EB_O_DPP20_WA

ORG./INO. CHEMISTRY

TARGET : JEE (Main) 2017 NO. 20

Course : AADHAAR(EB) AADHAAR(EB)

This DPP is to be discussed in the week (28-09-2015 to 03-10-2015)

DPP No. # 20 (Revision DPP) Total Marks : 63 Singl Single e choi choice ce Objec Objectiv tive e (‘–1’ (‘–1’ nega negativ tive e mark markin ing) g) Q.1 Q.1 to Q.21 Q.21

1.

(3 mark marks, s, 2 min.) min.)

Max. Time : 42 min. [63, [63, 42] 42]

Which of the following conditions conditions is required for the formation of hydrogen bond bond :

gkbMª kbM ªkst u ca/k cukus cuku s ds fy, fuEu fuE u esa ls dkSulh l h ifjfLFkf ifjf LFkfr;k¡ r;k¡ vko';d vko ';d gSa % (1) Hydrogen atom should be bonded to a highly electronegative atom.

gkbMª kbM ªkst u ijek.kq ijek.k q] mPp fo|q fo| qr_.kh; _.k h; ijek.kq ijek .kq ls caf/kr / kr gksu k pkfg,A pkfg, A (2) The size of electronegative atom should be small.

fo|q r_.kh; r_.k h; ijek.kq ijek. kq dk vkdkj Nks Vk Vk gksuk pkfg,A pkf g,A (3) There should be a lone pair of electron on the electronegative electronegative atom.

fo|q r_.kh; r_.k h; ijek.kq ijek .kq ij ,d ,dkdh ,dk dh bysD VªkW u ;qXe mifLFkr mifL Fkr gks uk uk pkfg;s pkfg ;sA (4* (4 *) All of the above.

mijks Dr Dr lHkhA Sol.

Conditions for H–bonding : (1) Positive charge density on H–atom should be high. (2) Availability Availability of lone pair of EN atom should be high. (3) Size of EN atom should be small.

Sol.

gkbMª kbM ªkst u ca/k ds fy, ifjfLFkf ifjf LFkfr;k¡ r;k¡ (1) H– ijek.kq jek .kq ij /kukRed /kuk Red vkos 'k ?kuRo mPp gksrk gks rk gSA (2) fo|q r_.kh; r_.kh ; ijek.kq ijek.k q ij ,dkdh bys DVª DVªk Wu ;qXe dh mifLFkfr mifLF kfr mPp gksu h pkfg,A pkfg, A ijek .kq dk vkdkj vkdk j NksVk gks uk uk pkfg,A pkfg ,A (3) fo|q r _.kh; ijek.kq

2.

How many of the following molecules are polar?

fuEu esa ls fdrus fdrus v.kq /kqz oh; gSa\ (i) CO2

(ii) SO2

(iii) NO2

(iv) SOCl2

(v) COCl2

(vi) BeCl2(g)

(vii) TeCl4

(viii) CCl4

(2*) 6

(3) 7

(4) 8

(ix) ClO 2 (1) 5

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PAGE NO.- 1

Sol.

CO2

NO2

3.



=0

;

sp

= O

  0

;

SO2

SO2

  0

SOCl2

COCl2



0

;

BeCl2

TeCl4



0

;

CCl4

ClO2

  0

  0

Cl – Be – Cl

=

0

  0

How many among the following species contain P–P bond(s ) : fuEu esa ls fdruh Lih'kht esa P–P cU/k mifLFkr gS %

(1) 2

Sol.

O= C=O

(i) P (red) yky

(ii) H4P2O5

(iii) H4P2O7

(iv) (PO3 –)3

(v) (PO3 –)2

(vi) P4O10

(vii) P4S3

(viii) P4O6

(ix) P4

(2*) 3

(3) 4

(i)

(4) 5

(ii)

(iii)

O O

P O

(iv)

(vi) O P

(v)

O

P

O O P O O

O

(vii)

4.

(viii)

(ix)

How many species have bond order more than one but less than three

fuEu esa ls fdruh iztkfr;k¡ ,d ls vf/kd ysfdu rhu ls de cU/k Øe j[krs gSa A O2, O –2 , N 2, N2 , N –2 , N –22 , B2, H 2, B –2 , C 2, C –22 (1) 5 Sol.

(2) 6

(3*) 7

(4) 8

O2, O –2 , N2 , N –2 , N –22 , B2 –, C2

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

5.

How many of these species are paramagentic ?

fuEu esa ls fdruh Lih'kht vuqpqEcdh; gS \ O2, O2+, O2 –, O22–, C2, B2, Be2,Li2 (1) 2

(2) 3

(3*) 4

(4) 5

Sol.

O2, O2+, O2 –, B2

6.

In which of the following species, the O–O bond length is expected to be longest ?

fuEu esa ls fdl iztkfr esa] (1*) K2O2 Sol.

O–O caèk yEckbZ (2) RbO2

(3) O2

(4) O2[PF6]

O22– (M.O.T)  1s2 1s2 2s2 2s2 (2p)2 ( 2p)4(*2p)4 Bond order ( caèk

Øe ) (O22–)

=

1 (10 – 8) = 1 2

Similarly ( blh

iz dkj), Bond order ( caèk Øe ) (O2 –)

Similarly ( blh

iz dkj), Bond order ( caèk Øe ) (O2) =

Similarly ( blh

izdkj),

Bond order ( caèk

( Bond order 

( 7.

lokZf/kd gksxh \

caèk Øe

Øe ) (O2+)

1 (10 – 7) = 1.5 2

=

1 2 =

(10 – 6) = 2

1 (10 – 5) = 2.5 2

1 Bond length ) 1



caèk yEckb Z )

Which of the following leads to bonding? s-orbital

p-orbital + –

(1) p-orbital

(3)

(2*)

p-orbital

–

+

+

–

(4)

fuEufyf[kr es a ls dkSu ca/k cukus ds i{k esa gS&

8.

(1)

(2*)

(3)

(4)

How many of the following are planar ? XeF2, ClF3, H2O, [XeF5] –, I 3 –, BCl3, XeF4, SF4, PCl5, SF6, IF7.

fuEu esa ls fdruh Lih'kht leryh; gSa \

XeF2, ClF3, H2O, [XeF5] –, I 3 –, BCl3, XeF4, SF4, PCl5, SF6, IF7. (1) 5

(2) 6

(3*) 7

(4) 8


PAGE NO.- 3

Sol.

(i) Planar molecules : XeF 2, ClF3, H2O, [XeF5] –, I3 –, BCl3, XeF4. (ii) SF4 – See - Saw shape PCl5 – Trigonal bipyramidal SF6 – Square bipyramidal IF – Pentagonal bipyramidal 7 (i) leryh; v.kq : XeF2, ClF3, H2O, [XeF5] –, I3 –, BCl3, XeF4. (ii) SF4 – lh&lkW vkd`fr PCl5 – f=kdks.kh; f}fijkfefM; SF6 – oxkZdkj f}fijkfefM; IF7 – i apdks.kh; f}fijkfefM;

9.

Match list l with List II and select the correct answer using the codes given below the lists. List I List II (Compound) (Shape) (a) CS2 1. Bent (b) SO2 2. Linear (c) BF3 3. Trigonal planer (d) NH3 4. Tetrahedral 5. Trigonal pyramidal lwph l rFkk lwph lI dks lqesfyr dhft, rFkk lwph ds uhps fn, x;s dwVksa dk mi;ksx dj lgh mÙkj pqfu;sA I II ( ) ( ) (a) CS2 1. eqMk gqvk (b) SO2 2. js[kh; (c) BF3 3. lery f=kdks.kh; (d) NH3 4. prq "Qydh; 5. f=kdks.kh; fijkfeMh; : (a) (b) (c) (d) (a) (b) (c) (d) (1*) 2 1 3 5 (2) 1 2 3 5 (3) 2 1 5 4 (4) 1 2 5 4

Sol.

(1) S  C  S (linear )

(2)

(bent)

(4)

(trigonal pyramidal)

sp

(3)

(1) S

(3)

(trigonal planar)



(2)

C  S ( js[kh; ) sp

( lery

f=kdks.kh;)

(4)

( eqM+k

gqvk)

( f=kdks .kh; fijkfeMh;)


PAGE NO.- 4

10.

Select the incorrect order(s). (1*) PH3 < AsH3 < NH3 < SbH3 – bond angle (3) CO < CO2 < CO32– – bond length

xyr Øe vFkok Øeksa dk p;u dhft;s % (1*) PH3 < AsH3 < NH3 < SbH3 – ca/k dks.k (3) CO < CO 2 < CO32– – ca/k yEckbZ Sol.

(2) Cl2O > F2O rFkk F2O < H2O – ca/k

(4) NH4+ > NH3 > NH2 – – ca/k (1) NH3 > PH3 > AsH3 > SbH3 – bond angle (2) Cl2O > F2O rFkk F2O < H2O 111° 102° 102° 104°

(3) C

Sol.

(2) Cl2O > F2O and F2O < H2O – bond angle (4) NH4+ > NH3 > NH2 – – bond angle

dks.k

dks.k

O < O=C=O <

triple bond double bond partial double bond character due to resonance (1) NH3 > PH3 > AsH3 > SbH3 – ca/k dks.k (2) Cl2O > F2O rFkk F2O < H2O 111° 102° 102° 104°

(3) C

O < O=C=O <

f=kca/k 11.

f}ca/k

vuqukn dsdkj.k vkaf'kd f}ca/k vfHky{k.k

Among the following molecular orbitals, how many have only one nodal plane?

fuEu es a ls fdrus vkf.od d{kdksa esa dsoy ,d uksMy ry gSS \ 1s, 2s, *1s, 2px , *2py, 2py

(1) 2

Sol.

(2*) 3

(4) 5

*1s

uks My ry

2px or 2py

12.

(3) 4

*2px or *2py uks My

ry

How many antibonding electrons are there in O 2+ ? O2+ esa (1) 2

fdrus izfr vkca/kh bysDVªkWu mifLFkr gSA (2) 3

(3) 4

(4*) 5


PAGE NO.- 5

13

How many of the following species have all bonds of equal length ?

fuEu esa ls fdruh iztkfr;ksa dsf y, lHkh cU/kksa dh yEckbZ leku gksrh gS\ (a) ClO4 –

(b) NO3 –

(1) 2 14.

(c) AsO43–

(2) 3

(d) CO2

(e) SO32–

(3) 4

(4*) 5

The shape of SO24 ion is : (1) Square planar

(2*) Tetrahedral

(3) Trigonal bipyramidal (4) Hexagonal

SO24 vk;u

dh vkÑfr gS % (1) oxZleryh; (2*) prq "Qydh; 15.

(3) f=kHkq th; f}fijkfeMh;

(4) "kV~ dks .kh;

IUPAC name of the following compound is fuEu ;kSfxd dk IUPAC uke gS

(1) 3-(Prop-2-ynl)hexe-1,3,5-triene (3) 4-Ethenylhexa-4, 6-dien-1-yne s -2-vkbZ fuy) g Ds l-1,3,5- V kª bZ bu (1) 3-( i kz i

(2*) 4-Ethenylhepta-1, 3-dien-6-yne (4) 4-(Prop-2-ynyl) hexa-1,3,5-triene (2*) 4- , fs Fkukbygs IVk-1, 3- MkbZ u-6-vkbu

(3) 4- , fs Fkukbygs Dlk-4, 6- MkbZ u-1-vkbu

s -2-vkbZ fuy) g Ds lk-1,3,5- V kª bZ b uZ (4) 4-( i kz i

Sol.

16.

The IUPAC name of the

is :

(1) 1–Ethyl–4–chlorocyclopent–1,3–diene (2) 1–Chloro–4–ethylcyclopent–1,3–diene (3) 1–Ethyl–4–chlorocyclopenta–1,3–diene (4*) 1–Chloro–4–ethylcyclopenta–1,3–diene

dk

17.

IUPAC uke

gS %

(1) 1– ,fFky –4– Dyks jks lkbDyks i Us V –1,3– MkbbZ u

(2) 1– Dyks j –4– ks ,fFkylkbDyks i Us V –1,3– MkbbZ u

(3) 1– ,fFky –4– Dyks jks lkbDyks i Us Vk –1,3– MkbbZ u

(4*) 1– Dyks j –4– ks ,fFkylkbDyks i Us Vk –1,3– MkbbZ u

IUPAC name of

is

(1) N-Deutero-N-formylbenzenamine (3*) N-Deutero-N-phenylmethanamide

;kSfxd

Sol.

dk

IUPAC uke

(2) N-Phenylamino-N-deuteromethanal (4) N-Deuterobenzene carboxamide

gS

~ fw Vfj;ks-N- QkfeZ ycs Uthuvehu (1) N- M ;

~ fw Vfj;ks eFs ks u y s (2) N- Q fs uyvehuks-N- M ;

~ fw Vfj;ks-N- Q fs uyesFks ukekbM (3*) N- M ;

~ fw Vfj;ks c US thu dkcks DZ lkekbM (4) N- M ;

D | H – C – N – Ph || O N-Deutero-N-phenylmethanamide. ~ fw Vfj;ks-N- Q fs uyesFks ukekbM N- M ;


PAGE NO.- 6

18.

The IUPAC name of following compounds is HOOC  CH2  CH  CH2  CH2  COOH |

COOH

(1) 3-Carboxyhexane -1,6- dioic acid

(2*) Butane-1,2, 4-tricarboxylic acid

(3) 4-Carboxyhexane-1,6-dioic acid

(4) 1, 2-Dicarboxypentanoic acid

HOOC  CH2  CH  CH2  CH2  COOH dk IUPAC uke

fuEufyf[kr esa gSA

|

COOH

(1) 3- dkcks fZ Dlgs Dls u-1,6- MkbvkW bd

vEy (3) 4- dkcks fZ Dl gs Dls u-1,6- MkbvkWbd vEy 19.

(2*) C;q V su-1,2, 4- V kª bZdkcks l Z fs yd

vEy Z vEy (4) 1, 2- Mkbdkcks fZ Dlis UVsukW b d

The total number of lone pairs in chlorate ion is :

DyksjsV vk;u esd a qy ,dkdh bySDVªkWu ;qXeksa dh la[;k gS % (1) 5

(2) 6

(3) 7

(4*) 8

Sol. 20.

In how many of the following species the central atoms have two lone pairs of electrons ?

fuEu esas ls fdruh Lih'kht es]a dsUnzh; ijek.kqi j nks ,dkdh bysDVªkW u ;qXe mifLFkr gSa \ XeF4 ICl4 –

ClF3 SCl2

XeF3+

F2SeO2 XeOF2

(1) 5

(2) 6

NH2 –

ClOF3

(3*) 7

(4) 8

F

Sol.

XeF4

ClF3

F | Xe – F | F

XeF3+

F2SeO2

+ NH2 –

ClOF3

– Cl I

ICl4 –

Cl

21.

Cl

SCl2 Cl

XeOF2

S Cl

Cl

In how many of the given species there is no any lone pair on the central atom.

fn;s x;s ;kSfxdksa esa fdrus ij dsfUæ; ijek.kqi j ,dkdh bysDVªkWu ;qXe ugh gSa A (i) XeF4 (vi) XeOF4 (1*) 5

(ii) NH3 (vii) ICl3 (2) 6

(iii) SO2 (viii) IF7

(iv) NO3 – (ix) SO42–

(3) 7

(v) O3 (x) XeO3 (4) 8


PAGE NO.- 7

Sol.

(i) XeF4

(ii) NH3

(iv) NO3 –

(iii) SO2

(v) O3

(vi) XeOF4

(vii) ICl3

(ix) SO42–

(x) XeO3

(viii) IF7


PAGE NO.- 8

EB_O_DPP20_WA

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