XII - ISC BOARD - Rao IIT Board Paper... · 2015. 3. 23. · rao iit academy/ isc - board...

Rao IIT Academy/ ISC - Board _Chemistry_Solutions

1 1SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR |SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

PART IQuestion 1:(a)(i) positive , lesser(ii) efficient , 74(iii) decreases , common-ion effect(iv) Cannizzaro , no hydrogen (v) paramagnetic , dimagnetic(b)(i) (2) –0.372°C(ii) (3) 1155(iii) (2) Secondary Alcohol(iv) (4) Octahedral and sp3d3

(v) (1) +3 and + 4(c)(i) As temperature increases, concentration of H+ and OH– increases and hence pH value decreases.

At 298 K.14H OH 10

for neutral solution pH = 7and at high temperature pH will be less than 7.

(ii) 3 2Fe e Fe

for 1 mole of 3 ,Fe 1 mole of e is required

for 3 moles of 3 ,Fe 3 moles of e are required

3 3 3 96500e F C

Q I t

3 96500 2 t

3 965002 3600

t

40.2t hrs

XII - ISC BOARDDate: 20.03.2015 CHEMISTRY - SOLUTIONS



(iii) Mixture containing potassium cyanate and ammonium sulphate on heating gives ammonium cyanate whichfurther gives urea

4 2 4 4 2 4urea

2 ( ) 2KCNO NH SO NH CNO N H CO

(iv) Yes(v) 3 2 3 Strong AcidWeak Acid

CH COONa H O CH COOH NaOH

3 2 3CH COO Na H O CH COOH Na OH 1

3 2 3CH COO H O CH COOH OH

OH are left in the solution

Sodium acetate on hydrolysis gives basic solution.

(d) (i) ; ii (iii) iv vd c b e a

PART II

SECTION AQuestion 2:(a)(i) 2 0.5w g

1 100w g1

1 18 .M g mol

0 0.24 0.24fT C

22

1

1000 1000 1.86 0.5 38.75 /0.24 100

f

f

k wM g mol

T w

Calculated (theoritical) mol mass of 39 35.5KCl

174.5 g mol

Calculated molar mass 74.5 1.92Theobserved molar mass 38.75

i

1.92i

KCl K Cl

Initial 1 0 0Equillibrium 1-

KCl K Cl

total number of moles after dissociation = 1

= 1

1 or 11

i i

1.92 1 0.92 degree of dissociation = 0.92



(ii) Given:mass of sugar = 1.71 gmolar mass = 342 gvolume of solution = 500 mltemperature = 300 KFormula used

number of moles mass of sugarmolar mass

number of moles of soluteMolarity =volume of the solution in litres

Osmotic pressure CRT

Solution:

number of moles = 1.71342

0.005

0.005 1000Molarity C500

= 0.01 MOsmotic Pressure = CRT 0.01 0.083 300

0.249 bar

(iii) Given

Mass of solute = 20.70 g w

Mass of solvent = 32 g (w1)

0.25bT C

11.72 .bK K kg mol

To find = M2 (molecular mass of organic compound)

Formula 22

1

1000 b

b

K wMw T

21000 1.72 0.70

32 0.25M

2 150.5M g



(b)(i)

Order Molecularity 1. Order of a reaction is an

experimental quantity. It can be zero and even a fr action.

Molecularity of a reaction is number of species (atoms, ions or molecules) taking part in elementary reaction. It cannot be zero or non integer

2. Order is applicable to e lementary as well as complex reactions.

Molecularity is applicable only for elementary reactions.

3. For complex reaction, order is given by slowest step.

Molecularity of the slowest step is equal to the order of over all reaction.

(ii) Given : Following is a 1st order reaction.

1/ 2t half life of a reaction = 120 min

To Find :Time required to complete 90% of the reaction.Solution:Let the reaction constant be KInitial concentration [R0]Now calculating constant K

1/ 2

0.693Kt

0.693120

K

10.0058 minK

Now time required to complete 90% of the reaction

Final concentration 0 00.9R R

00.1 R

Now,

2.303 log0.1

o

o

RK

t R

2.303 1log

0.0058 0.1t

2.303

0.0058t

397.06 mint time required to complete 90% of reaction = 397 .06 min.

(c) Crystal structure of Cu metal is FCC or CCP[Face centered cubic or cubic closed packed]



Question 3:(a)(i) Given :

Crystal structure of Cr = BCCEdge length (a) = 287 pmTo findAtomic radius r

Density

Solution:For BCC crystal system

Edge length (a) 4

3r

34

ar

0.433 287 pm

124.27r pmNow density of a crystal :

3A

zMa N

Wherenumber of atom in BCC unit cell 2z

M = Molar mass of 52.99Cr g

33 287a pm

312287 10 m

310287 10 cm

23 32.37 10 cm

236.022 10AN atoms

3DensityA

zMa N

23 232 52.99

2.37 10 6.022 10

105.9814.27

Density of crystal = 7.42 g/cm3



(ii) NaCl show metal excess defect due to anionic vacancies. When crystals of NaCl are heated in an atmosphere

of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse to thesurface of the crystal and combine with Na atoms to give NaCl. This happens by loss of electrons by sodiumatoms to form Na+ ions. The released electrons diffuse into the crystal and occupy anionic sites. As a resultcrystal now has excess Na+ ions. The anionic sites occupied by unpaired electrons are called F–centres. Theyimpart yellow colour to the crystals of NaCl. The colour results by excitation of these electrons when theyabsorb energy from visible light falling on the crystal.

(iii) 2( ) 2 33 2g g gN H NH

K (equilibrium constant) = 6.0 × 10–2 1mol L

Temperature (T) = 715 K

12 0.25H mol L

13 0.06NH mol L

2 ?N

23

32 2

NHK

N H

23

2 32

NHN

K H

2

32

0.06

6.0 10 0.25

20.0036

6.0 10 0.015625

12 3.84N mol L

(iv) 2 16.2 10H mol L 1410H OH

Now putting the value of H

14

210

6.2 10OH

13 11.61 10OH mol L (v) Le–Chatelier’s principle : The principle states that if an external stress is applied to a reacting system at

equilibrium, the system will adjust itself in such a way that the effect of the stress is reduced.



(b)(i) NaCl crystallises in FCC lattice

(ii) Each Na+ ion is surrounded by six Cl ions and each Cl ion is surrounded by six Na ions.(iii) Number of Cl ions = 4

Number of Na ions 4

(iv) In the NaCl crystal, Cl ions occupy all the FCC positions and Na+ ions occupy the octahedral voids.(c)(i) Increase in temperature – forward reaction

Decrease in temperature – backward reaction(ii) Increase in pressure – forward reaction

Decrease in pressure – backward reaction(iii) Increase in conc. of [N2O4] – forward reaction

Decrease in conc. of [N2O4] – backward reaction(iv) Removal of NO2 causes decrease in [NO2] – forward reactionQuestion 4:

(a) 4 1 11.65 10K cm

3

0 1 2 1

0 1 2 1

349.1

40.9H

CH COO

cm mol

cm mol

0.01 .C M

(i)4

1 2 12

1000 1000 1.65 10 16.510

K cm molC

1 2 116.5 cm mol

(ii) 3 349.1 40.9 390.0H CH COO

16.5 0.0423390

0.0423

(iii)2 2(16.5) 0.01

( ) 390 (390 16.5)CK

51.86 10K



(b)

(i) 2 2Mg Cu Mg Cu

2 3

4210 1010

MgQ

Cu

2n

010

0.0592 logcell cellE E Qn

100.05922.71 log 10

2

2.71 0.0296

2.68cellE V

(ii) In pure water2 2

4spK Ba SO

9 21.5 10 S

91.5 10S

1015 10

53.87 10S M

in 0.1 M 22BaCl Ba 0.1x

Cl x

22K Ba Clsp

9 21.5 10 0.1x x

9 21.5 10 0.1 is very smallx x

2 81.5 10x

81.5 10x

41.224 10x M



(c)

(i) 4 4NH OH NH OH

4 4NH Cl NH Cl

33Fe 3OH Fe OH

22Ca 2OH Ca OH

sp3 2K Fe OH is very less compared to K Ca OHsp hence 3Fe OH will be precipitated

and 2Ca OH remains in solution.

(ii) 22H S 2H S when strong acid is added HCl H Cl . Due to common ion effect,

concentration of H+ is suppressed. Therefore dissociation of 2H S suppressed.

SECTION B

Question 5:(a) IUPAC Name

(i) 3 2 34 2Cr NH H O Cl Tetra ammine diaqua chromium (III) chloride

(ii) 2 3 44Pt Cl NH PtCl Tetraamminedichloro platinum (IV) tetrachloro platinate (II)

(b) 3

6Fe CN

hybridisation = d2sp3

magnetic property = paramagnetic(c)

(i) 2 2

3 4 3 45 5Co NH Br SO Co NH Br SO

1

3 4 3 45 5Co NH SO Br Co NH SO Br

Both are ionisation isomers.

3 4 2 45 red violet ppt ofCo NH Br SO BaCl BaSO

3 4 35 red ppt ofCo NH SO Br AgNO AgBr



(ii) 2 2Co(en) Cl

Co Co

Cl ClCl Cl

enen

en en

+ +

Cis form

Non super imposable mirror image structure

Question 6:(a)

(i) 2 2 22 2 2NaOH F F O NaF H O

(ii) 24 2 2 2 22 6 5 2 8 5MnO H H O Mn H O O

(iii) 2 4 2 2 22H SO H S H O SO S

(b) 4XeF

Xe

F

FF

F

Hybridisation = 3 2sp dGeometry = square planar

Question 7:(a) Silver :

Argentite or silver glance 2Ag SLeaching Process:Steps:

2 22Ag S 4 NaCN 2 Na Ag CN Na S

2AgCl + 2 NaCN Na Ag CN NaCl

2 2 2 4Na S 2O Na SO

2 2 2 42Ag S + 4 NaCN + 2O 2 Na Ag CN + Na SO

22 4ppt

2 Na Ag CN + Zn 2 Ag + Na Zn CN



(b)(i) Due to small size of transition metal atoms or their cations and high effective nuclear charge, they have

a high positive charge density on them. This high positive charge density makes the atoms or cations toattract the lone pairs of e– from the ligands.

The transition metal cation or atoms have vacant d-orbital in which they can accomodate the lone pairs

of e– donated by ligands and thus can form L M coordinate bonds.(ii) Paramagnetic character increases with the increase in the number of unpaired electrons. The larger is the

number of unpaired e– in central atom or ion greater is the paramagnetic character.

Hence from Sc to Mn number of unpaired e– increases and then from Fe to Zn decreases. Hence paramagneticcharacter also increases from Sc to Mn and then decreases from Fe to Zn.

SECTION CQuestion 8:(a)(i) Glycerol to formic acid

HO OHOH

3H– C – OHFormic

acid

Pd(I) CatalystH O2 2

Glycerol

O

(ii) Chlorobenzene to phenolCl OH

+ OH–

Chlorobenzene Phenol

NaOHpressure, H+

(iii) Diethyl ether to ethanol

3 2 2 3 2 5 2 5Diethyl ether Ethanol

CH CH O CH CH HI C H OH C H I

(iv) Phenol to anilineOH NH2

+ NH3

Phenol Aniline

ZnCl /2

High pressure + H2O

(b)

(i) 3 2 2 3 2Ethanol Iodoform

CH CH OH 4I 6 NaOH CHI HCOONa 5 NaI 5H O

(ii) Chlorobenzene is heated with sodium metal in presence of dry ether gives diphenyl compound.

+ 2Na2

Cl

dryether + 2 NaCl

Chlorobenzene Diphenyl Compound



(c) Identifying compounds :

CH COCH3 3conc. HNO3

SOCl2

(O)CH C OH3

– – CH – C – Cl3

ONH3 CH – C – NH3 2

O

LiAlH4

CH CH NH3 2 2HNO2CH COCl3 C H OH2 5C H – O – C –CH2 5 3

O

O

(A) (B) C

(D)(E)(F)

Question 9:(a)(i) Reimer – Tiemann reaction :

OH ONa OH

+ CHCl3

Phenol

NaOH, 340 K –H O2

CHCl2 CH(OH)22 NaOH

ONa

CHO

– H O2

dil. HCl– NaCl

OH

CHO

Hydroxybenzaldehyde

(ii) Rosemund Reaction :

CH – C – Cl + H3 2 CH – C – H + HCl3

O OPd , BaSO4

Boiling xyleneAcetyl chloride Acetaldehyde

(iii) Hoffmann’s degradation reaction :

CH – C – NH3 2 + Br + 4NaOH2 CH NH + 2 NaBr + Na CO + 2H O3 2 2 3 2

O

Acetamide Methylamine



(b)(i) Ethylamine and diethylamine

Ethylamine reacts with Nitrous acid in cold condition to form unstable diazonium salt and which decomposesto give alcohol with liberation of N2 gas.

2 5 2 2 2 5 2 2C H NH + NaNO + 2HCl C H N Cl NaCl + 2H O +

2 5 2 2 5 2C H N Cl H O C H OH + HCl N diethylamine reacting with Nitrous acid give yellow oily N–Nitrosoamines

22 5 2 5 2 5 2

N nitroso diethylamineC H NH C H HO N = O C H N N = ONaNO

HCl

(ii) Acetaldehyde and benzaldehyde

Acetaldehyde reacts as reducing agent which reduces Fehling solution. The aldehyde gets oxidized tocarboxylate ion and Cu2+ is reduced to Cuprous oxide

23 3 2

Carboxylateion

Acetaldehyde

O|

CH C H + 2Cu 5OH CH C = O Cu O||O

Benzaldehyde

CHO(aromatic aldehyde) are not oxidised by Fehling solution.

(c)(i) Compounds in ascending order of their basic strength:

diethylether < Aniline < Methylamine < ethylamineIn aliphatic amines, due to presence of electron donating alkyl group, electron density on nitrogen increases.Hence lone pair is easily available for protonation.In aniline the lone pair of nitrogen is in conjugation with benzene and less available for protonation henceaniline is less basic than aliphatic amine and ethers are inert in nature due to stable C O C bond .

(ii) (a) Polyester :Monomer used :Ethylene Glycol 2 2OH CH CH OH

Dimethyl terephthalate H C – O – C3 C – O – CH3

O OType of Polymerization : Condensation polymerization.

(b) Bakelite :Monomer used :Phenol :

OH

Formaldehyde – HCOOH

Type of Polymerization : Condensation polymerization.



Question 10:

(a) 3 2 2A CH CH NH

3 2 2 2 3 2 2 2(B)(A)

CH CH NH HNO CH CH OH N H O

3 2 3 2(C)

Controlled

(B)CH CH OH (O) CH CHO H O

3 3 2 3 3 2(D)(C)

CH CHO 2[Ag(NH ) ] + 3OH CH COOH 2Ag 4NH 2H O

CH COOH + HO – CH – CH3 2 3 CH – COO – C H + H O3 2 5 2Con. H SO2 4

dil. H SO2 4

(E)Ethyl acetate

CH – CHO – NH3 3 CH – CH – OH3 CH – CH = NH3

NH2

Acetaldamine(C)

(b)

(i)Cold

3 2 2 2 3 2 2 2Ethylamine

CH CH NH HNO CH CH OH N H O

72 2

2 43 2 3 2

K Cr Odil. H SO ,CH CH OH (O) CH COOH H O

CH – COOH + NH3 3 CH – C – O NH3 – +

4

O

CH – C – O NH Br + 2KOH 3 2 – +

4 +

O CH – NH + 2KBr + 2H O + CO3 2 2 2

Methylamine

(ii) When globular proteins undergo coagulation or precipitation to give fibrous proteins, due to coagulation thenative shape of protein is destroyed and biological activity is lost. That is why coagulated proteins so formedare called denaturated proteins.During denaturation, secondary and tertiary structures are destroyed but primary structure remains intact.

(iii)

Adenine (A) Guanine (G)

NH2

N

N

N

N

O

N

N

N

NH N2

Thymine (T)

O

N

N

CH3

O

Cytosine (C)

NH2

N

NO



(c)

(i)

NH2 N2Cl

+ HNO + HCl2273

278 k

+ –

+ 2 H2O

(ii) CH – CO – Cl + HO – C H3 2 5 CH – COO – C H3 2 5 + HClPyridine

(iii)3 2 6 4 2

O||

6 H C H 4 NH (CH ) N 6H O

XII - ISC BOARD - Rao IIT Board Paper... · 2015. 3. 23. · rao iit academy/ isc - board...

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Transcript of XII - ISC BOARD - Rao IIT Board Paper... · 2015. 3. 23. · rao iit academy/ isc - board...