Jp Xii Physical&inorganic Chemistry (12) - Prev Chaps + Electrochemistry.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 Course : VIJETA (JP)

NO. 18 & 19

Date : 15-06-2015 TEST IN FORMATI ON

Date : 28.06.2015

PART TEST (PT) - 2 (ADVANCED)

Syllabus : Coordination Compounds, Electrochemistry (Upto Nernst equation, Concentration cell), Gaseous state, Chemical Bonding, p-Block (Boron, Carbon Family. (All ChemInfos and Handouts till date) This DPP is to be discussed in the week (15-06-2015 to 20-06-2015)

DPP No. # 18 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.20

1.

(3 marks, 2 min.)

[60, 40]

For the two reactions given below : H2(g) + ½O2(g)  H2O () + x1 kJ mol–1 H2(g) + ½O2(g)  H2O (g) + x2 kJ mol–1 Compare the magnitude of x1 and x2. (x1 and x2 are the heat released in above two process.)

uhps nh xbZ nks vfHkfØ;kvksa ds fy, % H2(g) + ½O2(g)  H2O () + x1 kJ mol–1 H2(g) + ½O2(g)  H2O (g) + x2 kJ mol–1

Sol.

x1 rFkk x2 ds ifjek.k dh rqyuk dhft,A (x1 rFkk x2 mijksDr nksuksa izØeksa (A*) x1 > x2 (B) x1 < x2 (C) x1 = x2 Some of the heat is used to vaporise the H2O () (H2O () dks ok"ihd`r djus ds fy, dqN ek=kk esa Å"ek iz;qDr gksrh gSA )  x1 > x2

esa fu"dkflr Å"ek gSA)

2.

Heat of hydrogenation of ethene is x 1 and that of benzene is x2. Hence, resonance energy of benzene is :

(D) x2 = 2x1

,Fkhu dh gkbMªkstuhdj.k Å"ek x1 rFkk csUthu dh x2 gS blfy, csUthu dh vuqukn ÅtkZ gSA Sol.

Sol.

(A) x1 – x2 (B) x1 + x2 (C*) 3x1 – x2 CH2 = CH2 + H2 CH3–CH3 H = X1 So, Hydrogenation energy of benzene sholud be 3x 1 H calculated = 3x1 So, Resonance energy = [ 3x1 – x2] CH2 = CH2 + H2 CH3–CH3 H = X1 blfy, cSathu dh gkbMªkstuhdj.k dh ÅtkZ 3x1 gksuh pkfg,A

(D) x1 – 3x2

H x.kfur = 3x1

blfy,] vuquknh ÅtkZ = [ 3x1 – x2 ]

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3.

For a real gas with very large value of molar volume, which of the following equation can most suitably be applied:

vR;f/kd eksyj vk;ru okyh ,d okLrfod xSl ds fy, fuEu esa ls dkSulh lehdj.k lokZf/kd mi;qDr :i ls iz;q Dr dh tk ldrh gS % a (A) Z = 1 – V RT m Sol.

(B*) PVm = RT

(C) Z = 1 +

Pb RT

a (D) PVm – RT = V m

For very large value of molar volume (Vm) a Vm and b can be neglected, so gas behaves as Ideal

 PVm = RT

vR;f/kd eksyj vk;ru (Vm) ds fy, a Vm

rFkk b dks ux.; ekuk tk ldrk gS vr% xSl vkn'kZ O;ogkj djrh gSA

 PVm = RT 4.

Given that the critical temperature of oxygen is 154K and its critical pressure is 50 atm, which of the following statements is are true : I. In a closed container at 154K and 50 atm, the solid, liquid, and gaseous phase of oxygen are in equilibrium. II. Oxygen gas can be liquefied at room temperature. III. It can be reasoned that ammonia has a critical temperature above 154 K. (A) I is true (B) II and III are true (C*) III is true (D) I and III are true vkWDlhtu dk Økafrd rki 154K rFkk Økafrd nkc 50 atm fn;k x;k gSA fuEu esa ls dkSuls dFku lR; gSa % I. 154K rFkk 50 atm ij

,d cUn ik=k esa, vkWDlhtu dh Bksl] nzo rFkk xSl voLFkkvksa esa lkE;koLFkk gksxhA xSl dks dejs ds rki ij nzfor fd;k tk ldrk gSA III. ;g dgk tk ldrk gS fd veksfu;k dk ØkafUrd rki 154 K ls vf/kd gSA (A) I lR; gS (B) II rFkk III lR; gS (C*) III lR; gSA (D) I rFkk III lR; gSA II. vkWDlhtu

Sol.

Refer class notes.

5.

Among the following compounds the one that is polar and has central atom with sp 3 hybridisation is fuEu esa ls dkSu lk v.kq /kzqoh; gS rFkk mldk dsUnzh; ijek.kq sp3 ladfjr voLFkk esa gSA (A) H2CO3 (B) SiF4 (C) NaBF4 (D*) HClO2 ; Cl is sp3 hybridised and molecule is polar

Sol.

; Cl , sp3 ladzfjr 6.

Sol.

gS vkSj v.kq /kzqoh; gS

Molecule XF4 has non-zero value of dipole moment. Then X is : (A) Xe (B) Si (C*) S v.kq XF4 ds f}/kqzo vk?kw.kZ dk eku v'kwU; gS] rc X gS :

(D) none

(A) Xe

(B) Si

(C*) S

(D) buesa

(A)

(B)

(C)

dksbZ ugha

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7.

Property of all the alkaline earth metals that increases with their atomic number is : (A) ionisation energy (B*) solubility of their hydroxides (C) solubility of their sulphate (D) electronegativity

{kkjh; e`nk /kkrqvksa dk dkSulk xq.k] ijek.kq Øekad c<+us ds lkFk c<+rk gS % (A) vk;uu ÅtkZ (B*) buds gkbMªkWDlkbMksa dh foys;rk (C) buds lYQsVks dh foys;rk (D) fo|qr_.krk Sol.

g y-

Down a group, for M(OH)2, ionic character increases, hence solubility of their hydroxide increases. (M = 2nd group element). oxZ esa uhps tkus ij] M(OH)2 ds fy, vk;fud vfHky{k.k c<+rs gS vr% buds gkbMªkWDlkbMksa dh foys;rk c<+rh gSA (M = 2nd oxZ

8.

Sol.

ds rRo)

A substance absorbs CO 2 and violently reacts with water. The substance is : (A) CaCO3 (B*) CaO (C) H2SO4 (D) ZnO ,d inkFkZ CO2 dks vo'kksf"kr djrk gS rFkk 'kh?kzrk ls ty ds lkFk vfHkfØ;k djrk gSA inkFkZ (A) CaCO3 (B*) CaO (C) H2SO4 (D) ZnO

gS

His sin g sound

 Ca(OH)2 + Heat CaO + H2O    CaO + CO2  CaCO3

9.

Given the following reaction at equilibrium N2(g) + 3H2 (g) 2NH3 (g) Some inert gas is added at constant volume. Predict which of the following facts will be affected ? (A) More of NH3(g) is produced. (B) Less of NH3(g) is produced (C*) No affect on the degree of advancement of the reaction at equilibrium. (D) Kp of the reaction is increased.

fuEu vfHkfØ;k lkE; ij nh x;h gS % N2(g) + 3H2 (g)

2NH3 (g)

fu;r vk;ru ij blesa dqN vfØ; xSl feyk;h x;hA fuEu esa ls] mu rF;ksa dks crkb;s] tks dh çHkkfor gksax s % (A) NH3(g) vf/kd mRiUu gksrh gSA (B) NH3(g) de mRiUu gksrh gSA (C*) lkE; ij vfHkfØ;k o`f) dh ek=kk (degree of advancement of the reaction) çHkkfor ugha gksrh gSA (D) vfHkfØ;k ds Kp esa o`f) gksrh gSA Sol.

At constant volume addition of inert gas do not change concentration of any of the species involved in reaction So, equilibrium is unaffected.

10.

The osmotic pressure of equimolar solutions of BaCl2, NaCl and glucose will be in the order (A) glucose > NaCl > BaCl2 (B*) BaCl2 > NaCl > glucose (C) NaCl > BaCl2 > glucose (D) NaCl > glucose > BaCl2 BaCl2, NaCl ,oe~ Xywdksl ds leeksyj foy;ukas ds ijklj.k nkc dk Øe gSA

Sol.

Sol.

(A) Xywdksl > NaCl > BaCl2

(B*) BaCl2 > NaCl > Xywdksl

(C) NaCl > BaCl2 > Xywdksl (b)  No. of particles/ions. BaCl2 = 3, NaCl = 2 glucose = 1 So. order of  = BaCl2 > NaCl > glucose (b)  d.kksa@vk;uksa dh la[;k

(D) NaCl > Xywdksl > BaCl2

BaCl2 = 3, NaCl = 2

Xywdksl = 1 vr%  dk Øe= BaCl2 > NaCl > Xywdksl 11.

Which one of the following complexes will have four or more than four isomers ? (A) [Co(en)(NH3)2Cl2]Cl (B) [Co(PPh3)2(NH3)2Cl2]Cl (C) [Co(en)3]Cl3 (D*) [Co(en)2Cl2]Br (en = ethylenediamine) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 3

fuEu esa ls fdl ladqy ds fy, pkj vFkok vf/kd leko;oh lEHko gS \

Sol.

12.

(A) [Co(en)(NH3)2Cl2]Cl (B) [Co(PPh3)2(NH3)2Cl2]Cl (C) [Co(en)3]Cl3 (D*) [Co(en)2Cl2]Br (en = ,fFkyhu Mkb,ehu) The four isomers of [Co(en)2Cl2]Br are trans isomer, cis-isomer (d-and -isomers) and ionization isomer [Co (en) BrCl]Cl. [Co(en)2Cl2]Br ds pkj leko;oh gksrs gS] foi{k leko;oh] lei{k leko;oh (d-rFkk -leko;oh) rFkk vk;uu leko;oh [Co (en) BrCl]Cl. Which of the following will exhibit maximum ionic conductivity ? (A*) K4[Fe(CN)6] (B) [Co(NH3)6]Cl3 (C) [Cu(NH3)4]Cl2

(D) [Ni(CO)4]

fuEu esa ls dkSu vf/kdre vk;fud pkydrk çnf'kZr djrk gS \ Sol.

(A*) K4[Fe(CN)6] (B) [Co(NH3)6]Cl3 (C) [Cu(NH3)4]Cl2 (D) [Ni(CO)4] Ionic conductivity depends upon the number of ions produced in aqueous solution. K4[Fe(CN)6] produces maximum number of ions, i.e., 5. vk;fud pkydrk tyh; foy;u esa mifLFkr vk;uksa dh la[;k ij fuHkZj djrh gSA K4[Fe(CN)6] vk;uksa dh vf/kdre la[;k

vFkkZr 5 gksrh gSA 13.

Which of the following will give maximum number of isomers ?

fuEu esa ls dkSulk ladqy lcls T;knk leko;oh j[krk gS \ Sol.

14.

Sol.

Sol.

(A) [Co(NH3)4Cl2] (B) [Ni (en)(NH3)4]2+ (C) [Ni(C2O4)en2]2– + [Cr(SCN)2 (NH3)4] shows linkage, geometrical and optical isomerism. [Cr(SCN)2 (NH3)4]+ fyadst] T;kferh; rFkk izdkf'kd leko;ork n'kkZrk gSA

(D*) [Cr(SCN)2(NH3)4]+

Atomic number of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic with the spin of electron? Cr rFkk Fe ds ijek.kq Øekad Øe'k% %24 o 26 gSaA bysDVªkWu ds çpØ.k ds lkFk dkSulk vuqpqEcdh; gS \ (A) [Cr(CO)6] (B) [Fe(CO)5] (C) [Fe(CN)6]4– (D*) [Cr(NH3)6]3+ Atoms, ions or molecules having unpaired electrons are paramagnetic. In [Cr(NH 3)6]3+, Cr is present as Cr (III). Cr3+ = 1s2, 2s2 2p6, 3s2 3p6 3d3

Number of unpaired electrons = 3, so it is paramagnetic while rest of the species are diamagnetic. v;qfXer bySDVªkWu j[kus okys ijek.kq] vk;u vFkok v.kq vuqpqEcdh; gksrsa gSaA [Cr(NH3)6]3+ esa Cr (III) ds :i esa Cr mifLFkr

gksrk gSA Cr3+ = 1s2, 2s2 2p6, 3s2 3p6 3d3

v;qfXer bySDVªkWuksa dh la[;k = 3, blfy, ;g vuqpqEcdh; gS tcfd ckdh Lih'kht ¼iztkfr½ izfrpqEcdh; gksrh gSA 15.

According to IUPAC nomenclature, sodium nitroprusside is named as : (A) Sodium nitroferricyanide (B) Sodium nitroferrocyanide (C*) Sodium pentacyanidonitrosoniumferratae (II) (D) Sodium pentacyano nitrosylferrate (III) IUPAC ukedj.k ds vuqlkj] lksfM;e ukbVªksizwlkbM dk uke gS : (A) lksfM;e

Sol.

ukbVªksQSfjlk;ukbM

(B) lksfM;e

ukbVªksQSjkslk;ukbM

(C*) lksfM;e isUVklk;ukbMksukbVªkslksfu;eQSjsV(II) (D) lksfM;e isUVklk;uks ukbVªksflyQSjsV (III) Sodium nitroprusside, Na2[Fe(CN)5NO], is named as sodium pentacyano nitrosyl ferrate(III). In this compound, NO is present as NO + and Fe has +2 charge. lksfM;e ukbVªksizwlkbM Na2[Fe(CN)5NO] dk uke gS % lksfM;eisUVklk;uksukbVªksfly QSjsV (III), bl ;kSfxd esa NO, NO+ ds

:i esa mifLFkr gS rFkk Fe ij +2 vkos'k gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

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16.

Which of the following coordination compounds would exhibit optical isomerism ? (A) Pentaamminenitrocobalt (III) iodide (B) Diamminedichloroplatinum (II) (C) Trans-dicyanobis (ethylenediamine chromium (III) chloride) (D*) Tris-(ethylenediamine) cobalt (III) bromide

fuEu esa ls dkSulk ;kSfxd çdk'kh; leko;rk çnf'kZr djsxk \ (A) isUVk,EehuukbVªksdksckWYV (III) vk;ksMkbM (B) Mkb,EehuMkbDyksjksIysfVue (II) (C) foi{k&Mkblk;uksfcl~ ¼,fFkyhu Mkb,ehu Øksfe;e (III) DyksjkbM) (D*) fVªl&¼,fFkyhuMkb,ehu½ dksckYV (III) czksekbM Sol.

Tris-(ethylenediamine) cobalt (III) bromide [Co(en)3]Br3 exhibits optical isomerism : fVªl - (,fFkyhuMkb,ehu) dksckYV (III) czksekbM [Co(en)3]Br3 izdkf'kd leko;rk j[krk gSA

17.

Which of the following complexes exhibits the highest paramagnetic behaviour ? where, gly = glycine, en = ethylenediamine and bpy = bipyridyl moities) (At no Ti = 22, V = 23, Fe = 26, Co = 27) (A) [V(gly)2(OH)2(NH3)2]+ (B) [Fe(en)(bpy)(NH3)2]2+ (C*) [Co(OX)2(OH)2]– (D) [Ti(NH3)6]3+

fuEu esa ls dkSulk ladqy mPpre vuqpqEcdh; O;ogkj çnf'kZr djrk gS \

Sol.

where, gly = glycine, en = ethylenediamine and bpy = bipyridyl moities) (At no Ti = 22, V = 23, Fe = 26, Co = 27) (A) [V(gly)2(OH)2(NH3)2]+ (B) [Fe(en)(bpy)(NH3)2]2+ (C*) [Co(OX)2(OH)2]– (D) [Ti(NH3)6]3+ The electron configuration of V (23) = [Ar], 4s2, 3d3 Let in [V(gly)2(OH)2(NH3)2]+ oxidation state of V is x. x + (– 1) × 2 (– 1)2 + (0 × 2) = + 1 V5+ = [Ar], 4s0, 3d0 (no unpaired electrons) The electronic configuration of Fe(26) = [Ar] 4s2, 3d6 Let the oxidation state of Fe in [Fe(en)(ppy)(NH3)2]2+ is x. [x + (0) + (0) + (0) × 2] = + 2 x=+2 Fe2+ = [Ar] 4s0, 3d6 ( no unpaired electron) but, bpy, en and NH 3 all are strong field ligands, so pairing occurs and thus, Fe2+ contains no unpaired electron. The electronic configuration of Co(27) = [Ar] 4s2, 3d7 Oxidation state of Co in [Co (OX)2 (OH)2]– x + (– 2) × 2 + (– 1) × 2 = – 1 x=+5 Co5+ = [Ar], 3d4 [4 unpaired electrons] OX and OH are weak field ligands. The electronic configuration of Ti (22) = [Ar] 4s2, 3d2 Oxidation state of Ti in [Ti(NH3)6]3+ is 3. Ti3+ = [Ar], 3d1 (one unpaired electron) Hence, complex [Co(OX)2(OH) 2]– has maximum number of unpaired electrons, thus show maximum paramagnetic.

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18.

Which of the following complex ions is not expected to absorb visible light ?

fuEu esa ls dkSulk ladqy vk;u n`'; {ks=k dks vo'kksf"kr ugha djrk gS\ Sol.

gy % 19.

(A*) [Ni(CN)4]2– (B) [Cr(NH3)6]3+ (C) [Fe(H2O)6]2+ There is no unpaired electrons in [Ni(CN)4]2– due to strong field ligand. [Ni(CN)4]2– esa izcy {ks=k fyxs.M ds dkj.k dksbZ v;qfXer bySDVªkWu ugha gSA

(D) [Ni(H2O)6]2+

In which of the following pairs, both the complexes show optical isomerism ? fuEu esa ls dkSulk ladqy ;qXe izdkf'kd leko;ork n'kkZrk gS ? (A) cis-[Cr(C2O4)2Cl2]3– , cis-[Co(NH3)4Cl2] (B*) [Co(en)3] Cl3 ,cis-[Co(en)2Cl2]Cl (C) [PtCl(dien)Cl] , [NiCl2Br2]2– (D) [Co(NO3)3(NH3)3], cis-[Pt(en)2Cl2] 3– (A) lei{k-[Cr(C2O4)2Cl2] , lei{k-[Co(NH3)4Cl2] (B*) [Co(en)3]Cl3 ,lei{k-[Co(en)2Cl2]Cl (C) [PtCl(dien)Cl] , [NiCl2Br2]2–

(D) [Co(NO3)3(NH3)3], lei{k-[Pt(en)2Cl2]

Sol.

20.

The correct order for the wavelength of absorption in the visible region is :

n`'; {ks=k esa vo'kksf"kr rjax}S/;Z ds fy, lgh Øe fuEu gS % Sol.

g y-

(A*) [Ni(NO2)6]4– < [Ni(NH3)6]2+ < [Ni(H2O)6]2+ (B) [Ni(NO2)6]4– < [Ni(H2O)6]2+ < [Ni(NH 3)6]2+ (C) [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4+ (D) [Ni(NH3)6]2+ < [Ni(H2O)6]2+ < [Ni(NO2)6]4– As all are octahedral complexes of the same metal(Ni), absorption will depend only on the nature of the ligand. From spectrochemical series, the CFSE of the given ligands are in the order : H 2O < NH3 < NO2– Hence, excitation energies absorbed to show particular colour will be in the order : [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4– The wavelength absorbed will be in the opposite order. pw¡fd ladqyksa esa dsUnzh; /kkrq ijek.kq (Ni) leku gS rFkk lHkh v"VQydh; ladqy gS] vo'kks"k.k dh ek=kk dsoy fyxs.M dh izd`fr

ij gh fuHkZj gksrh gSA LisDVªksjklk;fud Js.kh esa] fn;s x;s fyxs.M dh CFSE dk Øe fuEu gS % : H2O < NH3 < NO2– vr% ,d fuf'pr jax iznf'kZr djus ds fy;s vo'kksf"kr mÙksftr ÅtkZ dk Øe fuEu gksxk : [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4–

vo'kksf"kr rjaxnS/;Z blds foifjr Øe esa gksxhA

DPP No. # 19 (JEE-ADVANCED) Total Marks : 62

Max. Time : 42 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.14 Multiple choice objective ('–1' negative marking) Q.15 to Q.16 Subjective Questions ('–1' negative marking) Q.17 to Q.18 ChemINFO : 1 Questions ('–1' negative marking) Q.19

1.

(3 (4 (4 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 2 min.)

[42, [08, [08, [04,

28] 06] 06] 02]

Predict which of the following facts for the equilibrium reaction 2NH3(g) N2(g) + 3H2 (g) holds good ? (A) Kp of the reaction is changed with increase in pressure of the system. (B*) Kp of the reaction remains unaffected with increase in pressure of the system. (C) More of NH3(g) is decreased with increase in pressure. (D) Less of H2(g) is formed as compared to N2(g).

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PAGE NO.- 6

fuEu esa ls mu rF;ksa dks crkb;s] tks dh fuEu lkE; dh vfHkfØ;k ds fy, lgh gSa % 2NH3(g) (A) ra=k ds

N2(g) + 3H2 (g)

nkc esa o`f) ds lkFk vfHkfØ;k dk Kp ifjofrZr gksrk gSA (B*) ra=k ds nkc esa o`f) ds lkFk] vfHkfØ;k dk Kp vçHkkfor jgrk gSA (C) nkc esa o`f) ds lkFk NH3(g) dh ek=kk esa deh vkrh gSA (D) N2(g) dh rqyuk esa H2(g) dk fuekZ.k de gksrk gSA Sol.

KP depends on temperature. At constant temperature, Kp remains constant irrespective of the pressure.

2.

A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The apparent percentage dissociation of Na2SO4 is leku rki ij Na2SO4 dk 0.004 M foy;u Xywdksl ds 0.010 M foy;u ds lkFk leijkljh gSA Na2SO4 dk fdruk

izfr'kr Hkkx fo;ksftr gksxk \ (A) 25% Sol.

(B) 50% +

Na2SO4

2Na + SO4

a (1 – ) i=

2a

(C*) 75%

(D) 85%

2–

a

a(1  2α) = 1 + 2 a

Where  is degree of dissociation of Na2SO4 Solution of Na2SO4 & glucose are isotonic So



π Na2 SO 4 = π glu cos e i × 0.004 × R × T = 0.010 × R × T



(1+ 2) 1 + 2 

 3.

% dissociation = 75 %

The concentration of free HN3 in a 0.01 M solution of KN3 if Ka = 1.9 × 10–5 is KN3 ds 0.01 M foy;u (A*) 2.3 mol/ML

4.

Sol.

10 4 = 2.5 = 0.75

=

esa eqDr HN3 dh lkUnzrk D;k gS ;fn Ka = 1.9 × 10–5 gSa % (B) 3.3 mol/ML

(C) 4.3 mol/ML

(D) 5.3 mol/ML

The number of possible isomers of an octahedral complex [Co(C2O4)2(NH3)2]– is : v"VQydh; ladqy [Co(C2O4)2(NH3)2]– ds laHko leko;oh dh la[;k gS % (A) 1 (B) 2 (C*) 3 (D) 4 [Co(C2O4)2(NH3)] has 3 isomers (A) Cis-isomer (B) Trans-isomer It forms two optically isomeric forms [Co(C2O4)2(NH3)] 3 leko;oh j[krs gSaA (A) lei{k&leko;oh

(B) foi{k&leko;oh ;g nks izdk'kh; :i ls leko;oh :i dks cukrh gSA

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PAGE NO.- 7

5.

An excess of AgNO3 is added to 100 mL of a 0.01M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be : 0.01 eksyj MkbZDyksjksVsVªk,DokØksfe;e (III) DyksjkbM ds 100 mL foy;u esa AgNO3 dh vf/kdre ek=kk feykbZ xbZ gSA AgCl

ds vo{ksfir gksus okys eksyksa dh la[;k gksxh % (A) 0.002

(B) 0.003

(C) 0.01

(D*) 0.001

Sol.

10  10 3 1000 So mole of AgCl = 0.001 mole = 0.01 ×

Sol.

eksy = 0.01 ×

10  10 3 1000

vr% AgCl ds eksy = 0.001 6.

E° for F2 + 2e–

2F¯ is 2.8 V, E° for ½ F2 + e– = F¯ is -

Sol.

F2 + 2e– 2F¯ ds fy, E° 2.8 V gS( ½ F2 + e– = F¯ ds fy, E° fuEu (A*) 2.8 V (B) 1.4 V (C) –2.8 V Eº is intensive property.

7.

0 0 0 +2 stable to disproportionation in E Fe E Fe E Fe 3 3 / Fe  2 = +0.77 V ; / Fe = – 0.036 V. What is / Fe  2 and is Fe

gksxk& (D) –1.4 V

aqueous solution under standard conditions

Eº Fe / Fe2  dk eku D;k gS \ D;k Fe2+ ekud ifjfLFkfr;ksa esa tyh; foy;u esa fo"kekuqikru ds fy, LFkk;h gSa\ [

0 0 E Fe ;fn E Fe 3 3 / Fe  2 = +0.77 V ; / Fe = – 0.036 V].

(A*) +0.44 V, yes Sol.

8.

(B) – 0.44 V, No

(C) + 0.44 V, No

(D) – 0.44 V, yes

(A*) +0.44 V, gk¡ (B) – 0.44 V, ugha (C) + 0.44 V, ugha (D) – 0.44 V, gk¡ 3+ – 2+ Fe + e Fe (A) Eº1 = 0.77V G1º = – 1 × F × 0.77 Fe3+ + 3e– Fe (B) Eº2 = –0.036V G2º = 3 × F × 0.036 Fe Fe2+ + 2e– (C) E3º = G3º = –2 × F × E3º (A) – (B) = 3 G1º – G2º = G3º  –F × 0.77 – 3F – 0.036 = –2 × F × E3º  (0.77 × 0.108) = 2 × E3º  E3º ~ 0.44V Ans. Disproportionation reaction. 3Fe2+ Fe + 2Fe3+ (D) Fe2+ + 2e– Fe Eº = –0.44V Gº = –2 × F × (–0.44) (5) Fe2+ Fe3+ +e– Eº = –0.77V Gº = –1 × F × (–0.77) (6) (5) + 2 × (6) = (D) Gº4 = –2 × F × (–0.44) + 2 (–1 × F × (–0.77)) = +ve  G4º > 0,  Reaction is not spontaneous. Hence, Fe2+ is stable to disproportionation. º Given that EFe 2 

Fe

º = – 0.44 V ; EFe3 

(A) Fe3+ increases (C) Fe2+ / Fe3+ remains unchanged

Fe 2 

= 0.77 V if Fe2+, Fe3+ and Fe solid are kept together then (B*) Fe3+ decreases (D) Fe2+ decreases

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PAGE NO.- 8

º º EFe 2 3 fn;k x;k gS EFe Fe = – 0.44 V ;

= 0.77 V gS

Fe 2 

;fn Fe2+, Fe3+ rFkk Fe Bksl dks ,d lkFk j[kk tkrk gS]

rks (A) Fe3+ c<+rk gSA (C) Fe2+ / Fe3+ vifjofrZr Sol.

(B*) Fe3+ ?kVrk gSA (D) Fe2+ ?kVrk gSA

jgrk gSA

From the given data, we see that Fe3+ reduces to Fe2+ as E o

(Fe3  / Fe 2 )

Eo

(Fe2 / Fe )

is positive and Fe oxidises to Fe2+ are

is negative.

So, Fe3+ decreases, Fe2+ increases. 9.

The spontaneous redox reaction/s among the following is/are

fuEu es ls dkSulh Lor% vkWDlhdj.k vip;u ¼jsMkWDl½ vfHkfØ;k,a gS & (a) 2Fe3+ + Fe  3Fe++ (b) Hg2++  Hg++ + Hg (c) 3 AgCl + NO + 2H2O  3 Ag + 3 Cl– + NO3– + 4H+ Given that  EFe 

10.

= 0.77 V

 EFe 

Fe

= – 0.44 V

 EHg 

Hg

= 0.85 V

 EHg 

Hg2 

E AgCl

Ag

= 0.22 V

 ENO 

NO

2

Sol.

Fe  

fn;k gS &

3

= 0.92 V

= 0.96 V

(A*) a (B) a, b, c (C) a, b (D) a, c Only for this reaction E0 will come out to be positive, calculate using relation G° = G10 + G20 and G0 = – nFE°cell 0 dsoy bl vfHkfØ;k ds fy, E /kukRed vkrk gS fuEu laca/k dk mi;ksx djds x.kuk djrs gSaA G° = G10 + G20 ,oa G0 = – nFE°cell Ag(s) + H+ + Cl¯ is –21.52 kJ. G° of 2AgCl(s) +

G° of the cell reaction AgCl(s) + ½ H2(g) H2(g) = 2Ag(s) + 2H+ + 2Cl¯ is AgCl(s) + ½ H2(g) Ag(s) lsy vfHkfØ;k ds fy, G° gSA (A) –21.52 kJ (B) –10.76 kJ

ds fy, G° –21.52 kJ gS 2AgCl(s) + H2(g) = 2Ag(s) + 2H+ + 2Cl¯ (C*) –43.04 kJ

(D) 43.04 kJ

1 H (g) Ag (s) + H+ + Cl–. 2 2 Gº = – nF Eºcell = – 21.52 KJ = – 1 × F Eºcell . 2AgCl (s) + H2 (g) 2Ag (s) + 2H+ + 2Cl–. Gº = – 2 × F × Eºcell = – 21.52 × 2 = – 43.04 kJ.

Sol.

AgCl (s) +

11.

The de Broglie wavelength associated with a ball of mass 1 kg having kinetic energy 0.5 J is : 1 kg nzO;eku dh ,d xsan] ftldh xfrt ÅtkZ 0.5 J gS] ds lkFk la;ksftr ns czkWXyh rjax&nS/;Z fuEu gS % (A*) 6.626  10–34 m (B) 13.20  10–34m (C) 10.38  10–21 m (D) 6.626  10–34 Å

1 mv 2 = 0.5 J 2

Sol.

1 × 1 kg × v2 = 0.5 J 2

or

v2 = 1 or v = 1 ms–1

–34



2 –1

6.626  10 kgm s h = mv 1 kg 1 m s –1

= 6.626 × 10–34 m.

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PAGE NO.- 9

12.

Plot at Boyle’s temperature for a real gas will be :

ckW;y rki ij ,d okLrfod xSl dk vkjs[k gksxk %

(A)

(B)

(C*)

(D)

Sol.

At Boyle's temperature, for low pressure regions, Z = 1. However, for high pressure regions, Z > 1. ckW;y rki ij, fuEu nkc okys {ks=k esa, Z = 1. tcfd mPp nkc okys {ks=k esa , Z > 1.

13.

The oxide that gives hydrogen peroxide on treatment with a dilute acids is : (A) PbO2 (B*) Na2O2 (C) MnO2 (D) TiO2

fuEu esa ls dkSulk vkWDlkbM ruq vEyksa ds lkFk vfHkfØ;k dj gkbMªkstu ijkWDlkbM nsrk gS % Ans. Sol. 14.

(A) PbO2 (B*) Na2O2 (C) MnO2 (B) Na2O2 + H2SO4 (20% ice cold)  Na2SO4 + H2O2

(D) TiO2

S1 : Hybridisation state of central atom of cation of solid PBr 5 is sp2. S2 : Shape of both I3– and N3– are linear S3 : He2 does not exist but He2+ does exist. S4 : In PCl5 , the axial P–Cl bonds are stronger than the equitorial P–Cl bonds. S1 : Bksl PBr5 ds /kuk;u ds dsUnzh; ijek.kq dh ladj.k voLFkk sp2 gSA S2 : I3– rFkk N3– nksuksa

dk vkdkj js[kh; gSA S3 : He2 ugha ik;k tkrk gS] ysfdu He2+ ik;k tkrk gSA S4 : PCl5 esa] v{kh; (axial) P–Cl ca/k] fo"kqorh; (equitorial) P–Cl ca/kksa dh rqyuk esa izcy gksrk gSA (A) TFTT

15.*

(B*) FTTF

(C) TTFT

(D) TFTF

The density of a hypothetical gas X (mol. wt = 8 amu) has been observed to be less than

1 g/L at STP.. 2. 8

Then, which of the following statements is/are correct for gas X under above conditions : (A*) Repulsive tendencies are dominant among molecules of gas X. (B) The value of TB (Boyle's temperature) for gas X is greater than 273.15 K. (C) The value of Vm (molar volume) for gas X is less than 22.4 L (D*) Gas X is less compressible with respect to an ideal gas. 1

,d dkYifud xSl X (v.kqHkkj = 8 amu) dk ?kuRo STP ij 2.8 g/L ls de izf{kr fd;k x;k gS] rks mijksDr ifjfLFkfr;ksa esa xSl X ds fy, fuEu esa ls dkSulk@dkSuls dFku lR; gSa % (A*) xSl X ds v.kqvksa esa izfrd"kZ.k izo`fÙk izHkkoh gSA (B) xSl X ds fy, TB (okW;y rki) dk eku 273.15 K ls vf/kd gSA (C) xSl X ds fy, Vm (eksyj vk;ru) dk eku 22.4 L ls de gSA (D*) vkn'kZ xSl dh rqyuk esa xSl X dh laihM~;rk de gSA Sol.

dgas <



1 g/L 2. 8

8 1 < Vm 2. 8

 Vm > 22.4 L Thus, repulsive tendencies are dominant among gas molecules and gas is less compressible with respect to an ideal gas. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 10

dxSl <

g y-

1 g/L 2 .8



8 1 Vm < 2.8



Vm > 22.4 L

vr% xSl v.kqvksa ds e/; izfrd"kZ.k cy izHkkoh gS rFkk xSl] vkn'kZ xSl dh rqyuk esa de lEihM+; gSA 16.*

Which is true about NH2–, NH3 ,NH4+ ? (A*) Hybridization of N is same. (B) No. of lone pair of electron on N are same. (C*) Molecular geometry (i.e. shape) is different. (D) Bond angle is same. NH2–, NH3 ,NH4+ ds fy, D;k lR; gS \ (A*) N dk

ladj.k leku gSA (C*) vk.kfod T;kfefr (vFkkZr vkd`fr) fHkUu gSA

(B) N ij

,dkdh bySDVªkWu ;qXe dh la[;k leku gSA (D) cU/k dks.k leku gSA

Sol.

(A)

gy-

(A)

17.

Calculate the heat of formation of HCl at 348K from the following data . 0.5 H2 (g) + 0.5 Cl2 (g)  HCl (g) ; Hº298 =  22060 Cal The mean heat capacities over this temperature range are , H2 (g) ; CP = 6.82 Cal mol 1 degree1 ; Cl2 (g) ; CP = 7.71 Cal mol 1 degree 1 HCl (g) ; CP = 6.81 Cal mol 1 degree 1 .

fuEufyf[kr vk¡dM+ks ds }kjk 348K rki ij HCl ds laHkou dh Å"ek Kkr dhft,A 0.5 H2 (g) + 0.5 Cl2 (g)  HCl (g) ; Hº298 =  22060 Cal

bl rki ijkl ij ek/; Å"ek /kkfjrk,sa fuEu gSa % Ans. Sol.

18.

Ans.

Sol.

H2 (g) ; CP = 6.82 Cal mol 1 degree1 ; Cl2 (g) ; CP = 7.71 Cal mol 1 degree 1 HCl (g) ; CP = 6.81 Cal mol 1 degree 1 .  22083 Cal (Cp) reaction = (Cp) products – (Cp) reactants = 6.81 – [0.5 × 6.82 – 0.5 × 7.71] Hº (at 348 K) = Hº (at 298 T) + (CP) reaction T The difference in the number of  and  bonds in trimer of SO3 i.e. S3O9 is : (consider no co-ordinate bond to be present) SO3 ds f=kyd (S3O9) esa  rFkk  ca/kksa dh la[;k esa vUrj gS : (blesa milgla;ksth ca/k dks mifLFkr ugha ekurs gq, ) 6

 = 12 and  = 6  difference (vUrj) = 12 – 6 = 6

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PAGE NO.- 11

ChemINFO-1.14

COORDINATION COMPOUNDS Applications of Coordination Compounds

Daily Self-Study Dosage for mastering Chemistry

Coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of biological systems, medicine, qualitative analysis, metallurgy, industrial catalysts, electroplating and textile dyeing. These are described below: 1. Biological Systems a) The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of Mg. b) Haemoglobin, the red pigment of blood which acts as oxygen carrier is a coordination compound of Fe. c) Vitamin B12 (cyanocobalamine), the anti–pernicious anaemia factor, is a coordination compound of Co. d) Enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems) contain coordinated metal ions. 2. Softening of Hard Water Hardness of water is estimated by simple titration with Na2EDTA. The Ca2+ and Mg2+ ions in hard water form stable complexes with disodium salt of EDTA. The selective estimation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes. 3. Extractive Metallurgy a) Cyanide process for the extraction of Gold and Silver b) van-Arkel process for extraction of ultra-pure Ti and Zr c) Mond (carbonyl) process for extraction of Ni 4. Industrial Catalysts a) Wilkinson catalyst: Rhodium complex, [RhCl(Ph3P)3], is a homogenous catalyst used for the hydrogenation of alkenes to alkanes. 5. Photographic plates In black and white photography, the developed film is fixed by washing with hypo i.e. Na 2(S2O3) solution which dissolves the undecomposed AgBr to form a complex ion [Ag(S 2O3)2]3–. 2Na2S2O3 + AgBr (s)  Na3[Ag(S2O3)2] + NaBr 6. Medicine a) Chelate Therapy: EDTA is used in the treatment of lead poisoning. b) Chelate Therapy: excess of copper and iron are removed by the chelating ligands D–penicillamine and desferrioxime B via the formation of coordination compounds. c) Some coordination compounds of platinum effectively inhibit the growth of cancer tumours. e.g. cis–platin i.e. cis-[Pt(NH3)2Cl2] is used in chemotherapy for cancer treatment. Usage of chemotherapy is controversial as it Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 19.

Sol:

Match the following: Column I (i) EDTA (ii) Wilkinson's Catalyst of Rh (iii) Complex of Iron (iv) D–penicillamine (v) Complex of Co (vi) cis-platin (vii) Sodium bis(thiosulphato)argentate(I) (viii) Coordination compound of Mg (i)- b, f, h

(ii)- e

(iii)- a

(iv)- f

(v)- c

Column II (a) haemoglobin (b) Treatment of Lead Poisoning (c) vitamin B12 (d) cancer treatment (e) Hydrogenation of Alkenes (f) Chelate Therapy (g) Chlorophyll (h) Softening of hard water (i) photography (vi)- d (vii)- i (viii)- g

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PAGE NO.- 12

ChemINFO-1.14

COORDINATION COMPOUNDS milgla;kstu ;kSfxdksa ds vuqiz;ksx

Daily Self-Study Dosage for mastering Chemistry

milgla;kstu ;kSfxd cgqr egRo ds gSaA ;s ;kSfxd [kfutksa] isM&ikS/kksa o tho txr esa O;kid :i ls ik;s tkrs gSa rFkk tSfod iz.kkyh] vkS"kf/k] xq.kkRed fo'ys"k.k] /kkrqdeZ] vkS|ksfxd mRizsjd] oS|qr ysiu] VsDlVkb jatd (textile dyeing) ds {ks=k esa budh egRoiw.kZ Hkwfedk,a gSaA bldk o.kZu uhps fd;k x;k gS & 1. tSfod iz.kkyh a) izdk'k la'ys"k.k ds fy, mRrjnk;h o.kZd] DyksjksfQy Mg dk ,d milgla;kstu ;kSfxd gSA b) jDr dk yky o.kZd gheksXyksfcu] tks fd vkWDlhtu dk okgd gS] Fe dk ,d milgla;kstu ;kSfxd gSA c) foVkfeu B12 (lk;ukdkscky,ehu),,d izfriz.kkyh vjDrkdkjd (anti–pernicious anaemia factor) Co dk ,d milgla;kstu ;kSfxd gSA d) ,Utkbe tSls dkcksDZ lhisfIVMst (carboxypeptidase A) rFkk dkcksZfud ,ugkbMªst (carbonic anhydrase) milgla;ksftr /kkrq vk;u j[krs gSaA 2. dBksj ty dk e`nqdj.k ty dh dBksjrk dk vkdyu Na2EDTA ds lkFk lkekU; vuqekiu }kjk fd;k tkrk gSA dBksj ty esa Ca2+ rFkk Mg2+ vk;u EDTA ds MkbZlksfM;e yo.k ds lkFk LFkk;h ladqy cukrs gSaA bu vk;uksa dk p;ukRed vkdyu fd;k tk ldrk gS D;ks afd dSfYl;e rFkk eSXuhf'k;e ds ladqyksa ds LFkkf;Ro fLFkjkad eas vUrj gksrk gSA 3. fu"df"kZr /kkrqdeZ a) xksYM rFkk flYoj ds fu"d"kZ.k ds fy, lk;ukbM izØe b) vfr'kq) (ultra-pure) Ti rFkk Zr ds fu"d"kZ.k ds fy, okWu vkjdSy izØe c) Ni ds fu"d"kZ.k ds fy, ekUWM ¼dkcksZfuy½ izØe 4. vkS|ksfxd mRizsjd a) foyfdulu mRizsjd % jksfM;e ladqy [RhCl(Ph3P)3] ,d lekaxh mRiszjd gS tks ,Ydhuksa ls ,Ydsuksa esa gkbMªkstuu ds fy, iz;qDr gksrk gSA 5. QksVksxzkfQd IysV 'osr&';ke QksVksxzkQh esa] fodflr dh gqbZ fQYe dk LFkk;hdj.k gkbiksfoy;u Na2(S2O3) esa /kksdj fd;k tkrk gS tks AgBr ls ladqy vk;u [Ag(S2O3)2]3–cukdj ty esa ?kksy ysrk gSA 2Na2S2O3 + AgBr (s)  Na3[Ag(S2O3)2] + NaBr 6. vkS"kf/k a) fdysV fpfdRlk % EDTA dks ysM dh fo"kkDrk ds mipkj esa iz;qDr fd;k tkrk gSA b) fdysV fpfdRlk % dkWij rFkk vk;ju dh vf/kdrk dks D–iSfufly,ehu rFkk MslQsjhvkWfDle B fyxs.Mksa ds lkFk milgla;kstu ;kSfxd cukdj nwj fd;k tk ldrk gSA c) IySfVue ds dqN milgla;kstu ;kSfxd dSUlj V;wej o`f) dks izHkkoh :i ls jksdrs gSaA mnk- lei{k & IysfVu vFkkZr cis[Pt(NH3)2Cl2] dk mi;ksx dSalj ds mipkj ds fy, dheksFkSjsih esa fd;k tkrk gSA dheksFkSjsih dk mi;ksx fooknkLin gS D;ksafd ;g laØfer dksf'kdkvksa ds Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 19.

fuEu dk feyku dhft,& dkWye - I

dkWye - II (a) gheksXyksfcu (b) ysM fo"kkDrk dk mipkj (c) foVkfeu B12 (d) dSalj mipkj (e) ,Ydhuksa dk gkbMªkstuu (f) dhVsy FkSjsih (g) DyksjksfQy (h) dBksj ty dk e`nqdj.k (i) QksVksFkSji s h

(i) EDTA (ii) Rh ds foyfdulu dk mRizsjd (iii) vk;u dk ladqy (iv) D–isfufly,ehu (v) Co dk ladqy (vi) lei{k & IysfVu (vii) lksfM;e fol (Fkk;kslYQsVks) vtsZUVsV (I) (viii) Mg dk milgla;kstu ;kSfxd Sol:

(i)- b, f, h

(ii)- e

(iii)- a

(iv)- f

(v)- c

(vi)- d

(vii)- i

(viii)- g

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PAGE NO.- 13