All India Board Paper 2007

8/2/2019 All India Board Paper 2007

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Section A

Question 1 ( 1.0 marks) What is the coordination number in a rock-salt type structure?

Solution:In rock-salt, every Na

+

ion is surrounded by 6 Cl−

ions and each Cl−

ion is surrounded by 6 Na+

ions. Thus, the coordinationnumber of each type of ion in rock salt is 6.

Question 2 ( 1.0 marks) State Raoult’s law for a binary solution containing volatile components.

Solution:Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution isdirectly proportional to its mole fraction.

Question 3 ( 1.0 marks) What is meant by order of a reaction being zero?

Solution:For a zero-order reaction, the sum of the powers to which the concentration terms are raised is zero. Thus, the reaction isindependent of concentration.

Question 4 ( 1.0 marks) Write the IUPAC name of the following compound:

Solution:

The IUPAC name of the given compound is 3, 3-dimethyl-butanoic acid.

Question 5 ( 1.0 marks) Mention one commercial use of N, N-Dimethyl aniline (DMA).

Solution:DMA is used as a promoter in the curing of polyesters and vinyl ester resins. It is also used for the preparation of severalorganic compounds.

Section B

Question 6 ( 2.0 marks) State as a mathematical formula the de Broglie relationship for moving particles. What experimental evidence is availablefor this concept?

OR

Specify the ranges of values for quantum numbers m l and m s for an electron in an atom when the n quantum numbervalue for it is 2. What is the significance of these values for the orbitals?

Solution:According to De Broglie relationship for moving particles,



Where, λ is the wavelength of the wave associated with mass ‘ m ’ moving with velocity ‘ v ’

This concept was experimentally verified by Nickel Crystal with the help of the diffraction pattern of electrons.

OR

When Principal Quantum number, n = 2,

l = 0, 1

When l = 0,

This signifies that the electron is present in the s sub-shell, and can have two orientations.

When l = 1,

This signifies that the electron is present in the p sub-shell, which has 3 orbitals. Electrons in each orbital can have twoorientations.

Question 7 ( 2.0 marks) When can an endothermic process be spontaneous? Give an example of such a process.

Solution:Endothermic reactions can be made spontaneous by increasing the entropy of the system.

For example −

It is an endothermic reaction, but spontaneous.

Question 8 ( 2.0 marks) Write balanced chemical equations for the following reactions:

(i)

(ii)

Solution:(i)



(ii)

Question 9 ( 2.0 marks) Explain any one of the following statements:

(i) The transition metals are well known for the formation of interstitial compounds.

(ii) The largest number of oxidation states are exhibited by manganese in the first series of transition elements.

Solution:(i) Small atoms like hydrogen, carbon, boron and nitrogen occupy the interstitial sites in the lattices of transition metals.This gives rise to their interstitial compounds. This is mainly due to the vacant d orbital in transition metals and variableoxidation states.

(ii) In each group of transition metals, oxidation states increase with increase in atomic number, reach a maximum in themiddle, and then, start decreasing. The increase in oxidation states is due to the increase in the number of valenceelectrons with the increase in atomic number. As manganese lies in the middle, it has the maximum number of valenceelectrons. Hence, it shows maximum oxidation states.

Question 10 ( 2.0 marks) Draw the three-dimensional representations of (R) - and (S) - butan- 2- ol.

Solution:

Question 11 ( 2.0 marks) Write chemical reaction equations to illustrate the following reactions:

(i) Williamson synthesis of ethers

(ii) Reimer−Tiemann reaction

Solution:(i) Williamson synthesis of ethers :

It is used for preparing both symmetrical and unsymmetrical ethers. It involves the treatment of alkyl halide with sodiumalkoxide or sodium phenoxide to give ether and a sodium salt of halide.

(ii) Reimer−Tiemann reaction :



It involves the reaction between phenol and chloroform in the presence of an aqueous alkali at 333 K − 343 K to givesalicylaldehyde.

If the reaction is carried out with carbon tetrachloride, salicylic acid is produced.

Question 12 ( 2.0 marks) Distinguish between addition polymers and condensation polymers and give one example of each class.

Solution:

Addition polymer Condensation polymer

(i) It is formed when the monomer unitsare repeatedly added to form longchains.

(i) It is formed when two or more monomerunits react together.

(ii) No elimination of any by-product

takes place.

(ii) Elimination of simple compounds like H 2O,

R−OH takes place. (iii) For example:

Polythene is addition polymer of ethylene.

(iii) For example:

Nylon (6, 6) is a condensation polymerobtained from hexamethylene diamine andadipic acid.

Section C

Question 13 ( 3.0 marks)

Answer the following in the light of MO theory:

(a) Which has a higher bond order, C 2 or ?

(b) Which species is not likely to exist, Li 2 or Be 2?

OR

(a) Compare the structural shapes of the following species:



SF 6 and SF 4

(b) What type of intermolecular forces exists between Cl 2 and CBr 4 present in a mutual solution?

Solution:(a) In C 2 ,

Number of electrons = 12

Electronic configuration =

Now, bond order =

In ,

Number of electrons = 12+ 2 = 14


Therefore, C 22- has the higher bond order.

(b) For Li 2:



For Be 2:





Therefore, Be 2 is unstable and is not likely to exist.

OR

(a) SF 6 molecule has perfect octahedral geometry, while SF 4 has distorted trigonal bipyramidal geometry.

(b) Cl 2 and CBr 4 are both non-polar molecules. These molecules have no permanent dipoles. Hence, dispersion forces arethe only forces that operate between these molecules.

Question 14 ( 3.0 marks) (a) Name an element with which silicon can be doped to give an n -type semiconductor.

(b) Which type of crystals exhibits piezoelectricity?

Solution:(a) When Si is doped with group-15 elements, i.e., phosphorus, arsenic, etc., n -type semi-conductor is formed.

(b) Piezoelectricity is exhibited by the crystals which produce voltage when pressure or stress is applied.

Example − Rochelle salt, Barium titanate

Question 15 ( 3.0 marks) The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a one molal solution of a non-volatilenon-ionic solute in water.

Solution:The molality of the solution is 1.

Therefore, 1 mol of solute is present in 1000 g of water.

Hence, number of moles of water in 1000 g =

= 55.55

Total number of moles in the solution = 1 + 55.55 = 56.55

∴ Mole fraction of H 2O =

Now, according to Raoult’s law,



Question 16 ( 3.0 marks) Using the values of Δ f HΘ and Δr S Θ, given herein, calculate the standard molar Gibbs energy of formation, (Δ f GΘ) for CS 2 ( l ).

Given:

Solution:ΔH = 89.70 kJ mol −1 = 89.70 × 10 3 J mol −1

T = 298 K

∴ Standard molar Gibbs energy of formation, G

= 89.70 × 10 3 J mol −1 − 298 K × 151.34 J K −1 mol −1

= 89.70 × 10 3 − 45.10 × 10 3 J mol −1

= 44.6 × 10 3 J mol −1

Question 17 ( 3.0 marks) The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate activation energy of such a reaction.

(R = 8.314 J mol −1 K−1 , log 2 = 0.3010)

Solution:

Where,

k 1 , k 2 are rate constants

E a is activation energy

R is universal gas constant = 8.314 J mol −1 K−1

T 1 = 298 K, T 2 = 308 K

On substituting the above values, we get



⇒ E a = 52897.77

= 52.9 × 10 3 J

≅ 53 kJ

Question 18 ( 3.0 marks) State what is observed when

(i) the electrodes connected to a battery are dipped into a sol.

(ii) an electrolyte solution is added to a sol.

(iii) an emulsion is subjected to high-speed centrifugation.

Solution:(i) When the electrodes connected to a battery are dipped in a solution, the ions produced by the dissociation of theelectrolyte move towards the oppositely charged electrodes. The cations move towards the negatively charged electrodewhile the anions move towards the positively charged electrode.

(ii) The addition of an electrolyte causes a colloidal solution to coagulate.

(iii) On high-speed centrifugation, an emulsion decomposes into its constituent liquids.

Question 19 ( 3.0 marks) Answer the following questions:

(i) Which element in the first series of transition elements does not exhibit variable oxidation states and why?

(ii) What happens when a solution of copper (II) sulphate is saturated with ammonia?

(iii) Why do actinoids, in general, exhibit a greater range of oxidation states than lanthanoids?

Solution:(i) Scandium, w ith the atomic number ‘21’, does not show variable oxidation states. After losing three electrons, all theorbitals in scandium are fully filled.

Thus, scandium exhibits only +3 oxidation state.

(ii) In the presence of ammonia, copper (II) sulphate splits into two sets having slightly different energies.Thus, d−d transition occurs and it acquires blue colour.

In this reaction, [Cu(NH 3) 4] 2+ is produced.

(iii) Actinoids exhibit greater range of oxidation states than lanthanoids. This is because 5 f , 6 d and 7 s sub-shells inactinoids are of comparable energies.

Question 20 ( 3.0 marks) (a) Illustrate the following with an example each.

(i) Linkage isomerism



(ii) Coordination isomerism

(b) Why is paramagnetic? (Ni = 28)

Solution:(a)

(i) Linkage isomerism results when there are two ways in which a ligand can get attached to the central atom.

For example, NO 2 group can bind to a metal, either through a nitrogen atom or an oxygen atom, as indicated:

(ii) Coordination isomerism is shown by the compounds in which the cation and the anion are complexes.

For example,

(b) In [NiCl 4] 2− , the oxidation state of nickel is +2. Thus, Ni has 8 electrons in the 3 d orbital and no electron in the4 s orbital. Chlorine, being a weak-field ligand, forms sp 3 hybridisation. As there are 2 unpaired electrons in the 3 d orbitalof Ni, the complex is paramagnetic.

Question 21 ( 3.0 marks) Write the nuclear reactions for the following radioactive changes:

(i) undergoes α-decay

(ii) undergoes β-decay

(iii) undergoes K-decay

(You can put ‘X’ for the symbol which is not correctly known)

Solution:

(i)

(ii)

(iii)

Question 22 ( 3.0 marks) Explain the mechanism of nucleophilic addition to a carbonyl group and give one example of such addition reactions.

Solution:The nucleophilic addition to a carbonyl groups proceeds via the following mechanism.



Example:

The addition of hydrogen cyanide proceeds as follows:

Question 23 ( 3.0 marks) Write the chemical equations for the following chemical reactions:

(a) 1-Nitropropene is prepared from acetaldehyde

(b) Benzonitrile is converted into acetophenone

(c) A primary amine is prepared from a primary alkyl halide

Solution:(a)

(b)

(c)



Question 24 ( 3.0 marks) Mention one important use of each of the following:

(i) Equanil

(ii) Sucralose

(iii) Carbon fibres

Solution:(i) Equanil is used in the treatment of depression & hypertension.

(ii) Sucralose is used as an artificial sweetening agent.

(iii) Carbon fibres are used for preparing strong fibres such as carbon fibre-reinforced plastics.

Section D

Question 25 ( 5.0 marks) (a) Write the formulation for the galvanic cell in which the reaction,

takes place.

Identify the cathode and the anode reactions in it.

(b) Write Nernst equation and calculate the emf of the following cell:

OR

(a) Explain with one example each the terms weak and strong electrolytes.

(b) Write the Nernst equation and calculate the emf of the following cell:

Solution:(a) The galvanic cell is represented as



The reaction at the anode is given by

The reaction at the cathode is given by

(b) The Nernst equation for the given cell is

Here, Eº cell = Eº cathode − Eº anode

= 0 − (−0.14)

= 0.14 V

= 0.14 − 0.059

= 0.081 V

OR

(a) Strong electrolytes completely ionise when dissolved in water.

For example, sodium chloride dissociates as Na + & Cl− .

Weak electrolytes do not ionise completely when dissolved in water.

For example, carbonic acid dissociates as H + &

(b) The Nernst equation for the given cell is

Here,



Question 26 ( 5.0 marks) (a) How would you account for any two of the following:

(i) PbO 2 is a stronger oxidising agent than SnO 2

(ii) H 3PO 2 acts as a monobasic acid.

(iii) The pKa value for HOCl is higher than that of HOClO

(b) Draw the structures of the following species:

(i) Peroxodisulphuric acid, H 2S 2O8

(ii) Xenon tetrafluoride, XeF 4

OR

(a) Assign reasons for any two of the following observations:

(i) The lower oxidation state becomes more stable with increasing atomic number in Group 13.

(ii) Hydrogen iodide is a stronger acid than hydrogen fluoride in aqueous solution.

(iii) The basic character among the hydrides of Group-15 elements decreases with increasing atomic numbers.

(b) Draw the structural formula for XeOF 4 .

Solution:(a)

(i) PbO 2 is stronger oxidising agent than SnO 2 . This is because of the presence of poor shielding f orbitals in Pb. Thus, theeffective nuclear charge on the outer shell electrons increases, and this makes PbO 2 a strong oxidising agent.

(ii) The structure of H 3PO 2 is



Though it has three hydrogen atoms, only one of these is attached to an oxygen atom. Only those hydrogen atoms areacidic which are connected to oxygen atoms; therefore, H 3PO 2 acts as a monobasic acid.

(iii) The p Ka value is determined by acidic strength. The conjugate base of HOCl is less stable than the conjugate base of HOClO. Thus, HO 2Cl is a stronger acid. Also, increase in the number of oxygen atoms increases the stability of theconjugate base.

(b)

(i)

(ii)

OR

(a)

(i) The outer electronic configuration of group-13 elements is ns 2 np 1 . On moving down the group, due to inert pair effect,there is a decrease in the tendency of the s electrons of the valence shell to participate in bond formation. This is due tothe poor shielding of ns 2 electrons by the intervening d orbital. Thus, the lower oxidation state becomes more stable withincreasing atomic number.

(ii) Hydrogen iodide is a strong acid. This is because the small size of hydrogen and the big size of iodine make the H−Ibond very weak. On the other hand, due to the simil ar sizes of H and F, the H−F bond is quite strong. Thus, HI is astronger acid than HF in aqueous solution.

(iii) Among the group-15 elements, nitrogen has a small size and its density per unit volume is more. On moving down thegroup, the size of the central metal increases; thus, electron density decreases. As a result, the electron-donor capacity orthe basic strength of group-15 hydrides decreases with increasing atomic numbers.

(b)

Question 27 ( 5.0 marks) (a) Name the three major classes of carbohydrates and give the distinctive characteristic of each class.



(b) What are nucleotides? Name two classes of nitrogen-containing bases found amongst nucleotides.

OR

(a) Describe the classification of lipids based on their chemical compositions. Mention the chief chemical characteristic of each class.

(b) Explain the term ‘mutarotation’.

Solution:(a) Based on their behaviour on hydrolysis, carbohydrates are classified into three major groups.

(i) Monosaccharides − They cannot be hydrolysed to still simpler carbohydrates.

Example: Glucose

(ii) Oligosaccharides − They give two to ten units of monosaccharides on hydrolysis.

Example: Sucrose

(iii) Polysaccharides − They are polymeric molecules and give large number of monosaccharide units on hydrolysis.

Example: Starch

(b) The monomeric units of nucleic acids are called nucleotides. It consists of three parts − a phosphate group, a five -carbon sugar and a nitrogen heterocyclic base.

The nitrogen- containing bases are derivatives of two classes − purine and pyrimidine.

OR

(a) Out of current syllabus

(b) Out of current syllabus

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All India Board Paper 2007

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