Jp Xii Physical&inorganic Chemistry (26) - Prev Chaps + Inorg. Chem.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 41 & 42

Course : VIJETA (JP)

This DPP is to be discussed in the week (21.09.2015 to 26.09.2015)

DPP No. # 41 (JEE-MAIN) ANSWERKEY DPP No. # 41 (JEE-MAIN) 1. 8. 15.

(D) (B) (B)

2. 9. 16.

(A) (B) (D)

3. 10. 17.

(D) (D) (B)

4. 11. 18.

(A) (D) (C)

5. 12. 19.

(B) (D) (C)

6. 13. 20.

(A) (C) (C)

7. 14.

(C) (B)

3.

(B)

4.

(D)

5.

(D)

6.

(C)

7.*

(AD)

(B)

11.

(A)

12.

06

DPP No. # 42 (JEE-ADVANCED) 1.

(C)

2.

(D)

8.*

(AD)

9.*

(ACD) 10.

13.

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

(D) p,s,q

Total Marks : 60

Max. Time : 40 min.

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

(3 marks, 2 min.)

[60, 40]

1.

Incorrect statement about interhalogen compound is : (A) These compound can be used as non aqueous solvent. (B) Interhalogen compound are very useful fluorinating agents. (C) These all are covalent & dimagnetic in nature. (D*) Generally interhalogen compound are less reactive then halogens. vUrjgSykstu ;kSfxd ds fy, vlR; dFku gS : (A) bu ;kSfxdksa dks vtyh;&foyk;d ds :i esa iz;qDr fd;k tkrk gSA (B) vUrjgSykstu ;kSfxd cgqr mi;ksxh ¶yksjhuhd`r ;kSfxd gksrs gSA (C) ;g lHkh izd`fr esa lgla;ksth rFkk izfrpqEcdh; gksrs gSA (D*) lkekU;r% vUrjgSykstu] gSykstu dh vis{kk de fØ;k'khy gksrs gSaA

2.

Gold (atomic radius = 0.144 nm) crystallises in face - centred unit cell. What is the length of a side of the cell? Qyd&dsfUnzr bdkbZ dksf"Bdk esa xksYM ¼ijek.oh; f=kT;k = 0.144 nm) fØLVyhd`r fd;k tkrk gSA dksf"Bdk ds fdukjs dh yEckbZ D;k gS ? (A*) 0.407nm (B) 0.288nm (C) 0.332 nm (D) 0.432nm

Sol.

FCC

= 2 2R = 2 2 × 0.144 nm.

= 0.407nm.

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PAGE NO.- 1

3.

What is the order of freezing point of aqueous solution if same mole of acetic acid, trichloroacetic acid and tri fluoroacetic acid are added in given amount of water (A) Acetic acid < Trichloroacetic acid < Tri fluoroacetic acid (B) Tri fluoroacetic acid > Acetic acid > Trichloroacetic acid (C) Trichloroacetic acid > Tri fluoroacetic acid > Acetic acid (D*) Acetic acid > Trichloroacetic acid > Tri fluoroacetic acid

tyh; foy;uksa ds fgekad dk Øe D;k gksxkA ;fn fn;s x;s ty esa ,flfVd vEy] VªkbDyksjks,flfVd vEy rFkk Vªkb¶yksjks,flfVd vEy ds leku eksy feyk;s tkrs gks (A) ,flfVd vEy < VªkbDyksjks,flfVd vEy < Vªkb¶yksjks,flfVd vEy (B) Vªkb¶yksjks,flfVd vEy > ,flfVd vEy > VªkbDyksjks,flfVd vEy (C) VªkbDyksjks,flfVd vEy > Vªkb¶yksjks,flfVd vEy > ,flfVd vEy (D*) ,flfVd vEy > VªkbDyksjks,flfVd vEy > Vªkb¶yksjks,flfVd vEy Sol.

Order of degree of dissociation is Tri fluoroacetic acid > Trichloroacetic acid > acetic acid therefore number of particles are maximum in trifluoroacetic acid and minimum in acetic acid.

fo;kstu dh ek=kk dk Øe fuEu gS Vªkb¶yksjks,flfVd vEy > VªkbDyksjks,flfVd vEy > ,flfVd vEy blfy, d.kksa dh la[;k Vªkb¶yksjks,flfVd vEy esa vf/kdre gS rFkk ,flfVd vEy esa U;wure gSA 4.

Sol.

Determine the amount of CaCl2 dissolve in 2.5 lit of water such that its osmatic pressure is 0.75 atm. at 27ºC. If degree of dissociation of CaCl2 is 75%. 2.5 yhVj ty esa foy; CaCl2 dh ek=kk Kkr dhft, ;fn 27ºC ij bl foy;u dk ijklj.k nkc 0.75 ok;qe.Myh; gS rFkk CaCl2 ds fo;kstu dh ek=kk 75% gSA (A*) 3.38gm (B) 5.27gm (C) 4.25 gm (D) 52.7gm i = 1 +  (n–1) =1+

3 (3 – 1) 4

0.75 = 2.5 × 5.

=1+

3 = 2.5 2

1 w × × 0.0821 × 300 111 2.5

w = 3.38gm.

Hybridization and magnetic behaviour of complex K 4[Mn(CN)6] is : (A) d2sp3, diamagnetic (B*) d2sp3, paramagnetic 3 2 (C) sp d , diamagnetic (D) sp3d2, paramagnetic ladqy K4[Mn(CN)6] ds fy, ladj.k o pqEcdh; O;ogkj fuEu gS : (A) d2sp3, izfrpqEcdh; (B*) d2sp3, vuqpqEcdh; (C) sp3d2, izfrpqEcdh; (D) sp3d2 ]

vuqpqEcdh;

6.

A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amp what is the mass of copper deposite at cathode (atomic weight of copper = 63) : CuSO4 ds ,d foy;u dk 1.5 ,fEi;j /kkjk ds lkFk 10 fefuV ds fy, oS|qr vi?kVu fd;k tkrk gSA dSFkksM+ ij fu{ksfir dkWij dk nzO;eku D;k gS \ (dkWij dk ijek.oh; nzO;eku = 63) : (A*) 0.29gm (B) 0.35gm (C) 0.0048gm (D) 0.7gm

Sol.

w=

7.

In which of the following ‘N’ atom is not sp2 hybridised fuEu eas ls fdl esa ‘N’ ijek.kq sp2 ladfjr ugha gSA

1 63 × × 1.5 × 600 96500 2 = 0.293 gm.

(A) HNO4

Sol.

(B) FNO3

(C*)

‘N’ atom is sp3 ‘N’ ijek.kq sp3

(D) B3N3H6

gSA

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PAGE NO.- 2

8.

Sol.

The first three successive ionisation energies of an element Z are 900, 1757 and 14850 kJ mol–1 respectively. The element Z belongs to ,d rRo Z dh izFke rhu Øekxr vk;uu ÅtkZ;s Øe'k% 900, 1757 rFkk 14850 kJ mol–1 gSA rRo Z lEcfU/kr gS % (A) group-1 (B*) group-2 (C) group-15 (D) group-17 (A) oxZ- 1 ls (B*) oxZ-2 ls (C) oxZ-15 ls (D) oxZ-17 ls As .E1 is not very high, element Z cannot belong to group-15 or group-17. As successive .E. increases therefore element belong to group-. tSlk dh iz'u ds vuqlkj .E1 cgqr vf/kd ugha gS] vr% rRo Z, oxZ-15 vFkok oxZ-17 ls lacaf/kr ugha gks ldrk gSA tSls fd Øekxr :i ls .E. esa o`f) gksrh tkrh gS] oSls gh rRo oxZ -2 ls lEcfU/kr gks tkrk gSA

9.

If 40 ml of 0.2 M CH3COOH is titrated with 0.2 M NaOH. How many ml of base must be added to form a buffer solution with greatest buffering capacity ;fn 0.2 M CH3COOH ds 40 ml dks 0.2 M NaOH ds lkFk vuqekfir fd;k tkrk gSA rc lokZf/kd cQj {kerk] okys cQj foy;u dks cukus ds fy, {kkj ds fdrus mL feykus pkfg;s \ (A) 10 ml (B*) 20 ml (C) 30 ml (D) 40 ml

Sol.

Quantity of acid initially =

40  0.2 = 8 × 10–3 mole 1000

[Salt ] 1 [ Acid]

Buffer capacity will be maximum when

 pH = pKa  quantity of salt = quantity of acid = 4 × 10–3 mole This happens when acid is hald neutralized i.e. Now, quantity of salt formed = quantity of NaOH added = 4 × 10 –3 mole so molarity of NaOH = 0.2 M 

nNaOH =

...(1)

V  0. 2 1000

....(ii)

from (1) & (2) V = 20 ml Sol.

40  0.2

izkjEHk esa vEy dh ek=kk = 1000 = 8 × 10–3 eksy foy;u dh cQj {kerk vf/kdre gksxh tc [ [

] 1 ]



pH = pKa

yo.k dh ek=kk = vEy dh ek=kk = 4 × 10–3 eksy ;g rc gksxk] tc vEy mnklhu j[kk tkrk gS] vFkkZr~ vc] fufeZr yo.k dh ek=kk = feyk;s x;s NaOH dh ek=kk = 4 × 10–3 eksy blfy, NaOH dh eksyjrk = 0.2 M 



nNaOH =

V  0. 2 1000

....(ii)

lehdj.k (1) rFkk (2) ls 10.

...(1)

V = 20 ml

Suppose that a hypothetical atom gives a red, green, blue and violet line spectrum . Which jump according to figure would give off the red spectral line.

ekuk fd ,d dkYifud ijek.kq ,d yky] gjk] uhyk rFkk cSaxuh js[kh; LisDVªe nsrk gSA fp=k ds vuqlkj dkSulk laØe.k yky LisDVªy js[kk nsxkA

(A) 3  1

(B) 2  1

(C) 4  1

(D*) 3  2

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PAGE NO.- 3

Sol. Sol.

Red photon will be of minimum energy & 3  2 is transition is of minimum energy so will be of red photon. yky QksVksu U;wure ÅtkZ dk gksxk] rFkk 3  2 dk U;wure ÅtkZ okyk laØe.k gksrk gSA bl izdkj ;g gh yky QksVksu

nsxkA 11.

Calculate the potential of hydrogen electrode in contact with a solution of 5 × 10 –5M of Ba(OH)2 at 1 atm pressure and 298K temperature. 1 atm nkc o 298K rki ij 5 × 10–5M Ba(OH)2 ds lkFk lEidZ esa gkbMªkstu bySDVªkWM dk foHko ifjdfyr dhft,A (A) –0.059V (B) 0.059V (C) 0.59V (D*) – 0.59V

Sol.

H+ + e– 

E red

=

1 H 2 2

E0red

(PH2 )1/ 2 1 0.0591 0.0591 – log =0– log   1 1 H [H ]

= – 0.0591 × pH = –0.0591 × 10 = – 0.591V pOH = –log10–2 [OH–] = 2 × 0.5 × 10–3 = 10–2 [pOH = 2 pH = 10 12.

H and S for a reaction are +30.558 kJ mol–1 and 0.066 kJ mol–1 at 1 atm pressure. The temperature at which free energy is equal to zero and the nature of reaction below this temperature are (A) 483 K, spontaneous (B) 443 K, non-spontaneous (C) 443 K, spontaneous (D*) 463 K, non-spontaneous 1 ok;qe.Myh; nkc ij ,d vfHkfØ;k ds fy, H rFkk S Øe'k% +30.558 kJ mol–1 rFkk 0.066 kJ mol–1 gSA og rki

ftl ij eqDr ÅtkZ 'kwU; ds cjkcj gks tkrh gS rFkk bl rki ds uhps vfHkfØ;k dh izd`fr fuEu gS& (A) 483 K, Lor% izØe (B) 443 K, vLor% izØe (C) 443 K, Lor% izØe (D*) 463 K, vLor% izØe Ans.

G = H – TS = 0 H 30 .558 = = 463 K S 0.066 G = H – TS = 30.558 – T × 0.066 at T < 463 K, 0.066 T < 463 × 0.066 or 0 < 30.558 – 0.066 T or 0 < G Thus at T < 463 K, G > 0 i.e. process is non spontaneous. bl izdkj T < 463 K, G > 0 vr% izØe vLor% gksrk gSA 

13.

T=

White phosphorus when boiled with strong solution of caustic soda produces : (A) sodium phosphide (B) sodium phosphate (C*) sodium hypophosphite (D) red phosphourus

dkWfLVd lksM+s ds izcy foy;u ds lkFk tc 'osr QkLQksjl dks xeZ fd;k tkrk gS rks mRiUu gksxk& (A) lksfM;e QkLQkbM (B) lksfM;e QkLQsV (C*) lksfM;e gkbiksQkLQkbZV (D) yky QkLQksjl Sol.

3NaOH +

P4 +

3H2O 

PH3

+ 3NaH2PO2 Phosphine Sodium Hypophosphite

QkWLQhu 14.

lksfM;e gkbiksQkLQkbZV

Which one of the following is the correct statement ? (A) Beryllium exhibits coordination number of six (B*) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase (C) B2H6.2NH3 is known as 'inorganic benzene' (D) Boric acid is a protonic acid fuEu esa ls dkSulk dFku lR; gS ? (A) csfjfy;e milgla;kstu la[;k N% iznf'kZr djrk gSA (B*) csfjfy;e rFkk ,syqfefu;e nksuksa ds DyksjkbMksa dh Bksl voLFkk esa lsrq DyksjkbM lajpuk,¡ gksrh gSa A (C) B2H6.2NH3 dks vdkcZfud csUt+hu’ dgrs gSaA (D) cksfjd vEy ,d izksVkWfud vEy gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 4

Sol.

They are diagonally related with each other and chlorides of both beryllium and aluminium have bridged chloride structures in solid phase.

gy

pw¡fd buds e/; fod.kZ lac/k gksrk gS] vr% csfjfy;e ,oe~ ,Y;qfefu;e nksukas ds DyksjkbM Bksl izkoLFkk esa lsrq DyksjkbM lajpuk ds :i esa ik;s tkrs gSaA

15.

Calculate value of 'n Keq' for the reaction at 250 K. 250K rki ij fuEu vfHkfØ;k ds fy,] 'n Keq' dk eku Kkr N2O4 (g) Given fn;k

Sol.

16.

dhft;sA

2NO2 (g)

gS : Hºƒ (NO2)g = + 40.407 kJ / mol

Hºƒ (N2O4)g = + 70 kJ / mol Sºr = 10 JK–1 (A) 4 (B*) – 4 Gº = Hº – TSº Hrº = (ƒHº)p – (ƒHº)R = 2 × (+ 40.407) – (+70) = + 10.814 kJ Gº = + 10.814 kJ – 250 × 10 J/K = + 10.814 kJ – 2500 J = 8314 J Gº = – RT n K 8314 = – 8.314 × 250 n K

(C) 1.2

(D) – 1.2

n K = – 4.

The INCORRECT statement is : (A) In metallurgy, flux is a substance used to convert infusible impurities to fusible mass. (B) Cryolite is Na3AlF6 and is used in the electrolysis of alumina for lowering the melting point of alumina. (C) Extraction of iron metal from iron oxide ore is carried out by heating with coke and lime stone. (D*) Haematite, Cassiterite and argentite are oxide ores.

vlR; dFku gS % (A) /kkrqdeZ esa] xkyd og inkFkZ gS ftldk mi;ksx vxyuh; v'kqf);ksa dks] xyuh; nzO;eku esa ifjofrZr djus ds fy,] fd;k tkrk gSA (B) Na3AlF6, Øk;ksykbV gS bldk mi;ksx] ,Y;qqfeuk ds oS|qr vi?kVu esa] ,Y;qfeuk ds xyukad dks de djus ds fy, fd;k tkrk gSA (C) vk;ju vkWDlkbM ls vk;ju /kkrq dk fu"d"kZ.k dksd o ykbeLVksu ds lkFk xeZ djds fd;k tkrk gSA (D*) gsesVkbV] dsflfVjkbV rFkk vtsZUVkbV vkWDlkbM v;Ld gSA Sol.

g y-

Argentitie is sulphide ore (Ag2S) vtsZUVkbV ,d lYQkbM v;Ld (Ag2S) gSA

17.

How many millilitres of a 9 N H2SO4 solution will be required to neutralize completely 20 mL of a 3.6 N NaOH solution? 3.6 N NaOH foy;u ds 20 ml dks iw.kZr;k mnklhu djus ds fy, 9 N H2SO4 foy;u ds fdrus feyh yhVj vko';d gksxsa & (A) 18.0 mL (B*) 8.0 mL (C) 16.0 mL (D) 80.0 mL

18.

What is the normality of the H2SO4 solution, 18.6 mL of which neutralizes 30.0 mL of a 1.55 N KOH solution? H2SO4 foy;u dh ukeZyrk D;k gksxh ftlds 18.6 mL ds mnklhuhdj.k ds fy, 1.55N KOH ds 30.0 mL dh vko';drk

gSa \ (A) 5.0 N 19.

(B) 1.25 N

(C*) 2.5 N

(D) 3.5 N

A sample of 1.0 g of calcite (pure CaCO3) required 39.5 mL of HCl complete reaction. Calculate the normality of the acid. 1.0 g dsfYl,sV ¼'kq) CaCO3) ds uewus dh vfHkfØ;k iw.kZ djus ds fy, HCl ds 39.5 mL vko';d gSA vEy dh ukeZyrk

dh x.kuk djksA (A) 0.41 N 20.

Hint:

(B) 0.61 N

Concentrated HNO 3 reacts with 2 to give : lkUnz HNO3 , 2 ds lkFk fØ;k djds nsrk gSA (A) H (B) HO

(C*) 0.51 N

(C*) HO3

(D) 0.15 N

(D) HOO3

2 + HNO3  2HO3 + 4H2O + 10 NO2 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

DPP No. # 42 (JEE-ADVANCED) Total Marks : 48

Max. Time : 31 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.6 Multiple choice objective ('–1' negative marking) Q.7 to Q.9 Comprehension ('–1' negative marking) Q.10 to Q.11 Single Integer type Questions ('–1' negative marking) Q.12 Match the Following (no negative marking) Q.13

1.

(3 (4 (3 (4 (8

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

[18, [12, [06, [04, [08,

12] 06] 04] 03] 06]

NH4HS (s) NH3 (g) + H2S (g) Kp = 2.25 atm2 Initially 2 moles of only NH4HS(s) is present in 2 L container, enclosed with piston at 300 K. If system is expanded by reversible isothermal process then at what volume of container, pressure of container starts to decrease continuously ? NH4HS (s) NH3 (g) + H2S (g) Kp = 2.25 atm2 ;fn 300 K ij] ,d fiLVu ;qDr 2 L ds cUn ik=k esa izkjEHk esa dsoy NH4HS(s) ds 2 eksy j[ks tkrs gSaA ;fn ra=k dks

mRØe.kh;] lerkih; izØe }kjk izlkfjr fd;k tkrk gS rc ik=k ds fdrus vk;ru ij] ik=k ds nkc esa] yxkrkj deh izkjEHk gks tkrh gS \ (A) 2 lit Sol.

NH4HS (s) at equi.

(B) 4.8 lit (C*) 32.8 lit (D) 63.6 lit NH3(g) + H2S(g) p p p2 = 2.25, pequ = 1.5 1.5atm 1.5atm

ptotal = 3atm If system expand by increase in volume at constant temperature then to maintain the equilibrium condition, pressure of container remain constant but after complete dissociation of 2 mole NH4HS(s), pressure get start decreasing. At that condition

nNH3 = 2 Sol.

nH2S = 2

;

NH4HS (s)

Vcontainer =

NH3(g) + H2S(g) p p 1.5atm 1.5atm

lkE; ij

4  0.082  300 nRT = = 4 × 8.2 = 32.8 lit. 3 P

p2 = 2.25, pequ = 1.5

pdqy = 3atm

;fn fu;r rki ij] ra=k ds izlkj }kjk vk;ru esa o`f) gksrh gS] rc lkE; voLFkk ds LFkkfir jgus ds dkj.k] ik=k dk nkc fu;r cuk jgrk gS ijUrq NH4HS(s) ds 2 eksy ds iw.kZr% fo;ksftr gksus ds i'pkr~ nkc] esa deh gksuk izkjEHk gks tkrk gSA bl ifjfLFkfr ij] nNH3 = 2 2.

nH2S = 2

Vik=k =

4  0.082  300 nRT = = 4 × 8.2 = 32.8 lit. 3 P

5 g of H2O2 solution, containing x % H2O2 by weight, requires x ml of KMnO 4 for complete oxidation in acidic medium. What is the normality of KMnO 4 solution : x % H2O2 (Hkkj dk), ;qDr 5 g H2O2 foy;u dks vEyh; ek/;e esa iw.kZ vkWDlhdj.k ds fy, KMnO4 ds x ml vko';d gSA KMnO4 foy;u (A) 0.588

Sol.

;

dh ukWeZyrk D;k gS % (B) 1.47

(C) 0.312

(D*) 2.94

H2O2 + KMnO4  Mn+2 + O2 v.f. = 2 equivalent of H2O2 = equivalent of KMnO4 mass NV mol. mass × v.f. = 1000 5x N x ×2= 100  34 1000

N=

50 = 2.94 N 17 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

g y-

H2O2 + KMnO4  Mn+2 + O2 v.f. = 2 H2O2 ds

rqY;kad = KMnO4 ds rqY;kad × v.f. =

NV 1000

5x N x ×2= 100  34 1000

N= 3.

50 = 2.94 N 17

Select the correct option : (A) Number of subshells in 3rd orbit of hydrogen atom are 9. (B*) The energy of an electron in a hydrogen atom depends only upon the principal quantum number and not on the other quantum numbers. (C) Wave length of de-Broglie wave associated with electron in 2nd orbit is more than that in 3rd orbit. (D) Orbital angular momentum of 3p orbital is zero. (E) Number of electrons in Fe3+ ion for n +  = 4 are five.

lgh fodYi dk p;u dhft,A (A) gkbMªkstu ijek.kq ds 3rd d{kk esa midks'kksa dh la[;k 9 gSA (B*) ,d gkbMªkstu ijek.kq esa ,d bySDVªkWu dh ÅtkZ dsoy eq[; DokaVe la[;k ij fuHkZj djrh gS rFkk vU; Doka Ve la[;k ij ughaA (C) 2nd d{kk esa bySDVªkWu ds lkFk lEcfU/kr Mh&czkWXyh rjax dh rjaxnS/;Z] 3rd d{kk dh rqyuk esa vf/kd gksrh gSA (D) 3p d{kd dk d{kd dks.kh; vk?kw.kZ 'kwU; gksrk gSA (E) Fe3+ vk;u esa n +  = 4 ds fy, bysDVªkWu dh la[;k ik¡p gSaA Sol.

(A) Number of subshells = orbit number = 3. (B) Energy of electron in H-like atom depends only on ‘n’. (C)  =

(D)  =

Sol.

h 1 and   . mv n  (   1)

2

 n

h 2

(E) For 3p orbital n +  = 4 and number of electrons in Fe3+ in 3p are six. (A) midks'kksa dh la[;k = d{kkvksa dh la[;k = 3. (B) H-leku (C)  =

(D)  =

ijek.kq esa bySDVªkWu dh ÅtkZ dsoy ‘n’ ij fuHkZj djrh gSA

h 1 rFkk  × . mv n  (   1)

(E) 3p d{kd 4.

h = 2



h = 2

 2

× n

h 2

ds fy, n +  = 4 gS rFkk Fe3+ esa mifLFkr 3p bysDVªkWu dh la[;k N% gSA

100 mL of 0.1N I2 oxidizes Na2S2O3 in 50 ml solution to Na2S4O6. The normality of this hypo solution against KMnO4 (which oxidizes it to Na2SO4) would be 0.1N I2 dk 100 mL, Na2S2O3 dks Na2S4O6 ds 50 ml foy;u esa vkWDlhd`r dj nsrk gSA KMnO4 (tks bls Na2SO4 vkDlhd`r djsa) ds lkis{k bl gkbiks foy;u dh ukeZyrk gksxh (A) 0.1 (B) 0.2 (C) 1.0 (D*) 1.6.

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esa

5.

Sol.

0.185 g of an iron wire containing 99.8% iron is dissolved in an acid to form ferrous ions. The solution requires 29.3 mL of K2Cr2O7 solution for complete reaction. The normality of the K2Cr2O7 solution is : vk;ju rkj dk 0.185 g tks 99.8% vk;ju j[krk gS] dks Qsjl vk;u cukus ds fy, vEy esa ?kksyrs gaSA vfHkfØ;k dks iw.kZ djus ds fy, K2Cr2O7 foy;u ds 29.3 mL vko';d gaSA K2Cr2O7 foy;u dh ukeZyrk gksxh % (A) 0.05 (B) 0.02 (C) 0.20 (D*) 0.10 Equivalent of Fe = Equivalent of K2Cr2O7 Fe ds rqY;kad = K2Cr2O7 ds rqY;kad 0.185  99 .8 × 1 = 29.3 × 10–3 × N 56  100

N = 0.1 6.

Moles of O2 used in second reaction to completely consume the MnO 2 produced from first reaction are : izFke vfHkfØ;k ls izkIr MnO2 dks iw.kZ :i ls dke esa ysus ds fy, f}rh; vfHkfØ;k esa iz;qDr O2 ds eksy fuEu gSa % (A) 0.05 (B) 0.2 (C*) 0.1 (D) 0.4

Sol.

Balanced equation : 2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O 

Sol.

1 1 × nMnO 2 produced = × 0.2 = 0.1 mole 2 2

lUrqfyr vfHkfØ;k : 2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O 

7.*

nO 2 used =

nO 2

iz;qDr =

1 × nMnO 2 2

mRikfnr =

1 × 0.2 = 0.1 mole 2

For a hypothetical H like atom which follows Bohr's model, some spectral lines were observed as shown in figure. If it is known that line 'E' belongs to the visible region, then the lines possibly belonging to ultraviolet region will be : [Given : n1, n2,n3 and n4 are consecutive orbits & n1 = 3] (A*) D (B) C (C) B (D*) A H ln`'; ,d dkYifud ijek.kq] tks fd cksj

ekWMy dh ikyuk djrk gS] blds fy, dqN LiSDVªe js[kk,sa izsf{kr gksrh gSa] ftUgas fp=k esa iznf'kZr fd;k x;k gSA ;fn 'E' js[kk n`'; {ks=k ls lEcfUèkr gS] rks ijkcSaxuh {ks=k ls lEcfU/kr js[kk,sa lEHkor% dkSulh gksaxh % [fn;k x;k gS % n1, n2,n3 o n4 Øekxr d{kk,sa gSa rFkk n1 = 3] Sol. Sol.

(A*) D (B) C (C) B (D*) A Line should have higher energy than the energy of line E. EA (6  3) and ED ( 4) will be greater than energy of line E (4  3). js[kk dh ÅtkZ] E js[kk dh ÅtkZ ls vf/kd gksuh pkfg,A EA (6  3) rFkk ED ( 4) dh

8.*

ÅtkZ] E (4  3) js[kk dh ÅtkZ ls vf/kd gksxhA

Which of the following sample of reducing agents is/are chemically equivalent to 25ml of 0.2 N KMnO 4 to be reduced to Mn2+ and H2O? (A*) 25 mL of 0.2M FeSO4 to be oxidized to Fe3+ (B) 50 mL of 0.1M H3AsO3 to be oxidized to H3AsO4 (C) 25 mL of 0.2M H2O2 to be oxidized to H+ and O2 (D*) 25 mL of 0.1 M SnCl2 to be oxidized to Sn4+ vipk;d vfHkdeZd dk fuEu esa ls dkSulk @ dkSuls uewuk] 0.2 N KMnO4 ds 25ml dks Mn2+ ,oa H2O esa vipkf;r djus

ds jklk;fud rqY;kdksa ds leku gS \ (A*) 0.2M FeSO4 ds 25 mL dks Fe3+ esa vkWDlhd`r djus ds (B) 0.1M H3AsO3 ds 50 mL dks H3AsO4 esa vkWDlhd`r djus ds (C) 0.2M H2O2 ds 25 mL dks H+ vkSj O2 esa vkWDlhd`r djus ds (D*) 0.1 M SnCl2 ds 25 mL dks Sn4+ esa vkWDlhd`r djus ds 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

9.*

Which of the following statement(s) is (are) correct ? (A*) The coordination number of each type of ion in CsCl is 8 (B) A metal that crystallizes in bcc structure has a coordination number of 12 (C*) A unit cell of an ionic crystal shares some of its ions with other unit cells (D*) The length of the unit cell in NaCl is 552 pm. (rNa+ = 95pm; rCl– = 181 pm) fuEu esa ls dkSulk@dkSuls dFku lgh gS@gSa ? (A*) CsCl esa

izR;sd izdkj ds vk;u dh leUo; la[;k 8 gSA (B) ,d /kkrq tks fd bcc lajpuk esa fØLVyhd`r gksrk gS] bldh leUo; la[;k 12 gSA (C*) ,d vk;fud fØLVy dh bdkbZ dksf"Bdk bldk dqN vk;uksa dks vU; bdkbZ dksf"Bdk ds lkFk lkf>r djrh gSA (D*) NaCl esa bdkbZ dksf"Bdk dh yEckbZ 552 pm. (rNa+ = 95pm; rCl– = 181 pm) Paragraph for Question Nos. 10 to 11 0.6 mole of K2MnO4 is hydrolysed by dilute H2SO4 to produce KMnO4 along with dark brown precipitate of MnO2, according to following reaction : MnO42– + H+  MnO4– + MnO2 + H2O The MnO2 obtained is fused with KOH in presence of O 2 to give K2MnO4 according to the reaction : MnO2 + KOH + O2  K2MnO4 + H2O Now, answer the following questions :

iz'u

10

ls

11

ds fy, vuqPNsn

ruq H2SO4 }kjk 0.6 eksy K2MnO4 dks ty vi?kfVr dj fuEu vfHkfØ;k ds vuqlkj MnO2 ds xgjs Hkwjs vo{ksi ds lkFk KMnO4 dks mRikfnr fd;k tkrk gS % MnO42– + H+  MnO4– + MnO2 + H2O

vc] fuEu vfHkfØ;k ds vuqlkj O2 dh mifLFkfr esa KOH ds lkFk izkIr MnO2 dks xfyr dj K2MnO4 fn;k tkrk gS % MnO2 + KOH + O2  K2MnO4 + H2O

vc] fuEu iz'uksa ds mÙkj nhft, % 10.

Sol.

Number of moles of MnO2 produced according to first reaction is : izFke vfHkfØ;k ds vuqlkj izkIr MnO2 ds eksyksa dh la[;k fuEu gS % (A) 0.3 (B*) 0.2 (C) 0.1 6

7

4

v.f.=1

v.f.=2

(D) 0.4

MnO 24– + H+  MnO –4 + MnO 2 Partial balancing vkaf'kd

larqyu

3MnO42–  2MnO4– + MnO2 0.6mole 11.

0.2mole

Volume of H2S gas at STP required to completely react wtih KMnO4 produced in first reaction in acidic medium to get converted into elemental sulphur is : rkfRod lYQj esa iw.kZ:i ls ifjofrZr gkssus ds fy, vEyh; ek/;e esa izFke vfHkfØ;k esa mRikfnr KMnO4 ds lkFk iw.kZ :i

ls fØ;k djus ds fy, STP ij vko';d H2S dk vk;ru fuEu gS % Sol.

(A*) 22.4 L (B) 11.2 L eq. of H2S = eq. of KMnO 4

(C) 5.6 L

(D) 2.8 L

V × 2 = 0.4 × 5 22.4



V = 22.4 L

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PAGE NO.- 9

12.

How many of the following molecules are polar?

fuEu esa ls fdrus v.kq /kqzoh; gSa\ Ans.

(i) CO2 (vi) BeCl2(g) 06

Sol.

CO2

Ans.

(iii) NO2 (vii) TeCl4

O= C=O

= 0

(viii) CCl4 ;

sp

NO2

13.

(ii) SO2

=O

 0

;

SO2

(iv) SOCl2 (ix) ClO2

SO2

SOCl2

COCl2

 0

;

BeCl2

TeCl4

 0

;

CCl4

ClO2



Cl – Be – Cl

(v) COCl2



0



0

= 0



0

0

1 mole of A is added in 1 kg water. Volume is nearly 1L. To this mixture , small amount of B is added such that it act as limiting reagent. Match the effect of addition of B in Column I with entries in column II Column-I A B (A) KCN + AgCN(s) (B) CuS(s) + Conc. HNO3 (C) CH3COOH + Ba(OH)2 ( < 0.1) (D) HgI2 + KI (A) r,q ; (B) p,s,q (C) p,s,q (D) p,s,q

Column-II (p) (q) (r)

Boiling point increase Freezing point does not increase Osomotic pressure is unchanged

(s)

Vapor pressure decreases

A

ds 1 eksy 1 kg ty eas feyk;s tkrs gSA bldk vk;ru yxHkx 1L gSA bl feJ.k esa B dh dqN ek=kk bruh feykrs gS fd ;g lhekar vfHkdeZd dh rjg dk;Z djsA dkWye&I esa B dks feykus ij gksus okys izHkko dk dkWye II esa izfof"B;ksa ds lkFk feyku dhft,A dkWy e-I dkW ye -II (A)

A KCN

(B)

CuS(s)

(C)

Ans. Sol. (A)

(B)

(C)

+ +

B AgCN(s)

(p)

Conc. HNO3

(q)

CH3COOH + Ba(OH)2 ( < 0.1) (D) HgI2 + KI (A) r,q ; (B) p,s,q (C) p,s,q (D) p,s,q

(r)

DoFkukad c
(s)

ok"i nkc ?kVrk gSA

KCN + AgCN  K[Ag(CN)2](aq) Two particles reacting (K+ and CN–) and two particles are producing (K+ and [Ag(CN)]–) Hence no change in colligative properties CuS + Conc. HNO3  Cu(NO3)2 (aq) + H2S  + NO2 + H2O Cus is insoluble. Addition of HNO3 produces ions Cu2+ , (NO3)– . Hence solute particles increases so boiling point increases and vapor pressure decreses. 2 CH3COOH + Ba(OH)2  Ba(CH3COO)2 + 2H2O 1 – 2x 0 x Hence 2x × (1 + ) are giving 3x moles of solute and  < 0.1. Hence, moles of solute are increasing. So, boiling point increases and vapor pressure decreases.

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PAGE NO.- 10

(D)

(A)

HgI2 + 2KI  K2[HgI4] HgI2 is insoluble. Adding KI produces [K2(HgI4)]. Hence, ions increase. So boiling point increases and vapor pressure decreases. KCN + AgCN  K[Ag(CN)2](aq)

nks v.kq (K+ rFkk CN–) fØ;k djrs gS rFkk nks v.kq (K+ rFkk[Ag(CN)]–) curs gSA vr% v.kqla[;d xq.k/keZ esa ifjorZu ugh gksrk gSA (B)

CuS + Conc. HNO3  Cu(NO3)2 (aq) + H2S  + NO2 + H2O Cus vfoys; gksrk gSA HNO3 ds ;ksx ij Cu2+ , (NO3)– vk;u curs vr%

foys; ds d.k c
gS rFkk ok"i nkc ?kVrk gSA (C)

2 CH3COOH + Ba(OH)2  Ba(CH3COO)2 + 2H2O 1 – 2x 0 x vr% 2x × (1 + ) foys; ds 3x eksy nsrs gS rFkk  < 0.1 vr%

foys; ds eksy c
?kVrk gSA (D)

HgI2 + 2KI  K2[HgI4] HgI2 vfoys; gSA KI ds ;ksx ij [K2(HgI4)]

curk gSA vr% vk;u c
gSA

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PAGE NO.- 11