DPP-41 to 42 With Answer

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 41 to 42 Class : XIII Course : DPP No.1 Total Marks : 48 Max. Time : 48 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] Multiple choice objective ('–1' negative marking) Q.6 (4 marks 4 min.) [4, 4] Comprehension ('–1' negative marking) Q.7 to Q.8 (3 marks 3 min.) [6, 6] Subjective Questions ('–1' negative marking) Q.9 to Q.10 (4 marks 4 min.) [8, 8] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.11 to Q.12 (3 marks 3 min.) [6, 6] Integer Types Questions Short Subjective Questions ('–1' negative marking) Q.13[Q.(i) to Q(iii)] (3 marks 3 min.) [9, 9] 1. Spin only magnetic moment of a complex having CFSE = – 0.6  be: and surrounded by weak field ligands can (1) 1.73 BM (2) 4.9 BM (3*) both (1) & (2) (4) None of these ,dladqy]ftldkCFSE=–0.60gSrFk tksnqcZy{ks=kfyxs.Mksalsf?kjkgS]dkdsoypØ.kpqEcdh;vk?kq.kZgksldrk gSA (1) 1.73 BM (2) 4.9 BM (3*)(1)rFk (2)nksuks (4)buesalsdksbZugha Sol. The options can give CFSE = – 0.6  with weak field ligands  d4 and d9. 2. Which of the following statements is not correct? (a) [Ni(H O) ]2+ and [Ni(NH ) ]2+ have same value of CFSE 2 6 3 6 (b) [Ni(H O) ]2+ and [Ni(NH ) ]2+ have same value of magnetic moment 2 6 3 6 (1*) Only a (2) Only b (3) Both a and b (4) None of these fuEuesalsdkSulkdFkulghughagSA (a) [Ni(H O) ]2+ o[Ni(NH ) ]2+ dsa CFSE ds eku leku gSA 2 6 3 6 (b)[Ni(H O) ]2+ and[Ni(NH ) ]2+ dsapqEcdh;vk?kw.kZds ekuleku gSA 2 6 3 6 (1*) dsoy a (2) dsoy b (3)aobnksukas (4)buesalsdksbZugh Sol. Ammonia is a stronger field ligand than water. 3. Nitroprusside ion is : A : [FeII(CN) NO+]2– and not B : [FeII(CO) NO]2+. A and B can be differentiated by. (1) estimating the concentration of iron (2) measuring the concentration of CN– (3*) measuring the magnetic moment (4) thermally decomposing the compound. ukbVªksizqlkbM vk;u A=[FeII(CN) NO+]2– uk dh B=[FeII(CO) NO]2+ 5 5 ArFk BdksfuEuds}kjkfoHksfnrfd;ktkldrkgS (1)vk;judhlkanzrkdsifjdyu}kjk (2)CN–dhlkanzrkdsekiu}kjk (3*)pqEcdh;vk?kw.kZdsekiu}kjk (4);kSfxddsÅ"eh;fo;kstu}kjk 4. It is given that a complex formed by one Ni2+ ion and some Cl– ions and some PPh3 molecules does not show geometrical isomerism and its solution does not show electrical conductance. Then which is correct about the complex : (1) It is square planar (2*) It is tetrahedral (3) It is diamagnetic (4) none of the above is correct fn;kx;kgSfd,dNi2+vk;urFkkdqNCl–vk;urFkkdqNPPh v.kqlscuk,dladqyT;kferh;leko;orkiznf'kZr ughadjrkgSrFk bldkfoy;ufo|qrpkydrkughan'k ZrkgSArcladqydsckjsesadkSulkdFkulghgSA (1);glery oxkZdkjgSA (2*);g prq"Qydh; gSA (3);gizfrpqEcdh;gSA (4)mijksDresalsdksbZlghughagSA Sol. It is not showing geometrical isomerism tetrahedral. 5. The green coloured complex K2[Cr(CN)4(NH3) (NO)] is paramagnetic and its paramagnetic moment (spin only) is 1.73 B.M. Which of the following is correct about it : (i) Its IUPAC name is Potassium amminetetracyanonitrosylchromate () (ii) It IUPAC name is Potassium amminetetracyanonitrosoniumchromate () (iii) Hybridisation state of chromium is sp3d2 (iv) It cannot show geometrical isomerism (v) Hybridisation state of chromium is d2sp3 (vi) It can show linkage isomerism gjsjaxdkladqyK2[Cr(CN)4(NH3)(NO)]vuqpqacdh;gSrFk bldkvuqpqacdh;vk?kw.kZ¼dsoypØ.k½1.73B.M.gSAfuEu esalsbldslanHkZesaD;klghgSA (i)bldkIUPACukeiksVsf'k;e,EehuVsVªklk;uksukbVªksflyØksesV()gSA (i)bldkIUPACukeiksVsf'k;e,EehuVsVªklk;uksukbVªksflfy;eØksesV()gSA (iii)Øksfe;edhladj.kvoLFkksp3d2 gSA (iv);gT;kferh;leko;orkiznf'kZrughadjldrkgSA (v)Øksfe;edhladj.kvoLFkkd2sp3gSA (vi);gca/kuhleko;orkiznf'kZrdjldrkgSA (1) (ii), (iii),(iv) (2) (i),(iii), (vi) (3) (i), (v) (4*) (ii), (v), (vi) Sol. (4) The complex is actually  ()   k Cr(CN)4 (NH3 )(NO) in which Cr() is d2sp3 hybridised with one unpaired electron. 2   6. Which are correct statements ? (1*) [Ag(NH ) ]+ is linear with sp hybridised Ag+ ion. (2*) NiCl 2– , VO 3– and MnO – have tetrahedral geometry 4 4 4 (3*) [Cu(NH ) ]2+ , [Pt(NH ) ]2+ and [Ni(CN) ]2– have dsp2 hybridisation of the metal ion 3 4 3 4 4 (4*) Fe(CO)5 have bipyramidal structure with dsp hybridised iron. 3 dkSulslghdFkugS? (1*) [Ag(NH ) ]+ ,spladjhr Ag+ vk;u ds lkFk js[kh; vkd`frdk gS (2*) NiCl 2– ,VO 3– rFkk MnO – leprq"Qydh; T;kfefr j[krs gS 4 4 4 (3*) [Cu(NH ) ]2+ ,[Pt(NH ) ]2+ rFkk [Ni(CN) ]2– , /kkrq vk;u dk ladj.k dsp2 j[krs gS 3 4 3 4 4 (4*)Fe(CO) ,dsp3ladjhrvk;ju dslkFk f}fijkfeMylajpuk j[krkgS Passage :(Q.7 to Q.8) Read the following passage based on Applications of crystal field theory to explains magnetic and spectral properties of complexes carefully and answer the questions. With the help of CFT number of unpaired electron in a compound can be calculated and we can calculate its paramagnetic moment(due to spin only), by the formula : µ = Bohr magneton (BM). where n is the number of unpaired electron in the complex. For spectral analysis the separation between t and e orbitals, called ligand field splitting. Parameter  (for octahedral complexes) should be known to us, which can be easily calculated by observing the absorption spectrum of one electron complex. Figure shows the optical absorption spectrum of the d1 hexaaquatitanium(III) ion [Ti(H O) ]3+. The CFT assigns the first absorption maximum at 20,300 cm–1 to the transition e  t . For 2 6 g 2g multielectronic(d2 to d10) system, the calculation of  by absorption spectrum is not that easy as the absorption spectrum will also be affected by electron-electron repulsions. vuqPNsn : (Q.7 to Q.8) fØLVy{ks=kfl)kUrdsvuqiz;ksxijfuEuvuqPNsndksif<+;srFk /;kuiwoZdladqydhpqEcdh;rFk LisDVªexq.ksadhO;k[;k dhft,rFk iz'uksadsmÙkjnhft,A ,d;kSfxdesav;qfXerbySDVªksudksCFTla[;kdhlgk;rklsifjdfyrfd;ktkldrkgSarFk gefuEulw=k}kjkbldk vuqpqEcdh;vk?kw.kZ¼dsoypØ.kdsdkj.k½ifjdfyrdjldrsgSA µ= cksjeSXusVkWu(BM)tgk¡nladqyesav;qfXerbySDVªksudhla[;kgSA LisDVªyfo'ys"k.kdsfy,t2grFk eg d{kddschpi`Fkd~dj.kdksfyxsaM{ks=kfoikVudgrsgSAizkpy0(v"VQydh;ladqy dsfy,)gesKkrgksukpkfg,]tksfd,dbySDVªkWuladqydsvo'ks"k.kLisDVªelsizsf{krfd;ktkldrkgSaAfn;kx;kfp=k] d1 gsDlk,DokVkbVsfu;e (III) vk;u [Ti(H O) ]3+ dk izdk'kh; vo'kks"k.k LisDVªe n'kkZrk gSA CFT laØe.k 2 6 e  t dsfy,20,300cm–1ijizFkevo'k s"k.kvf/kdrecrkrkgSAcgqbySDVªksfud(d2tod10)fudk;dsfy,]vo'k s"k.k g 2g LisDVªeds}kjk0dkifjdyudjukbrukvklkuughagS]D;ksafdbySDVªkWu&bySDVªkWuizfrd"kZ.k}kjkvo'k s"k.kLisDVªe izHkforgksrkgSA 7. The crystal field stabilization energy (CFSE) for complex given in the passage, [Ti(H O) ]3+ will be 2 6 (in kJ/mol) mijksDrvuqPNsnesaladqy[Ti(H O) ]3+ dsfy,nhxbZfØLVy{ks=kLFk ;hdj.kÅtkZ(CFSE)(kJ/molesa) gksxh& 2 6 (1) 243 kJ/mole (2*) 97 kJ/mole (3) 194 kJ/mole (4) 143 kJ/mole J Sol. Ti+3 is 3d1 system, 0= 6.63 × 10–34 × 3 × 108 × 20300 × 102 ion kJ = 6.63 × 10–34 × 3 × 108 × 20300 × 102 × 10–3 × 6.02 × 1023 mol kJ Now, CFSE = 0.4 × 0 = 0.4 × 243 = 97.2 mol kJ = 243 mol 8. The magnetic moments of following, arranged in increasing order will be (atomic number of Co = 27) fuEudspqEcdh;vk?kw.kZdkvkjksghØe(Codkijek.kqØekad=27)gksxk& (1) Co3+ (octahedral complex with a strong field ligand) (2) Co3+ (octahedral complex with a weak field ligand) (3) Co2+ (tetrahedral complex) (4) Co2+ (square planar complex) (1) Co3+ (,dizcy{ks=kfyxsaMdslkFkv"VQydh;ladqy) (2)Co3+ (,dnqcZy{ks=kfyxsaMdslkFkv"VQydh;ladqy) (3) Co2+ (prq"Qydh; ladqy) (4) Co2+ (oxZ leryh; ladqy) (1) 1 > 2 > 3 > 4 (2*) 2 > 3 > 4 > 1 (3) 3 > 2 > 4 > 1 (4) 2 > 4 > 3 > 1 Sol. (1) D (0) , (2) P (4) , (3) P(3), 3d 4s 4p (4) P(1) dsp2 9. In each of the following pair of complexes, choose the one that absorbs light at a longer wave length. fuEuizR;sdladqyksads;qXeesa]mlladqydkspqfu;s]tksfdyEchrjaxnS/;Zdkvo'ks"k.kdjrkgSA (1) [Co(NH ) ]2+ , [Co(H O) ]2+ (2) [FeF ]3– , [Fe(CN) ]3– (3) [Cu(NH ) ]2+ , [CuCl ]2– 3 6 2 6 6 6 3 4 4 Sol. Stronger field ligand will split more so, more energy is required to transition of electrons from t2g to eg so, smaller wavelength light is required. 10. Observed (expermiental) lattice energies (in KJ/mol) of octahedrally coordinated crystals of VO and FeO are – 3917 and – 3923 respectively. The lattice energies of these crystals in the absence of CFSE are – 3691 and – 3856 kJ/mol respectively. Assuming that oxide ion (O2–) is a weak field ligand, calculate CFSE value of V2+ and Fe2+ ion. v"VQydh; milgla;ksthfØLVyVOrFkkFeOdhizsf{kr¼izk;ksfxd½ tkydÅtkZØe'k%–3917o–3923gSACFSE dh vuqifLFkfresa bufØLVyksa dh tkydÅtkZ Øe'k%–3691rFkk –3856kJ/molgSA;g ekudjfd vkWDlkbMvk;u (O2–),dnqcZy{ks=kfyxsaMgSarksV2+ rFkkFe2+vk;udsfy,CFSEdkekuifjdfyrdhft,A Ans. – 226, – 67. Sol. CFSE of Fe2+ = observed L.E. – L.E. (in absence of CFSE). = – 3923 – (– 3856) = – 67 kJ/mol. CFSE of V2+ = – 3917 – (– 3691) = – 226 kJ/mol. BooSt YoUr PreViouS ConCept 11. Ca3(PO4)2(s) and H3PO3(s) contains same number of 'P' atom then the ratio of oxygen atom in the two compounds respectively is : Ca3(PO4)2(s)vkSjH3PO3(s)nksuksaghlekula[;kesa'P'ijek.kqj[krsgSarcbunks;kSfxdksaesavkWDlhtuijek.kqdkvuqikr Øe'k%gS: 8 (1) 3 2 (2) 3 4 (3) 3 (4*) 3 Sol. Let Ca3(PO4)2 is x-mole H3PO3 is y-mole x 1  2x = y  y = 2  moles of 'O' Ca3(PO4 )2 moles of ' O' in H3PO3 8x = 3y 4 = 3 . 12. 17 g of AgNO3 solution is mixed with 400 ml of 0.2 M HCl solution then how many gram of AgCl will be precipitated : [Ag = 108, Cl = 35.5, N = 14, O = 16] 17g]AgNO3dsfoy;udks0.2MHClfoy;uds400mldslkFkfeyk;ktkrkgSrcfdrusxzkeAgCldsvo{ksfirgksaxs [Ag = 108, Cl = 35.5, N = 14, O = 16] (1) 16.7 (2) 14.35 (3*) 11.48 (4) 36.5 Sol. AgNO3 + HCl 17  AgCl  + HNO m moles 170 x 1000 80  moles = 80 1000 x 143.5 = 11.48. Integer Answer Type 13. This section contains 3 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. iw.k±dmÙkjizdkj bl[k.Mesa3iz'ugSaAizR;sdiz'udkmÙkj0ls9 rdiw.k ±dgSA (i) Find out the number of 3d electrons occupied in t orbitals of hydrated Cr3+ ion (octahedral). gkbMªsVsVCr3+ vk;u¼v"VQydh;½dst d{kdesamifLFkr3dbysDVªkWuksadhla[;kKkrdjsaA Ans. 3 Sol. Cr3+ ; 3d3 n = 3 (ii) How many Cr—O bonds are equivalent in dichromate anion ? MkbØksesVds_.k ;uesafdrusCr—Oca/krqY;¼leku½gksrsgSaA Ans. 6 Sol. Six Cr—O bonds are equivalent due to resonance and two bridged Cr–O bonds are equivalent (no resoance) gy. vuqukndsdkj.kN%Cr—OcU/krqY;gksrsgSrFk nksCr–OlsrqcU/krqY;gksrsgSaA¼buesavuquknughagksrkgS½A (iii) CrO42– + 2H+ + 2H2O2  CrO5 + 3H2O How many peroxide linkages are found in the structure of CrO5 ? CrO42– + 2H+ + 2H2O2  CrO5 + 3H2O CrO5dhlajpukesafdrusijkWDlkbMla;kstuik;stkrsgSa? Ans. 2 Sol. DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 42 Class : XIII Course : DPP No.2 Total Marks : 45 Max. Time : 45 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] Multiple choice objective ('–1' negative marking) Q.6 to Q.8 (4 marks 4 min.) [12, 12] Comprehension ('–1' negative marking) Q.9 to Q.11 (3 marks 3 min.) [9, 9] BooSt YoUr PreViouS ConCept Comprehension ('–1' negative marking) Q.9 to Q.11 (3 marks 3 min.) [9, 9] 1. Which of the following is true : (1) [Zn(Cl)2 (NH3)2] will exist in cis and trans forms (2) [Pt(Br) (Cl) (NH3) (Py)] is an optically active compound (3*) The brown ring complex [Fe(H O) NO+]2+ is paramagnetic 2 5 (4) All the above are true fuEuesalsdkSulklR;gS& (1)[Zn(Cl)2 (NH3)2]lei{kofoi{kizdkjksaesavfLrRoj[ksxkA (2)[Pt(Br)(Cl)(NH3)(Py)] ,d izdkf'k; :i ls lfØ; ;kSfxd gSA (3*)Hkwjhoy;ladqy[Fe(H O) NO+]2+ vuqpqacdh;gSA 2 5 (4)mijksDrlHkhlR;gSA Sol. (3) (1) is tetrahedral (2) is square planar [Pt() complex] (3) is Fe() complex, contains three unpaired e. 2. Which of the following is true about the complex [PtCl2(NH3)(OH2)] ; [Atomic no. of Pt = 78] (i) It will have two geometrical isomeric forms, cis and trans (ii) The hybridisation state of Pt() is sp3 (iii) It is a square planar complex (iv) It is a diamagnetic complex (v) It can show hydrate isomerism (vi) It is a tetrahedral complex ladqy[PtCl2(NH3)(OH2)]dslanHkZesafuEuesalsdkSulklR;gSA[Ptdkijek.kqØekad=78] (i)bldslei{kofoi{knksT;kferh;leko;ohgksaxsA (ii) Pt()dh ladj.k voLFkksp3gSA (iii) ;g ,d lery oxkZdkj ladqy gSA (iv);g,dizfrpqEcdh;ladqygSA (v);ggkbMªsVleko;orkiznf'kZrdjldrkgSA (vi);g ,d prq"Qydh; ladqy gSA (1*) (i), (iii),(iv) (2) (ii),(iv),(v) (3) (ii),(v),(vi) (4) (i),(v),(vi) Sol. (1) Pt() is 5d8, forms square planar complex which is diamagnetic. [Pt Cl (NH ) (OH )] will show geometrical isomerism. 2 3 2 3. The octahedral complex [Rh(NO ) (SCN) (en) ]+ can exist in a total number of isomeric forms including 2 2 stereoisomers : v"VQydh;ladqy[Rh(NO )(SCN)(en) ]+f=kfoeleko;ohlfgrdqyfdrusleko;ohizdkjksaesavfLrRoj[kldrkgSA 2 2 (1) 2 (2) 4 (3) 8 (4*) 12 Sol. (4) (1) NO2 / SCN (5) NO2 / SCN (9) (2) ONO / SCN (6) ONO / SCN (10) (3) NO2 / NCS (7) NO2 / NCS (11) (4) ONO NCS (8) ONO / NCS (12) * Only cis isomer is optical active. 4. Total number of geometrical isomers of Ma3 b3 type of octahedral complex are (1*) Two (2) four (3) Six (4) Zero Ma3b3izdkjdsv"VQydh;ladqydsT;kferh;leko;fo;ksadhdqyla[;kgSA (1*)nks (2)pkj (3) N% (4)'kwU; 5. STATEMENT -1 : Tetrahedral complexes do not show geometrical isomerism STATEMENT -2 : All the bond angles in tetrahedral geometry are 109.5°. (1*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True (5) Statement-1 is False, Statement-2 is False dFku-1:leprq"Qyfd;ladqy(Tetrahedralcomplexe)]T;kferh;leko;ork(geometricalisomerism)ughan'k ZrsgSaA dFku -2:lHkhleprq"Qyfd;T;kfefr(tetrahedralgeometry) esa ca/kdks.k109.5°gksrkgSA (1*)dFku&1lR; gS] dFku&2lR; gS ;dFku&2,dFku&1 dk lgh Li"Vhdj.k gSA (2)dFku&1lR; gS] dFku&2lR; gS ; dFku&2,dFku&1 dk lgh Li"Vhdj.k ugha gSA (3) dFku&1lR; gS] dFku&2vlR; gSA (4) dFku&1vlR; gS] dFku&2lR; gSA (5) dFku&1vlR; gS] dFku&2vlR; gSA 6. Which is/are correct statement (s) ? (1*) [Co(en)3] [Cr(CN)6] will display coordination isomerism. (2*) [Mn(CO)5(SCN)] will display linkage isomerism. (3*) [Co(NH3)5(NO3)]SO4 will display ionisation isomerism (4) None is correct. dkSulk@dkSulsdFkulghgS? (1*)[Co(en)3][Cr(CN)6]milgl;kasthleko;orkn'kkZ;sxsa (2*)[Mn(CO)5(SCN)]fyadstleko;orkn'kkZ;sxsa (3*)[Co(NH3)5(NO3)]SO4 vk;uuleko;orkn'kkZ;sxsa (4)dksbZlghughagSA 7. Which of the following statement(s) is(are) correct ? (1*) [Fe(CN) ]3– and [Co(C O ) ]3– are inner orbital complexes. 6 2 4 3 (2*) [FeF ]3– and [CoF ]3– are outer orbital complexes. 6 6 (3*) [Ni(NH ) ]2+ is an outer orbital complex. (4*) [Fe(H O) ]3+ is outer orbital complex with five unpaired electrons. 2 6 fuEuesalsdkSuls@dkSulkdFkulghgSa\ (1*) [Fe(CN) ]3– rFkk[Co(C O ) ]3– vkarfjd d{kd ladqy gaSA 6 2 4 3 (2*)[FeF ]3– rFkk[CoF ]3– ckgjhd{kd ladqygSaA 6 6 (3*)[Ni(NH ) ]2+ ,dvkarfjd d{kd ladqy gSaA (4*)[Fe(H O) ]3+ ikWpv;qfXerbysDVªkWudslkFkckgjhd{kdladqygSaA 2 6 Sol. Complexes in which inner d-orbital (3d) is used in hybridisation is called an inner orbital or low spin or spin paired complexes. Complexes in which outer d-orbital (4d) is used in hybridisation is called outer orbital or high spin or spin free complexes. gy- ladqy ftuesa vkarfjd d-d{kd (3d) ladj.k esa ç;qDr gksrs gS vkarfjd d{kd ladqy vFkok U;wu pØ.k vFkok pØ.k ;qfXer ladqy dgykrs gSA ladqy ftuesa ckgjh d-d{kd (4d) ladj.k esa ç;qDr gksrs gS ckgjh d{kd laadqy vFkok mPp pØ.k vFkok pØ.k eqDr ladqy dgykrs gSA 8. Which are correct statements ? (1*) [Ag(NH ) ]+ is linear with sp hybridised Ag+ ion. (2*) NiCl 2– , VO 3– and MnO – have tetrahedral geometry 4 4 4 (3*) [Cu(NH ) ]2+ , [Pt(NH ) ]2+ and [Ni(CN) ]2– have dsp2 hybridisation of the metal ion 3 4 3 4 4 (4*) Fe(CO)5 have bipyramidal structure with dsp hybridised iron. 3 dkSulslghdFkugS? (1*) [Ag(NH ) ]+ ,spladfjr Ag+ vk;u ds lkFk js[kh; vkd`frdk gS (2*) NiCl 2– ,VO 3– rFkk MnO – leprq"Qydh; T;kfefr j[krs gS 4 4 4 (3*) [Cu(NH ) ]2+ ,[Pt(NH ) ]2+ rFkk [Ni(CN) ]2– , /kkrq vk;u dk ladj.k dsp2 j[krs gS 3 4 3 4 4 (4*)Fe(CO) ,dsp3ladfjrvk;ju dslkFk f}fijkfeMylajpuk j[krkgS Sol. (1) [NH3 – Ag – NH ] linear shape + (2) Cl–, O–2 are weak ligand, so Ni2+  3d8 no paring V+5  3d0 no paring Mn+7  3d0 no paring (3) Cu2+  3d9 NH3 is strong ligand so one unpaired electron jumps from d orbital to p orbital. Pt2+  4d8 Pairing the electron Ni2+  3d8 CN– strong ligand, pairing the electron (4) Fe(CO)5 has zero dipole moment so Fe0  [Ar] 4s2, 3d6 CO is strong ligand, hybridisation dsp2 BooSt YoUr PreViouS ConCept Comperhension : (Q.9 to Q.11) Many times two different solutions are mixed together to get desired concentration in different volume ratio. To 100 ml of 5 M NaOH solution (density 1.2 g/ml) were added 200/3 ml of another NaOH solution which has a density of 1.5 g/ml and contains 20 mass percent of NaOH. Aluminium reacts with this (final) solution according to the following reaction. Al + NaOH + H O  NaAlO + H (At.wt. Al = 27, Na = 23, O = 16, H = 1) vuqPNsn % (Q.9to Q.11) dbZckjnksfHkUufoy;uksadksfofHkUuvk;rudsvuqikresa,dlkFkfeykdjmi;qDrlkUnzrkizkIrdjldrsgSA 5MNaOHfoy;u ¼?kuRo 1.2g/ml)ds 100ml dks vU; NaOHds200/3mlds lkFk feyk;k tkrkgS] ftldk ?kuRo 1.5 g/mlgSarFk blesa20%(Hk jls)NaOH mifLFkrgSA,Y;qfefu;ebl(ifj.k eh)foy;udslkFkfØ;kfuEufyf[krvfHkfØ;k dsvuqlkjdjrkgS& Al + NaOH + H O  NaAlO + H (At.wt. Al = 27, Na = 23, O = 16, H = 1) 9. What is the concentration of resulting NaOH solution obtained by mixing above two NaOH solutions : mijksDrnksNaOHfoy;udksfeykuslsizkIrifj.k ehNaOHfoy;udhlkUnzrkD;kgksxhA (1) 5 M (2) 3 M (3*) 6 M (4) 2 M Sol. Mole of NaOH in 1st solution = 0.5 moles 200 1.5  0.2 moles of NaOH addded = 40 x 3 = 0.5 moles of NaOH in the final solution = 0.5 + 0.5 = 1 moles  concentration of resulting solution = 1 500 / 3 x 1000 = 6 M gy- izFke foy;uesa NaOHdseksy =0.5moles 200 1.5  0.2 feyk;s x;s NaOHds eksy = 40 x 3 =0.5 vfUre foy;u esa NaOHds eksy = 0.5+0.5 =1eksy 1  ifj.kkeh foy;u dh lkUnzrk = 500 / 3 x 1000= 6M 10. What is the maximum volume of H2(g) liberated at STP when Al reacts with the complete solution as in the above reaction : STPijH2(g)dkvf/kdrevk;ruD;kgksxktcmijksDrvfHkfØ;kesaiw.kZfoy;udhAllsfØ;kgkstk;sA (1) 22.4 L (2) 44.8 L (3) 67.2 L (4*) 33.6 L Sol. Al + NaOH + H O  NaAlO + 3/2 H moles of H2 produced from 1 moles of NaOH = 3 moles 2 volume of H2 produced at STP = 3 × 22.4 = 33.6 litre 2 gy- Al + NaOH + H2O  NaAlO2 + 3/2 H2 1 eksy NaOHls mRikfnr H2ds eksy 3 = 2 moles STP ij mRikfnr H2 dk vk;ru = 3 × 22.4 = 33.6 litre 11. How much volume of the above solution is needed for the complete reaction with 54 g of Al : mijksDrfoy;udh,Y;qfefu;eds54gdslkFkiw.kZvfHkfØ;kdjokusdsfy,fdrusvk;rudhvko';drkiM+sxhA 2500 (1) 3 500 ml (2) 3 1000 ml (3*) 3 ml (4)NoneofthesebuesalsdksbZughaA Sol. Al + NaOH + H2O  NaAlO + 3/2 H2 2 mole M × V M × V = 2 2 1 V = 6 = 3 Lt.