Jp Xii Physical&inorganic Chemistry (6) - Prev Class Xi Chaps + Solution & Colligative Properties +solid State.pdf

PHY./INO. CHEMISTRY

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

NO. 09

Date : 04-05-2015 TEST IN FORMATI ON

DATE : 17.05.2015

PART TEST (PT) - 01 (ADVANCED)

Syllabus : Solution & Colligative Properties, Solid State, Thermodynamics, Ionic equilibrium, s-Block Element. (All ChemInfos and Handouts till date)

This DPP is to be discussed in the week (04-05-2015 to 09-05-2015)

DPP No. # 09 (JEE-MAIN) Total Marks : 64

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.16 ChemINFO : 4 Questions ('–1' negative marking) Q.17 to Q.20

1.

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

[48, 32] [16, 08]

Arrange the alkalline earth metals in decreasing order of size of their hydrated ion :

fuEu esa ls dkSulk {kkjh; e`nk /kkrqvksa ds ty;ksftr vk;u dh f=kT;k ds ?kVrs gq, lgh Øe dks n'kkZrk gS%

Sol.

(A*) Be2+ (aq) > Mg2+ (aq) > Ca2+ (aq) > Sr2+ (aq) > Ba2+ (aq) (B) Ba2+ (aq) > Sr2+ (aq) > Ca2+ (aq) > Mg2+ (aq) > Be2+ (aq) (C) Be2+ (aq) > Mg2+ (aq) > Ca2+ (aq) > Ba2+ (aq) > Sr2+ (aq) (D) Ba2+ (aq) > Sr2+ (aq) > Mg2+ (aq) > Ca2+ (aq) > Be2+ (aq) Greater the charge density on cation, greater will be the hydration, and greater will be the hydrated radii.

/kuk;u ij vkos'k ?kuRo ftruk T;knk gksxk] ty;kstu Hkh vis{kkd`r mruk gh T;knk gksxk rFkk bl izdkj ty;ksftr f=kT;k Hkh T;knk gksxhA 2.

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

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

Refer class notes.

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PAGE NO.- 1

3.

The following diagram shows arrangement of lattice point with a = b = c and  =  =  = 90º.

Choose the correct options. (A*) The arrangement is SC with each lattice point surrounded by 6 nearest neighbours. (B) The arrangement is SC with each lattice point surrounded by 8 nearest neighbours. (C) The arrangement is FCC with each lattice point surrounded by 12 nearest neighbours. (D) The arrangement in BCC with each lattice point surrounded by 8 nearest neighbours fuEu vkjs[k tkyd fcUnq dh O;oLFkk n'kkZrk gS ftlds a = b = c vkSj  =  =  = 90º gaSA lgh oDrO; (A*) O;oLFkk

igpkuksA

ljy ?kuh; gS ftlesa izR;sd tkyd fcUnq 6 fudVre d.kksa ls f?kjk jgrk gSA

(B) O;oLFkk

ljy ?kuh; gS ftlesa izR;sd tkyd fcUnq 8 fudVre d.kksa ls f?kjk jgrk gSA (C) O;oLFkk FCC gS ftlesa izR;sd tkyd fcUnq 12 fudVre d.kksa ls f?kjk jgrk gSA (D) O;oLFkk BCC gS ftlesa izR;sd tkyd fcUnq fudVre 8 d.kksa ls f?kjk jgrk gSA

g y-

. According to figure, it shows a simple cubic lattice. Now observe the center atom , its has 6 nearest neighbours fp=kkuqlkj] ;g ,d ljy ?kuh; tkyd gS] ftlesa dsfUnz; ijek.kq ds 6 fudVre iM+kslh gSA

4.

A body centred cubic arrangement is show :

Sol.

O is the body centre ; A, B, C, D, .........., H are the corners. What is the magnitude of the angle AOB? (A) 120° (B) 109° 28 (C) 104° 31 (D*) 70° 32 ,d dk; dsUnzh; ?kuh; O;oLFkk fuEu izdkj n'kkZ;h x;h gS :

O dk; dsUnz (A) 120°

Sol.

gS] A, B, C, D, .........., H dksus gS] dks.k AOB dk eki D;k gS ? (B) 109° 28

(C) 104° 31

(D*) 70° 32

3a 2 3a 2   a2 1 4 4  cos  = 2  3 a   3 a  3 = 70º32'  2  2    

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5.

The radius of metal atom can be expressed in terms of the length of a unit cell is : (A) it is a/2 for simple cubic lattice (B) it is (3a/4) for b.c.c. lattice (C) it is (a/22) for F.C.C. lattice (D*) All of the above.

bdkbZ dks"Bhdk dh dksj yEckbZ ds in esa /kkrq ijek.kq dh f=kT;k fuEu esa ls fdl izdkj O;Dr dh tk ldrh gSA (A) ljyre ?kuh; tkyd ds fy, ;g a/2 gSA (B) b.c.c. tkyd ds fy, ;g (3a/4) gSA (C) F.C.C. tkyd ds fy, ;g (a/22) gSA (D*) mDr lHkh lgh gSA 6.

The correct order of hybridisation of the central atom in the following species NF 3, BF3, PF5 and [SiF6]2– is: fuEufyf[kr iztkfr;ksa] NF3, BF3, PF5 rFkk [SiF6]2– esa dsUnzh; ijek.kq ds ladj.k dk lgh Øe gS : (A) sp2, sp3, sp3d2, sp3d (B*) sp3, sp2, sp3d, sp3d2 2 3 3 3 2 (C) sp , sp , sp d, sp d (D) sp3, sp2, sp3d2, sp3d3

Sol.

7.

The numbers of tetrahedral and octahedral holes in a ccp array of 100 atoms are respectively (A*) 200 and 100 (B) 100 and 200 (C) 200 and 200 (D) 100 and 100 100 ijek.kqvksa dh ccp O;oLFkk esa leprq"Qydh; rFkk v"BQydh; fNnzksa dh la[;k Øe'k% gSA (A*) 200 rFkk 100

8.

(B) 100 rFkk 200

(C) 200 rFkk 200

(D) 100 rFkk 100

In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids?

/kkfRod ijek.kqvksa ds Qyd dsUnzh; ?kuh; lajpuk esa] leprq"Qydh; rFkk v"BQydh; fjfDr;ksa ds vkdkj dk vkisf{kd vuqikr D;k gS & (A*) 0.543 Sol.

(B) 0.732

(C) 0.414

(D) 0.637

rtetrahedra l 0.225 R roctahedral = 0.414 R = 0.543.

9.

CsBr has b.c.c. structure with edge length 4.3 A. The shortest inter ionic distance in between Cs + and Br– is: CsBr dksj yEckbZ 4.3 Å ds lkFk b.c.c. lajpuk j[krk gS rc Cs+ rFkk Br– ds e/; U;wure varj vk;fud nwjh gS: (A*) 3.72 (B) 1.86 (C) 7.44 (D) 4.3

Sol.

r+ + r– =

10.

Select the INCORRECT order from the following according to the property mentioned along with : (A) F– > Cl– > Br– > I– ; ionic mobility in water (B*) CsCl > RbCl > KCl > NaCl > LiCl ; lattice energy. (C) Li2CO3 < Na2CO3< K2CO3 < Rb2CO3 ; solubility in water. (D) CsF > CsCl > CsBr > CsI ; melting point.

3a = 2

3  4.3 = 3.72 Å 2

v|ksfyf[kr esa ls dkSulk Øe fn;s x;s xq.k ds vk/kkj ij vlR; gS % (A) F– > Cl– > Br– > I– ; ty esa vk;fud xfr'khyrk (B*) CsCl > RbCl > KCl > NaCl > LiCl; tkyd ÅtkZ (C) Li2CO3 < Na2CO3< K2CO3 < Rb2CO3; ty esa foys;rk (D) CsF > CsCl > CsBr > CsI; xyukad fcUnq Sol.

(B) Correct order is LiCl > NaCl > KCl > RbCl > CsCl. lattice energy 

(B) lgh

Øe gS % LiCl > NaCl > KCl > RbCl > CsCl. tkyd ÅtkZ 

1 size of cation 1

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11.

A real gas most closely approaches the behaviour of an ideal gas at : (A) 15 atm and 200 K (B) 1 atm and 273 K (C*) 0.5 atm and 500 K (D) 15 atm and 500 K

,d okLrfod xSl] ,d vkn'kZ xSl O;ogkj ds utnhd fuEu esa ls fdu ifjfLFkfr;ksa esa igq¡prh gS % (A) 15 atm rFkk 200 K (B) 1 atm rFkk 273 K (C*) 0.5 atm rFkk 500 K (D) 15 atm rFkk 500 K Sol.

A real gas behaves idealy under conditions of low pressure and high temperature .

fuEu nkc rFkk mPp rki ij ,d okLrfod xSl] vkn'kZ xSl dh rjg O;ogkj djrh gSA 12.

If a gas X (having molar mass 41 g/mol) has a density of 1.33 g/L at 1 atm pressure and 300 K temperature, then which of the following statements is correct for gas X : (A*) Repulsive tendencies are dominant over attractive tendencies among the gas molecules. (B) Gas is more compressible with respect to an ideal gas. (C) Gas molecules are found to behave ideally under the given set of conditions. (D) Both (A) and (B). ;fn 1 atm nkc rFkk 300 K rki ij] ,d xSl X dk (eksyj nzO;eku 41 g/mol) ?kuRo 1.33 g/L gS] rc xSl X ds fy, fuEu

esa ls dkSulk dFku lgh gS % (A*) xSl v.kqvksa ds e/;] vkd"kZ.k izo`fÙk dh rqyuk esa] izfrd"kZ.k izo`fr izHkkoh gSA (B) ,d vkn'kZ xSl dh rqyuk esa] xSl vf/kd lEihM~; gSA (C) nh xbZ ifjfLFkfr;ksa esa] xSl v.kqvksa dk O;ogkj] vkn'kZ ik;k tkrk gSA (D) (A) rFkk (B) nksuksa 1 41 PM = = 1.66 g/L 0.082  300 RT  dgas < dideal  Vreal > Videal  Repulsive tendencies are more dominant & gas is less compressible wrt an ideal gas.

Sol.

dideal =

Sol.

dvkn'kZ =

1 41 PM = = 1.66 g/L 0.082  300 RT

 dxSl < dvkn'kZ  VokLrfod > Vvkn'kZ  vr% izfrd"kZ.k izo`fr vf/kd izHkkoh gSa 13.

rFkk ,d vkn'kZ xSl dh rqyuk esa] xSl de lEihM~; gSA

In hexagonal close packing of sphere in three dimensions.

(A) In one unit cell there are 12 octahedral voids and all are completely inside the unit cell. (B*) In one unit cell there are six octahedral voids and all are completely inside the unit cell. (C) In one unit cell there are six octahedral void and of which three are completely inside the unit cell and other three are partially inside the unit cell. (D) In one unit cell there are 12 tetrahedral voids, all are completely inside the unit cell.

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,d xksys ds f=kfoeh; "kVdks.kh; fufcM ladqyu esa %

(A) ,d

,dy dksf"Bdk esa 12 v"VQydh; fNnz gaS rFkk lHkh iw.kZr;k ,dy dksf"Bdk ds vUnj gSaA (B*) ,d ,dy dksf"Bdk esa 6 v"VQydh; fNnz gSa rFkk lHkh iw.kZr;k ,dy dksf"Bdk ds vUnj gSaA (C) ,d ,dy dksf"Bdk esa 6 v"VQydh; fNnz gSa ftuesa ls rhu iw.kZr;k ,dy dksf"Bdk ds vUnj rFkk vU; rhu vkaf'kd :i ls bdkbZ dksf'kdk ds vUnj gSaA (D) ,dy dksf"Vdk esa 12 prq"dQydh; fNnz gaS tks fd iw.kZr;k ,dy dksf"Bdk ds vUnj gaSA Sol.

gy

HCP = AB AB AB ...............pattern repeat For calculating voids between two layers A and B. Total tetrahedral voids = 12 (represented by dots) out of which 8 are completely inside but rest are shared by other unit cells. Total octahedral voids = 6 (represented by cross). All are completely inside. HCP = AB AB AB ...............rjhds dh iqujko`fÙk gksrh gSA

nksuksa ijrksa A o B ds e/; fNnzksa dh x.kuk djus ds fy,µ dqy prq"Qydh; fNnzksa dh la[;k = 12. (fcUnq }kjk çnf'kZr) ftuesa ls 8 iw.kZr;k vUnj gSa tcfd cps gq, 4 vU; ,dy dksf"Bdkvksa ls Hkkxhnkjh dj jgs gaSA dqy v"VQydh; fNnz = 6 ¼ØkWl (x) ds }kjk iznf'kZr½ blfy, lHkh iw.kZr;k vUnj gSaA 14.

Select the incorrect statement. (A*) Br2 and ICl have the same number of electrons and thus both have nearly the same boiling points. (B) N2H4 is pyramidal about each N-atom. (C) In P4S3 molecule, there are six P-S bonds, three P-P bonds and ten lone pairs of electrons (on all atoms). (D) In ClO4–, all Cl–O bonds are identical and there is strong p–d bonding between chlorine and oxygen atoms.

vlR; dFku dks pqfu;sA (A*) Br2 rFkk ICl esas bysDVªksuksa dh la[;k leku gksrh gS blfy, nksuksa dk yxHkx leku DoFkukad gksrk gSA (B) izR;sd N-ijek.kq ij N2H4 fijSfeMh; gksrk gSA (C) P4 S3 v.kq esa] N% P - S cU/k] rhu P - P cU/k rFkk bysDVªksuksa ds nl ,dkdh ;qXe ¼lHkh ijek.kqvksa ij½ gksrs gSaA (D) ClO4–, esa lHkh Cl–O cU/k leku gksrs gSa rFkk ;gka Dyksjhu rFkk vkWDlhtu ijek.kqvksa ds chp izcy p–d cU/ku gksrk gSA Sol.

(A) Boiling point of ICl > Br2 as ICI is polar and Br2 is non-polar in nature. (B)

(C)

gy .

(A) izd`fr

pyramidal about each N-atom.

(D)

identical due to resonance.

esa ICl /kzqoh; rFkk Br2 v/kzqoh; gksus ls] ICl dk DoFkukad > Br2 gksrk gSA

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PAGE NO.- 5

15.

(B)

izR;sd N-ijek.kq dh vksj fijkfeMh;

(C)

(D)

vuqukn ds dkj.k leferA

At 900ºC Kp = 1.04 atm for the reaction CaCO3(s)  CaO(s) + CO2(g). At low temperature 58.4 g CO2, 655 g CaO and 95 g CaCO3 are introduce in 50 L reaction chamber. The temperature is raised to 900ºC amount of CaO will …….. (A) increase (B*) decrease (C) remain same (D) none fuEu vfHkfØ;k ds fy, 900ºC ij Kp = 1.04 atm CaCO3(s)  CaO(s) + CO2(g). fuEu

rki ij 58.4 g CO2, 655 g CaO rFkk 95 g CaCO3 dks 50 L vfHkfØ;k ik=k esa ysrs gSA rkieku 900ºC rd c<+rk gSA CaO dh ek=kk — gksxh % (A) c<s+xh (B*) ?kVsxh (C) leku jgsxh (D) buesa ls dksbZ ugha Sol.

nCO2 =

PV 1.04  50 = = 0.53 mol RT 0.0821 1173

WCO2

= 23.75 g

QP > KP hence reaction goes to back. QP > KP vr% vfHkfØ;k izrhi fn’kk esa tkrh 16.

gSA

Borax on heating strongly above its melting point melts to a liquid, which then solidifies to a transparent mass commonly known as borax-bead. The transparent glassy mass consists of : (A) sodium pyroborate (B) boric anhydride (C) sodium meta-borate (D*) boric anhydride and sodium metaborate

cksjsDl blds xyukad ls Åij izcy :i ls xeZ djus ij nzo esa ifjofrZr gks tkrk gS] tks fd Bksl ikjn'khZ nzO;eku tks fd lkekU;r% cksjsDl eudk ds :i esa tkuk tkrk gS] esa ifjofrZr gks tkrk gSA ikjn'khZ dk¡p tSlk nzO;eku fuEu dk cuk gksrk gS& (A) lksfM;e ikbjkscksjV s (B) cksfjd ,ugkbMªkbM (C) lksfM;e esVkcksjsV (D*) cksfjd ,ugkbMªkbM o lksfM;e esVkcksjsV Sol.

  Na2B4O7 10H2O   Na2B4O7  2NaBO2 + B2O3

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ChemINFO-1.6

Solid State

Ionic solid AB Type The lattices in ionic crystals consists of alternative positive and negative ions in equivalent amount arranged in an order so that the potential energy of the ions in the lattice is minimum and such crystal normally found in ionic compound.

Daily Self-Study Dosage for mastering Chemistry

NaCl  Na+ + Cl – Limiting radius ratio + –

r /r

C.N

Shape

<0.155 0.155 –0.225 0.225 – 0.414

2 3 4

linear planar triangle Tetrahedral

0.414 – 0.732

4

Square planar

0.414 – 0.732

6

Octahedral

0.732 – 0.999

8

Body- centered cubic

NaCl Type IIn NaCl

r r–

value is 0.52 This value suggests it to be octahedral in shape. In this each Na+ ion surrounded by six Cl–

ion and each CI– is surrounded by six Na+ ions. In the octahedral structure of NaCl, Cl– ion may be regarded as having c.c.p arrangement in which all the octahedral holes are filled by Na + ions. Other examples of this structure are halide of Li. Na, K, Rb and Ag (except AgF) oxide and sulphides of Mg, Sr, Ba.

ZnS Type In ZnS the radius of cation to anion is 0.40 which suggests it to be tetrahedral in shape. Each Zn2+ ion is tetrahedrally surrounded by four S2–ions and each S2– is tetrahedrally surrounded by four Zn2+ ions. i.e.,coordination no. of ZnS. is 4. Thus this structure is termed as 4:4 arrangement it should be noted that zinc blend (ZnS) has ccp structure while wurtzuite (ZnS) has hcp structure. Other example are CuCI, CuBr, CuI, and BeS. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 17.

In the zinc blende structure, zinc ions occupy alternate tetrahedral voids and S 2– ions exist as ccp. The radii of Zn2+ and S2– ions are 0.83 Å and 1.74 Å respectively. The edge length of the ZnS unit cell is (A) 2.57 Å (B) 5.14 Å (C*) 5.93 Å (D) 4.86 Å

18.

If the distance between Na+ and Cl – ions in NaCl crystal is ‘a’ pm, what is the length of the cell edge? (A*) 2a

(B)

a 2

(C) 4a

(D)

19.

Which of the following statement is incorrect about rock salt type structure ? (A) It has FCC arrangement of Na+. (B) Na+ & Cl– ions have co-ordination number of 6 :6. (C) A unit cell fo NaCl consists of four NaCl units (D*) All halides of alkali metals have rock salt type structure.

20.

In the zinc blende structure S2– adopt ccp arragement and Zn2+ occupy. (A) Octahedral sites (B) Hexagonal sites (C*) Tetrahedral sites (D) None of these

a 4

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ChemINFO-1.6

Solid State Ionic solid AB Type

Daily Self-Study Dosage for mastering Chemistry

vk;fud fØLVy esa tkyd esa rqY;kad ek=kk esa ,d /kuk;uksa o _.kk;uksa esa bl çdkj ls Øe esa O;ofLFkr fd;k tkrk gS fd tkyd esa vk;uksa dh fLFkfrt ÅtkZ U;wUkre gksa o bl çdkj ds fØLVy lkekU; :i ls vk;fud ;kSfxd esa çkIr gksr s gSA such crystal normally found in ionic compound. NaCl  Na+ + Cl – Limiting radius ratio + –

r /r

NaCl NaCl

C.N

<0.155 0.155 –0.225 0.225 – 0.414

2 3 4

0.414 – 0.732

4

0.414 – 0.732

6

0.732 – 0.999

8

çdkj r r–

dk eku 0.52 gSA ;g çdkj v"VQydh; vkÑfÙk dk crkrk gSA blesa çR;sd Na+ vk;u ds pkjksa vksj N% Cl– vk;uksa

}kjk f?kjk jgrk gS o çR;sd CI– vk;u N % Na+ vk;uksa }kjk f?kjk jgrk gSA NaCldh v"VQydh; lajpuk esa Cl– vk;u c.c.p O;oLFkk j[krk gSA ftlesa lHkh v"VQydh; fNæksa dks Na+ vk;uksa }kjk Hkjk tkrk gSA bl lajpuk ds vU; mnkgj.k Li. Na, K, Rb o Ag (AgF ds flok;) ds gSykbM o Mg, Sr, Ba ds lYQkbM gSA

ZnS çdkj ZnS esa /kuk;u

o _.kk;u dh f=T;k 0.40 gS tks bls vkd`fÙk dksa prq"Qydh; crkrh gSA çR;sd Zn2+ vk;u pkj S2– vk;uksa }kjk prq"Qydh; :i ls f?kjk gksrk gS o çR;sd S2– vk;u pkj Zn2+ vk;uksa }kjk prq"Qydh; :i ls f?kjk gksrk gSA vFkkZr ZnS dh leUo; la[;k 4 gSA bl çdkj ;g lajpuk 4: 4 O;oLFkk j[krh gS] ;g /;ku nsus ;ksX; gS fd ftUd Cys.M (ZnS) dh ccp ljapuk j[krk gSA tcfd oqVZtkbV (ZnS) hcp j[krk gSA vU; mnkgj.k CuCI, CuBr, CuI, o BeS gSA Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 17.

ftUd Cys.M ljapuk esa ftUd vk;u ,dkUrfjr] prq"Qydh; fjfDr;ksa ls f?kjk gksrk gS o S2– vk;u ccp dsa :i esa gksrk gSA Zn2+ o S2– vk;uksa dh f=kT;k Øe'k% 0.83 Å o 1.74 Å gaSA ZnS ,dd~ dksf"Bdk ds fy, dksj yEckbZ fuEu gSA (A) 2.57 Å

18.

(B) 5.14 Å

(C*) 5.93 Å

(D) 4.86 Å

;fn NaCl fØLVYk esa Na+ o Cl – vk;uksa ds fy, nwjh ‘a’ pm gS] rks dksf"Bdk dh dksj yEckbZ D;k gS \ (A*) 2a

(B)

a 2

(C) 4a

(D)

19.

jkWd lkYV çdkj ljapuk ds fy, fuEu esa ls dkSulk dFku xyr gS \ (A) ;g Na+ vk;uksa ds fy, FCC O;oLFkk gS (B) ;g Na+ o Cl– vk;uksa ds fy, leUo; la[;k 6 :6 gSA (C) NaCl ds fy, ,dd~ dksf"Bdk pkj NaCl bdkbZ j[krk gSA (D*) {kkj /kkrqvksa ds lHkh gSykbM jkWd lkWYV çdkj dh ljapuk j[krs gSA

20.

ftUd Cys.M ljapuk esa S2– ccp O;oLFkk o Zn2+ vk;u (A) v"VQydh; O;oLFkk j[krs gSA (B) "kV~dks.kh; O;oLFkk j[krs gSA (C*) prq"Qydh; O;oLFkk j[krs gSA (D) dksbZ ugha j[krs gSA

a 4

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