1/2 + 1/2 Q.3 2-Hydroxy-3,3-dimethylbutanoic acid 1. Q.4 Due to +I effect of alkyl group 1

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1 CODE No:050/S Total Pages:07 KENDRIYA VIDYALAYA SANGATHAN, GUWAHATI REGION nd PREBOARD EXAMINATION 067 MARKING SCHEME SUB : CHEMISTRY Q. Due to similar size of cations and anions. Q. Chloromethylpropane Because it is a tertiary halide. / / Q.3 Hydroxy3,3dimethylbutanoic acid Q.4 Due to I effect of alkyl group Q.5 A catalyst whose catalytic action depends upon its pore structure and molecular sizes of the reactants as well as the products is known as shape selective catalysis and the catalytic action is called shape selective catalysis. Q.6 In the case of CH 3 COOH, which is a weak electrolyte, the number of ions increases on dilution due to an increase in degree of dissociation. In the case of strong electrolyte the number of ions remains the same but the inter ionic attraction decreases. Q.7 A reaction which looks like of higher order but follows the kinetics of first order under special condition is called pseudo first order reaction. Example, CH 3 COOC H 5 H O H CH 3 COOH C H 5 OH Rate equation of the above reaction is: Rate = k[ch 3 COOC H 5 ] (a)total number of collision per unit volume per unit time is called collision frequency. (b)the minimum energy that the reacting molecules must possess in order to undergo effective collision to form product. Q.8 a) H S O 8 O O Q.9 S S O O O O OH OH b) Total number of electron pair around central atom I =6; B.P=4 and L.P=, so the compound should be square planer.therefore the noble gas compound will be XeF 4. Q.0

2 Q. z M d = a 3 N A d=6.3g/cm 3, M=60g/mol, N A = a=400pm =4 0 8 cm z = (4 0 8 ) For fcc, z = 4 The unit cell is fcc. r = 4.4pm Q. Oxidation Half Mg Mg e Reduction Half Cu e Cu Cell Reaction Mg Cu Mg Cu Here number of moles of electrons (n) = o o o E cell E E = 0 34 ( 37) = 7 V Cu Cu Mg Mg o [Mg ] The Nernst equation for the cell : Ecell Ecell log [Cu ] E cell log = log 0 = = 6805 V Q.3 (a) zero order reaction (b) unit of rate constant is mol l s (c) When [R] = [R 0 ], t = t / For a zero order reaction, k = { [R 0 ] [R]}/ Substituting the above values in integrated rate equation of zero order reaction, we get t / = { [R 0 ] [R 0 ]/} k t / = [R 0 ]/K t / [R 0 ] Q.4 a) (i) SiO

3 (ii) NaCN b) (i) Mond s process : It is a method of purification of Ni by using vapour phase refining. Ni 4CO [Ni(CO) 4 ] Ni 4 CO (ii) Zone refining : It is a method of obtaining a metal in very pure state. It is based on the principal that impurities are more soluble in molten state of metal than solidified state. In this method, a rod of impure metal is moved slowly over circular heater. The portion of the metal being heated melts & forms the molten zone. As this portion of the rod moves out of heater, it solidified while the impurities pass into molten zone. The process is repeated to obtain ultrapure metal and end of rod containing impure metal cutoff. a) it lowers the melting point of the mixture and it increase the electrical conductivity of the mixture b) Al O 3 Al 3 3 O Cathode: Al 3 3e Al Anode: C O CO e C O CO 4 e Overall reaction Al O 3 3C 4Al 3CO Q.5 i) 5SO MnO 4 H O 5SO 4 ii) 4H 3 PO 3 PH 3 3H 3 PO 4 iii) XeF 6 3H O XeO 3 6HF 4H Mn Q.6 i) Absorption of radiations in visible region by halogen atoms, results in the excitation of outer electrons to higher energy level. By absorbing radiation of different wavelength, they display different colours. ii) )PP bond is stronger than NN bond. iii) H SO 4 is very strong acid in water mainly because its first ionization to H and HSO 4 is very fast then ionization of HSO 4 to SO 4 is very small. Q.7 a)pentaaminechloridocobalt(iii) sulphate b)though both [NiCl 4 ] and [Ni(CO) 4 ] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl is a weak field ligand and does not cause the pairing of unpaired 3d electrons. Hence, [NiCl 4 ] is paramagnetic. In Ni(CO) 4, Ni is in the zero oxidation state i.e., it has a configuration of 3d 8 4s. But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp 3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO) 4 ] is diamagnetic.

4 Q.8 (a) CHBrCH 3 i) C 6 H 5 CH CH 3 Br /UV light ii) CH 3 CH=C(CH 3 ) HBr Peroxide CH 3 CHBrCH(CH 3 ) b) SP carbon of chlorobenzene is more electronegative than the SP 3 carbon of cyclohexyl chloride. Q.9 a) It involves the reaction of bromine with an acid amide in the presence of an alkali. It results in the formation of a primary amine with one carbon less than the parent compound. Here, the alkyl group migrates from carbonyl, with the elimination of CO. For example: b)nitrobenzene to,4,6tribromo aniline NO NH SnCl HCl Br /H O,4,6tribromo aniline Q.0 When extent of adsorption is x/m is plotted against pressure at a constant temperature, curve thus obtained is known as adsorption isotherm. Freundlich adsorption isotherm gives a relation between magnitude of adsorption (x/m) and pressure (p) at constant temperature.it can be expressed as (x/m) = kp /n (n > ).(i) Where x is the amount of gas adsorbed by m grams of adsorbent at pressure p, k and n are constants. Taking logarithm of equation (i) log(x/m) = (/n) logp logk A plot of log(x/m) vs logp is a straight line whose slope gives the value of (/n) and intercept is logk. /n can have any value between 0 and. i) When (/n) =, (x/m) = kp, adsorption varies log(x/m) slope=/n directly as pressure. ii) When (/n) = 0, (x/m) = k, adsorption is independent of pressure. logp Q. a) Neoprene<Polyvinyl chloride<nylon6 b) Name: Isoprene(Methyl,3Butadiene) Structure: CH =C(CH 3 )CH=CH c) Those polymers which are remoulded into our desire shape on heating and cooling are called thermoplastic.eg.pvc,teflon. Those polymers which are not remoulded into our desire shape on heating and cooling are called thermoplastic.eg.bakelite,melamine.

5 Q. i)tanquilizer: Drug which act on CNS to help in reducing anxiety. Example : Equanil ii)antifertility drug: Drugs which are used to prevent unwanted pregnancies in women. Example : Novestrol iii)disinfectant: Chemicals that kill microorganism or stop their growth but are harmful for living tissues. Example : % phenol Q.3 i)because they are good sources of vitamins which is required for the development of foetus. ii) VitaminB iii) Neha is a helping and caring person with scientific attitude. Q.4 a)osmosis : The process of flow of solvent molecules from pure solvent to solution or from solution of lower concentration of solution of higher concentration through a semi permeable membrane is called osmosis. Osmotic pressure : The pressure required to just stop the flow of solvent due to osmosis is called osmotic pressure of the solution. Yes, the osmotic pressure of a solution is colligative property. The osmotic pressure is expressed as. n = number of moles of solute,v = volume of solution,t = temperature From the equation, it is clear that 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) = C RT = C RT n /V = n /V W /M = W /M 5/34 = 3/M M = 05. a)when a nonvolatile solute is added to a volatile solvent the vapour pressure of pure solvent decreases because a part of the surfaces occupied by nonvolatile solute which can t volatilise. As a result, the vapour pressure of solution decreases and hence, the solution requires a comparatively higher temperature to boil causing an elevation of boiling point.

6 b) Given, K b = 0.5 k kg mol w = 5.00 g w = 50.0 g M = g Using the formula, = = = 0.53 Now, T b = T o = = K Q.5 a)pyrolusite is an ore of manganese having formula MnO.There are two steps involved in the preparation of KMnO 4. st step : oxidation of pyrolusite MnO 4KOH (aq) O K MnO 4 H O nd step : electrolysis of manganate to obtain permanganate MnO 4 MnO 4 b) (i) 5 Fe MnO 4 8 H Mn 4H O 5Fe 3 (ii) MnO 4 5C O 4 (aq) 6 H Mn 8H O 0 CO a) (i) A large number of unpaired electrons take part in bonding so they have very strong metallic bonds and hence high enthalpy of atomization (ii)because they have variable oxidation states and hence can form different intermediates. They also provide large surface area. (iii)e 0 value for Cr 3 /Cr is negative but that of Mn 3 /Mn is positive so Cr can lose electron to form Cr 3 while Mn 3 accepts electron to form Mn. In case of Cr d 4 to d 3 occurs for Cr to Cr 3. d 3 is stable. b) (i) Cr O 7 6I 4H Cr 3 3I 7 H O (ii) Cr O 7 3S 4 H Cr 3 3S 7 H O Q.6 (a) i) X = CH 3 CH(OH)CH CHO (Aldol) ii) Y = PCC iii) Z = CH 3 COCH CH 3 (b) i) Propanal and Benzaldehyde : Propanal will give Fehling s test(red ppt.) but Benzaldehyde do not. ii)phenol and Benzoic acid : Phenol will react with FeCl 3 and give violet colour but

7 Benzoic acid will not. a) i) Ethanal and Acetone: Ehanal will give Tollen s test(silver mirror) but Acetone will not. ii) Acetophenone and Benzophenone : Acetophenone will give iodoform test(yellow ppt.) but Benzophenone will not. b)ch 3 CH CH COOCH CH CH CH 3 H O CH 3 CH CH COOH CH 3 CH CH CH OH Dil HSO 4 (A) (B) (C) CH 3 CH CH CH OH (C) CrO3 CH 3 CH CH COOH (B) CH 3 CH CH CH OH Conc HSO4 CH 3 CH CH=CH (C)