SOLUTION OF CBSE MODEL PAPER OF CHEMISTRY 2018

SOLUTION OF THE PAPER CBSE MODEL
                                                                             (In brief)






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Metal excess or anionic vacancies or F-centres
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Catalysis by zeolites which depends on the shape and size of the reactants and the products as compared to those of the pores and cavitites of zeolites.
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Neopentane or 2,2-Dimethylpropane
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3-Chloropropanamine, CH3CH(Cl)CH2NH2
½, ½

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Three ions [Co(NH3)6]2+, 2Cl
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No, the elevation in boiling point is not the same.
Elevation in boiling point is a colligative property which depends on the number of particles. NaCl is an ionic compound which dissociates in solution to give more number of particles whereas sugar is made up of molecules and thus does not dissociate.
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(a) As seen from the graph, electrolyte A is a strong electrolyte which is completely ionised in solution. With dilution, the ions are far apart from each other and hence the molar conductivity increases.
(b) To determine the value of limiting molar conductivity for electrolyte B, indirect method based upon Kohlrausch law of independent migration of ions is used.
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(a) Scandium (Sc)
(b) KMnO4 or any other suitable example
(c) Cerium (Ce) or any other example.
(d) Chromite ore
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(a) I2 < F2 < Br2 < Cl2
(b) BiH3 < SbH3 < AsH3 < PH3 < NH3
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(a) 1-Methylcyclohexene
(b) C6H6—H+àC6H5SO3H-NaOH-àC6H5SO3Na-H+àC6H5ONa

OR
Any two isomers out of the following:
(i) CH3−CH2−CH2−CH(OH)− CH3 Pentan-2-ol
(ii) CH3−CH2−CH(OH)- CH2− CH3 Pentan-3-ol
(iii) CH3−CH(CH3) −CH(OH) −CH3 3-Methylbutan-2-ol
(iv) CH3−CH2−C(CH3)(OH) − CH3 2-Methylbutan-2-ol
½+½
½+½

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For the bcc structure, z = 2
Density
By mole concept,
51.8 g of the element contains 6.022 × 1023 atoms
208 g of the element will contain atoms 8 .51208 1
= 24.17 × 1023 atoms.
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Molar mass of KCl = 39+35.5 = 74.5 g mol-1
As KCl dissociates completely, number of ions produced are 2.
Therefore, van’t Hoff factor, i=2
Mass of KCl solution = 1000 × 1.04 = 1040 g
Mass of solvent = 1040 – 74.5 = 965.5 g = 0.9655 kg
Molality of the solution:
Therefore, boiling point of solution = 100 + 1.078 = 101.078 oC

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Anode reaction:
Cathode reaction:
Cell representation:
According to Nernst equation: ECEll=E0cell-0.0591/n{log(oxid)/Red)
0.6094V


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(a) A reddish brown coloured colloidal solution is obtained.
(b) Stability of lyophilic sols is due to:

(i) same charge on all the colloidal particles.
(ii) solvation of the colloidal particles.
(c) At high pressures, amount of gas adsorbed (x/m) becomes independent of pressure
½+½
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The feasibility of thermal reduction can be predicted on the basis of Ellingham diagram. Metals for which the standard free energy of formation ( ) is more negative can reduce those metals for which is less negative. At a given temperature, any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram.
(a) Below the temperature approx 1623K), corresponding to the point of intersection of Al2O3 and MgO curves, Mg can reduce alumina.
(b) At temperatures below 1073K, the CO,CO2 line lies Fe, FeO line, thus CO is a better reducing agent.
At temperatures above 1073K, Coke will reduce FeO and itself get oxidised to CO.
                                                           OR
(a) Entropy is higher when a metal is in the liquid state than when it is in the solid state. Thus increases, thus becomes more negative and the reduction becomes easier..do it self
(b) Limestone provides the flux (CaO) which combines with the impurities (SiO2) to form slag (CaSiO3). Thus it helps in the removal of impurities.
(c) Pine oil (Collector) enhances the non wettability of the ore particles, which become lighter and hence rise to the surface along with the froth.
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(i)(a) Cr3+/Cr2+ has a negative reduction potential. Hence Cr3+ cannot be reduced to Cr2+. Cr3+ is most stable. Mn3+/Mn2+ have large positive Eo values. Hence Mn3+ can be easily reduced to Mn2+. Thus Mn3+ is least stable. Fe3+/Fe2+ couple has a positive Eo value but small. Thus the stability of Fe3+ is more than Mn3+ but less stable than Cr3+.
(b) If we compare the reduction potential values, Mn2+/Mn has the most negative value i.e its oxidation potential value is most positive. Thus its most easily oxidised. Thus the decreasing order for their ease of oxidation is Mn > Cr >Fe.
(ii) K4[Mn(CN)6]    Mn is in +2 oxidation state. Magnetic moment 2.2 indicates that it has one unpaired electron and hence forms inner orbital or low spin complex. In presence of 
CN− is a strong ligand, hybridisation involved is d2sp3 (octahedral complex)

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(i) Fe exists as Fe2+. There are 4 unpaired electrons. Water is a weak ligand. Thus the hybridisation involved is sp3d2. It is an octahedral outer orbital complex.
(marks to be granted if hybridisation is depicted diagrammatically)
(ii) The ionisation isomer is [Co(NH3)5SO4]Br. The IUPAC name is Pentaamminesulphatocobalt(III)bromide.
The isomer [Co(NH3)5 Br] SO4 gives a white precipitate of BaSO4 with BaCl2 solution whereas the isomer [Co(NH3)5 SO4]Br does not form this precipitate. (or any other relevant test)

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(a) Due to greater s-character, a sp2 hybrid carbon is more electronegative than a sp3 hybrid carbon. Therefore, the sp2 hybrid carbon of C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than a sp3 hybrid carbon of cyclohexyl chloride.
( marks to be granted if shown with the help of a figure)
(b) Since the alkyl halide reacts with KOH to form a racemic mixture, it must be a 3o alkyl halide and the reaction will follow SN1 mechanism


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(a) A – C6H5NH2 B – C6H5N2+Cl− C – C6H5-N2 –C6H4 -OH
(b) A - C6H5CN B - C6H5COOH C - C6H5CONH2
½+½+½
½+½+½

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(a) A – Sucrose (C12H22O11)
The mixture of D-(+)- glucose and D-(-)-Fructose is known as invert sugar. The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic linkage.
(b) The amino acids exist as dipolar zwitter ion. Due to this dipolar salt like character they have strong dipole dipole attractions Thus their melting points are higher than the corresponding haloacids which do not exist as zwitter ions.
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(a) Phenacetin is an antipyretic, while the rest are tranquilizers.
(b) 0.2% solution of phenol acts as antiseptic whereas 1% solution of phenol acts as disinfectant.
(c) Carbohydrates, proteins, nucleic acids, lipids (any two)
½+½
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(i) Since the alkyl halide is a 3o halide and C2H5ONa is a strong base, therefore elimination occurs preferably. The product obtained is 2- Methylprop-1-ene. CH3−C(CH3)=CH2
(ii) To prepare t-Butyl ethyl ether, the alkyl halide should be 1o i.e. chloroethane and the nucleophile should be sodium t-butoxide because the 3o nucleophile is able to attack 1o alkyl halide.

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½+½

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(a) The class of polymers is Biodegradable polymers.
(b) One example of biodegradable polymers is PHBV (Poly- - hydroxybutyrate-co- -hydroxyvalerate). The names of its monomers are: 3-hydroxybutanoic acid and 3- hydroxypentanoic acid
(c) Care for environment, concern for the health of the people or any other two relevant points.
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½+½

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(a) (i) The delocalisation of benzene electrons contributes little towards the stability of phenoxide ion. The carboxylate ion is much more resonance stabilized than phenoxide ion.
(ii) Semicarbazide has two –NH2 groups. One of them, which is directly attached to C=O is involved in resonance. Thus electron density on this group decreases and it does not act as a nucleophile. In contrast, the lone pair of electrons on the other –NH2 group is available for nucleophilic attack.

                                OR
(a) Ethanal and propanal can be distinguished by Iodoform test.
Ethanal gives a yellow precipitate of iodoform with an alkaline solution of NaOH. Propanal does not give this test. The name of the reaction is Hell Vohlard Zelinsky reaction
C(i)

(ii) (iii)


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(a) For a first order reaction
where [R]o = initial concentration, [R] = conc. after time
When half of the reaction is completed, [R] = [R]o/2. Representing, the time taken for half of the reaction to be completed, by t1/2, equation becomes:
Do it Self 
The above equation shows that half life first order reaction is independent of the initial concentration of the reactant.
(b) For a first order reaction
Self
                                                      OR
(a) r = k[R]n
When concentration is increased three times, [R] = 3a
27r = k(3a)n
or 27 = 3n or 3
n = 3
(b) According to Arrhenius equation,
RT E A k a 303 .2log log
For uncatalysed reaction
(i)self.For catalysed reaction
(ii) A is equal for both the reactions.
Subtracting equation(i) from equation(ii)  self
= 2.45×104
Rate of reaction increases by 2.45×104 times.
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(a) (i)self
(ii) self
(b) XeO3 is isostructural with BrO3−. (pyramidal structure)
(c) The bond dissociation enthalpy of F-F bond is lower than that of Cl-Cl bond and hydration enthalpy of F− ion is much higher than that of Cl− ion.
These two factors more than compensate the less negative electron gain
enthalpy of F2. Thus, F2 is a stronger oxidizing agent than Cl2.
(d) H2SO4 ionises in two stages and hence has two dissociation constants.
a2 a1 K K
.
This is because the negatively charged HSO4
− ion has much less tendency to
donate a proton to H2O as compared to neutral H2SO4.
OR
(a) Due to stronger H-F bond than HCl bond, HF ionises less readily than HCl
in aqueous solution to give H+ ions. Therefore HF is a weaker acid than HCl.
(b) In solid state, PCl5 consists of ions [PCl4]+[PCl6]− . On melting these ions
become free to move and hence conducts electricity in the molten state.
(c) In SF6, S is sterically protected by six F atoms and hence does not allow H2O molecules to attack the S molecule. Also, F does not have d-orbitals to accept the electrons donated by H2O molecules.
(d) In the structure of H3PO3 , it contains only two ionisable H-atoms which are present as -OH groups, thus it behaves as a dibasic acid.
(e) Except radon which is radioactive, Xenon has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with O2 and F2.
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