Download CBSE-12th/2011/CHEMISTRY

CBSE-12th/2011/CHEMISTRY
S.No
Q.1
Questions
'Crystalline solids are anisotropic in nature'. What does this statement mean?
Q.2
Express the relation between conductivity and molar conductivity of a solution
held in a cell.
Q.3
Define 'electrophoresis'.
Q.4
Draw the structure of XeF2 molecule.
Q.5
Write the IUPAC name of the following compound:
(CH3)3CCH2Br
Q.6
Draw the structure of 3-methylbutanal.
Q.7
Arrange the following compounds in an increasing order of their solubility in
water: C6H5NH2, (C2H5)2NH, C2H5NH2
Answers
Ans.1 The physical properties of crystalline solids such as electrical resistance or
refractive index show different values when measured along different directions
in the same crystals. This is the means of the above statement.
Ans.2
Molar conductivity of a solution is given by-
Where m=Molar Conductivity,k=Conductivity
Ans.3 The movement of charged particles in a fluid or gel under the influence of an
electric field.
Ans.4
Ans.5
2,2 –dimethylbromopropane
Ans.6
Ans.7 The more extensive the H−bonding, the higher is the solubility. C2H5NH2
contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2
undergoes more extensive H−bonding than (C2H5)2NH. Hence, the solubility in water
of C2H5NH2 is more than that of (C2H5)2NH. Further, the solubility of amines
decreases with increase in the molecular mass. This is because the molecular mass of
amines increases with an increase in the size of the hydrophobic part. The molecular
mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH.
Hence, the increasing order of their solubility in water is as follows:
C6H5NH2 < (C2H5)2NH < C2H5NH2
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Q.8
What are biodegradable polymers?
Q.9
The chemistry of corrosion of iron is essentially an electrochemical
phenomenon. Explain the reactions occurring during the corrosion of iron in
the atmosphere.
Ans.8 Biodegradable polymers are a specific type of polymer that breaks down after its
intended purpose to result in natural byproducts such as gases (CO2, N2), water, biomass,
and inorganic salts.
Ans.9 Corrosion is the deterioration of a metal as a result of chemical reactions between
it and the surrounding environment. The reaction at the anode is can be written as
follows.
Anodic reaction:
Electrons released at the anodic spot move through the metallic object to reach another
spot of object. There, in the presence of H+ ions, the electrons reduce molecular oxygen.
This spot behaves as the cathode. There H+ ions come either from H2CO3 , which are
formed due to the dissolution of carbon dioxide from air into water or from the
dissolution of other acidic oxides from the atmosphere in water.
Cathodic reaction is as follows:
Thus, the overall reaction is:
Q.10
Determine the values of equilibrium constant (Kc) and o ∆G for the following
reactions:
Q.11
Distinguish between 'rate expression' and 'rate constant' of a reaction.
Also, ferrous ions are further oxidized by atmosphere oxygen to ferric ions. These ferric
ions combine with moisture, present in the surroundings, to form hydrated ferric oxide
(Fe2O3, xH2O) i.e., rust.
Ans.10 Ni(s) +2Ag+(aq)
Ni+2 (aq) +2Ag(s)
E0=1.05V
The galvanic cell of given cell can be calculated as
Ni(s)/Ni+2 Ag +/Ag(s)
F=96500Cmol-1
∆rG0=-nFE0cell
=-2×96500×1.05V
=-202650Jmol-1
=-2.02.65KJmol-1
Ans.11
RATE OF REACTION
Rate of reaction is the change in concentration of a reactant or product per unit time.
The rate of reaction at any instant of time depends upon the molar concentrations of the
reactants at that time.
Rate of reaction is the change in concentration of a reactant or product per unit time.
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The rate of reaction at any instant of time depends upon the molar concentrations of the
reactants at that time.
A reaction is said to be of the zero order, first order, second order, and third order if some
of concentration terms is equal to 0, 1, 2 and 3 respectively
REACTION RATE CONSTANT
It is a constant of proportionality in the rate law equation and is equal to the rate of
reaction when the molar concentration of each of the reaction is unity.
Q.12
State reasons for each of the following:
(i) The N - O bond isNO2- is shorter than N-O bond in NO3(ii)SF6 is kinetically an inert substance
OR
State reason for the each of the following:
(i)
All the P-Cl bonds in PCl5 molecule are not equivalent.
(ii)
Sulphur has greater tendency for catenation than oxygen.
The rate constant is constant for a particular reaction at a particular temperature and does
not depend upon the concentrations of the reactants.Its units are always moles
litre<sub>-1</sub> time<sup>-1</sup>It units depend upon the order or reaction.
It is a constant of proportionality in the rate law equation and is equal to the rate of
reaction when the molar concentration of each of the reaction is unity.
The rate constant is constant for a particular reaction at a particular temperature and does
not depend upon the concentrations of the reactants.Its units are always moles
litre<sub>-1</sub> time<sup>-1</sup>It units depend upon the order or reaction.
Ans.12(i) Bond length increases with decreasing bond order. The nitrogen–oxygen bond
order in NO3- is less than the nitrogen–oxygen bond order in NO2- (1.3 versus 1.5,
respectively). Therefore, NO3- has longer nitrogen–oxygen bonds than does NO2-.
(ii) SF6 is chemically inert due the reason that 6F atom protects S atom from attack by
reagent to such an extent that even thermodynamically most favourable reaction like
hydrolysis do not occur as shown in fig.
OR
(i) There are two bonds in PCl5 .One is Equatorial and other is Axial, because of the
repulsion factor one is longer and the other shorter. The bond nature is trigonal
bipyramid. Two axial bonds of same length are at angle 90. Three equatorial bonds of
same length are at bond angle of 120. This is due to the minimum repulsion of charge
between the Cl atoms.
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Q.13
Q.14
Assign reasons for the following:
(i) Copper (I) ion is not known in aqueous solution.
(ii) Actinoids exhibits greater range of oxidation states than Lanthanoids
Explain the following giving one example for each:
(i)Reimer-Tiemann reaction.
(ii)Friedel Craft's acetylation of anisole.
(ii)O=O is a much stronger bond than O-O (about 3 times). Also, O has a small size. S is
larger in size. so lp repulsion is less significant. Also, S-S bond is stronger than O-O
bond & S=S is less strong(less than 2 S-S bonds). This is also affected by the fact that O
forms strong bonds with mostly other elements than itself.
Ans.13
(i)In aqueous solution, Cu+ ion undergoes oxidation to Cu2+ ion. The relative stability of
different oxidation states can be seen from their electrode potentials. Due to more
reduction electrode potential value of Cu+, it undergoes oxidation reaction quite feasibly.
Hence, Copper (I) ion is not known in aqueous solution
(ii)Because actinoids have 5f, 6d and 7s energy levels which are comparable energies.
Therefore all these three sub shells can participate. But the most common oxidation state
of actinoids is +3.
Ans.14
(i)The Reimer-Tiemann reaction is an organic reaction used to convert a phenol to an ohydroxy benzalde-hyde using chloroform, a base, and acid work-up. The mechanism
begins with abstraction of the proton from chloroform with the base to form a
trichlorocarbanion which spontaneously loses a chloride ion to form a neutral
dichlorocarbene. The base also deprotonates the phenol reagent which then attacks the
carbene. A series of steps and a final acid work-up result in the o-hydroxy benzaldehyde
product.
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(ii) This electrophilic aromatic substitution allows the synthesis of monoacylated
products from the reaction between arenes and acyl chlorides or anhydrides. The
products are deactivated, and do not undergo a second substitution. Normally, a
stoichiometric amount of the Lewis acid catalyst is required, because both the substrate
and the product form complexes.
Q.15
How would you obtain:
(i) Picric acid (2, 4, 6-trinitrophenol) from phenol
(ii) 2-Methylpropene from 2-methylpropanol?
Ans.15(i)
(ii)
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Q.16
What is essentially the difference between α-form of glucose and β-form of
glucose? Explain.
Ans.16 The difference between alpha and beta glucose is nothing more than the position
of one of the four –OH groups. The carbon to the right of the oxygen atom in the
hexagonal ring is called the anomeric carbon. If the –OH group attached to it is below the
ring, the molecule is alpha glucose. If the –OH group is above the ring, the molecule is
beta glucose. Since the linear and cyclic forms of glucose inter-convert with each other,
alpha glucose can turn into beta glucose and vice versa.
Q.17
Describe what do you understand by primary structure and secondary structure
of proteins.
Ans.17
Primary structure: the linear rearrangement of amino acids in a protein and the location
of covalent linkages such as disulfide bonds between amino acids.
Secondary structure: areas of folding or coiling within a protein; examples include
alpha helices and pleated sheets, which are stabilized by hydrogen bonding.
Q.18
Mention two important uses of each of the following:
(i) Bakelite
Ans.18 (i) Bakelite
In its industrial applications, Bakelite was particularly suitable for the emerging
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(ii) Nylon 6
Q.19
Q.20
. Silver crystallizes in face-centred cubic unit cell. Each side of this unit cell
has a length of 400 pm. Calculate the radius of the silver atom. (Assume the
atoms just touch each other on the diagonal across the face of the unit cell.
That is each face atom is touching the four corner atoms.)
Nitrogen pentaoxide decomposes according to equation:
electrical and automobile industries because of its extraordinarily high resistance – not
only to electricity, but to heat and chemical action. It was soon used for all
nonconducting parts of radios and other electrical devices, such as bases and sockets for
light bulbs and vacuum tubes, supports for electrical components, automobile
distributor caps and other insulators.
Bakelite is used today for wire insulation, brake pads and related automotive
components, and industrial electrical-related applications.
(ii)Nylon 6
Nylon 6 finds application in a broad range of products requiring materials of high
strength. It is widely used for gears, fittings, and bearings, in automotive industry for
under-the-hood parts, and as a material for power tools housings.Nylon 6 is used as
thread in bristles for toothbrushes, surgical sutures, and strings for acoustic and
classical musical instruments, including guitars, sitars, violins, violas, and cellos.
Ans.19 Silver crystallizes in face-centred cubic unit cell hence-
Given
unit cell has a length(a) =400 pm=400×10-10 cm
so r=400×10-10/2×1.41
=141.4 pm
Ans.20(a) Plot of [N2O5 ] vs time(min.)
This first order reaction was allowed to proceed at 40 oC and the data below
were collected:
[N2O5](M)
Time (min.
Log [N2O5]
0.400
0.0
-0.3979
0.289
20.0
-0.5391
0.209
40.0
-0.6798
(a) Calculate the rate constant. Include units with your answer.
0.151
60.0
-0.8210
(b) What will be the concentration of N2O5 after 100 minutes?
0.109
80.0
-0.9625
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(c) Calculate the initial rate of reaction.
From the plot
(a)Slop=-0.70-(-0.60)/40-20
=-0.70+0.60/20
=-0.10/20
Slop of line plot=-K/2.303=0.10/20
Or K=0.1/20x2.303=0.01151
Or K=1.15x10-3min-1
(b)After 100min.
K=2.303/t log [N2O5]0/[N2O5]t
So after 100min
0.01151=2.303/100log[0.400]0/[N2O5]t
[N2O5]t=0.357M
©Initial rate of reaction
R=K[N2O5]
=1.15x10-3x0.400
=4.6x10-4S-1
Q.21
Explain how the phenomenon of adsorption finds applications in each of the
following processes:
Ans.21
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(i) Production of high vacuum
(ii) Heterogeneous catalysis
(iii)Froth floatation process
OR
Define each of the following terms:
(i) Micelles
(ii) Peptization
(iii)Desorption
Q.22
Describe the principle behind each of the following processes:
(i) Vapour phase refining of a metal.
(ii) Electrolytic refining of a metal.
(iii)Recovery of silver after silver ore was leached with NaCN.
(i)Production of high vacuum: Traces of air can be adsorbed by charcoal from a vessel,
evacuated by a vacuum pump to give a very high vacuum.
(ii) Heterogeneous catalysis: The gaseous reactants are adsorbed on the surface of the
solid catalysts. As a result, the concentration of the reactants increases on the surface and
hence the rate of the reaction increases.
(iii)Froth floatation process: This process is used to remove gangue from sulphide ores.
The basic principle involved in this process is adsorption. In this process, a mixture of
water pine oil is taken in tank. The impure powdered sulphide ore is dropped in through
hopper and the compressed air is blown in through the agitator and rotator is rotated
several times. As a result, froth is formed and the sulphide ores get adsorbed in the froth.
The impurities settled down and are let out through an outlet at the bottom.
OR
(i)Micelles –An aggregate of molecules in a colloidal solution, such as those formed by
detergents.
(ii) Peptization – Peptization is the process responsible for the formation of stable
dispersion of colloidal particles in dispersion medium. In other words it may
be defined as a process of converting a precipitate into colloidal sol by shaking it with
dispersion medium in the presence of small amount of electrolyte.
(iii)Desorption -Desorption is a phenomenon whereby a substance is released from or
through a surface. The process is the opposite of sorption (that is, either adsorption or
absorption).
Ans.22 (i) Vapour phase refining of metal:
Vapour phase refining is the process of refining metal by converting it into its volatile
compound and then, decomposing it to obtain a pure metal. To carry out this process,
the metal should form a volatile compound with an available reagent, and the volatile
compound should be easily decomposable so that the metal can be easily
recovered.Nickel, zirconium, and titanium are refined using this method.
(ii) Electrolytic refining;
Electrolytic refining is the process of refining impure metals by using electricity. In this
process, impure metal is made the anode and a strip of pure metal is made the cathode. A
solution of a soluble salt of the same metal is taken as the electrolyte. When an electric
current is passed, metal ions from the electrolyte are deposited at the cathode as pure
metal and the impure metal from the anode dissolves into the electrolyte in the form of
ions. The impurities present in the impure metal gets collected below the anode.
This is known as anode mud.
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(iii) Recovery of silver after silver ore was leached with NaCN.
In the process of leaching, the finely divided silver is treated with dilute solution of
sodium cyanide while a current of air is continuously passed. As a result, silver pass into
the solution forming solution dicyanoargenate(I) while the impurities remain unaffected
which are filtered off.
Ag S
Q. 23 Complete the following chemical equations:
Q.24
Write the name, stereochemistry and magnetic behaviour of the following: (At.
Nos. Mn = 25, Co = 27, Ni = 28)
(i) K4[Mn(CN)6]
(ii) [Co(NH3)5Cl] Cl2
(iii) K2[Ni(CN)4]
+
4NaCN
2Na[Ag (CN)2 ] + Na2 S
Sodium dicyanoargenate(I)
Ans.23
Ans.24
(i) K4[Mn(CN)6]
Name: Potassium hexacyanomanganate(II)
Stereochemistry – Does not show geometric or optical isomerism
Magnetic 10ehavior – Paramagnetic
(ii) [Co(NH3)5Cl] Cl2
Name: Pentaamminedchloridocobalt (III) chloride
Stereochemistry – Does not show geometric isomerism but is optically active
Magnetic 10ehavior – Paramagnetic
(iii) K2[Ni(CN)4]
Name: Potassium tetracyanoinickelate (II)
Stereochemistry – Does not show geometric or optically isomerism
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Q.25
Answer the following:
(i) Haloalkanes easily dissolve in organic solvents, why?
(ii) What is known as a racemic mixture? Give an example.
(iii)Of the two bromo derivatives, C6H5CH(CH3)Br and C6H5CH(C6H5)Br,
which one is more reactive in SN1 substitution reaction and why?
Magnetic 11ehavior – Diamagnetic
Ans.25(i)
When haloalkanes are dissolved in organic (non polar) solvents the new forces of
attraction set up between haloalkane and solvent molecules are of the same strength as
the forces of attraction being broken,i.e., existing between haloalkane and solventsolvent molecules. Thus, it can easily get dissolved in organic solvents.
(ii)A solution in which both enantiomers of a compound are present in equal amounts is
called a racemic mixture, or racemate. Racemic mixtures can be symbolized by a (d/l)- or
()- prefix in front of the substance’s name. Since enantiomers have equal and opposite
specific rotations, a racemic mixture exhibits no optical activity.
Eg.-
(iii)The SN1 substitution reaction involves the formation of carbocation, which is not
affected by the presence of bulky groups. Thus, C6H5CH(C6H5)Br will be more reactive
towards SN1 substitution reaction forming racemic mixture.
Q.26
(a) Explain why an alkylamine is more basic than ammonia.
(b) How would you convert: (i) Aniline to nitrobenzene (ii) Aniline to
iodobenzene?
Ans.26 (a) Alkyl Amine is more basic than ammonia because of the +I effect of the
Alkyl group present. It donates its electron cloud to the nitrogen group therefore nitrogen
has a greater tendency to donate it lone pair of electrons. This makes it more basic than
ammonia where such a positive inductive effect is abscent.
(b)(i) Aniline to nitrobenzene
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(ii) Aniline to iodobenzene
Q.27
Describe the following giving one example for each:
(i)Detergents
(ii)Food preservatives
(iii)Antacids
Q.28
(a) Difference between molarity and molality for a solution. How does a
change in temperature influence their values?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of
MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86
K kg mol-1)
OR
(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a
solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of
NaCl to 250.0 g of water.(Kb for water = 0.512 K kg mol-1), Molar mass of
NaCl = 58.44 g.
Ans.27(i) Detergent -a water-soluble cleansing agent which combines with impurities
and dirt to make them more soluble, and differs from soap in not forming a scum with
the salts in hard water.eg. Sodium dodecylbenzenesulfonate
(ii)Food preservatives- A preservative is a substance that is added to products such
as foods, pharmaceuticals, paints, biological samples, wood, etc. to prevent
decomposition by microbial growth or by undesirable chemical changes. In general
preservation is implemented in two modes, chemical and physical.eg. sodium benzoate
(C6H3COONa)
(iii) Antacids- Antacids are medicines that neutralize stomach acid.eg. Alka-Seltzer –
NaHCO3 and/or KHCO3
Ans.28
(a) Molarity , also known as molar concentration , is the number of moles of a
substance per liter of solution . Molality is a property of a solution and
is defined as the number of moles of solute per kilogram of solvent. Molarity
decreases with an increase in temperature whereas molality is independent of
temperature. This happens because molality involves mass, which does not
change with a change in temperature, while molarity involves volume, which is
temperature dependent.
(b) Given w2 = 10.50 g w1 = 200g Molar mass of MgBr2 (M2) = 184 g
Formula for this:
Tf=1000 x Kf x W2/ W1 x M2
= 1000 x 1.86 x 10.50 /200 x 184
=0.53
Tf=T0 - ∆Tf
=273 -.053 =272.47K
OR
(a) Osmosis-A process by which molecules of a solvent tend to pass through a
semipermeable membrane from a less concentrated solution into a more
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Q.29
(a) Give the chemical test to distinguish between:
(i)Propanal and propanone, (ii) Benzaldehyde and acetophenone.
(b) How would you obtain (i) But-2-enal from ethanal, (ii) Butanoic acid from
butanol,
(iii) Benzoic acid from ethylbenzene
OR
(a) Describe the following giving linked chemical equations:
(i) Cannizzaro reaction
(ii) Decarboxylation
(b) Complete the following chemical equations:
concentrated one.
Osmotic pressure- The pressure that would have to be applied to a pure solvent to
prevent it from passing into a given solution by osmosis, often used to express the
concentration of the solution.
Π=iRT=n/VRT
Where n=no.of moles, Π=osmotic pressure,T=temperature
osmotic pressure depends upon the number of moles of solute ‘n’ irrespective of the
nature of the solute. Hence, osmotic pressure is a colligative property.
(b) Given, Kb = 0.512 k kg mol-1 w2 = 15.00 g w1 = 250.0 g M2 = 58.44 g
By formula
∆Tb=1000 x Kb x w2 / w1 x M2
=1000 x0.512 x 15.00 / 250.0 x 58.44
= 0.52
Tb = To + ∆Tb
=373+ 0.53
=373.53 K
Ans.29(a)
(i) Propanal and Propanone :- Propanal and Propanone can be distinguished
by iodoform test, Tollens reagent test and Fehlings solution test.
Iodoform test:- Propanal does not give iodoform when heated with I2 and NaOH
while propanone give iodoforms when heated with I2 and NaOH.
(ii)Benzaldehyde and Acetophenone
By Tollen’s reagent Test
R=C6H5-
Acetophenone does not give this test.
(b)(i) But-2-enal from ethanol
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(ii)Butanoic acid from butanol
(iii)
Benzoic acid from ethylbenzene
OR
(a)(i) Cannizaro reaction: In this reaction, the aldehydes which do not have an hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on
treatment with a concentrated alkali. Example:
(b)Decarboxilation
In decarboxylation,the –COOH or –COONa group is removed and replaced with a
hydrogen atom.
(b) Complete reactions:
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Q.30
OR
(a) Account for the following:
(i) The acidic strength decreases in the order: HCl>H2S>PH3
(ii)Tendency to form pentahalides decreases down the group in group 15 of
the periodic table. (b) Complete the following chemical equations:
Ans.30(a)
(i) N≡N + 3X2 → 3NX3 ΔHrxn +ve(endothermic) or -ve(exothermic)
Approximately ΔHrxn = 3ΔHbond(NX) - ΔHbond(N2) - 3ΔHbond(X2)
The NN bond energy is huge (-ve) and formation will only be thermodynamically
allowed if X-X is low -ve (the F-F bond strength is anomalously small) and the N-X
bond energy is high (-ve). The N-F covalent bond is much stronger than the N-Cl bond
due to better 2p-2p overlap.
(ii) Due to the small size of F atom, the three lone pair of electrons on each F atom of F F molecule repels the bond pair. As a result, F - F is most reactive of all the four common
halogens.
(b)(i)
C + 2H2SO4
2SO2 + CO2 + 2H2O
Sulphur dioxide
∆CO2
(ii) P4 + 3NaOH + 3H2O
PH3 + 3NaH2PR2 + 2 ,CO
Phosphine
(iii)Cl2 + 3F2
2ClF3 (excess)
Chlorine trifluoride
OR
(a)(i) In a period, the electro negativity decreases in the order Cl > S > P. As a result, the
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loss of H+ ions decreases. Thus, the acidic strength of the hydrides decreases in the
following order: HCl > H2S > PH3
(ii) The tendency to form pentahalides decreases down the group 15 due to inert pair
effect i.e., in Bi the s-electrons remain inert and do not take part in bonding.
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