CBSE ANNUAL PAPER - 1999

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.

Give the maximum number of electrons which can be accommodated in a set of (i) p-orbitals and (ii) d-orbitals.

There can be 6 electrons in a set of p-orbitals and 10 electrons in a set of d-orbitals.

What is an 'orbital' in an atom ? What does the angular momentum quantum number tell about on orbital ?

An atomic orbital refers to three- dimensional regions around the nucleuis in which there is high probability of electron of given energy being found is maximum.

Angular quantum number (l) describes the shape of the electron cloud or the shape of the orbital.

When any of the particles of a crystal lattice is either missing or is dislocated from its original position called interstitial site we get point defect schottky or Frioendkel defect.

Name any two metals, which can be used for cathodic protection of iron.

Magnesium and zinc metals can be used for cathodic protection of iron from rusting.

Two liquids A and B from an ideal solution if A-A and B-B interaction forces are the same and there is no net change in molecular interaction forces A-B due to identical structure and polarity of molecules of A and B. An ideal solution will obey Raoults law.

Selenium finds use in xerography, in photoelectric cells and as a rectifier in semi conductor devices.

Fusion reactions require high temperature (> 106K) to overcome electrostatic repulsions between nuclei when they come together to fuse. Hence nuclear fusion reactions are called thermonuclelar reactions.

Lipids are esters of long chain fatty acids and alcohols. They are found in all living organisms. Lipids are classified as (a) simple lipids or triglyeerides and waxes which on hydrolysis produce fatty acids, alcohols, glycerol etc.) and (b) complex lipids (Phospholipids) which produce glycerol, fatty acids, phosphoric acid and choline etc - on hydrolysis example lecithin.

What are the two most important oxidation states of Group 6 elements of the periodic table ?

Group 6 elements show important oxidation states of 3 and 6 example Cr³+ IN Cr₂ (So₄)₃ and Cr⁶+ CrO₃.

(i) CO2 and SiO2 are not isostructural.

(ii) O2 is paramagnetic but O2/2- is not.

CO₂ is a linear molecule with zero dipole moment. It is represented as O = C = O. This is due to the small size of carbon atom which can form C = O by p

p-pp lateral overlap. In silicon dioxide (SiO2) silicon and oxygen atoms are held together by strongcovalent bond forces forming a three dimensional net - work structure. Each silicon atom is tetrahedrally bonded to four oxygen atoms.

Larger size silicon atoms cannot form bonds unlike carbon atoms. It cannot form O = Si = O structure.

Distance between K+ AND F- in KF = 269 pm

(Atomic masses : K = 39 and F = 19 amu)

Edge length (a) = 2 x 269 = 538 pm.

= 538 x 10-10 cm

Mass of K+ F- unit = 39+19=58=(M)

Density of KF = 2.48 g cm^-3

(z) no of K+ F- units in KF crystals = 4k+ F- = 4

N₀ = ?

r Z x M / a³ x N₀ N₀ = Z x M / a3 x r

= 4 x 58 / (538 x 10^-10)³ x 2.48

= 6.02 x 10²³ atoms mol^-1

(i) What is osmotic pressure and how is it related to the molecular mass of a non-volatile substance ?

(ii) What advantage the osmotic pressure method has over the elevation of boiling point method for determining molecular masses ?

Osmotic pressure is the measure of the external pressure which has to be applied on the surface of the solution to prevent osmosis of the solvent molecules through the semi permeable membrane which separates the solvent and the solution.

formula to be inserted.

(b) ADVANTAGE OF OSMOTIC PRESSURE METHOD OVER THE ELEVATION OF BOILING POINT METHOD FOR DETERMINING MOLECULAR MASSES.

(i) Elevation of boiling point for measuring the molecular masses of macromolecules like protein can not be used because they are not stable at high temperature. On the other hand the Osmotic pressure measurements can be taken at room temperatures.

(ii) For very dilute solution osmotic pressures are of the order of atmosphere. This makes accurate measurements of osmotic pressure quite convenient for dilute solution.

TB values are of the order of 10^-5 which are too small to be measure accurately by any instrument.

Calculate the molarity and molarity of a 13% solution (by weight) of sulphuric acid ? Its density is 1.020 g cm-3.

[ Atomic masses : H=!, O=16, S=32 amu)

mole of H₂SO₄ = 13/98 mole

Volume of H₂SO₄ solution + Weight of soluton / Density = 100/1.02 ml

= 100 / 1.02 x 1/1000 = 1/1.02 litre

Molarity of H₂SO₄ solution = mole of H₂SO₄ / Volume of solution

= 13/98 / 1/1.02 = 10.2 x 13/98 = 1.353 M

Molality of H₂SO₄ solution = mole of H₂SO₄ / wt of water in kg. = 13/98 / 87/1000 = 13/98 x 1000/87

= 1.525 mol kg^-1 (mole Kg^-1)

NO₂ (g) + CO (g) -- CO₂ (g) + NO (g)

the proposed mechanism is as below :
(i) NO₂ + NO₂ -- NO + NO3 (slow)

(ii) NO₃ + CO -- CO₂ + NO₂ (fast)

What is the rate law for the reaction ?

NO₂ (g) + CO (g) -- CO₂ (g) + NO(g)

Rate mechanism is (i) NO₂ + NO₂ --NO + NO₃ (slow)

(g) (g) (g) (g)

(ii) NO₃ + CO -- CO + NO₂ (g)

(g) (gas)

This shows that (i) is the slow step in he reaction. This is the rate determining step. Rate mechanism of the reaction shows that rate law is Rate = K [ NO₂] [ NO₂] = K [ NO₂]²

At constant temperature and pressure heat is given out of the surroundings

DS surrounding = -Qp /T = - DH / T ( Since Qp = DH at constant pressure)

= heat lost by surrounding)

From (1) DS_total + DS_system = DH / T

Multiplying both sides by T

TDS_total = TDS(_system) - DH ......(3)

or - TDS _total = DH - TDS .......(4)

Also free energy change DG = DH - TDS

Hence for a change conducted at constant temperature and pressure

From (3) and (4) DG = -TDS_total

For t_2/3 = let initial concentration = 1 mol L^-1

Final concentration = 1 - 2/3 = 1/3 mol 1^-1

By the relation t _2/3 = 2.303 / K log a/a-x

= 2.303/ 5.4 x 10_-14 x log 1/ 1/3

= 2.303 / 5.4 x 10_-14 x log 3

= 2.303 x 0.4771 / 5.4 x 10_-14

= 2303 x 4771 / 54 x 10⁸

= 2 x 10¹³ seconds

What are transition elements? Account for the following

(i) Transition elements have high enthalpies of atomisation

(ii) Most of the compounds formed by transition elements are coloured.

(iii) Transition metals form complex compounds.

Transition metals are elements of d-block which form at least one stable ion which has partially filled d-orbitals. Example Cu²⁺ (3d⁹ 4S⁰) therefore copper is a transition element.

(i) In transition metals both as electrons and unpaired electrons in (n-1) d orbitals take part in d-d metal bonding. Since in these elements metal bonding is stronger than other elements of S and P block the transition elements have high enthalpies of atomisation.

(ii) In transition metal ions the unpaired electrons in (n-1)d orbitals absorb radiations from the visible range in d-d transition. They emit coloured radiations complementary to the coloured radiations absorbed by them in under-going transitions. Examples Cu²+ ions are blue coloured. Ni²⁺ salts are green. Sc³+ (3d⁰ 4S⁰) ions are colourless as there is no unpaired electron in 3d orbitals.

(iii) In transition metal ions there are empty or vacated d-orbitals of appropriate energy which can accept Ione pair of electrons from neutral or negatively charged ligands resulting in the formation of complexes. Examples : [ Ni (CN₄]²-, [Cu(NH₃)₄]²+ and [ Fe (CN)₆]^4-

Write the complete reactions involved in (a) the isocyanide test, and in (b) the iodoform test.

Only 1⁰ ammines having - NH₂ group give foul smelling isocyanides.

(ii) IODOFORM TEST FOR ALCHOL having CH₃ --- CH --- function

|
OH

and for aldehydes and ketones having CH₃ ---C--(R or H)

||
O

CH₃ -- CH --- R + 3I₂ + 4NaOH +D CHI₃ + RCOONa + 3Nal + 4H₂O)

| (Yellow iodoform)
OH

CH₃ ----C---R+ 2I₂ + 4NaOH +D CHI + RCOONa + 3 Nal + 3 H₂O

||

O

What are nucleotides ? Name two classes of nitrogen containing bases found in nucleotides.

(i) Nucleotides are basic components of nucleic acid. They contain II. a phosphate ---sugar ---base ---unit. sugar may be ribose or deoxribose. The bases - adenine, guanine, cytosine, uracil or thymine are attached to sugar molecule at No. 1 carbon atom.

(ii) Two classes of nitrogen containing bases found in nucleotides. are (a) Purines (Adenine and guanine and (b) Pyrimidines ( uracil, thymine and cytosine).

Write the equations for the synthesis of (i) Neoprene and (ii) Glyptal. Which one of the two is a condensation polymer ?

Giving a suitable example describe the importance of the formation of complex compounds in

(a) The estimation of hardness of water, and

(b) the extraction of a particular metal from its natural source.

(a) Importance of Co-ordination (complex) compounds in (i) estimation of hardness of water. Hard water contains Ca²⁺ and Mg²⁺ ions which form high stability complexes with hexadentate ligand EDTA. The selective estimation of Ca²⁺ and Mg²⁺ ions in hard water can be done because there is a large difference in the stability constants of Ca²⁺ and Mg²⁺ with EDTA.

(b) Extraction of metals like gold and silver from their ores by formation of their complex compounds (soluble in water) whereas their ores are insoluble in water.

Ag₂S(ore) | 4NaCN -------> 2NalAg[CN₂] Na₂S

(insoluble) (Soluble complex)

From these complexes the metals can be precipitated by adding Zinc dust (powder).

What do you understand by 'broad spectrum antibiotics'? Is penicillin a broad spectrum antibiotic ? Name a place in India where penicillin is manufactured.

(i) Chloramphenicol, tetracycline like antibiotics which can be used for curing a large number of infections are called broad spectrum antibiotics.

(ii) Penicillin is a narrow spectrum antibiotic.

(iii) It is manufactured in India at Hindustan Antibiotics in Pimpri and I.D.P.L. at Rishikesh (U.P.).

(i) What is 'photosynthesis' and where does it occur in plants?

(ii) Name the two products of photosythesis which are needed by most chemotrophs for their survival.

(i) The process of conversion of Solar energy into chemical energy in which plants produce organic molecules like carbohydrates and amino acids 6H₂O + 6CO₃ ---> C₆H₁₂O₆+6O₂ is called photosynthesis. It takes place in sunlight.

The phenomenon of photosynthesize takes place in chloroplasts of plant cells which contain green pigment called chlorophyl.

(ii) Two important products of photosynthesis are carbohydrates - glucose, starch. etc and amino acids which the chemotrophs need for their survival.

(a) 238/92 U undergoes a-decay.

(b) 234/91 Pa undergoes b-decay.

(b) 234/91 Pa --------------> 0/-1e + 234/92 U

(i) Using the data given below calculate the value of equilibrium constant for the reaction at 298 K.

Large value of K shows that this process can be recommended as a practical method for preparing benzene from acetylene.

Write the structure of the main product of each of the following reactions :

(i) Chlorination of benzene in the presence of ultra violet light.

(ii) Action of excess of bromine on phenol.

(iii) Action of alkaline potassium permanganate on ethane.

Account for the following :

(i) Chloroacetic acid has lower pKa value than acetic acid.

(ii) Electrophic substitution in benzoic acid takes place at meta position.

(iii) Carboxylic acids have higher boiling points than alcohols of comparable molecular masses.

(i) Due to greater stability of chloroacetate ion than acetate ion owing to

CH₂ --COOH is > K_A

electron with drawl effect of -----------cl group. KA of Cl

of acetic acid. Hence PK_A of chloroacetic acid is less than PK_A of acetic acid.

(ii) -COOH group in benzoic acid is an election with drawing group. By inductive effect -COOH group. Hence electrophilic substitution in benzoic acid takes place at meta position.

(iii) Carboxylic acid have higher boiling points than alcohols of comparable molecular masses because they form dimers on intermolecular hydrogen bonding CH₃COOH has higher boiling point than C₃H₅OH. In dimers two carboxylic acid molecules are held through Hydrogen bonds.

Write chemical equations for the conversion of :

(1) chromate ore to sodium chromate.

(2) Pyrolustile to potassium manganate.

(3) Potassium permanganate to manganese dioxide.

Chromite + D Sodium chromate

(ii) 2MnO₂ + 4 KOH O₂ -------> 2 K₂ MnO₄ + 2H₂O

Pyrolusite ore (Potassium manganate)

(iii) 4 K MnO4 + 2H₂O -->4 KOH + 4 Mno₂ + 3O₂

( Pot permanganate)

PT BR₂ (I) IBr^-1 (0.01 M) || H⁺ (0.03 M) IH₂ (1 atm|Pt

Given E⁰ Br|br - = 41.08 V.

The cell notation is wrong because the lower value of E^O of electrode will serve as anode. hence.

(i) Explain the meaning of the statement, "adsorption is a surface phenomenon."

(ii) State two features of chemical adsorption which are not found with physical adsorption.

(i) In adsorption the molecules of a liquid or a gas get concentrated only on the surface of the solid. The surface of the solid possesses particles with unbalanced forces like weak vanderwall forces or chemical bond forces. On breaking a piece into two parts two new surfaces are formed which attract more molecules of liquid or gases. Adsorption in creases with increase in surface area. Hence adsorption is a surface phenomenon.

(ii) (a) Chemical adsorption occurs at comparatively high temperature. It first increases due to requirement of activation energy. (like all chemical reactions.) in initial stages. Physical adsorption. takes place at low temperature. It decreases on increasing temperature.

(b) Chemical adsorption is highly specific in nature but physical adsorption is not as specific. Charcoal adsorbs all gases physically but tungsten adsorbs oxygen chemically.

(a) Account for the following :

(i) Like ammonia, amines are good nucleophiles.

(ii) In contrast to arenas, aliphatic hydrocarbons do not undergo nitration easily.

(iii) Alkylamines are stronger bases than arylamines.

(b) Illustrate the following reactions with an example in each case :

(i) Sandmeyer reaction

(ii) Coupling reaction.

(i) Like ammonia all amines primary, secondary or tertiary contain lone pair of electrons on nitrogen atoms. Hence amines are good nucleophiles. Like ammonia they react with alkylhalides undergoing nucleophilic substitution reactions.

C₂H₅Cl+CH₃ --NH₂ ------> C₂H₅NH ---CH₃. They displace weaker nueophile like Cl.

(ii) Aliphateic hydrocarbons undergo nitration in drastic conditions. With fuming nitric acid in the vapour phase at 423 to 673 K under pressure alkanes undergo under go nitration.

CH₄ + HNO₃ (funning) 673 K CH₃ NO₂ + H₂O

This is becaue nitration of alkanes occurs by free radical mechanism. In arenes nitration is through electrophilic substitution which is easier.

(iii) In Alkyl animes the alkl groups are strong electron pushing. By their _1 effect the electron density on N-atom of R-- NH₂ is increased but the benzene ring in C₆H₅ --- NH₃ is electron attracting. C₆H₅ -- NH₂ group decreases electron density on N-atom. Hence alkyl amines are stronger bases than aryl amines.

(b) (i) Sandmeyer's reaction

(1) Structure of beryllium chloride and aluminium chloride.

(2) Proton affinities of NH3 and PH3.

(3) Action of water on CCl4 and SiCl4.

(4) Physical states of nitrogen and phosphorus.

(5) Shapes of ClO2 and Cl2O

(i) Beryllium Chloride has a bridged structure with three centre bonding in the solid state. It is polymeric chain structure.

In vapour phase Be Cl₂ has liner geometry.

Alluminium chloride exists as a simer Al₂ Cl₆ in which each atom forms one co-ordinate bond by accepting tone pair from the chlorine atom of another AlCl3 molecule. This makes Aluminum to attain complete octet.

(ii) In ammonia : (NH₃) the lone pair has more affinity for proton H+ due to higher electrogegativity and small size of N atom. Due to the lower electronegativity and large size of phosphorous atom (compared to nitrogen) the lone pair on Phosphorous atom in : PH₃ has lower affinity for H⁺ (proton). NH³ is more basic than PH³.

Si Cl₄ + 4 H₂O ---> Si (OH)₄ + 4HCl.

Silicicx acid

(iv) Nitrogen is a gas it exists as diatomic molecules. N₂ with weak vander Wall forces between them. Phosphorous has P₄ molecules or P₄ chains with stronger intermolecular forces. Phosphorous is a solid at ordinary temperature.

There is larger bond angle in O - Cl - O than in ClO. Both are angular in shape.