CBSE ANNUAL PAPER - 2000

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.

What is the physical significance of the lines in the following depictions of atomic orbital :

For an S- orbital the radial electron probability is spherically symmetrical. For a p-orbital the electron probability takes the form of dumbell along along a coordinate axis.

An ionic compound AB₂ possesses CaF₂ type crystal structure. Write the co-ordination numbers of A²⁺ and B^- ions in crystals of AB₂.

How may the state of a thermodynamic system be defined ?

State of a thermodyanamic system is described by its measurable properties like Pressure, Temperature, Volume and Moles. they are cattled variables of state.

State Kohlrausch's law for electrical conductance of an electrolyte at infinite dilution.

Formalin (HCHO) is used in the preparation of phenol formaldehyde resin (bakelite). It is also used as a preservative for biological specimens.

For an amine RNH₂ write the expression for K_b to indicate its base strength.

What is responsible for the blue colour of a solution of an alkali metal in liquid ammonia ?

A solutions of alkali metal in liquid ammonia has blue colour due to the solvated electron present in the solution.

2M + 2NH₃ --- 2M NH₂ + H₂ ( M is alkali metal)

e^-1 + X NH₃ -- [ e - (NH₃)_x]^-1

Colloidal solutions are heterogeneous in nature. They exhibit Tyndall effect due to scattering of light by the particles of the dispersed phase ( which are colloids).

Write any one characteristic feature of enzymatic catalysts.

Enzyme catalysts are highly specific in their action on substrate molecules and each enzyme catalyses only a specific type of reaction.

Account for the following :

(i) Silicone is an insulator but silicon doped with phospphorus acts as a semi - conductor.

(ii) Some of the glass objects recovered from ancient monuments look milky instead of being transparent.

(i) In pure silicon all its four valencies are used up in producing diamond like structure. When silicon is doped with phosphorous ( with five electrons in its valence shell). Out of these 5 electron four from bonds and the fifth electron can convey electric current readily. Hence, silicon doped with phosphorus acts as a semi-conductor.

(ii) some of the glass objects from ancient monuments look milky instead of being transparent because of some crystallisation occurring after a long exposure to the atmosphere. They become a bit opaque.

Give one example each of miscible liquid pairs showing positive and negative deviations from Raoult's law. Give one reason each for such deviations.

Non - ideal solutions do not obey Raolt's law. In them the solute-solvent intermolecular forces are either weaker or stronger than solvent-solvent intermolecuilar forces. Example solution of cyclohexane in ethanol.

The solution of cyclohexane in ethanol shows positive deviation from Raoult's law. Ethanol molecules are associated through hydrogen bonds. On adding cyclohexane some hydrogen bonds break. The intermolecular forces cyclohexane - ethanol become weaker than ethanol - ethanol

intractions. The boiling point of the solutions is less than the boiling point of ethanol alone. Vapour pressure of this solution is more than the vapour pressure of pure ethanol.

The solution of acetone and chloroform is an example of negative deviation from Raoult law. There are no hydrogen bonds in between chloroform - chloroform or between acetone - acetone molecules but when they form solution hydrogen bonding takes place as shown.

Due to this intermolecular forces acetone - chloroform are stronger than chloroform -chloroform- or acetone - acetone intermoleculr forces. Solution boils at higher temperature than the b.p. of the solvent and vapour pressure of the solution is less than the vapour pressure of pure solvent.

For final stage H₂ =E₂+PV@ AND FOR INITIAL STAGE H₁ = E₁ +PV,

Substracting H₂ - H₁ = (E₂ + PV₂) - (E₂ + PV₁)

or H₂ - H₁ = DH = (E₂-E₁) +P (V₂-V₁)

or DH = DE = PDV .....(1)

Also DE + PDV = q-PDV

or DE + PDV = qp

From (1) and (2) DH = qp

CH₄ (g) + 2O₂ (g) --CO₂ (g) + 2H₂O (l)

Calculate the enthalpy of formation of CH₄ (g) at 298 K.

Enthalpy of reaction = -890.5 kJ

Enthalpy of formation of CO₂ (g) = -393.5 kJ mol^-1

Enthalpy of formation of H₂O(1) =-286.0 kJ mol^-1

(I) An ether would possess a dipole moment even if the alkyl groups present in it are identical.

(ii) Towards nucleophilic reagents aldehydes are more reactive than ketones.

(a) In any ether R--O--R the bond angle R--O--r is 110⁰ due to the presence of two Ione pairs on central oxygen atom.

Therefore, ether R--O--R will have dipole moment even if alkyl groups are identical.

(b)

For the following conversion reactions write the chemical equations :

(i) Ethyl isocyanide to ethylamine

(ii) Aniline to N-phenylethanamide

(i) ammonolysis of alkyl halides does not give a coresponding amine in pure state.

(ii) If --No₂ or --- COOH group is attached to a carbon of benezene ring, electrophilic substitution becomes difficult.

(a) The reaction of ammonia with alkyl halides called ammonia alysis produces a mixture of primary, secondary and tertiary amines. Since this mixture of amines cannot be separated easily. We cannot get pure amines by ammonialysis

(b) If NO₂ or --COOH group is attached to the carbon atom of benzene ring, the benzene ring is deactivated by reasonance as follows :

In these structures there is +ve charge in ortho and para positions. Therefore, they are unreactive to the attacking electrophile which is positively charged. Electrophilic substitution becomes difficult.

The sums of first and second ionization energies and those of third and fourth ionization energies of nickel and platinum are given below : -

formula to be inserted :

(i) The most common oxidation state for Ni and Pt and its reasons.

(ii) The name of the metal (Ni or Pt) which can form compounds in +4 oxidation state more easily and why.

(b) If reach one of the above ionic species is in turn placed in a magnetic field, how will it respond and why ?

a) AG⁺ (4d¹⁰ 5s⁰) and Ti⁴⁺ (3d⁰ 4s⁰) have no unpaired electrons but CO²⁺ (3d⁷ 4s⁰) has three unpaired electrons. Hence only CO²⁺ will be coloured in acqueous solutions. Others Ag⁺ and Ti⁴⁺ will be colourless.

(b) Out nof Ag⁺ (4d¹⁰ s0) Co²+ (3d⁷4s⁰) and Ti⁴⁺ (3d⁰s⁰) only CO²⁺ has three unpaired electrons. When placed in a magnetic field CO²⁺ will be attracted because it is paramagnetic. Other ionic species Ag⁺, Ti⁴⁺ have no unpaired electrons. They are diamagnetic. They will be fecbly repelled by the magnetic field.

less stable stable

state state

(i) Chain - growth polymerization

(ii) Step - growth polymerization

In chain growth polymerisation an organic perioxide is used as initiator and the monomers are added to the freeradical in chain fashion. The polymer chain is grown by adding small quantities of indicator to produce reactive free radicals and adding monomers at each addition of initiator.

In step growth polymerization the condensation polymer like backelite is produced from phenol and formaldehyde by condensation reaction step by step with or without elimination of smaller molecules. The polymer chain is grown step by step.

Nylon 66 is synthesised by the condensation of hexamethyline diamine and adipicacid. this produces polymamide linkages with elimination of water molecule in each step.

Giving an example of each, differentiate between multimolecular and macromolecular colloids.

Multimolecular - the colloidal particles of these colloidal solutions consist of aggregates of atmosor small molecules with diameter less than 1 nm. Example gold sol consists of particles of various sizes having several atoms. The atoms are held together by weak Vander Waal forces.

Macromolecular Colloids - In macromolecular colloids the dispersed particles are themselves large molecules called macromolecules of more than 1 nm diameter. They have high molecular masses (1000 to 10⁶). Naturally occurring macromolecules are starch, cellulose, gum and protein.

What arepolysachharides? Name one of them. How is it important for us ?

Polysachharides are formed when large number of monosachharide molecules get condensed with elimination of H2O molecules. Example cellolose and starch. Starch is a food reserve in plants. On hydrolysis starch forms maltose. One further hydrolysis maltose forms glucose. Glucose is a chief source of energy in living systems as its oxidation in the body provides 38 molecules of A.T.P.

What do you understand by the statement : "ATP molecules are the currency of energy metabolism in a cell."

A.T.P. is an energy source in human system. The molecule has four negative oxygen atoms. Due to the repulsion of these oxygen atoms ATP hreases readily and repulsive forces are reduced.

ATP -----------> ADP + P(D G < 0)

ADP -----------> AMP + P(D G < 0)

Hydrolysis of ATP releases a large amount of energy required for the human system.

Name a broad spectrum antibiotic and state two diseases for which it is prescribed.

Chloramphenicol is a broad spectrum antibiotic. It can be used for the treatment of disentry, typhoid and urinary infections.

The ionization energy of hydrogen atom in ground state is 1312 kJ mol^-1. Calculate the wavelength of radiation emitted when the electron in this atom makes a transition from n = 2 state to n = 1 state. [ Planck constant, h = 6.63 X 10^-34 Js, velocity of light, c=3X10¹⁰ cm S^-1 and Avogadro's number, N_A = 6.02X10²³ mol^-1)

Energy of electron in hydrogen (E_n) = -1312/n2 kJ mol^-1

E₂ (energy in x = 2) = -1312/ (2)² = -328 kJ mol^-1

E₁ (in n = 1) = -1312 / (1)² = -1312 kJ mol^-1

DE = Energy emitted in transition form n = 2 to n =1

= -328 - (-1312) = 984 kJ mol^-1

Also E = hc /l or l = hc/E

Wavelength of radiation emitted = (l)

= 6.6 x 10^-34 x 3 x10⁸ / 984 x1000 x 6.02 x 10²³

l = 6.6 x 10^-34 x 3 x 10⁸ x 6.02 x 10²³ / 984 x 1000 metre x 1/10^-10

= 1207 A⁰

On dissolving 3.24 of sulphur in 40 g. of benzene boiling point of solution was higher than that of benzene by 0.81 K. K1 value for benzene is 2.53 K kg. mol^-1. What is the molecular formula of sulphur? (Atomic mass of sulphur = 32 g mol^-1).

Wt. of benzene (W_A) = 40 g

DT_b = 0.81 K

K_b = 2.53 K kg mol^-1

Molecular mass of Sulphur = K_h / DT_b x W_B/ W_A X 1000

= 2.53 / 0.81 x 3.24/40 x 1000

= 253 g mol^-1

The no. of atoms of sulphur in its molecule = 253 / 32 = 7.9

= 8 approx.

Molecular formula of Sulphur is S₈.

Zn(s) | Zn²+ (0.0004 M) || Cd²+ (0.2 M) | CD (s)

E⁰ values at 25⁰C : Zn²⁺ / Zn = - 0.763V ; Cd²⁺/Cd = - 0.403 V ;

F = 96500 C mol^-1 ; R = 8.314 JK^-1 mol^-1.

= - 0.36 - 0.0296 log 2 x 10^-3

= 0.36 - (0.0296) x [ 0.2010 - 3]

= 0.36 - (0.0296) x (-2.6990)

= 0.44 V

D G = -n FE⁰ = -2x 96500 x 0.36

= 69.48 kj

answer to be inserted.

In general it is observed that the rate of a chemical reaction doubles with every 10⁰ rise in temperature. If this generalization holds for a reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for this reaction ? ( R = 8.314 K K^-1 mol^-1).

T₂ = 305 K

K₂ / K₁ = 2 ( rate of reaction doubles)

By the relation log K₂ / K₁ = E_AC / 2.303 x R x [ 1/T₁ - 1/T₂]

or, E_AC ( activation energy ) = log K₂ / K₁ x 2.303 x R X T₂ x T₁ / T₂ - T₁

= log 2 x 2.303 x 8.314 x 305 x 295 / 10

= 0.3010 x 2.303 x 8.314 x 305 x 295 / 10

= 5139998.255 Joule

(i) Optical isomerism

(ii) Magnetic behaviour of [ Ni (CN)₄]² - ion. ( At. no. of Ni = 28)

(iii) Preparation of tetrabutyl tin.

(i) Optical isomerism in Co-ordination compounds is exhibited when the given compounds possesses no plane of symmetry. The two isomers are mirror image of each other and they rotate the plane of polarished light in opposite directions. It is found that octahedral complexes having bidentate ligands NH₂--CH₂--CH₂--NH₂ and oxalate COO - ion exhibit optical isomerism. E.g. Co (en)₃ has two optical isomers.

(ii) [Ni (CN)4]2- . Oxidation state of Ni is =2 in this complex.

There is no unpaired electron in [Ni (CN)4}@_ ION HENCE [Ni (CN)4]2- is diamagnetic.

(iii) Preparation of tetrabutyl tin

SnCl4 + 4 Bu Li -- Bu4 Sn = 4 LiCl)

( Tetrabutyl tin)

(a) What is a nuclear fission reaction ?

(b) Describe the basic principle of a nuclear reactor.

(a) In Nuclear fission a heavier nucleous like 235/92 U breaks into a number of fragments by suitable bombardment with slow neutrons. It is a chain reaction because neutrons produced strike more particles of 235/92 U again and again Example : 235/92 u + 1/0 n ----- 139/56 Ba + 93/36 Kr + 3 1/0 n

(b) Principle of nuclear reactor : In a nuclear reactor fission reaction is made to take place under controlled conditions. Energy released is used for the production of electricity. In order to make the speed of neutrons slow a moderator ( graphite or heavy water) is used. Control rods made up of boron, steel or cadmium are used to capture some of the neutrons so that chain reaction does not become violent.

(a) Write chemical equations and reaction conditions for the conversion of :

(i) Propene to 1-bromo-propane.

(ii) Chlorobenzene to phenol.

(iii) 2-proppanol to iodoform.

(b) How is glyeerol obtained commercially. State its two uses.

Uses of glycerol : (i) glycerol is used as a sweetening agent in medicines like cough syrup.

(ii) It is used in making cosmetics because glycerol has the property of retaining moisture.

(i) In several chemical properties lithium resembles magnesium.

(ii) + 1 oxidation state of thallium ( atomic no. 81) is more stable than its + 3 oxidation state.

(iii) Boron chloride exists as a monomer while in the same group anhydrous aluminium chloride exists as a dimmer.

(b) Complete and balance the following chemical equations :

(i) NH₃ -- NaOCI --

(ii) XeF₄ + SbF₅ --

(i) Important functions of nucleic acids are :

(a) REPLICATION : DNA has double strand helical structure. It unwinds itself and each strand synthesises its complimentory template so that two new molecules of DNA closely resem,bling the parent molecule are produced.

(b) PROTEIN SYNTHESIS : involves two steps (a) Transcription in which DNA synthesises m-RNa which have base sequence complimentary to base sequence in DNA.

(c) TRANSLATION : iN WHICH M-RNA provides code corresponding to base triplelt on its nucleotide sequence to t-RNA which carries the specific amino acid to the site of protein synthesis. r-RNA attaches the amino acids in the sequence provided by t-RNA.

(ii) Thailium (6s²6P¹) has + 1 stable state because of Inert pair effect. Two electrons in 6s-orbital do not take part in bond formation. This pair remains inert. Hence Thallium has + 1 oxidation state more stable than + 3 oxidation state.

(iii) The small size boron atom cannot accommodate four large sized Cl ions and B cl₃ exists as monomeric unit. In the case of larger sized Aluminum atom used its vacent P-orbial in its valence shell to complete its octet by forming dimer. Hence, AlCl₃ forms dimer in anhydrous state.