Jp Xii Physical&inorganic Chemistry (36) - Prev Chaps + Surface Chemistry + Inorg. Chem (2).pdf

TARGET : JEE (Main + Advanced) 2016 Course : VIJETA (JP) This DPP is to be discussed in the week (14.12.2015 to 19.12.2015)

DPP No. # 56 (JEE-MAIN) Total Marks : 54

Max. Time : 36 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.18

(3 marks, 2 min.)

[54, 36]

ANSWER KEY DPP No. # 56 (JEE-MAIN) 1.

(C)

2.

(D)

3.

(B)

4.

(A)

5.

(D)

6.

(A)

7.

(A)

8. 15.

(D) (B)

9. 16.

(A) (A)

10. 17.

(D) (C)

11. 18.

(B) (D)

12.

(D)

13.

(D)

14.

(D)

(B) 4. (B) 5. (D) (ABCD) 11.* (BD) 12. (C) (A – r), (B – s), (C – p), (D – q)

6.* 13.

(ABD) 7.* (A) 14.

DPP No. # 57 (JEE-ADVANCED) 1. 8.* 15.

(A) (B) 4

2. 9.* 16.

1.

K 2[Hg 4] detects the ion/group – K 2[Hg 4] (A)

2.

(B) (AD) 3

3. 10.* 17.

(ABC) (B)

fuEu vk;u@lewg dh igpku djrk gS –

NH 2–

(B) NO–

(C*) NH4+

(D) Cl–

The ion that cannot be precipitated by both HCl and H2S is

vk;u tks] HCl rFkk H2S nksuksa ls vo{ksfir ugha gksrk gSA Sol.

(A) Pb2+ (B) Cu+ (C) Ag+ (D*) Sn2+ (A) (B) (C) can be precipitated by HCl as well as H2S as their insoluble chlorides and sulphides respectively. (D) Sn2+ + 2HCl SnCl2 (soluble) + 2H+. It can be precipitated only by H2S as its insoluble sulphide (brown). (A) (B) (C), HCl ds 2+

(D) Sn + 2HCl 3.

lkFk&lkFk H2S }kjk Hkh Øe'k% buds DyksjkbM rFkk lYQkbM ds :i esa vo{ksfir gks ldrs gSA SnCl2 (foys;) + 2H+

A metal nitrate reacts with KI to give a black precipitate which on addition of excess of KI is converted into orange colour solution. The cation of the metal nitrate is : ,d /kkrq ukbVªsV] KI ds lkFk fØ;kdj dkyk vo{ksi nsrk gSa tks fd KI dk vkf/kD; feykus ij ukjaxh jax ds foy;u

esa ifjofrZr gks tkrk gS] /kkrq ukbVªsV dk /kuk;u fuEu gS& (A) Hg2+ Sol.

Bi(NO3)2 + K Bi 3 + K

4.

(B*) Bi3+ KNO3 + BiI3

(C) Pb2+

(D) Cu+

(black) ¼dkyk½

K [Bi 4] (orange) ¼ukjaxh½

Black precipitate of copper dissolves in : (A*) KCN solution (C) sodium hydroxide

(B) sodium sulphide solution (D) boiling dilute (M) sulphuric acid

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dkWij dk dkyk vo{ksi fuEu esa ls fdlesa foys; gS % (A*) KCN foy;u (B) lks f M;e lYQkbM foy;u (C) lksf M;e gkbMªk WDlkbM (D) ruq (M) lY¶;wf jd vEy ds lkFk mckyus ij + 8 CN–

2 [Cu(CN)4]3– + S22–

Sol.

2 CuS

5.

Except which all reagent are used for dissolving the green precipitate of Cr(OH) 3 ?

fuEufyf[kr esa ls fdlds vfrfjDr lHkh dks Cr(OH)3 ds gjs vo{ksi dks foys; djus ds fy, mi;ksx esa yk;k tkrk gS \ (A) Alkaline sodium perborate (C) Sodium hydroxide in excess

(B) Bromine water in alkaline solution (D*) Ammonium hydroxide

(A) {kkjh;

(B) {kkjh;

lksfM;e ijcksjsV (C) vkf/kD; esa NaOH Sol.

6.

foy;u esa czksehu ty (D*) veksfu;e gkbMªkWDlkbM

(A)

BO3– + 2OH–

BO33– + H2O2

(B)

2Cr3+ + 3OBr– + 10OH–

(C) (D)

Cr(OH)3 + OH– [Cr(OH)4]– Only slightly soluble in ammonia solution. dsoy

Cr ( OH )3

CrO42–

2CrO42– + 3Br– + 5H2O

veksfu;k foy;u esa vkaf'kd foy;A

In the following redox equation, xUO2+ + Cr2O2–7 + yH+ aUO2+2 + zCr3+ + bH2O the values of coefficients x, y and z are, respectively : (A*) 3, 8, 2 (B) 3, 8, 7 (C) 3, 2, 4

(D) 3, 1, 8

fuEu jsMkWDl vfHkfØ;k esa xUO2+ + Cr2O2–7 + yH+ aUO2+2 + zCr3+ + bH2O xq.kkad x, y o z ds eku Øe'k% fuEu gaS : (A*) 3, 8, 2 (B) 3, 8, 7 7.

(C) 3, 2, 4

(D) 3, 1, 8

The e/m ratio for cathode rays (A*) is constant (B) varies as the atomic number of the element forming the cathode in the discharge tube changes (C) varies as the atomic number of the gas in the discharge tube varies (D) has the smallest value when the discharge tube is filled with hydrogen. dSFkksM fdj.k ds fy, e/m vuqikr (A*) fu;r

gksrk gSa (B) ifjofrZr gksrk gSa tSls gh foltZd ufydk esa cuk;s x;s dSFkksM es mifLFkr rRo dk ijek.kq Øekad ifjofrZr gksrk gSA (C) ifjofrZr gksrk gS tSls gh foltZu ufydk esa xSl dk ijek.kq Øekad ifjofrZr gksrk gSA (D) lcls de eku gksrk gSa tc foltZu ufydk dks gkbMªkstu ds lkFk Hkjk tkrk gSA 8.

In diborane, boron involves Mkbcksjsu (A) no hybridization dksbZ (C)

sp2

hybridization

sp2

ladj.k ugha ladj.k

esa] cksjkWu esa gksrk gSa& (B) sp hybridization sp ladj.k (D*) sp3 hybridization sp3 ladj.k

Sol.

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9.

In the formation of HBr from H2 & Br2, following mechanism is observed. (A) Br2 2Br Equilibrium step (B) H2 + Br

HBr + H

Slow step

(C) H + Br2 HBr + Br Fast step Calculate the rate of reaction, if concentration of hydrogen is twice that of bromine and the rate constant is equal to 1 M–1/2 Sec –1. Concentration of bromine is 1 M. H2 rFkk Br2 ls HBr ds (A) Br2

fuekZ.k esa] fuEu fØ;kfof/k izsf{kr dh x;h & 2Br lkE; in

(B) H2 + Br

HBr + H

/khek in

(C) H + Br2

HBr + Br rhoz in vfHkfØ;k dh nj@osx ifjdfyr dhft,] ;fn H2 dh lkUnzrk czksehu dh rqyuk esa nqxquh gSa rFkk nj fu;rkad 1 M–1/2 Sec–1 ds cjkcj gSA czksehu dh lkUnzrk 1 M gSA Sol.

Sol.

10.

(A*) 2 M Sec–1 (B) 3 M Sec–1 (A) Rate = K[H2]1 [Br2]1/2 Rate = 1 [2]1 [1]1/2 = 2 M Sec–1 (A) nj = K[H2]1 [Br2]1/2 nj = 1 [2]1 [1]1/2 = 2 M Sec–1

(C) 4 M Sec–1

(D) 5 M Sec–1

Which one is incorrect statement among the following ?

d bond is present in SO 2 molecule. (A) PH5 , SCl6 and FCl3 do not exist. (B) p (C) 12 P–O bonds are present in P4O6 molecule. (D*) Bond angle in SiH4 less than that in CH 4.

fuEu esa ls dkSulk dFku xyr ugha gS \ (A) PH5 , SCl6 rFkk FCl3 dk vfLrRo ugha gksrk gSA (C) P4O6 v.kq esa 12 P–O ca/k mifLFkr gksrs gSaA Sol.

esa p d ca/k mifLFkr gksrk gSA (D*) SiH4 dk cU/k dks.k CH4 ds cU/k dks.k ls de gSA (B) SO2 v.kq

(A) PH5 does not exist as there is large difference in energies of s, p and d orbitals and hence it does not undergo sp3d hybridisation. In SCl6, smaller S cannnot accomodate six larger Cl– ions. Fluorine can not expand its octet because it does not have vacant d-orbitals. So FCl3 does not exist. (A) PH5 dk vfLrRo ugha gksrk gS D;ksafd ;gka s, p rFkk d d{kdksa dh ÅtkZ esa vUrj vf/kd gksrk gS] rFkk bl izdkj ;g sp3d

ladj.k esa Hkkx ugha ysrs gSAA SCl6 esa NksVs vkdkj dk S ijek.kq cM+s vkdkj ds N% Cl– vk;uksa dks vfHkxzfgr ugha djrk gSA ¶yksjhu dk v"Vd izlkj ugha gks ldrk gS D;ksafd ¶yksjhu esa fjDr d-d{kd ugha gksrs gSaA vr% FCl3 dk Hkh vfLrRo ugha gksrk gSA

(B)

11.

(C) P4O6

For a reaction Fe0.9O

vfHkfØ;k Fe0.9O (A)

Sol.

M 0.8

n-factor (n dkjd )= 3

Fe2O3 . What is the equivalent weight of Fe0.9O [molar mass = M] Fe2O3 (B*)

2 0.9

ds fy, Fe0.9O dk rqY;kadh Hkkj D;k gSA [eksyj nzO;eku = M] M 0.7

0.9

(C)

0.7

M 0.9

(D)

eq.wt

M 0.85

M 0.7

M 0.7

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12.

Which statement is incorrect about the following graph :

(A) It is an exothermic reaction (C) Number of transition state = 4

(B) Number of reaction intermediate = 3 (D*) None of these

fuEu vkjs[k ds ckjs esa dkSulk dFku xyr gS \

(A) ;g ,d (C) laØe.k 13.

Å"ek{ksih vfHkfØ;k gS voLFkk dh la[;k = 4

(B) vfHkfØ;k e/;orhZ dh (D*) buesa ls dksbZ ugh

la[;k = 3

200 mL of an aqueous solution of a protein contains its 1.26 g. The Osmotic pressure of this solution at 300 K is found to be 2.57 × 10–3 bar. The molar mass of protein will be (R = 0.083 L bar mol–1 K–1) : ,d izksVhu ds 200 mL tyh; foy;u esa bldk 1.26 g gSA bl foy;u dk 300 K ij ijklj.kh nkc 2.57 × 10–3 ckj ik;k

x;kA izksVhu dk eksyj nzO;eku gksxkA (R = 0.083 L bar mol–1 K–1) : Ans. Sol.

(A) 51022 g mol –1 (D) = CRT =

(B) 122044 g mol –1

(C) 31011 g mol –1

(D*) 61038 g mol –1

wt 1000 RT GMM V

2.57 × 10–3 =

1.26 1000 × 0.083 × 300 GMM 200

GMM = 61038 g 14.

Which of the following statement is INCORRECT : (A) In general, adsorption of a gas on a solid involves H < 0 and S < 0. (B) Chemisorption is much more specific than physisorption. (C) Among common protective colloids, gelatin has the least gold number and thus, maximum protective power. (D*) If critical temperature (T C) of gas increases then extent of adsorption decreases.

fuEu esa ls dkSulk dFku gS \ (A) lkekU;r;k] ,d Bksl ij xSl dks vf/k'kks"k.k ds fy, H < 0 rFkk S < 0 gksrk gSA (B) jklk;fud vf/k'kks"k.k] HkkSfrd vf/k'kks"k.k dh rqyuk esa cgqr T;knk fof'k"V gksrk gSA (C) lkekU; laj{kh dksykbMksa esa ls] ftysfVu dh Lo.kZ la[;k lcls de rFkk laj{kh {kerk lcls T;knk gksrh gSA (D*) ;fn xSl dk Økafrd rki (TC) esa o`f) gksrh gS rc vf/k'kks"k.k dh ek=kk esa deh gksrh gSA Sol.

If critical temperature (T C) of gas increases then extent of adsorption increases. ;fn xSl dk Økafrd rki (TC) esa o`f) gksrh gS rc vf/k'kks"k.k dh ek=kk esa o`f) gksrh gSA

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15.

For the gaseous equilibrium, A2 (g) + B2(g) 2AB(g), the equilibrium constant at a certain temperature is 49. If an equimolar mixture of A 2 and B2 is heated in a closed container to this temperature, the mole fraction of B2 in the equilibrium mixture is xSlh; lkE;, A2 (g) + B2(g) 2AB(g) ds fy, fuf'pr rki ij lkE; fu;rkad 49 gSA ;fn bl rki ij ,d cUn ik=k

esa A2 o B2 ds leeksyj feJ.k dks xeZ fd;k x;k rks lkE; feJ.k esa B2 dk eksy fHkUu D;k gksxkA (A) Sol.

2 9

A2 1 1–x

(B*) +

1 9

B2 1 1–x

(C)

3 8

(D)

1 8

2AB ; K = 49 0 2x

2x =7 1 x 2x = 7 – 7x 9x = 7 x=

7 9

XB2 = 16.

2/9 2

=

1 9

For the chemical equilibrium, CaCO3(s) CaO(s) + CO2(g) H = +ve Which of the following curve is correct

jklk;fud lkE; ds fy,] CaCO3(s)

CaO(s) + CO2(g) H = +ve

dkSu lk oØ lgh gS &

(A*)

Sol.

(B)

CaCO3(s)

(C)

(D)

CaO(s) + CO2(g)

Kp

PCO2

log Kp = log A –

log PCO2 = log A –

H r0 2.303 RT

H 0r l 2.303 R T

......... (i)

Graph (a) represents (i) and its slope will be used to determine the heat of the reaction. vkjs[k (a) leh- (i) dks iznf'kZr djrk gS o
50 ml 0.1 M KOH is mixed with 50 ml 0.1 M HCOOH, pH of the final solution is (Given Ka (HCOOH) = 2 × 10–4) (A) 2.35 (B) 2.5 (C*) 8.2 (D) 8.5 50 ml 0.1 M HCOOH ds lkFk 50 ml 0.1 M KOH foy;u dks feyk;k x;k gS rks vfUre foy;u (fn;k x;k (A) 2.35

Sol.

pH =

gS Ka (HCOOH) = 2 ×

dh pH gksxhA

10–4)

(B) 2.5

(C*) 8.2

(D) 8.5

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18.

In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (Molecular mass = 283) the equilibrium which sets in is AgIO3(s) Ag+(aq) + O–3(aq) If the solubility product constant Ksp of Ag O3 at a given temperature is 1.0 × 10–8, what is the mass of Ag O3 contained in 100 ml of its saturated solution?

vkaf'kd foys; izcy oS|qr&vi?kV~; AgIO3 (v.kqHkkj = 283) ds ,d lar`Ir foy;u esa fuEufyf[kr lkE; LFkkfir gksrk gS % AgIO3(s)

;fn fn, x, rki ij dk Hkkj gksxk \ Sol.

Ag+(aq) + O–3(aq) Ag O3 dk foys;rk xq.kuQy

fLFkjkad (Ksp) 1.0 × 10–8 gS] rks lar`Ir foy;u ds 100 ml esa Ag O3

(A) 1.0 × 10–7 g (B) 1.0 × 10–4 g (C) 28.3 × 10–2 g (D*) 2.83 × 10–3 g AgIO3(s) Ag+(aq) + O–3(aq) [s = Solubility] [s = foys;rk] 2 Ksp = s or s = 1.0 × 10–4 mol/lit = 1.0 × 10–4 × 283 g/lt = 2.83 × 10–3 gm/100 ml.

DPP No. # 57 (JEE-ADVANCED) Total Marks : 64

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.6 to Q.11 Comprehension ('–1' negative marking) Q.12 to Q.14 Integer type Questions ('–1' negative marking) Q.15 to Q.16 Match the Following (no negative marking) Q.17

1.

(3 (4 (3 (4 (8

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

[15, [24, [09, [08, [08,

10] 12] 06] 06] 06]

Sodium carbonate extract of a metal salt is acidified with dilute acetic acid. To this is added concentrated solution of cobalt chloride followed by the addition of potassium or ammonium chloride. On warming and keeping, a yellow precipitate is obtained. The metal anion is :

/kkrq yo.k ds lksfM;e dkcksZusV fu"d"kZ dks ruq ,flfVd vEy ds lkFk vEyhd`r~ fd;k tkrk gSa bl foy;u esa lkanz dksckYV DyksjkbM foy;u feyk;k tkrk gS rFkk blds i'pkr~ iksVsf'k;e ;k veksfu;e DyksjkbM dks feyk;k tkrk gSA bls xeZ djus rFkk j[kus ij ihyk vo{ksi izkIr gksrk gSA /kkrq _.kk;u gSA Hint.

(A*) NO2– (B) SO32– CoCl2 + 7NaNO2 + 2CH3COOH Na3[Co(NO2)6] + K+ or NH4+

(C) NO32– (D) CO32– Na3 [Co(NO2)6] + H2O + 2NaCl + 2CH3COONa + NO K3 or (NH4)3 [Co(NO2)6] yellow ppt.

2.

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm –3. The molar mass of the metal is : (NA Avogadro's constant = 6.02 1023 mol–1) ,d /kkrq fcc tkyd gSA ,dd dksf"Bdk dh dksj yEckbZ 404 pm gSA /kkrq dk ?kuRo 2.72 g cm –3 gSA /kkrq dk eksyj nzO;eku gksxk % (NA vkoksxknzks fLFkjkad = 6.02 1023 mol–1) (A) 30 g mol–1 (B*) 27 g mol–1 (C) 20 g mol–1 (D) 40 g mol–1

Sol.

d

ZM NA a3 4 M

2.72 6.02 10

M

3.

23

( 404 10 10 )3

2.72 6.02 (404 )3 4 10 7

= 26.99 = 27 gm mole–1

Which of the following statement(s) is correct ? (A) The pH of 1.0 × 10–8 M solution of HCl is 8 (B*) The conjugate base of H2PO–4 is HPO42– (C) Autoprotolysis constant of water decreases with temperature. (D) Basic strength of NaOH, KOH and CsOH can be compared in water.

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fuEu esa ls dkSuls dFku lgh gS \ (A) HCl ds 1.0 × 10–8 foy;u dh pH 8 gSA (B*) H2PO–4 dk la;qXeh{kkj HPO42– gSA (C) ty dk Lo izksVksuh vi?kV~u fu;rkad esa rkieku ds lkFk deh gksrh gSA (D) NaOH, KOH o CsOH ds {kkjh; lkeF;Z dh ty esa rqyuk dh tk ldrh gSA 4.

Three test tubes A,B,C contain Pb 2+, Hg22+ and Ag+ (but unknown). To each aqueous solution, NaOH is added in excess. Following changes occur. A : Black ppt ; B : Brown ppt ; C : W hite ppt but dissolves in excess of NaOH A,B,and C contain respectively.

rhu ij[kufy;k¡ A,B,C ftuesa Pb2+, Hg22+ rFkk Ag+ vk;u ds foy;u gS ¼ysfdu vKkr gSaA½ çR;sd tyh; foy;u esa NaOH dks vkf/kD; esa feyk;k tkrk gS] rks fuEu ifjorZu izkIr gksrk gS – A : dkyk vo{ksi ; B : Hkwjk vo{ksi ; C : lQsn vo{ksi ysfdu NaOH ds vkf/kD; esa foys; rc A,B, rFkk C ij[kufy;k¡ Øe'k% vk;u j[krh gSaA (A) Pb2+, Hg22+ , Ag+ Sol.

5.

(B*) Hg22+ , Ag+, Pb2+

(C) Ag+ , Pb2+, Hg2+

(D) Ag+ ,Hg22+, Pb22+

Hg22+ + 2NaOH

Hg2O

(black) ¼dkyk½ + 2Na+ + H 2O

2Ag+ + 2NaOH

Ag2O

(brown but turns black when dried) + 2Na+ + H 2O

2Ag+ + 2NaOH

Ag2O

¼Hkwjk ysfdu 'kq" d djus ij dkyk½ + 2Na+ + H2O

Pb2+ + 4NaOH

Na2PbO2 (soluble) ¼foy;'khy½ + 2Na+ + 2H 2O

Ammonium salts on heating with slaked lime liberates a colourless gas (X). Identify the correct statement for gas (X) and ammonium salt. (A) It turns red litmus blue and gives yellow ppt. with Na3[Co(NO2)6] (B) It turns filter paper moistened with mercurous nitrate black and gives intense blue coloured solution with CuSO4(aq) (C) It when passed through Nessler reagent produces a brown colour ppt. (D*) All of these

veksfu;e yo.k] cq>s gq, pwus ¼LysDM ykbe½ (slaked lime) ds lkFk xeZ djus ij] jaxghu xSl (X) nsrk gSA veksfu;e yo.k rFkk xSl (X) ds fy, lgh dFku dh igpku dhft,& (A) ;g yky fyVel dks uhyk dj nsrh gS rFkk Na3[Co(NO2)6] ds lkFk ihyk vo{ksi cukrh gSA (B) ;g ejD;wjl ukbVªsV ds foy;u esa Hkhxs fQYVj isij dks dkyk dj nsrh gS rFkk CuSO4(tyh;) ds lkFk xgjs uhys jax dk foy;u cukrh gSA (C) bls tc uSlyj vfHkdeZd esa ls izokfgr fd;k tkrk gS] rc Hkwjs jax dk vo{ksi mRiUu djrk gS (D*) mijksDr lHkh Sol.

(A) NH3, alkaline in nature turns red litmus blue; 3NH4Cl + Na3[Co(NO2)6] (B) 2Hg NO3 + 2NH3

(NH4)3 [Co(NO2)6]

Hg(NH2 )NO3

(yellow) + 3NaCl.

Hg + NH4NO3 ; CuSO4 + 4NH3

[Cu(NH3)4] SO4(intense blue)

black

(C) 2K2 (Hg 4) + NH3 + 3KOH

HgOHgNH2

(brown) + 7K + 2H2O

(A) NH3, izd`fr

esa {kkjh; gS vr% ;g yky fyVel dks uhyk dj nsrh gSA 3NH4Cl + Na3[Co(NO2)6] (NH4)3 [Co(NO2)6] (ihyk) + 3NaCl. (B) 2Hg NO3 + 2NH3

Hg(NH2 )NO3

Hg + NH4NO3 ; CuSO4 + 4NH3

[Cu(NH3)4] SO4(xgjk uhyk)

black

(C) 2K2 (Hg 4) + NH3 + 3KOH

HgOHgNH2

(Hkwjk) + 7K + 2H2O

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6.*

Sol.

An inorganic lewis acid [X] gives gelatinous white precipitate with NH4OH in presence of NH4Cl and its aqueous solution in water gives white precipitate with AgNO3 solution. [X] will respond to which of the following characteristics ? (A*) X fumes in moist air. (B*) X on heating with solid K2Cr2O7 and conc. H2SO4 gives deep red fumes. (C) X on addition of excess NaOH gives white precipitate. (D*) X on heating with Na2CO3 and cobalt nitrate gives a blue bead in oxidising flame. ,d vdkcZfud yqb Zl vEy [X], NH4Cl dh mifLFkfr esa] NH4OH ds lkFk lQsn ftysVhuh vo{ksi nsrk gS rFkk ty esa bldk tyh; foy;u AgNO3 foy;u ds lkFk lQsn vo{ksi nsrk gSA [X] fuEu esa ls dkSuls xq.kksa ds fy, mÙkjnk;h gS& (A*) X ueok;q esa /kw ez (fumes) cukrk gSA (B*) X dks Bks l K2Cr2O7 rFkk lkUnz H2SO4 ds lkFk xeZ djus ij xgjs yky ja x ds /kw ez cukrk gS A (C) X , NaOH ds vkf/kD; esa lQs n vo{ksi nsr k gSA (D*) X, Na2CO3 rFkk dks ckYVukbVªs V ds lkFk vkWDlhdkjd Tokyk esa xeZ djus ij uhyk eudk cukrh gS A (A, B, D) (A) AlCl3 + 3H2O

Al(OH)3 + 3HCl

(B) 4AlCl3 + 3K2Cr2O7 + 9H2SO4 + 9H2O (D) Gives blue bead of Al2O3 . CoO (A) AlCl3 + 3H2O

2Al2(SO4)3 + 3K2SO4 + 6CrO2Cl2

Al(OH)3 + 3HCl

(B) 4AlCl3 + 3K2Cr2O7 + 9H2SO4 + 9H2O (D) Al2O3 . CoO dh uhyh eudk ns rk 7.*

(fumes) (deep red or orange red vapour)

(/kw ez) 2Al2(SO4)3 + 3K2SO4 + 6CrO2Cl2

(xgjh

yky ;k ukjaxh yky ok"i )

gSA

W hich of the following statement(s) is / are correct ? (A*) From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl (B*) Thiourea test is given by NO2– and not by SO32– (C*) On boiling a solution having K +, Ca2+ and HCO 3– ions we get a precipitate of CaCO 3 (D) Ag2SO3 is insoluble in dilute HNO3 .

fuEu esa ls dkSulk okD; lgh gS \ (A*) AgCl

o AgI ds fefJr vo{ksi esa ls] veksfu;k foy;u dsoy AgCl dks foys; djrk gSA (B*) }kjk Fkk;ks;wfj;k ijh{k.k fn;k tkrk gS ysfdu SO32– }kjk ughaA (C*) o HCO3– ;qDr foy;u dks mckyus ij CaCO 3 dk vo{ksi çkIr gksrk gSA (D) Ag2SO3, ruq HNO3 esa vfoys; jgrk gSA NO2– K+, Ca2+

Hint :

Ag2SO3 + H+

2 Ag+ + SO2 + H2O

8.*

S1 : Silver iodide is fairly soluble in hypo solution. S2 : Heavy metal chlorides like AgCl, HgCl2 etc. also respond to Chromyl chloride test. S3 : Bromine reacts with K liberating violet vapours of iodine. S4 : Diphenylamine reagent test is also given by nitrites chlorates, bromates, iodates, etc. in addition to NO3– S1 : flYoj

vk;ksMkbM] gkbiks foy;u esa iw.kZ foy; gSA S2 : Hkkjh /kkrq DyksjkbM tSls AgCl, HgCl2 vkfn Hkh Øksfey DyksjkbM ifj{k.k nsrs gSA S3 : czksehu] K ls vfHkd`r gksdj vk;ksMhu dh cSaxuh ok"i mRiUu djrh gSA S4 : MkbZfQukby ,ehu vfHkdeZd ifj{k.k NO3– ds lkFk&lkFk ukbVªkbV] DyksjsV] czksesV] vk;ksMsV] }kjk Hkh fn;k tkrk gSA Hint :

(A) T T T F (B*) T F T T (C) T T F F S2 : do not respond as they are partially dissociated.

Hint :

S2 : ;g

(D) F F T T

ifj{k.k ugha nsrs gSA D;ksafd ;g vkaf'kd :i ls fo;ksftr gksrs gSA

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9.*

Which of the folllowing is correct process for the separation of given ions ? (A*) Cu2+ from the mixture of Cu2+ and Cd2+ in aqueous solution, Cu2+ + Cd2+

Add excess

pass H2S

KCN

& filter

Cu2+ in the filtrate.

(B) Cu2+ from the mixture of Cu2+ and Cd2+ in aqueous solution, Cu2+ + Cd2+

Add excess

pass H2S

KCN

& filter

Cu2+ in the precipitate.

(C) Zn2+ from the mixture of Zn2+ and Cu2+ in aqueous solution, H S

dil. HCl

Filter

2 Zn2+ + Cu2+ Zn2+ in the precipitate. 3+ 2+ 3+ (D*) Fe from the mixture of Fe and Fe in aqueous solution,

Fe2+ + Fe3+

NH4Cl

NH3 solution and filter

Fe3+ in the precipiate.

fn;s x;s vk;uksa ds i`FkDdj.k ds fy,] fuEu esa ls dkSulk @ dkSuls izØe lgh gS \ (A*) tyh; foy;u esa Cu2+ rFkk Cd2+ ds feJ.k ls] Cu2+ dk i`FkDdj.k] KCN

Cu2+ + Cd2+ (B) tyh;

foy;u esa

Cu2+

rFkk

ds feJ.k ls]

KCN

Cu2+ + Cd2+ (C) tyh;

Cd2+

H2S

Cu2+

Nfu=k esa Cu2+ dk i`FkDdj.k]

H2S

vo{ksi esa Cu2+

foy;u esa Zn2+ rFkk Cu2+ ds feJ.k ls] Zn2+ dk i`FkDdj.k]

Zn2+ + Cu2+

H2S

Fe2+ + Fe3+

NH4Cl

HCl

vo{ksi esa Zn2+ (D*) tyh; foy;u esa Fe2+ rFkk Fe3+ ds feJ.k ls] Fe3+ dk i`FkDdj.k]

10.*

NH3

vo{ksi esa Fe3+

W hich of the following statement(s) is/are true. (A*) Only strontium sulphate being insoluble is precipitated by the addition of ammonium sulphate ; CaSO4 dissolves in (NH 4) 2 SO4 forming soluble complex (B*) Barium chromate is insoluble in dil. acetic acid (C*) Cr(OH) 3 is soluble in NaOH and Br 2 water while Fe(OH) 3 is insoluble (D*) Cu and Cd separation is based upon the fact that in presence of KCN, only Cd is precipitated as sulphide on passing H 2S

fuEu esa ls dkSulk@dkSuls dFku lR; gS \ (A*) veksfu;e lYQsV feykus ij] dsoy LVªkaf'k;e lYQsV vfoys; gksus ds dkj.k vo{ksfir gksrk gS ( CaSO4, veks fu;e lYQsV ds lkFk ladqy fuekZ.k dj foys ; gks tkrk gS A (B*) ruq ,lhfVd vEy esa csjh;e ØksesV v?kqyu'khy gSA (C*) Cr(OH) 3 ; NaOH rFkk Br 2 ty es a foys ; gS A tcfd Fe(OH) 3 vfoys; gS A (D*) Cu rFkk Cd dk i`FkDdj.k bl rF; ij vk/kkfjr gS fd KCN dh mifLFkfr es a H 2S xSl izo kfgr djus ij dsoy Cd dk CdS ds :i esa vo{ksi.k gksrk gSA Ans.

ABC&D

Sol.

(A) CaSO4 + (NH4)2SO4 (NH 4) 2 [Ca(SO 4) 2] (soluble) (B) BaCrO4 is insoluble in dil. acetic acid & thus gets ppt. with K 2CrO 4. (C) Cr(OH) 3 is soluble in NaOH & Br 2 water forming sodium chromate while Fe(OH) 3 is insoluble. (D) [Cu(CN) 4] 3– more stable than [Cd(CN) 4]2– ; So[Cd(CN) 4]2– gives yellow ppt. with H 2S. (A) CaSO4 + (NH4) 2SO4 (NH4)2 [Ca(SO4)2] ( foys;'khy ) (B) BaCrO 4 ruq ,lhfVd vEy esa v?kq yu'khy gS rFkk blfy, K 2CrO4 ds lkFk vo{ks i (C) Cr(OH) 3 , NaOH rFkk Br 2 ty ds lkFk lksf M;e Øks esV cukdj foys;'khy gks

cukrk gSA tkrk gSA tcfd Fe(OH)3

v?kqyu'khy jgrk gSA (D) [Cu(CN) 4]3– , [Cd(CN) 4]2–

dh rqyuk esa vf/kd LFkk;h gS bl izdkj [Cd(CN)4]2–, H2S ds lkFk ihyk vo{ksi nsrk

gSA

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11.*

An aqueous solution of ZnCl2 with conc. H2SO4 is electrolysed using zinc electrodes at anode and cathode then which of following options are correct ? (A) Cl2 gas can evolve at anode due to over voltage conditions (B*) H2 gas evolve at cathode so pH of solution get increases (C) Zinc will oxidise at anode but conc. of Zn+2 of solution remain constant. (D*) conc. of Zn+2 in electrolyte increases but conc. of anion remain same lkUnz H2SO4 ds lkFk ZnCl2 ds tyh; foy;u dks ,uksM rFkk dSFkksM ij ftad bysDVªkWM dk iz;ksx djds fo|qr vi?kfVr fd;k

tkrk gS rc fuEu esa ls dkSuls fodYi lgh gSa \ (A) Cl2 xSl vf/kd oksYVst ifjfLFkfr;ks ds dkj.k fu””"dkflr gks ldrh gSA (B*) H2 xSl dSFkksM ij fu””"dkflr gksrh gS blfy, foy;u dh pH c<+rh gSA (C) ftad ,uksM ij vkWDlhd`‘r gksxh ysfdu foy;u ds Zn+2 dh lkUnzrk fu;r jgrh gSA (D*) fo|qr vi?kV; esa Zn+2 dh lkUnzrk c
Electrolyte ZnCl2 and conc. H2SO4 electrode Zn rod Zn+2 + 2e–

Anode reaction Zn

Cathode reaction 2H+ + 2e– +2

H2

–

So conc. of [Zn ]

, conc. of Cl and SO

2– 4

remain same and pH

fo|qr vi?kV~; ZnCl2 rFkk lkUnz H2SO4 bysDVªkWM Zn NM+ gSA ,uksM vfHkfØ;k Zn Zn+2 + 2e– dSFkksM vfHkfØ;k 2H+ + 2e–

H2

blfy;s [Zn+2] dh lkUnzrk

, Cl–

rFkk SO42– dh lkUnzrk leku jgrh gS rFkk pH

Comprehension # (Q.12 to Q.14) # (Q.12 Q.14) (A) + NaCl

(B)

(White ppt.)

(B) + KI

(C)

(C)

(D) (Colourless) + (E)

+ KI

(C)

(Green ppt)

Boiling with water

F

+ (E)

12.

Compound (A) and (B) are respectively : (A) AgNO3 and AgCl (B) Pb(NO3)2 and PbCl2 (C*) Hg2(NO3)2 and Hg2Cl2 (D) Cu2(NO3)2 and CuCl ;kSfxd (A) rFkk (B) +Øe'k% fuEufyf[kr esa ls dkSu gS % (A) AgNO3 rFkk AgCl (B) Pb(NO3)2 rFkk PbCl2 (C*) Hg2(NO3)2 rFkk Hg2Cl2 (D) Cu2(NO3)2 rFkk CuCl

13.

Compound (E) is : ;kSfxd (E) fuEufyf[kr (A*) Hg

esa ls dkSu gS % (B) Ag2O

(C) HgCl2

(D) AgCl

14.

When compound A reacts with Na2CrO4 solution, the colour of the compound formed is almost similar to tc ;kSfxd A , Na2CrO4 foy;u ds lkFk fØ;k djrk gS rc cuus okys ;kSfxd dk jax fuEu esa ls fdlds leku gSA (A) Hg (B*) Hgl2 (C) PbCrO4 (D) BaCrO4

Sol.

Hg2(NO3)2 + 2NaCl

Hg2Cl2 (White ppt) (B) + 2NaNO3

Hg2Cl2 (B) + 2KI

Hg2l2 (Green ppt) (C) + 2KCl

Hg2l2 (C) + 2 KI (excess) Hg2l2 (C)

K2[Hgl4] (D) + (Hg)

Boiling with water

Hg2+2 + CrO42–

HgI2

Hg2CrO4

(E)

(F) red . + Hg (E) black.

red.

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15.

A solution of Hg2+ ion on treatment with a solution of cobalt(II) thiocyanate gives rise to a deep blue crystalline precipitate. The coordination number of mercury in the deep blue coloured compound is. tc Hg2+ vk;u ;qDr foy;u dh vfHkfØ;k dksckYV (II) Fkk;kslk;usV foy;u ds lkFk djokrs gSa] rks ,d xgjs uhys jax dk

fØLVyh; vo{ksi izkIr gksrk gSA bl xgjs uhys jax ds ;kSfxd esa edZjh dh milgla;kstu la[;k@leUo;oh la[;k fdruh gksxh\ Ans. Sol.

4 Hg2+ + Co(SCN)2 Hg2+ + Co(SCN)2

16.

Ans.

How many of the following reagents are able to bring about the reduction of copper(II) salt into copper (I) salt in aqueous solution ? KCN, NaOH, NH3, KI, K4[Fe(CN)6], KSCN. fuEu esa ls fdrus vfHkdeZd tyh; foy;u esa dkWij(II) yo.k dks dkWij(I) esa vipf;r dj ldrs gS \ KCN, NaOH, NH3, KI, K4[Fe(CN)6], KSCN. 3

Sol.

Cu + 2CN

2+

2+

–

Cu(CN)2

–

2Cu + 5I 2+

Co[Hg(SCN)4] Co[Hg(SCN)4]

2CuI –

Cu + 2SCN 17.

+ I3

;

(deep blue) (xgjk uhyk)

2Cu(CN)2

2CuCN

+ (CN)2

–

Cu(SCN)2

;

2Cu (SCN)2

Match the following : Column A (A) Na + Cl2 (water) + CCl 4 - shake (B) CH 3COO¯ + FeCl3 + H 2O - boil (C) CoCl2 + KNO2 + CH 3COOH - warm (D) FeCl3 + NO2– + CH 3COOH + CS(NH2)2

2CuSCN

+ (SCN)2

Column B (p) yellow ppt (q) deep red colouration (r) purple colouration (s) red brown ppt (t) green colouration

fuEu dks lqesfyr dhft;s& A (A) Na + Cl2 ( ty) + CCl4 - fgykus ij (B) CH 3COO¯ + FeCl3 + H 2O - mckyuk (C) CoCl2 + KNO2 + CH 3COOH - xeZ djuk (D) FeCl3 + NO2– + CH 3COOH + CS(NH2)2 Ans.

(A – r), (B – s), (C – p), (D – q)

Sol.

(A) I¯ + Cl2 2Cl¯ + (B) 2FeCl3 + CH 3COO¯ Fe(CH 3 COO) 3 + 2H 2O

2

jax½

(purple colour in CCl4) Fe (CH3COO) 3 + 3Cl¯

shake

boil

(C) CoCl2 + 6KNO2 + CH 3COOH (D) FeCl3 + 3HSCN

B (p) ihyk vo{ksi (q) xgjk yky ja x (r) cSa xuh jax ¼tkewu h (s) yky Hkwj k ja x (t) gjk jax

CH3 COO (OH) 2 Fe warm

( Red brown ppt) + 2CH 3COOH

K 3 [Co(NO 2)6]

(yellow) + 2KCl + CH 3COOK

Fe (SCN) 3 (blood red colouration) + 3HCl

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TARGET : JEE (Main + Advanced) 2016 Course : VIJETA (JP) This DPP is to be discussed in the week (14.12.2015 to 19.12.2015)

DPP No. # 58 (JEE-ADVANCED) Total Marks : 66

Max. Time : 44 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.11 Multiple choice objective ('–1' negative marking) Q.12 to Q.13 Comprehension ('–1' negative marking) Q.14 to Q.16 Integer type Questions ('–1' negative marking) Q.17 to Q.18 Match the Following (no negative marking) Q.19

(3 (4 (3 (4 (8

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

[33, [08, [09, [08, [08,

22] 04] 06] 06] 06]

ANSWER KEY DPP No. # 58 (JEE-ADVANCED) 1. 8. 15.

(C) (C) (D)

1.

H 2S would separate the following when pH of the solution will be less than 7. tc foy;u dh pH, 7 ls de gks rc H2S xSl fuEu esa ls dkSuls vk;uksa dks i`Fkd djsxh \ (A) Zn2+, Co2+ (B) Cu2+, Cd2+ (C*) Cu2+, Cr3+ (D) Cu2+, As 3+ 2+ 3+ Cu – black ppt ; Cr no ppt remains in solution. Cu2+ – dkyk vo{ksi ; Cr 3+ foy;u esa dksbZ vo{ksi 'ks"k ugha jgrk gSA

Sol.

2.

2. 9. 16.

(C) (A) (C)

3. 10. 17.

(B) (D) 1.

4. 11. 18.

(C) (B) 3.

5. 12.* 19.

(B) 6. (A) 7. (ABCD) 13.* (BD) 14. (A - p,s,t ; B - q ; C - r ; D - p,t)

(D) (B)

Among the following, the coloured commound is :

fuEufyf[kr ;kSfxdksa esa jaxhu ;kSfxd gS % Sol.

3.

(A) Cu2Cl2 (B) K3 [Cu(CN)4] (C*) CuF2 (D) [Cu(CH3NH2)4]BF4 CuF2 contains Cu2+, having d9 configuration, therefore, there is one unpaired electron which undergoes d–d transition in visible region. CuF2 in crystalline form is blue in colour. CuF2 ; Cu2+ j[krk gS] ftldk bysDVªkWfud foU;kl d9 gksrk gS] vr% ;gk¡ ij ,d v;qfXer bysDVªkWu gS tks fd n`'; {ks=k esa d–d laØe.k djrk gSA CuF2 fØLVyh; voLFkk esa uhys jax dk gksrk gSA The increasing order of paramagnetism of :

vuqpqEcdRo dk c<+rk gqvk Øe gS %

Sol. Sol.

4.

(I) MnSO4.4H2O (II) FeSO4.7H2O (III) NiSO4.6H2O (IV) CuSO4.5H2O (A) I < II < III < IV (B*) IV < III < II < I (C) III < IV < II < I Mn2+ n = 5 ; Fe2+ n = 4 ; Ni2+ n = 2 and Cu2+ n = 1 n = number of unpaired electron and n paramagnetism. Mn2+ n = 5 ; Fe2+ n = 4 ; Ni2+ n = 2 rFkk Cu2+ n = 1 n = v;qfXer bySDVªkWuksa dh la[;k rFkk n vuqpqEcdRo

(D) III < IV < I < II

(I) V2O5, Cr2O3 are amphoteric oxides. (II) Interstitial compounds are very reactive (III) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine. Correct statements amongs the following are(A) I, II (B) II, III (C*) I, III (D) I, II

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(I) V2O5, Cr2O3 mHk;/kehZ vkWDlkbM gksrs gSaA (II) vUrjkdk'kh ;kSfxd cgqr fØ;k'khy gksrs gSA (III) eSaxuht viuh mPp vkWDlhdj.k voLFkkvksa esa

vkWDlhtu vkSj ¶yksjhu ds lkFk LFkk;h ;kSfxd cukrk gSA

fuEu esa ls lgh dFku gS& Sol.

(A) I, II (B) II, III Interstitial compounds are chemically inert

(C*) I, III

(D) I, II

vUrjkdk'kh ;kSfxd jklk;fud :i ls vfØ; gksrs gSA 5.

Sol.

Na2S2O3. 5H2O Sodium thiosulphate is used in photography to : (A) remove reduced silver (B*) remove undecomposed AgBr as soluble silver thiosulphate complex (C) convert the metallic silver to silver salt (D) reduce the silver bromide grains to metallic silver QksVksxzkQh esa lksfM;e Fkk;kslYQsV Na2S2O3. 5H2O dk iz;ksx djrs gS % (A) vipf;r flYoj dks gVkus ds fy,A (B*) vfo?kfVr AgBr dks foy;'khy flYoj Fkk;kslYQsV ladqy ds :i esa gVkus ds (C) /kkfRod flYoj dks flYoj yo.k esa :ikUrfjr djus esa (D) flYoj czksekbM d.kksa dks /kkfRod flYoj esa vipf;r djus ds fy,A 2Na2S2O3 +

fy,A

Na3 [Ag(S2O3)2] + NaBr

AgBr Un exp osed

This property is used for fixing in photography. Sol.

2Na2S2O3 + AgBr

Na3 [Ag(S2O3)2] + NaBr

;g xq.k QksVksxzkQh esa LFkk;hdj.k ds fy, iz;qDr gksrk gSA 6.

Ferrous sulphate on heating gives : (A*) SO2 and SO3 (B) SO2 only

(C) SO3 only

(D) S

Qsjl lYQsV xeZ djus ij nsrk gS % (A*) SO2 o SO3 (B) dsoy SO2

(C) dsoy SO3

(D) S

Heat

Fe2O3 + SO3 + SO2.

Sol.

FeSO4

7.

The number of unpaired electrons in d6, low spin, octahedral complex is : d6, fuEu pØ.k] v"VQydh; ladqy esa v;qfXer bysDVªkWuksa dh la[;k gSSA (A) 4 (B) 2 (C) 1 (D*) 0 In d6, 'low spin' octahedral complex all electron will be paired because of higher CFSE. fuEu pØ.k v"VQydh; ladqy ftlesa bysDVªkWu foU;kl d6 gksrk gS] lkjs bysDVªkWu ;qfXer gksrs gS D;ksafd CFSE dk

Sol.

eku

mPp gksrk gSA 8.

Sol.

9.

Sol.

The yellow colour solution of Na2CrO4 changes to orange red on adding of H2SO4 due to the formation of : Na2CrO4 dk ihys jax dk foy;u H2SO4 dks feykus ij fuEu ds fuekZ.k ds dkj.k ukjaxh yky jax esa cny tkrk gSA (A) CrO5 (B) CrO3 (C*) Na2Cr2O7 (D) Na3CrO8. + 2– 2– + 2– CO2 + H2O H2CO3 2H + CO3 ; CrO4 + 2H Cr2O7 (orange red) + H 2O CO2 + H2O H2CO3 2H+ + CO32– ; CrO42– + 2H+ Cr2O72– (ukjaxh yky) + H 2O Large quantity of KMnO 4, on treatment with cold conc. H2SO4 forms a compound (X) which decomposed explosively, on heating forming (Y). The (X) and (Y) respectively are KMnO4 dh vf/kd ek=kk] B.Ms lkanz H2SO4 ds lkFk mipkfjr djus ij] ,d ;kSfxd (X) nsrh gS tks fd xeZ djus ij foLQksV ds lkFk fo;ksftr gksdj (Y) cukrk gSA (X) rFkk (Y) Øe'k% gS % (A*) Mn2O7, MnO2 (B) Mn2O7, Mn2O3 (C) MnSO4, Mn2O3 (D) Mn2O3, MnO2 (A).

2KMnO 4

H2 SO4

Mn2 O7

K 2SO 4

H2O

;

2Mn2 O7

4MnO2 3O2

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10.

Consider the given sequence of reaction and identify P and Q. Na2B4O7 . 10H2O

P + NaBO2 + H2O

Cr2 O3

Q (Green coloured)

vfHkfØ;kvksa ds fn;s x;s fuEu Øeksa ij fopkj dhft, rFkk P rFkk Q dh igpku dhft,A Na2B4O7 . 10H2O

P + NaBO2 + H2O

(A) P = Na3BO3 , Q = Cr(BO2)3 (C) P = Na2B4O7 , Q = Cr(BO2)3 11.

Cr2O3

Q (gjk

jax)

(B) P = B2O3 , Q = CrBO3 (D*) P = B2O3 , Q = Cr(BO2)3

When CS2 layer containing both Br2 and I2 is shaken with excess of Cl2 water, the violet colour due to I2 disappears and a pale yellow colour appears in the solution. The disappearance of violet colour and appearance of pale yellow colour is due to the formation of : – (A) I3 and Br2 respectively. (B*) HIO3 and BrCl respectively. – – (C) ICl and BrCl respectively. (D) I and Br respectively. tc CS2 ijr Br2 rFkk I2 nksuksa ;qDr gks rFkk Cl2 ty ds vkf/kD; esa ijr dks fgyk;k tkrk gS] rks foy;u esa I2 ds dkj.k mRiUu

cSaxuh jax foyqIr gks tkrk gS rFkk ,d gYdk ihyk jax foy;u esa fn[kkbZ nsrk gSA cSaxuh jax dk foyqIr gksuk rFkk gYds ihyk jax dk fn[kkbZ nsuk fuEu ds fuekZ.k ds dkj.k gksrk gS & – – – (A) Øe'k% I3 rFkk Br2 (B*) Øe'k% HIO3 rFkk BrCl (C) Øe'k% ICl rFkk BrCl (D) Øe'k% I rFkk Br Sol.

5Cl2 + I2 + 6H2O

Sol.

Br2 + Cl2 2BrCl (pale yellow) 5Cl2 + I2 + 6H2O 2HIO3 (jaxghu) + 10HCl Br2 + Cl2

12.*

2BrCl (gYdk

ihyk)

Hg22+ and Hg2+ salt can be distinguished by : (A*) Aq NH 3 (B*) KI (not in excess) (C*) SnCl2 Hg22+ rFkk Hg2+ dks (A*) tyh; NH3

Sol.

2HIO3 (colourless) + 10HCl

(D*) Dilute HCl

fuEu esa ls fdlds }kjk foHksfnr fd;k tk ldrk gS % (B*) KI ( vkf/kD; esa ugha) (C*) SnCl2 (D*) ruq HCl

(A) Hg2Cl2 + NH 3

Hg + Hg(NH 2) Cl + NH 4Cl |_______ Black ________|

HgCl2 + 2NH 3 Hg(NH 2)Cl (White) + NH 4Cl 2+ (B) Hg2 gives green precipitate whereas Hg2+ gives red precipitate (C) Hg22+ gives directly black precipitate. Hg 2Cl2 + SnCl2 Hg2+ gives first white silky precipitate and then black. HgCl2 + SnCl2

Hg2Cl2 Hg2Cl2

(A) Hg2Cl2 + NH 3

Hg + Hg(NH 2) Cl

HgCl2 + 2NH 3 (B) Hg22+ gjk vo{ksi (C) Hg22+ lh/ks Hg2+ igys

(Black) + SnCl 4

(White) + SnCl4 ; Hg2Cl2 + SnCl2

(D) Hg22+ + 2 Cl–

|_______

Hg

Hg

(Black) + SnCl 4

(W hite) + NH 4Cl

dkyk________|

Hg(NH 2)Cl ( lQsn ) + NH 4Cl nsrk gS tcfd Hg2+ yky vo{ksi nsrk gSA

gh dkyk vo{ksi nsrk gSA Hg2Cl2 + SnCl2 2Hg ( dkyk) + SnCl4 flYd leku lQsn vo{ksi rFkk fQj dkyk vo{ksi nsrk gSA

HgCl2 + SnCl2 (D) Hg22+ + 2 Cl–

Hg2Cl2

( lQsn) + SnCl4 ; Hg2Cl2 + SnCl 2

Hg2Cl2

( lQsn)

2Hg

( dkyk) + SnCl 4

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13.*

W hich of the following statement(s) is/are incorrect with reference to ammonium ion ? (A) Alkaline solution of ammonium salt on heating with sodium hydroxide liberates a gas which produces dense white fumes with dilute HCl. (B*) It with potassium cobaltinitrite solution gives a yellow precipitate. (C) It with alkaline K 2Hg 4 gives brown precipitate. (D*) It with chloroplatinic acid gives a red precipitate.

veksfu;e vk;u ds lanHkZ esa fuEu esa ls dkSulk @ dkSuls dFku lgh ugha gSa \ (A) veks fu;e yo.k dk {kkjh; foy;u dks lksfM;e gkbMªkW DlkbM ds lkFk xeZ djus ij ,d xS l eqDr gks rh gS tks fd ruq HCl ds lkFk lQsn xgjs /kwez cukrh gSA (B*) ;g] iks Vs f 'k;e dks ckYVh ukbVª k bV ds foy;u ds lkFk ihyk vo{ksi ns rs gSA (C) ;g] {kkjh; K 2Hg 4 ds lkFk Hkw jk vo{ksi nsr s gSA (D*) ;g] Dyks jks I ysfVfud vEy ds lkFk yky vo{ks i ns rs gS A Sol.

NH3 + HCl

NH 4Cl (white)

Potassium cobaltinitrite is a yellow coloured compound insoluble in water.

K 2Hg 4 + 2NH3

2NH 4Cl + H2PtCl6

(NH 4) 2 PtCl6

(yellow) + 2HCl

NH 3 + HCl

NH 4Cl ( lQsn) iksVsf'k;e dksckYVh ukbVªkbV ,d ihys jax dk ;kSfxd gS] tks fd ty esa vfoys; gSA K 2Hg 4 + 2NH3

2NH4Cl + H 2PtCl6

(NH4) 2 PtCl6

( ihyk) + 2HCl

Comprehension # Read the following comprehension carefully and answer the following questions. (Q.14 to Q.16)

(a) (b) (c) (d)

In qualitative analysis, group includes Fe3+, Al3+ and Cr3+ . The group reagent is ammonium hydroxide in presence of ammonium chloride. NH 4Cl is added to suppress the ionisation of NH4OH so that only the group radicals are precipitated as their insoluble hydroxide and not the ivth and vth group radicals as the solubility products of their hydroxides are much higher. Before adding group reagent to the filterate of II group, one drop of concentrated HNO3 (oxidising agent) is added and the content is boiled for 2-3 minutes and then cool because the solubility product of Fe(OH)2 is very high as compared to Fe(OH)3. When a light bluish green crystalline compound containing cations of III group and zero group is analysed, it responds to following tests /reactions. Its aqeous solution gives a reddish brown precipitate with alkaline solution of potassium tetraiodomercurate (II) Its aqueous solution after boiling with conc. HNO3 gives a reddish brown ppt. with sodium hydroxide. Solution of reddish brown ppt. in dil. HCl gives blood red colouration with ammonium sulphocyanide and prussian blue with potassium ferrocyanide. Its very dilute solution gives white ppt. with (CH3COO)2 Pb solution and white ppt. is only soluble in hot amonium acetate.

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# (Q.14 to Q.16)

(a) (b) (c) (d) 14.

xq.kkRed fo'kys"k.k esa] lewg esa Fe3+, Al3+ rFkk Cr3+ vkrs gSA bldk lewg vfHkdeZd NH4Cl dh mifLFkfr esa NH4OH foy;u gSA NH4Cl dks NH4OH ¼tyh;½ ds vk;uu dks ean (suppress) djus ds fy, Mkyk tkrk gS rkfd dsoy lewg ds ewyd gh muds vfoys;'khy gkbMªkWDlkbM ds :i esa vo{ksfir gks lds rFkk ivth vkSj vth lewg ds ewyd ugha] D;ksafd bu lewg ds ewydksa ds gkbMªkWDlkbM rqyukRed :i ls vf/kd foys;rk xq.kuQy j[krs gSaA II lewg ds Nfu=k esa lewg vfHkdeZd feykus ls igys ,d cawn lkUnz HNO3 (vkWDlhdkjd inkFkZ) dh feyk;h tkrh gSa rFkk vo;oksa dks 23 feuV rd mckyk tkrk gS rFkk fQj B.Mk fd;k tkrk gS D;ks af d Fe(OH)2 dk foys;rk xq.kuQy Fe(OH)3 dh rqyuk esa cgqr vf/kd gksrk gSA tc ,d gYds uhys gjs fØLVyh; ;kSfxd ftlesa III lewg rFkk 'kwU; lewg ds /kuk;u mifLFkr gSa] dk fo'ys" k.k fd;k tkrk gS rks og fuEu vfHkfØ;k rFkk ijh{k.k nsrs gaS& bldk tyh; foy;u] iksVsf'k;e VsVªk vk;ksMkseDZ;wjsV (II) ds {kkjh; foy;u ds lkFk xgjk yky Hkwjk vo{ksi nsrk gSA bldk tyh; foy;u lkUnz HNO3 ds lkFk xeZ djus ds i'pkr] NaOH ds lkFk xgjk yky Hkwjk vo{ksi nsrk gSA yky Hkwjs vo{ksi dk ruq HCl esa foy;u] veksfu;e lYQkslk;ukbM ds lkFk jDryky jax rFkk iksVsf'k;e QsjkslkbukbM ds lkFk izqf'k;u Cyw jax nsrk gSA bldk cgqr ruq foy;u (CH3COO)2 Pb ds lkFk lQsn vo{ksi nsrk gS rFkk ;g lQsn vo{ksi dsoy xeZ veksfu;e ,lhVsV esa ?kqyu'khy gSA Why one ml conc. HNO3 is added to IInd group filterate before proceeding for IIIrd group radicals ? (A) Because it oxidises the dissolved H2S to colloidal sulphur (B) Because it oxidises Fe2+ to Fe3+ so that Fe3+ can be completely precipitated as its hydroxide (C) Because precipitation of cations belonging to IIIrd group by group reagent takes place only in presence of conc. HNO3 (D) None of these IIIrd lewg ds ewydka s ij tkus ls igys IInd lewg ds Nfu=k esa ,d feyhyhVj lkUnz HNO3 D;ksa (A) D;ksaf d ;g ?kq f yr H2S dks dksykW bMy lYQj es a vkWDlhd`r dj nsrk gS A (B) D;ks af d ;g Fe2+ dks Fe3+ es a vkWDlhd`r dj ns rk gS rkfd Fe3+ blds gkbMªk W DlkbM ds :i

feyk;k tkrk gSA esa iw.kZr% vo{ksfir gks

tkrk gSA (C) D;ksaf d III lewg ls lEcfU/kr /kuk;uksa dk vo{ks i.k] buds lew g vfHkdeZ d }kjk dsoy lkUnz HNO3 dh mifLFkfr esa gksrk gSA (D) buesa ls dks bZ ugha A Sol.

(B) (B) As complete precipitation of Fe(OH)2 (Ksp very high) does not take place in presence of NH 4Cl so it is oxidised to Fe3+ by HNO3 and Fe(OH)3 has lower ksp which is completely separated as reddish brown ppt. (B) Fe(OH)2 dk iw .kZ vo{ks i.k (Ksp cgqr vf/kd gS) NH4Cl dh mifLFkfr esa ugha gks rk gS A vr% Fe2+ vk;u HNO3 }kjk Fe+3 esa vkWDlhd`r gks tkrk gSA Fe(OH)3 dk Ksp de gksrk gSA vr% yky Hkwjs vo{ksi ds :i esa iw.kZr% i`Fkd gks tkrk gSA

15.

Identifies the correct statement : (A) NH4Cl is added along with NH4OH so that only IIIrd group cations can be precipitated as their hydroxides (B) In place of NH4Cl, (NH4)2 SO4 can not be used as barium (vth group radical) will also be precipitated as BaSO4 along with Al+3 , Fe+3 & Cr+3 (C) Aqueous solution of ammonium sulphate also produces white ppt. with BaCl2 solution which is insoluble in conc. HCl. (D*) All of these

lgh dFku igpkfu;s& (A) NH4Cl dks NH4OH ds lkFk blfy, feyk;k tkrk gS ftlls fd IIIrd lew g ds /kuk;u buds gkbMª k WDlkbM ds :i esa vo{ksfir fd;s tk ldrs gSA (B) NH4Cl ds LFkku ij (NH4)2 SO4 dk mi;ksx ugha dj ldrs gSa D;ksafd Al+3 , Fe+3 rFkk Cr+3 ds lkFk csf j;e (vth lewg ewyd½ Hkh BaSO4 ds :i esa buds lkFk vo{ksfir gks tk;sxkA (C) veksfu;e lYQsV dk tyh; foy;u BaCl2 ds foy;u ds lkFk Hkh lQsn vo{ksi mRiUu djrk gS] tks fd lkUnz HCl esa vfoys; gSA (D*) mijksDr lHkhA 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]

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Sol.

(A) As Ksp of hydroxides of Al3+, Fe3+ & Cr3+ are low and NH4Cl suppresses the ionisation of NH4OH (B) (NH4)2SO4 + Ba2+

BaSO4

(C) SO42– + Ba2+

(white) + 2NH4+

(white) as it contains SO42– as anion.

BaSO4

(A) Al3+, Fe3+, Cr3+ dk Ksp de (B) (NH4)2SO4 + Ba

gSa rFkk NH4Cl , NH4OH ds vk;uu dks eUn ¼ suppress ½ djrk gSA BaSO4 (lQs n) + 2NH4+ BaSO4 (lQsn) D;ksaf d ;g SO42– dks _.kk;u ds :i esa j[krk gSA

2+

(C) SO42– + Ba2+ 16.

Sol. 17.

The reddish brown precipitate formed in (a) is of (A) Mercury amido iodide (B) Mercury iodide (C*) Oxydimercuric amido iodide (D) Mercury amido iodide and mercury (a) esa cuk yky Hkwj k vo{ks i fuEUk dk curk gS& (A) edZ jh ,ekbMks vk;ksM kbM (B) edZ jh vk;ks MkbM (C*) vkWDlh MkbZ eD;wZf jd vehMks vk;ksM kbM (D) edZj h ,ekbMks vk;ksM kbM rFkk edZj h 2K2(Hg 4) + NH3 + 3KOH HgOHgNH2 + 7KI + 2H2O. 2+

–

2+

–

Cu (aq) + CN (aq) complex – Aqueous solution of CN is taken in excess. What is the oxidation state of copper in the complex.

Ans.

Cu (aq) + CN (aq) – CN ds tyh; foy;u dks 1.

Sol.

Cu (aq) + 2CN (aq)

Cu(CN)2

2Cu(CN)2

(white) + (CN)2

2+

CuCN 18.

ladqy vkf/kD; esa ysus ij cuus okys ladqy esa dkWij dh vkWDlhdj.k voLFkk gksxhA

–

2CuCN

(yellow)

1

–

[Cu(CN)4 ] 3

+ 3CN (aq)

In how many of the following reactions, one of the products is obtained as a yellow precipitate ? Ba2+ + CrO42– Product Hg2+ + Co2+ + SCN– Product NH4+ + [PtCl6]2– Product Ag+ + CrO42– Product NH4+ + [Co(NO2)6]3– Product

fuEu vfHkfØ;kvksa esa ls ,slh vfHkfØ;kvksa dh la[;k fdruh gS] ftuesa de ls de ,d mRikn ihys jax ds vo{ksi ds :i esa izkIr gksrk gS \ Ba2+ + CrO42– mRikn 2+ 2+ – Hg + Co + SCN mRikn + 2– NH4 + [PtCl6] mRikn Ag+ + CrO42– mRikn NH4+ + [Co(NO2)6]3– mRikn Ans. Sol.

3. Ba2+ + CrO42– 2+

2+

BaCrO4 –

Hg + Co + SCN NH4+ + [PtCl6]2–

Sol.

Co[Hg(SCN)4] (NH4)2 [PtCl6]

Ag+ + CrO42– +

(yellow)

Ag2CrO4

(deep blue)

(yellow)

(brick red)

3–

NH4 + [Co(NO2)6] (NH4)3 [Co(NO2)6] (yellow) Ba2+ + CrO42– BaCrO4 (ihyk) Hg2+ + Co2+ + SCN– Co[Hg(SCN)4] (xgjk uhyk) NH4+ + [PtCl6]2– (NH4)2 [PtCl6] (ihyk) Ag+ + CrO42– Ag2CrO4 (b±V ds leku yky) NH4+ + [Co(NO2)6]3–

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19.

Match the products of the reactions listed in column-I with the suitable colour of the precipitate(s) listed in column-II. Column I Column II boiling

(A) Hg2I2 (green) (B) BiI3 (black)

(p) One of product is grey / black precipitate

H2 O

(q) One of product is orange precipitate

H2 O boiling

(C) FeCl3 + CH3COO– (salt)

water

(r) One of product is reddish-brown precipitate

boiling

(D) Ag2SO3 (white)

(s) One of product is red precipitate

water

(t) Redox reaction

LrEHk -I esa nh xbZ vfHkfØ;k dks LrEHk -II esa fn;s x;s mi;qDr vo{ksi (mRikn/xq.k) ds jax ds lkFk lqesfyr djksA . I

II

(A) Hg2I2 (gjk) (B) BiI3 (dkyk)

(p) ,d

mRikn LysVh @ dkyk vo{ksi gSA

(q) ,d

mRikn ukjaxh vo{ksi gSA

(r) ,d

mRikn yky Hkwjk vo{ksi gSA

(s) ,d

mRikn yky vo{ksi gSA

H2O

H2 O

(C) FeCl3 + CH3COO– (yo.k)

(D) Ag2SO3 ('osr)

H2O

H2O

(t) mikip;h Ans.

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

Sol.

(A)

Hg2I2 (green )

(B)

BiI3 (black) + H2O

boiling

vfHkfØ;k ¼Redox reaction½

HgI2 (red) + Hg (black)

H2 O

BiOI (orange) + 2H+ + 2I–

(C) FeCl3+CH3COO– (salt)

[Fe3(OH)2(CH3COO)6]+ + H2O

boiling

3Fe(OH)2CH3COO + 3CH3COOH + H+

(deep-red colouration)

Sol.

(D)

Ag2SO3 (white)

(A)

Hg2I2 (gjk )

(B)

boiling water

2Ag (grey) + SO42– + 2H+ HgI2 (yky) + Hg (dkyk)

H2O

BiOI (ukjaxh) + 2H+ + 2I–

BiI3 (dkyk) + H2O

(C) FeCl3+CH3COO– (yo.k)

[Fe3(OH)2(CH3COO)6]+ + H2O (xgjk

(D)

Ag2SO3 ('osr)

(reddish brown ppt.)

H2O

H2O

3Fe(OH)2CH3COO + 3CH3COOH + H+

yky jax)

(yky

Hkwjk vo{ksi)

2Ag (LysVh ) + SO42– + 2H +.

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