Chemistry 2009 Set 1 Sol

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Subjective Test

(i) All questions are compulsory.(ii) This question paper consists of four sections A, B, C and D. Section A contains 8 questions of one mark each. Section B is of 10 questions of two marks each. Section C is of 9 questions of three marks each and Section D is of 3 questions of five marks each.(iii) There is no overall choice. However, an internal choice has been provided.(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.

Section A

Question 1 ( 1.0 marks)

Which point defect in crystals does not affect the density of the relevant solid?

Solution:

Frenkel defect in crystals does not affect the density of the relevant solid.


Define the term ‘Tyndall effect’.

Solution:

Tyndall effect is observed when a fine beam of light enters a room through a small hole. It happensbecause of the scattering of light by particles of colloidal and suspension solutions.


Why is the froth flotation method selected for the concentration of sulphide ores?

Solution:

Froth flotation process is selected for the concentration of sulphide ores as in this process, sulphideore particles are preferentially wetted by oil whereas gangue particles are wetted by water.


Why is Bi (V) a stronger oxidant than Sb (V)?

Solution:

Bismuth and antimony both belong to the nitrogen family and exhibit the +5 oxidation state.However, on moving down the group, i.e., from antimony to bismuth, the stability of the +5 oxidationstate decreases. This is due to the inert pair effect. Thus, Bi (V) is a stronger oxidant than Sb (V).


Give the IUPAC name of the following compound:

Solution:

2 − Bromo-3-methyl-but-2-ene-1-ol.

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Write the structure of 3-oxopentanal.

Solution:


Why is an alkylamine more basic than ammonia?

Solution:

An alkylamine is more basic than ammonia because of inductive effect (+I effect). Alkyl group or ‘R’has an electron-releasing effect, which increases electron density over nitrogen atom. This increasesits basicity.


Give an example of elastomer.

Solution:

Natural rubber is an example of elastomer.

Section B


A reaction is of second order with respect to a reactant. How will the rate of reaction be affected ifthe concentration of this reactant is

(i) Doubled,

(ii) Reduced to half?

Solution:

(i) When the concentration of the reactant is doubled, the rate of reaction will become four times.

(ii) When the concentration of the reactant is reduced to half, the rate of reaction will become one-fourth.


Explain the role of

(i) Cryolite in the electrolytic reduction of alumina.

(ii) Carbon monoxide in the purification of nickel.

Solution:

(i) Cryolite is used in the electrolytic reduction of alumina so as to reduce its melting point and makeit a good conductor of electricity.

(ii) Carbon monoxide is used in the purification of nickel because it reacts with nickel to give a volatilecomplex called nickel tetracarbonyl, which on heating, decomposes to gives pure nickel metal.


Draw the structures of the following molecules:

(i) XeF4

(ii) BrF3

Solution:




Complete the following chemical reaction equations:

(i)

(ii)

Solution:

(i)

(ii)


Differentiate between molality and molarity of a solution. What is the effect of change in temperatureof a solution on its molality and molarity?

Solution:

Molarity of a solution is defined as the number of gram moles of solute present in l L of solution,while molality of a solution is defined as the number of gram moles of solute present in 1 kg ofsolvent.

Molality of a solution decreases with increase in temperature, while molality of a solution is notaffected by temperature.


Which ones in the following pairs of substances undergoes SN2 substitution reaction faster and

why?

(i)

(ii)

or

Solution:

(i) undergoes SN2 substitution reaction faster than . This is because the alkyl group

present in benzyl chloride increases its basicity due to +I effect. Stronger the base, lesser is its

leaving ability. So, reacts faster.

(ii) Iodide is a weaker base than chloride. Weaker the base, greater is its leaving ability. So,

undergoes SN2 substitution reaction faster.


Complete the following reaction equations:

(i)



(ii)

Solution:

(i)

(ii)


Explain what is meant by

(i) A peptide linkage

(ii) A glycosidic linkage

Solution:

(i) A peptide linkage (−CO−NH−) holds together amino acid units in proteins. It is an amide bondformed between −COOH of one amino acid and −NH2 group of another amino acid by the loss of

water molecule.

(ii) The linkage formed by the reaction of the −OH group of anomeric carbon of a monosaccharidewith the −OH group of other monosaccharide is called glycosidic linkage.


Name two water soluble vitamins, their sources and the diseases caused due to their deficiency indiet.

Solution:

Thiamine (vitamin B1) & riboflavin (vitamin B2) are soluble in water.

Thiamine is found in unpolished rice, whole cereals, yeast, egg yolk, milk, green vegetables, etc. Thedeficiency of thiamine causes beriberi and loss of appetite.

Riboflavin is found in egg yolk, liver, milk and green leafy vegetables. The deficiency of riboflavincauses cracked lips, sore tongue and skin disorders.


Draw the structures of the monomers of the following polymers:

(i) Teflon

(ii) Polythene

OR

What is the repeating unit in the condensation polymer obtained by combining HO2CCH2CH2CO2H

(succinic acid) and H2NCH2CH2NH2 (ethylene diamine)?

Solution:

(a) The monomer present in Teflon is Tetrafloroethylene. The structure of the monomer of teflon is

(b) The monomer present in polythene is ethene. The structure of the monomer of polythene is

OR

The repeating unit of the condensation polymer obtained by combining succinic acid and ethylene



The repeating unit of the condensation polymer obtained by combining succinic acid and ethylenediamine is

Section C


Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g

cm−3. Use this information to calculate Avogadro’s number.

(At. Mass of Fe = 56 g mol−1)

Solution:

In a body-centred cubic unit cell, number of atoms present = 2

At mass of iron = 56 g mol−1

Density of iron = 7.87 g cm−3

Mass of iron = 7.87 × Volume

Volume in BCC = (a)3

= (286.65)3 pm

= 2.34 × 10−23 cm

Mass = 7.87 × 2.34 × 10−23 g

∴ Avogadro’s number = 6.022 × 1023


100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solutionhas an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?

(R = 0.0821 L atm mol−1 K−1 and 760 mm Hg = 1 atm)

Solution:

We know that

Osmotic pressure (π)

π = 13.3 mm Hg

V = 10 mL = 10/1000 L = 0.01 L

R = 0.0821 L atom mol−1 K−1

T = 25ºC = 273 + 25 K = 298 K

W = 100 mg = 100/1000 g = 0.1 g

On putting the above values in the formula, we get



Thus, the molar mass of the protein is .


A first-order reaction has a rate constant of 0.0051 min−1. If we begin with 0.10 M concentration ofthe reactant, what concentration of reactant will remain in the solution after 3 hours?

Solution:

The integrated rate equation of the first-order reaction is

Here,

A0 = Initial concentration = 0.10 M

k = 0.0051 min-1

t = 3 hours = 3 × 60 min

Log 0.1 − log A = 0.4

Log A = −1 − 0.4

A= antilog (−1 − 0.4)

A = 0.0398


How are the following colloids different from each other with respect to dispersion medium anddispersed phase? Give one example of each type.

(i) An aerosol (ii) A hydrosol (iii) An emulsion

Solution:

(i) An aerosol is a suspension of a solid or a liquid in a gas.

Example: smog, smoke

(ii) Hydrosol is a colloidal suspension of essential oils. It is obtained from steam distillation ofaromatic plants,

Example: rose water used as facial toner

(iii) An emulsion contains one liquid dispersed in another liquid.

Example: butter and margarine; milk and cream


Account for the following:

(i) NH3 is a stronger base than PH3.

(ii) Sulphur has a greater tendency for catenation than oxygen.

(iii) Bond dissociation energy of F2 is less than that of Cl2.

OR

Explain the following situations:

(i) In the structure of HNO3 molecule, the N − O bond (121 pm) is shorter than the N − OH bond (140

pm).

(ii) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.

(iii) XeF2 has a straight linear structure and not a bent angular structure.

Solution:



(i) On moving from nitrogen to phosphorus, i.e., down the group, the atomic size increases. As thesize of the central atom increases, the lone pair of electrons occupy a larger volume. Consequently,the electron density, and hence, the basic strength decrease. Thus, NH3 is a stronger base than

PH3.

(ii) The tendency for catenation depends upon the bond energy. The bond energy of a sulphurmolecule is more than that of an oxygen molecule. Thus, the sulphur − sulphur bond strength ishigher, and as a result, the tendency of catenation is also higher. Sulphur shows catenation up toeight atoms.

(iii) Bond dissociation energy of fluorine is less than that of chlorine. It is due to the low value ofelectron affinity of small-sized fluorine. Also, the value of enthalpy of hydration of fluorine is muchhigher than that of chlorine.

OR

(i) The structure of nitric acid is

The N − O bond has a double-bond character. On the other hand, N − OH bond is a single bond.Since a double bond is shorter than a single bond, the N − O bond (121 pm) is shorter than the N −OH bond (140 pm).

(ii) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed. SF6 is an inert, non- toxic gas. This

is due to the presence of stearically protected six fluorine atoms. As a result, reactions like hydrolysisare not allowed to take place.

(iii)

According to VSEPR theory, XeF2 has trigonal bipyramidal geometry with sp3d hybridisation. It has

two bond pairs and three lone pairs of electrons. The bond pairs occupy axial positions and the lonepairs occupy the equatorial positions. This is the most favourable arrangement as far as stability isconcerned. The two fluorine atoms if placed at the equatorial positions will face minimum repulsion,and this would consequently increase the stability. Hence, the structure is linear.


For the complex [Fe(en)2Cl2] Cl, (en = ethylene diamine), identify

(i) The oxidation number of iron,

(ii) The hybrid orbitals and the shape of the complex,

(iii) The magnetic behaviour of the complex,

(iv) The number of geometrical isomers,

(v) Whether there is an optical isomer also, and

(vi) Name of the complex. (At. No. of Fe = 26)

Solution:

(i) Let the oxidation number of Fe in be x.

The oxidation state of Fe can be calculated as follows:

x + 2 (0) + 2 (-1) = + 1

Or, x − 2 = 1

Or, x = 3



Or, x = 3

(ii) The electronic configuration of Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5

So, hybridisation of Fe3+ is sp3d2, i.e., one s, three p and two d orbitals hybridise. It has octahedralgeometry

(iii) The complex is para magnetic due to the presence of 5 unpaired electrons.

(iv) This complex exists as cis-trans isomers. Thus, it has 2 geometrical isomers.

(v) The cis isomer exhibits optical activity.

(vi) The name of the complex is Dichlorobis-(ethylenediamine) iron (III) chloride.


Explain the mechanism of the following reactions:

(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed byhydrolysis.

(ii) Acid catalysed dehydration of an alcohol forming an alkene.

(iii) Acid catalysed hydration of an alkene forming an alcohol.

Solution:

(i) Grignard’s reagent is an alkyl magnesium halide. The alkyl group has a partial negative charge,whereas the magnesium group has a partial positive charge. The alkyl group attacks the carbon ofthe carbonyl group to form an addition compound.

Grignard’s reagent acts as a nucleophilic agent and attacks electrophilic carbon atoms to yield acarbon − carbon bond. The addition to the nucleophile is an irreversible process due to the high pkavalue of the alkyl group.

(ii) When heated with concentrated sulphuric acid, phosphoric acid or boric acid, alcohols undergodehydration to form alkenes. The mechanism of this reaction involves the protonation of alcohol,followed by loss of a water molecule and a proton.

(a)

(b)



(c)

During the dehydration of alcohol, the intermediate carbocation may undergo re-arrangement,resulting in the formation of a stable carbocation.

(iii) Some reactive alkenes like 2 − methyl propene undergo direct hydration in the presence ofmineral acids which act as catalysts. The addition of water to the double bond takes place inaccordance with Markonikoff’s rule.


Giving an example for each, describe the following reactions:

(i) Hofmann’s bromamide reaction

(ii) Gatterman reaction

(iii) A coupling reaction

Solution:

(i) Hofmann’s bromamide reaction: It involves the reaction of bromine with an acid amide in thepresence of an alkali. It results in the formation of a primary amine with one carbon less than theparent compound. Here, the alkyl group migrates from carbonyl, with the elimination of CO2. For

example:

(ii) Gatterman reaction: This is a modification of Sandmeyer reaction in which benzenediazoniumchloride is treated with copper powder and halogen acid to form aryl halides.

(iii) Coupling reaction: It is the reaction of diazonium salts with phenols and aromatic amines to formazo compounds of the general formula Ar − N = N − Ar. The coupling of phenol takes place in a mildlyalkaline medium.


Explain the following types of substances with one suitable example, for each case:

(i) Cationic detergents

(ii) Food preservatives

(iii) Analgesics

Solution:

(i) Cationic detergents are quaternary ammonium salts such as chlorides and acetates.

Example − Cetyl trimethyl ammonium chloride



These detergents are very good cleansing agents and are used as germicidals.

(ii) Food preservatives are chemical substances used for inhibiting the growth of micro-organisms infood materials so as to prevent their spoilage.

Example − benzoic acid and sulphur dioxide

Benzoic acid is used for preserving fruits, fruit juices, jams, etc.,as it is soluble in water, while sulphurdioxide is used for the preservation of colourless food materials.

(iii) Analgesics are the chemical substances used for relieving pain. They are also used for alleviatingfever.

Example − Aspirin, Analgin, Novalgin

Section D


(a) Define molar conductivity of a substance and describe how for weak and strong electrolytes,molar conductivity changes with concentration of solute. How is such change explained?

(b) A voltaic cell is set up at 25°C with the following half cells:

Ag+ (0.001 M) | Ag and Cu2+ (0.10 M) | Cu

What would be the voltage of this cell?

( )

OR

(a) State the relationship amongst the cell constant of a cell, the resistance of the solution in the celland the conductivity of the solution. How is molar conductivity of a solute related to the conductivityof its solution?

(b) A voltaic cell is set up at 25°C with the following half-cells:

Al | Al3+ (0.001 M) and Ni | Ni2+ (0.50 M)

Calculate the cell voltage.

Solution:

(a) Molar conductivity is defined as the conductance of a solution containing 1 g molecule or 1 mol ofelectrolyte such that the entire solution is placed between two electrodes one cm apart. It isdenoted as ^m. In the case of strong electrolytes, when the concentration is zero, molarconductivity attains a definite value known as limiting molar conductivity. It is denoted by ^m, whichcan be calculated.

For weak electrolytes, when concentration reaches zero, the graph of ^m vs. becomes

parallel to the y-axis. Thus, the limiting molar conductivity cannot be calculated.

(b)

OR

1. The conductivity ( ) of the solution in a cell is the reciprocal of its resistivity.

The quantity is cell constant.



l → Distance between 2 electrodes

a → Area of cross section

R → Resistance

Also, conductivity,

Thus, ^m =

(b) For the given cell,

Eº cell = Eº right − Eº left

Also, Ecell = Eºcell +

The net reaction is

∴ n = 6


(a) Complete the following chemical reaction equations:

(i)

(ii)

(b) Explain the following observations about the transition/inner transition elements:

(i) There is in general an increase in density of element from titanium (Z = 22) to copper (Z = 29).

(ii) There occurs much more frequent metal−metal bonding in compounds of heavy transition

elements (3rd series).

(iii) The members of the actinoid series exhibit a larger number of oxidation states than thecorresponding members of the lanthanoid series.

OR

(a) Complete the following chemical equations for reactions:

(i)

(ii)

(b) Give an explanation for each of the following observations:

(i) The gradual decrease in size (actinoid contractions) from element to element is greater among theactinoids than among the lanthanoids (lanthanoid contraction).

(ii) The greatest number of oxidation states are exhibited by the members in the middle of atransition series.

(iii) With the same d-orbital configuration (d4) Cr2+ ion is a reducing agent but Mn3+ ion is anoxidising agent.



Solution:

(a)

(i)

(ii)

(b)

(i) On moving from titanium (Z = 22) to copper (Z = 29), electrons get added to the 3d orbital. So, themass per unit volume increases. As a result, density also increases.

(ii) In the third transition series, due to the introduction of the d orbital, the shielding effectdecreases. Therefore, the effective nuclear charge increases. Consequently, the atomic volumeincreases. Due to an increase in mass by volume ratio, density increases.

(iii) The lanthanoids and actinoids exhibit a principal oxidation state of +3. Some lanthanoids alsoexhibit +4 and +2 oxidation states. However, actinoids exhibit oxidation states of +2, +4 ,+5, +6 and+7. They exhibit greater range of oxidation states because the 5f, 6d and 7s sub-shells present inactinoids are of comparable energy. Thus, they can take part in bonding, giving rise to differentoxidation states.

OR

(a)

(i)

(ii)

(b)

(i) The gradual decrease in size (actinoid contraction) from element to element is greater among theantinoids than among the lanthanoids. This is because of the poor shielding effect of electrons in the5f orbital of actinoids

(ii) The elements at the beginning of a series exhibit fewer oxidation states because they have lessnumber of electrons which they can lose or contribute for bond formation. The elements at the endof a series exhibit fewer oxidation states because they have too many d electrons, and hence, fewervacant d orbitals that can be involved for bonding.

(iii) For Cr, the +3 oxidation state is more stable than the +2 state. Thus, Cr2+ changes to Cr3+ andit behaves as a strong reducing agent. However, for Mn, the +2 state is more stable than the +3

state. So, Mn3+changes into Mn2+ and it behaves as a strong oxidising agent.


(a) Illustrate the following name reactions by giving example:

(i) Cannizzaro’s reaction

(ii) Clemmensen reduction

(b) An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. Themolecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an additioncompound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation itgives ethanoic and propanoic acids. Derive the possible structure of compound A.

OR

(a) How are the following obtained?

(i) Benzoic acid from ethyl benzene

(ii) Benzaldehyde from toluene

(b) Complete each synthesis by giving the missing material, reagent or products:

(i)

(ii)

(iii)



Solution:

(a)

(i) In Cannizzaro’s reaction, aldehydes which do not have α-H atom undergo self oxidation andreduction on treatment with a concentrated alkali, an alcohol and an acid.

Thus, 1 molecule is oxidised to alcohol and the other is reduced to carboxylic acid salt.

(ii) In Clemmenson’s reduction, the carbonyl group of aldehydes and ketones is reduced to CH2group on treatment with zinc amalgam and conc. HCl.

(b) Since the compound does not reduce Tollen’s reagent, it has a ketonic group.

Also, as it forms an addition compound with NaHSO3 and gives positive iodoform test, the presence

of methyl ketone is confirmed.

On oxidation, it gives ethanoic acid and propanoic acid.

So, the compound can be

This is because in unsymmetrical ketone, the point of cleavage is such that the keto group stays withthe smaller alkyl group.

The compound is pentan − 2 − one, with molecular mass 86.

(a)

(i)

(ii)

(b)

(i)

(ii)

(iii)





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Chemistry 2009 Set 1 Sol