4F-(aq) + O2(g)
1x3=3
23.
(i) The standard Gibbs free energy of formation of ZnO from Zn
is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is
not extracted from ZnO through reduction using CO.
(ii) Copper matte contains Cu2S and FeS. Copper matte is put in a silica-lined converter to
remove the remaining FeO and FeS present in the matte as slag (FeSiO3). Also, some silica
is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the
remaining FeS and FeO are converted to iron silicate (FeSiO3) and Cu2S is converted into
metallic copper.
(assign marks for either correct description or only
reactions)
(iii) Zinc being above iron in the electrochemical series (more reactive
metal is zinc), the reduction will be faster in case zinc scraps are
used. But zinc is costlier metal than iron so using iron scraps will be
advisable and advantageous.
24.
(i)It is because in aniline the –NH2 group is attached directly to the benzene ring.It results
in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and
thus making it less available for protonation.
(or any other suitable reason)
(ii) Methyl amine in water gives OH- ions which react with FeCl3 to give precipitate of ferric
hydroxide/ or
+
1x3=3
CH3NH2 + H2O
Fe3+ + 3OH-
CH3NH3OH-
CH3NH+3 +OH-
Fe (OH)3
(iii)Aniline does not undergo Friedel-Crafts reaction due to salt formation with aluminium
chloride, the Lewis acid.
1X 3=3
25.
(i) The monomeric repeating unit of nylon 6 is
Caprolactam.
, which is derived from
The monomeric repeating unit of nylon 6, 6 is
which is derived from hexamethylene diamine and adipic acid.
,
1
1
(ii) Natural rubber though useful has some problems associated with its use. These
limitations are discussed below:
1. Natural rubber is quite soft and sticky at room temperature. At elevated temperatures (>
335 K), it becomes even softer. At low temperatures (< 283 K), it becomes brittle. Thus, to
maintain its elasticity, natural rubber is generally used in the temperature range of 283 K-335
1
K.
2. It has the capacity to absorb large amounts of water.
3. It has low tensile strength and low resistance to abrasion.
4. It is soluble in non-polar solvents.
5. It is easily attacked by oxidizing agents.
Vulcanization of natural rubber is done to improve upon all these properties. In this process,
a mixture of raw rubber with sulphur and appropriate additive is heated at a temperature
range between 373 K and 415 K.
26.
(i) If propene is allowed to react with water in the presence of an acid as a catalyst, then
propan-2-ol is obtained.
(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is
produced.
(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced
which gives propan-1-ol on hydrolysis.
1x3=3
27.
(a) Tetrahedral complexes do not show geometrical isomerism because
the relative positions of the unidentate ligands attached to the central
metal atom are the same with respect to each other.
(b) In [Ni(H2O)6]2+,
is a weak field ligand. Therefore, there are unpaired electrons in
Ni2+. In this complex, the d electrons from the lower energy level can be excited to the
higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H2O)6]2+ is
coloured.
In [Ni(CN)4]2−, the electrons are all paired as CN- is a strong field ligand. Therefore, dd transition is not possible in [Ni(CN)4]2−. Hence, it is colourless.
28.
(a)Factors affecting rate of chemical reaction are:
(i)Concentration of reactants
(ii)Temperature
(iii)Presence of catalyst
(iv)Surface Area
(v)Activation energy
1
1
1
1/2x4=2
(any four)
(b) k
=
0.693
t1/2
k = 0.693
5730 y
K = 1.21 x 10-4 y-1
t = 2.303
k
log [ A0 ]
[A]
k = 2.303
log 100
-4 -1
1.21 x 10 y
80
k=
2.303
log 1.25
1.21 x 10-4 y-1
k=
2.303
x 0.0969
1.21 x 10-4 y-1
½
½
1
= 1845 years
1
OR
(a) The half-life of a reaction is the time in which the concentration of a
reactant is reduced to one half of its initial concentration. It is
represented as t1/2.
1
It can be seen that for a first order reaction, half-life period is
constant, i.e., it is independent of initial concentration of the reacting
species.
(b) For a first order reaction, the time required for 99% completion is
1
½
For a first order reaction, the time required for 90% completion is
½
29.
Therefore, t1 = 2t2
Hence, the time required for 99% completion of a first order reaction is twice the time
required for the completion of 90% of the reaction.
1
1
(a) (i)
1
1
(ii)
(iii)
1
(b)
(i)
1+1=2
(ii)
OR
(a)
(i)Because the stability of conjuguate base of monochloroethanoic acid is less due to
presence of one electron withdrawing -Cl group than in dichloroethanoic acid.
(ii)This is because of greater electronegativity of sp2 hybridised carbon to which carboxyl
carbon is attached.
( or any other suitable reason)
1+1=2
1+1+1=3
(b)
30.
(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the
presence of unpaired electrons with each electron having a magnetic moment associated with
its spin angular momentum and orbital angular momentum. However, in the first transition
series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is
only because of the unpaired electron.
(ii) Transition elements have high effective nuclear charge and a large number of valence
electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of
atomization of transition metals is high.
(iii) Most of the complexes of transition metals are coloured. This is because of the
absorption of radiation from visible light region to promote an electron from one of
the d−orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of
orbitals having different energies. Therefore, the transition of electrons can take place from
one set toanother. The energy required for these transitions is quite small and falls in the
visible region of radiation. The ions of transition metals absorb the radiation of a particular
wavelength and the rest is reflected, imparting colour to the solution.
(iv) The catalytic activity of the transition elements can be explained by two basic facts.
(v) Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing
agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be
written as
configuration, which is a more stable configuration. In the case of Mn3+ (d4), it
acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled dorbital and is highly stable.
OR
(a)
(i)
1x5=5
4FeCr2O4 + 8Na2CO3 + 7O2
8 Na2CrO4 + 2 Fe2O3 + 8 CO2
1
(ii)
2MnO2 + 4KOH + O2
2K2MnO4 + 2H2O
1
(b) 3d34s2(Vanadium): Oxidation states +2,+3,+4,+5
Stable oxidation state: +4 as VO2+ ,+5 as VO43-
1
3d54s2(Manganese): Oxidation states +2,+3,+4,+5,+6,+7
Stable oxidation states: +2 as Mn2+ ,+7 as MnO-4
1
3d64s2(Iron): Oxidation states +2,+3
Stable oxidation state: +2 in acidic medium, +3 in neutral or in alkaline medium.
1
BLUE PRINT
CLASS XII – CHEMISTRY
S.NO.
UNIT
VSA
(1 MARK)
1
2
Solid State
1(1)
3
4
5
6
Electrochemistry
7
8
9
10
p-Block elements
11
12
SA I
(2 MARKS)
Solutions
1(1)
SA II
(3 MARKS)
4
3(1)
5
2(1)
4(2)
5
5(1)
1(1)
General principles and
processes of Isolation of
Elements
1(1)
4(2)
Haloalkanes &
Haloarenes
Alcohols,Phenols &
Ethers
Aldehydes,Ketones &
Carboxylic Acids
4
3(1)
3
3(1)
8
5(1)
3(1)
13
Organic Compounds
containing Nitrogen
14
15
16
Biomolecules
4
3(1)
1(1)
4
5(1)
1(1)
3(1)
4
3(1)
Chemistry in Everyday
Life
1(1)
2(1)
TOTAL
8(8)
20(10)
6
4
4(2)
Polymers
5
3
4(2)
1(1)
5
3(1)
d- & f- Block Elements
Coordination Compounds
TOTAL
3(1)
Chemical Kinetics
Surface Chemistry
LA
(5 MARKS)
3
3
27(9)
15(3)
70(30)