Jp Xii Physical&inorganic Chemistry (22) - Prev Chaps + Inorg. Chem.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 35 & 36

Course : VIJETA (JP)

This DPP is to be discussed in the week (24.08.2015 tob 29.08.2015)

DPP No. # 35 (JEE-ADVANCED) Total Marks : 91

Max. Time : 54 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.6 to Q.12 Single Integer type Questions ('–1' negative marking) Q.13 to Q.16 Match the Following (no negative marking) Q.17 ChemINFO : 5 Questions ('–1' negative marking) Q.18 to Q.23

1.

(3 (4 (4 (8 (4

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

[15, [28, [16, [08, [24,

10] 14] 12] 06] 12]

The number of atoms in 50 g of an FCC crystal with density d = 10 gcm –3 and cell edge of 200 pm is equal to ,d FCC fØLVy ftldk ?kuRo d = 10 gcm–3 o ,dd dksf"Bdk dh dksj yEckbZ 200 pm gS] ds 50 g esa ijek.kqvksa dh la[;k

gksxh % Sol.

(A) 3 × 1025 (B) 5 × 1024 a = 200 pm = 200 × 10–10 cm = 2 × 10–8 cm volume (vk;ru) = (2 × 10–8)3 No. of atoms (ijek.kqvksa

2.

The rms velocity of hydrogen is

(A) T(H2 )  T(N2 ) Given that : So

gy .

fn;k x;k gSA fd :

7 times the rms velocity of nitrogen. If T is the temperature of the gas, then :

blfy,

(B) T(H2 )  T(N2 ) 3RTH2 2 3RTH2 2 3RTH2 2 3RTH2 2

(D*) 2.5 × 1024

dh la[;k) = = = 2.5 × 1024

gkbMªkstu dk rms osx] ukbVªkstu ds rms osx ls

Sol.

(C) 1.25 × 1024

= =7×

= =7×

7

3RTN2 28

3  RTN2 28 7

3RTN2 28

3  RTN2 28

7

xquk gSA ;fn xSl dk rkieku T gS] rks % (C*) T(H2 )  T(N2 ) (because Urms = 

3RT ) M

TN2 = 2TH2

(D;ksafd Urms = 

(D) T(H2 )  7 T(N2 )

or

TN2 > TH2 .

;k

TN2 > TH2

3RT ) M

TN2 = 2TH2

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

3.

The preparation of SO3(g) by reaction SO2 (g) +

1 O (g) 2 2

SO3 (g)

is an exothermic reaction. If the preparation follows the following temperature-pressure relationship for its % yield, K1 , K2 and K3 are the equilibrium constant for the given reaction at temperature T1, T2 and T3 respectively. The correct option is :

(A) T3 > T2 > T1 (B*) T3 < T2 < T1 (C) K1 = K2 = K3 (D) Nothing could be predicted about temperature through given information. SO2 (g) +

1 O (g) 2 2

SO3 (g)

vfHkfØ;k ds }kjk SO3(g) dk la'ys"k.k gksrk gS tks fd m"ek{ksih vfHkfØ;k gS bldh % yC/kh ds fy, rkieku nkc lEcUèk ds vuqlkj vfHkfØ;k ds fy, rkieku T1, T2 rFkk T3 ij Øe'k% lkE;koLFkk fu;rkad K1 , K2 rFkk K3 gS rks lgh fodYi gksxkA

(A) T3 > T2 > T1 (B*) T3 < T2 < T1 (C) K1 = K2 = K3 (D) nh xbZ lwpuk ds 4.

vk/kkj ij rkieku dk vuqeku ugha yxk;k tk ldrk gSA

A barium salt [X] insoluble in dilute HCl on heating in a reducing flame with charcoal and sodium carbonate gives a salt. An aqueous solution of this salt produces yellow precipitate with cadmium acetate solution. The anion of [X] is : ruq HCl esa vfoys; ,d csfj;e yo.k [X] dks pkjdksy rFkk lksfM;e dkckZsusV ds lkFk vipk;d Tokyk esa xeZ djus ij ,d

yo.k nsrk gSA bl yo.k dk tyh; foy;u dSMfe;e ,lhVsV foy;u ds lkFk ihyk vo{ksi nsrk gSA [X] dk _.kk;u gS % (A) S2– (B) SO32– (C*) SO42– (D) All of these mijksDr lHkh  Hints : BaSO4 (insoluble in dilute HCl) + Na2CO3 + C  Na2S Na2S + (CH3COO)2Cd  CdS  (yellow) + 2 CH3COONa

5.

Sol.

When a nitrate is boiled with aluminium turnings and con. NaOH solution, the gas liberated is: tc ukbVªsV dks] ,Y;wfefu;e dh Nhyu rFkk lkanz NaOH foy;u ds lkFk mckyk tkrk gS rc mRiUu xSl gSA (A) NO2 (B*) NH3 (C) NO (D) N2O Al + H2O + NaOH  NaAlO2 + 3H ; NO3¯ + 8H  NH3 + OH¯ + 2H 2O 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.- 2

6.*

Which of the following is/are correctly matched ? (A*) Brown ring test – NO 3– (B) Thiourea test – SO32– 2– (C*) Starch iodate test – S (D) All of these

fuEu esa ls dkSulk ;k dkSuls lgh lqesfyr gSA (A*) Hkwjh oy; ijh{k.k – NO3– (C*) LVkpZ vk;ksMsV ijh{k.k – S2–

(B) Fkk;ks;wfj;k ijh{k.k – SO32– (D) mijksDr lHkh

Ans.

Thiourea test  For NO2–

7.*

When CO2 is passed through lime water , it first turns milky and then it becomes clear. The turbidity reappears if : (A) Solution of MgCl2 is added. (B) More CO 2 bubbled through it. (C*) The clear solution is boiled. (D*) Fresh lime water is added. tc CO2 dks pwus ds ikuh esa ls izokfgr fd;k tkrk gS] rc ;g igys nqf/k;k gks tkrk gS rFkk fQj ;g jaxghu gks tkrk gSA xanykiu (turbidity) iqu% mRiUu gksrk gS ;fn % (A) MgCl2 dk foy;u feyk;k tkrk gSA (B) vf/kd CO2 cqycqfyr dh tkrh gSA (C*) tc LoPN foy;u dks mckyk tkrk gSA (D*) rktk cus pwus ds ikuh dks feyk;k tkrk gSA

Hint :

 Ca(HCO3)2  CaCO3 + H2O + CO2

8.*

Which of the following statement(s) is/are true ? (A) Soluble bicarbonates gives white precipitate with MgCl2 in cold. (B*) Soluble calcium bicarbonate gives white precipitate with dilute ammonia solution. (C*) All bicarbonates are generally soluble in water (D*) Hg(II) chloride forms a reddish brown precipitate in a solution of soluble carbonate.

Sol.

;

Ca(HCO 3)2 + Ca(OH)2  2 CaCO3  + 2 H2O

fuEu esa ls dkSulk@dkSuls dFku lR; gS@gSa& (A) foy;'khy ckbdkckZsusV] MgCl2 ds B.Ms foy;u ds lkFk lQsn vo{ksi nsrk gSA (B*) foy;'khy dSfY'k;e ckbdkckZsusV ruq veksfu;k foy;u ds lkFk lQsn vo{ksi nsrk gSA (C*) lHkh ckbdkckZsusV lkekU;r% ty esa foy;'khy gksrs gSaA (D*) Hg(II) DyksjkbM] foy;'khy dkckZsusV ds foy;u esa yky&Hkwjk vo{ksi cukrk gSA Cold  (A) HCO3– + Mg2+   MgHCO3(soluble) foys;  MgCO3  + CO2 + H2O (B) Ca(HCO3)2 + 2NH3  (NH4)2CO3 + CaCO3 (C) True (D) CO32– + 4Hg2+ + 3H2O  Hg4O3CO3  (reddish brown) ykyHkwjk + 6H+ basic mercury (II) carbonate {kkjh; edZjh (II) dkcksZusV

9.*

Which of the following statement(s) is/are correct ? (A*) Sulphide gives black precipitate with AgNO 3 solution and is soluble in hot concentrated nitric acid. (B*) Sulphide gives no precipitate with barium chloride solution. (C*) Sulphide ions react with sodium nitroprusside giving a purple colouration. (D) Free H2S gas form white precipitate with tetrahydroxo plumbate (II) solution.

fuEufyf[kr esa ls dkSulk@dkSuls dFku lgh gS@gSa \& (A*) lYQkbM] AgNO3 foy;u ds lkFk dkyk vo{ksi nsrs gSa tksfd xeZ] rFkk lkanz ukbVªhd vEy esa foy; gSA (B*) lYQkbM] csfj;e DyksjkbM foy;u ds lkFk dksbZ vo{ksi ugha nsrk gSA (C*) lYQkbM vk;u] lksfM;e ukbVªksçqlkbV ds lkFk tkequh jax dk foy;u nsrk gSA (D) eqDr H2S xSl] VªsVkgkbMªksDlksIyEcsV (II) foy;u ds lkFk lQsn vo{ksi cukrh gSA Sol.

[Pb(OH)4]2– + H2S  PbS  (black) dkyk + 2OH– + 2H2O.

10.*

For the equilibrium CuSO 4.5H 2O(s) CuSO 4.H 2O(s) + 4H 2O(g), the equilibrium constant –10 4 Kp = 2.56 × 10 atm at 27ºC. Now, if an air sample 40% saturated with water vapour is exposed to the above reaction at equilibrium, which of the following statement(s) are correct : Given : Saturated vapour pressure of water at 27ºC is 12.5 torr. (A*) Mass of CuSO4.5H2O will increase. (B) Mass of CuSO 4.5H2O will decrease. (C) Mass of CuSO4.H2O will increase. (D*) Mass of CuSO4.H2O will decrease. 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

27ºC

rki ij ,d lkE; CuSO 4.5H2O(s) CuSO 4.H 2O(s) + 4H 2O(g) ds fy,] lkE; fu;rka d –10 4 Kp = 2.56 × 10 atm gSA vc] ;fn tyok"i ls 40% lar`Ir ok;q ds ,d izkn'kZ dks mDr vfHkfØ;k ds lkE; ij ls izokfgr fd;k tk,] rc fuEu esa ls dkSulk@dkSuls dFku lgh gS@gSa : fn;k gS : 27ºC ij] ty dk lar`Ir ok"i nkc = 12.5 VkWj gSA (A*) CuSO4.5H2O ds nzO;eku esa o`f) gksrh gSA (B) CuSO4.5H2O ds nzO;eku esa deh gksrh gSA (C) CuSO4.H2O ds nzO;eku esa o`f) gksrh gSA (D*) CuSO4.H2O ds nzO;eku esa deh gksrh gSA Sol. 

Kp = (pH2O)4 = 2.56 × 10–10 atm 4 pH2O = 4 × 10–3 atm = 4 × 10–3 × 760 = 3.04 torr.

Partial pressure of water vapour in air =

Sol.

40 × 12.5 = 5 100

So, the amount of water vapour in air should decrease to decrease value of partial pressure of water vapour from 5 torr to the equilibrium value (3.04 torr). so, mass of CuSO4.5H2O will increase and mass of CuSO4.H2O will decrease. Kp = (pH2O)4 = 2.56 × 10–10 atm 4  pH2O = 4 × 10–3 atm = 4 × 10–3 × 760 = 3.04 torr.

ok;q esa] ty ok"i dk vkaf'kd nkc =

40 × 12.5 = 5 100

vr% ty ok"i ds vkaf'kd nkc ds eku 5 VkWj ls lkE; eku] (3.04 VkWj) rd de djus ds fy, ok;q esa] ty ok"i dh ek=kk esa deh gksuh pkfg,A blfy,] CuSO4.5H2O ds nzO;eku esa o`f) gksxh rFkk CuSO4.H2O ds nzO;eku esa deh gksxhA 11.*

To a 200 ml 0.1 M aq. Na3PO4 solution, 50 ml 0.2 M aq. NaH2PO4 solution is added. select correct statements: [K1 = 10–4, K2 = 10–7, K3 = 10–12 for H3PO4] (A) Resulting solution will be acidic (B*) Resulting solution will act as buffer solution (C*) Resulting solution will be basic (D) Concentration of PO34– in resulting solution will be 0.1 M

;fn 200 ml, 0.1 M tyh; Na3PO4 foy;u esa] 50 ml 0.2 M tyh; NaH2PO4 foy;u feyk;k tkrk gS rks fuEu esa ls dkSulk dFku lgh gksxkA [K1 = 10–4, K2 = 10–7, K3 = 10–12 for H3PO4] (A) ifj.kkeh foy;u vEyh; gksxkA (B*) ifj.kkeh foy;u cQj foy;u gksxkA (C*) ifj.kkeh foy;u {kkjh; gksxkA (D) ifj.kkeh foy;u esa PO 34– dh lkUærk 0.1 M gksxhA Sol.

Na3PO 4( aq ) 20 – 10  10 m mol

+ NaH2PO 4( aq )  2Na HPO 2 4 50  0.2 O

 10 m mol (L.R.)

20 m mol

Resulting solution has : Na 3PO 4 + Na 2PO 4 10 m mol PH = 12 + log [PO3–4]f = 12.*

20 m mol

10 = 11.7 20

10 1 = = 0.04 M. 250 25

Which of the following statement(s) is (are) correct when a mixture of NaCl and K 2Cr2O7 is gently warmed with concentrated H2SO4? (A*) A deep red vapour is evolved (B*) The vapour when passed into NaOH solution gives a yellow solution of Na2CrO4 (C) Chlorine gas is evolved (D) Intensity of deep red vapours increases if dil H2SO4 is used instead of concentrated H2SO4.

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.- 4

fuEu esa ls dkSulk@dkSuls dFku lgh gS@gSa] tc NaCl o K2Cr2O7 ds feJ.k dks lkanz H2SO4 ds lkFk xeZ fd;k tkrk gS& (A*) xgjh yky ok"i fudyrh gSA (B*) tc ok"i dks NaOH foy;u esa ls izokfgr djrs gS rc Na2CrO4 dk ihyk foy;u izkIr gksrk gSaA (C) Dyksjhu xSl eqDr gksrh gSA (D) ;fn lkUnz H2SO4 ds LFkku ij ruq H2SO4 feyk;k tk;s rc xgjh yky ok"i dh l?kurk esa o`f) gksrh gSA Sol.

13.

Ans. Sol.

14.

Chromyl chloride confirmatory test for ionic chlorides which forms CrO2Cl2 (deep red). Øksfey DyksjkbM ijh{k.k vk;fud DyksjkbMksa dk fu'p;kRed ijh{k.k gSA tks CrO2Cl2 (xgjk yky) cukrk

gSA

Two solids AB and CB2 are simultaneously heated in a closed vessel to attain equilibrium. AB(s) A(g) + B(g) ... (1) CB2(s) C(s) + 2B(g) ... (2) –6 KP for (1) and (2) reaction are 4.8 × 10 and 5.76 × 10–6. If partial pressure of A at equilibrium is 2 × 10–x, find x. ,d cUn ik=k esa nks Bksl AB rFkk CB2 lkE; izkIr djus ds fy, ,d lkFk xeZ fd;s tkrs gSaA AB(s) A(g) + B(g) ... (1) CB2(s) C(s) + 2B(g) ... (2) (1) o (2) vfHkfØ;k ds fy, KP Øe'k% 4.8 × 10–6 o 5.76 × 10–6 gSaA ;fn lkE; ij A dk vkaf'kd nkc 2 × 10–x gS] rks x Kkr dhft;s) 3 AB(s) A(g) + B(g) y 2x+y KP = 4.8 × 10–6 CB2(s) C(s) + 2B(g) 2x+y KP = 5.76 × 10–6 2 –6 (2x + y) = 5.76 × 10 (2x + y) = 2.4 × 10–3  2.4 × 10–3 × y = 4.8 × 10–6 y = 2 × 10–3 How many of the following compounds do not impart characteristic colour to the Bunsen flame : NaCl , BeCl2 , KOH , BaSO4 , MgCl2 , CsCl , Na2SO4 , Mg(OH)2 , K2CO3

fuEu ;kSfxdksa esa ls fdrus ;kSfxd] cqulsu Tokyk dks ykf{.kd jax çnku ugha djrs gSa %

g y-

NaCl , BeCl2 , KOH , BaSO4 , MgCl2 , CsCl , Na2SO4 , Mg(OH)2 , K2CO3 3 Compounds of Be and Mg do not impart any colour to Bunsen flame. Be rFkk Mg ds ;kSfxd] cqulsu Tokyk dks dksbZ jax çnku ugha djrs gaSA

15.

1.0 gram of a monobasic acid HA in 100 gram H2O lower the freezing point by 0.155 K. 0.45 gram of same

Ans. Sol.

1 M KOH solution for complete neutralisation. If the degree of dissociation of acid is , 5 then value of ‘20’ is : (Kf for H2O = 1.86 K.Kg/mole) 100 g H2O esa ,d ,dy {kkjh; vEy HA dk 1.0 g, fgekad fcUnq esa 0.155 K dk voueu ¼deh½ djrk gSA leku vEy acid require 15 ml of

1

ds 0.45g ds iw.kZ mnklhuhdj.k ds fy;s 5 M KOH ds 15 ml dh vko';drk gksrh gSA ;fn vEy ds fo;kstu dh ek=kk  gS rc ‘20’ dk eku gS: (H2O ds fy;s Kf = 1.86 K.Kg/mole) Ans. Sol.

5

 Wacid  1000   TF = KF × m = KF ×  M  acid  H2O  (Macid)exp = 120 Also N1V1 = N2V2 0.45 1 1 (Macid )thor = 5 × 15 × 1000 '

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.- 5

(Macid)thor =



Sol.

0.45  5  1000 = 150 15

 Mthor  150   i = M = = 1.25  120  exp   = (i – 1) = 1.25 – 1 = 0.25. 20 = 5

W TF = KF × m = KF ×  M  Also

(MvEy)izk;ksfxd = 120 N1V1 = N2V2 (M

0.45 )

(MvEy)lS)kfUrd =



16.

 1000    H2O 

=

1 1 × 15 × ' 5 1000

0.45  5  1000 = 150 15

M  150   i= M  = 120 = 1.25    = (i – 1) = 1.25 – 1 = 0.25. 20 = 5

Ans.

How many of the following liberate coloured vapours/gas with concentrated H2SO4? fuEu esa ls fdrus ;kSfxd lkUnz H2SO4 ds lkFk jaxhu ok"i@xSl nsrs gS ? (i) KBr (s) + K2Cr2O7 (s) (ii) KNO2 (s) (iii) KI (s) (iv) K2S (s) (v) KCl (s) (vi) KBr (s) + MnO2 (s) (vii) KNO3 (s) (viii) KCI(s) + MnO2 (s) (ix) K2SO3 (s) 6.

Sol.

(i)

6KBr + K2Cr2O7 + 7H2SO4  3Br2  (reddish-brown) + 2Cr3+ + 8K+ + 7SO42– + 7H 2O

(ii)

NO2– + H+  HNO2

;

3HNO2  HNO3 + NO  + H 2O

2NO  + O2   NO2  (brown)

Sol.

(iii)

3I– + 2H2SO4  I3–  (violet) + SO42– + 2H2O + SO2

(iv)

K2S + H2SO4  H2S  (colourless) + K2SO4

(v)

– Cl– + H2SO4  HCl  (colourless) + HSO 4

(vi)

2KBr + MnO2 + 2H2SO4  Br2  (reddish-brown) + 2K+ + Mn2+ + 2SO42– + 2H 2O

(vii)

4NO3– + 2H2SO4  Br2  (reddish-brown) + 2K+ + Mn2+ + 2SO42– + 2H2O

(viii)

KCl + MnO(OH)2 + 2H2SO4 + 2Cl–  Mn2+ + Cl2  (yellowish green) + 2SO 42– + 3H2O

(ix)

SO32– + 2H+  SO2  (colourless) + H2O

(i)

6KBr + K2Cr2O7 + 7H2SO4  3Br2  (xgjk

(ii)

NO2– + H+  HNO2

;

yky Hkwjk) + 2Cr3+ + 8K+ + 7SO42– + 7H2O

3HNO2  HNO3 + NO  + H 2O

2NO  + O2   NO2  (Hkwjk) (iii)

3I– + 2H2SO4  I3–  (cSaxuh) + SO42– + 2H2O + SO2

(iv)

K2S + H2SO4  H2S  (jaxghu) + K2SO4

(v)

– Cl– + H2SO4  HCl  (jaxghu) + HSO4

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.- 6

17.

yky Hkwjk) + 2K+ + Mn2+ + 2SO42– + 2H2O

(vi)

2KBr + MnO2 + 2H2SO4  Br2  (xgjk

(vii)

4NO3– + 2H2SO4  Br2  (xgjk

(viii)

KCl + MnO(OH)2 + 2H2SO4 + 2Cl–  Mn2+ + Cl2  (ihyk

(ix)

SO32– + 2H+  SO2  (jaxghu) + H2O

yky Hkwjk) + 2K+ + Mn2+ + 2SO42– + 2H2O

Match the following (one or more than one). Column – I (Reagent) (A) Silver nitrate solution (B) Barium chloride solution (C) Lead nitrate solution (D) Acidified potassium permangnate solution

fuEu dks lqesfyr dhft, ¼,d vFkok ,d ls vf/kd½ % dkWy e-I (vfHkdkjd) (A) flYoj ukbVsªV foy;u (B) csfj;e DyksjkbM foy;u (C) ysM ukbVsªV foy;u (D) vEyhd`r ikSVsf'k;e ijeSXusV foy;u Ans.

gjk) + 2SO42– + 3H2O

(p) (q) (r) (s)

Column – II (Anion) CO32– SO32– S2– NO2–

(p) (q) (r) (s)

dkW ye -II (_.kk;u) CO32– SO32– S2– NO2–

(A – p, q, r, s) ; (B – p, q) ; (C – p, q, r) ; (D – q, r, s)

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.- 7

ChemINFO-1.19

p-Block Element

Daily Self-Study Dosage for mastering Chemistry

Comparision between white and Red phosphorus

Property

White prosphorus

Red prosphorus

(1) Physical state

Soft waxy solid

Brittle powder

(2) Colour

White when pure Attains, yellow

Red

colour on standing (3) Odour

Garlic

Odourless

(4) Solubility in H2O

Insoluble

Insoluable

(5) Solubility in CS2

Soluable

Insoluable

(6) Physiological Action

Poisonous

Non poisonous

(7) Chemical activity

Very Active

Less active

(8) Stability

Unstable

Stable

(9) Phosphorescence

Glows in dark

Does not glow in dark

(10) Reaction with NaOH

Evolves phosphine

No action

(11) Molecular Formula

P4

Complex polymer

Note : Order of stability : Black phosphorus > R.P. > W.P. Order of Reactivity : B.P. < R.P. < W.P. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 18.

19.

White phosphorus is : (A) A monoatomic gas

(B*) P4 a tetrahedral solid

(C) Pgas a crown

(D) A linear diatomic molecule

White phosphorus when boiled with strong solution of caustic soda produces. (A*) Phosphine

20.

22.

23.

(C) Phosphorus Acid

(D) No reaction

(C*) Black

(D) All stable

Which of the following phosphorus is most stable. (A) Red

21.

(B) Phosphoric acid

(B) White

Which of the following properties of white phosphorus are shared by red phosphorus. (A) It show phosphorescenes in air

(B) It reacts with hot aqueous NaOH to give phosphine.

(C) It dissolve in carbon disulphide

(D*) It not dissolve in H2O.

White phosphorus (P4) has : (A*) Six P–P single bond

(B) Four P–P single bond

(C*) Four lone pair of electron

(D*) PPP angle of 60º

Phosphorus is used in : (A) Photography

(B) Cement Industry

(C) Rubber industry

(D*) Match industry

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.- 8

ChemINFO-1.19

p-Block Element

Daily Self-Study Dosage for mastering Chemistry

xq.kèkeZ (1) HkkSfrd voLFkk (2) jax

lQsan o yky QkLQksj l ds eè; rqyuk

lQsan QkLQkWjsl ekse tSlk eqyk;e Bksl 'kq) voLFkk esa lQsan gksrk gS rFkk dqN nsj j[kus ij ihyk gks tkrk gSA yglwu tSlh vfoys; foys; fo"kSyk vfèkd lfØ; vLFkk;h vaèksjs esa pedrk gSA QkWLQhu eqDr djrk gSA

(3) xaèk (4) H2O esa

foys;rk (5) CS2 esa foys;rk (6) 'kkfjjhd fØ;k (7) jlk;fud lfØ;rk (8) LFkkf;Ro (9) LQqjnhfIr 'khyrk (10) NaOH ds lkFk vfHkfØ;k (11) v.kqlw=k

yky QkLQkWjsl Hkaxwj pw.kZ yky xaèkghu vfoys; vfoys; vfo"kSyk de lfØ; LFkk;h vaèksjs esa ugh pedrk gSA dksbZ fØ;k ughaA ladqy cgqydA

P4

uksV : LFkkf;Ro dk Øe : dkyk QkLQksjl > yky QkLQksjl > lQsn QkLQksjl fØ;k'khyrk dk Øe : dkyk QkLQksjl < yky QkLQksjl < lQsn QkLQksjl Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 18.

lQsan QkLQksjl gS % (A) ,d ,dy ijekf.o; xSl (C) PxSl rkt (crown) uqek gksrh gSA

(B*) P4 ,d

prq"Qydh; Bksl (D) js[kh; f}ijek.oh; v.kq

19.

lQsan QkWLQksjl dks tc dkfLVd lksMk ds izcy foy;u ds lkFk xeZ fd;k tkrk gS] rks curk gS % (A*) QksLQhu (B) QkLQksfjd vEy (C) QkLQksjl vEy (D) dkbZ vfHkfØ;k ugha

20.

fuEu esa ls dkSulk QkLQksjl vfèkd LFkk;h gSA (A) yky (B) lQsn

(C*) dkyk

(D) lHkh LFkk;h

21.

lQsan QkLQksjl dk fuEu esa ls dkSulk xq.k yky QkLQksjl ls feyrk gSA (A) ;g ok;q esa LQqjnhfIr n'kkZrk gSA (B) ;g xeZ tyh; NaOH ds lkFk fØ;k djds QkWLQhu curk gSA (C) ;g dkcZu MkbZ lYQkbM esa ?kqyrk gSA (D*) ;g H2O esa ugha ?kqyrk gSA

22.

lQsan QkWLQksjl (P4) j[krk gS % (A*) N% P–P ,dy caèk (C*) bysDVªkWu ds pkj ,dkdh ;qXe

23.

QkLQksjl dk iz;ksx fdlesa gksrk gSA (A) QksVksxzkQh (B) lhesUV m+|ksx

(B) pkj P–P ,dy

caèk (D*) PPP dks.k 60º dk gksrk gSA

(C) jcj

m|ksx

(D*) ekfpl

m|ksx

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.- 9

DPP No. # 36 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

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

1.

[60, 40]

NaOH + N2O5  Product of this reaction are NaOH + N2O5 

bl vfHkfØ;k esa mRikn gS %

(A) NaNO2, H2O 2.

(3 marks, 2 min.)

(B*) NaNO3, H2O

(C) HNO3, Na2O

(D) HNO2, Na2O

In the process of extraction of silver,

Identify the complexes [P] and [Q]

flYoj ds fu"d"kZ.k izØe esa]

ladqy [P] o [Q] dks igpkfu;s \

Sol.

(A) P = [Ag(CN)4]–, Q = [Zn(CN)4]2–

(B) P = [Ag(CN)4]–, Q = [Zn(CN)6]4–

(C) P = [Ag(CN)4]3–, Q = [Zn(CN)4]2–

(D*) P = [Ag(CN)2]–, Q = [Zn(CN)4]2–

2Ag + 4CN– + H2O + 1/2 O2  2[Ag(CN)2]– + 2OH– 2[Ag(CN)2]– + Zn  [Zn(CN)4]2– + 2Au

3.

At 727 C, N2(g) + 3H2(g)  2NH3(g);  H = –123.77 KJmol–1 Substance N2 H2 NH3 Cp 3.5R 3.5R 4R The heat of formation of ammonia at in KJ mol–1 at 27 C is : (A) + 44.42 (B*) – 44.42 (C) + 88.85 727 C ij N2(g) + 3H2(g)  2NH3(g) ds

N2 H2 Cp 3.5R 3.5R 27 C ij veksfu;k ds lEHkou dh Å"ek KJ mol–1 esa (A) + 44.42 (B*) – 44.42 Using Kirchhoff's equation,

inkFkZ

Sol.

(D) – 88.85

fy,  H = –123.77 KJmol–1 NH3 4R

gS % (C) + 88.85

(D) – 88.85

fdjpkWQ lehdj.k ls H2(1000 K) = H1 (300 K) + Cp(1000 – 300) Here (;gk¡) H2(1000 K) = – 123.77 KJ mol–1 H1(300 K) = ? Cp = 2Cp(NH3) – [Cp(N2) + 3Cp(H2)] = – 6R = – 6  8.314  10–3 KJmol–1 K–1  –123.77 = H1(300 K) – 6  8.314  10–3  700  H1(300 K) = – 88.85 KJ for two moles of NH3  

NH3 ds nks eksy ds fy,H1(300 K) = – 88.85 KJ H1 = – 44.42 KJ mol–1

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

4.

How many elements would be in the IInd period of the Periodic Table if the spin quantum numbers could have the value +

1 1 , 0, – ? 2 2 1 2

1 2

;fn pØ.k DokUVe la[;k dk eku + , 0, – gS rc vkorZ lkj.kh ds IInd vkorZ esa fdrus rRo gksxs \ Sol.

(A) 18 For IInd period (IInd n = 2 hence l m

(B*) 12

(C) 10

(D) 8

vkorZ ds fy,) (n = 2 vr%½ s

0

0

+

1 1 , 0, – 2 2

1

–1

+

1 1 , 0, – 2 2

0

+

1 1 , 0, – 2 2

+1

+

1 1 , 0, – 2 2

Hence , total number of elements (total value of spin quantum number) = 12

vr% rRoksa dh dqy la[;k ¼pØ.k DokUVe la[;k dk dqy eku½ = 12 5.

1 mol CH3COOH is added in 250 g benzene. Acetic acid dimerises in benzene due to hydrogen bond. Kb of benzene is 2 K kgmol–1. The boiling point has increased by 6.4K. % dimerisation of acetic acid is : 1 eksy CH3COOH dks 250 g csathu

esa feykrs gSA gkbMªkstu ca/k ds dkj.k ,flfVd vEy cst a hu esa f}ydhd`r gks tkrk gSA csathu ds Kb dk eku 2 K kgmol gSA DoFkukad esa 6.4K dh o`f) gksrh gSA ,flfVd vEy dk % f}ydhdj.k (dimerisation) gS % –1

Sol.

(A) 50 Tb = i Kb .m Given molality (fn;k

(B*) 40

x;k gS eksyyrk) =

For dimerisation (f}ydhdj.k

(C) 30

(D) 20

1 1000 = 4m, 6.4 = i  2  4 or i = 0.8 250

ds fy,)

 –  0.8 – 1 = or  = 0.4  40% 2 2 H2S reacts with lead acetate forming a black compound which reacts with H2O2 to form another compound. The colour of the compound is : (A) pink (B) black (C) yellow (D*) white H2S, ySM ,lhVsV ds lkFk vfHkfØ;k dj ,d dkyk ;kSfxd cukrk gS tks H2O2 ds lkFk vfHkfØ;k dj ,d vU; ;kSfxd cukrk i=1–

6.

gSA ;kSfxd dk jax gS % (A) xqykch Sol.

(B) dkyk

(C) ihyk

(D*) 'osr

H2S  (CH3COO)2 Pb  PbS  2CH3COOH black

PbS + 4H2O2  PbSO 4  4H 2O white

7.

(I) [ MnCl6 ]3 – , [ FeF6 ]3 – and [ CoF6]3 – are paramagnetic having four, five and four unpaired electrons respectively. (II) Valence bond theory gives a quantitative interpretation of the thermodynamic stabilities of coordination compounds. (III) The crystal field splitting o , depends upon the field produced by the ligand and charge on the metal ion. Amongs the following correct statements are : 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.- 11

(I) [ MnCl6 ]3 – , [ FeF6 ]3 –

rFkk [ CoF6]3 – vuqpqEcdh; gksrs gaSA buesa Øe'k% pkj] ik¡p rFkk pkj v;qfXer bysDVªkWu gSaA (II) la;kstdrk ca/k fl)kUr] milgla;kstd ;kSfxdksa ds m"ekxfrdh LFkkf;Ro dh ek=kkRed O;k[;k djrh gSA (III fØLVy {ks=k foikVu o fyxs.M }kjk mRiUu {ks=k rFkk /kkrq vk;u ds vkos'k ij fuHkZj djrk gSA fuEu esa ls lgh dFku gSa % Sol.

gy .

(A) I, II (B*) I, III (C) I, II, III (D) II, III I : As all are weak field ligands therefore all will have same number of unpaired electrons as in central metal ion. II : V.B.T. does not give any interpretation about the relative thermodynamic stabilities of various complexes. This is one of the limitation of V.B.T. III : is correct statement. I : D;ksafd lHkh nqcZy {ks=k fyxsaM gS vr% buesa mrus gh v;qfXer bysDVªkWu gksaxs ftrus fd dsUnzh; /kkrq vk;u esa gSA II : V.B.T. fdlh

Hkh ladqy dh vkisf{kd Å"ekxfrdh LFkkf;Ro ds ckjs esa dqN Hkh ugha crkrh gSA ;g V.B.T. dh ,d deh

gSA III % ,d lR; dFku gSA 8.

In the synthesis of sodium carbonate by Solvay (or ammonia soda) process the recovery of ammonia is done by treating NH4Cl with Ca(OH)2. The by-product obtained in this process is (A*) CaCl2 (B) NaCl (C) NaOH (D) NaHCO3 lkWYos izØe (;k veksfu;k lksMk izØe) }kjk lksfM;e dkcksZusV ds fuekZ.k esa NH4Cl dks Ca(OH)2 ls fØ;k djkdj NH3 dks

iqu% çkIr fd;k tkrk gSA bl çØe esa çkIr lgmRikn gS & (A*) CaCl2 Sol.

(B) NaCl

(C) NaOH

(D) NaHCO3

NH3 + CO2 + H2O  NH4HCO3 NaCl + NH4HCO3  NaHCO3 + NH4Cl 150 º C 2NaHCO3    Na2CO 3 + CO2 + H2O

2NH4Cl + Ca(OH)2  2NH3 + H2O + CaCl2 (byproduct) (lg 9.

mRikn).

Which of the following statement is correct ? (A*) S–S bond is present in H2S2O6. (B) In peroxosulphuric acid (H2SO5) sulphur is in +8 oxidation state. (C) Copper powder along with Al2O3 and K2O is used as a catalyst in the preparation of NH 3 by Haber’s process. (D) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation of SO 2.

fuEu esa ls dkSulk dFku lgh gS % (A*) H2S2O6 esa S–S ca/k mifLFkr gksrk gSA (B) ijvkWDlkslY¶;wfjd vEy (H2SO5) esa lYQj +8 vkWDlhdj.k voLFkk esa gksrh gSA (C) gscj izØe ds }kjk NH3 ds fuekZ.k esa Al2O3 o K2O ds lkFk dkWij pw.kZ mRizsjd dk dk;Z djrk gSA (D) SO2 ds mRiszjdh; vkWDlhdj.k ds }kjk SO3 fuekZ.k esa ,UFksYih esa ifjorZu /kukRed gksrk gSA Sol.

In H2SO5 the oxidation number of O is +6 as two O are having peroxide bond. H2SO5 esa O dk vkWDlhdj.k vad +6 gS D;ksafd nks O ijkWDlkbM ca/k ls caf/kr gSA 10.

1 g sample of alkaline earth metal react completely with 4.08 g H2SO4 and yield an ionic product MSO 4. Then find out the atomic mass of Alkaline earth metal (M) ?

,d {kkjh; e`nk /kkrq dk 1 g çkn'kZ 4.08 g H2SO4 ds lkFk iw.kZr% vfHkd`r gksrk gS rFkk ,d vk;fud mRikn MSO4 cukrk gS rc {kkjh; e`nk /kkrq M dk ijek.kq nzO;eku crkb;sa \ Sol.

(A) 9 (B*) 24 M + H2SO4  MSO4 + H2

(C) 40

(D) 87

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.- 12

Mole of H2SO 4 mole of M = 1 1 1 4.08 = a 98 a=

98 = 24.01 4.08

This is atomic weight of M (M dk 11.

ijek.kq nzO;eku) = 24.

Pt (H2) | pH = 2 | pH = 3 | Pt (H2) 1 atm 1 atm The cell reaction for the given cell is (A) spontaneous (B*) non-spontaneous Pt (H2) | pH = 2 | pH = 3 | Pt (H2) 1 atm 1 atm

fn;s x;s lsy ds fy, lsy vfHkfØ;k gS % (A) Lor% (B*) vLor%

(C) equilibrium

(D) None of these

(C) lkE;

(D) buesa

ij

ls dksbZ ughaA

[H ]anode 0.0591 0.0591 10 –2 log10  =– log10 = – ve (non spontaneous). (vLor%) 1 1 [H ]cathode 10 – 3

Sol.

Ecell = E0cell –

12.

The edge lengths of the unit cells in terms of the radius of spheres constituting fcc,bcc and simple cubic unit cell are respectively fcc, bcc o ljy ?kuh; ,dd dksf"Bdkvksa esa dksj yEckbZ (edge length) dks xksys dh f=kT;k ds inksa esa fuEu izdkj ls fy[kk

tk ldrk gS& (A) 2r,

4r 3

,2 2r

(B)

4r 3

,2 2r, 2r

3a  4r ; fcc :

(C) 2r,2 2r,

4r 3

(D*) 2 2r,

4r ,2r 3

Sol.

scc : a = 2r ; bcc :

13.

At constant P of 0.0821 atm log V v/s log T graph is plotted for 3 samples of ideal gas as shown. Value of moles of gas in these three samples is respectively : P = 0.0821 atm ds ,d fu;r nkc ij vkn'kZ xSl ds rhu izkn'kkZs ds fy, log V v/s log T vkjs[k uhps n'kkZ;k x;k gSA

2a  4r

bu rhuksa izkn'kks esa xSl ds eksyksa ds eku Øe'k% fuEu gS : (A*) 3, 1,

(B)

1 , 1, 3 2

(C) 1,

(D)

Sol.

1 2

1 ,3 2

1 , 3, 1 2

V  cos nst(k ) T log V = log T + log k where K =

nR P

for 2 graph

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.- 13

log

nR = 0 = log 1 P

nR  1 (R = P) or n = 1 mol P for 1 graph nR = log 3 or n = 3 mol P for 3 graph log

log

nR 1 = – log 2 = log P 2 1 mol 2

n= 14.

The atomic weights of two elements A and B are 40 u and 80 u respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B : nks rRoksa A rFkk B ds ijek.kq Hkkj Øe'k% 40 u rFkk 80 u gSA ;fn A ds x g esa y ijek.kq mifLFkr gSa] rks B ds 2x g esa fdrus

ijek.kq mifLFkr gkasxs % (A)

Sol.

y 2

(B)

y 4

(C*) y

Number of moles of A =

x 40

Number of atoms of A =

x × Avogadro no. = y (say) 40

(D) 2y

40 y Or x = Avogadro no. Number of moles of B =

2x 80

Number of atoms of B =

g y-

2x 2 40 y  × Av. no = × Av. no. = y.. 80 80 Av.no. x

A dh

eksy la[;k = 40

A ds

ijek.kqvksa dh la[;k = 40 × vkoksxknzks la[;k = y (say)

;k x =

x

40 y

2x 80

B

dh eksy la[;k =

B

ds ijek.kqvksa dh la[;k

=

2x × vkoksxknzks 80

la[;k =

2  80

40 y

× vkoksxknzks

la[;k = y..

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.- 14

15.

Calculate the lattice energy of LiF(s), given that : Hsub for Lithium is 155.2 kJ mol–1

(i)

(iii) I.E of Lithium = 520 kJ mol–1 (v) Hf for LiF(s) = – 594.1 kJ mol-1. (A*) – 1011.6 kJ mol–1. (B) + 2056.6 kJ mol–1. LiF (s) tkyd ÅtkZ dh x.kuk djksA fn;k x;k gS % (i) yhfFk;e

ds fy, Hsub 155.2 kJ mol–1 gSA

(C) – 906.6 kJ mol–1.

(ii) F2 ds

(A*) – 1011.6 kJ mol–1.

(C) – 906.6 kJ mol–1.

Li(s)

(B) + 2056.6 kJ mol–1.

1 F (g) 2 2

+

H

f  

(D) – 573.6 kJ mol–1.

1 eksy dk Hdiss 75.3 kJ gSA 2

(iv) ¶yksjhu

dh bysDVªkWu cU/kqrk (E.A.) = 333 kJ mol–1 (D) – 573.6 kJ mol–1.

LiF(s)

1 Hdiss. + I.E. + E.A. + HLE 2 HLE= – 1011.6 kJ/mol Hf = Hsub. +

  16.

(iv)

1 mole of F2 (g) is 75.3 kJ 2 E.A. of Fluorine = 333 kJ mol–1

Hdiss of

dh vk;uu ÅtkZ (I.E) = 520 kJ mol–1 (v) LiF(s) ds fy, Hf = – 594.1 kJ mol-1 gSA (iii) yhfFk;e

Sol.

(ii)

Which bond angle  would result in maximum dipole moment for the triatomic molecule XY2 shown below :

uhps fn[kk;s x;s fp=k ds vuqlkj f=kijek.oh; v.kq XY2 esa cU/k dks.k dk dkSulk eku] vf/kdre f}/kqzo vk?kw.kZ nsxk%

(A*)  = 90°

(B)  = 120°

(C)  = 150°

(D)  = 180°

P 2  Q 2  2 PQ cos  ; as  increases from 90º, cos becomes – ve. So R .

Sol.

R=

17.

When one mole NO 3 is converted into 1 mole NO 2, 0.5 mole N 2 and 0.5 mole N2O respectively, it accepts x, y and z mole of electrons. Then, x , y and z are respectively : tc ,d eksy NO3– dks Øe'k% 1 eksy NO2, 0.5 eksy N2 vkSj 0.5 eksy N2O esa cnyrs gSa] rks ;g bysDVªkWuksa ds x, y vkSj

–

z eksy xzg.k (A*) 1, 5, 4 Sol.

– NO3

djrk gSA rc] x , y vkSj z Øe'k% D;k gSa % (B) 1 ,2 , 3

(C) 2 , 1 , 3

+

–

–

+

–

1 N + 3H2O. 2 2

Y = 5.

–

+

–

5 H O. 2 2

Z = 4.

+ 2H + 1e  NO2 + H2O.

NO3 + 6H + 5e 

NO3 + 5H + 4e  N2O +

(D) 2, 3, 4

X = 1.

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.- 15

18.

Two moles of HI were heated in a sealed tube at 440°C till the given equilibrium was reached. HI was found to be 20% decomposed. The equilibrium constant for dissociation is : 2H I (g) H 2 (g) + I2 (g) 440°C ij HI ds

nks eksyksa dks ,d cUn ufydk esa rc rd xeZ djrs gSa] tc rd fuEu lkE;koLFkk u vk tk;asA ;g ik;k x;k fd HI dk 20% fo?kVu gksrk gSA fo;kstu ds fy;s lkE;koLFkk fu;rkad fuEu gS % 2HI (g) H2 (g) + I2 (g) (A)

1 16

Sol. Initial

(B) 2HI 2moles

20 ×2 100 = 2 – 0.4 = 1.6

At eqm. 2 –

K=

19.

1 32

(C*)

H2 +

I2

0.2

0.2

1 64

(D)

1 128

1 [H2 ] [ I 2 ] 0.2  0.2 = = . 2 64 [HI] (1. 6 )

The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of 3% in its 0.1M solution at 250C. Given that the ionic product of water is 1014 at this temperature, what is the dissociation constant of the acid? 250C ij ,d nqcZy ,d{kkjh; dkcZfud vEy dk lksfM;e yo.k blds 0.1 M foy;u esa 3% rd tyvi?kfV~r gksrk gS] fd

bl rki ij ty dk vk;fud xq.kuQy 1014 gS] rks vEy dk fo;kstu fu;rkad D;k gS \ Sol.

(A*)  1 x 1010 h = .03 Kh = Ch2 = 9 × 10–5 Kh =

20.

Sol.

gy .

Kw Ka

= 9 × 10–5

(B)  1 x 109 C = 0.1 M



Ka =

(C) 3.33 x 109

10 14 9  10 – 5

(D) 3.33 x 1010

= 1.11 × 10–10  1 × 10–10

A complex containing K+, Pt(IV) and Cl¯ is 100% ionised giving i = 3. Thus, complex is : ,d ladqy K+, Pt(IV) rFkk Cl¯ ;qDr 100% vk;fur gksdj i = 3 nsrk gS rks ladqy gksxk % (A) K2[PtCl4] (B*) K2[PtCl6] (C) K3[PtCl5] (D) K[PtCl3] i = van’t Hoff factor = 1 + (y – 1) x (y = number of ions, x = degree of ionisation) y i = 1 + (y – 1) x (a) K2PtCl4 2K+ + PtCl42– 3 1+2=3 (b) K2PtCl6 2K+ + PtCl62– 3 3 In (a), oxidation number of Pt = 2 In (b) oxidation number of Pt = 4 i = okWVgkQ dkjd = 1 + (y – 1) x (y = vk;uksa

dh la[;k, x = vk;uu dh ek=kk)

(a) K2PtCl4 (b) K2PtCl6 (a) esa Pt dk

2K+ + PtCl42– 2K+ + PtCl62– vkWDlhdj.k vad = 2

(b) esa Pt dk

vkWDlhdj.k vad = 4

y 3 3

i = 1 + (y – 1) x 1+2=3 3

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.- 16