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CHEMISTRY(043) CLASS XI:2019-20

Marking Scheme for: UNIT TEST-II EXAMINATION

TIME:3 Hrs M.M.:70

SECTION-A 1. Electrophilic addition reaction.[or any other example] 1 2. Because of resonance. 1 3. Tertiary carbocation> secondary carbocation> primary carbocation

[reason is not required]

1

4. Presence of electron releasing groups[+I & +R groups] neutralize the +ve charge and thus stabilizes the carbocation.

1

5. Greater the number of α-hydrogens , greater is the extent of hyperconjugation & greater is the stability.

1

6. CaSO4.2H2O 1 7. Three 1 8. Lithium 1 9.

1

10.

1

11. (d) 1 12. (a) 1 13. (c) 1 14. (c) 1 15. (a) 1 16. (c) 17. (b) 1 18. (a) 1 19. (c) 1 20. (a) 1 SECTION-B 21. In the process of chemical reduction, oxides of metals are reduced using a stronger

reducing agent. Alkali metals and alkaline earth metals are among the strongest reducing agents and the reducing agents that are stronger than them are not available.

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Therefore, they cannot be obtained by chemical reduction of their oxides. OR

½ +½ +½ +½

22. (i) 2 2

4

.

2 2 2 2 2

Br NaNHalc KOH

CCl heatH C CH BrCH CH Br BrCH CH HC CH 1

(ii)

1

OR The peroxide effect is not observed in addition of HCl and HI. This may be due to the fact

that the H–Cl bond being stronger (430.5 kJ mol–1) than H–Br bond (363.7 kJ mol–1), is not cleaved by the free radical, whereas the H–I bond is weaker (296.8 kJ mol–1) and iodine free radicals combine to form iodine molecules instead of adding to the double bond.

2

23. Greater the number of α-hydrogens greater is the extent of hyperconjugation. Therefore the stability order is 3°>2°>1°.

1+1

24. (i)

1

(ii)

The dipole moment of cis-compound is a sum of the dipole moments of C–CH3 and C– CH2CH3 bonds acting in the same direction. The dipole moment of trans-compound is the resultant of the dipole moments of C–CH3 and C–CH2CH3 bonds acting in opposite directions. Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.

1

25. Benzene is a hybrid of resonating structures given as:

The six π electrons are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized π-electrons stabilize benzene.

1+1

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26. (i) (a)

(b)

½ +½

(ii) NO2 group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO2 group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O2NCH2CH2O– is expected to be more stable than CH3CH2O–.

1

27. 27

(i) A redox couple is defined as having together the oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

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(ii) Br2 > Ag > I2> Cu 1 SECTION- C 28. (i) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated

sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate. The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.

2

(ii) Kjeldahl’s method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

1

29. (a) (i) 2,4-dinitrochlorobenzene (ii) propane-1,2,3-tricarbaldehyde 1+1

(b) Anhydrous AlCl3 and BF3 1 30. (i) In this structure two identical groups are attached to one of the doubly bonded carbon

atom therefore geometrical isomerism is not possible. 1

(ii)

since OH- ions are liberated at cathode therefore pH increases.

1

(iii)

1

31. (i) With increase in number of branched chains, the molecule attains the shape of a sphere. This results in smaller area of contact and therefore weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperatures.

1

(ii) The ease of nitration depends on the presence of electron density on the compound to form nitrates. Nitration reactions are examples of electrophilic substitution reactions

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where an electron-rich species is attacked by a nitronium ion (NO2+). Now, CH3– group is electron donating and NO2– is electron withdrawing. Therefore, toluene will have the maximum electron density among the three compounds followed by benzene. On the other hand, m– Dinitrobenzene will have the least electron density. Hence, it will undergo nitration with difficulty. Hence, the increasing order of nitration is as follows:

(iii)

Note:-Marks to be awarded only on names of these compounds; structures are not required

Or any other suitable example

½ +½

OR

(i) Acidic character of a species is defined on the basis of ease with which it can lose its H– atoms. The hybridization state of carbon in the given compound is:

As the s–character increases, the electronegativity of carbon increases and the electrons of C–H bond pair lie closer to the carbon atom. As a result, partial positive charge of H– atom increases and H+ ions are set free. The s–character increases in the order: sp3 < sp2 < sp Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

1

(ii) For preparation of alkanes with odd number of carbon atoms two different alkyl halides need to be taken which then gives a mixture of products which is difficult to separate. For example if we have to prepare propane then the starting halides should be methyl halide and ethyl halide but the undergo reaction to form ethane, butane alongwith some propane

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(iii) Newmann projection for staggered conformation of ethane

Sawhorse projection for staggered conformation of ethane

½ +½

32. (i) Because of similar polarizing power. 1 (ii) Electrode potential (E°) of any M2+/M electrode depends upon three factors:

(i) Ionisation enthalpy (ii) Enthalpy of hydration (iii) Enthalpy of vaporisation The combined effect of these factors is approximately the same for Ca, Sr, and Ba. Hence, their electrode potentials are nearly constant.

1

(iii) Because of low ionisation enthalpy values and more negative reduction electrode potential values.

1

OR (i) The alkali metals and their salts impart characteristic colour to an oxidizing flame. This is

because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.

1

(ii) When sodium carbonate is added to water, it hydrolyses to give sodium bicarbonate and sodium hydroxide (a strong base). As a result, the solution becomes alkaline.

1

(iii) Halides of Be are covalent due to small size of Be+2 and are therefore soluble in organic solvents while the halides of Ba are largely ionic.

1

33. (i) 2 5 4 2

2 2 5 4

. 2, 6, 7 & 4O S are BaO CrO MnO CO

BaO CO CrO MnO

½ +½ +½ +½

(ii) Oxygen is oxidized and chlorine is reduced. ½ +½

34. (i) CCl4 will not give the white precipitate of AgCl on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in CCl4. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne’s extract of CCl4.

1

(ii) Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test.

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(iii) Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This is called “Lassaigne’s test”. The chemical equations involved in the test are

1

SECTION -D 35. (i) 0, +4 1 (ii) Because fluorine is the most electronegative element therefore can not be oxidized. 1 (iii)

Note:-Give marks on the steps.

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OR (i) An example of redox reaction occurs between the titration of potassium permanganate

and Mohr’s salt where permanganate ion acts as oxidizing agent and Fe+2 ion acts as reducing agent. Permanganate ion acts as self indicator in this reaction in the redox titration The visible end point in this case is achieved after the last of the reductant Fe2+ is oxidised and the first lasting tinge of pink colour appears at MnO4 –concentration as low as 10–6 mol dm–3 (10–6 mol L–1). This ensures a minimal ‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.

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(ii)

Note:-Give marks on the steps.

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36. (i) CH+

tropylium Or any other suitable example

1

(ii) [A]=cis-but-2-ene and [B]=trans-but-2-ene 2 (iii)

Nitronium ion acts as the electrophile

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E+=NO2+

OR (i)

cyclooctatetraene Or any other suitable example

1

(ii) Hex-3-ene is more stable due to greater number of α-hydrogens. 2 (iii)

3 3

aq.3 3

CH H CH H| | | |

CH C C H + H I CH C C H| |I H

Mechanism:

[morestablecarbocation]

Step II

I

3 3

aq.+3 3

3 3

aq.3 3

-

Step - I

CH H CH H| | | |

CH C C H + H CH C C H+ | H

CH H CH H| | | |

CH C C H + I CH C C H| | H

2

37. (i) The superoxide O2– is paramagnetic because of one unpaired electron in π*2p molecular orbital.

1

(ii) BeCl2 has a chain structure in the solid state

In the vapour phase BeCl2 tends to form a chloro-bridged dimer which dissociates into the linear

monomer at high temperatures of the order of 1200 K.

2

(iii) On moving down the alkali group, the ionic and atomic sizes of the metals increase. The given alkali metal ions can be arranged in the increasing order of their ionic sizes as: Li+ < Na+ < K+ < Rb+ < Cs+ Smaller the size of an ion, the more highly is it hydrated. Since Li+ is the smallest, it gets heavily hydrated in an aqueous solution. On the other hand, Cs+ is the largest and so it is the least hydrated. The given alkali metal ions can be arranged

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in the decreasing order of their hydrations as: Li+ > Na+ > K+ > Rb+ > Cs+ Greater the mass of a hydrated ion, the lower is its ionic mobility. Therefore, hydrated Li+ is the least mobile and hydrated Cs+ is the most mobile. Thus, the given alkali metal ions can be arranged in the increasing order of their mobilities as: Li+ < Na+ < K+ < Rb+ < Cs+

OR (i) Alkali metals are paramagnetic because they contain 1 unpaired electron where as their

compounds are in +1 O.S state and contain no unpaired electron. 1

(ii) When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution.

The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is blue in colour. At a higher concentration (3 M), clusters of metal ions are formed. This causes the solution to attain a copper–bronze colour and a characteristic metallic lustre.

2

(iii) Lithium ion and Fluoride ion are the two small sized ions and therefore pack effectively in the lattice and thus the lattice of LiF is extremely hard and therefore has lowest solubility in water.

2