DPP-53 to 54 Physical Chemistry With Answer

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 53 to 54 Class : XIII Course : DPP No.1 Total Marks : 41 Max. Time : 41 min. Single choice Objective ('–1' negative marking) Q.1 to Q.11 (3 marks 3 min.) [33, 33] Subjective Questions ('–1' negative marking) Q.12 to Q.13 (4 marks 4 min.) [8, 8] ANSWER KEY DPP No.-53 1. 1 2. 4 3. 1 4. 2 6. 4 7. 2 8. 2 9. 4 10. 4 11. 3 12. (a) No. ; (b) Yes 13. (a) F (b) F (c) F (d) F (e) F (f) F (g) T (h) T (i) T (j) F (k) T (l) T (m) T 1. Which statement about standard reduction potentials is correct (1*) E  2 (2) E  = Zero at all temperature = zero at 298 K D / D2 (3) A redox reaction is feasible if sum of SRP of oxidant and that of reductant is a positive quantity (4) K2Cr2O7 (acid) is stronger oxidising agent than KMnO4 (acid) [Given : E – 2 = 1.51 V ; E 2 – 3  = 1.33 V] MnO4 / Mn Cr2O7 / Cr ekudvip;u foHkods fy,dkSulk dFkulgh gS& (1*) E  (2) E  / H2 / D2 = fdlhHkhrkiij'kwU; = 298 K ij'kwU; (3) ,djsMkDlvfHkfØ;kgksrhgS;fnvkWDlhdkjddsSRP rFkkvipk;ddsSRP dk;ksx,d/kukRedek=k gS (4) K2Cr2O7 (vEy) ( KMnO4 (vEy) dh rqyuk esa izcyre vkWDlhdkjd vfHkdeZd gSA [fn;k x;k gS : E – 2  = 1.51 V; E 2 – 3  = 1.33 V] MnO 4 / Mn Cr2O7 / Cr Sol. (1) It is a convention that SRP of standard hydrogen electrode is zero at all tempertures (2) is incorect (3) the difference should be +ve quantity (4) KMnO4 is stronger oxidising agent (SRP values) 2. A solution containing NH Cl and NH OH has a hydroxide ion concentration of 10 6 mol lit 1. Which of the 4 4 following hydroxides could be precipitated when this solution is added in equal volume to a solution containing 0.1 M of metal ions . [4 marks, 4 min.] NH ClrFk NH OH;qDr,dfoy;uesagkbMªksDlkWbMvk;udhlkUnzrk106eksy@yhVjgSA0.1M/k rqvk;u;qDrfoy;u 4 4 esablfoy;udslekuvk;rudksfeykusijfuEuesalsdkSulkgkbMªksDlkWbMvo{ksfirgksldrkgSA (1) AgOH (Ksp = 5  10 ) (2) Cd(OH) (K = 8  10 ) 3 6 sp (3) Mg(OH)2 (Ksp = 3  10 ) (4*) Fe(OH) (K = 8  10 ) 11 16 sp 3. The solubility products of Al(OH)3 and Zn(OH)2 are 8.5 × 10 and 1.8 × 10 respectively. If NH OH is added –23 –14 to a solution containing Al3+ and Zn2+ ions, then substance precipitated that is : [4 marks, 4 min.] (1*) Al(OH)3 (2) Zn(OH)2 (3) Both together (4) None of these Al(OH) rFkk Zn(OH) dk foys;rk xq.kuQy Øe'k% 8.5×10–23 rFkk1.8×10–14 gSA ;fn Al3+ rFkk Zn2+ vk;u ;qDr 3 2 foy;uesaNH4OHdksfeyk;ktkrkgSrksinkFkZvo{ksfirgkstkrkgSA (1*) Al(OH)3 (2) Zn(OH)2 (3)nksukslkFk&lkFk (4)buesalsdksbZugh 4. E° for some half cell reactions are given below dqNv)ZlsyvfHkfØ;kvksadsfy,E°uhpsfn;sx;sgSaA Sn+4 + 2e–  Sn2+ ; E° = 0.151 v 2Hg2+ + 2e  Hg2+2 ; E° = 0.92 v PbO2 + 4H+ + 2e–  Pb2+ + 2H2O ; E° = 1.45 v based on the given data which statement is correct. fn;sx;svkdMksadsvk/k jijdkSulkdFkulghgSA (1) Sn4+ is a stronger oxidising agent than Pb+4 (2*) Sn2+ is a stronger reducing agent than Hg 2+ (3) Hg2+ is a stronger oxidising agent than Pb4+ (4) Pb+2 is a stronger reducing agent than Sn2+ (1)Sn4+,Pb+4 fdrqyukesaizcyvkWDlhdkjdvfHkdeZdgSA (2*)Sn2+ ,Hg 2+ fd rqyuk esa izcy vipk;d vfHkdeZd gSA (3) Hg2+ ,Pb4+fdrqyukesaizcyvkWDlhdkjdvfHkdeZdgSA (4) Pb+2 ,Sn2+ fd rqyukesa izcy vipk;d vfHkdeZdgSA Sol. Sn2+  Sn+4 + 2e– E° = –0.15 v Hg 2+  2Hg2+ + 2e– E° = –0.92 v Sn2+ ] Hg 2+ Sn2+]Hg 2+ fdrqyukesaizcyvipk;dvfHkdeZdgSA 6. Which of the following equations is not involved in the Solvay process ? fuEuesalsdkSulhvfHkfØ;klkYosizØeesaughagksrhgS\ (1) CaCO heat  CaO + CO (2) NaCl + NH + H O + CO  NH Cl + NaHCO (3) CaO + 2NH Cl  2NH + H O + CaCl (4*) Na CO + CO + H O  2NaHCO 4 3 2 2 2 3 2 2 3 7. The composition of Sorel's cement is : lksjsy lhesUV (Sorel'scement) dk la?kVugSA (1) MgCl2 .5H2O (2*) MgCl2 .5MgO . nH2O (3) MgCO3 .5H2O (4) MgCO3 .5H2O . nH2O 8. Select the most ionic and most covalent compounds respectively from the following. fuEuesalslokZf/kdvk;fudolokZf/kdlgla;ksth;kSfxddkp;udhft,A CrO5, Mn2O7, PbO, P4O10, SnO2 (1) CrO5, Mn2O7 (2*) PbO, Mn2O7 (3) CrO5, P4O10 (4) SnO2, CrO5 Sol. As anion is same, the covalent character in ionic compound will depend upon the size and charge on the cation. According to Fajan's rule, PbO (most ionic) = Pb2+, Mn O (most covalent) = Mn+7. gy. tc_.k ;ulekugksrkgS]rcvk;fud;kSfxdesalgla;ksthvfHky{k.k/kuk;udsvkdkjovkos'kijfuHkZjdjrsgSaAQk;ku fu;edsvuqlkj,PbO(lokZf/kdvk;fud)=Pb2+,Mn O (lokZf/kdlgla;ksth)=Mn+7 gksrkgSA 2 7 9. In which of the following sets the central atom of each member involves sp3 hybridisation ? fuEuesalsdkSulsleqPp;esa]izR;sddkdsfUnz;ijek.kq(centralatom)sp3ladfjrvoLFk esagSa\ (1) O – , Cl – , F + (2) XeO , XeO , XeF 4 4 4 3 4 4 (3) SO3, SO 2– , SO 2– (4*) PCl + , BF – , ClO – 10. The BF3 is a planar molecule where as NF3 is pyramidal because : (1) B – F bond is more polar than N – F bond. (2) boron atom is bigger than nitrogen atom. (3) nitrogen is more electronegative than boron. (4*) BF3 has no lone pair but NF3 has a lone pair of electrons. BF3,dleryh;v.kqgStcfdNF3fijsfeMyv.kqgS]D;ksafd (1)B–Fca/k]N–Fca/kdhrqyukesavf/kd/kqzfo;gksrkgSA (2)cksjksuijek.kqukbVªkstuijek.kqdhrqyukesacM+kgSA (3)ukbVªkstucksjksudhrqyukesavf/kdfo|qr_.khgSA (4*)BF3v.kqesadksbZ,dkdhbysDVªkWu;qXeughagS]fdUrqNF3esa,d,dkdhbysDVªkWu;qXegSA 11. Which species has the maximum number of lone pairs of electrons on the central atom? fuEuesalsdkSulhLih'khtesadsfUnz;ijek.kqij,dkdhbysDVªkWu;qXedhvf/kdrela[;kgSA (1) XeOF (2) IF + (3*) XeF (4) BrF 4 4 2 3 12.(a) Equal volumes of 0.02 M CaCl2 and 0.0004 M Na2SO4 are mixed . Will a precipitate form . Ksp for CaSO4 = 2.4  10 5 . 0.02MCaCl2rFk 0.0004M Na2SO4dks lekuvk;ruksaeasfeyk;ktkrkgSrksD;kvo{ksicusxk\CaSO4dsfy, Ksp = 2.4  105 gSA Ans. No. (b) Will a precipitate form when 0.150 L of 0.10 M Pb(NO3)2 and 0.100 L of 0.20 M NaCl are mixed? KSP (PbCl ) = 1.2 × 10–5 tc 0.10MPb(NO3)2ds 0.150LrFkk 0.20MNaClds 0.100Ldks fefJr fd;k tkrk gSa rks D;k vo{ksi cusxk \ KSP (PbCl ) = 1.2 × 10–5 . Ans. Yes.gk¡A 13. State whether each of the following is true or false. fuEuesalsdkSulkdFkulR;;kvlR;gSA (a) MgCl2 . 6H2O on heating gives anhydrous MgCl2. MgCl2 . 6H2Odks xeZ djus ij futZyh; MgCl2 nsrk gSA (b) Aluminium chloride (AlCl3) is a Lewis acid because it can donate electrons. ,Y;qfefu;eDyksjkbZM(AlCl3)yqbZlvEygSD;ksafd;gbysDVªkWunkudjrkgSA (c) All the Al–Cl bonds in Al2Cl6 are equivalent. Al2Cl6 esalHkhAl–ClcU/klekugksrsgSA (d) Sodium monoxide can be prepared by burning sodium in air. lksfM;eeksuksavkWDlkbMdkla'ys"k.kok;qesalksfM;edkstykusds}kjkfd;ktkrkgSA (e) Potassium superoxide is diamagnetic in nature. iksVsf'k;elqijvkWDlkbMizd`fresaizfrpqEcdh;gSA (f) The thermal stability of hydroxides of group 1 decreases on moving down the group. oxZ1stdsgkbMªksDlkbMdkrkih;LFk ;hRoo.kZesauhpsdhvksj?kVrkgSA (g) Potassium carbonate cannot be prepared by Solvay process as potassium bicarbonate is highly soluble in water. lkYosizØeds}kjkiksVsf'k;edkcksZusVdkla'ys"k.kughafd;ktkldrkgSaD;ksafdiksVsf'k;eckbZdkcksZusVtyesa vR;fèkdfoys;gksrkgSA (h) All chlorides of Group 1 elements are ionic except LiCl. vioknLo:iLiCldksNksM+djoxZ1dslHkhrRoksadsDyksjkbZMvk;fudgksrsgSA (i) The metallic oxides of group 1 elements becomes more basic on going down the group. oxZ1dsrRoksads/k fRodvkDlkbMoxZesauhpsdhvksjtkusijvf/kd{k jh;gkstkrsgSA (j) The oxides of alkaline earth metals are more basic then the oxides of alkali metals. {kjh;/krqvksadsvkWDlkbMdhrqyukesa{kjh;e`nk/krqvksadsvkWDlkbMvf/kd{kjh;gksrsgSA (k) The bicarbonates of Mg and Ca are not stable in the crystalline salt. MgrFk CadsckbZdkcksZusVfØLVyh;yo.kesaLFk;hughgksrsgSaA (l) Unlike magnesium chloride, calcium chloride can be obtained by heating CaCl2 . 6H2O. CaCl2.6H2OdksxeZdjusijdSfY'k;eDyksjkbZMizkIrgksrkgSysfdueSXuhf'k;eDyksjkbZMizkIrughagksrkgSA (m) The decrease in the solbuility from BeSO4 to BaSO4 is primarily due to decrease in the hydration energy as one moves from Be2+ to Ba2+. BeSO lsBaSO rd foys;rk?kVrh gSD;ksafd Be2+ ls Ba2+ rdtkus ijty;kstu ÅtkZ?kVrhgSA 4 4 Ans. (a) F (b) F (c) F (d) F (e) F (f) F (g) T (h) T (i) T (j) F (k) T (l) T (m) T DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 54 Class : XIII Course : DPP No.2 Total Marks : 45 Max. Time : 45 min. Single choice Objective ('–1' negative marking) Q.1 to Q.11 (3 marks 3 min.) [33, 33] Multiple choice objective ('–1' negative marking) Q.12 to Q.14 (4 marks 4 min.) [12, 12] 1. The temperature defining the standard electrode potential is (1) 298 K (2) 273 K (3) 373 K (4*) any temperature can be selected but it must remain constant and species must be in their standard states. ekudbysDVªkWMfoHkodksifjHk f"krdjusdsfy,rkigSA (1) 298 K (2) 273 K (3) 373 K (4*)fdlh Hkh rkieku dk p;u fd;k tk ldrk gS ysfdu ;g fu;r jguk pkfg, rFkk Lih'kht viuh ekud voLFkk esagksuhpkfg,A Sol. Any temperature can be selected. 2. E° for F2 + 2e– 2F¯ is 2.8 V, E° for ½ F2 + e– = F¯ is - F + 2e– 2F¯ ds fy, E° 2.8V gS( ½ F + e– = F¯ ds fy, E° fuEu gksxk& 2 2 (1*) 2.8 V (2) 1.4 V (3) –2.8 V (4) –1.4 V Sol. E°isintensiveproperties.(E°ek=kkLora=k xq.kgSA½ 3. The standard reduction potential of Cu2+/Cu and Cu2+/Cu+ are 0.337 and 0.153 respectively. The standard electrode potential of Cu+/Cu half-cell is - Cu2+/Cu rFkk Cu2+/Cu+ ds fy, ekud vip;u foHko Øe'k% 0.337rFkk 0.153gSA Cu+/Cuv)Z lsy ds fy, ekud bysDVªkWMfoHkogSA (1) 0.184 V (2) 0.827 V (3*) 0.521 V (4) 0.490 V Sol. (i) Cu2+ + 2e–  Cu Eº = 0.337 V G º = – 2F × 0.337. (ii) Cu2+ + e–  Cu+ Eº = 0.153 V G º = – 1F × 0.153. (iii) Cu2+ + e–  Cu E0 = ?  G0 = – F E° eq. (iii) = (i) – (ii)   G0 =  G0 □–  G0 1 3 ;  G0 = –2 F  0.337 – (–F  0.153) 3 1 2 3 E0 = 0.153 – 2 (0.337) = 0.521 V. 4. E0 3  / Fe2 = +0.77 V ; E0  3 / Fe = – 0.036 V. What is E0 2 and is Fe+2 stable to disproportionation in aqueous solution under standard conditions (1*) +0.44 V, yes (2) – 0.44 V, No (3) + 0.44 V, No (4) – 0.44 V, yes EºFe/ Fe2 dkekuD;kgS\rFk D;kFe2+ ekudifjfLFkfr;ksaesatyh;foy;uesafo"kekuqikrudsfy,LFk ;hgSa\ 0 Fe / Fe 2 rFkk Fe+2 D;k gSA [ ;fn E0 3 / Fe 2 = +0.77 V ; E0  3 / Fe = – 0.036 V]. (1*) +0.44V, gk¡ (2) – 0.44 V, ugha (3) + 0.44 V, ugha (4)– 0.44 V, gk¡ Sol. (i) Fe3+ + e–  Fe+2 G º = – F × 0.77 (ii) Fe3+ + 3e–  Fe G º = – 3F (– 0.036) (iii) Fe  Fe2+ + 2e–  G0 = – 2F E° eq. (iii) = (i) – (ii)   G0 =  G0 □–  G0 3 1 2 –2FE° = – F (0.77) – (3F  (0.036) E0 = + 0.44 V. 5. Given that Eº 2 = – 0.44 V ; Eº 3 2 = 0.77 V if Fe2+, Fe3+ and Fe solid are kept together then Fe Fe Fe Fe (1) Fe3+ increases (2*) Fe3+ decreases (3) Fe2+ / Fe3+ remains unchanged (4) Fe2+ decreases fn;k x;kgS Eº 2 =–0.44V; Eº 3 2 =0.77VgS;fn Fe2+,Fe3+ rFkkFeBksl dks,d lkFkj[kk tkrkgS]rks (1)Fe3+ c<+rkgSA (2*)Fe3+ ?kVrkgSA (3)Fe2+ /Fe3+ vifjofrZrjgrk gSA (4)Fe2+ ?kVrkgSA Sol. Fe3+ + e–  Fe+2 E° = +ve \ it is spontaneous i.e. Fe+3  and Fe2+  6. Which of the following are isoelectronic and isostructural ? (1) NO +, CO (2) NO –, BF (3) NH , CH – (4*) all of these. 2 2 3 3 3 3 fuEueslsdkSulebySDVªkWfudolelajpukRedgS\ (1) NO +, CO (2) NO –, BF (3) NH , CH – (4*)mijksDrlHkh 2 2 3 3 3 3 7. There is no S–S bond in : fuEuesalsfdlesaS–Sca/kughagS% (1) S O 2– (2) S O 2– (3) S O 2– (4*) S O 2– 8. What is the correct order from the weakest to the strongest carbon-oxygen bond for the following species ? CO, CO , CO 2– , CH OH. 2 3 3 fuEuLih'khtdsfy,nqcZylsizcydkcZu&vkWDlhtuca/kdklghØegS\ CO, CO , CO 2– , CH OH. 2 3 3 (1*) CH OH < CO 2– < CO < CO (2) CO < CO < CO 2– < CH OH 3 3 2 2 3 3 (3) CH OH < CO 2– < CO < CO (4) CH OH < CO < CO 2– CO 3 3 2 3 2 3 Sol. C – O = 1.43 Å ; C = O 1.23 Å ; C  O 1.09 Å. 9. The spontaneous redox reaction/s among the following is/are fuEueslsdkSulhLor%vkWDlhdj.kvip;u¼jsMkWDl½vfHkfØ;k,agS& (1) 2Fe3+ + Fe  3Fe++ (2) Hg ++  Hg++ + Hg (3) 3 AgCl + NO + 2H O  3 Ag + 3 Cl– + NO – + 4H+ Given that fn;kgS&  Fe Fe = 0.77 V  Fe Fe = – 0.44 V E  = 0.85 V E   = 0.92 V Hg2 E Hg = 0.22 V Hg E  Hg2 = 0.96 V AgCl Ag NO3 NO (1*) a (2) a, b, c (3) a, b (4) a, c Sol. Only for this reaction E0 will come out to be positive, calculate using relation G° = G 0 + G 0 and G0 = – nFE° 1 2 cell dsoyblvfHkfØ;kdsfy,E0/kukRedvkrkgSfuEulaca/kdkmi;ksxdjdsx.kukdjrsgSaA G° = G 0 + G 0 ,oa G0 = – nFE° 1 2 cell 10. G° of the cell reaction AgCl(s) + ½ H (g) Ag(s) + H+ + Cl¯ is –21.52 kJ. G° of 2AgCl(s) + H2(g)  2Ag(s) + 2H + 2Cl¯ is - + AgCl(s) + ½ H2(g) Ag(s) lsy vfHkfØ;k ds fy, G° –21.52 kJ gS 2AgCl(s) + H2(g)  2Ag(s) + 2H+ + 2Cl¯ ds fy, G° gSA (1) –21.52 kJ (2) –10.76 kJ (3*) –43.04 kJ (4) 43.04 kJ Sol. (i) AgCl (s) + 1 H (g) Ag (s) + H+ + Cl–. G ° 2 2 1 (ii) 2AgCl (s) + H2 (g) 2Ag (s) + 2H+ + 2Cl– G2° eq. (i)  2 = (ii) G1º  2 = G2° G2° = –21.52  2 = – 43.04 kJ 11. Which of the following species are correctly matched with their geometries according to the VSEPR theory? (1*) BrF +  octahedral. (2*) SnCl –  trigonal bipyramidal. 6 5 (3*) ClF –  linear. (4*) IF +  see – saw. 2 4 VSEPR fl)kUr ds vuqlkj] fuEu esa ls dkSulh Lih'kht mudh T;kfefr ds lkFk lqesfyr gS \ (1*) BrF +  v"VQydh; (2*)SnCl –f=kdks.kh;f}fijsfefM; 6 5 (3*)ClF – js[kh; (4*) IF + lh&lkW 2 4 12. For the cell TI | TI+ (0.001 M) | | Cu2+ (0.1 M) | Cu. E cell at 25°C is 0.83 V, which can be increased- (1*) by increasing [Cu2+] (2) by increasing [Tl+] (3) by decreasing [Cu2+] (4*) by decreasing [Tl+] 25ºC ij TI| TI+ (0.001M) | |Cu2+ (0.1M) | Culsy ds fy, E 0.83 V gS rks bls fdl çdkj c<+k;k tk ldrk gSA (1*)[Cu2+]c<+kusls (2)[Tl+]c<+kusls (3)[Cu2+]?kVkusls (4*)[Tl+]?kVkusls 0.059 Sol. Ecell = E°cell – 2 [Tl ]2 log [Cu2 ] 13. The standard reduction potentials of some half cell reactions are given below : PbO + 4H+ + 2e– Pb2+ + 2H O E0 = 1.455 V 2 2 MnO – + 8H+ + 5e– Mn2+ + 4H O E0 = 1.51 V 4 2 Ce4+ + e– Ce3+ E0 = 1.61 V H2O2 + 2H + 2e 2H O E = 1.71 V + – 0 Pick out the correct statement : (1*) Ce+4 will oxidise Pb2+ to PbO (2*) MnO- will oxidise Pb2+ to PbO (3*) H2O2 will oxidise Mn+2 to MnO- (4) PbO2 will oxidise Mn+2 to MnO- dqNv)ZlsyvfHkfØ;kvksadsekudvip;ufoHkouhpsfn,gSa& PbO + 4H+ + 2e– Pb2+ + 2H O E0 = 1.455 V 2 2 MnO – + 8H+ + 5e– Mn2+ + 4H O E0 = 1.51 V 4 2 Ce4+ + e– Ce3+ E0 = 1.61 V H2O2 + 2H+ + 2e– 2H O E0 = 1.71 V lghdFkudkspqfu;s& (1*)Ce+4,Pb2+dksPbO esavkWDlhd`rdjsxk (2*) MnO- ,Pb2+dksPbO esavkWDlhd`rdjsxk 2 4 2 (3*)H O ,Mn+2dks MnO- esavkWDlhd`rdjsxk (4)PbO ,Mn+2dks MnO- esavkWDlhd`rdjsxk 2 2 4 2 4 Sol. The species with more SRP undergo reduction (forward) and with low SRP value undergoes oxidation (backward).