DPP-39 to 40 Faculty Copy

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 39 to 40 Class : XIII Course : DPP No.1 Total Marks : 41 Max. Time : 41 min. Single choice Objective ('–1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21] Subjective Questions ('–1' negative marking) Q.8 to Q.12 (4 marks 4 min.) [20, 20] 1. A complex with the molecular formula CrCl3.6H2O is such that 1/3 of the total chloride is precipitated by adding AgNO3 to its aqueous solution. Then which of the following is its best representation : CrCl3.6H2O vkf.odlw=k okykladqy blizdkj gSfd bldstyh; foy;uesa AgNO3Mkyusij dqyDyksjkbM dk1/3 Hk xvo{ksfirgkstkrkgSrcfuEuesalsdkSulklw=kblslclslghiznf'kZrdjrkgSA (1) CrCl3.6H2O (2) [Cr(H2O)3 Cl3].3H2O (3*) [CrCl2(H2O)4]Cl.2H2O (4) [CrCl(H2O)5]Cl2.H2O 1 Sol. [Cr Cl2 (H2O)4] Cl.2H2O will liberate 3 of the total chloride ions for precipitation. 2. Match list  with list  and select the correct answer : lwphrFk lwph dkslqesfyrdhft,rFk lghmÙkjpqfu;sA List (Ι) lwph (Ι) (Equiv. conductance at infinite dilution) ¼vuUrruqrkijrqY;kadhpkydrk½ List (ΙΙ) lwph (ΙΙ) (Formula) ¼lw=k½ (A) 229 (B) 97 (C) 404 (D) 523 (a) [Pt(NH3)5Cl]Cl3 (b) [Pt(NH3)3Cl3]Cl (c) [Pt(NH3)4Cl2]Cl2 (d) [Pt(NH3)6]Cl4 Thecode:¼dwVfuEugS½% A B C D (1) e a b d (2) a c d b (3) a d c b (4*) c b a d Ans. Eq. conductance at infinite dilution depends on no. of ions produced in the sol. [Pt(NH3)6] CI4 > [Pt(NH3)3 CI]CI3 > [Pt(NH3)4 CI2] CI2 > [Pt (NH3)3 CI3] CI  no. of ions produced is sol. 3. Which of the following complex ions obeys Sidgwick’s effective atomic number (EAN) rule ? fuEuladqyvk;uksaesalsdkSuflM~fodizHk ohijek.kqØekad(EAN)fu;edkikyudjrkgS\ (1) [Fe(CN) ]3– (2*) [Fe(CN) ]4– (3) [Cr(NH ) ]3+ (4) [Ni(en) ]2+ 6 6 3 6 3 Sol. (2) EAN = 26 – 2 + 2  6 = 24 + 12 = 36. 4. In which of the following complexes the effective atomic number is not equal to the atomic number of a inert gas : fuEuesalsfdlladqyesaizHk ohijek.kqØekadvfØ;xSldsijek.kqØekaddslekuughagksrkgSA (1) Ni(CO) (2) [Co(NH ) ]3+ (3) [Fe(CN) ]4– (4*) [CuCl ]– 4 3 6 6 2 Sol. In [CuCl ]– EAN = 28 + 4 = 32 5. A coordination compound of cobalt has the molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three moles of ions in an aqueous solution. The aqeuous solution on treatment with an excess of AgNO3 gives two moles of AgCl as a precipitate the formula of this complex would be dksckYVdk,dmilgla;ksth;kSfxdftlesa],ddksckYVijek.kqdsfy,ik¡pveksfu;kv.kq],dukbVªkslewgrFk nksDyksjhu ijek.kqokykv.kqlw=kj[krkgSA,dtyh;foy;uesabl;kSfxddk,deksyvk;uksadsrhueksynsrkgSAAgNO3ds vkfèkD;dslkFktyh;foy;udksmipkfjrdjusij]AgCldsvo{ksids:iesa nkseksynsrkgSaAblladqydklw=k gksxk& (1) Co[NH3)4NO2Cl][NH3Cl] (2) [Co(NH3)Cl] [ClNO2] (3*) [Co(NH3)5NO2]Cl2 (4) [Co(NH3)5] [(NO2)2Cl2] Ans. Co (NH3)5 NO2 CI2 1 mole of this complex gives 3 moles of ions in aq. solution. 1 mole of this complex gives 2 mole of AgCI  2CI– outside the coordination sphere.  [Co(NH ) Co (NH3)5 NO2 CI2 NO2] CI2 bl ladqydk 1eksy tyh;foy;uesa3eksyvk;u nsrkgSA blladqydk1eksyAgCl ds2eksynsrkgSmilgla;ksthe.Myesa2CI– vk;ugSaA  [Co(NH ) NO ] CI 6. Which of the following complex ions obeys Sidgwick’s effective atomic number (EAN) rule ? fuEuladqyvk;uksaesalsdkSuflM~fodizHk ohijek.kqØekad(EAN)fu;edkikyudjrkgS\ (1) [Fe(CN) ]3– (2*) [Fe(CN) ]4– (3) [Cr(NH ) ]3+ (4) [Ni(en) ]2+ 6 6 3 6 3 Sol. (2) EAN = 26 – 2 + 2  6 = 24 + 12 = 36. 7. In which of the following complexes the effective atomic number is not equal to the atomic number of a inert gas : fuEuesalsfdlladqyesaizHk ohijek.kqØekadvfØ;xSldsijek.kqØekaddslekuughagksrkgSA (1) Ni(CO) (2) [Co(NH ) ]3+ (3) [Fe(CN) ]4– (4*) [CuCl ]– 4 3 6 6 2 Sol. In [CuCl ]– EAN = 28 + 4 = 32 8. Two compounds have the molecular formula, Co(H2O)4(NO2)3. In aqueous solution one of these compounds does not conduct electricity while the other does. Write the possible structures of these two compounds. nks;kSfxdftudkv.kqlw=kCo(H2O)4(NO2)3gSAtyh;foy;uesabuesals,d;kSfxdfo|qrdkpkydughagSatcfdnwljk pkydgSAbunks;kSfxdksadhlaHkolajpukfyf[k,A Ans. Co (H2O)4 (NO2)3 [Co(H2O)3 (NO2)3]. H2O does not conduct electricity because no ion is generated in aq. sol. [Co (H2O)4 (NO2)2] NO2 will conduct electricity. Co(H2O)4(NO2)3 [Co(H2O)3(NO2)3].H2OoS|qrpkyd ugh gS D;ksafdtyh; foy;uesa dksbZvk;u mRiUuugh djrk gS [Co(H2O)4 (NO2)2]NO2 oS|qrh; pkyd gksxkA 9. A coordination compound has the formula CoCl3.4NH3. It does not liberate NH3 but precipitates Cl– ions as AgCl. Give the IUPAC name of the compound and write its structural formula. ,dmilgla;ksth;kSfxddklw=kCoCl3.4NH3gSA;gNH3eqDrughadjrkgSysfduAgClds:iesaCl–vk;uvo{ksfir djrkgSaA;kSfxddkIUPACukefyf[k,rFk bldhlajpuklw=kdksfyf[k,A Sol. [Co(NH3)4Cl2]Cl Tetramminedichloridocobalt(III) chloride. [Co(NH3)4Cl2]Cl VsVªk,EehuMkbDyksjkbMksdksckYV(II)DyksjkbMA 10. Explain the following, giving appropriate reasons. (i) Out of K4 [Fe(CN)6] and K3 [Fe(NH3)6] solutions, the former has higher value of molar conductivity. (ii) [Pt(NH3)2 Cl2] and [Pt(NH3)6] Cl4 differ in their electrolytic conductance (iii) The value of molar conductivity of the aqueous solution of [CoCl3 (NH3)3] is zero. mi;qDrdkj.knsdjfuEudhO;k[;kdhft,A (i)K4[Fe(CN)6]rFkkK3[Fe(NH3)6]foy;uksaesals]igysokysdheksyjpkydrkdkekumPpgksrkgSA (ii)[Pt(NH3)2Cl2]rFkk[Pt(NH3)6]Cl4oS|qrvi?kV~;pkydrkesafHkUugksrsgSA (iii)[CoCl3(NH3)3]ds tyh; foy;u dh eksyj pkydrk dk eku 'kwU; gksrk gSA Sol. (i) K4 [Fe(CN)6] will produce 5 ions while K3 [Fe(CN)6] will produce 4 ions in aq. sol. so, higher molar conductivity. (ii) [Pt(NH3)2CI2] produce no ions but [Pt(NH3)6]CI4 produce 5 ions in aqueous sol. so, higher molar conductivity. (iii) [CoCI3(NH3)3] will not produce any ion in aq. sol. so, conductivity. is zero. (i) K4 [Fe(CN)6] 5vk;u nsrk gS tcfd K3 [Fe(CN)6] 4vk;u nsrk gSA blfy, mPp pkydrk gSA (ii) tyh; foy;u esa [Pt(NH3)2CI2]dksbZ vk;u ugha nsrk gS tcfd [Pt(NH3)6]CI4 5vk;u nsrk gSA blfy, mPp eksyj pkydrkgSA (iii) tyh;foy;u esa[CoCI3(NH3)3]dksbZvk;uugh nsxk]blfy,pkydrk'kwU; gSA 11. Arrange the following complexes in the increasing order of their electrical conductivity : [Co(NH3)3 Cl3], [Co(NH3)6] Cl3 and [Co(NH3)5 Cl]Cl. fuEuladqyksadksmudsoS|qrpkydrkdsvkjksghØeesaO;ofLFkrdhft,A [Co(NH3)3 Cl3], [Co(NH3)6] Cl3 rFkk [Co(NH3)5 Cl]Cl. Ans. [Co(NH3)6] CI3 > [Co(NH3)5 CI]CI > [Co(NH3)3 CI3] More no. of ion present in aqueous solution quantivity increase. tyh;foy;uesavk;uksadhla[;kesao`f)gksusijpkydrkc<+rhgSA 12. The compound CoCl3.4NH3 contains only one Cl– ion that is precipitated immediately on the addition of Ag+ ions. Draw the structure of the compound on the basis of Werner’s coordination theory. ;kSfxdCoCl3.4NH3dsoy,dCl– vk;u;qDrgksrkgSatksfdAg+vk;udksfeykusijrqjUrvo{ksfirgkstkrkgSAouZj milgla;ksthfl)kUrdsvk/k jij;kSfxddhlajpukcukb;saA Ans. CoCl3.4NH3 Since only one CI– ion is precipiteated by Ag+ ion. This implies that only one CI– ion outside coordination sphere.  [Co(NH3)4CI2]CI CoCl3.4NH3 Ag+ vk;u ds }kjk dsoy ,d Cl– vk;u vo{ksfirgksrkgSA vFk Zr~dsoy,dCl–vk;u milgla;ksthoxZ(e.My)dsckgj mifLFkr gSA [Co(NH3)4CI2]CI H3N H3N H3N H3N CI CI Trans CI CI NH3 NH3 NH3 NH3 + CI H3N NH3 NH3 Cis + CI H3N NH3 NH3 + CI NH3 + CI NH3 Trans Cis DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 38 Class : XIII Course : DPP No.2 Total Marks : 33 Max. Time : 33 min. Single choice Objective ('–1' negative marking) Q.1 to Q.8 (3 marks 3 min.) [24, 24] Comprehension ('–1' negative marking) Q.9 to Q.11 (3 marks 3 min.) [9, 9] 1. It is experimentally found that the compound K3[Ni(CN)5] shows an decrease in its weight when placed in a magnetic balance and four metal–ligand bond lengths are equal but the rest is different. Then which of the following set of informations is correct : (i) The transition metal is sp3d hybridised (ii) The net dipole moment of complex is  zero (iii) The transition metal is dsp3 hybridised (iv) The net dipole moment of the complex is zero (v) The complex ion is trigonal bipyramidal (vi) The complex ion is square pyramidal izk;ksfxd:ilsik;kx;kfd;kSfxdK3[Ni(CN)5]dkstcpqEcdh;rqykesaj[k tkrkgSrks;gbldsHk jesadehiznf'kZr djrkgSrFk pkj/krq&fyxs.Mca/kyEckbZ;k¡leku]ijUrq'ks"kfHkUugksrhgSaArcfuEuesalslwpukvksadkdkSulkleqPp; lghgS& (i) laØe.k/k rqsp3dladfjrgSA (i)ladqydkdqyf}/kzqovk/kw.kZ'kwU;ughagksrkgSA (ii) laØe.k/k rqdsp3ladfjrgSA (iv)ladqydkdqyf}/kzqovk/kw.kZ'kwU;gksrkgSA (v) ladqyvk;uf=kHkqth;f}fijsfefM;gSA (vi)ladqyvk;uoxZfijsfeMh;gSA (1) (i),(ii),(v) (2*) (ii),(iii),(vi) (3) (iii),(iv),(v) (4)noneofthesebuesalsdksbZugha Sol. According to the question K3 [Ni(CN)5] is diamagnetic and square pyramidal with non-zero dipole moment. bliz'udsvuqlkjK3[Ni(CN)5]izfrpqEcdh;gSrFk v'kwU;f}/kzqovk?kq.kZdslkFkoxkZdkjfijkfeMhgSA 2. When the complex K6 [(CN)5 Co–O–O–Co(CN)5] is oxidised by bromine into K5[(CN)5 Co–O–O–Co(CN)5]. Then which of the following statements will be true about this change: (In both compound Co is in + 3 oxidation state) (1) Co() is oxidised in Co() (2) The O–O bond length will increase (3*) The O–O bond length will decrease (4) ‘A’ & ‘B’ both are correct tc ladqy K6 [(CN)5 Co–O–O–Co(CN)5]czksehu }kjk K5[(CN)5 Co–O–O–Co(CN)5]esavkWDlhd`r gksrk gS] rksbl ifjorZudslanHkZesafuEuesalsdkSulkdFkulghgS& (1)Co(),Co()esavkWDlhd`rgksrkgSA (2)O–Oca/kyEckbZesao`f)gksxhA (3*)O–Oca/kyEckbZesadehgksxhA (4)‘A’o‘B’nksukslghgSaA Sol. In the first complex ligand is O 2– which is oxidised into O 1– hence O – O bond length decreases. 2 2 izFkeladqyesaO 2–fyxs.MgStksO 1–esavkWDlhd`rgksrkgSblizdkjO–OcU/kyEckbZ?kVrhgSA 2 2 3. The crystal field-splitting for Cr3+ ion in octahedral field increases for ligands I–, H O, NH , CN– and the order 2 3 is : v"VQydh;{ks=kesaCr3+ vk;udsfy;sfØLVy&{ks=kfoikVu]I–,H O,NH ,CN–fyxs.Mksadsfy;sc Q > R > S (2*) Q > R > S > P (3) S > R > P > Q (4) R > Q > P > S Sol. (2) On the basis of nature of ligands the correct order is Q > R > S > P. (2) fyxs.M dh izd`frds vk/kkjij lghØe Q>R>S>PgSA 8. In the reaction : [Ag(CN) ]– + Zn  the complex formed will be : fuEuvfHkfØ;kesa] [Ag(CN) ]– + Zn  cu sokykladqygksxk& (1*) Tetrahedral (2) square planar (3) octahedral (4) triangal bipyramidal (1*)prq"Qydh; (2) oxkZdkj lery (3) v"VQydh; (4)f=kdks.kh;f}fijkfeMh; Sol. 2[Ag(CN) ]– + Zn  2Ag + [Zn(CN) ]2–. 2 4 Shape of [Zn(CN) ]2– (3d)10 is tetrahedral. 2[Ag(CN) ]– + Zn  2Ag + [Zn(CN) ]2–. [Zn(CN) ]2– esa(3d)10 dsdkj.kvkd`fÙkprq"Qydh;gSA Comprehension : (Q.9 to Q.11) Werner performed two experiments : Expt-1 : He prepared a compound X by reacting KCl with PtCl4. The compound X didn't give any ppt. with AgNO3 but gave electrical conductance corresponding to 3 ions. Expt-2 : He took 0.319 g of CrCl3.6H2O and passed through a cation exchange resin & the acid coming out required 28.5 ml of 0.125 M NaOH. Hence vuqPNsn % (Q.9lsQ.11) ouZjusnksvuqiz;ksxfd;s& iz;ksx-1:ouZjusKCldslkFkPtCl4dhvfHkfØ;kdjkdj;kSfxdXcuk;kA;kSfxdX,AgNO3dslkFkdksbZvo{ksiughansrk gSAysfdu3vk;udslUnHkZesafo|qrpkydrkn'k ZrkgSA iz;ksx-2 : ouZj us CrCl3.6H2O dk 0.319 g fy;k rFkk bls ,d /kuk;u fofue; jsft+u (cation exchange resin) ls izokfgr fd;k rFkk ik;k fd cuus okys vEy ds fy, 0.125 M NaOH ds 28.5 ml vko';d gSA blizdkj& 9. The formula of the compound X is - ;kSfxdXdklw=kgS%& (1) [KPtCl]Cl2 (2) K2[PtCl4] (3*) K2[PtCl6] (4) K[PtCl4] Sol. PtCl4 + 2 KCl  K2[PtCl6] 10. The hybridization in K2[PtCl6] is - K2[PtCl6]esaladj.kgS% (1) sp3 (2*) d2sp3 (3) sp3d2 (4) dsp3 Sol. Pt+4 = 5d6 11. The complex CrCl3.6H2O can be rightly represented as : ladqyCrCl3.6H2Odkslghiznf'kZrfd;ktkldrkgS%& (1) [Cr(H2O)4Cl2]Cl (2*) [Cr(H2O)6]Cl3 (3) [Cr(H2O)3Cl3]3H2O (4) [Cr(H2O)5Cl]Cl2 0.319 Sol. 266.5  V.F.  1000 = 0.125  28.5 V.F. = 3 So complex is [Cr(H2O)6]Cl3

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