CBSE ANNUAL PAPER - 1998

CHEMISTRY

(SET-I)

Time allowed : 3 hours

Maximum Marks : 70

General Instructions :
(i) All questions are compulsory.

(ii) Marks for each question are indicated against it.

(iii) Question numbers 1 to 10 are very short-answer questions each of 1 mark. Answer them in about one sentence each.

(iv) Question numbers 11 to 26 are short - answer question of 2 marks each. Answer them in not more than 30 words each.

(v) Question numbers 27 to 32 are short - answer questions of 3 marks each. Answer them in not more than 40 words each.

(vi) Question numbers 33 and 34 are long-answer questions of 5 marks each. Answer them in not more than 70 words each.

(vii) Use Log Tables, if necessary.

Why are all the P.F. bonds in PF₅ molecule not of the same length ?

The molecule of PF₅ has trigonal bypyrimidal geometry with three equatorial fluorine atoms close to the central phosphorous atom than the two vertical fluorine atoms. Hence all the five P-F bonds are of different lengths.

Name the type of structure processed by a unit cell of C_sCL.

Unit cell of cezium chloride has body centered cubic crystal lattice structure.

Unit cell of cezium chloride has body centered cubic crystal lattice structure.

23/11 Na.... ------> 23/12 Mg + )0/1 n

CH₃ --CO--CH--CH₂-->CH₂Cl

CH₃

CH₃

_{5 chloro 3 methyl 2 pentanone.}

Al, B,Ca, tl, Be.

[Co(ONO) (NH₃)₅] Cl₂

What type of substance is phenylalanine bydroxlase ? What is the importance for us ?

Phenylalanine hydroxylase is an enzyme whose deficiency causes congenital disease phenyketone urea.

What is the role of dimethyl and monomethyl hydrazines in rockets ?

Dimethyl and monomethyl hydrazines along with liquid N₂O₄ are used as liquid propellents in rockets.

Alizarine is an anthraquinone dye, it is also a morodant dye.

In the compound AX, the radius of A+ ion is 95 pm and that of Ion is 181 pm. Predict the crystal structure of AX and write the coordination numbers of each of the ions.

Radius A+ / X- = 95/181 = 0.5 Which lies between 0.414 - 0.732 A + X- has octahedral crystallattice with co-ordination numbers A + : X- as 6:6 each.

Differentiate between molarity and molarity of solution. When and why is molarity preferred over molarity in handling solutions in chemistry ?

Molarity of a solution equals number of moles of solute in one litre of the solution but molarity of a solution refers to the number of moles of solute dissolved in one kg. of the solvent.

Molarity of a solution is preferred for expressing the concentration in procedures where temperature changes may occur because molarity unlike molarity does not change with temperature. Colligative properties like depression of ferrzing point,

Define the term 'entropy'. How does TDS determine the spontaneity of process ?

Entrophy is the property of a system which describes the disorder in it. The more the disorder in the system greater its entropy.

For any spontaneous reaction DG should be negative By the relation DG = DH - TDS if DH is + ve, this reaction will be spontaneous only when DS is positive and TDS is greater than DH. Then DG will be -ve.

(i) rate of reaction and rate constant of a reaction.

(ii) order of a reaction and its molecularity.

Calculate the energy change in J/mol for the nuclear reaction

0/1 n -----> 1/1 p + 0/1 e

Decrease in mass = 1.00867 - 1.00728

= 0.00139 amu.

Energy change per atom = mc ²

= 0.00139 x 1.66 x 10^-27 x (3 x 10⁸ ms)²

Energy change per mole = 0.00139 x 1.66 x 10^-27 x (3x10⁸)² x 6.02 x 10²³

= 1.24 x 10^-27 Jmol^-1

(i) Haloarenes are insoluble in water but are soluble in benzene.

(ii) Phenols are move acidic than alcohols.

Haloarenes are insoluble in water because they cannot break hydrogen bounds between water molecules. They are however soluble in benzene due to similar structure of haloarene and benzene.

(ii) Phenols are more acidic than alcohols because the phenoxide ion C6H5O is more stablised by resonance than phenol.

Anorganic compound (A) which has a characteristic odour, on treatment with NaOH forms two compounds (B) and (C). Compound (B) hs the molecular formula. C₇H₈O which on oxidation gives back compound (A). Compound (c) is the sodium salt of an acid. (C) when heated with sodalime yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D).

Add iodine and dil NaOH solution and warm no reaction.

Propanone (CH₃COCH₃)

On boiling with I2 and NaOH (aq) a yellow crystalline substance iodoform is produced.

CH₃C-CH₃ ^I₂^+3NaoH

|| CH I₃ + CH₃ CooNa

O

(ii) Acetaldehyde ( CH₃ CHO)

Add Fehling reagent Cuso₄ + Naoh and warm. A reddish precipitate is formed

2CU ^z+ 50H + CH₃ CHO -- CH₃COO + 3H₂O + Cu₂O (red)

Benzaldehyde ( C₆H₅CHO)

on warning with fehling solution C₆H₅CHO does not give any reddish colour.

C₆H₅CHO + 50H + 2Cu 2+ -- No reaction

What is meant by 'Catenation'? How does the catenation tendency for elements of group 14 vary ?

Catenation is the property of interlinking of atoms of the same element to from long, branched or cyclic chains.

This phenomenon is shown by carbon, sulphur and phosphorous.

• Tendency for contenation decreases in going down group 14 C-C>Si-Si>Ge-Ge>Sn-Sn>Pb-Pb

This is due to decrease in the strength of M-M bond C-C>Si-Si>Ge - Ge > Sn-Sn > Pb - Pb

Using the Valence Bond approach, explain the shape and magnetic behaviour of [Ni (NH₃)₆]²⁺. Given : At. No. of Ni = 28.

[ Ni(NH₃)₆]²⁺ has octahedral geometry because the six pairs of electrons donated by 6 NH₃ ligands are accommodated in six sp³ d² hyberdised orbitals of Ni²⁺.

There are two upaired electrons in the complex (in 3d orbitals) It is paramagnetic.

Calculate the standard free energy change for the reaction occurring in the cell :

Zn(s) | Zn²⁺ (!M) || Cu²⁺ (1M) | Cu(s) [ Given E⁰ Zn²⁺ / Zn = -0.76V, E⁰ Cu²⁺ /Cu = +0.34 V; F + (^%)) Cmol^-1]

(1M) (1M)

E⁰ cell = C⁰ Cathode - E⁰ Anode = +0.34 -(-0.76) = 1.10 Volt.

D G0 = nFE⁰ ( Cell reactuib us Zn(s) + Cu²+ (aq) --- Zn²+ (aq) + CU⁰ (s)

= - 2x 96500 x 1.10

= - 212.3 K Joule.

Equilibrium constant K is related to free energy change for the cell reaction by the relation.

D G⁰ = -2.303 x RT log K = - nFE⁰

The decomposition of N2 in CCl4 solution follows the first order rate law. The concentrations of N2O5; measured at different time intervals are given below :

formula to be inserted.

Calculate its rate constant at t = 410 s and t = 1130s. What do these results show ?

k = 2.303 / t log a/a-x ( ist order reaction)

K1 = 2.303 /410 log 5.5/4.0 = 7.708 x 10^-3

k2 = 2.303 / 1130 log 5.5 / 2.4 = 7.401 x 10^-3

Write equations for the synthesis of the following polymers : (a) Glypptal (b) Teflon

(b) Synthesis of Teflon --------->

n(CF₂ -- CF₂) n ----------> (CF₂--CF₂) n

(Tetrafluro-ethylene) Teflon

Explain the term CONDON in relation to mRNA. What are nucleotides ?

(i) Condon : mRNA has Adenine, guanine, cytosin and uracil bases attached to ribose sugar. Base triplet on m RNA chain which codes for a specific ammino acid is called a condon.

(ii) Nucleotides : A nucleotide consists of three components.

(a) A five carbon atom sugar Ribox/deoxyribox.

(b) A phosphate - PO₄

(c) A nitrogenous base - adenine, guanine, cytosine.

base

Uracil and thiamine :Sugar --Phosphate Re[resemts miceptode/

What is genetic engineering ? In what way are biotechnological products useful to human beings? Illustrate with two examples.

Genetic engineering is a biotechnology involving enzymes which help in inserting a section of a DNA molecule into an organism ( bacterium) to synthesise newer proteins with altered aminoacid sequences.

Biotechmoligal products are helpful to human beings.

(i) Insulin is used for treatment of diabetes.

(ii) interferon is anti viral agent.

In what way Heisenberg's uncertainty principle contradicts the concept of stationary orbit for electrons as suggested by Bohr ?

An electron has a speed of 500 ms^-1 with an uncertainty of 0.02 %. What is the uncertainty in locating its position ?

[ Give mass of electron = 9.1 x 10^-31 kg, h = 6.6 x 10^-34 Js]

(a) The concept of stationary orbits for electron suggested by Bohr assigns a fixed position to an electron moving round the nucleus whereas heisen - burg "it is impossible to determine simultaneously and accurately the position and momentum of the moving electron. The position and momentum of the moving electron. The position and momentum of the moving electron in uncertain.

(b)D V = 0.02 / 100 x 500 = 0.1ms^-1

m = 9.1 x 10-31kg

m DV x D x = h /4p x 1 /mDV

= 6.6 x 10^-34 x 7 / 4 x 22 x 1 / 9.1 x 10^-31x0.1

= 66 x 7 / 4x22x9113 x 10^-34x10⁺³¹ x x 10

3/52 x 10^-2 = 5.77 x 10^-4

Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00g. of Glauber's salt, Na₂SO₄. 10H2O in 0.100 kg of H₂O. [ Given for water, Kr = 1.86 K kg mol-1 Atomic masses : Na = 23, S = 32, O = 16, H = 1 amu]

Molar mass of Na₂ SO₄ - 10H₂O = 44 + 32 + 64 + 10 x 18 = 322.

Wa = Mass of H₂O = 0.1 kg = 100 g

WB = Mass of the solute = 6.0 g

Kf = 1.86 k kg. mol^-1 and i (Vont Hof factor) = 3

By the relation DTf = kf / MB x 1WB/WA x 1000 x i

= 1.86 / 322 x 6.0 / 100 x 1000 x 3 = 0.346k

Freezing point of solution = 273 - 0.346 k = .654 k.

Calculate the standard free energy change for the following reaction at 27⁰ C. f

H₂ (g) + I₂(g) -- 2HI (g) ; D H0 = + 51.9 kJ

[Given S⁰ H2 = 130.6JK^-1 mol^-1, S⁰ I₂ = 116.7 JK^-1

mol^-1 S⁰ HI = 206.3 JK^-1 mol^-1

= S⁰ (2HI) - [S⁰(H2) + S⁰ (I₂)]

= 2 x 206.3 - (130.6 + 116.7)

= 412.6 - 247.3 = 165.3Jk^-1

DH⁰ = 51.9 kJ

= 51900 J

DG⁰ = DH0 - TDS⁰

= 51900 - 300 x 165.3

= 51900 - 49590 = 2310 Joule

Since DG0 is +ve the reaction is not feasible at 300 k.

(a) Compare physical adsorption and chemical adsorption in terms of rate and prevailing temperature.

(b) Show graphically how the amount of a gas adsorbed on a solid in physical adsorption varies with (i) pressure and (ii) temperature ?

How are the following conversions carried out ? (Write reactions and conditions only)

(i) 1-Bromopropane to 2-bromopropane

(ii) propanone to iodoform

(iii) phenol to salicylic acid

(i) NH₃ has higher boiling point than PH3

(ii) H₃PO₃ is a diprotic acid

(iii) OF₂ should be called oxygen difluoride and not fluorine oxide.

(a) Due to high electronegativity of nitrogen atom of NH₃ its molecules are associated through hydrogen bonds

H H H

| | |

H---N-----H---N-----H-----N--H

| |

H H

It has a higher boiling point than PH₃. Phosphorous is less electrogengative than nitrogen. No hydrogen bonds exist between PH₃ molecules.

(b) Experiments show that H₃PO₃ has reducing properties. One of its three hydrogen atoms is directly attached to P-atom as in the structure

H

|

HO-----O---OH

||

O

This H-atom is not ionised while two other hydrogens are ionisable they exist as - OH groups. This explains why H₃PO₃ is diprotic.

(c) in OF₂ oxygen is electropositive with respect to fluorine OF₂ is called oxygen fluride and not fluorine oxide.

What is vapour phase nitration ? Describe the preparation of nitroalkanes and alkyl nitrites from alkyl halides. Explain why the boiling points of nitroalkanes are higher than those of corresponding alkanes ?

How can the following compounds be prepared :

(i0 Aniline from nitrobenzene

(ii) Ethylamine from nitroethane

(i) Vapour phase nitration : Saturday by hydrocarbons react with nitric acid in vapour phase between 420 - 700 K when a mixture of nitro derivatives is formed.

(iii) Boiling points of nitro compounds are higher than the corresponding hydrocarbons (alkanes ) because alkanes are nonpolar whereas nitroalkanes are polar compounds.

(iv) Conversions (i) aniline from nitrobenzene

(a) Describe with the help of chemical equations the preparation of potassium dichromate from chromite ore.

(b) Give balanced chemical equations for what happens when acidified potassium dichromate solutions reacts with :

(i) ferrous sulphate solution

(ii) hydrogen sulphide gas

(c) Draw the structure of Cr2O2/7- ion.

(i) Fusing chromite ore with Na₂  CO₃ and lime as flux in excess of air

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ -- 8Na₂ CrO₄ + 2Fe O₃ + 8CO₂

(ii) Reacting Na₂ CrO₄ solution with conc H₂SO₄ to get Na₂, Cr₂, O₇

2Na₂ CrO₄ + H₂ SO₄ -- Na₂ Cr₂ O₇ + Na₂SO₄ + H₂O

(iii) Addition of KCI solution

Na₂ Cr₂ O₇+ 2KCI -- K₂Cr₂ O₇ + 2NaCl

( Less soluble) (More soluble)

On cooling the hot concentrated solution obtained in step (iii) the less soluble K₂Cr₂ O₇ separates as orange crystals.