Jp Xii Physical&inorganic Chemistry (14) - Prev Chaps + Inorg. Chem.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 22 & 23

Course : VIJETA (JP)

This DPP is to be discussed in the week (29.06.2015 to 04.07.2015)

DPP No. # 22 (JEE-MAIN) Total Marks : 61

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.19 Multiple choice objective ('–1' negative marking) Q.20

1.

(3 marks, 2 min.) (4 marks, 2 min.)

[57,38] [04,02]

Values of heats of formation for SiO2 and MgO are – 48.4 and – 34.7 kJ respectively. The heat of the reaction 2 Mg + SiO2  2MgO + Si is SiO2 rFkk MgO dh

lEHkou Å"ekvksa ds eku Øe'k% – 48.4 rFkk – 34.7 kJ gSA 2 Mg + SiO2  2MgO + Si vfHkfØ;k

dh Å"ek gksxh % (A) 21.16 kJ 2.

(B*) – 21.00 kJ

(C) – 13.62 kJ

(D) 13.6 kJ

In the reaction CO 2(g) + H2(g) CO(g) + H2O(g); H = 2.8 kJ H represents (A*) heat of reaction (B) heat of combustion (C) heat of formation (D) heat of solution CO2(g) + H2(g) CO(g) + H2O(g) vfHkfØ;k (A*) vfHkfØ;k

3.

esa ; H = 2.8 kJ, H iznf'kZr djrk gSA (B) ngu dh Å"ek (C) lEHkou Å"ek (D) foy;u dh Å"ek

dh Å"ek

Heat of neutralisation of a strong acid by a strong base is equal to H of -

,d izcy {kkj }kjk ,d izcy vEy dh mnklhuhdj.k dh Å"ek dk eku fuEu esa ls fdlds H ds cjkcj gS % (A*) H+ + OH– = H2O (C) 2H2 + O2 = 2H2O 4.

(B) H2O + H+ = H3O+ (D) CH3COOH + NaOH = CH3COONa + H2O

Which of the following relations hold good (where symbols have their usual meaning) :

 TC2   (A)  a  P  C  

 TC  (B)  b  P  C  

(C*) A and B both

(D) None of these

fuEu esa ls dkSulk lEcU/k lgh gSA ¼tgk¡ fn;s x;s izrhd budk lkekU; vFkZ j[krs gSa ½ 2    a  TC  (A)  PC  

27R 2 64

 TC   (B)  b  PC  

 TC2  R    PC  , b = 8  

 TC  P  C

(C*) A rFkk B nksuksa

(D) buesa

ls dksbZ ugha

   

Sol.

a=

5.

The correct relationship between critical temperature (T C) and Boyle's temperature (T B) for CO2 gas is : CO2 xSl ds fy, ØkfUrd rki (TC) o ckW;y rki (TB) ds e/; lgh lEcU/k gSA (A*) TB > TC (B) TB < TC (C) TB = TC (D) Nothing can be predicted.

vuqeku ugh yxk ldrs gSaA Sol.

TB =

a , bR

TC =

8  a   TB > TC 27  bR  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.- 1

6.

Sol.

Arrange the following in increasing order of magnitude of lattice enthalpy : Al2O3, Na2O, MgO fuEufyf[kr ;kSfxdksa Al2O3, Na2O, MgO dks tkyd ,UFkSYih ds ifjek.k ds c<+rs gq, Øe esa O;ofLFkr dhft, (A) Al2O3 < Na2O < MgO (B) MgO < Al2O3 < Na2O (C*) Na2O < MgO < Al2O3 (D) Na2O < Al2O3 < MgO As charge on cation increases, magnitude of lattice enthalpy increases.

gy

tSls&tSls /kuk;u ij vkos'k c<+rk gS] oSls&oSls tkyd ,UFkSYih dk ifjek.k Hkh c<+rk gSA

7.

K[IBrCl]  X + Y X & Y are respectively : (Answer according to lattice energy) (A) KI and BrCl (B) KBr and ICl (C*) KCl and IBr

%



(D) None of these



Sol.

g y-

K[IBrCl]  X + Y X rFkk Y Øe'k% gS (mÙkj tkyd ÅtkZ ds vk/kkj ij nhft, ) (A) KI rFkk BrCl (B) KBr rFkk ICl (C*) KCl rFkk IBr (D) buesa ls dksbZ ugha For the same cation, lesser the anionic size, greater will be the lattice energy. So, the product formed will be KCl, corresponding to maximum lattice energy. leku /kuk;u gksus ij vis{kkd`r NksVs _.kk;u ds fy, tkyd ÅtkZ dk eku vf/kd gksxkA blfy,] fufeZr mRikn KCl gksxk]

bldh tkyd ÅtkZ vf/kdre gksxhA 8.

Which of the following statements are correct ? (1) The number of sigma bonds in CH2 = C = C = CH2 is 7. (2) All the hydrogen atoms in CH2 = C = C = CH2 lie in the same plane. (A) Only (1) (B) Only (2) (C*) Both (1) and (2) fuEu dFkuksa esa ls dkSulk lgh gS ? (1) CH2 = C = C = CH2 esa

(D) Neither (1) nor (2)

flXek ca/kksa dh la[;k 7 gSA lHkh gkbMªkstu ijek.kq ,d gh ry esa gSA (B) dsoy (2) (C*) (1) rFkk (2) nksuksa

(2) CH2 = C = C = CH2 esa Sol.

g y-

(A) dsoy (1) (D) uk rks (1), uk gh (2) (1) Number of sigma bonds is 7. (2) Odd number of consecutive double bonds. So, all the hydrogen atoms lie in same plane. (1) flXek ca/kks dh la[;k 7 gSA (2) la;qXeh

9.

f}ca/kks dh fo"ke la[;kA blfy, lHkh gkbMªkstu ijek.kq ,d gh ry esa mifLFkr gSaA

The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. 1 and 2 will be : nks fofdj.kksa dh ÅtkZ,¡ E1 rFkk E2 Øe'k% 25 eV rFkk 50 eV gSa] muds rjaxnS/;Z] vFkkZr~ 1 rFkk 2 ds chp lEcU/k gksxk% (A) 1 = 2

Sol.

E1 = 25 eV,

hc E1 =  , 1 10.

(B*) 1 = 22

(C) 1 = 42

(D) 1 =

2 25 =  50 1

1 = 22

1  2 2

E2 = 50 eV

hc E2 =  2



The orbital angular momentum of a p-electron is given as :

,d p-bysDVªkWu dk vkWfcZVy dks.kh; laosx bl izdkj fn;k tkrk gS % h (A*) Sol.

(B)

2

Orbital angular momentum =

d{kd dks.kh; losax = = 1

h 2

3

h 2

h 2

3h 2

(C)

(D)

6.

h 2

(  1)

(   1)

So =

h 2

2

=

h 2

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PAGE NO.- 2

11.

At very low temperature, oxygen O 2, freezes and forms a crystalline solid. Which term best describes the solid ? (A) Covalent network (B*) Molecular crystals (C) Metallic (D) Ionic vfrU;wu rkieku ij] vkWDlhtu O2 te tkrh gS rFkk ,d fØLVyh; Bksl curk gSA bl izdkj ds Bksl ds fy, mi;qDr in

gS \ (A) lgla;kstd vFkok usVodZ Bksl (C) /kkfRod Bksl Sol.

(B*) vkf.od Bksl (D) vk;fud O2 solid is made up of O 2 molecules held together by van der Walls forces. Therefore, it is molecular

12.

Which of the following is incorrect

fuEu esa ls dkSulk dFku lgh ugha gSA (A) The defect is known as schottky defect

iznf'kZr =kqfV 'kkWVdh =kqVh dgykrh gSA (B) Density of compound in the defect decreases

bl =kqfV esa ;kSfxd ds ?kuRo esa deh vkrh gSA (C) NaCl(s) is example which generally shows this defect NaCl(s) og mnkgj.k gS] tks lkekU;r% ;g =kqfV n'kZrk gSA (D*) Stoichoimetry of compound will change slightly.

;kSfxd fd jllehdj.kfefr FkksM+h lh ifjofrZr gks tkrh gSA 13.

If this represents FCC of CaF 2, then shaded plane passes through the centre of how many fluoride and calcium ions respectively : ;fn ;g CaF2 ds FCC dks iznf'kZr djrk gks] rks Nk;kafdr ry fdrus ¶yksjkbM vk;u o dSfY'k;e vk;u ds dsUnz

esa ls Øe'k% xqtjrk gS \

(A) 4, 4 14.

Sol:

15.

(B) 6, 6

(C) 2, 4

(D*) 4, 6

The correct name of [Pt(NH3)4Cl2][PtCl4] is : (A) tetraamminedichloridoplatinum(II) tetrachloridoplatinate (IV). (B) dichloridotetraammineplantinum(IV) tetrachloridoplatinate(II). (C*) tetraamminedichloridoplatinum(IV) tetrachloridoplatinate(II). (D) tetrachloridoplantinum(II) dichloridotetraammineplatinate(IV). [Pt(NH3)4Cl2][PtCl4] dk lgh uke gS : (A) VSVªk,EehuMkbDyksjkbMksIysfVue(II) VSVªkDyksjkbMksIysfVusV(IV)A (B) MkbDyksjkbMksVSVªk,EehuIySfVue(IV) VSVªkDyksjkbMksIysfVusV(II)A (C*) VSVªk,EehuMkbDyksjkbMksIysfVue(IV) VSVªkDyksjkbMksIysfVusV(II)A (D) VSVªDyksjkbMksIysfVue(II) MkbDyksjkbMksVSVªk,EehuIySfVusV(IV)A According to IUPAC nomenclature [Pt(NH 3) 4 Cl 2 ][PtCl 4] is tetraamminedichloridoplatinum(IV) tetrachloridoplatinate(II). Where oxidation number of Pt are (IV) and (II) respectively IUPAC ukedj.k ds vuqlkj [Pt(NH3)4Cl2][PtCl4] VSVªk,EehuMkbDyksjkbMksIysfVue(IV) VSVªkDyksjkbMksIysfVusV(II) gS] tgka Pt ds fy, vkWDlhdj.k la[;k Øe'k% (IV) o (II) gSA In a mixture of trichloromethane and ethyl ethanoate there are strong intermolecular forces between the two types of molecule. Would you expect the mixture to give a positive or negative deviation from Raoult's law ? (A) Positive (B*) Negative (C) no deviation (D) Can't say

VªkbDyksjksesFksu rFkk ,FksuksbV ds ,d feJ.k esa nks izdkj ds v.kqvksa ds e/; ,d izcy vUrjkf.kod cy gSA vki D;k vis{kk djksxs fd feJ.k jkWmYV fu;e ls ,d /kukRed ;k _.kkRed fopyu nsxk \ (A) /kukRed (B*) _.kkRed (C) dksbZ fopyu ugha (D) dqN dgk ugha tk ldrk gSA Sol.

Negative. The bonding in the mixture will prevent the molecules escaping easily into the vapour.

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PAGE NO.- 3

16.

The following two graphs are plotted between the temperature and mole fraction of the components of two different azeotropic mixture. (X1 = mole fraction of solvent).

Choose the correct options (A*) graph A represents minimum boiling point and maximum vapour pressure (B) graph B represents maximum boiling point and maximum vapour pressure (C) graph A represents maximum boiling point and minimum vapour pressure (D) graph B represents minimum boiling point and maximum vapour pressure

nks fHkUu fLFkjDokFkh feJ.k ds ?kVdks ds rki rFkk eksy izHkkt ds e/; fuEu nks vkjs[k cuk, x, gSA (X1 = foyk;d dk eksy izHkkt ).

lgh fodYi pqfu, % (A*) vkjs[k A fuEure DoFkukad rFkk mPpre ok"i nkc dks iznkf'kZr djrk gSA (B) vkjs[k B mPpre DoFkukad rFkk mPpre ok"i nkc dks iznkf'kZr djrk gSA (C) vkjs[k A mPpre DoFkukad rFkk fuEure ok"i nkc dks iznkf'kZr djrk gSA (D) vkjs[k B fuEure DoFkukad rFkk mPpre ok"i nkc dks iznkf'kZr djrk gSA 17.

18.

Sol.

Which of the following is not a function of salt bridge in an electrochemical cell ? (A) To maintain electrical neutrality of the solution (B) To complete the circuit so that current can flow. (C) To supply ions to cathodic and anodic compartment. (D*) To increase cell voltage oS|qr jlk;fud lsy esa mifLFkr yo.k lsrq (Salt bridge) dk fuEufyf[kr esa ls dkSulk dk;Z (A) foy;u dh oS|qr mnklhurk dks cuk;s j[kuk (B) oS|qr ifjiFk (electrical circuit) dks iwjk djuk] ftlls fo|qr dk izokg gks ldsA (C) dSFkksM o ,uksfMd d{k dh vksj vk;uksa dks izokfgr djukA (D*) lsy foHko esa o`f) djukA A given cell reaction is spontaneous when : (A) E°red is negative (B) E°red is positive

lSy vfHkfØ;k Lor% gksrh gS tc % (A) E°vipk;d _.kkRed gS (B) E°vip;u /kukRed gS G = – nFE; if E > 0 ; G < 0



(C*) Ecell is positive

(C*) Elsy /kukRed Spontaneous reaction.

gS

ugha gS \

(D)G = 0 (D)G = 0

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PAGE NO.- 4

Note : If Eºcell > 0, then the reaction will be spontaneous at standard conditions, not necessarily at all condition or given condition. gy % G = – nFE; if E > 0 ; G < 0  Lor% vfHkfØ;k uksV : ;fn Eºlsy > 0 gks rks ekud ifjfLFkfr;ksa ij vfHkfØ;k Lor% gksxh] ysfdu nh xbZ ifjfLFkfr;ksa ;k lHkh ifjfLFkfr;ksa ij Lor%

gksuk vko';d ugha gSA 19.

What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together : ;fn 0.1 M AgNO3 rFkk 0.1 M NaCl ds leku vk;ru ,d lkFk fefJr fd;s tkrs gSa] rks ukbVªsV vk;uksa dh lkUnzrk D;k gksxh: (A) 0.1 M (B) 0.2 M (C*) 0.05 M (D) 0.25 M (E) None of these buesa ls dksbZ ugha

Sol.

[NO3–] =

20.*

Sol.

0.1V  0 2V

0 .1 = 2 = 0.05 M

In which of the following reactions is Kp < Kc ? fuEu esa ls fdl vfHkfØ;k ds fy;s Kp < Kc gksxk ? (A*) CO(g) + Cl2(g) COCl2 (g) (C) 2BrCl(g) Cl2(g) + Br2(g) kp = kc (RT) ng So, if ng = – Ve, kp < kC So, anwers are (A) & (B)

(B*) CO(g) + 3H 2(g) (D) I2(g) 2(g).

CH4(g) + H2O(g)

DPP No. # 23 (JEE-ADVANCED) Total Marks : 78

Max. Time : 49 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.7 Multiple choice objective ('–1' negative marking) Q.8 to Q.10 Subjective Questions ('–1' negative marking) Q.11 to Q.13 Comprehension ('–1' negative marking) Q.14 to Q.16 Match the Following (no negative marking) Q.17 ChemINFO : 4 Questions ('–1' negative marking) Q.18 to Q.21

1.

(3 (4 (4 (3 (8 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 2 min.) marks, 6 min.) marks, 2 min.)

[21, [12, [12, [09, [08, [16,

14] 06] 09] 06] 06] 08]

From the figure representing enthalpy change of various transitions as indicated, a substance has maximum enthalpy when present in -

fuEu fp=k esa inkFkksZ ds fofHkUu laØe.kksa dh ,UFkSYih dks fuEukuqlkj iznf'kZr fd;k x;k gS] fuEu es a ls fdl voLFkk esa mifLFkr gksus ij inkFkZ dh ,UFkSYih dk eku vf/kdre gksrk gSA

2.

(A) solid state

(B) liquid state

(C*) gaseous state

(D) equal in all states

(A) Bksl

(B) nzo

(C*) xSl

(D) lHkh

voLFkk

voLFkk

voLFkk

voLFkkvksa esa cjkcj

Standared enthalpy of given reaction H2(g) + Br2(g)  2HBr(g) is H01 and standard enthalpy of condensation of bromine is H02, standard enthalpy of formation of HBr(g) at 250C is

ds fy, ,UFkSYih ifjorZu] H01 gS rFkk czksehu ds la?kuu dh ekud ,UFkSYih dk eku H02 gS] rks 250C rki ij HBr(g) dh ekud laHkou ,UFkSYih dk eku gksxk % H2(g) + Br2(g)  2HBr(g) vfHkfØ;k

Sol.

(A) H01 / 2 (B) H01 / 2 + H02 (C) H01 / 2  H02 0 H2(g) + Br2(g)  2HBr(g) ; H 1 Br2()  Br2(g) ; H0 = – H02 H2(g) + Br2()  2HBr(g)

;

2 × H0f = H01 – H02

(D*) (H01H02) / 2

H0f =

Hº1 Hº 2 2

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PAGE NO.- 5

3.

If the above plot is replotted at temperature T2 (where T2 > T1) , then which of the following plots may show the correct behaviour at T2 .

;fn mijksDr vkjs[k dks T2 rki (tgk¡ T2 > T1) ij iqu% vkjsf[kr fd;k tk;as] rks fuEu esa ls dkSulk vkjs[k T2 rki ij lgh O;ogkj iznf'kZr djsxk %

(A)

Sol.

(B)

(C*)

(D)

As temperature increases, graph moves upwards.

rki c<+kus ij vkjs[k Åij dh vksj foLFkkfir gksus yxrk gSA 4.

Calculate the compressibility factor for CO2, if one mole of it occupies 0.4 litre at 300 K and 40 atm. Comment on the result. (A) 0.40, CO2 is more compressible than ideal gas (B*) 0.65, CO2 is more compressible than ideal gas (C) 0.55, CO2 is more compressible than ideal gas (D) 0.62, CO2 is more compressible than ideal gas

ds fy, laihMh;rk xq.kkad ifjdfyr dhft,] ;fn bldk ,d eksy 300 K rFkk 40 atm ij 0.4 yhVj ?ksjrk gS, ifj.kke ij oDrO; nhft,A (A) 0.40, CO2 vkn'kZ xSl dh vis{kk vf/kd laihM+hr gksrh gSA (B*) 0.65, CO 2 vkn'kZ xSl dh vis{kk vf/kd laihM+hr gksrh gSA (C) 0.55, CO2 vkn'kZ xSl dh vis{kk vf/kd laihM+hr gksrh gSA (D) 0.62, CO2 vkn'kZ xSl dh vis{kk vf/kd laihM+hr gksrh gSA CO2

Sol.

5.

PV real Z = PV ideal W hen 0.1 mole HCl gas is added in 1lt of 0.1 M CH 3COOH(aq) then which statement is wrong? (Ka = 2 × 10–5, log 2 = 0.3) (A) degree of dissociation of CH3COOH decreases sharply. (B) change in pH would be 1.85 (C*) conc of [Cl–] = 0.1 M, [CH3COOH] = 0.1 M, [H+] = 0.2 M in final solution (D) on addition of HCl, Ka of CH3COOH does not change. tc 0.1 eksy Bksl HCl dks 0.1 M tyh; CH3COOH ds 1 yhVj esa feyk;k tkrk gS] rc dkSulk dFku xyr gSA (Ka = 2 × 10–5 , log 2 = 0.3) (A) CH3COOH ds fo;kstu dh ek=kk esa rhozrk ls deh gksrh gSA (B) pH esa ifjorZu 1.85 gksxkA (C*) ifj.kkeh foy;u esa [Cl–] dh lkUnzrk = 0.1 M, [CH3COOH] = 0.1 M, [H+] = 0.2 M. (D) HCl ds feykus ls CH3COOH ds Ka ds eku esa dksbZ ifjorZu ugha gksrk gSA

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PAGE NO.- 6

6.

E ºAl3  / Al = – 1.66 V and KSP of Al(OH)3 = 1.0 × 10–33. Reduction potential of the above couple at pH = 14 is: E ºAl3  / Al = – 1.66 V rFkk Al(OH)3

Sol.

(A*) – 2.31 V pH = 14

dk KSP = 1.0 × 10–33 gSaA pH = 14 ij mijksDr ;qXe ds fy, vip;u foHko gSa µ

(B) + 2.31 [CH–] = 1 M



(C) – 1.01 V

(D) + 1.01 V.

Ksp

[AI3+] = [OH– ]3 = 10–33 Al3+ + 3e–  Al 1 0.0591 log [ AI3 ] . 3

Ecell = E ºAI3  / Al – 7.

Sol.

At 298K the standard free energy of formation of H 2O() is –257.20 kJ/mole while that of its ionisation into H + ion and hydroxyl ions is 80.35 kJ/mole, then the emf of the following cell at 298 K will be(take F = 96500 C] H 2(g,1 bar) | H+ (1M) || OH¯ (1M) | O 2 (g, 1bar) 298K rki ij H 2O() ds laHkou dh ekud eqDr ÅtkZ –257.20 kJ/mole gS] tcfd blds H + vk;u rFkk gkbMªksfDly (OH¯) vk;u esa vk;uu dh ÅtkZ 80.35 kJ/mole gSA rks 298 K rki ij fuEu lsy dk emf gksxk (F = 96500 C] H 2(g,1 bar) | H+ (1M) || OH¯ (1M) | O 2 (g, 1bar) (A) 0.40 V (B*) 0.50 V (C) 1.23 V (D) – 0.40 V Cell reaction lsy vfHkfØ;k Cathode Anode

dSFkksM+ :

1 O (g) + H 2(g)  2H+(aq.) + 2OH¯(aq.) 2 2

Also we have gekjs H 2(g) +

1 O (g) + 2e –  2OH¯(aq.) 2 2

H 2(g)  2H + (aq.) + 2e –

,uksM :

H 2O() +

H 2O() +

ikl

1 O (g)  H 2O() 2 2

H 2O()  H +(aq.) + OH¯ (aq.) Hence for cell reaction

G° f = –257.2 kJ/mole G° = 80.35 kJ/mole

bl izdkj lsy vfHkfØ;k ds fy, G° = –96.50 kJ/mole

So vr% , E° = – 8.*

96500 G = 2  96500 = 0.50 V nF

The correct statement(s) regarding defects solids is (are) (A*) Schottky defect is usually favoured by small difference in the sizes of cation and anion. (B*) Schottky defect lowers the density of solids. (C) Compounds having F-centres are diamagnetic. (D*) Frenkel defect is a dislocation defect.

Bkslks esa =kqfV;ksa ds lanHkZ esa dkSulk@dkSuls dFku lgh gS % (A*) lkekU;r% 'kkSVdh =kqfV] /kuk;u rFkk _.kk;uksa ds vkdkj esa cgqr de vUrj ds dkj.k gksrh gSA (B*) 'kkSVdh =kqfV ds dkj.k Bkslksa ds ?kuRo esa deh vkrh gSA (C) ;kSfxd ftuesa F-dsUnz mifLFkr gksrs gSa] og izfrpqEcdh; gksrs gSA (D*) Ýsady =kqfV ,d izdkj dh LFkku ifjorZu (dislocation) =kqfV gSA

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PAGE NO.- 7

9.*

Which of the following factors do not favour electrovalency ? (A) Low charge on ions (B*) High charge on ions (C) Large cation and small anion (D*) Small cation and large anion

fuEu esa ls dkSuls dkjd oS|qr la;kstdrk dk leFkZu ugha djrs gSa \ (A) vk;u ij fuEu vkos'k (B*) vk;u ij mPp vkos'k (C) cM+k /kuk;u rFkk NksVk _.kk;u (D*) NksVk /kuk;u rFkk cM+k _.kk;u Sol.

(B) and (D) favour the covalent bond formation according to Fajan's rule.

Sol.

Qtku fu;e ds vuqlkj (B) rFkk (D) lgla;ksth ca/k ds fy, mfpr dkj.k gSA

10.*

Most ionic compounds have : (A) high melting points and low boiling points (B*) high melting points and non-directional bonds (C*) high solubilities in polar solvents and low solubilities in nonpolar solvents (D*) three-dimensional crystal structures, and are good conductors of electricity in the molten state

vf/kdrj vk;fud ;kSfxd dk gksrk gS % (A) mPp xyukad rFkk fuEu DoFkukad (B*) mPp xyukad rFkk vfn'kkRed cU/k (C*) /kzqoh; foyk;d esa mPp foys;rk rFkk v/kqoh; foyk;d esa fuEu foys;rk (D*) f=kfofe; fØLVyh; lajpuk rFkk xfyr voLFkk esa oS|qr dk vPNk pkyd gSA Sol.

(A) is incorrect, as ionic compounds have higher melting points as well as higher boiling points due to strong electrostatic force of attraction between the ions. (B), (C) and (D) are properties of ionic compounds. (A) lgh

ugha gS D;ksafd vk;u ds e/; vkd"kZ.k cy ds dkj.k vk;fud ;kSfxd dk xyukad rFkk DoFkukad vf/kd gks rk gS] vk;fud ;kSfxd dh (B), (C) rFkk (D) xq.k gSA 11.

How will you prepare bleaching powder from slaked lime ?

vki cq>s gq;s pwus ls fojatd pw.kZ dSls cuk ldrs gS \ Ans. Sol.

3Ca(OH)2 + 2CI2

Ca(OCI)2.CaCI2.Ca(OH)2.H2O + H2O

Slaked Lime

12.

Bleaching Powder

Work out the following using chemical equations : Chlorination of calcium hydroxide produces bleaching powder.

fuEufyf[kr jklk;fud vfHkfØ;k dk mi;ksx djrs gq;s jklk;fud lehdj.k fyf[k, % dsfY'k;e gkbMªkWDlkbM ds Dyksjhuhdj.k ls fojtad pw.kZ izkIr djukA Sol.

3 Ca(OH)2 + 2CI2 

Ca(OCI)2. Ca(OH)2.CaCI2.2H2O Bleaching powder

(fojatd pw.kZ )

(a mixture of Ca(OCI)2 and basic chloride)

13.

For a cell Mg(s) |Mg2+ (aq) || Ag+ (aq)| Ag, Calculate the equilibrium constant at 25°C. Also find the maximum work that can be obtained by operating the cell. E°(Mg2+ / Mg) = – 2.37V, E° (Ag+/Ag) = 0.8 V. ,d lsy Mg(s) |Mg2+ (aq) || Ag+ (aq)| Ag ds fy, 25°C ij lkE;koLFkk fLFkjkad dh x.kuk dhft, \ rFkk vf/kdre dk;Z

dks Kkr dhft,] tks fd lSy ds lapkyu ls izkIr gksrk gSA Ans.

E°(Mg2+ / Mg) = – 2.37V, E° (Ag+/Ag) = 0.8 V. Kc = 1.864 × 10107 , G° = – 611.8 kJ

Sol.

Ecell = 3.11 –

0.0591 log Keq. 2

Ecell = 0.

3.11  2 0.0591 Keq. = 1.864 × 10107 Gº = – 2.303 RT log 1.864 × 10107 = – 2.303 × 8.314 × 298 log 1.86 × 10107 log Keq. =

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Comprehension # (Read the following passage and answer the questions numbered 14 to 16. They have only one correct option) Standard reduction potentials (SRP) for different systems can be used to decide the spontaneity of a reaction e.g. E Zn 2  = – 0.76V, hence for the reaction Zn + 2H +  Zn2+ + H 2, G° is negative. It has Zn

been found experimentally that if (SRP of an oxidant – SRP of a reductant) is more than 1.7V, then their combination may lead to explosion (though it may be prevented by kinetic factors). Now go through the following data and answer the questions. Data :

E Ag / Ag = 0.80 V

;

E N2 / N3 ¯ = – 3.09 V

E ClO ¯ / ClO  = 1.23 V

;

 = – 2.71 V E Na  / Na

 = 0.77 V EFe 3 / Fe2 

;

E O 2 / H2 O 2 = – 1.03 V

E H 2O 2 / H2 O = 1.76 V

;

E O3 / O2 = 2.07 V

 EMnO – 2  = 1.54 V 4 / Mn

;

E Cr O 2 – / Cr 3  = 1.33 V 2 7

4

3

vuqPNsn % fuEu vuqPNsn dks i<+dj rFkk iz'u 14 ls 16 rd ds mÙkj nhft,A ;g dso y ,d lgh fodYi j[krs gSA fofHkUu fudk;ksa ds eku vip;u foHko (SRP) dks ,d vfHkfØ;k dh Lorrk% dks fu/kkZfjr djus esa iz;qDr fd;k tk ldrk gS mnkgj.k ds fy, E Zn 2 Zn = – 0.76V vr% Zn + 2H+  Zn2+ + H2 vfHkfØ;k ds fy, G° _.kkRed gS] ;g iz;ksfxd :i ls ik;k x;k gS fd tc (,d vkWDlhdkjd dk SRP– ,d vipk;d dk SRP) 1.7 V ls vf/kd gS] rc ;g la;kstu foLQksfVr gks ldrs gSA ¼gkyk¡fd bls xfrdh dkjdks }kjk jksdk tk ldrk gS½ fuEu vk¡dM+kas ds vk/kkj ij iz'uksa ds mÙkj nhft, \ E N / N ¯ = – 3.09 V E Ag / Ag = 0.80 V vk¡dM+s : ; 2

E ClO ¯ / ClO  4 3

14.

3

= 1.23 V

;

 = E Na  / Na

 = 0.77 V EFe 3 / Fe2 

;

E O 2 / H2 O 2 = – 1.03 V

E H 2O 2 / H2 O = 1.76 V

;

E O3 / O2 = 2.07 V

 EMnO – 2  = 1.54 V 4 / Mn

;

E Cr O 2 – / Cr 3  = 1.33 V 2 7

– 2.71 V

W hich of the following ionic combinations may lead to the formation of explosive substance. (A) Sodium ion and azide ion (B) Silver ion and perchlorate ion (C*) Silver ion and azide ion (D) All the above

fuEu esa ls dkSulk vk;fud la;kstu foLQksVd inkFkZ ds fuekZ.k esa iz;qDr gksrs gS \ (A) lksfM;e vk;u rFkk ,tkbM vk;u (B) flYoj vk;u rFkk ijDyksjsV vk;u (C*) flYoj vk;u rFkk ,tkbM vk;u (D) mijksDr lHkh Sol.

For the given options

fn;s x;s fodYi ds fy, (A) Sodium ion and azide ion ( lksfM;e

vk;u rFkk ,tkbM vk;u)

Na+ + N 3–1  Na + N 2 (B) Silver ion and perchlorate ion

E° = – 2.71 + 3.09 < 1.7 V

flYoj vk;u rFkk ijDyksjsV vk;u ClO 4– + Ag  Ag+ + ClO 3– (C) Silver ion and azide ion

E° = 1.23 – 0.80 < 1.7 V

flYoj vk;u rFkk ,tkbM vk;u Ag+ + N 3–  Ag + N 2

E° = 0.80 + 3.09 = 3.89 > 1.7 V

So will lead to explosion

blfy, foLQksVu gksxkA

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PAGE NO.- 9

15.

Sol.

16.

Sol.

17.

Ans.

W hich of the following ions will be capable of causing catalytic decomposition of H 2O 2. H 2O2 ds mRizsfd; vi?kVu ds fy, fuEu esa ls dkSulk vk;u mi;qDr gksxk \ (A) Fe 3+ (B) Fe 2+ (C*) both nksu ksa (D) None of these bues a ls dks bZ 3+ If Fe is present, then Fe3+ + H 2O2  Fe2+ + O 2 E° = 0.77 + 1.03 = 1.8 V > 0 so spontaneous Fe2+ + H 2O 2  Fe3+ + H 2O E° = 1.76 + 0.77 = 0.99 > 0 Hence Fe 2+ and Fe 3+ both are capable of decomposing H 2O 2. bl izdkj Fe2+ rFkk Fe3+ nksuks H2O 2 dks fo;ksftr djus leFkZ gSA

ugha

W hich is correct about the reaction between H 2O2 and O 3 (A) It is a case of mutual reduction (B*) O3 will oxidise H 2O 2 into O 2 (C) It is not a redox reaction (D) H 2O2 being a stronger oxidising agent will decompose ozone into oxygen H 2O2 rFkk O3 ds e/; vfHkfØ;k ds fy, dkSulk lgh gS (A) ;g vU;ksU; vip;u dh fLFkfr gS (B*) O3 ; H 2O 2 dks O 2 esa vkWDlhd`r djrh gS (C) ;g jsMkWDl vfHkfØ;k ugha gS (D) pwafd H 2O2 ,d izcy vkWDlhdkjd gS] vr% ;g vkstksu dks vkWDlhtu From SRP values it is clear that O 3 will oxidise H 2O 2 into O2. SRP ekuks ls ;g Li"V gS] O3, H 2O2 dks O2 esa vkWDlhd`r djsxkA Column – I dkWye– I (A) H+ (aq) (tyh;) (B) H (gas) (xSl) (C) H2 (gas) (xSl) (D) C (s, diamond) (s, ghjk) (A) – p,r ; (B) – q,s ; (C) – p,r

esa fo?kfVr djrk gSA

Column – II

dkWye– II (p) (q) (r) (s)

Hf0 = 0 Hf0  0 Gf0 = 0 Hf0 > 0

; (D) – q,s

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PAGE NO.- 10

ChemINFO-1.16

ELECTROCHEMISTRY

Daily Self-Study Dosage for mastering Chemistry

Lead Storage Batterty (Lead-Acid Battery)

The lead storage battery used in cars was first demonstrated to the French Academy of Sciences by Gaston plante in 1860. It contained nine cells in parallel and was able to deliver extraordinarily large currents. The cells were constructed from lead plates separated by layers of flannel that were immersed in a 10% H 2SO4 solution. As might be expected, the system soon became known as the Lead-acid battery Lead storage cell consists of lead grid filled with spongy lead which acts as the anode and a lead grid packed with lead oxide that acts as the cathode. A number of (generally six) such electrode pairs are dipped in sulphuric acid solution (20%) which acts as the electrolyte and these are separated by inert and porous partitions. When the battery discharges, PbO2 is reduced to PbSO 4 . When it is charged, the reaction is reversed. The standard electrode potential for the half-reaction is 1.682 V. The cell reactions are as follows: At the anode: At the cathode: At the anode:

Pb(s) + SO42– (aq)  PbSO4(s) + 2e– PbO 2(s) + 4H+ (aq) + SO 42– (aq) + 2e–  PbSO4(s) + 2H2O(I) Pb(s) + PbO2(s) + 2H2SO4 (aq)  2PbSO4(s) + 2H2O(I)

The cell reaction shows that the electrolyte H2SO4 is consumed during the reaction, so as the reaction proceeds, the concentration of H2SO4 decreases. Once the concentration of H2SO4 has fallen to a certain permissible level, the battery needs to be recharged by reversing the discharge reaction. For this, an external potential higher than the cell potential is applied. The cell then operates like an electrolyte cell and the PbSO 4 deposited on the electrodes is reverted to Pb and PbO2 by the following reactions: At the anode: At the cathode: At the anode:

PbSO 4(s) + 2H 2O(I)  PbO2(s) + 4H+ (aq) + SO 42– (aq) + 2e– PbSO 4(s) + 2e–  Pb(s) + SO42–(s) 2PbSO 4(s) + 2H2O(I)  Pb(s) + PbO 2(s) + 4H+ (aq) + 2SO 42–(aq)

The cell reactions show that the concentration of H 2SO4 increases during recharging of the cell. The potential developed by each electrode pair is about 2V and higher potential can be obtained by connecting a number of electrode pairs in series. These batteries find extensive use as automobile batteries, besides being used for electrical supply in trains, hospitals, etc.

The Lead Storage battery Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 18. Th cathode reaction during the charging of a lead-acid battery leads to the (A) Formation of PbSO4 (B*) Reduction of Pb 2+ to Pb (C) Formation of PbO 2 (D) Deposition of Pb at the anode 19. In a lead storage battery (A) Pb is oxidized to PbSO 4 at the anode (B) PbO2 is reduced to PbSO4 at the cathode (C) Both electrodes are immersed in the same aqueous solution of H 2SO4 (D*) All the above are true

20. Which of the following reactions occurs at the anode during the recharging of lead storage battery ? (A) Pb + SO42–  PbSO 4 + 2e – (B) Pb + PbO 2 + H2SO 4  2PbSO 4 + 2H2O (C*) PbSO 4 + 2e –  Pb + SO 42– (D) PbSO4 + 2H 2O  Pb + PbO 2 + 2H 2SO4 + 2e – 21. When a lead storage battery is discharged, (A) SO2 is evolved (B) Lead is formed (C) Lead sulphate is consumed (D*) Sulphuric acid is consumed

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PAGE NO.- 11

ChemINFO-1.16

fo|qr jlk;u

Daily Self-Study Dosage for mastering Chemistry

lhlk lapk;d lSy (lhlk&vEy cSVjh )

dkjksa esa iz;qDr lhlk lapk;d cSVjh dks loZizFke xSLVu IykUV }kjk foKku dh ÝSap vdsMeh }kjk 1860 esa iznf'kZr fd;k x;kA blesa ukS lSy lekukUrj ry esa gksrs gSa rFkk ;g vk'p;Ztud :i ls vf/kd /kkjk igqapkus esa l{ke gksrk gSA lSy dks ¶ysuy] tks 10% H2SO4 foy;u esa Mwch gqbZ gS] dh ijrksa }kjk i`Fkd ysM ;k lhlk dh IysVksa ls cuk;k x;kA vis{kkd`r :i ls fudk; vc lhlk&vEy cSVjh dgykrk gSA lhlk lapk;d lSy esa LikWUth ysM ls Hkjh gqbZ ysM fxzM gksrh gS] tks ,uksM ds :i esa dk;Z djrh gS rFkk ysM vkWDlkbM ls ladqfyr ysM fxzM dSFkksM ds :i esa dk;Z djrh gSA bl izdkj ds bysDVªkWM ;qXeksa dh la[;k (lkekU;r% N%) dks lY¶;wfjd vEy foy;u (20%) esa Mwcksrs gSa] tks fo|qr vi?kV~; ds leku dk;Z djrh gS rFkk ;g vfØ; o fNfnzr Hkkxks }kjk i`Fkd gksr s gSaA tc cSVjh dks fujkosf'kr djrs gSa rks PbO2 , PbSO4 esa vipf;r gksrk gSA tc bls vkosf'kr djrs gSa rks vfHkfØ;k O;qRØe gksrh gSA v)Z vfHkfØ;k ds fy, ekud bysDVªkWM foHko 1.682 V gSA lSy vfHkfØ;k,a fuEu izdkj gS % ,uksM ij : Pb(s) + SO42– (aq)  PbSO4(s) + 2e– dSFkksM ij : PbO2(s) + 4H+ (aq) + SO42– (aq) + 2e–  PbSO4(s) + 2H2O(I) ,uksM ij : Pb(s) + PbO2(s) + 2H2SO4 (aq)  2PbSO4(s) + 2H2O(I) lSy vfHkfØ;k n'kkZrh gS fd fo|qr vi?kV~; H2SO4 dk vfHkfØ;k ds nkSjku miHkksx gksrk gS vr% vfHkfØ;k lEiUu gksrh gS] H2SO4 dh lkUnzrk ?kVrh gSA H2SO4 dh lkUnzrk fuf'pr Lohd`r Lrj rd de gksrh gSA cSVjh dks fujkos'ku vfHkfØ;k ds O;qRØe }kjk iqu% vkosf'kr djus dh vko';drk gksrh gSA blds fy, lSy foHko ls mPp ckáre foHko iz;qDr fd;k tkrk gSA ckn esa lSy ,d oS|qr vi?kV~; lSy ds leku lapkfyr gksrk gS rFkk bysDVªkWM ij fu{ksfir PbSO4 fuEu vfHkfØ;kvksa }kjk Pb rFkk PbO2 eas cnyrk gS : ,uksM ij : PbSO4(s) + 2H2O(I)  PbO2(s) + 4H+ (aq) + SO42– (aq) + 2e– dSFkksM ij : PbSO4(s) + 2e–  Pb(s) + SO42–(s) ,uksM ij : 2PbSO4(s) + 2H2O(I)  Pb(s) + PbO2(s) + 4H+ (aq) + 2SO42–(aq) lSy vfHkfØ;k,¡ n'kkZrh gS fd H2SO4 dh lkUnzrk lSy ds iqu%vkos'ku (recharging) ds nkSjku c
The Lead Storage battery Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice.

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PAGE NO.- 12

18.

Th cathode reaction during the charging of a lead-acid battery leads to the (A) Formation of PbSO 4 (B*) Reduction of Pb2+ to Pb (C) Formation of PbO2 (D) Deposition of Pb at the anode

19.

In a lead storage battery (A) Pb is oxidized to PbSO 4 at the anode (B) PbO2 is reduced to PbSO4 at the cathode (C) Both electrodes are immersed in the same aqueous solution of H 2SO4 (D*) All the above are true

20.

Which of the following reactions occurs at the anode during the recharging of lead storage battery ?

21.

(A) Pb + SO42–  PbSO4 + 2e–

(B) Pb + PbO2 + H2SO4  2PbSO4 + 2H 2O

(C*) PbSO4 + 2e–  Pb + SO42–

(D) PbSO 4 + 2H2O  Pb + PbO2 + 2H2SO4 + 2e–

When a lead storage battery is discharged, (A) SO2 is evolved (C) Lead sulphate is consumed

(B) Lead is formed (D*) Sulphuric acid is consumed

18.

ysM vEy cSVjh ds vkos'ku ds nkSjku dSFkksM+ vfHkfØ;k esa gksrh gSA (A) PbSO4 dk fuekZ.k (B*) Pb2+ dk Pb es vip;u (C) PbO2 dk fuekZ.k (D) ,uksM ij Pb dk fu{ksi.k

19.

lhlk lapk;d lsy es (A) ,uksM+ ij Pb, PbSO4 es vkWDlhd`r gksrk gSA (B) dFkkSM+ ij PbO2, PbSO4 es vipf;r gksrk gSA (C) nksuks bysDVªkWM H2SO4 ds leku tyh; foy;u es Mwcs jgrs gSA (D*) mijksDr lHkh lR; gSA

20.

lhlk lapk;d cSVjh ds iqu% vkos'ku ds nkSjku ,uksM+ ij fuEUk esa ls dkSulh vfHkfØ;k gksrh gSA

21.

(A) Pb + SO42–  PbSO4 + 2e–

(B) Pb + PbO 2 + H2SO4  2PbSO 4 + 2H2O

(C*) PbSO4 + 2e–  Pb + SO42–

(D) PbSO4 + 2H 2O  Pb + PbO 2 + 2H2SO 4 + 2e–

tc lhlk lapk;d lsy fujkosf'kr gksrk gS%& (A) SO2 eqDr gksrh gSA (C) ysM lYQkbM iz;qDRk gksrk gSA

(B)

ysM curk gSA (D*) lY¶;wfjd vEy iz;qDRk gksrk gSA

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PAGE NO.- 13