<p> PRE-BOARD-II EXAMINATION SESSION 2016-17 Model Answers SUBJECT: CHEMISTRY (043) Q.NO. ANSWERS The trapping of anionic sites by the electrons in metal excess non-stoichiometric crystal 1 defects are called F-centres. 3+ [Fe(OH)3]Fe ….. is a positively charged sol hence it will be neutralized by the negative part of the an electrolyte. According to Hardy-Schulze rule the flocculation power is 2 directly proportional to charge density of the ion. + 4- K4[Fe(CN)6] 4K + [Fe(CN)6] (aq) 1 0 3 3 (a) ,because 2 – chlorobutane when undergoes SN reaction , produces 2 carbocation whereas 1-chlorobutane gives 10carbocation. Since 20 carbocation is more stable than 10 1 carbocation, therefore 2-chlorobutane undergoes SN reaction faster than 1- chlobutane. </p><p>4 2-Acetoxy benzoic acid. </p><p>5 C2H5NH2 < (C2H5)3N < (C2H5)NH</p><p>6 a) The reaction at cathode in a dry cell : + - MnO2+ NH4 + e →MnO(OH) + NH3 ZnCl2 combines with NH3 produced to form complex [Zn(NH3)2Cl2], otherwise the pressure developed due to NH3 would crack the seal of the cell. + - b) AgNO3 Ag (aq) + NO3 (aq) + - H2O<------ H +OH At cathode: Ag+ + e- → Ag(s) discharge potential of Ag+ is greater than H+ Ag+ ions will be deposited as silver. At anode: Ag(s)→ Ag+ + e- - + Ag electrode is attacked by NO3 ions, Ag anode will dissolve to form Ag ions in the solution. </p><p>7 a)(i)The sum of powers of the concentration of the reactants in the rate law expression is called the order of reaction whereas the number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity. (ii) Order is experimental whereas molecularity is theoretical (iii) Order may be zero or fractional but molecularityis always integral eg CH3CHO → CH4 + CO 3/2 Rate = K[CH3CHO] Order= 3/2 Molecularity= 1 .(or any other example).</p><p> b) Inversion of cane sugar/ hydrolysis of ester/ SN1reaction .(or any other example). eg Ester + H2O → Acid + Alcohol Rate = K[Ester]1[Water]0 Order = 1 ,Seems to follow 2nd order kinetics but actually follows 1st order kinetics due to its independence to the conc. of water which is taken in large excess 8 (i) Preparation of Sodium Chromate</p><p>4FeO.Cr2O3 + 8Na2CO3+7O2 →8Na2CrO4+2Fe2O3+8CO2 (ii) Conversion of Sodium Chromate into Sodium Dichromate</p><p>2Na2CrO4+ H2SO4 →Na2Cr2O7+ Na2SO4+ H2O (iii) Conversion of Sodium Dichromate into Potassium Dichromate Na2Cr2O7 + 2KCl → K2Cr2O7+ 2NaCl </p><p>H Effect of increasing p on aqueous solution of K2Cr2O7</p><p>2- 2- + Cr2O7 + H2O → 2CrO4 + 2H ←</p><p>OR</p><p>(i) Electronic Configuration – The general electronic configuration of lanthanoids is [Xe]54 4f1-14 5d0-1 6s2 whereas that of actinoids is [Rn]54 5f1-14 6d0-1 7s2 ( 1) (ii) Oxidation State – Lanthanoids show limited oxidation states ( +2,+3, +4) but +3 is the common oxidation state whereas actinoids show large number of oxidation states (+3, +4, +5, +6, +7). </p><p>9 (i) The major product is (CH3)2C=CH2 , C2H5OH and NaBr -1 + (ii) (CH3)3CO Na + C2H5Br → (CH3)3COC2H5 + NaBr </p><p>10 Mechanism of hydration of ethene. Step I : Protonation of Alkene + + CH2 = CH2 + H CH3 - CH2 </p><p>Step II : Nucleophile attacks on carbocation</p><p>+ + CH3 - CH2 CH3 - CH2 - OH2</p><p>Step III : Deprotonation + CH3 - CH2 - OH2 CH3-CH2-OH </p><p>11 For fcc structure, z= 4 a= 4.07 x 10-8cm d=10.5gm/cc 23 N0= 6.022 x 10 3 d= Z x M/a x N0 3 M= d x a x N0/z = 10.5 X (4.07)3 x10-24 x 6.022 x 1023/4 = 106.55g 12 Half cell reaction At Anode: Zn → Zn2+ + 2e- </p><p>+ At cathode: 2H + 2e- → H2/Pt ______</p><p>+ -2 2+ -3 Zn(s) + 2H (10 M) →Zn (10 M) + H2/Pt</p><p>0 E Zn2+/Zn = -0.76V</p><p>0 E H+/H2= 0.00V</p><p>0 2+ + 2 E cell =E cell- 0.059/n log [Zn ] / [H ] Substituting values</p><p>-3 -4 E cell = 0.76 – 0.0591/2 log 10 /10</p><p>E cell = 0.76 – 0.0295 log 10 (log 10=1) = 0.7305 V</p><p>SO2Cl2 (g) → SO2(g)+ Cl2(g) k = (2.303/t) log p0/(2p0-pt) t= 100s k= (2.303/100) log 0.5/2 x 0.5- 0.6 13 =2.303/100 x log1.25 = 2.303/100 x 0.0969 = 2.23 x 10-3 s-1 When Pt = p0 + p = 0.65 atm , p = 0.65 – p0 = 0.65 – 0.50 = 0.15 atm Pressure of SO2Cl2= p0 - p =50- 0.15 = 0.35 atm 3 -5 -1 Rate =k x P SO2Cl2= 2.2316x 10- x 0.35 = 7.8 x 10 atms (a) Breaking of emulsions by heating, freezing, centrifugating etc. Examples - Epoxy resins, poly amines, polyols, siloxanes etc. 14 (b) The catalytic reaction that depends on the pore structure of catalyst and size of reactant and product molecules is called shape selective catalysis. Eg alcohols gasoline (a) In froth floatation process, the role of the depressant is to prevent one type of sulphide ore particles from forming thr froth with air bubbles. For example, NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. Because NaCN forms a complex with ZnS, Na [Zn(CN) ], but 15 2 4 PbS forms froth. (b) The standard free energy of formation of CO2 from CO is higher than that of formation of ZnO from Zn. (c) The impurities are more soluble in the molten state than in the solid state of the metal. </p><p>(a) P4+3NaOH+3H2O → PH3 + 3Na2H2PO2 3+ 2+ 2- 16 (b) 2Fe + SO2 +2H2O →2Fe + SO4 +4H+ (c) 4NaCl + MnO2 +4H2SO4 → MnCl2 +4NaHSO4 +2H2O +Cl2 (a) (i) Oxygen stabilizes the highest oxidation state even more than fluorine, e.g, highest fluoride of Mn is MnF4 whereas highest oxide is Mn2O7. The reason for this is the ability of oxygen to form multiple bonds with metal atoms (ii) Transition metals and their compounds show catalytic properties because of: 17 . Variable oxidation state . Unpaired electrons in their incomplete d – orbitals . Large surface area 2- + -1 (b) 3MnO4 + 4H → 2MnO4 + MnO2 + 2H2O </p><p>(a)Correct structure of cis and trans isomer 4 0 (b) t2g eg (c) dsp2, diamagnetic OR 18 3 [Ni(CO)4] – tetrahedral geometry, diamagnetic, sp hybridisation 3- 3 2 [CoF6] - octahedral geometry, paramagnetic, sp d hybridisation 3+ 2 3 [Co(NH3)6] - octahedral geometry, diamagnetic, d sp hybridisation </p><p>. a) CH3 – CH = CH2 + HBr CH3CHCH2Br CH3CH2NO2</p><p> b) CH – CH – OH CH CH Cl CH CH F 19 3 2 3 2 3 2 c) +2 Na + 2NaBr</p><p>A) Aniline doesn’t undergoes friedel crafts reaction as aniline is a Lewis base it reacts with catalyst AlCl3 which is Lewis acid & forms a complex which deactivates the benzene ring towards electrophilic substitutions reactions.</p><p>20 B) i) CH3CONH2+Br2+4NaOH→ CH3 – NH2 + 2NaBr +Na2CO3 + H2O </p><p> ii) C6H5N2Cl + C6H5OH C6H – N = N – C6H5OH + HCl</p><p>22 a) Tranquilizers are chemical compounds which are used to cure stress, anxiety, mild or severe mental diseases For eg: Derivatives of Barbiturates, luminal, equanil etc.</p><p> b) The use of aspartame is limited to cold food and drinks because it decomposes on heating and loses its sweetening properties.</p><p> c) Antidepressant drugs should not be taken without consulting a doctor as it may increase aggression, anxiety, and can cause hypertention. a) Fat soluble vitamins b) Concern for health, awareness about vitamins, helpful.</p><p> c) Vitamin B12 can be stored in our body. 23 d) Deficiency of Vitamin A – Night Blindness Deficiency of Vitamin D – Rickets</p><p>(a)As solubility of gases decreases with increase of temperature, less oxygen is available in summer in the lakes / as cold water contains more oxygen dissolved. (b)0.91% saline water is isotonic to human blood cells. When placed in a hypertonic solution (conc.> 0.91% NaClsoln) they will shrink due to exo-osmosis. </p><p>(c) P0- Ps / P0= XB </p><p>P0- Ps / P0= WB x MA / MB x WA </p><p>1-Ps/P0 = 15 x 18/180 x 150 24 = 1/100</p><p>Ps/P0 = 1- 1/100</p><p>Ps/P0 = 99/100</p><p>Ps = P0 x 99/100 = 17.5 x99/100 mm Hg = 17.325 mm Hg 25 (i) Nitrogen shows less tendency for catenation than phosphorous because of its small size and more repulsion due to lone pair of electrons on the two N – atoms whereas in phosphorous there is less repulsion on P –P bond due to its comparatively larger size. * (ii) Due to presence of two unpaired electrons in the π antibonding orbitalof S2 molecule. (1) (iii) Bond dissociation enthalpy of fluorine is less than chlorine because of small size of fluorine atom, there is more repulsion between lone pair of electrons of F- atoms </p><p>(b) (i) XeF6 + 3H2O → XeO3 + 6HF (1) (ii) 4H3PO3 → 3H3PO4 + PH3 (1) OR (a) Any of the two poisonous gases like phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2CH2Cl), etc. </p><p>(b) (i) High pressure (ii) low temperature (iii) use of catalyst </p><p>(c) When O3is treated with excess of KI solution buffered with borate buffer, I2 is liberated. - - 2I (aq) + H2O (l) + O3 (g) → 2OH (aq) +I2 (s) + O2 (g) I2 so liberated is titrated against standard solution of sodium thiosulphate using starch as indicator.(1)</p><p>(d) Correct structure </p><p>(e) Halogens are colored because there molecules absorb light in the visible region which excite their electrons to the higher energy levels whereas the remaining light is transmitted. The color of halogens is the color of its transmitted light that is halogens have complementary colors. The amount of energy required for excitation decreases from fluorine to iodine as the size of the atom increases. Conversely , the energy of transmitted light increases progressively i.e. deepening of colour of halogens from fluorine to iodine. </p><p>26 a) i) Aldol Condensation</p><p>H H NaOH CH3CHO + CH3CHO CH3–C–C–CHO</p><p>OH H</p><p>CH3–CH=CHCHO But-2enal ii) Cannizzaro Reaction</p><p>KOH HCHO + HCHO CH3OH + HCOOK b) i) CH3COCH3 and CH3CHO</p><p>These can be distinguished by Fehling solution test or Tollen’s Regent test. CH3CHO will give silver mirror with Tollen’s Reagent or red ppt with Fehling solution. ii) Ethanal and Ethanoic acid can be distinguished by Tollen’s reagent test or </p><p>NaHCO3 test Ethanoic acid will give brisk effervscences with NaHCO3 due to </p><p> evolution of CO2 gas.</p><p>- c) In 4 Nitrobenzoic acid NO2 group has -I effect so it decrease e density on benzene and hence removal of H+ becomes easier and is more acidic whereas in 4-Methyl benzoic acid methoxy gp has +I effect and increase e- density on benzene and hence removal of H+ becomes difficult making it less acidic.</p>