Chemistry 2007 Set 1.3 Sol

Chemistry 2007 Set 1 Close

Subjective Test

(i) All questions are compulsory.(ii) This question paper consists of four sections A, B, C and D. Section A contains 5 questions of one mark each. Section B is of 7 questions of two marks each. Section C is of 12 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)

Find out the number of atoms per unit cell in a face-centred cubic structure having only single atomsat its lattice points.

Solution:

Calculation of the number of atoms per unit cell in a face-centred cubic structure

Contribution by atoms at the corners =

Contribution by atoms on the faces =

∴ Number of atoms present in the unit cell = 1 + 3 = 4


State the condition resulting in reverse osmosis.

Solution:

Reverse osmosis will occur if a pressure higher than the osmotic pressure is applied on the solution.


Express the rate of the following reaction in terms of disappearance of hydrogen in the reaction.

Solution:

In terms of disappearance of hydrogen in the reaction , the rate of reaction

is .


Name the following compound according to IUPAC system:

Solution:

2/4/2011 Subjective Test Paper - Chemistry - Meri…

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The IUPAC name of is pentan−2, 4−dione.


Why do amines react as nucleophiles?

Solution:

A nucleophile is an electron-rich species seeking electron-deficient sites. Amines are nucleophilic dueto the presence of unshared or lone pair of electrons.

Section B


(a) Write the mathematical expression for the relationship of the wavelength (λ) of a moving particleand its momentum (P)

(b) What physical meaning is attributed to the square of the absolute value of wave function, ?

OR

State the Heisenberg Uncertainty Principle and explain as to why it is not of real consequence whenapplied to a macroscopic object, like a cricket ball.

Solution:

(a) The mathematical expression for the relationship of the wavelength (λ) of a moving particle andits momentum (P), as given by De Broglie, is

Where, h is Planck’s constant, equivalent to 6.63 × 10−34 Js

(b) The square of the absolute value of wave function, gives the probability of finding an electron

in the different regions of space around the nucleus.

OR

Heisenberg Uncertainty Principle states that it is impossible to measure simultaneously the positionand momentum of a microscopic object with absolute accuracy or certainty.

The product of uncertainty in the position (∆x) and uncertainty in the momentum (∆P = m∆v; where,m is the mass of the particle and ∆v is the uncertainty in velocity) is always constant and equal to orgreater than h/4π (where, h is Planck’s constant), i.e.,

For macroscopic objects, like a cricket ball, the value of momentum (P) is very large. The position andmomentum of a macroscopic object can be measured with good precision.


Define Conductivity and molar conductivity for the solution of an electrolyte.

Solution:

Conductivity is the reciprocal of resistivity. It is denoted by k (Kappa). If k is the conductivity and C isthe conductance of a solution, then

and

Or,



Molar conductivity of a solution is defined as the conductance of all the ions produced from one mole

of an electrolyte dissolved in V cm3 of solution, when the electrodes are one cm apart and the areaof the electrodes is so large that the entire solution is contained between them. It is represented ByΛm.


How would you account for the following?

(i) Sulphur hexafluoride is less reactive than sulphur tetrafluoride.

(ii) Of the noble gases only xenon forms known chemical compounds.

Solution:

(i) Sulphur hexafluoride (SF6) is chemically inert as 6 F atoms protect the sulphur atom from attack by

reagents to such an extent that even thermodynamically most favourable reactions like hydrolysis donot occur. In contrast, sulphur tetrafluoride (SF4) is less sterically hindered. Hence, it undergoes

reactions easily.

(ii) All noble gases have full s and p outer electron shells, so they do not form chemical compoundseasily. However, in heavier members like Xe, the outermost electrons experience a shielding effectfrom the inner electrons. As a result, they are comparatively easily ionised. Hence, first ionisationenergy is roughly equivalent to molecular oxygen. Xe reacts with electronegative elements likefluorine and oxygen to form stable compounds.


On the basis of the standard electrode potential values stated for acid solution, predict whether Ti4+

species may be used to oxidise FeII to FeIII

Solution:

The reduction electrode potential of Fe3+ is more than that of Ti4+. So, Fe3+ will get reduced to Fe2+

more easily. In other words, Fe3+ is acting as an oxidising agent. So, Ti4+ cannot be used for

oxidising Fe2+ to Fe3+.


What are chiral objects? Indicate the presence of centre of chirality, if any, in the molecules of 3-bromopent-1-ene.

Solution:

The objects which are non-superimposable on their mirror images are called chiral objects.

3-bromopent-1-ene is represented as

The centre of chirality is designated as *. It is due to the presence of four different groups.


How may the following conversions be carried out?

(i) Propene to propan-2-ol

(ii) Anisole to phenol

(Write the reaction only).

Solution:

(i)



(ii)


Write formulae of the monomers of polythene and Teflon.

Solution:

The monomer of polythene is ethane or CH2 = CH2.

The monomer of Teflon is tetra-fluoro ethylene or CF2 = CF2.

Section C


Define bond order in a diatomic molecule. Find the bond order in O2 molecule. State and explain

magnetic character of molecular oxygen.

Solution:

Bond order is defined as the number of chemical bonds between a pair of atoms.

Mathematically, it is represented as

Where, Nb is the number of electrons present in the bonding molecular orbitals and Na is the number

of electrons present in the anti-bonding molecular orbitals.

Number of electrons in an oxygen molecule (O2) = 2 × 8 = 16

Electronic configuration of O2 =

∴Bond order =

Oxygen molecule has 2 unpaired electrons; it is therefore paramagnetic in nature.


Assign reasons for the following:

(i) Phosphorus-doped silicon is a semiconductor.

(ii) Schottky defect lowers the density of a solid.

(iii) Some of the very old glass objects appear slightly milky instead of being transparent.

Solution:

(i) When silicon is doped with phosphorus (having 5 electrons), four valence electrons of phosphorusare involved in bond formation with the neighbouring silicon atoms, while the fifth valence electron isleft free to conduct electricity. This type of conduction which arises due to the availability of excesselectrons is called n-type conduction.

(ii) Schottky defect arises when equal number of cations and anions are missing from their latticesites. As the number of ions decreases due to this defect, the mass decreases, but the volumeremains the same. As a result, the density of the solid decreases.

(iii) Some of the very old glass objects appear slightly milky instead of being transparent because ofsome crystallisation at that point.




A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol−1) has a freezing point of271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an

aqueous solution containing 5 g of glucose (mol. mass = 180 g mol−1) per 100 g of solution?

Solution:

Molality of cane sugar solution = 0.1539 molal

Depression at freezing point, ∆Tf = 273.15 K − 271 K = 2.15 K

∆Tf= Kf m

Or

The freezing point of an aqueous solution containing 5 g of glucose is calculated as follows:

Amount of glucose in 100 g of solution = 5 g

Amount of water in 100 g of solution = 100 g − 5 g = 95 g = 0.095 kg

∴Freezing point of glucose solution = 273.15 K − 4.08 K = 269.07 K


Calculate the standard cell potential of the galvanic cell in which the following reaction takes place:

Also calculate the ∆rG Θ value of the reaction.

(Given and F = 96500 C mol−1)

Solution:

For the cell reaction ,

The electrochemical reactions taking place at the cathode and anode are as follows:

Number of electrons involved = 6


The rate constant for a first-order reaction is 60 s−1. How much time will it take to reduce the

concentration of the reactant to 1/10th of its initial value?



concentration of the reactant to 1/10th of its initial value?

Solution:

Let [A]0 = 1.0 M, [A]t = 1 ×1/10 M

= 1/10 M

On substituting the values in the first-order rate equation, we get


Describe the following types of colloids, giving an example for each:

(i) Multimolecular colloids

(ii) Macromolecular colloids

OR

Explain the following terms with a suitable example in each case:

(i) Shape-selective catalysis

(ii) Dialysis

Solution:

(i) Multimolecular colloids are large aggregates of atoms or small molecules with diameters less than1 nm. The particles are held together by van der Waals force. An example is gold sol. The dispersedparticles consist of aggregates of different sizes of atoms or molecules.

(ii) In macromolecular colloids, the dispersed particles are large molecules (usually polymers). Thesizes of these particles lie within the colloidal range. Thus, their dispersions behave like colloidalsolutions. An example is synthetic rubber.

OR

(i) A catalytic reaction which depends upon the pore structure of the catalyst and the size of themolecules of the reactants and products is known as shape-selective catalysis.

(ii) The process of separating a soluble crystalloid from a colloid is called dialysis.


How would you account for the following?

(i) The transition elements have high enthalpies of atomisation.

(ii) The transition metals and their compounds are found to be good catalysts in many processes.

Solution:

(i) The atoms of transition elements are held together by strong metallic bonds which involve theparticipation of both 4s and 3d electrons. The greater the number of unpaired d electrons, thestronger is the metallic bond. It is due to these strong metallic bonds that the transition elementshave high enthalpies of atomisation.

(ii) The transition metals can exhibit multiple oxidation states. Hence, they are found to be goodcatalysts in many processes.


Describe for any two of the following complex ions, the type of hybridisation, shape and magneticproperty:

(i)



(i)

(ii)

(iii)

Solution:

(i) Let the oxidation state of Fe in [Fe(H2O)6]2+ be x.

x + 6(0) = +2

Or, x = +2

Since H2O is a weak-field ligand, outer orbital complex is formed, i.e., the hybridisation is sp3 d2.

Hybridisation : sp3 d2

Shape : Octahedral

Magnetic property : Paramagnetic (due to the presence of unpaired electrons)

(ii) Let the oxidation state of Co in [Co(NH3)6]3+ be x.

x + 6(0) = + 3

Or, x = +3

Since NH3 is a weak-field ligand, outer orbital complex is formed.

Hybridisation : sp3 d2

Shape : Octahedral

Magnetic property : Paramagnetic (due to the presence of unpaired electrons)

(iii) Let the oxidation state of Ni in [NiCl4]2− be x.

x + 4(− 1) = − 2

x − 4 = − 2

Or, x = +2

Since Cl− is a weak-field ligand, outer orbital complex is formed.



Hybridisation : sp3

Shape : Tetrahedral

Magnetic property : Paramagnetism (due to the presence of unpaired electrons)


Complete the following statements for nuclear reactions:

(i)

(ii)

(iii)

(Note: You may use ‘X’ as symbol if the correct symbol in a reaction is not known)

Solution:

(i)

(ii)

(iii)


Write one chemical equation for each, to illustrate the following reactions:

(i) Rosenmund reduction

(ii) Cannizzaro reaction

(iii) Fischer esterification

Solution:

(i) Rosenmund reduction

(ii) Cannizzaro reaction

(iii) Fischer esterification


Account for any two of the following

(a) Amines are basic substances while amides are neutral.

(b) Nitro compounds have higher boiling points than the hydrocarbons having almost the samemolecular mass.

(c) Aromatic amines are weaker bases than aliphatic amines.

Solution:

(a) Amines are basic because of the presence of unpaired or lone pair of electrons.

Amides, on the other hand, are neutral species because the lone pair of electrons is involved in



Amides, on the other hand, are neutral species because the lone pair of electrons is involved inresonance. Hence, no electron pair remains for donation.

(b) Nitro compounds are polar compounds whereas hydrocarbons are non-polar. Due to theirpolarity, nitro compounds have higher boiling points than the hydrocarbons having almost samemolecular mass.

(c) Aromatic amines are weaker bases than aliphatic amines. This is because in aromatic amines,owing to resonance, the lone pair of electrons on the nitrogen atom is less available for protonation.The partial positive charge on the nitrogen atom tends to repel a proton.


(a) Describe and illustrate with an example each, a mordant dye and a detergent,

(b) Give an example of a liquid propellant.

Solution:

(a) Mordant dyes: These dyes do not dye the fabric directly, but require a mordant. The mordant, infact, acts as the binding agent between the fibre and the dye.

Detergent:

These are chemical substances which concentrate at the surface of the solution or interfaces, formsurface films, reduce surface tension of the solution, and help in removing dirt and dust byemulsifying grease. Detergents do not form curdy white precipitate with hard water. They can beused in acidic conditions.

For example − Sodium lauryl sulphate,

(b) One example of a liquid propellant is liquid nitrogen tetroxide (N2O4)

Section D


(a) Prove that ∆Gsystem = − T∆Stotal for a system which is not isolated.

(b) The decomposition of Fe2O3 is a non-spontaneous process

Show that the reduction of Fe2O3 by CO can be made spontaneous by coupling with the following

reaction:

OR

(a) Define the following terms:

(i) Entropy

(ii) A spontaneous process

(b) Given below are the standard Gibbs energy changes for two reactions at 1773 K:

Discuss the possibility of reducing Al2O3 with carbon at this temperature. Given that:



Solution:

(a) …(i)

If the heat lost by the surroundings is represented by qp, then by definition

Further, we know that at constant pressure

qp = ∆H

Or,

On substituting this value in equation (i), we get

On multiplying throughout with T, we get

… (ii)

But for a system which is not isolated,

∆G = ∆H − T∆S … (iii)

From equations (ii) and (iii), we get

(b) Given that

On multiplying equation (ii) with 3, we get

On adding equations (i) and (iii), we get

Hence, the reduction of Fe2O3 by CO can be made spontaneous by coupling with CO.

OR

(a)

(i) Entropy is the measure of randomness or disorder of a system.

(ii) A process which under some given conditions may take place by itself or by initiation independentof the rate is called a spontaneous process.

(b) Given that

On multiplying equations (i) and (ii) with 3, we get

On subtracting equation (iii) from equations (v) and (vi), we get



Since for both the above reactions is positive, i.e., greater than zero, Al2O3 cannot be reduced

by carbon.


(a) Assign reasons for the following:

(i) PbO2 is a stronger oxidising agent than SnO2

(ii) In solid state PCl5 behaves as an ionic species,

(iii) Aluminium chloride (AlCl3) is very often used as a catalyst.

(b) What is the structural difference between orthosilicates and pyrosilicates?

OR

(a) Assign reasons for the following:

(i) The acid strengths of acids increase in the order

HF < HCl < HBr < HI

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

(iii) H3PO2 behaves as a monoprotic acid.

(b) Draw the structures of the following compounds:

(i) SF4

(ii) XeF2

Solution:

(a)

(i) Due to inert pair effect, +2 oxidation state of Pb is the most stable. Therefore, among dioxides ofgroup 14, PbO2 is a powerful oxidising agent.

(ii) In solid state, phosphorus pentachloride (PCl5) exists as [PCl4]+ [PCl6]

+. So, solid phosphorus

pentachloride exhibits some ionic character. The cation [PCl4]+ has tetrahedral structure and the

anion [PCl6]− has octahedral structure.

(iii) Aluminium is electron deficient, and hence, it acts as a Lewis acid catalyst. For example, it acts asa catalyst in Friedel crafts alkylation and helps in generating electrophiles.

(b)

Orthosilicates Pyrosilicates

These are simple silicates containing

discrete tetrahedra that do not share

corners with one another.

When two tetrahedra share one corner (i.e.,

oxygen atom), anions are formed. These

are called pyroilicates.

OR

(a) (i) The strength of an acid depends upon its degree of ionisation, which in turn depends uponbond strength.

Higher the bond dissociation energy, lower is the degree of ionisation, and hence, weaker is theacid. The bond dissociation energies of the halogen acids increase in the order:



HI < HBr < HCl < HF

Therefore, the strengths of the acids increase in the reverse direction, i.e.,

HF < HCl < HBr < HI

(ii) The outer electronic configuration of group-13 elements is ns2np1. On moving down the group,due to inert pair effect, there is a decrease in the tendency of s electrons of the valence shell to

participate in bond formation. This is due to the poor shielding of the ns2 electrons of the valenceshell by the intervening d or f electrons. So, only the p electrons can participate in bonding. Hence,the oxidation state is restricted to +1.

(iii) The structure of H3PO2 is

Since, it has only one ionisable hydrogen, it behaves as a monoprotic acid.

(b)

(i) SF4

(ii) XeF2


(a) Answer the following question briefly:

(i) What are reducing sugars?

(ii) What is meant by denaturation of a protein?

(iii) How is oxygen replenished in our atmosphere?

(b) Define enzymes.

OR

(a) Answer the following questions briefly:

(i) What are any two good sources of vitamin A?

(ii) What are nucleotides?

(iii) Give an example of simple lipids.

(b) How are carbohydrates classified?

Solution:

(a)

(i) The carbohydrates which reduce Fehling’s solution, Tollen’s reagent, etc., and contain a freealdehyde or ketonic group are referred to as reducing sugars.

(ii) Denaturation of a protein refers to the loss of biological activity of a protein due to change in itssecondary and tertiary structures. The change occurs due to heat, pH change, presence of salts, etc.

(iii) Plants release oxygen by the process of photosynthesis. The following reaction takes placeduring this process.

Animals, including human beings, take up this oxygen during the process of respiration and releaseCO .



CO2.

(b) Enzymes are proteins that catalyse biological reactions.

OR

(a)

(i) Two good sources of vitamin A are green vegetables and milk.

(ii) Nucleotides are a combination of a base, a sugar and a phosphate moiety.

There are two nucleotides − RNA (Ribonucleic acid) and DNA (Deoxyribonucleic acid).

(iii) Palmitic acid (C15 H31 COOH) is a simple lipid.

(b) Carbohydrates are classified on the basis of the number of products formed on hydrolysis.

(i) Monosaccharides: These are the carbohydrates which cannot be hydrolysed to simplercarbohydrates.

Example: Glucose

(ii) Oligosaccharides: These are the carbohydrates which on hydrolysis yield 2 − 10 molecules ofmonosaccharides.

Example: Sucrose

(iii) Polysaccharides: These are the carbohydrates which yield many molecules of monosaccharideson hydrolysis.

Example: Starch



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Chemistry 2007 Set 1.3 Sol