700000584 cbse 12 chemistry boardpaper2006 set1solution

CBSE XII | Chemistry

Board Paper – 2006

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CBSE Board

Class XII Chemistry – Set 1

Board Paper – 2006 (Solution)

Time allowed: 3 hours Maximum Marks: 70

Sol. 1 The atom at the body centre makes a contribution of 1 to the unit cell, while the atom

at the corner makes a contribution of 1/8 to the unit cell.

Thus, number of atoms Y per unit cell

= Number of atoms × Contribution per unit cell

= 8 (at the corners) × 1/8 atoms per unit cell

= 1

Thus, number of atoms X per unit cell

= Number of atoms × contribution per unit cell

= 1 (at the body centre) × 1

= 1

Thus, the formula of the given compound is XY.

Sol. 2 Liquid A has higher vapour pressure at 80 oC because of its lower boiling point.

Sol. 3 Let r = k[A]n

Then 27r = k[3A]n [According to question] n

3 n

n

k 3A27ror 3 3

r k A

Order of reaction, n = 3.

Sol. 4 3-Bromo - N-methyl - butanamide.

Sol. 5 When warmed with chloroform in the presence of alc. KOH, aniline gives offensive

smell of isocyanides while N-methyl aniline does not give this test.

C6 H5 NH2 + CHCl3 + 3KOH C6H5NC + 3KCl + 3H2O

Aniline Phenylisocyanide

Sol. 6

(a) XeF2 has a linear structure.

F - Xe - F

(b) The outer electronic configuration of Cr atom is 3d5 4s1.

Sol. 7 Not in current syllabus

Sol. 8

3 2 2 3 2a Ca P 6H O 2PH 3Ca OH

6 2 3b XeF 3H O XeO 6HF




Sol. 9 Preparation of potassium permanganate: Potassium permanganate is prepared

by the fusion of MnO2 (pyrolusite) with potassium hydroxide and an oxidizing agent like

KNO3 to form potassium manganate (green mass), which disproportionate in a neutral or

acidic solution to form permanganate. heat

2 2 2 4 2

2 4 4 2 2

2MnO 4KOH O 2K MnO 2H O

potassium

manganate(greenmass)

3K MnO 4HCl 2KMnO MnO 2H O 4KCl

potassium

permanganate


Sol. 11

H O2 2

3 2 3 3 2 2 3 2 2 3 32 3 OH ,H O2

1a 3CH CH CH BH CH CH CH B 3CH CH CH OH H BO

2

Pr opene Diborane Tripropylborane 1 Pr opanol

(b)

Sol. 12

(a) Tetrafluoroethylene : CF2 = CF2

(b) Methyl methacrylate

Or

(a) Heating rubber with sulphur causes cross-linking of polymer chains through disulphide

bonds. This makes rubber hard stiff. It prevents the intermolecular movement of rubber

springs resulting in change of physical character of rubber.

(b) Nylon - 6 is obtained from the monomer caprolactum which contains 6 carbon atoms.

Nylon - 66 is condensation polymer of hexamethylene - diamine and adipic acid, the two

monomers have 6 carbon atoms each.





Sol. 14 Volume of unit cell = (288 pm)3 = (288 10-10 cm)3

= 2.389 10-23 cm3

Volume of 208 g of the element = 3

3

208gMass28.89cm

Density 7.2gcm

Number of unit cells = 3

23 3

Total volume 28.89cm

Volume of a unit cell 2.389 10 cm

= 12.09 1023

For a b. c. c. structure, number of atoms per unit cell = 2

Number of atoms present in 208 g

= No. of atoms per unit cell No. of unit cells

= 2 12.09 1023 = 24.18 1023 = 2.418 1024.

Sol. 15

(a) A part of the water surface is occupied by non - volatile glucose molecules. This

decreases the effective surface area for the vaporization of water molecules. Consequently,

the vapour pressure of solution of glucose in water is lower than that of water.

(b) A 6.90 M solution contains 6.90 mol of KOH in 1000 cm3 of the solution.

Mass of KOH present in 1000 cm3 of solution = 56 x 6.90 = 386.4 g

A 30% solution contains 30g of KOH present in = 100 g of solution

386.4 g of KOH is present in = 100 386.4

g30

= 1288 g of solution

Density of KOH solution = 3

Mass 1288g

Volume 1000cm

= 1.288 g cm-3.


Sol. 17 T1 = 273 + 50 = 323 K, T2 = 273 + 100 = 373 K,

K2 = 3k1

a a2 1

1 1 2 1 1 2

a1 1

a1

a

1

E Ek 3k1 1 1 1log logk 2.303R T T k 2.303R T T

E 1 1log3

323K 373K2.303 8.31Jmol K

E 500.4771

323 3732.303 8.314 Jmol

0.477 2.303 8.314 323 373E

50

22011.76 Jmol




Sol. 18

(a) The path of light becomes visible when passed through the colloidal solution while it is

not visible in case of true solution. This is because of Tyndall effect caused by the scattering

of light by colloidal particles.

(b) Applications of adsorption:

(i) Activated charcoal is used in gas masks to remove poisonous gases such as CH1, CO, etc.

(ii) Animal charcoal is used as decolouriser in the manufacture of sugar.

(iii) Silica is used for removing moisture.

(iv) The ion exchange resins are used for removing hardness of water.

Or

(a) A lyophilic sol is stable due to the charge and the hydration of sol particles. Such a sol

can only be coagulated by removing the water and adding solvents like alcohol, acetone,

etc. and then an electrolyte.

On the other hand, a lyophobic sol is stable due to the charge only and hence can be easily

coagulated by adding small amount of an electrolyte.

(b) The colloidal particles get precipitated i.e., ferric hydroxide is precipitated. When an

excess of electrolyte (e.g., NaCl solution) is added to the colloidal solution of ferric

hydroxide. This is because of the fact that colloids interact with the ions carrying the charge

opposite (Cl−) to the one present on them. This causes neutralisation, which leads to their

coagulation.

(c) The atmospheric particles of colloidal range scatter blue component of the white

sunlight preferentially. That is why the sky appears blue.

Sol. 19 Argentite, which is a sulphide ore, is the chief source of silver.

It is extracted by the process of leaching with a dilute solution of sodium cyanide.

Silver dissolves and forms a complex, argentocyanide. It is further treated with scrap zinc,

which displaces silver from the complex. This is called displacement method or

hydrometallurgy. The chemical reactions involved are given below.




Sol. 20

(a) The oxidation number of chromium (Cr) in [Cr (NH3)6]3+ ion is +3. Electronic

configuration of Cr is [Ar] 3d54s1.

Since the resulting complex [Cr(NH3)6]3+ involves d2sp3 hybridisation, it is octahedral in

shape. The presence of three unpaired electrons in the complex makes the complex

paramagnetic.

(b) IUPAC name of [Pt (NH3)2Cl2] is diamminedichloridoplatinum (II).


Sol. 22

(a) (i) Cannizzaro reaction: Aldehydes having no α-hydrogen atoms undergo self oxidation

and reduction (disproportionation) reactions on treatment with a concentrated alkali. In

such reactions, one molecule of aldehyde gets oxidised to form an acid and the other

molecule of aldehyde gets reduced to form an alcohol.

For example, two molecules of formaldehyde, in the presence of concentrated NaOH,

produce methanol and sodium formate.

NaOH

3HCHO HCHO CH OH HCOONa

Methanal Methanol Sodium formate

(ii) Aldol condensation: In the presence of a dilute alkali, aldehydes and ketones (having at

least one α-hydrogen atom) produce β-hydroxyl aldehyde (aldol) and β-hydroxyl ketone

respectively. The aldol and ketol then readily lose water to give α, β-unsaturated carboxyl

compounds. Such reactions are called aldol condensation.

For example, in the presence of dil. NaOH, ethanal (aldehyde) produces 3-hydroxybutanal

(aldol), which then readily loses water to produce but-2-enal.

(b) When warmed with iodine and sodium hydroxide solution, ethanal gives yellow

crystals of iodoform.

Propanal does not give this iodoform test




3 2 3 2CH CHO 4NaOH 3I CH I HCOONa 3NaI 3H O

Ethanal Iodoform

Sol. 23

(i) Benzene ring in aromatic amines is highly activated. This is due to the displacement of

lone pair of nitrogen towards the ring, which results in the increase in the electron density

on the ring. This facilitates the electrophilic attack on the ring.

(ii) In CH3CONH2, the lone pair of electrons on nitrogen atom is involved in resonance with

the carbonyl group. So the electron pair of nitrogen is not easily available for protonation.

Hence CH3CONH2 is a weaker base than CH3CH2NH2.

(iii) Nitro compounds are polar compounds whereas hydrocarbons are non-polar. Due to

their polarity, nitro compounds have higher boiling points than the hydrocarbons having

almost same molecular mass.

Sol. 24

(a) Tranquillisers are the neurologically active drugs. They are the class of chemical

compounds used for the treatment of stress, anxiety, and mild or severe mental

diseases. They are essential components of sleeping pills.

(b) Not in current syllabus

(c) Not in current syllabus

Sol. 25

(a) At cathode, the following reduction reactions can take place:

A reduction reaction with higher reduction potential is preferred. Therefore, the reaction at

the cathode during electrolysis is

This is why electrolysis of aqueous solution of NaCl gives H2 at the cathode.

At anode, the following oxidation reactions can take place:

At anode, the reaction with lower value of E0 is preferred. But, due to overvoltage,

oxidation of chloride ion occurs and chlorine gas is obtained. Hence, the reaction at the

anode during electrolysis is

This is why electrolysis of aqueous solution of NaCl gives Cl2 at the anode.

The overall cell reaction is given below.




.

(a) Emf of the given cell = Eocathode - Eoanode

= [- 0.40 - (- 0.76)] V

= [-0.40 + 0.76] V = 0.36 V

Thus, the emf of the given cell is 0.36 V.

or

(a) (i) The alkaline medium prevents the availability of H+ ions. This in turn reduces rate of

oxidation of Fe to Fe2+. Thus, the rusting of iron is inhibited.

(ii) Even if the zinc coating is broken in an iron pipe, the remaining zinc layer undergoes

oxidation in preference to iron because of its more electropositive nature than iron. Hence

iron does not rust.

(b) (i) Since E0Ag+/Ag is greater than EoCu2+/Cu so reduction will occur at silver electrode. The

cell representation is 2Cu(s)|Cu Ag | Ag(s)

(ii) The cell reaction is:

Cu + 2Ag+ (aq) Cu2+ (Aq) + 2Ag(s)

Clearly, n = 2 0 0cell anodeE E 0.80 0.34 0.46V

At 25oC, Ecell = E0cell -

2

cell2

Cu0.059

log given E 02

Ag

2

0.059 0.10 0.46V log

2Ag

2

15

2

217

15

9

0.1 0.46 2log 15.6

0.059Ag

0.1Antilog 15.6 3.98 10

Ag

0.1Ag 2.51 10

3.98 10

Ag 5.00 10 M.

Sol. 26

(a) Aluminium bromide exists as a dimer, Al2Br6. In this structure, each aluminium atom

forms one coordinate bond by accepting a lone pair of electrons from the bromine atom of

another aluminum bromide molecule and thus completes an octet of electrons. Due to the

lack of free electrons, molten aluminum bromide is a poor conductor of electricity.

(b) Nitric oxide reacts with air to get oxidized into NO2 which has brown yellow vapours.

2NO + O2 2NO2




(c) In solid state, PCl5 exists as [PCl4]+ [PCl6]- and hence it is ionic in nature. Due to ionic

nature, it conducts current on fusion.

(d) N atom in ammonia has lone pair of electrons which can coordinate with other atoms or

cations needing electron pair for stability.

(e) Not in current syllabus

Or

Not in current syllabus

Sol. 27

(a) Differences between globular proteins and fibrous proteins

Globular proteins Fibrous proteins

They are cross-linked

condensation polymers of

acidic and basic amino

acids.

They are linear

condensation products.

They are usually soluble in

water.

They are generally insoluble

in water.

(b) Not in current syllabus

Or

(a) Hormones are molecules that transfer information from one group of cells to distant

tissue or organ and thus control the metabolism. So they act as chemical

messengers.

(b)

Vitamins Deficiency disease Sources

Vitamin A Harding of cornea of eye Fish liver oil, butter,

milk

Vitamin B6 Serve dermatitis,

convulsions

Yeast, milk, egg yolk

Vitamin E Sterility. Vegetable oils

700000584 cbse 12 chemistry boardpaper2006 set1solution

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