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56/3 1 [P.T.O.

¸üÖê»Ö ®ÖÓ.

Roll No.

¸ü ÃÖ Ö µÖ ®Ö × ¾Ö – Ö Ö ®Ö (ÃÖî ü Ö × ®ŸÖ Û ú ) CHEMISTRY (Theory)

×®Ö¬ÖÖÔ×¸üŸÖ ÃÖ´ÖµÖ : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú : 70

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ÃÖÖ´ÖÖ®µ Ö ×®Ö¤ìü¿Ö ::::

(i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü …

(ii) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 1 ÃÖê 5 ŸÖÛú †×ŸÖ »Ö‘Öã-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü †Öî ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 1 †ÓÛú Æîü …

(iii) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 6 ÃÖê 10 ŸÖÛú »Ö‘Öã-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü †Öî ü ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 2 †ÓÛú Æïü …

(iv) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 11 ÃÖê 22 ŸÖÛú ³Öß »Ö‘Öã-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü †Öî ü ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 3 †ÓÛ Æïü …

(v) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 23 ´Öæ»µÖÖ¬ÖÖ×¸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî ü ‡ÃÖÛêú ×»Ö‹ 4 †ÓÛú Æïü …

(vi) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 24 ÃÖê 26 ŸÖÛú ¤üß‘ÖÔ-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü †Öî ü ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 5 †ÓÛú Æïüü …

(vii) µÖ×¤ü †Ö¾Ö¿µÖÛúŸÖÖ ÆüÖê, ŸÖÖê »ÖÖòÝÖ ™êü²Ö»ÖÖë ÛúÖ ¯ÖÏµÖÖêÝÖ Ûú¸ëü … Ûîú»ÖÛãú»Öê™ü¸üÖë Ûêú ˆ¯ÖµÖÖêÝÖ Ûúß †®Öã Ö×ŸÖ ®ÖÆüà Æîü …

Series : SSO/C 56/3

• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´Öã×¦üŸÖ ¯ÖéÂšü 12 Æïü … • ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê ü ×¤ü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ÃŸÖÛúÖ Ûêú ´ÖãÜÖ-¯ÖéÂšü ¯Ö¸ü ×»ÖÜÖë … • Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü … • Ûéú¯Öµ ÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢ Ö¸ü ×»ÖÜÖ®ÖÖ ¿Ö ãºþ Ûú¸ü®Öê ÃÖ ê ¯ÖÆü»Ö ê, ¯Ö Ï¿®Ö ÛúÖ ÛÎú ´ÖÖÓÛú †¾Ö¿µ Ö ×»ÖÜÖë … • ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê

×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî üÖ®Ö ¾Öê ˆ¢Ö¸ü-¯Öã×ÃŸÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …

• Please check that this question paper contains 12 printed pages.

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• Please check that this question paper contains 26 questions.

• Please write down the Serial Number of the question before attempting it.

• 15 minutes time has been allotted to read this question paper. The question paper will be

distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the

question paper only and will not write any answer on the answer-book during this period.

ÛúÖê› ü ®ÖÓ. Code No.

¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ÃŸÖÛúÖ Ûêú ´ÖãÜÖ-¯ÖéÂšü ¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … Candidates must write the Code on

the title page of the answer-book.

SET – 3

56/3 2

General Instructions :

(i) All questions are compulsory.

(ii) Question number 1 to 5 are very short answer questions and carry 1 mark each.

(iii) Question number 6 to 10 are short answer questions and carry 2 marks each.

(iv) Question number 11 to 22 are also short answer questions and carry 3 marks each.

(v) Question number 23 is a value based question and carry 4 marks.

(vi) Question number 24 to 26 are long answer questions and carry 5 marks each.

(vii) Use log tables, if necessary. Use of calculators is not allowed.

1. ‘Ûúß»Öê™’ü ¯ÖÏ³ÖÖ¾Ö ÛúÖ ŒµÖÖ ŸÖÖŸ¯ÖµÖÔ ÆüÖêŸÖÖ Æîü ?

What is meant by chelate effect ?

2. ×®Ö´®Ö ÛúÖ †Ö‡Ô µÖæ ¯Öß ‹ ÃÖß (IUPAC) ®ÖÖ´Ö ×»Ö×ÜÖ‹ :

CH3 – CH2 – CHO

Write the IUPAC name of the following :

CH3 – CH2 – CHO

3. ×®Ö´®Ö ÛúÖê õÖÖ¸üßµÖ õÖ´ÖŸÖÖ Ûêú ²ÖœÌüŸÖê ÛÎú´Ö ´Öë ¾µÖ¾Ö×Ã£ÖŸÖ Ûúß×•Ö‹ :

‹ê×®Ö»Öß®Ö, p-®ÖÖ‡™ÒüÖê‹ê×®Ö»Öß®Ö †Öî ü p-™üÖ»Öã‡›Ìüß®Ö

Arrange the following in increasing order of basic strength :

Aniline, p-Nitroaniline and p-Toluidine

4. AgCl ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ÛúÖ Ã™üÖê‡×ÛúµÖÖê Öß™Òüß ¤üÖêÂÖ ¤ü¿ÖÖÔŸÖÖ Æîü ?

What type of stoichiometric defect is shown by AgCl ?

5. ‡´Ö»¿Ö®Ö ŒµÖÖ ÆüÖêŸÖê Æïü ? ‹Ûú ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ …

What are emulsions ? Give an example.

56/3 3 [P.T.O.

6. ¯ÖÖê™îü×¿ÖµÖ´Ö ¯Ö¸ü´ÖïÝÖ®Öê™ü Ûúß ×®Ö´ÖÖÔÞÖ ×¾Ö×¬Ö ÛúÖ ¾ÖÞÖÔ®Ö Ûúß×•Ö‹ … †´»ÖßÛéúŸÖ ¯Ö¸ü´ÖîÓÝÖ®Öê™ü †ÖòŒÃÖî×»ÖÛú †´»Ö Ûêú ÃÖÖ£Ö ÛîúÃÖê

†×³Ö×ÛÎúµÖÖ Ûú¸üŸÖÖ Æîü ? †×³Ö×ÛÎúµÖÖ Ûêú ×»ÖµÖê †ÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖ ×»Ö×ÜÖ‹ …

†£Ö¾ÖÖ

¯ÖÖê™îü×¿ÖµÖ´Ö ›üÖ‡ÛÎúÖê Öê™ü Ûúß †ÖòŒÃÖßÛú¸üÞÖ ×ÛÎúµÖÖ ÛúÖ ¾ÖÞÖÔ®Ö Ûúß×•Ö‹ †Öî ü ‡ÃÖÛúß (i) †ÖµÖÖê›üÖ‡›ü (ii) H2S Ûêú ÃÖÖ£Ö

ÆüÖê®Öê ¾ÖÖ»Öß †×³Ö×ÛÎúµÖÖ†Öë Ûêú ×»ÖµÖê †ÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖÖë ÛúÖê ×»Ö×ÜÖ‹ …

Describe the preparation of potassium permanganate. How does the acidified

permanganate solution react with oxalic acid ? Write the ionic equations for the

reactions.

OR

Describe the oxidising action of potassium dichromate and write the ionic equations

for its reaction with (i) an iodide (ii) H2S.

7. ‹£Ö®ÖÖò»Ö ÃÖê ‹£Öß®Ö ²Ö®Ö®Öê ´Öë †´»Ö ×®Ö•ÖÔ»ÖßÛú¸üÞÖ Ûúß ×ÛÎúµÖÖ×¾Ö×¬Ö ×»Ö×ÜÖ‹ …

Write the mechanism of acid dehydration of ethanol to yield ethene.

8. ×®Ö´®Ö ¯Ö¤üÖë Ûúß ¯Ö×¸ü³ÖÖÂÖÖ‹Ñ ×»Ö×ÜÖµÖê :

(i) ´ÖÖê»ÖÖÓ¿Ö (x)

(ii) ‹Ûú ×¾Ö»ÖµÖ®Ö Ûúß ´ÖÖê»Ö»ÖŸÖÖ (m)

Define the following terms :

(i) Mole fraction (x)

(ii) Molality of a solution (m)

56/3 4

9. •Öß¸üÖê †Öò›Ôü¸ü †Öî ü ×«üŸÖßµÖ †Öò›Ôü¸ü †×³Ö×ÛÎúµÖÖ†Öë Ûêú ×»ÖµÖê ¤ü¸ü ×Ã£Ö¸üÖÓÛúÖë Ûêú µÖæ×®Ö™ü ×»Ö×ÜÖ‹ ˆÃÖ ×Ã£Ö×ŸÖ ´Öë •Ö²Ö ÃÖÖÓ¦üŸÖÖ

ÛúÖê mol L–1 †Öî ü ÃÖ´ÖµÖ ÛúÖê ÃÖêÛúÞ›üÖë ´Öë ×»ÖÜÖÖ ÝÖµÖÖ ÆüÖê …

Write units of rate constants for zero order and for the second order reactions if the

concentration is expressed in mol L–1 and time in second.

10. ×®Ö´®Ö Ûú£Ö®ÖÖë Ûúß ¾µÖÖÜµÖÖ Ûúß×•Ö‹ :

(i) ±úÖòÃ±úÖê üÃÖ Ûúß †¯ÖêõÖÖ ®ÖÖ‡™ÒüÖê•Ö®Ö ²ÖÆãüŸÖ Ûú´Ö ÃÖ×ÛÎúµÖ Æîü …

(ii) NF3 ‹Ûú ‰úÂ´ÖÖõÖê Öß ¯Ö¤üÖ£ÖÔ Æîü ¯Ö¸ü®ŸÖã NCl3 ‰úÂ´ÖÖ¿ÖÖêÂÖß ¯Ö¤üÖ£ÖÔ Æîü …

Explain the following :

(i) Nitrogen is much less reactive than phosphorus.

(ii) NF3 is an exothermic compound but NCl3 is an endothermic compound.

11. ‘†®ÖÖ®Öã ÖÖŸÖ®Ö’ ÃÖê ŒµÖÖ ŸÖÖŸ¯ÖµÖÔ ÆüÖêŸÖÖ Æîü ? •Ö»ÖßµÖ ×¾Ö»ÖµÖ®Ö ´Öë †®ÖÖ®Öã ÖÖŸÖ®Ö †×³Ö×ÛÎúµÖÖ†Öë ÛúÖ ‹Ûú ˆ¤üÖÆü üÞÖ ¤üß×•Ö‹ …

What is meant by ‘disproportionation’ ? Give one example of disproportionation

reaction in aqueous solutions.

12. ×®Ö´®Ö×»Ö×ÜÖŸÖ Ûêú IUPAC ®ÖÖ´Ö ×»Ö×ÜÖ‹ :

(i) [Co(NH3)6]Cl3

(ii) [NiCl4]2–

(iii) K3[Fe(CN)6]


(i) [Co(NH3)6]Cl3

(ii) [NiCl4]2–

(iii) K3[Fe(CN)6]

56/3 5 [P.T.O.

13. ×®Ö´®Ö µÖÖî×ÝÖÛúÖë Ûêú †Ö‡Ô µÖæ ¯Öß ‹ ÃÖß (IUPAC) ®ÖÖ´ÖÖë ÛúÖê ¤üß×•Ö‹ :

(i) CH3 – CH –| Br

CH2 – CH3

(ii)

Br

Br

(iii) CH2 = CH – CH2 – Cl

Give the IUPAC names of the following compounds :

(i) CH3 – CH –| Br

CH2 – CH3

(ii)

Br

Br


14. ×®Ö´®Ö ºþ¯ÖÖÓŸÖ¸üÞÖ ÛîúÃÖê ×ÛúµÖê •ÖÖŸÖê Æïü ?

(i) ²Öê×®•Ö»Ö Œ»ÖÖê üÖ‡›ü ÛúÖ ²Öê×®•Ö»Ö ‹ê»ÛúÖêÆüÖò»Ö ´Öë

(ii) ‹×£Ö»Ö ´ÖîÝ®Öß×¿ÖµÖ´Ö Œ»ÖÖê üÖ‡›ü ÛúÖ ¯ÖÏÖê Öê®Ö-1-†Öò»Ö ´Öë

(iii) ¯ÖÏÖê Öß®Ö ÛúÖê ¯ÖÏÖê Öê®Ö-2-†Öò»Ö ´Öë

How are the following conversions carried out ?

(i) Benzyl chloride to Benzyl alcohol

(ii) Ethyl magnesium chloride to Propan-1-ol

(iii) Propene to Propan-2-ol

56/3 6

15. ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë ´Öë ´ÖãÜµÖ ˆŸ¯ÖÖ¤ü ×»Ö×ÜÖ‹ :

(i) CH3 – CH2OH –––––––––→PCl

5 ?

(ii)

OH

+ CH3– Cl ––––––––––––––––––→anhyd. AlCl

3 ?

(iii) CH3 – Cl + CH3CH2 – ONa → ?

Write the major product in the following equations :

(i) CH3 – CH2OH –––––––––→PCl

5 ?

(ii)

OH

+ CH3– Cl ––––––––––––––––––→anhyd. AlCl

3 ?


16. ¯ÖÏÖê™üß®Ö ÃÖê ÃÖÓ²Ö×®¬ÖŸÖ ×®Ö´®Ö ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ :

(i) ¯Öê ™üÖ‡›ü Ø»ÖÛêú•Ö

(ii) ¯ÖÏÖ‡´Ö¸üß ÃÖÓ ü“Ö®ÖÖ

(iii) ›üß®Öî“Öã êü¿Ö®Ö

Define the following as related to proteins :

(i) Peptide linkage

(ii) Primary structure

(iii) Denaturation

17. ‘ÛúÖê ÖÖò»Öß´Ö¸üÖ‡•Öê¿Ö®Ö’ ¯Ö¤ü Ûúß ¾µÖÖÜµÖÖ Ûúß×•Ö‹ †Öî ü ‘ÛúÖê ÖÖò»Öß´Ö¸üÖ‡•Öê¿Ö®Ö’ Ûêú ¤üÖê ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ …

Explain the term ‘copolymerization’ and give two examples of copolymerization.

56/3 7 [P.T.O.

18. ×ÃÖ»¾Ö¸ü fcc •ÖÖ»ÖÛú ´Öë ×ÛÎúÃ™ü×»ÖŸÖ ÆüÖêŸÖÖ Æîü … µÖ×¤ü µÖæ×®Ö™ü ÃÖê»Ö Ûêú ÛúÖê ü Ûúß »Ö´²ÖÖ‡Ô 4.077 × 10–8 cm ÆüÖê, ŸÖÖê

×ÃÖ»¾Ö¸ü ÛúÖ †¬ÖÔ¾µÖÖÃÖ (r) ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ …

Silver crystallises in fcc lattice. If edge length of the unit cell is 4.077 × 10–8 cm, then

calculate the radius of silver atom.

19. Ûêú®Ö-¿ÖãÝÖ¸ü (M.W. 342) ÛúÖ 5 ¯ÖÏ×ŸÖ¿ÖŸÖ ‘ÖÖê»Ö (¦ü¾µÖ´ÖÖ®Ö †Ö¬ÖÖ¸ü ¯Ö¸ü) ‹Ûú ¯Ö¤üÖ£ÖÔ X Ûêú 0.877% ‘ÖÖê»Ö Ûêú ÃÖÖ£Ö

†Ö‡ÃÖÖê™üÖê×®ÖÛú Æîü … X ÛúÖ †ÖÞÖ×¾ÖÛú ³ÖÖ¸ü ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ …

A 5 percent solution (by mass) of cane-sugar (M.W. 342) is isotonic with 0.877%

solution of substance X. Find the molecular weight of X.

20. ‹Ûú ¯ÖÏ£Ö´Ö ÛúÖê×™ü †×³Ö×ÛÎúµÖÖ Ûêú ×»ÖµÖê ¤ü¸ü ×Ã£Ö¸üÖÓÛú 60 s–1 Æîü … †×³ÖÛúÖ¸üÛú Ûêú ¯ÖÏÖ¸ü×´³ÖÛú ÃÖÖÓ¦üÞÖ ÛúÖê ‡ÃÖÛêú 1/10

ŸÖÛú ‘Ö™®Öêê ´Öë ×ÛúŸÖ®ÖÖ ÃÖ´ÖµÖ »ÖÝÖêÝÖÖ ?

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

reduce the initial concentration of the reactant to its 1/10th

value ?

21. ×®Ö´®Ö ¯ÖÏÛÎú´ÖÖë Ûúß ¾µÖÖÜµÖÖ Ûúß×•Ö‹ :

(i) ›üÖµÖ×»Ö×ÃÖÃÖ

(ii) ‡»ÖêŒ™ÒüÖê±úÖê êü×ÃÖÃÖ

(iii) ×™üÞ›ü»Ö ¯ÖÏ³ÖÖ¾Ö

Describe the following processes :

(i) Dialysis

(ii) Electrophoresis

(iii) Tyndall effect

56/3 8

22. ×®Ö´®Ö Ûêú ˆ¢Ö¸ü ¤üß×•Ö‹ :

(i) ‹ê»Öã×´Ö×®ÖµÖ´Ö Ûêú ¬ÖÖŸÖãÛú´ÖÔ ´Öë ÛÎúÖ‡µÖÖê»ÖÖ‡™ü Ûúß ŒµÖÖ ³Öæ×´ÖÛúÖ ÆüÖêŸÖß Æîü ?

(ii) ³Ö•ÖÔ®Ö ×ÛÎúµÖÖ †Öî ü ×®ÖÃŸÖÖ¯Ö®Ö ´Öë †ÓŸÖ¸ü Ûúß×•Ö‹ …

(iii) ‘ÛÎúÖê Öîê™üÖêÝÖÏî±úß’ ¯Ö¤ü ÃÖê ŒµÖÖ ŸÖÖŸ¯ÖµÖÔ ÆüÖêŸÖÖ Æîü ?

†£Ö¾ÖÖ

»ÖÖêÆüÖ ²Ö®ÖÖ®Öê Ûêú ×»Ö‹ ²»ÖÖÃ™ü ±ú®ÖìÃÖ Ûêú ×¾Ö×³Ö®®Ö ³ÖÖÝÖÖë ´Öë ÆüÖê®Öê ¾ÖÖ»Öß †×³Ö×ÛÎúµÖÖ‹Ñ ×»Ö×ÜÖ‹ …

Answer the following :

(i) What is the role of cryolite in the metallurgy of aluminium ?

(ii) Differentiate between roasting and calcination.

(iii) What is meant by the term ‘chromatography’ ?

OR

Write the reactions taking place in different zones of the blast furnace to obtain Iron.

23. ®Öß¸ü•Ö ×›ü ÖÖ™Ôü´Öê®™ü»Ö Ã™üÖê ü ´Öë ‘Ö¸ü Ûêú ÃÖÖ´ÖÖ®Ö ÜÖ¸üß¤ü®Öê Ûêú ×»ÖµÖê ÝÖµÖÖ … ‹Ûú ÜÖÖ®Öê ´Öë ¾ÖÆü Ûãú”û ¿ÖãÝÖ¸ü¸ü×ÆüŸÖ ×™ü×ÛúµÖÖÑ

¤êüÜÖÖ … ¾ÖÆü Ûãú”û ‹êÃÖß ×™ü×ÛúµÖÖ ÜÖ¸üß¤ü®Öê ÛúÖ ×®Ö¿“ÖµÖ ×ÛúµÖÖ •ÖÖê ˆÃÖÛêú ¤üÖ¤üÖ Ûêú ×»ÖµÖê ˆ¯ÖµÖÖêÝÖß £Öß, ŒµÖÖë×Ûú ˆÃÖÛêú

¤üÖ¤üÖ ¿ÖãÝÖ¸ü Ûêú ´Ö¸üß•Ö £Öê … ¾ÖÆüÖÑ ŸÖß®Ö ¯ÖÏÛúÖ¸ü Ûúß ¿ÖãÝÖ¸ü¸ü×ÆüŸÖ ×™ü×ÛúµÖÖÑ £ÖßÓ … ˆÃÖ®Öê ×®ÖÞÖÔµÖ ×ÛúµÖÖ ¾ÖÆü ÃÖãÛÎúÖê»ÖÖêÃÖ ÜÖ¸üß¤êü

•ÖÖê ˆÃÖÛêú ¤üÖ¤üÖ Ûêú ×»ÖµÖê ˆ¯ÖµÖÖêÝÖß £Öß …

(i) ‹Ûú †®µÖ ¿ÖãÝÖ¸ü ¸ü×ÆüŸÖ ÛúÖ ®ÖÖ´Ö ¤üß×•Ö‹ •ÖÖê ®Öß¸ü•Ö ®Öê ®ÖÆüà ÜÖ¸üß¤üÖ …

(ii) ŒµÖÖ ›üÖŒ™ü¸ü Ûêú ¯Ö“Öá Ûêú ×²Ö®ÖÖ ‹êÃÖß ¤ü¾ÖÖ ÜÖ¸üß¤ü®ÖÖ ®Öß¸ü•Ö Ûêú ×»ÖµÖê ˆ×“ÖŸÖ £ÖÖ ?

(iii) ˆ¯Ö¸üÖêŒŸÖ ÃÖê ®Öß¸ü•Ö ÛúÖ ÛúÖî®Ö ÃÖÖ ÝÖãÞÖ ¯ÖÏ×ŸÖ»Ö×õÖŸÖ ÆüÖêŸÖÖ Æîü ?

Neeraj went to the departmental store to purchase groceries. On one of the shelves he

noticed sugar free tablets. He decided to buy them for his grandfather who was a

diabetic. There were three types of sugar free tablets. He decided to buy sucrolose

which was good for his grandfather’s health.

(i) Name another sugar free tablet which Neeraj did not purchase.

(ii) Was it right to purchase such medicines without doctor’s prescription ?

(iii) What quality of Neeraj is reflected above ?

56/3 9 [P.T.O.

24. (a) ×®Ö´®ÖÖë Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ :

(i) p-´Öê×£Ö»Ö²Öï•ÖÌ‹×»›üÆüÖ‡›ü

(ii) 4-´Öê×£Ö»Ö¯Öî®™ü-3-‡Ô®Ö-2-†Öò®Ö

(b) ×®Ö´®Ö µÖÖî×ÝÖÛú µÖãÝ´ÖÖë ´Öë †ÓŸÖ¸ü Ûú¸ü®Öê Ûêú ×»ÖµÖê ¸üÖÃÖÖµÖ×®ÖÛú •ÖÖÑ“ÖÖë ÛúÖê ¤üß×•Ö‹ :

(i) ²Öê®•ÖÌÖê‡Ûú ‹ê×ÃÖ›ü †Öî ü ‹×£Ö»Ö²Öê®•ÖÌÖê‹™ü …

(ii) ²Öê®•Öî×»›üÆüÖ‡›ü †Öî ü ‹êÃÖß™üÖê±úß®ÖÖê®Ö

(iii) ±úß®ÖÖò»Ö †Öî ü ²Öê®•ÖÌÖê‡Ûú ‹ê×ÃÖ›ü

†£Ö¾ÖÖ

(a) ×®Ö´®Ö ¾µÖãŸ¯Ö®®ÖÖë Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ †Ö¸êü×ÜÖŸÖ Ûúß×•Ö‹ :

(i) ¯ÖÏÖê Öê®ÖÖê®Ö †Öò×ŒÃÖ´Ö

(ii) CH3CHO ÛúÖ ÃÖê ÖßÛúÖ²Öì•ÖÖê®Ö

(b) ‹£Öî®ÖÖò»Ö ÛúÖê †Ö¯Ö ×®Ö´®Ö µÖÖî×ÝÖÛúÖë ´Öë ÛîúÃÖê ºþ¯ÖÖÓŸÖ×¸üŸÖ Ûú ëüÝÖê ? ÃÖ´Ö²Ö¨ü ¸üÖÃÖÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖÖë ÛúÖê ¤üß×•Ö‹ …

(i) CH3 – CH3

(ii) CH3 – CH –|

OH

CH2 – CHO

(iii) CH3CH2OH

(a) Draw the structures of the following :

(i) p-Methylbenzaldehyde

(ii) 4-Methylpent-3-en-2-one

(b) Give chemical tests to distinguish between the following pairs of compounds :

(i) Benzoic acid and Ethyl benzoate.

(ii) Benzaldehyde and Acetophenone.

(iii) Phenol and Benzoic acid.

OR

56/3 10

(a) Draw the structures of the following derivatives :

(i) Propanone oxime

(ii) Semicarbazone of CH3CHO

(b) How will you convert ethanal into the following compounds ? Give the chemical

equations involved.

(i) CH3 – CH3

(ii) CH3 – CH –|

OH

CH2 – CHO

(iii) CH3CH2OH

25. (a) ÝÖÏã Ö 16 Ûêú ŸÖ¢¾Ö ÃÖÖ¬ÖÖ¸üÞÖŸÖµÖÖ ¯ÖÏ£Ö´Ö †ÖµÖ®Ö®Ö ‹®£Öî»¯Öß ŸÖŸÃÖ´²Ö®¬Öß †Ö¾ÖŸÖÔ ¾ÖÖ»Öê ÝÖÏã Ö 15 Ûêú ŸÖ¢¾ÖÖë Ûúß

ŸÖã»Ö®ÖÖ ´Öë Ûú´Ö ´ÖÖ®Ö ¤ü¿ÖÖÔŸÖê Æïü … ‹êÃÖÖ ŒµÖÖë Æîü ?

(b) ŒµÖÖ ÆüÖêŸÖÖ Æîü •Ö²Ö –

(i) ÃÖÖÓ¦ü H2SO4 ÛúÖê CaF2 ¯Ö¸ü ›üÖ»ÖÖ •ÖÖŸÖÖ Æîü ?

(ii) ÃÖ»±Ìú¸ü ›üÖ‡†ÖòŒÃÖÖ‡›ü “ÖÖ¸üÛúÖê»Ö Ûúß ˆ¯Ö×Ã£Ö×ŸÖ ´Öë Œ»ÖÖê üß®Ö ÃÖê †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖß Æîü ?

(iii) †´ÖÖê×®ÖµÖ´Ö Œ»ÖÖê üÖ‡›ü ÛúÖê Ca(OH)2 Ûêú ÃÖÖ£Ö ˆ¯Ö“ÖÖ×¸üŸÖ ×ÛúµÖÖ •ÖÖŸÖÖ Æîü ?

†£Ö¾ÖÖ

(a) ×®Ö´®ÖÖë Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ :

(i) BrF3

(ii) XeO3

(b) ×®Ö´®Ö ¯ÖÏ¿®ÖÖë Ûêú ˆ¢Ö¸ü ¤üß×•Ö‹ :

(i) PH3Ûúß †¯ÖêõÖÖ NH3 ŒµÖÖë †×¬ÖÛú õÖÖ¸üßµÖ ÆüÖêŸÖÖ Æîü ?

(ii) Æîü»ÖÖê•Ö®Ö ¯ÖÏ²Ö»Ö †ÖòŒÃÖßÛúÖ¸üÛú ŒµÖÖë ÆüÖêŸÖê Æïü ?

(iii) XeOF4 Ûúß ÃÖÓ ü“Ö®ÖÖ †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ …

56/3 11 [P.T.O.

(a) Elements of Gr. 16 generally show lower value of first ionization enthalpy

compared to the corresponding periods of Gr. 15. Why ?

(b) What happens when

(i) concentrated H2SO4 is added to CaF2 ?

(ii) sulphur dioxide reacts with chlorine in the presence of charcoal ?

(iii) ammonium chloride is treated with Ca(OH)2 ?

OR

(a) Draw the structure of the following :

(i) BrF3

(ii) XeO3

(b) Answer the following :

(i) Why is NH3 more basic than PH3 ?

(ii) Why are halogens strong oxidising agents ?

(iii) Draw the structure of XeOF4.

26. ∆rG° †Öî ü e.m.f.(E) ÛúÖê ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ •ÖÖê 25 °C ¯Ö¸ü Ã™ïü›ü›Ôü ×Ã£Ö×ŸÖ ´Öë ×®Ö´®Ö ÃÖê»Ö ÃÖê ¯ÖÏÖ¯ŸÖ ÆüÖêŸÖÖ Æîü :

Zn(s) | Zn2+(aq) || Sn2+(aq) | Sn(s)

×¤üµÖÖ ÝÖµÖÖ : E°Zn

2+/Zn

= – 0.76 V; E°Sn

2+/Sn

= – 0.14 V

†Öî ü F = 96500 C mol–1

†£Ö¾ÖÖ

(a) ‹Ûú ×¾ÖªãŸÖË-†¯Ö‘Ö™ü¶ Ûêú ×¾Ö»ÖµÖ®Ö Ûêú ×»ÖµÖê “ÖÖ»ÖÛúŸÖÖ †Öî ü ´ÖÖê»Ö¸ü “ÖÖ»ÖÛúŸÖÖ ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ …

ÃÖÖÓ¦üŸÖÖ Ûêú ÃÖÖ£Ö ˆ®ÖÛêú ¯Ö×¸ü¾ÖŸÖÔ®Ö Ûúß ¾µÖÖÜµÖÖ Ûúß×•Ö‹ …

56/3 12

(b) ˆÃÖ ÝÖî»Ö¾ÖÖò×®ÖÛú ÃÖê»Ö Ûêú Ã™ïü›ü›Ôü ÃÖê»Ö ×¾Ö³Ö¾Ö ÛúÖê ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ ×•ÖÃÖ´Öë ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ ÆüÖêŸÖß Æîü :

Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

†×³Ö×ÛÎúµÖÖ ÛúÖ ∆rG° †Öî ü ŸÖã»µÖÖÓÛúß ×Ã£Ö¸üÖÓÛú ÛúÖ ¯Ö×¸üÛú»Ö®Ö ³Öß Ûúß×•Ö‹ …

(E°Ag

+/Ag

= 0.80 V; E°Fe

3+/Fe

2+ = 0.77 V)

Calculate ∆rG° and e.m.f. (E) that can be obtained from the following cell under the

standard conditions at 25 °C :

Zn(s) | Zn2+(aq) || Sn2+(aq) | Sn(s)

Given : E°Zn

2+/Zn

= – 0.76 V; E°Sn

2+/Sn

= – 0.14 V

and F = 96500 C mol–1.

OR

(a) Define conductivity and molar conductivity for the solution of an electrolyte.

Discuss their variation with concentration.

(b) Calculate the standard cell potential of the galvanic cell in which the following

reaction takes place :


Calculate the ∆rG° and equilibrium constant of the reaction also.

(E°Ag

+/Ag

= 0.80 V; E°Fe

3+/Fe

2+ = 0.77 V)

____________

1

Qu

es. Value points Marks

1 Formation of stable complex by polydentate ligand. 1

2 Propanal 1

3 p-Nitroaniline < Aniline < p-Toluidine 1

4 Frenkel defect 1

5 Emulsions are liquid – liquid colloidal systems.

For example – milk, cream (or any other one correct example)

½ + ½

6 Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an

oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a

neutral or acidic solution to give permanganate.

Oxalate ion or oxalic acid is oxidised at 333 K:

OR

1

1

6 i)

ii)

1

1

7

½

½

1

CHEMISTRY MARKING SCHEME

SET -56/3

Compt. July, 2015

2

8 i)

ii) Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the

solvent. Or

1

1

9 Zero order : mol L-1

s-1

Second order : L mol-1

s-1

1

1

10 i) Due to high bond dissociation enthalpy of N N

ii) Due to low bond dissociation enthalpy of F2 than Cl2 and strong bond formation

between N and F

1

1

11 Disproportionation : The reaction in which an element undergoes self-oxidation and self-

reduction simultaneously. For example –

2Cu+ (aq) Cu

2+ (aq) + Cu(s)

(Or any other correct equation)

1 ½

1 ½

12 i) Hexaamminecobalt(III) chloride

ii) Tetrachlorido nickelate(II)

iii) Potassium hexacyanoferrate(III)

1

1

1

13 i) 2-bromobutane

ii) 1, 3-dibromobenzene

iii) 3-choloropropene

1

1

1

14

i)

ii)

1

1

1

15

i)

1

3

ii)

iii)

1

1

16 i) Peptide linkage – in proteins, -amino acids are connected to each other by peptide

bond or peptide linkage (-CONH- bond).

ii) Primary structure - each polypeptide in a protein molecule having amino acids which

are linked with each other in a specific sequence.

iii) Denaturation - When a protein is subjected to physical change like change in

temperature or chemical change like change in pH, protein loses its biological activity.

1

1

1

17 Copolymerisation is a polymerisation reaction in which a mixture of more than one monomeric

species is allowed to polymerise and form a copolymer.

(or any other correct example)

1

1

1

18 r =

√

r=

r = 1.44 x cm

1

1

1

cane sugar = πXח 19

Therefore, ccane sugar = cX (where c is molar concentration)

=

=

MX =

gmol

-1

MX = 59.9 or 60 gmol-1

1

1

1

20 k=

log

1

4

60 s-1

=

log

t=

log 10

t=

s

t= 0.0384 s

1

1 21 i) It is a process of removing the dissolved substance from a colloidal solution by means

of diffusion through a semi - permeable membrane.

ii) The movement of colloidal particles under an applied electric potential towards

oppositely charged electrode is called electrophoresis.

iii) Colloidal particles scatter light in all directions in space. This scattering of light

illuminates the path of beam in the colloidal dispersion.

1

1

1 22 i) It lowers the melting point of alumina / acts as a solvent.

ii)

Roasting Calcination

Ore is heated in a regular supply of air Heating in a limited supply or

absence of air.

(Or with equation)

iii) It is a process of separation of different components of a mixture which are differently

adsorbed on a suitable adsorbent. OR

1

1

1

22

(any 6 correct equations)

6 x ½

= 3

23 i) Aspartame, Saccharin (any one)

ii) No

iii) Social concern, empathy, concern, social awareness (any 2 )

1

1

2 24 a) i)

ii)

b) i)Add NaHCO3, benzoic acid will give brisk effervescence of CO2 whereas ethylbenzoate

1

1

5

will not.

ii)Add NaOH and I2, acetophonone forms yellow ppt of iodoform on heating whereas

benzaldehyde will not.

iii)Add neutral FeCl3, phenol gives violet colouration whereas benzoic acid does not. (or any other correct test)

OR

1

1

1

24 a) i)

ii)

b) i)

ii)

iii)

1

1

1

1

1

25 a) Due to relatively stable half – filled p-orbitals of group 15 elements

b) i) CaF2 + H2SO4 CaSO4 + 2HF

ii)

iii)

OR

2

1

1

1

25

a) i)

1

6

ii)

b) i)Due to small size of nitrogen, the lone pair of electron on nitrogen is localized/ easily

available for donation.

ii)Because they need only one electron to attain stable/noble gas configuration.

iii)

1

1

1

1

26 E0cell = E

0Sn2+ / Sn - E

0Zn2+ / Zn

= - 0.14V –(- 0.76V)

= 0.62V

∆rG0 = -n F E

0cell

= - 2 x 96500 C mol-1

x 0.62 V

= - 119660 J mol-1

Ecell = E0

cell -

log

Ecell = 0.62 -

log

OR

1

1

1

1

1

26 a) The conductivity of a solution at any given concentration is the conductance of one unit

volume of solution kept between two platinum electrodes with unit area of cross section

and at a distance of unit length.

Molar conductivity of a solution at a given concentration is the conductance of the volume

V of solution containing one mole of electrolyte kept between two electrodes with area of

cross section A and distance of unit length.

Molar conductivity increases with decrease in concentration.

b)E0cell = E

0C - E

0A

= 0.80V – 0.77V

= 0.03V

∆rG0 = -n F E

0cell

= - 1 x 96500 C mol-1

x 0.03 V

= - 2895 J mol-1

Log Kc=

½

½

1

½

½

1

½

7

Dr. Sangeeta Bhatia Sh. S.K. Munjal Sh. D.A. Mishra

Ms. Garima Bhutani

Log Kc=

Log Kc= 0.508

½

56/1/3 1 [P.T.O.


Roll No.






(ii) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 1 ÃÖê 5 ŸÖÛú †×ŸÖ »Ö‘Öã-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü … ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 1 †ÓÛú ×®Ö¬ÖÖÔ×¸üŸÖ Æîü …

(iii) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 6 ÃÖê 10 ŸÖÛú »Ö‘Öã-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü … ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 2 †ÓÛú ×®Ö¬ÖÖÔ×¸üŸÖ Æïü …

(iv) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 11 ÃÖê 22 ŸÖÛú ³Öß »Ö‘Öã-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü … ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 3 †ÓÛú ×®Ö¬ÖÖÔ×¸üŸÖ Æïü …

(v) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 23 ´Öæ»µÖÖ¬ÖÖ×¸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî ü ‡ÃÖÛêú ×»Ö‹ 4 †ÓÛú ×®Ö¬ÖÖÔ×¸üŸÖ Æïü …

(vi) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 24 ÃÖê 26 ¤üß‘ÖÔ-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü … ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 5 †ÓÛú Æïüü …

(vii) µÖ×¤ü †Ö¾Ö¿µÖÛú ÆüÖê ŸÖÖê »ÖÖòÝÖ ™êü²Ö»Ö ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü … Ûîú »ÖÛãú»Öê™ü¸ü Ûêú ˆ¯ÖµÖÖêÝÖ Ûúß †®Öã Ö×ŸÖ ®ÖÆüà Æîü …

Series : SSO/1/C 56/1/3














SET – 3

56/1/3 2



(ii) Q. No. 1 to 5 are very short answer questions and carry 1 mark each.

(iii) Q. No. 6 to 10 are short answer questions and carry 2 marks each.

(iv) Q. No. 11 to 22 are also short answer questions and carry 3 marks each.

(v) Q. No. 23 is a value based question and carry 4 marks.

(vi) Q. No. 24 to 26 are long answer questions and carry 5 marks each.

(vii) Use log tables if necessary, use of calculator is not allowed.

1. ÛúÖò ¯»ÖêŒÃÖ [Ni(NH3)6]Cl2 ÛúÖ †Ö‡Ô µÖæ ¯Öß ‹ ÃÖß (IUPAC) ®ÖÖ´Ö ×»Ö×ÜÖ‹ …

What is the IUPAC name of the complex [Ni(NH3)6]Cl2 ?

2. ×®Ö´®Ö µÖÖî×ÝÖÛú Ûúß ÃÖÓ ü“Ö®ÖÖ †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ …

3-´Öê×£Ö»Ö¯Öê®™îü®Öî»Ö

Draw the structure of 3-methylpentanal.

3. ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ ÃÖ´ÖßÛú¸üÞÖ ÛúÖê ¯ÖæÞÖÔ Ûúß×•Ö‹ :

C6H5N2Cl + H3PO2 + H2O ––––→ - - -

Complete the following reaction equation :

C6H5N2Cl + H3PO2 + H2O ––––→ - - -

4. ‹Ûú †ÓŸÖ: Ûêú×®¦üŸÖ ‘Ö®ÖßµÖ ÃÖÓ ü“Ö®ÖÖ ´Öë ¯Ö¸ü´ÖÖÞÖã†Öë Ûúß ÃÖÓÜµÖÖ ¯ÖÏ×ŸÖ ‹ÛúÛú ÛúÖê×ÂšüÛúÖ (z) ŒµÖÖ ÆüÖêŸÖß Æîü ?

What is the no. of atoms per unit cell (z) in a body-centred cubic structure ?

5. ÃÖŸÖÆü ¸üÃÖÖµÖ®Ö Ûêú ÃÖÓ¤ü³ÖÔ ´Öë ›üÖµÖÖ×»Ö×ÃÖÃÖ ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ …

In reference to surface chemistry, define dialysis.

56/1/3 3 [P.T.O.

6. ‹£Öî®ÖÖò»Ö Ûêú ×®Ö•ÖÔ»ÖßÛú¸üÞÖ Ûúß ¯ÖÏ×ÛÎúµÖÖ Ûêú “Ö¸üÞÖÖë Ûúß ¾µÖÖÜµÖÖ Ûúß×•Ö‹ :-

CH3CH2OH H+

→443 K

CH2 = CH2 + H2O

Explain the mechanism of dehydration steps of ethanol :-

CH3CH2OH H+

→443 K

CH2 = CH2 + H2O

7. ×¾Ö»ÖµÖ®Ö Ûêú ¯Ö üÖÃÖ üÞÖß ¤üÖ²Ö ÛúÖê ¯Ö× ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ … ×¾Ö»ÖµÖ®Ö ´Öë ×¾Ö»ÖêµÖ Ûêú ÃÖÖÓ¦üÞÖ ÃÖê ¯Ö üÖÃÖ üÞÖß ¤üÖ²Ö ÛîúÃÖê ÃÖÓ²Ö×®¬ÖŸÖ Æîü ?

Define osmotic pressure of a solution. How is the osmotic pressure related to the

concentration of a solute in a solution ?

8. ×®Ö´®Ö×»Ö×ÜÖŸÖ ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ :

(i) †×³Ö×ÛÎúµÖÖ Ûúß †¬ÖÖÔµÖã (t½)

(ii) ¾ÖêÝÖ ×Ã£Ö¸üÖÓÛú (k)


(i) Half-life of a reaction (t½)

(ii) Rate constant (k)

9. ×®Ö´®ÖÖë Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ :

(i) H2SO4

(ii) XeF2

Draw the structures of the following :

(i) H2SO4

(ii) XeF2

10. ‘†ÃÖ´ÖÖ®ÖÖ®Öã ÖÖŸÖ®Ö’ ÛúÖ ŒµÖÖ ŸÖÖŸ¯ÖµÖÔ Æîü ? •Ö»ÖßµÖ ×¾Ö»ÖµÖ®Ö ´Öë †ÃÖ´ÖÖ®ÖÖ®Öã ÖÖŸÖ®Ö †×³Ö×ÛÎúµÖÖ ÛúÖ ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ …

†£Ö¾ÖÖ

ÃÖÓÛÎú´ÖÞÖ ¬ÖÖŸÖã ¸üÃÖÖµÖ®Ö Ûêú ×®Ö´®Ö »ÖõÖÞÖÖë Ûêú ×»ÖµÖê ÛúÖ¸üÞÖ ÃÖã—ÖÖ‡‹ :

(i) ÃÖÓÛÎú´ÖÞÖ ¬ÖÖŸÖã‹Ñ †Öî ü ˆ®ÖÛêú µÖÖî×ÝÖÛú ÃÖÖ´ÖÖ®µÖŸÖÖ †®Öã“Öã ²ÖÛúßµÖ ÆüÖêŸÖê Æïü …

(ii) ÃÖÓÛÎú´ÖÞÖ ¬ÖÖŸÖã‹Ñ ¯Ö×¸ü¾ÖŸÖÔ®Ö¿Öß»Ö ˆ¯Ö“ÖµÖ®Ö †¾ÖÃ£ÖÖ‹Ñ ¯ÖÏ¤üÙ¿ÖŸÖ Ûú¸üŸÖß Æïü …

What is meant by ‘disproportionation’ ? Give an example of a disproportionation

reaction in aqueous solution.

OR

Suggest reasons for the following features of transition metal chemistry :

(i) The transition metals and their compounds are usually paramagnetic.

(ii) The transition metals exhibit variable oxidation states.

56/1/3 4


(i) ¯ÖÏÖê Öß®Ö ÛúÖê ¯ÖÏÖê Öê®Ö-2-†Öò»Ö ´Öë …

(ii) ²Öê×®•ÖÌ»Ö Œ»ÖÖê üÖ‡›ü ÛúÖê ²Öê×®•ÖÌ»Ö ‹ê»ÛúÖêÆüÖò»Ö ´Öë …

(iii) ‹ê×®ÖÃÖÖê»Ö ÛúÖê p-²ÖÎÖê ÖÖê‹ê×®ÖÃÖÖê»Ö ´Öë …


(i) Propene to propane-2-ol

(ii) Benzyl chloride to Benzyl alcohol

(iii) Anisole to p-Bromoanisole

12. ‹Ûú ‹ê üÖê Öî×™üÛú µÖÖî×ÝÖÛú ‘A’ •Ö»ÖßµÖ †´ÖÖê×®ÖµÖÖ Ûêú ÃÖÖ£Ö ˆ¯Ö“ÖÖ×¸üŸÖ ÆüÖê®Öê †Öî ü ÝÖ´ÖÔ Ûú¸ü®Öê ¯Ö¸ü µÖÖî×ÝÖÛú ‘B’ ²Ö®ÖÖŸÖÖ Æîü

•ÖÖê Br2 †Öî ü KOH Ûêú ÃÖÖ£Ö ŸÖÖ×¯ÖŸÖ Ûú¸ü®Öê ¯Ö¸ü µÖÖî×ÝÖÛú ‘C’ ²Ö®ÖÖŸÖÖ Æîü … ‘C’ ÛúÖ †ÖÞÖ×¾ÖÛú ÃÖæ¡Ö C6H7N Æîü …

A, B †Öî ü C µÖÖî×ÝÖÛúÖë Ûêú †Ö‡Ô µÖæ ¯Öß ‹ ÃÖß (IUPAC) ®ÖÖ´ÖÖë ÛúÖê ×»Ö×ÜÖ‹ †Öî ü ˆ®ÖÛúß ÃÖÓ ü“Ö®ÖÖ‹Ñ †Ö êü×ÜÖŸÖ

Ûúß×•Ö‹ …

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms

compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of

molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B

and C.

13. ×¾Ö™üÖ×´Ö®Öë ÛîúÃÖê ¾ÖÝÖáÛéúŸÖ Ûúß •ÖÖŸÖß Æïü ? ¸üŒŸÖ Ûêú ÃÛÓú¤ü®Ö Ûêú •ÖÖê ×¾Ö™üÖ×´Ö®Ö ˆ¢Ö¸ü¤üÖµÖß ÆüÖêŸÖê Æïü ˆ®ÖÛêú ®ÖÖ´Ö ¤üß×•Ö‹ …

How are vitamins classified ? Name the vitamin responsible for the coagulation of

blood.

14. ×®Ö´®Ö ²ÖÆãü»ÖÛúÖë Ûêú ‹Ûú»ÖÛúÖë Ûêú ®ÖÖ´Ö †Öî ü ˆ®ÖÛúß ÃÖÓ ü“Ö®ÖÖ‹Ñ ×»Ö×ÜÖ‹ :

(i) ²Öæ®ÖÖ-S

(ii) ®Öß†Öê ÖÏß®Ö

(iii) ™êü°»ÖÖò®Ö

Write the names and structures of the monomers of the following polymers :

(i) Buna-S

(ii) Neoprene

(iii) Teflon

56/1/3 5 [P.T.O.

15. ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ :

(i) ¿ÖÖò™üÛúß ¤üÖêÂÖ

(ii) ±ÏëúÛêú»Ö ¤üÖêÂÖ

(iii) F-Ûëú¦ü

Define the following :

(i) Schottky defect

(ii) Frenkel defect

(iii) F-centre

16. ‹×£Ö»Öß®Ö Ý»ÖÖ‡ÛúÖê»Ö (C2H4O2) ÛúÖ 45 g •Ö»Ö Ûêú 600 g Ûêú ÃÖÖ£Ö ×´Ö»ÖÖµÖÖ ÝÖµÖÖ Æîü … ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ …

(i) ×Æü´ÖÖÓÛú ÛúÖ †¾Ö®Ö´Ö®Ö †Öî ü

(ii) ×¾Ö»ÖµÖ®Ö ÛúÖ ×Æü´ÖÖÓÛú

(×¤üµÖÖ ÝÖµÖÖ Æîü : Kf ÛúÖ ´ÖÖ®Ö ¯ÖÖ®Öß Ûêú ×»Ö‹ = 1.86 K kg mol–1)

45 g of ethylene glycol (C2H4O2) is mixed with 600 g of water. Calculate

(i) the freezing point depression and

(ii) the freezing point of the solution

(Given : Kf of water = 1.86 K kg mol–1)

17. 500 K †Öî ü 700 K ¯Ö¸ü ‹Ûú †×³Ö×ÛÎúµÖÖ ÛúÖ ¤ü¸ü ×Ã£Ö¸üÖÓÛú ÛÎú´Ö¿Ö: 0.02 s–1 †Öî ü 0.07 s–1 Æîü … ÃÖ×ÛÎúµÖÞÖ

‰ú•ÖÖÔ, Ea ÛúÖ ¯Ö×¸üÛú»Ö®Ö Ûúß×•Ö‹ … (R = 8.314 J K–1 mol–1)

The rate constants of a reaction at 500 K and 700 K are 0.02 s–1 and 0.07 s–1

respectively. Calculate the value of activation energy, Ea. (R = 8.314 J K–1 mol–1)

18. ×®Ö´®Ö ¯Ö¤üÖë ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ :

(i) ‡»ÖêŒ™ÒüÖê±úÖê êü×ÃÖÃÖ

(ii) †×¬Ö¿ÖÖêÂÖÞÖ

(iii) ¿Öê Ö-ÃÖê»Öê×Œ™ü¾Ö (†ÖÛéú×ŸÖ †Ö¬ÖÖ×¸üŸÖ) ˆŸ¯ÖÏê üÞÖ


(i) Electrophoresis

(ii) Adsorption

(iii) Shape selective catalysis

56/1/3 6

19. ×®Ö´®Ö ×¾Ö×¬ÖµÖÖë «üÖ¸üÖ ¬ÖÖŸÖã†Öë Ûêú ¯Ö×¸üÂÛú¸üÞÖ Ûêú †Ö¬ÖÖ¸ü ´Öæ»Ö ×ÃÖ¨üÖ®ŸÖ ×»Ö×ÜÖ‹ :

(i) †ÖÃÖ¾Ö®Ö

(ii) •ÖÖê®Ö ¯Ö×¸üÂÛú¸üÞÖ

(iii) ¾ÖîªãŸÖ †¯Ö‘Ö™ü®Ö

†£Ö¾ÖÖ

†ÖµÖ¸ü®Ö Ûêú ×®ÖÂÛúÂÖÔÞÖ Ûêú ÃÖ´ÖµÖ ²»ÖÖÃ™ü ±ú®ÖìÃÖ Ûêú ×¾Ö×³Ö®®Ö ³ÖÖÝÖÖë •ÖÖê †×³Ö×ÛÎúµÖÖ‹Ñ ÆüÖêŸÖß Æïü ˆ®Æëü ×»Ö×ÜÖ‹ … œü»Ö¾Öë

»ÖÖêÆêü ÃÖê Ûú““ÖÖ (Pig) »ÖÖêÆüÖ ÛîúÃÖê ×³Ö®®Ö ÆüÖêŸÖÖ Æîü ?

Outline the principles of refining of metals by the following methods :

(i) Distillation

(ii) Zone refining

(iii) Electrolysis

OR

Write down the reactions taking place in different zones in the blast furnace during the

extraction of iron. How is pig iron different from cast iron ?

20. »Öî®£Öê®ÖÖòµÖ›ü ÃÖÓÛãú“Ö®Ö ŒµÖÖ Æîü ? »Öî®£Öê®ÖÖòµÖ›ü ÃÖÓÛãú“Ö®Ö Ûêú ŒµÖÖ ¯Ö×¸üÞÖÖ´Ö ÆüÖêŸÖê Æïü ?

What is lanthanoid contraction ? What are the consequences of lanthanoid contraction ?

21. ×®Ö´®Ö ÛúÖò ¯»ÖêŒÃÖÖë «üÖ¸üÖ •ÖÖê ÃÖ´ÖÖ¾ÖµÖ¾ÖŸÖÖ Ûêú ¯ÖÏÛúÖ¸ü ¯ÖÏ¤üÙ¿ÖŸÖ ÆüÖêŸÖê Æïü ˆ®ÖÛúÖ ÃÖÓÛêúŸÖ Ûúß×•Ö‹ :

(i) [Co(NH3)5(NO2)]2+

(ii) [Co(en)3]Cl3 (en = ‹×£Ö»Öß®Ö ›üÖ‡‹ê Öß®Ö)

(iii) [Pt(NH3)2Cl2]

Indicate the types of isomerism exhibited by the following complexes :

(i) [Co(NH3)5(NO2)]2+

(ii) [Co(en)3]Cl3 (en = ethylene diamine)

(iii) [Pt(NH3)2Cl2]

56/1/3 7 [P.T.O.

22. ×®Ö´®Ö Ûêú †Ö‡Ô µÖæ ¯Öß ‹ ÃÖß (IUPAC) ®ÖÖ´Ö ¤üß×•Ö‹ :

(i) CH3 – CH –| OH

CH2 – CH3

(ii)

(iii) CH3

CH3|

– C –|

CH3

CH2 – Cl

Name the following according to IUPAC system :

(i) CH3 – CH –| OH

CH2 – CH3

(ii)

(iii) CH3

CH3|

– C –|

CH3

CH2 – Cl

23. ¸ü´Öê¿Ö ‹Ûú ×›ü¯ÖÖ™Ôü´Öê®™ü»Ö Ã™üÖê ü ´Öë ÝÖµÖÖ ¾ÖÆüÖÑ ˆÃÖê Ûãú”û ‘Ö¸ü Ûêú ×»ÖµÖê ÃÖÖ´ÖÖ®Ö ÜÖ¸üß¤ü®ÖÖ £ÖÖ … ‹Ûú ÜÖÖ®Öê ´Öë ˆÃÖ®Öê ¿ÖãÝÖ¸ü- ±Ïúß ×™ü×ÛúµÖÖÑ ¤êüÜÖß … ˆÃÖ®Öê ˆ®Æëü †¯Ö®Öê ¤üÖ¤üÖ Ûêú ×»ÖµÖê ÜÖ¸üß¤ü®Öê ÛúÖ ×®ÖÞÖÔµÖ ×ÛúµÖÖ •ÖÖê ¿ÖãÝÖ¸ü Ûêú ´Ö¸üß•Ö £Öê … ŸÖß®Ö ¯ÖÏÛúÖ¸ü Ûúß ¿ÖãÝÖ¸ü-±Ïúß ×™ü×ÛúµÖÖÑ ´ÖÖî•Öæ¤ü £Öà … ¸ü´Öê¿Ö ®Öê ÃÖãÛÎúÖê»ÖÖêÃÖ ÜÖ¸üß¤ü®Öê ÛúÖ ×®Ö¿“ÖµÖ ×ÛúµÖÖ •ÖÖê ˆÃÖÛêú ¤üÖ¤üÖ Ûêú Ã¾ÖÖÃ£µÖ Ûêú ×»ÖµÖê †“”ûß £Öà …

(i) ‹Ûú †®µÖ ¿ÖãÝÖ¸ü ±Ïúß ×™ü×ÛúµÖÖ ÛúÖ ˆ»»ÖêÜÖ Ûúß×•Ö‹ ×•ÖÃÖê ¸ü´Öê¿Ö ®Öê ®ÖÆüà ÜÖ¸üß¤üÖ …

(ii) ×²Ö®ÖÖ ›üÖòŒ™ü¸ü Ûúß ¯Ö“Öá Ûêú ‹êÃÖß ¤ü¾ÖÖ ÜÖ¸üß¤ü®ÖÖ ŒµÖÖ ¸ü´Öê¿Ö Ûêú ×»Ö‹ ˆ×“ÖŸÖ £ÖÖ ?

(iii) ˆ¯Ö¸üÖêŒŸÖ ×¾Ö¾ÖÞÖÔ ÃÖê ¸ü´Öê¿Ö ÛúÖ ÛúÖî®Ö ÃÖÖ ÝÖãÞÖ ¯ÖÏ¤üÙ¿ÖŸÖ ÆüÖêŸÖÖ Æîü ?

Ramesh went to a departmental store to purchase groceries. On one of shelves he

noticed sugar-free tablets. He decided to buy them for his grandfather who was a

diabetic. There were three types of sugar-free tablets. Ramesh decided to buy

sucrolose which was good for his grandfather’s health.

(i) Name another sugar free tablet which Ramesh did not buy.


(iii) What quality of Ramesh is reflected above ?

56/1/3 8

24. (a) ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë ÛúÖ ¸üÖÃÖÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖÖë ÛúÖê ¤êüŸÖê Æãü‹ ¾ÖÞÖÔ®Ö Ûúß×•Ö‹ :

(i) ›üßÛúÖ²ÖÖì×ŒÃÖ»ÖßÛú¸üÞÖ †×³Ö×ÛÎúµÖÖ

(ii) ±ÏúÖ‡›êü»Ö-ÛÎîú°™ü †×³Ö×ÛÎúµÖÖ

(b) †Ö¯Ö ×®Ö´®Ö ºþ¯ÖÖÓŸÖ¸üÞÖ ÛîúÃÖê Ûú¸ëüÝÖê ?

(i) ²Öê®•ÖÌÖê‡Ûú †´»Ö ÛúÖê ²Öê®•Öî×»›üÆüÖ‡›ü ´Öë

(ii) ²Öê®•ÖÌß®Ö ÛúÖê m-®ÖÖ‡™ÒüÖê‹ÃÖß™üÖ±úß®ÖÖê®Ö ´Öë

(iii) ‹£Öî®ÖÖò»Ö ÛúÖê 3-ÆüÖ‡›ÒüÖòŒÃÖß ²µÖæ™îü®Öî»Ö ´Öë

†£Ö¾ÖÖ

(a) ×®Ö´®Ö ×ÛÎúµÖÖ†Öë ÛúÖ ¾ÖÞÖÔ®Ö Ûúß×•Ö‹ :

(i) ‹ÃÖß×™ü»ÖßÛú¸üÞÖ

(ii) ‹ê»›üÖê»Ö ÃÖÓ‘Ö®Ö®Ö

(b) ×®Ö´®Ö×»Ö×ÜÖŸÖ †×³Ö×ÛÎúµÖÖ†Öë Ûêú ´ÖãÜµÖ ˆŸ¯ÖÖ¤ü ÛúÖê ×»Ö×ÜÖ‹ :

(i) CH3 – C –| |O

CH3 –––––––––––––→LiAlH

4 ?

(ii)

CHO

–––––––––––––––––––→HNO

3 / H

2SO

4

273 – 283 K

?

(iii) CH3 – COOH –––––––––––––→PCl

5 ?

(a) Describe the following giving chemical equations :

(i) De-carboxylation reaction

(ii) Friedel-Crafts reaction

(b) How will you bring about the following conversions ?

(i) Benzoic acid to Benzaldehyde

(ii) Benzene to m-Nitroacetophenone

(iii) Ethanol to 3-Hydroxybutanal

OR

56/1/3 9 [P.T.O.

(a) Describe the following actions :

(i) Acetylation (ii) Aldol condensation

(b) Write the main product in the following equations :

(i) CH3 – C –| |O

CH3 –––––––––––––→LiAlH

4 ?

(ii)

CHO

–––––––––––––––––––→HNO

3 / H

2SO

4

273 – 283 K

?

(iii) CH3 – COOH –––––––––––––→PCl

5 ?

25. (a) ×®Ö´®Ö ¸üÖÃÖÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖÖë ÛúÖê ¯ÖæÞÖÔ Ûúß×•Ö‹ :

(i) Cu + HNO3(ŸÖ®Öã) →

(ii) P4 + NaOH+ H2O →

(b) (i) ŒµÖÖë R3P = O ²Ö®ÖŸÖÖ Æîü ¯Ö¸ü®ŸÖã R3N = O ®ÖÆüà ²Ö®ÖŸÖÖ Æîü ? (R = ‹×»Ûú»Ö ÝÖÏã Ö)

(ii) ›üÖ‡†ÖòŒÃÖß•Ö®Ö ŒµÖÖë ‹Ûú ÝÖîÃÖ Æîü ¯Ö¸ü®ŸÖã ÃÖ»±ú¸ü ‹Ûú šüÖêÃÖ Æîü ?

(iii) Æîü»ÖÖê•Ö®Ö ŒµÖÖë ÓüÝÖµÖãŒŸÖ ÆüÖêŸÖê Æïü ?

†£Ö¾ÖÖ

(a) ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë Ûêú ×»Ö‹ ÃÖÓŸÖã×»ÖŸÖ ¸üÖÃÖÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖ ×»Ö×ÜÖ‹ :

(i) ²Öã—Öê “Öæ®Öê Ûêú ÃÖÖ£Ö Œ»ÖÖê üß®Ö †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖß Æîü …

(ii) ÛúÖ²ÖÔ®Ö ÃÖÖÓ¦ü H2SO4 ÃÖê †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖÖ Æîü …

(b) ÃÖ»°µÖæ×¸üÛú †´»Ö ÛúÖê ÛúÖÓ™îüŒ™ü ×¾Ö×¬Ö ÃÖê ²Ö®ÖÖ®Öê ÛúÖ ×®Ö´®Ö ÃÖÓ¤ü³ÖÖí, •ÖîÃÖê †×¬ÖÛúŸÖ´Ö ˆŸ¯ÖÖ¤ü, ˆŸ¯ÖÏê üÞÖ †Öî ü †®µÖ ×Ã£Ö×ŸÖ ´Öë ¾ÖÞÖÔ®Ö Ûúß×•Ö‹ …

(a) Complete the following chemical reaction equations :

(i) Cu + HNO3(dilute) →


56/1/3 10

(b) (i) Why does R3P = O exist but R3N = O does not ? (R = alkyl group)

(ii) Why is dioxygen a gas but sulphur a solid ?

(iii) Why are halogens coloured ?

OR

(a) Write balanced equations for the following reactions :

(i) Chlorine reacts with dry slaked lime.

(ii) Carbon reacts with concentrated H2SO4.

(b) Describe the contact process for the manufacture of sulphuric acid with special

reference to the reaction conditions, catalysts used and the yield in the process.

26. (a) ×®Ö´®Ö×»Ö×ÜÖŸÖ ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ :

(i) ´ÖÖê»Ö¸ü “ÖÖ»ÖÛúŸÖÖ (^m)

(ii) ÃÖÓ“ÖÖµÖÛú ²Öî™ü×¸üµÖÖÑ

(iii) ‡Õ¬Ö®Ö ÃÖê»Ö

(b) ×®Ö´®Ö×»Ö×ÜÖŸÖ ×®ÖµÖ´ÖÖë ÛúÖê ×»Ö×ÜÖ‹ :

(i) ±îú¸üÖ›êü Ûêú ¾ÖîªãŸÖ†¯Ö‘Ö™ü®Ö ÛúÖ ¯ÖÏ£Ö´Ö ×®ÖµÖ´Ö

(ii) ÛúÖê»Ö¸üÖ‰ú¿Ö Ûêú †ÖµÖ®ÖÖë Ûêú Ã¾ÖŸÖÓ¡Ö †×³ÖÝÖ´Ö®Ö ÛúÖ ×®ÖµÖ´Ö

†£Ö¾ÖÖ

(a) ×¾ÖµÖÖê•Ö®Ö Ûúß ×›üÝÖÏß ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ … ‹Ûú ¾µÖÓ•ÖÛú ×»Ö×ÜÖ‹ •ÖÖê ¤ãü²ÖÔ»Ö ×¾ÖªãŸÖË-†¯Ö‘Ö™ü¶ Ûúß ´ÖÖê»Ö¸ü

“ÖÖ»ÖÛúŸÖÖ ÛúÖê ‡ÃÖÛêú ×¾ÖµÖÖê•Ö®Ö Ûúß ×›üÝÖÏß ÃÖê ÃÖÓ²Ö×®¬ÖŸÖ ÆüÖêŸÖÖ Æîü …

(b) ÃÖê»Ö †×³Ö×ÛÎúµÖÖ

Ni(s) | Ni2+(aq) || Ag+

(aq) | Ag(s)

Ûêú ×»ÖµÖê 25 °C ¯Ö¸ü ŸÖã»µÖ ×Ã£Ö¸üÖÓÛú ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ … ‡ÃÖ ÃÖê»Ö Ûêú ÛúÖ´Ö Ûú¸ü®Öê ¯Ö¸ü †×¬ÖÛúŸÖ´Ö ×ÛúŸÖ®ÖÖ

ÛúÖµÖÔ ¯ÖÏÖ¯ŸÖ ÆüÖêŸÖÖ Æîü ?

E°Ni2+/Ni

= 0.25 V, E°Ag+/Ag

= 0.80 V.

56/1/3 11 [P.T.O.

(a) Define the following terms :

(i) Molar conductivity (^m)

(ii) Secondary batteries

(iii) Fuel cell

(b) State the following laws :

(i) Faraday first law of electrolysis

(ii) Kohlrausch’s law of independent migration of ions

OR

(a) Define the term degree of dissociation. Write an expression that relates the molar

conductivity of a weak electrolyte to its degree of dissociation.

(b) For the cell reaction

Ni(s) | Ni2+(aq) || Ag+

(aq) | Ag(s)

Calculate the equilibrium constant at 25 °C. How much maximum work would

be obtained by operation of this cell ?

E°Ni2+/Ni

= 0.25 V and E°Ag+/Ag

= 0.80 V.

___________

56/1/3 12

1

Qu


1 Hexaamninenickel (II) chloride 1

2

1

3

(where Ar is C6H5)

1

4 2 1

5 It is a process of removing a dissolved substance from a colloidal solution by means of diffusion

through a suitable membrane.

1

6

½

½

1

7 The external pressure which is applied on solution side to stop the flow of solvent across the

semi-permeable membrane.

The osmotic pressure is directly proportional to concentration of the solution. / = CRT

1

1

8 The half-life of a reaction is the time in which the concentration of a reactant is reduced to one-

half of its initial concentration.

Rate constant is the rate of reaction when the concentration of the reactant is unity.

1

1


SET -56/1/3

Compt. July, 2015

2

9

i) ii)

1+1



2Cu+ (aq) Cu

2+ (aq) + Cu(s)


OR

1

1

10 i) Due to presence of unpaired electrons in d-orbitals.

ii) Due to incomplete filling of d-orbitals.

1

1

11 i)

ii)

iii)

1

1

1

12

A – Benzoic acid

B – Benzamide

½ +

½

½ +

½

3

C - Aniline

½ +

½

13 Fat soluble vitamin- Vitamin A, D

Water soluble vitamin-Vitamin B,C

Vitamin K

½+½

½+½

1

14 i)

ii)

iii)

½ +

½

½ +

½

½ +

½

15 i) The defect in which equal number of cations and anions are missing from the lattice.

ii) Due to dislocation of smaller ion from its normal site to an interstitial site.

iii) Anionic vacancies are occupied by unpaired electron.

1

1

1

16 i) ∆Tf = Kf m

∆Tf = Kf

∆Tf =

∆Tf =2.325K or 2.3250

C

ii) Tf0- Tf = 2.325

0 C

O0C - Tf = 2.325

0 C

Tf = - 2.3250

C or 270.675 K

½

½

1

1

17

1

1

1

18 i) The movement of colloidal particles under an applied electric potential towards oppositely

charged electrode is called electrophoresis.

ii) The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid

1

1

4

is termed adsorption.

iii) The catalytic reaction that depends upon the pore structure of the catalyst and the size of the

reactant and product molecules is called shape-selective catalysis.

1

19 i) The impure metal is evaporated to obtain the pure metal as distillate.

ii) This method is based on the principle that the impurities are more soluble in the melt than in

the solid state of the metal.

iii) The impure metal is made to act as anode. A strip of the same metal in pure form is used as

cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal.

The more basic metal remains in the solution and the less basic ones go to the anode mud.

OR

1

1

1

19

( any four correct equations) Cast iron has lower carbon content (about 3%) than pig iron / cast iron is hard & brittle whereas pig iron is soft.

½ x 4

= 2

1

20 The steady decrease in atomic radii from La to Lu due to imperfect shielding of 4f – orbital.

Consequences –

i) Members of third transition series have almost identical radii as coresponding members

of second transition series.

ii) Difficulty in separation.

1

1+1

21 a) Linkage isomerism

b) Optical isomerism

c) Cis - trans / Geometrical isomerism

1

1

1

22 a) Butan – 2 – ol

b) 2 – bromotoluene

c) 2, 2-dimethylchlorpropane

1

1

1


ii) No


1

1

2

24 a) i) Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are

heated with sodalime (NaOH and CaO).

ii) When the alkyl / acyl group is introduced at ortho and para positions by reaction

with alkyl halide / acyl halide in the presence of anhydrous aluminium chloride (a Lewis

acid) as catalyst.

1

5

(Note : Award full marks if correct equation is given )

b) i)

ii)

iii)

(or any other correct method)

OR

1

1

1

1

24 a) i) When the acyl groups are introduced at ortho and para positions by reaction with acyl halide in the

presence of anhydrous aluminium chloride (a Lewis acid) as catalyst.

ii) Aldehydes and ketones having at least one -hydrogen undergo a reaction in the presence of

dilute alkali as catalyst to form -hydroxy aldehydes (aldol) or hydroxy ketones (ketol),

respectively.


b)i)

1

1

1

6

ii)

iii) CH3COCl

1

1

25 a) i)

ii)

b) i) Due to absence of d-orbital, nitrogen cannot expand its valency beyond four.

ii) Because of pπ – pπ multiple bonding in dioxygen which is absent in sulphur.

iii) Due to excitation of electron by absorption of radiation from visible region. OR

1

1

1

1

1

25 a) i)

ii)

b) It is manufactured by Contact Process which involves following steps:

i) burning of sulphur or sulphide ores in air to generate SO2.

ii) conversion of SO2 to SO3 by the reaction with oxygen in the presence of a catalyst (V2O5)

iii) absorption of SO3 in H2SO4 to give Oleum (H2S2O7). The oleum obtained is diluted to give

sulphuric acid

Reaction condition – pressure of 2 bar and temperature of 720 K

Catalyst used is V2O5

Yield – 96 – 98% pure

1

1

1

1

1

26 a)i)Molar conductivity of a solution at a given concentration is the conductance of the volume V

of solution containing one mole of electrolyte kept between two electrodes with area of cross

section A and distance of unit length.

ii) Secondary battery- can be recharged by passing current through it in opposite direction so that

it can be used again.

iii) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen,

methane, methanol, etc. directly into electrical energy are called fuel cells.

b)i) The amount of chemical reaction which occurs at any electrode during electrolysis by a

current is proportional to the quantity of electricity passed through the electrolyte (solution or

melt).

ii) Limiting molar conductivity of an electrolyte can be represented as the sum of the individual

contributions of the anion and cation of the electrolyte.

OR

1

1

1

1

1

26 a) Degree of dissociation is the extent to which electrolyte gets dissociated into its constituent

ions.

b) E

0cell = E

0Ag+ / Ag - E

0Ni2+ / Ni

= 0.80V – 0.25V

1

1

7


Ms. Garima Bhutani

= 0.55V

1og Kc =

=

log Kc = 18.644

∆G0 = - nFE

0cell

= -2x96500 Cmol-1

x 0.55V

= -106,150 Jmol-1

Max.work =+106150 Jmol-1

or 106.150k Jmol-1

½

½

½

½

1

56/2 1 [P.T.O.


Roll No.












Series : SSO/C 56/2














SET – 2

56/2 2














CH3 – CH2 – CHO


CH3 – CH2 – CHO







56/2 3 [P.T.O.









†£Ö¾ÖÖ





reactions.

OR



8. ‹£Öî®ÖÖò»Ö ÃÖê ‹£Öß®Ö ²Ö®Ö®Öê ´Öë †´»Ö ×®Ö•ÖÔ»ÖßÛú¸üÞÖ Ûúß ×ÛÎúµÖÖ×¾Ö×¬Ö ×»Ö×ÜÖ‹ …








56/2 4





11. ×®Ö´®Ö Ûêú ˆ¢Ö¸ü ¤üß×•Ö‹ :




†£Ö¾ÖÖ






OR






(i) [Co(NH3)6]Cl3

(ii) [NiCl4]2–

(iii) K3[Fe(CN)6]


(i) [Co(NH3)6]Cl3

(ii) [NiCl4]2–

(iii) K3[Fe(CN)6]

56/2 5 [P.T.O.


(i) CH3 – CH –| Br

CH2 – CH3

(ii)

Br

Br



(i) CH3 – CH –| Br

CH2 – CH3

(ii)

Br

Br










56/2 6


(i) CH3 – CH2OH –––––––––→PCl

5 ?

(ii)

OH

+ CH3– Cl ––––––––––––––––––→anhyd. AlCl

3 ?



(i) CH3 – CH2OH –––––––––→PCl

5 ?

(ii)

OH

+ CH3– Cl ––––––––––––––––––→anhyd. AlCl

3 ?







(i) Peptide linkage


(iii) Denaturation



56/2 7 [P.T.O.













value ?






(i) Dialysis



56/2 8





















†£Ö¾ÖÖ

56/2 9 [P.T.O.


(i) BrF3

(ii) XeO3











OR


(i) BrF3

(ii) XeO3





56/2 10



(ii) 4-´Öê×£Ö»Ö¯Öî®™ü-3-‡Ô®Ö-2-†Öò®Ö





†£Ö¾ÖÖ





(i) CH3 – CH3

(ii) CH3 – CH –|

OH

CH2 – CHO

(iii) CH3CH2OH








OR

56/2 11 [P.T.O.


(i) Propanone oxime



equations involved.

(i) CH3 – CH3

(ii) CH3 – CH –|

OH

CH2 – CHO

(iii) CH3CH2OH


Zn(s) | Zn2+(aq) || Sn2+(aq) | Sn(s)


2+/Zn

= – 0.76 V; E°Sn

2+/Sn

= – 0.14 V

†Öî ü F = 96500 C mol–1

†£Ö¾ÖÖ






(E°Ag

+/Ag

= 0.80 V; E°Fe

3+/Fe

2+ = 0.77 V)

56/2 12



Zn(s) | Zn2+(aq) || Sn2+(aq) | Sn(s)

Given : E°Zn

2+/Zn

= – 0.76 V; E°Sn

2+/Sn

= – 0.14 V

and F = 96500 C mol–1.

OR







(E°Ag

+/Ag

= 0.80 V; E°Fe

3+/Fe

2+ = 0.77 V)

____________

1

Qu




½ + ½


3 Propanal 1


5 Frenkel defect 1



between N and F

1

1





OR

1

1

7 i)

ii)

1

1

8

½

½

1


SET -56/2

Compt. July, 2015

2

9 i)


solvent. Or

1

1


s-1


s-1

1

1

11 i) It lowers the melting point of alumina / acts as a solvent.

ii)



absence of air.

(Or with equation)



1

1

1

11


6 x ½

= 3



2Cu+ (aq) Cu

2+ (aq) + Cu(s)


1 ½

1 ½




1

1

1

3

14 i) 2-bromobutane



1

1

1

15

i)

ii)

1

1

1

16

i)

ii)

iii)

1

1

1







1

1

1



1

1

4


1

19 r =

√

r=

r = 1.44 x cm

1

1

1



=

=

MX =

gmol

-1

MX = 59.9 or 60 gmol-1

1

1

1

21 k=

log

60 s-1

=

log

t=

log 10

t=

s

t= 0.0384 s

1

1

1

22 i) It is a process of removing the dissolved substance from a colloidal solution by means






1

1

1 23 i) Aspartame, Saccharin (any one)

ii) No


1

1

2 24 a) Due to relatively stable half – filled p-orbitals of group 15 elements


ii)

iii)

OR

2

1

1

1

5

24 a) i)

ii)




iii)

1

1

1

1

1

25 a) i)

ii)


will not.



iii)Add neutral FeCl3, phenol gives violet colouration whereas benzoic acid does not.

1

1

1

1

6

(or any other correct test)

OR 1

25 a) i)

ii)

b) i)

ii)

iii)

1

1

1

1

1

26 E0cell = E

0Sn2+ / Sn - E

0Zn2+ / Zn

= - 0.14V –(- 0.76V)

= 0.62V

∆rG0 = -n F E

0cell

= - 2 x 96500 C mol-1

x 0.62 V

= - 119660 J mol-1

Ecell = E0

cell -

log

Ecell = 0.62 -

log

OR

1

1

1

1

1








½

½

1

7


Ms. Garima Bhutani

b)E0cell = E

0C - E

0A

= 0.80V – 0.77V

= 0.03V

∆rG0 = -n F E

0cell

= - 1 x 96500 C mol-1

x 0.03 V

= - 2895 J mol-1

Log Kc=

Log Kc=

Log Kc= 0.508

½

½

1

½

½

56/1/2 1 [P.T.O.


Roll No.


























SET – 2

56/1/2 2

















C6H5N2Cl + H3PO2 + H2O ––––→ - - -


C6H5N2Cl + H3PO2 + H2O ––––→ - - -



56/1/2 3 [P.T.O.


†£Ö¾ÖÖ






OR





CH3CH2OH H+

→443 K

CH2 = CH2 + H2O


CH3CH2OH H+

→443 K

CH2 = CH2 + H2O










56/1/2 4


(i) H2SO4

(ii) XeF2


(i) H2SO4

(ii) XeF2


(i) [Co(NH3)5(NO2)]2+


(iii) [Pt(NH3)2Cl2]


(i) [Co(NH3)5(NO2)]2+


(iii) [Pt(NH3)2Cl2]


(i) CH3 – CH –| OH

CH2 – CH3

(ii)

(iii) CH3

CH3|

– C –|

CH3

CH2 – Cl

56/1/2 5 [P.T.O.


(i) CH3 – CH –| OH

CH2 – CH3

(ii)

(iii) CH3

CH3|

– C –|

CH3

CH2 – Cl
















and C.

56/1/2 6



blood.


(i) ²Öæ®ÖÖ-S




(i) Buna-S

(ii) Neoprene

(iii) Teflon




(iii) F-Ûëú¦ü


(i) Schottky defect

(ii) Frenkel defect

(iii) F-centre









56/1/2 7 [P.T.O.










(i) Electrophoresis

(ii) Adsorption






†£Ö¾ÖÖ




(i) Distillation

(ii) Zone refining

(iii) Electrolysis

OR



56/1/2 8



23. ¸ü´Öê¿Ö ‹Ûú ×›ü¯ÖÖ™Ôü´Öê®™ü»Ö Ã™üÖê ü ´Öë ÝÖµÖÖ ¾ÖÆüÖÑ ˆÃÖê Ûãú”û ‘Ö¸ü Ûêú ×»ÖµÖê ÃÖÖ´ÖÖ®Ö ÜÖ¸üß¤ü®ÖÖ £ÖÖ … ‹Ûú ÜÖÖ®Öê ´Öë ˆÃÖ®Öê ¿ÖãÝÖ¸ü-

±Ïúß ×™ü×ÛúµÖÖÑ ¤êüÜÖß … ˆÃÖ®Öê ˆ®Æëü †¯Ö®Öê ¤üÖ¤üÖ Ûêú ×»ÖµÖê ÜÖ¸üß¤ü®Öê ÛúÖ ×®ÖÞÖÔµÖ ×ÛúµÖÖ •ÖÖê ¿ÖãÝÖ¸ü Ûêú ´Ö¸üß•Ö £Öê … ŸÖß®Ö ¯ÖÏÛúÖ¸ü

Ûúß ¿ÖãÝÖ¸ü-±Ïúß ×™ü×ÛúµÖÖÑ ´ÖÖî•Öæ¤ü £Öà … ¸ü´Öê¿Ö ®Öê ÃÖãÛÎúÖê»ÖÖêÃÖ ÜÖ¸üß¤ü®Öê ÛúÖ ×®Ö¿“ÖµÖ ×ÛúµÖÖ •ÖÖê ˆÃÖÛêú ¤üÖ¤üÖ Ûêú Ã¾ÖÖÃ£µÖ Ûêú

×»ÖµÖê †“”ûß £Öà …

















†£Ö¾ÖÖ

56/1/2 9 [P.T.O.

(a) ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë Ûêú ×»Ö‹ ÃÖÓŸÖã×»ÖŸÖ ¸üÖÃÖÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖ ×»Ö×ÜÖ‹ : (i) ²Öã—Öê “Öæ®Öê Ûêú ÃÖÖ£Ö Œ»ÖÖê üß®Ö †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖß Æîü … (ii) ÛúÖ²ÖÔ®Ö ÃÖÖÓ¦ü H2SO4 ÃÖê †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖÖ Æîü …








OR













†£Ö¾ÖÖ

56/1/2 10





(i) CH3 – C –| |O

CH3 –––––––––––––→LiAlH

4 ?

(ii)

CHO

–––––––––––––––––––→HNO

3 / H

2SO

4

273 – 283 K

?

(iii) CH3 – COOH –––––––––––––→PCl

5 ?








OR




(i) CH3 – C –| |O

CH3 –––––––––––––→LiAlH

4 ?

(ii)

CHO

–––––––––––––––––––→HNO

3 / H

2SO

4

273 – 283 K

?

(iii) CH3 – COOH –––––––––––––→PCl

5 ?

56/1/2 11 [P.T.O.








†£Ö¾ÖÖ




Ni(s) | Ni2+(aq) || Ag+

(aq) | Ag(s)



E°Ni2+/Ni

= 0.25 V, E°Ag+/Ag

= 0.80 V.




(iii) Fuel cell




OR




Ni(s) | Ni2+(aq) || Ag+

(aq) | Ag(s)



E°Ni2+/Ni


= 0.80 V.

___________

56/1/2 12

1

Qu




1


3

1

4

(where Ar is C6H5)

1

5 2 1



2Cu+ (aq) Cu

2+ (aq) + Cu(s)


OR

1

1



1

1

7

½

½

1

8 The external pressure which is applied on solution side to stop the flow of solvent across the



1

1


SET -56/1/2

Compt. July, 2015

2

9 The half-life of a reaction is the time in which the concentration of a reactant is reduced to one-



1

1

10

i) ii)

1+1




1

1

1




1

1

1

13 i)

ii)

iii)

1

1

1

14

A – Benzoic acid

B – Benzamide

C - Aniline

½ +

½

½ +

½

3

½ +

½



Vitamin K

½+½

½+½

1

16 i)

ii)

iii)

½ +

½

½ +

½

½ +

½




1

1

1

18 i) ∆Tf = Kf m

∆Tf = Kf

∆Tf =

∆Tf =2.325K or 2.3250

C

ii) Tf0- Tf = 2.325

0 C

O0C - Tf = 2.325

0 C

Tf = - 2.3250

C or 270.675 K

½

½

1

1

19

1

1

1







1

1

1

4







OR

1

1

1


½ x 4

= 2

1


Consequences –




1

1+1


ii) No


1

1

2

24 a) i)

ii)




1

1

1

1

1

a) i)

ii)





sulphuric acid



1

1

1

1

1

5






acid) as catalyst.


b) i)

ii)

iii)


OR

1

1

1

1

1





respectively.

1

1

6


b)i)

ii)

iii) CH3COCl

1

1

1










melt).



OR

1

1

1

1

1


ions.

b) E

0cell = E

0Ag+ / Ag - E

0Ni2+ / Ni

= 0.80V – 0.25V

= 0.55V

1og Kc =

=

log Kc = 18.644

∆G0 = - nFE

0cell

= -2x96500 Cmol-1

x 0.55V

= -106,150 Jmol-1


or 106.150k Jmol-1

1

1

½

½

½

½

1

7


Ms. Garima Bhutani

56/1 1 [P.T.O.


Roll No.












Series : SSO/C 56/1














SET – 1

56/1 2
















CH3 – CH2 – CHO


CH3 – CH2 – CHO





56/1 3 [P.T.O.



















†£Ö¾ÖÖ



56/1 4



reactions.

OR



10. ‹£Öî®ÖÖò»Ö ÃÖê ‹£Öß®Ö ²Ö®Ö®Öê ´Öë †´»Ö ×®Ö•ÖÔ»ÖßÛú¸üÞÖ Ûúß ×ÛÎúµÖÖ×¾Ö×¬Ö ×»Ö×ÜÖ‹ …














value ?

56/1 5 [P.T.O.






(i) Dialysis



15. ×®Ö´®Ö Ûêú ˆ¢Ö¸ü ¤üß×•Ö‹ :




†£Ö¾ÖÖ






OR





56/1 6


(i) [Co(NH3)6]Cl3

(ii) [NiCl4]2–

(iii) K3[Fe(CN)6]


(i) [Co(NH3)6]Cl3

(ii) [NiCl4]2–

(iii) K3[Fe(CN)6]


(i) CH3 – CH –| Br

CH2 – CH3

(ii)

Br

Br



(i) CH3 – CH –| Br

CH2 – CH3

(ii)

Br

Br


56/1 7 [P.T.O.










(i) CH3 – CH2OH –––––––––→PCl

5 ?

(ii)

OH

+ CH3– Cl ––––––––––––––––––→anhyd. AlCl

3 ?



(i) CH3 – CH2OH –––––––––→PCl

5 ?

(ii)

OH

+ CH3– Cl ––––––––––––––––––→anhyd. AlCl

3 ?







(i) Peptide linkage


(iii) Denaturation

56/1 8


















Zn(s) | Zn2+(aq) || Sn2+(aq) | Sn(s)


2+/Zn

= – 0.76 V; E°Sn

2+/Sn

= – 0.14 V

†Öî ü F = 96500 C mol–1

†£Ö¾ÖÖ

56/1 9 [P.T.O.






(E°Ag

+/Ag

= 0.80 V; E°Fe

3+/Fe

2+ = 0.77 V)



Zn(s) | Zn2+(aq) || Sn2+(aq) | Sn(s)

Given : E°Zn

2+/Zn

= – 0.76 V; E°Sn

2+/Sn

= – 0.14 V

and F = 96500 C mol–1.

OR







(E°Ag

+/Ag

= 0.80 V; E°Fe

3+/Fe

2+ = 0.77 V)

56/1 10







†£Ö¾ÖÖ


(i) BrF3

(ii) XeO3











OR

56/1 11 [P.T.O.


(i) BrF3

(ii) XeO3







(ii) 4-´Öê×£Ö»Ö¯Öî®™ü-3-‡Ô®Ö-2-†Öò®Ö





†£Ö¾ÖÖ





(i) CH3 – CH3

(ii) CH3 – CH –|

OH

CH2 – CHO

(iii) CH3CH2OH

56/1 12








OR


(i) Propanone oxime



equations involved.

(i) CH3 – CH3

(ii) CH3 – CH –|

OH

CH2 – CHO

(iii) CH3CH2OH

_________

1

Qu


1 Frenkel defect 1



½ + ½


4 Propanal 1


6 i)


solvent. Or

1

1


s-1


s-1

1

1



between N and F

1

1





OR

1

1

9 i)

ii)

1

1


SET -56/1

Compt. July, 2015

2

10

½

½

1

11 r =

√

r=

r = 1.44 x cm

1

1

1



=

=

MX =

gmol

-1

MX = 59.9 or 60 gmol-1

1

1

1

13 k=

log

60 s-1

=

log

t=

log 10

t=

s

t= 0.0384 s

1

1

1

14 i) It is a process of removing the dissolved substance from a colloidal solution by means




1

1

3



1 15 i) It lowers the melting point of alumina / acts as a solvent.

ii)



absence of air.

(Or with equation)



1

1

1

15


6 x ½

= 3

16

Disproportionation : The reaction in which an element undergoes self-oxidation and self-


2Cu+ (aq) Cu

2+ (aq) + Cu(s)


1 ½

1 ½




1

1

1

18 i) 2-bromobutane



1

1

1

19

i)

ii)

1

1

4

1

20

i)

ii)

iii)

1

1

1







1

1

1




1

1

1


ii) No


1

1

2

24 E0cell = E

0Sn2+ / Sn - E

0Zn2+ / Zn

= - 0.14V –(- 0.76V)

= 0.62V

∆rG0 = -n F E

0cell

= - 2 x 96500 C mol-1

x 0.62 V

= - 119660 J mol-1

1

1

1

1

5

Ecell = E0

cell -

log

Ecell = 0.62 -

log

OR

1








b)E0cell = E

0C - E

0A

= 0.80V – 0.77V

= 0.03V

∆rG0 = -n F E

0cell

= - 1 x 96500 C mol-1

x 0.03 V

= - 2895 J mol-1

Log Kc=

Log Kc=

Log Kc= 0.508

½

½

1

½

½

1

½

½

25 a) Due to relatively stable half – filled p-orbitals of group 15 elements


ii)

iii)

OR

2

1

1

1

25 a) i)

ii)


1

1

1

6



iii)

1

1

26 a) i)

ii)


will not.



iii)Add neutral FeCl3, phenol gives violet colouration whereas benzoic acid does not. (or any other correct test)

OR

1

1

1

1

1

26 a) i)

ii)

b) i)

ii)

1

1

1

1

7


Ms. Garima Bhutani

iii)

1

56/1/1 1 [P.T.O.


Roll No.


























SET – 1

56/1/1 2



















C6H5N2Cl + H3PO2 + H2O ––––→ - - -


C6H5N2Cl + H3PO2 + H2O ––––→ - - -

56/1/1 3 [P.T.O.











(i) H2SO4

(ii) XeF2


(i) H2SO4

(ii) XeF2


†£Ö¾ÖÖ






OR




56/1/1 4


CH3CH2OH H+

→443 K

CH2 = CH2 + H2O


CH3CH2OH H+

→443 K

CH2 = CH2 + H2O




(iii) F-Ûëú¦ü


(i) Schottky defect

(ii) Frenkel defect

(iii) F-centre













56/1/1 5 [P.T.O.






(i) Electrophoresis

(ii) Adsorption






†£Ö¾ÖÖ




(i) Distillation

(ii) Zone refining

(iii) Electrolysis

OR





56/1/1 7 [P.T.O.
















and C.



blood.


(i) ²Öæ®ÖÖ-S




(i) Buna-S

(ii) Neoprene

(iii) Teflon

56/1/1 8

23. ¸ü´Öê¿Ö ‹Ûú ×›ü¯ÖÖ™Ôü´Öê®™ü»Ö Ã™üÖê ü ´Öë ÝÖµÖÖ ¾ÖÆüÖÑ ˆÃÖê Ûãú”û ‘Ö¸ü Ûêú ×»ÖµÖê ÃÖÖ´ÖÖ®Ö ÜÖ¸üß¤ü®ÖÖ £ÖÖ … ‹Ûú ÜÖÖ®Öê ´Öë ˆÃÖ®Öê ¿ÖãÝÖ¸ü-

±Ïúß ×™ü×ÛúµÖÖÑ ¤êüÜÖß … ˆÃÖ®Öê ˆ®Æëü †¯Ö®Öê ¤üÖ¤üÖ Ûêú ×»ÖµÖê ÜÖ¸üß¤ü®Öê ÛúÖ ×®ÖÞÖÔµÖ ×ÛúµÖÖ •ÖÖê ¿ÖãÝÖ¸ü Ûêú ´Ö¸üß•Ö £Öê … ŸÖß®Ö ¯ÖÏÛúÖ¸ü

Ûúß ¿ÖãÝÖ¸ü-±Ïúß ×™ü×ÛúµÖÖÑ ´ÖÖî•Öæ¤ü £Öà … ¸ü´Öê¿Ö ®Öê ÃÖãÛÎúÖê»ÖÖêÃÖ ÜÖ¸üß¤ü®Öê ÛúÖ ×®Ö¿“ÖµÖ ×ÛúµÖÖ •ÖÖê ˆÃÖÛêú ¤üÖ¤üÖ Ûêú Ã¾ÖÖÃ£µÖ Ûêú

×»ÖµÖê †“”ûß £Öà …


















†£Ö¾ÖÖ




Ni(s) | Ni2+(aq) || Ag+

(aq) | Ag(s)



E°Ni2+/Ni

= 0.25 V, E°Ag+/Ag

= 0.80 V.

56/1/1 9 [P.T.O.




(iii) Fuel cell




OR




Ni(s) | Ni2+(aq) || Ag+

(aq) | Ag(s)



E°Ni2+/Ni


= 0.80 V.







†£Ö¾ÖÖ

(a) ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë Ûêú ×»Ö‹ ÃÖÓŸÖã×»ÖŸÖ ¸üÖÃÖÖµÖ×®ÖÛú ÃÖ´ÖßÛú¸üÞÖ ×»Ö×ÜÖ‹ :

(i) ²Öã—Öê “Öæ®Öê Ûêú ÃÖÖ£Ö Œ»ÖÖê üß®Ö †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖß Æîü …

(ii) ÛúÖ²ÖÔ®Ö ÃÖÖÓ¦ü H2SO4 ÃÖê †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖÖ Æîü …


56/1/1 10







OR













†£Ö¾ÖÖ




56/1/1 11 [P.T.O.


(i) CH3 – C –| |O

CH3 –––––––––––––→LiAlH

4 ?

(ii)

CHO

–––––––––––––––––––→HNO

3 / H

2SO

4

273 – 283 K

?

(iii) CH3 – COOH –––––––––––––→PCl

5 ?








OR




(i) CH3 – C –| |O

CH3 –––––––––––––→LiAlH

4 ?

(ii)

CHO

–––––––––––––––––––→HNO

3 / H

2SO

4

273 – 283 K

?

(iii) CH3 – COOH –––––––––––––→PCl

5 ?

___________

56/1/1 12

1

Qu


1 2 1



1


4

1

5

(where Ar is C6H5)

1

6. The external pressure which is applied on solution side to stop the flow of solvent across the



1

1

7. The half-life of a reaction is the time in which the concentration of a reactant is reduced to one-



1

1

8.

i) ii)

1+1

9

Disproportionation : The reaction in which an element undergoes self-oxidation and self-


2Cu+ (aq) Cu

2+ (aq) + Cu(s)


OR

1

1



1

1

10

½


SET -56/1/1

2

½

1




1

1

1

12 i) ∆Tf = Kf m

∆Tf = Kf

∆Tf =

∆Tf =2.325K or 2.3250

C

ii) Tf0- Tf = 2.325

0 C

O0C - Tf = 2.325

0 C

Tf = - 2.3250

C or 270.675 K

½

½

1

1

13

1

1

1







1

1

1







OR

1

1

1

3

15


½ x 4

= 2

1


Consequences –




1

1+1




1

1

1




1

1

1

19 i)

ii)

iii)

1

1

1

20

A – Benzoic acid

½ +

½

4

B – Benzamide

C - Aniline

½ +

½

½ +

½



Vitamin K

½+½

½+½

1

22 i)

ii)

iii)

½ +

½

½ +

½

½ +

½


ii) No


1

1

2










melt).



OR

1

1

1

1

1

5


ions.

b) E

0cell = E

0Ag+ / Ag - E

0Ni2+ / Ni

= 0.80V – 0.25V

= 0.55V

1og Kc =

=

log Kc = 18.644

∆G0 = - nFE

0cell

= -2x96500 Cmol-1

x 0.55V

= -106,150 Jmol-1


or 106.150k Jmol-1

1

1

½

½

½

½

1

25 a) i)

ii)




1

1

1

1

1

25 a) i)

ii)





sulphuric acid




1

1

1

1

1





acid) as catalyst.

1

6


b) i)

ii)

iii)


OR

1

1

1

1





respectively.


b)i)

1

1

1

7

ii)

iii) CH3COCl

1

1

56/2/3/F 1 P.T.O.

narjmWu H$moS >H$mo CÎma-nwpñVH$m Ho$ _wI-n¥ð >na Adí` bIo§ & Candidates must write the Code on the

title page of the answer-book.

Series SSO/2 H$moS> Z§.

Code No.

amob Z§. Roll No.

agm`Z dkmZ (g¡ÕmpÝVH$)

CHEMISTRY (Theory)

ZYm©[aV g_` : 3 KÊQ>o AYH$V_ A§H$ : 70

Time allowed : 3 hours Maximum Marks : 70

H¥$n`m Om±M H$a b| H$ Bg àíZ-nÌ _o§ _wÐV n¥ð> 15 h¢ &

àíZ-nÌ _| XmhZo hmW H$s Amoa XE JE H$moS >Zå~a H$mo N>mÌ CÎma-nwpñVH$m Ho$ _wI-n¥ð> na bI| &

H¥$n`m Om±M H$a b| H$ Bg àíZ-nÌ _| >26 àíZ h¢ & H¥$n`m àíZ H$m CÎma bIZm ewê$ H$aZo go nhbo, àíZ H$m H«$_m§H$ Adí` bI| &

Bg àíZ-nÌ H$mo n‹T>Zo Ho$ bE 15 _ZQ >H$m g_` X`m J`m h¡ & àíZ-nÌ H$m dVaU nydm©• _| 10.15 ~Oo H$`m OmEJm & 10.15 ~Oo go 10.30 ~Oo VH$ N>mÌ Ho$db àíZ-nÌ H$mo n‹T>|Jo Am¡a Bg AdY Ho$ Xm¡amZ do CÎma-nwpñVH$m na H$moB© CÎma Zht bI|Jo &

Please check that this question paper contains 15 printed pages.

Code number given on the right hand side of the question paper should be

written on the title page of the answer-book by the candidate.

Please check that this question paper contains 26 questions.

Please write down the Serial Number of the question before

attempting it.

15 minute time has been allotted to read this question paper. The question

paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the

students will read the question paper only and will not write any answer on

the answer-book during this period.

56/2/3/F

SET-3

56/2/3/F 2

gm_mÝ` ZXe :

(i) g^r àíZ AZdm`© h¢ &

(ii) àíZ g§»`m 1 go 5 VH$ AV bKw-CÎmar` àíZ h¢ Am¡a àËòH$ àíZ Ho$ bE 1 A§H$ h¡ &

(iii) àíZ g§»`m 6 go 10 VH$ bKw-CÎmar` àíZ h¢ Am¡a àËòH$ àíZ Ho$ bE 2 A§H$ h¡§ &

(iv) àíZ g§»`m 11 go 22 VH$ ^r bKw-CÎmar` àíZ h¢ Am¡a àËòH$ àíZ Ho$ bE 3 A§H$ h¢ &

(v) àíZ g§»`m 23 _yë`mYm[aV àíZ h¡ Am¡a BgHo$ bE 4 A§H$ h¢ &

(vi) àíZ g§»`m 24 go 26 VH$ XrK©-CÎmar` àíZ h¢ Am¡a àËòH$ àíZ Ho$ bE 5 A§H$ h¢ &

(vii) `X Amdí`H$Vm hmo, Vmo bm°J Q>o~bm| H$m à`moJ H$a| & H¡$ëHw$boQ>am| Ho$ Cn`moJ H$s AZw_V Zht h¡ &



(ii) Questions number 1 to 5 are very short answer questions and carry

1 mark each.

(iii) Questions number 6 to 10 are short answer questions and carry 2 marks

each.

(iv) Questions number 11 to 22 are also short answer questions and carry

3 marks each.


(vi) Questions number 24 to 26 are long answer questions and carry 5 marks

each.


1. Cg `m¡JH$ H$m gyÌ Š`m h¡ Og_| VÎd Y ccp OmbH$ ~ZmVm h¡ Am¡a X Ho$ na_mUw

AîQ>\$bH$s` [a>º$ H$m 2/3dm± ^mJ KoaVo h¢ ? 1

What is the formula of a compound in which the element Y forms ccp

lattice and atoms of X occupy 2/3rd of octahedral voids ?

56/2/3/F 3 P.T.O.

2. XE JE `m¡JH$ H$m AmB©.`y.nr.E.gr. Zm_ bIE : 1

HO CH2 CH = C CH3

CH3

Write the IUPAC name of the given compound :

HO CH2 CH = C CH3

CH3

3. ^m¡VH$emofU CËH«$_Ur` h¡ O~H$ amgm`ZH$emofU AZwËH«$_Ur` hmoVm h¡ & Š`m| ? 1

Physisorption is reversible while chemisorption is irreversible. Why ?

4. ZåZbIV `w½_ _| H$m¡Z SN2 A^H«$`m AYH$ Vrd«Vm go H$aoJm Am¡a Š`m| ? 1

CH3 CH2 Br Am¡a CH3 CH2 I

Which would undergo SN2 reaction faster in the following pair and why ?

CH3 CH2 Br and CH3 CH2 I

5. gm_mÝ` Vmn_mZ na gë\$a H$m H$m¡Z-gm Anaê$n (EbmoQ´>mon) D$î_r` ê$n go ñWm`r h¡ ? 1

Which allotrope of sulphur is thermally stable at room temperature ?

6. (a) Obr` H$m°na(II) ŠbmoamBS> db`Z Ho$ dÚwV ²-AnKQ>Z Ho$ Xm¡amZ H¡$WmoS> na ZåZbIV A^H«$`mE± hmoVr h¢ :

Cu2+ (aq) + 2e Cu(s) E0 = + 0·34 V

H+ (aq) + e 2

1 H2(g) E

0 = 0·00 V

CZHo$ _mZH$ AnM`Z BboŠQ´>moS> d^d (E0) Ho$ _mZm| Ho$ AmYma na H¡$WmoS> na H$g

A^H«$`m H$s g§^mdZm (gwg§JVVm) h¡ Am¡a Š`m| ?

(b) Am`Zm| Ho$ ñdV§Ì A^J_Z Ho$ H$mobamD$e Z`_ H$m H$WZ H$sOE & BgH$m EH$ AZwà`moJ bIE & 2

56/2/3/F 4

(a) Following reactions occur at cathode during the electrolysis of

aqueous copper(II) chloride solution :

Cu2+ (aq) + 2e Cu(s) E0 = + 0·34 V

H+ (aq) + e 2

1 H2(g) E

0 = 0·00 V

On the basis of their standard reduction electrode potential (E0)

values, which reaction is feasible at the cathode and why ?

(b) State Kohlrausch law of independent migration of ions. Write its

one application.

7. g§H«$_U VÎd Š`m| n[adVu CnM`Z AdñWmAm§o H$mo àXe©V H$aVo h¢ ? 3d loUr _| (Sc go Zn)

H$m¡Z-gm VÎd gdm©YH$ CnM`Z AdñWmE± Xem©Vm h¡ Am¡a Š`m| ? 2

Why do transition elements show variable oxidation states ? In 3d series

(Sc to Zn), which element shows the maximum number of oxidation

states and why ?

8. (i) ZåZbIV H$m°åßboŠg H$m AmB©.`y.nr.E.gr. Zm_ bIE :

[Cr (en)3]Cl3

(ii) ZåZbIV H$m°åßboŠg H$m gyÌ bIE :

nmoQ>¡e`_ Q´>mB© Am°Šgb¡Q>mo H«$mo_oQ>(III) 2

(i) Write down the IUPAC name of the following complex :

[Cr (en)3]Cl3

(ii) Write the formula for the following complex :

Potassium tri oxalato chromate(III)

56/2/3/F 5 P.T.O.

9. ZåZbIV A^H«$`mAm| _| à`moJ AmZo dmbo A^H$maH$m| Ho$ Zm_ XrOE : 2

(i) CH3 CHO ?

CH3 CH CH3

|

OH

(ii) CH3 COOH ?

CH3 COCl

Name the reagents used in the following reactions :

(i) CH3 CHO ?

CH3 CH CH3 |

OH

(ii) CH3 COOH ?

CH3 COCl

10. amCëQ> Ho$ Z`_ go G$UmË_H$ dMbZ go Š`m VmËn`© h¡ ? EH$ CXmhaU XrOE & G$UmË_H$ dMbZ Ho$ bE ∆mixH H$m Š`m M• hmoVm h¡ ? 2>

AWdm EµOAmoQ´>mon H$mo n[a^mfV H$sOE & amCëQ> Ho$ Z`_ go G$UmË_H$ dMbZ Ûmam ~ZZo dmbm EµOAmoQ´>mon H$g àH$ma H$m hmoVm h¡ ? EH$ CXmhaU XrOE & 2

What is meant by negative deviation from Raoult’s law ? Give an

example. What is the sign of mixH for negative deviation ?

OR

Define azeotropes. What type of azeotrope is formed by negative

deviation from Raoult’s law ? Give an example.

11. (a) EpëH$b h¡bmBS>| Ob _| KwbZerb Zht h¢ & Š`m| ?

(b) ã`wQ>¡Z-1-Am°b àH$meH$s` ZpîH«$` (Y«wdU AKyU©H$) h¡ naÝVw ã`wQ>¡Z-2-Am°b

àH$meH$s` gH«$` (Y«wdU KyU©H$) h¡ & Š`m| ?

(c) `Ún ŠbmoarZ BboŠQ´>m°Z H$mo AmH$f©V H$aZo dmbm J«wn h¡ \$a ^r `h BboŠQ´>m°ZñZohr

Eamo_¡Q>H$ àVñWmnZ A^H«$`mAm| _| Am°Wm- VWm n¡am- ZXeH$ h¡ & Š`m| ? 3

(a) Why are alkyl halides insoluble in water ?

(b) Why is Butan-1-ol optically inactive but Butan-2-ol is optically

active ?

(c) Although chlorine is an electron withdrawing group, yet it is

ortho-, para- directing in electrophilic aromatic substitution

reactions. Why ?

56/2/3/F 6

12. ZåZbIV H$m ê$nm§VaU Amn H¡$go H$a|Jo : 3

(i) ~oÝµOmoBH$ EgS> H$mo ~oÝµO¡pëS>hmBS> _|

(ii) EWmB©Z H$mo EW¡Z¡b _|

(iii) EogrQ>H$ EgS> H$mo _rWoZ _|

AWdm

ZåZbIV A^H«$`mAm| go gå~pÝYV g_rH$aUm| H$mo bIE : 3

(i) ñQ>r\$Z A^H«$`m

(ii) dmoë\$-H$íZa AnM`Z

(iii) EQ>mS>© A^H«$`m

How do you convert the following :


(ii) Ethyne to Ethanal

(iii) Acetic acid to Methane

OR

Write the equations involved in the following reactions :

(i) Stephen reaction

(ii) Wolff-Kishner reduction

(iii) Etard reaction

13. 37·2 g Ob _| NaCl (_mob Ðì`_mZ = 58·5 g mol1) H$s H$VZr _mÌm KwbmB© OmE H$ h_m§H$ 2°C KQ> OmE, `h _mZVo hþE H$ NaCl nyU© ê$n go dKQ>V hmoVm h¡ ?

(Kf Ob Ho$ bE = 1·86 K kg mol1) 3

Calculate the mass of NaCl (molar mass = 58·5 g mol1) to be dissolved in

37·2 g of water to lower the freezing point by 2C, assuming that NaCl

undergoes complete dissociation. (Kf for water = 1·86 K kg mol1)

56/2/3/F 7 P.T.O.

14. ZåZbIV ~hþbH$m| Ho$ EH$bH$m| Ho$ Zm_ Am¡a CZH$s g§aMZmE± bIE : 3

(i) Q>oarbrZ (ii) ~¡Ho$bmBQ> (iii) ~wZm-S

Write the names and structures of the monomers of the following

polymers :

(i) Terylene

(ii) Bakelite

(iii) Buna-S

15. (i) ZåZbIV _§o H$m¡Z _moZmog¡Ho$amBS> h¡ : ñQ>mM©, _mëQ>mog, \«$ŠQ>mog, gobwbmog

(ii) Aåbr` Eo_Zmo EogS>m| Am¡a jmar` Eo_Zmo EogS>m| Ho$ ~rM$ Š`m A§Va hmoVm h¡ ?

(iii) Cg dQ>m_Z H$m Zm_ bIE OgH$s H$_r Ho$ H$maU _gy‹S>m| _| IyZ AmZo bJVm h¡ & 3

(i) Which one of the following is a monosaccharide :

starch, maltose, fructose, cellulose

(ii) What is the difference between acidic amino acids and basic amino

acids ?

(iii) Write the name of the vitamin whose deficiency causes bleeding of

gums.

16. (i) H$m°åßboŠg [Pt(en)2Cl2]2+ Ho$ Á`m_Vr` g_md`dm| H$mo AmaoIV H$sOE &

(ii) H«$ñQ>b \$sëS> gÕmÝV Ho$ AmYma na `X ∆o > P h¡, Vmo d4 Am`Z H$m BboŠQ´>m°ZH$ dÝ`mg bIE &

(iii) H$m°åßboŠg [Ni(CN)4]2 H$m g§H$aU àH$ma Am¡a Mwå~H$s` ì`dhma bIE & (Ni H$m na_mUw H«$_m§H$ = 28) 3

(i) Draw the geometrical isomers of complex [Pt(en)2Cl2]2+.

(ii) On the basis of crystal field theory, write the electronic

configuration for d4 ion, if o P.

(iii) Write the hybridization type and magnetic behaviour of the

complex [Ni(CN)4]2. (Atomic number of Ni = 28)

56/2/3/F 8

17. (i) ZH$b Ho$ n[aîH$aU _| H$m_ AmZo dmbr dY Ho$ nrN>o Omo gÕmÝV hmoVm h¡ CgH$m

C„oI H$sOE &

(ii) gmoZo Ho$ ZîH$f©U _| VZw NaCN H$s Š`m ^y_H$m hmoVr h¡ ?

(iii) ‘H$m°na _¡Q>o’ Š`m hmoVm h¡ ? 3

(i) Indicate the principle behind the method used for the refining of

Nickel.

(ii) What is the role of dilute NaCN in the extraction of gold ?

(iii) What is ‘copper matte’ ?

18. 25C na ZåZ gob H$m dÚwV²-dmhH$ ~b (B©.E_.E\$.) n[aH$bV H$sOE : 3

Zn | Zn2+ (0.001 M) | | H+ (0.01 M) | H2(g) (1 bar) | Pt(s)

0

)Zn/2Zn(E = 0.76 V, 0

)2

H/H(E = 0.00 V

Calculate the emf of the following cell at 25C :


0

)Zn/2Zn(E = 0.76 V, 0

)2

H/H(E = 0.00 V

19. ZåZbIV A^H«$`mAm| Ho$ CËnmXm| H$s àmJwº$ H$sOE : 3

(i) CH3 CH = CH2 –

22

62

OH/OH3)ii

HB)i ?

(ii) C6H5 OH )aq(Br2 ?

(iii) CH3CH2OH K573/Cu

?

Predict the products of the following reactions :

(i) CH3 CH = CH2 –

22

62

OH/OH3)ii

HB)i ?



?

56/2/3/F 9 P.T.O.

20. EH$ VÎd X (_moba Ðì`_mZ = 60 g mol1) H$m KZËd 6·23 g cm3 h¡ & `X `yZQ> gob Ho$ H$moa H$s bå~mB© 4 × 108 cm h¡, Vmo Š`y~H$ `yZQ> gob Ho$ àH$ma H$s Š`m nhMmZ hmoJr ? 3

An element X (molar mass = 60 g mol1) has a density of 6.23 g cm3.

Identify the type of cubic unit cell, if the edge length of the unit cell is

4 × 108 cm.

21. (a) ZåZbIV H$mo Amn H$maU XoVo hþE H¡$go g_PmE±Jo :

(i) Mn H$m CƒV_ âbwAmoamBS> MnF4 h¡ O~H$ CƒV_ Am°ŠgmBS> Mn2O7

h¡ &

(ii) g§H«$_U YmVwE± Am¡a CZHo$ `m¡JH$ CËàoaH$ JwUY_© Xem©Vo h¢ &

(b) ZåZbIV g_rH$aU H$mo nyU© H$sOE :

3–2

4MnO + 4H+ 3

(a) How would you account for the following :

(i) Highest fluoride of Mn is MnF4 whereas the highest oxide is

Mn2O7.

(ii) Transition metals and their compounds show catalytic

properties.

(b) Complete the following equation :

3–2

4MnO + 4H+

22. ZåZbIV AdbmoH$Zm| Ho$ bE H$maUm| H$mo XrOE : 3

(i) g_wÐr Ob Am¡a ZXr H$m Ob Ohm± _bVo h¢ dhm± EH$ S>oëQ>m ~Z OmVm h¡ &

(ii) MmaH$mob H$s gVh na N2 J¡g H$s Anojm NH3 J¡g AYH$ erK«Vm go AYemofV hmoVr h¡ &

(iii) MyU© H$E hþE nXmW© AYH$ à^mdembr AYemofH$ hmoVo h¢ &

56/2/3/F 10

Give reasons for the following observations :

(i) A delta is formed at the meeting point of sea water and river

water.

(ii) NH3 gas adsorbs more readily than N2 gas on the surface of

charcoal.

(iii) Powdered substances are more effective adsorbents.

23. ~ƒm| _| _Yw_oh Am¡a CXmgr Ho$ ~‹T>Vo Ho$gm| H$mo XoIZo Ho$ ~mX EH$ àgÕ ñHy$b Ho$ qàgnb

lr Mmon ‹S>m Zo EH$ go_Zma H$m Am`moOZ H$`m Og_| CÝhm|Zo ~ƒm| Ho$ A^^mdH$m| VWm AÝ`

ñHy$bm| Ho$ qàgnbm| H$mo Am_§ÌV H$`m & CÝhm|Zo ñHy$bm| _| g ‹S>o hþE ^moÁ` nXmWm] (O§H$ \y$S>) na àV~§Y bJmZo H$m ZU©` b`m, gmW hr `h ZU©` b`m H$ ñHy$bm| _| ñdmñÏ`dY©H$

nXmW© O¡go gyn, bñgr, XÿY AmX H¢$Q>rZm| _| CnbãY H$amB© OmE± & CÝhm|Zo `h ^r ZU©`

b`m H$ àmV:H$mbrZ Egoå~br Ho$ g_` ~ƒm| H$mo àVXZ AmYo K§Q>o H$s emar[aH$ H$gaV ^r H$amB© OmE & N>: _mh níMmV² lr Mmon ‹S>m Zo ~ƒm| Ho$ ñdmñÏ` H$m AYH$V_ dÚmb`m| _|

nwZ: ZarjU H$adm`m Am¡a ~ƒm| Ho$ ñdmñÏ` _| AZwn_ gwYma nm`m J`m &

Cn w©º$ àH$aU H$mo n‹T>Zo Ho$ ~mX, ZåZbIV àíZm| Ho$ CÎma XrOE : 4

(i) lr Mmon ‹S>m Ûmam H$Z _yë`m| (H$_-go-H$_ Xmo) H$mo Xem©`m J`m h¡ ?

(ii) EH$ dÚmWu Ho$ ê$n _|, Amn Bg df` _| H¡$go OmJê$H$Vm \¡$bmE±Jo ?

(iii) AdZ_Z-damoYr S´>J ~Zm S>m°ŠQ>a H$s gbmh Ho$ Š`m| Zht boZ o MmhE ?

(iv) H¥$Ì_ _YwaH$ Ho$ Xmo CXmhaU XrOE &

Seeing the growing cases of diabetes and depression among children,

Mr. Chopra, the principal of one reputed school organized a seminar in

which he invited parents and principals. They all resolved this issue by

strictly banning the junk food in schools and by introducing healthy

snacks and drinks like soup, lassi, milk etc. in school canteens. They also

decided to make compulsory half an hour physical activities for the

students in the morning assembly daily. After six months,

Mr. Chopra conducted the health survey in most of the schools and

discovered a tremendous improvement in the health of students.

56/2/3/F 11 P.T.O.

After reading the above passage, answer the following questions :

(i) What are the values (at least two) displayed by Mr. Chopra ?

(ii) As a student, how can you spread awareness about this issue ?

(iii) Why should antidepressant drugs not be taken without consulting

a doctor ?

(iv) Give two examples of artificial sweeteners.

24. (a) àË`oH$ Ho$ bE Cn wº$ CXmhaU XoVo hþE ZåZbIV A^H«$`mAm| H$mo àXe©V H$sOE :

(i) A_moZrH$aU

(ii) H$pßb¨J (`w½_Z) A^H«$`m

(iii) Eo_rZm| H$m EogrQ>brH$aU

(b) àmW_H$ (àmB_ar), ÛVr`H$ (goH$ÊS>ar) Am¡a V¥Vr`H$ (Q>e©`ar) E_rZm| H$s nhMmZ H$aZo Ho$ bE hÝg~J© dY H$m dU©Z H$sOE & gå~Õ A^H«$`mAm| Ho$ amgm`ZH$ g_rH$aUm| H$mo ^r bIE & 5

AWdm

(a) O~ ~oÝµOrZ S>mBEµOmoZ`_ ŠbmoamBS> (C6 H5–

2ClN ) ZåZbIV A^H$maH$m| go

A^H«$`m H$aVm h¡, V~ àmßV _w»` CËnmXm| H$s g§aMZmE± bIE :

(i) HBF4 /

(ii) Cu / HBr

(b) ZåZbIV A^H«$`mAm| _| A, B Am¡a C H$s g§aMZmE± bIE :

5

56/2/3/F 12

(a) Illustrate the following reactions giving suitable example in each

case :

(i) Ammonolysis

(ii) Coupling reaction

(iii) Acetylation of amines

(b) Describe Hinsberg method for the identification of primary,

secondary and tertiary amines. Also write the chemical equations

of the reactions involved.

OR

(a) Write the structures of main products when benzene diazonium

chloride (C6 H5–

2ClN ) reacts with the following reagents :

(i) HBF4 /

(ii) Cu / HBr

(b) Write the structures of A, B and C in the following reactions :

25. (a) ZåZbIV nXm| H$mo n[a^mfV H$sOE :

(i) gH«$`U D$Om©

(ii) Xa pñWam§H$

(b) 25% d`moOZ Ho$ bE EH$ àW_ H$moQ> H$s A^H«$`m 10 _ZQ> boVr h¡ & A^H«$`m Ho$ bE t

1/2 H$m n[aH$bZ H$sOE >& 5

(X`m J`m : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

AWdm

56/2/3/F 13 P.T.O.

(a) EH$ amgm`ZH$ A^H«$`m R P Ho$ bE gm§ÐU _| n[adV©Z ln [R] vs. g_` (s) ZrMo ßbm°Q> _| X`m J`m h¡ :

ln [R]

t (s)

(i) A^H«$`m H$s H$moQ> H$s àmJwº$ H$sOE &

(ii) dH«$ H$m T>bmZ Š`m h¡ ?

(iii) A^H«$`m Ho$ bE Xa pñWam§H$ H$s `yZQ> bIE &

(b) Xem©BE H$ 99% nyU© hmoZo _| Omo g_` bJVm h¡ dh Cg g_` H$m XwJwZm h¡ Omo

A^H«$`m Ho$ 90% nyU© hmoZo _| bJVm h¡ & 5


(i) Activation energy

(ii) Rate constant

(b) A first order reaction takes 10 minutes for 25% decomposition.

Calculate t1/2

for the reaction.

(Given : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

OR

(a) For a chemical reaction R P, the variation in the concentration,

ln [R] vs. time (s) plot is given as

ln [R]

t (s)

56/2/3/F 14

(i) Predict the order of the reaction.

(ii) What is the slope of the curve ?

(iii) Write the unit of rate constant for this reaction.

(b) Show that the time required for 99% completion is double of the

time required for the completion of 90% reaction.

26. (a) ZåZbIV Ho$ H$maU XoVo hþE ñnîQ> H$sOE :

(i)

4NH _| Omo Am~ÝY H$moU h¡ dh NH3 Ho$ H$moU go CƒVa h¡ &

(ii) H2O H$s Anojm H2S H$m ŠdWZm§H$ Ý`yZVa h¡ &

(iii) AnM`Z ì`dhma SO2 go TeO2 H$s Amoa KQ>Vm h¡ &

(b) ZåZbIV H$s g§aMZmE± AmaoIV H$sOE :

(i) H4P2O7 (nm`amo\$m°ñ\$mo[aH$ EogS>)

(ii) XeF2 5

AWdm

(a) ZåZbIV H$s g§aMZmE± AmaoIV H$sOE :

(i) XeF4

(ii) H2S2O7

(b) ZåZbIV Ho$ H$maU XrOE :

(i) HCl go A^H«$`m go Am`aZ FeCl2 ~ZmVm h¡ Z H$s FeCl3.

(ii) HClO H$s Anojm HClO4 à~bVa Aåb h¡ &

(iii) dJ© 15 Ho$ g^r hmBS´>mBS>m| _| BiH3 à~bV_ AnMm`H$ h¡ & 5

(a) Account for the following :

(i) Bond angle in

4NH is higher than NH3.

(ii) H2S has lower boiling point than H2O.

(iii) Reducing character decreases from SO2 to TeO2.

56/2/3/F 15 P.T.O.

(b) Draw the structures of the following :

(i) H4P2O7 (Pyrophosphoric acid)

(ii) XeF2

OR


(i) XeF4

(ii) H2S2O7

(b) Account for the following :

(i) Iron on reaction with HCl forms FeCl2 and not FeCl3.

(ii) HClO4 is a stronger acid than HClO.

(iii) BiH3 is the strongest reducing agent amongst all the

hydrides of group 15.

1

Qn Value points Marks

1 X2Y3 1

2 3-Methylbut-2-en-1-ol 1

3 Because of weak van der Waals’ forces in physisorption whereas there are strong chemical

forces in chemisorption.

1

4 CH3CH2I , because I is a better leaving group. ½ , ½

5 Rhombic sulphur 1

6 a) Cu2+

(aq) + 2 e Cu(s) because of high E0 value/ more negative ∆G

b) It states that limiting molar conductivity of an electrolyte is equal to the sum of the individual

contributions of cations and anions of the electrolyte.

It is used to calculate the Ʌm0 for weak electrolyte / It is used to calculate α and Kc

(Any one application)

½ , ½

1

1

7 a) Due to presence of unpaired d-electrons/ comparable energies of 3d and 4s orbitals.

b) Mn , due to involvement of 4s and 3d electrons/ presence of maximum unpaired d-

electrons.

1

½ ,½

8 i) tris-(ethane-1,2-diamine)chromium(III) chloride

ii) K3[ Cr(C2O4)3]

1

1

9 (i) CH3MgBr/ H3O+

(ii) PCl5/ PCl3 / SOCl2

1

1

10

10

When solute- solvent interaction is stronger than pure solvent or solute interaction.

Eg: chloroform and acetone (or any other correct eg)

∆mixH= negative

OR

Azeotropes –binary mixtures having same composition in liquid and vapour phase and boil at

constant temperature / is a liquid mixture which distills at constant temperature without

undergoing change in composition

1

½

½

1

½

CHEMISTRY MARKING SCHEME 2015

SET -56/2/3 F

2

Maximum boiling azeotropes

eg: HNO3 (68%) and H2O(32%) (or any other correct example)

½

11 a) Because they are unable to form H-bonds with water molecules.

b) Because of the presence of chiral carbon in butan-2-ol.

c) Due to dominating +R effect

1

1

1

12 i) C6H5COOH PCl5 C6H5COCl H2/Pd C6H5CHO

BaSO4

ii) CH≡CH + H2O Hg2+

/H2SO4 CH3CHO

iii) CH3COOH NaOH

CH3COONa NaOH + CaO , heat

CH4

OR

i)

ii)

iii)

1

1

1

1

1

1

13 ∆ Tf = i. Kf m

= i Kf wB x 1000

MB x wA

2K= 2 x 1.86K kg/mol x wB x 1000

58.5 g/mol x 37.2 g

wB = 1.17g

1

1

1

14

i)

ii)

Phenol and formaldehyde

1

1

3

iii)

(Note: half mark for structure/s and half mark for name/s)

1

15 i) Fructose

ii) Acidic amino acid has more number of acidic carboxylic group than basic amino

group whereas basic amino acid has more number of basic amino group.

iii) Vitamin C

1

1

1

16

i)

cis- isomer trans-isomer

ii) t2g4

iii) dsp 2

, diamagnetic

1

1

½ , ½

17 a) Impure Ni reacts with CO to form volatile Ni(CO)4 which when heated at higher

temperature decomposes to give pure Ni.

b) NaCN acts as a leaching agent to form a soluble complex with gold.

c) It is a mixture of Cu2S and FeS

1

1

1

18 E cell = E

0 cell –

log

E cell = 0.76 V -

V log

( )

E cell = 0.76 – 0.0295 V log 10

= 0.7305 V

1

1

1

19 i) CH3CH2CH2OH

ii)

1

1

4

iii) CH3CHO

1

20 d=

6.23 g cm-3

=

( )

z=4

fcc

½

½

1

1

21 i) Because oxygen stabilizes Mn more than F due to multiple bonding

ii) Because of their ability to show variable oxidation state(or any other correct reason)

iii) 3MnO42-

+ 4H+ 2MnO4

- + MnO2 + 2H2O

1

1

1

22 i) Due to coagulation of colloidal clay particles.

ii) Because NH3 is easily liquefiable than N2 due to its larger molecular size.

iii) Because of more surface area.

1

1

1

23 a) Concern for students health, Application of knowledge of chemistry to daily life, empathy

, caring or any other

b) Through posters, nukkad natak in community, social media, play in assembly (or any other

relevant answer)

c) Wrong choice and overdose may be harmful

d) Aspartame, saccharin (or any other correct example)

½ , ½

1

1

½+ ½

24

a) i)ammonolysis

ii)

1

1

5

24

(any one)

iii) (or any other correct reaction)

b)reaction of primary amine

(soluble in alkali)

Reaction of secondary amine

(insoluble in alkali)

Tertiary amine doesn’t react

OR

a) i)

ii)

1

1

1

1

1

½,½,

½

6

b) i) A- B- C-

ii) A- CH3CN B- CH3CH2NH2 C- CH3CH2OH

½ ,½,

½

25 a)i) Activation energy- Extra energy required by reactants to form activated complex.

ii) Rate constant- rate of reaction when the concentration of reactant is unity.

b)

k= 2.303 log [ A0 ]

t [A]

k = 2.303 log 100

10 min 75

k = 2.303 x 0.125

10 min

k = 0.02879 min-1

t1/2 =

=

t1/2 = 24.07min

OR

a) i)First order ii) -k iii) s-1

b)

t =

log

t99% =

log

t =

x 2

1

1

½

½

1

1

1,1,1

½

7

t90% =

log

=

t99% = 2 x t90%

½

1

26 a) i)Because of lone pair in NH3 , lone pair- bond pair repulsion decreases the bond angle

ii)Because of absence of H-bonding in H2S

iii)Because stability of +4 oxidation state increases from SO2 to TeO2

b)

OR

a)

b)i)Because iron on reaction with HCl produces H2(g) which prevents the formation of FeCl2 to

FeCl3 / Because HCl is a weak oxidising agent.

ii) Because of higher oxidation state of chlorine in HClO4

iii) Because of lower dissociation enthalpy of Bi-H bond.

1

1

1

1,1

1,1

1

1

1

56/1/1/D 1 [P.T.O.


Roll No.

¸üÃÖÖµÖ®Ö ×¾Ö–ÖÖ®Ö (ÃÖî üÖ×®ŸÖÛú) CHEMISTRY (Theory)

×®Ö¬ÖÖÔ×¸üŸÖ ÃÖ´ÖµÖ : 3 ‘ÖÞ™êüü] [†×¬ÖÛúŸÖ´Ö †ÓÛúú : 70


ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö :






(vi) ¯ÖÏ¿®Ö-ÃÖÓÜµÖÖ 24 ÃÖê 26 ¤üß‘ÖÔ-ˆ¢Ö¸üßµÖ ¯ÖÏ¿®Ö Æïü … ¯ÖÏŸµÖêÛú ¯ÖÏ¿®Ö Ûêú ×»Ö‹ 5 †ÓÛú Æïü …

(vii) µÖ×¤ü †Ö¾Ö¿µÖÛú ÆüÖê ŸÖÖê »ÖÖòÝÖ ™êü²Ö»Ö ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü … Ûîú»ÖÛãú»Öê™ü¸ü Ûêú ˆ¯ÖµÖÖêÝÖ Ûúß †®Öã Ö×ŸÖ ®ÖÆüà Æîü …

Series : SSO/1 ÛúÖê›ü ®ÖÓ. Code No.

56/1/1/D

• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´Öã×¦üŸÖ ¯ÖéÂšü 12 Æïü …

• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê ü ×¤ü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ÃŸÖÛúÖ Ûêú ´ÖãÜÖ-¯ÖéÂšü ¯Ö¸ü ×»ÖÜÖë …

• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …

• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …

• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê ×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ¯ÖœÌëüÝÖê †Öî ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî üÖ®Ö ¾Öê ˆ¢Ö¸ü-¯Öã×ÃŸÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …







distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will only read the

question paper and will not write any answer on the answer-book during this period.



SET – 1

56/1/1/D 2



(ii) Q. no. 1 to 5 are very short answer questions and carry 1 mark each.

(iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each.

(iv) Q. no. 11 to 22 are also short answer questions and carry 3 marks each.

(v) Q. no. 23 is a value based question and carry 4 marks.

(vi) Q. no. 24 to 26 are long answer questions and carry 5 marks each.

(vii) Use log tables if necessary, use of calculators is not allowed.

1. H3PO4 Ûúß õÖÖ¸üÛúŸÖÖ ŒµÖÖ Æîü ? 1

What is the basicity of H3PO4 ?

2. ×¤üµÖê ÝÖµÖê µÖÖî×ÝÖÛú ÛúÖ †Ö‡Ô.µÖæ.¯Öß.‹.ÃÖß. ®ÖÖ´Ö ×»Ö×ÜÖ‹ : 1


3. ×®Ö´®Ö µÖãÝ´Ö ´Öë ÛúÖî®Ö SN2 †×³Ö×ÛÎúµÖÖ †×¬ÖÛú ŸÖß¾ÖÎŸÖÖ ÃÖê Ûú êüÝÖÖ †Öî ü ŒµÖÖë ? 1

CH3 – CH2 – Br ŸÖ£ÖÖ CH3 –

C|

C|

B

H3

–

r

CH3


CH3 – CH2 – Br and CH3 –

C|

C|

B

H3

–

r

CH3

56/1/1/D 3 [P.T.O.

4. BaCl2 †Öî ü KCl ´Öë ÃÖê ÛúÖî®Ö ŠúÞÖÖŸ´ÖÛú “ÖÖ•ÖÔ ¾ÖÖ»Öê ÛúÖê»ÖÖ‡›üß ÃÖÖò»Ö ÛúÖ ÃÛÓú¤ü®Ö †×¬ÖÛú ¯ÖÏ³ÖÖ¾Ö¿ÖÖ»Öß œÓüÝÖ ÃÖê Ûú¸êüÝÖÖ ? ÛúÖ¸üÞÖ ¤üß×•Ö‹ … 1

Out of BaCl2 and KCl, which one is more effective in causing coagulation of a

negatively charged colloidal Sol ? Give reason.

5. ˆÃÖ µÖÖî×ÝÖÛú ÛúÖ ÃÖæ¡Ö ŒµÖÖ ÆüÖêÝÖÖ ×•ÖÃÖ´Öë Y ŸÖ¢¾Ö ccp •ÖÖ»ÖÛú ²Ö®ÖÖŸÖÖ Æîü †Öî ü X “ÖŸÖãÂ±ú»ÖÛúßµÖ ×¸ü×ŒŸÖ ÛúÖ 1/3¾ÖÖÑ ³ÖÖÝÖ ‘Öê üŸÖÖ Æîü ? 1

What is the formula of a compound in which the element Y forms ccp lattice and

atoms of X occupy 1/3rd

of tetrahedral voids ?

6. ÃÖÓÛÎú´ÖÞÖ ŸÖŸ¾Ö ŒµÖÖ Æïü ? ÃÖÓÛÎú´ÖÞÖ ŸÖŸ¾ÖÖë Ûúß ¤üÖê ×¾Ö¿ÖêÂÖŸÖÖ†Öë ÛúÖê ×»Ö×ÜÖ‹ … 2

What are the transition elements ? Write two characteristics of the transition elements.

7. (i) ÛúÖò ¯»ÖêŒÃÖ [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine) ÛúÖ †Ö‡Ô.µÖæ.¯Öß.‹.ÃÖß. ®ÖÖ´Ö ×»Ö×ÜÖµÖê …

(ii) ×®Ö´®Ö ÛúÖò ¯»ÖêŒÃÖ ÛúÖ ÃÖæ¡Ö ×»Ö×ÜÖ‹ :

¯Öê®™üÖ‹ê ÖÖ‡®Ö®ÖÖ‡™ÒüÖ‡™üÖê-o-ÛúÖê²ÖÖ»™ü (III). 2


[Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine)


Pentaamminenitrito-o-Cobalt (III).

8. ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë ´Öë ¯ÖÏµÖãŒŸÖ †×³ÖÛúÖ¸üÛúÖë Ûêú ®ÖÖ´Ö ¤üß×•Ö‹ : 2


(i) CH3 – CO – CH3 ?

→ CH3 – C|

O

H

H

– CH3

(ii) C6H5 – CH2 – CH3

?→ C6H5 – COO–K+

9. ¸üÖˆ»™ü ×®ÖµÖ´Ö ÃÖê ¬Ö®ÖÖŸ´ÖÛú ×¾Ö“Ö»Ö®Ö ÃÖê ŒµÖÖ ŸÖÖŸ¯ÖµÖÔ Æîü ? ‹Ûú ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ … ¬Ö®ÖÖŸ´ÖÛú ×¾Ö“Ö»Ö®Ö Ûêú ×»ÖµÖê ∆mixH ÛúÖ ×“ÖÅ®Öü ŒµÖÖ Æîü ? 2

What is meant by positive deviations from Raoult’s law ? Give an example. What is

the sign of ∆mixH for positive deviation ?

†£Ö¾ÖÖ/OR

56/1/1/D 4

9. ‹×•ÖµÖÖê™ÒüÖê ÃÖ ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ … ¸üÖˆ»™ü ×®ÖµÖ´Ö ÃÖê ¬Ö®ÖÖŸ´ÖÛú ×¾Ö“Ö»Ö®Ö ÃÖê ¯ÖÏÖ¯ŸÖ ‹×•ÖµÖÖê™ÒüÖê Ö ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ÛúÖ

ÆüÖêŸÖÖ Æîü ? ‹Ûú ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ … 2

Define azeotropes. What type of azeotrope is formed by positive deviation from

Raoult’s law ? Give an example.

10. (a) ×ÃÖ»¾Ö¸ü Œ»ÖÖê üÖ‡›ü Ûêú •Ö»ÖßµÖ ×¾Ö»ÖµÖ®Ö Ûêú ×¾ÖªãŸÖ †¯Ö‘Ö™ü®Ö ´Öë Ûîú£ÖÖê›ü ¯Ö¸ü ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ‹Ñ ÆüÖêŸÖß Æïü :

Ag+(aq) + e– → Ag(s) E° = +0.80 V

H+(aq) + e– → 1

2H2(g) E° = 0.00 V

ˆ®ÖÛêú ´ÖÖ®ÖÛú †¯Ö“ÖµÖ®Ö ‡»ÖêŒ™ÒüÖê›ü ×¾Ö³Ö¾Ö (E°) Ûêú ´ÖÖ®ÖÖë Ûêú †Ö¬ÖÖ¸ü ¯Ö¸ü Ûîú£ÖÖê›ü ¯Ö¸ü ×ÛúÃÖ †×³Ö×ÛÎúµÖÖ Ûúß

ÃÖÓ³ÖÖ¾Ö®ÖÖ Æîü †Öî ü ŒµÖÖë ?

(b) ÃÖß´ÖÖÓŸÖ ´ÖÖê»Ö¸ü “ÖÖ»ÖÛúŸÖÖ ÛúÖê ¯Ö×¸ü³ÖÖ×ÂÖŸÖ Ûúß×•Ö‹ … ×¾ÖªãŸÖË †¯Ö‘Ö™ËüµÖ Ûúß “ÖÖ»ÖÛúŸÖÖ ÃÖÖÓ¦üÞÖ Ûêú ‘Ö™ü®Öê Ûêú ÃÖÖ£Ö

ŒµÖÖë Ûú´Ö ÆüÖê®Öê »ÖÝÖŸÖß Æîü ? 2

(a) Following reactions occur at cathode during the electrolysis of aqueous silver

chloride solution :

Ag+(aq) + e– → Ag(s) E° = +0.80 V

H+(aq) + e– → 1

2H2(g) E° = 0.00 V

On the basis of their standard reduction electrode potential (E°) values, which

reaction is feasible at the cathode and why ?

(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution

decreases with the decrease in concentration ?

11. ²Öê®•Öß®Ö Ûêú 49 g ´Öë ²Öê®•ÖÖê‡Ûú †´»Ö ÛúÖ 3.9 g ‘Öã»Ö®Öê ¯Ö¸ü ×Æü´ÖÖÓÛú ´Öë 1.62 K ÛúÖ †¾Ö®Ö´Ö®Ö ÆüÖê •ÖÖŸÖÖ Æîü … ¾Öî®™ü ÆüÖ±ú ÛúÖ¸üÛú ÛúÖê ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ †Öî ü ×¾Ö»ÖêµÖ Ûêú Ã¾Ö³ÖÖ¾Ö Ûúß ¯ÖÏÖÝÖã×ŒŸÖ Ûúß×•Ö‹ (ÃÖÓÝÖã×ÞÖŸÖ µÖÖ ×¾Ö‘Ö×™üŸÖ) 3

(×¤üµÖÖ ÝÖµÖÖ : ²Öê®•ÖÖê‡Ûú ‹×ÃÖ›ü ÛúÖ ´ÖÖê»Ö ¦ü¾µÖ´ÖÖ®Ö = 122 g mol–1, ²Öê®•Öß®Ö Ûêú ×»ÖµÖê Kf = 4.9 K kg mol–1)

3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point

of 1.62 K. Calculate the van’t Hoff factor and predict the nature of solute (associated

or dissociated).

(Given : Molar mass of benzoic acid = 122 g mol–1, Kf for benzene = 4.9 K kg mol–1)

56/1/1/D 5 [P.T.O.

12. (i) Ø•ÖÛú Ûêú ¯Ö×¸üÂÛú¸üÞÖ Ûú¸ü®Öê Ûúß ×¾Ö×¬Ö Ûêú ¯Öß”êû •ÖÖê ×ÃÖ¨üÖ®ŸÖ ÆüÖêŸÖÖ Æîü ˆÃÖÛúÖ ÃÖÓÛêúŸÖ Ûúß×•Ö‹ …

(ii) ÛúÖò Ö¸ü Ûêú ×®ÖÂÛúÂÖÔÞÖ ´Öë ×ÃÖ×»ÖÛúÖ Ûúß ŒµÖÖ ³Öæ×´ÖÛúÖ ÆüÖêŸÖß Æîü ?

(iii) ¾µÖÖ¯ÖÖ¸üß »ÖÖêÆü ÛúÖ ×¾Ö¿Öã ü ºþ¯Ö »ÖÖêÆü ÛúÖ ÛúÖî®Ö ºþ¯Ö Æîü ? 3

(i) Indicate the principle behind the method used for the refining of zinc.

(ii) What is the role of silica in the extraction of copper ?

(iii) Which form of the iron is the purest form of commercial iron ?

13. ´ÖÖê»Ö¸ü ¦ü¾µÖ´ÖÖ®Ö 27 g mol–1 Ûêú ÃÖÖ£Ö ‹Ûú ŸÖŸ¾Ö ÛúÖê ü »Ö´²ÖÖ‡Ô 4.05 × 10–8 cm ¾ÖÖ»Öß ‹Ûú ŒµÖæ×²ÖÛú µÖæ×®Ö™ü ÃÖê»Ö ²Ö®ÖÖŸÖÖ Æîü … µÖ×¤ü ‘Ö®ÖŸ¾Ö 2.7 g cm–3 ÆüÖê, ŸÖÖê ŒµÖæ×²ÖÛú µÖæ×®Ö™ü ÃÖê»Ö ÛúÖ Ã¾Öºþ¯Ö ŒµÖÖ Æîü ? 3

An element with molar mass 27 g mol–1 forms a cubic unit cell with edge length

4.05 × 10–8 cm. If its density is 2.7 g cm–3, what is the nature of the cubic unit cell ?

14. (a) †Ö¯Ö ×®Ö´®Ö ÛúÖê ÛîúÃÖê ÃÖ´Ö—ÖÖ‹ÑÝÖê ?

(i) »Öï£Öî®ÖÖêµÖ›ü ÃÖÓÛãú“Ö®Ö Ûúß †¯ÖêõÖÖ ‹ê×Œ™ü®ÖÖêµÖ›ü ÃÖÓÛãú“Ö®Ö †×¬ÖÛú Æîü …

(ii) ÃÖÓÛÎú´ÖÞÖ ¬ÖÖŸÖã‹Ñ ¸ÓüÝÖß®Ö µÖÖî×ÝÖÛú ²Ö®ÖÖŸÖß Æïü …

(b) ×®Ö´®Ö ÃÖ´ÖßÛú¸üÞÖ ÛúÖê ¯ÖæÞÖÔ Ûúß×•Ö‹ : 3

2MnO–

4 + 6H+ + 5NO–

2 →


(i) Actinoid contraction is greater than lanthanoid contraction.

(ii) Transition metals form coloured compounds.


2MnO–

4 + 6H+ + 5NO–

2 →

15. (i) ÛúÖò ¯»ÖêŒÃÖ [Pt(NH3)2Cl2] Ûêú •µÖÖ×´ÖŸÖßµÖ ÃÖ´ÖÖ¾ÖµÖ¾Ö ÛúÖê †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ …

(ii) ×ÛÎúÃ™ü»Ö ±úß»›ü ×ÃÖ¨üÖ®ŸÖ Ûêú †Ö¬ÖÖ¸ü ¯Ö¸ü µÖ×¤ü ∆o < P ÆüÖê ŸÖÖê d4 ÛúÖ ‡»ÖêŒ™ÒüÖò®Ö ×¾Ö®µÖÖÃÖ ×»Ö×ÜÖ‹ …

(iii) ÛúÖò ¯»ÖêŒÃÖ [Ni(CO)4] ÛúÖ ÃÖÓÛú¸üÞÖ †Öî ü “Öã ²ÖÛúßµÖ ¯ÖÏÛéú×ŸÖ ×»Ö×ÜÖ‹ … (¯Ö.ÃÖÓ. Ni = 28) 3

(i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2].

(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion

if ∆o < P.

(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].

(At.no. of Ni = 28)

56/1/1/D 6

16. ×®Ö´®Ö ÃÖê»Ö Ûúß 25 °C ¯Ö¸ü emf ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ :

Fe | Fe2+(0.001 M) || H+(0.01 M) | H2(g) (1 bar) | Pt(s)

E°(Fe2+ | Fe) = –0.44 V E°(H+ | H2) = 0.00 V 3

Calculate emf of the following cell at 25 °C :


E°(Fe2+ | Fe) = –0.44 V E°(H+ | H2) = 0.00 V

17. ×®Ö´®Ö †¾Ö»ÖÖêÛú®ÖÖë Ûêú ×»ÖµÖê ÛúÖ¸üÞÖ ¤üß×•Ö‹ :

(i) “Ö´ÖÔ¿ÖÖê¬Ö®Ö Ûêú ²ÖÖ¤ü “Ö´Ö›ÌüÖ ÃÖÜŸÖ ÆüÖê •ÖÖŸÖÖ Æîü …

(ii) ¦ü¾Ö ×¾Ö¸üÖê¬Öß ÃÖÖò»Ö Ûúß †¯ÖêõÖÖ ¦ü¾ÖÃ®ÖêÆüß ÃÖÖò»Ö †×¬ÖÛú Ã£ÖÖµÖß ÆüÖêŸÖÖ Æîü …

(iii) •Ö²Ö Æîü²Ö¸ü ¯ÖÏÛÎú´Ö «üÖ¸üÖ †´ÖÖê×®ÖµÖÖ ²Ö®ÖÖµÖÖ •ÖÖŸÖÖ Æîü ŸÖ²Ö CO ÛúÖê ¤æü¸ü ¸üÜÖ®ÖÖ †Ö¾Ö¿µÖÛú ÆüÖêŸÖÖ Æîü … 3


(i) Leather gets hardened after tanning.

(ii) Lyophilic sol is more stable than lyophobic sol.

(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process.

18. ×®Ö´®Ö ²ÖÆãü»ÖÛúÖë Ûêú ‹Ûú»ÖÛúÖë Ûêú ®ÖÖ´Ö ¾Ö ˆ®ÖÛúß ÃÖÓ ü“Ö®ÖÖµÖë ×»Ö×ÜÖ‹ :

(i) ®ÖÖµÖ»ÖÖò®Ö-6, 6

(ii) PHBV

(iii) ®Öß†Öê ÖÏß®Ö 3


(i) Nylon-6, 6

(ii) PHBV

(iii) Neoprene

19. ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë Ûêú ˆŸ¯ÖÖ¤üÖë Ûúß ¯ÖÏÖÝÖã×ŒŸÖ Ûúß×•Ö‹ :

(i) CH3 – C|C

=

H3

O (i) H2N – NH3

→(ii) KOH/Glycol, ∆

?

(ii) C6H5 – CO – CH3 NaOH/I2

→ ? + ?

(iii) CH3 COONa NaOH / CaO

→∆

? 3

56/1/1/D 7 [P.T.O.


(i) CH3 – C|C

=

H3

O (i) H2N – NH3


?


→ ? + ?


→∆

?

20. ×®Ö´®Ö ºþ¯ÖÖÓŸÖ¸üÞÖ †Ö¯Ö ÛîúÃÖê Ûú¸ëüÝÖê ?

(i) ±úß®ÖÖò»Ö ÛúÖê ‹ê×®ÖÃÖÖò»Ö ´Öë

(ii) ¯ÖÏÖê Öî®Ö-2-†Öò»Ö ÛúÖê 2-´Öê×£Ö»Ö¯ÖÏÖê Öî®Ö-2-†Öò»Ö ´Öë

(iii) ‹ê×®Ö»Öß®Ö ÛúÖê ±úß®ÖÖò»Ö ´Öë 3


(i) Phenol to anisole

(ii) Propan-2-ol to 2-methylpropan-2-ol

(iii) Aniline to phenol

†£Ö¾ÖÖ/OR

20. (a) ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ Ûúß ×ÛÎúµÖÖ-×¾Ö×¬Ö ÛúÖê ×»Ö×ÜÖ‹ :

2CH3CH2OH H+

→ CH3CH2 – O – CH2CH3

(b) ÃÖî×»Ö×ÃÖ×»ÖÛú ‹ê×ÃÖ›ü Ûêú ‹êÃÖß™üß»ÖßÛú¸üÞÖ ÃÖê ÃÖ´²Ö¨ü ÃÖ´ÖßÛú¸üÞÖ ÛúÖê ×»Ö×ÜÖ‹ … 3

(a) Write the mechanism of the following reaction :

2CH3CH2OH H+


(b) Write the equation involved in the acetylation of Salicylic acid.

21. (i) ×®Ö´®Ö ´Öë ÃÖê ÛúÖî®Ö ›üÖ‡ÔÃÖîÛêú¸üÖ‡›ü Æîü : Ã™üÖ“ÖÔ, ´ÖÖ»™üÖêÃÖ, ±ÏúŒ™üÖêÃÖ, Ý»ÖæÛúÖêÃÖ ?

(ii) ¸êü¿Öê¤üÖ¸ü ¯ÖÏÖê™üß®Ö †Öî ü ÝÖÖê»ÖÖÛúÖ¸ü ¯ÖÏÖê™üß®Ö ´Öë ŒµÖÖ †ÓŸÖ¸ü Æîü ?

(iii) ²Ö““ÖÖë ´Öë ×ÛúÃÖ ×¾Ö™üÖ×´Ö®Ö Ûúß Ûú´Öß Ûêú ÛúÖ¸üÞÖ Æüøüß ´Öë Ûãúºþ¯ÖŸÖÖ ÆüÖê •ÖÖŸÖß Æîü ? 3

(i) Which one of the following is a disaccharide : Starch, Maltose, Fructose,

Glucose ?

(ii) What is the difference between fibrous protein and globular protein ?

(iii) Write the name of vitamin whose deficiency causes bone deformities in children.

56/1/1/D 8

22. ×®Ö´®Ö Ûêú ÛúÖ¸üÞÖ ¤üß×•Ö‹ :

(a) t-²µÖæ×™ü»Ö ²ÖÎÖê ÖÖ‡›üü Ûúß †¯ÖêõÖÖ n-²µÖæ×™ü»Ö ²ÖÎÖê ÖÖ‡›ü ÛúÖ Œ¾Ö£Ö®ÖÖÓÛú ˆ““ÖŸÖ¸ü ÆüÖêŸÖÖ Æîü …

(b) ¸îüÃÖê×´ÖÛú ×´ÖÁÖÞÖ ¯ÖÏÛúÖ¿ÖÛúßµÖ ×®Ö×ÂÛÎúµÖ Æïü …

(c) ®ÖÖ×³ÖÛúÃ®ÖêÆüß ¯ÖÏ×ŸÖÃ£ÖÖ¯Ö®Ö †×³Ö×ÛÎúµÖÖ†Öë Ûêú ¯ÖÏ×ŸÖ Æîü»ÖÖê‹¸üß®ÖËÃÖ Ûúß ÃÖ×ÛÎúµÖŸÖÖ ²ÖœÌü •ÖÖŸÖß Æîü µÖ×¤ü o/p ¯ÖÖê•Öß¿Ö®Ö ¯Ö¸ü ®ÖÖ‡™ÒüÖê ÝÖÏã Ö (–NO2) ˆ¯Ö×Ã£ÖŸÖ ÆüÖê … 3

Give reasons :

(a) n-Butyl bromide has higher boiling point than t-butyl bromide.

(b) Racemic mixture is optically inactive.

(c) The presence of nitro group (–NO2) at o/p positions increases the reactivity of

haloarenes towards nucleophilic substitution reactions.

23. ‹Ûú ¯ÖÏ×ÃÖ¨ü ÃÛæú»Ö Ûêú ×¯ÖÏØÃÖ¯Ö»Ö ÁÖß ¸üÖµÖ ®Öê ‹Ûú ÃÖê×´Ö®ÖÖ¸ü ÛúÖ †ÖµÖÖê•Ö®Ö ×ÛúµÖÖ †Öî ü ˆÃÖ´Öë ˆ®ÆüÖë®Öê ²Ö““ÖÖë Ûêú

†×³Ö³ÖÖ¾ÖÛúÖë ŸÖ£ÖÖ †Öî ü ÃÛæú»ÖÖë Ûêú ×¯ÖÏØÃÖ¯Ö»ÖÖë ÛúÖê †Ö´ÖÓ×¡ÖŸÖ ×ÛúµÖÖ †Öî ü ÃÖ²Ö®Öê ×´Ö»ÖÛú¸ü ²Ö““ÖÖë ´Öë ´Ö¬Öã ÖêÆü ŸÖ£ÖÖ

ˆ¤üÖÃÖß •ÖîÃÖß ²Öß´ÖÖ×¸üµÖÖë Ûêú ²ÖœÌü®Öê Ûêú ÝÖÓ³Öß¸ü ×¾ÖÂÖµÖ ¯Ö¸ü ×¾Ö“ÖÖ¸-ü×¾Ö´Ö¿ÖÔ ×ÛúµÖÖ … ˆ®ÆüÖë®Öê ×®ÖÞÖÔµÖ ×ÛúµÖÖ ×Ûú ÃÛæú»ÖÖë ´Öë

ÃÖ›Ìêü Æãü‹ ÜÖÖª ¯Ö¤üÖ£ÖÔ ¯Ö¸ü ¯ÖÏ×ŸÖ²Ö®¬Ö »ÖÝÖÖµÖÖ •ÖÖµÖê †Öî ü Ã¾ÖÖÃ£µÖ¾Ö¬ÖÔÛú ¯Ö¤üÖ£ÖÔ •ÖîÃÖê ÃÖæ Ö, »ÖÃÃÖß, ¤æü¬Ö †Ö×¤ü ˆ¯Ö»Ö²¬Ö

Ûú¸üÖµÖÖ •ÖÖµÖ … ÃÖÖ£Ö Æüß ÃÛæú»ÖÖë ´Öë ¯ÖÏÖŸÖ:ÛúÖ»Öß®Ö ‹ÃÖê ²Ö»Öß Ûêú ÃÖ´ÖµÖ ²Ö““ÖÖë ÛúÖê †Ö¬Öê ‘ÖÓ™êü Ûúß ¿ÖÖ¸üß×¸üÛú ÛúÃÖ¸üŸÖ

Ûú¸üÖ‡Ô •ÖÖµÖê … ”û: ´ÖÖÆü Ûêú ¯Ö¿“ÖÖŸÖË ÁÖß ¸üÖµÖ ®Öê ‹Ûú Ã¾ÖÖÃ£µÖ ×®Ö¸üßõÖÞÖ ¯Öã®Ö: Ûú¸üÖµÖÖ †Öî ü ¤êüÜÖÖ ÝÖµÖÖ ×Ûú ²Ö““ÖÖë Ûêú

Ã¾ÖÖÃ£µÖ ´Öë †®Öã Ö´Ö ÃÖã¬ÖÖ¸ü Æãü†Ö Æîü …

ˆ¯Ö¸üÖêŒŸÖ ÛúÖê ¯ÖœÌü®Öê Ûêú ²ÖÖ¤ü ×®Ö´®Ö Ûêú ˆ¢Ö¸ü ¤üß×•Ö‹ :

(i) ÁÖß ¸üÖµÖ «üÖ¸üÖ ×Ûú®Ö ´Öæ»µÖÖë (Ûú´Ö ÃÖê Ûú´Ö ¤üÖê) ÛúÖê ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü ?

(ii) ‹Ûú ×¾ÖªÖ£Öá Ûêú ºþ¯Ö ´Öë ‡ÃÖ ×¾ÖÂÖµÖ ´Öë †Ö¯Ö ÛîúÃÖê •ÖÖÝÖºþÛúŸÖÖ ±îú»ÖÖµÖëÝÖê ?

(iii) ¿ÖÖ×®ŸÖÛúÖ¸üÛú ŒµÖÖ Æïü ? ‹Ûú ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ …

(iv) ‹Ã¯Öî™ìü´Ö ÛúÖ ˆ¯ÖµÖÖêÝÖ ŒµÖÖë šÓü›êü ³ÖÖê•µÖ ¯Ö¤üÖ£ÖÔ †Öî ü ¯ÖêµÖ ´Öë Æüß ÃÖß×´ÖŸÖ ÆüÖêŸÖÖ Æîü ? 4

Mr. Roy, the principal of one reputed school organized a seminar in which he invited

parents and principals to discuss the serious issue of diabetes and depression in

students. They all resolved this issue by strictly banning the junk food in schools and

to introduce healthy snacks and drinks like soup, lassi, milk etc. in school canteens.

They also decided to make compulsory half an hour physical activities for the students

in the morning assembly daily. After six months, Mr. Roy conducted the health survey

in most of the schools and discovered a tremendous improvement in the health of

students.

After reading the above passage, answer the following :

(i) What are the values (at least two) displayed by Mr. Roy ?


(iii) What are tranquilizers ? Give an example.

(iv) Why is use of aspartame limited to cold foods and drinks ?

56/1/1/D 9 [P.T.O.

24. (a) ×®Ö´®Ö Ûêú ÛúÖ¸üÞÖ ¤êüŸÖê Æãü‹ Ã¯ÖÂ™ü Ûúß×•Ö‹ :

(i) HF ÃÖê HI Ûúß †Öê ü †´»ÖßµÖ Ã¾Ö³ÖÖ¾Ö ²ÖœÌüŸÖÖ Æîü …

(ii) †ÖòŒÃÖß•Ö®Ö †Öî ü ÃÖ»±ú¸ü Ûêú ×Æü´ÖÖÓÛú †Öî ü Œ¾Ö£Ö®ÖÖÓÛú Ûêú ²Öß“Ö ²Ö›ÌüÖ †ÓŸÖ¸ü Æîü …

(iii) ®ÖÖ‡™ÒüÖê•Ö®Ö ¯Öê®™üÖÆîü»ÖÖ‡›ëü ®ÖÆüà ²Ö®ÖÖŸÖÖ Æîü …

(b) ×®Ö´®Ö Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ †Ö êü×ÜÖŸÖ Ûúß×•Ö‹ :

(i) Cl F3

(ii) XeF4 5


(i) Acidic character increases from HF to HI.

(ii) There is large difference between the melting and boiling points of oxygen

and sulphur.

(iii) Nitrogen does not form pentahalide.


(i) Cl F3

(ii) XeF4

†£Ö¾ÖÖ/OR

24. (i) ±úÖòÃ±úÖê üÃÖ ÛúÖ ÛúÖî®Ö ‹»ÖÖê™ÒüÖò Ö †×¬ÖÛú ÃÖ×ÛÎúµÖ Æîü †Öî ü ŒµÖÖë ?

(ii) ÃÖã Ö¸üÃÖÖê×®ÖÛú •Öê™ü ¯»Öê®Ö †Öê•ÖÖê®Ö ¯ÖŸÖÔ Ûêú †¾ÖõÖµÖ Ûêú ×»Ö‹ ÛîúÃÖê ×•Ö´´Öê¤üÖ¸ü Æïü ?

(iii) Cl2 Ûúß †¯ÖêõÖÖ F2 Ûúß †Ö²Ö®¬Ö ×¾ÖµÖÖê•Ö®Ö ‹®£Öî»¯Öß Ûú´Ö ŒµÖÖë Æîü ?

(iv) ´ÖÖîÃÖ´Ö×¾Ö–ÖÖ®Ö ´Öë †¾Ö»ÖÖêÛú®Ö Ûêú ×»ÖµÖê ²Öî»Öæ®ÖÖë ´Öë ³Ö¸ü®Öê Ûêú ×»ÖµÖê ×ÛúÃÖ ˆŸÛéúÂ™ü ÝÖîÃÖ ÛúÖ ¯ÖÏµÖÖêÝÖ ×ÛúµÖÖ •ÖÖŸÖÖ

Æîü ?

(v) ×®Ö´®Ö ÃÖ´ÖßÛú¸üÞÖ ÛúÖê ¯Öæ üÖ Ûúß×•Ö‹ :

XeF2 + PF5 → 5

(i) Which allotrope of phosphorus is more reactive and why ?

(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone

layers ?

(iii) F2 has lower bond dissociation enthalpy than Cl2. Why ?

(iv) Which noble gas is used in filling balloons for meteorological observations ?

(v) Complete the equation :

XeF2 + PF5 →

56/1/1/D 10

25. †ÞÖã ÃÖæ¡Ö C7H7ON ÛúÖ ‹Ûú ‹¸üÖê Öê×™üÛú µÖÖî×ÝÖÛú ‘A’ ®Öß“Öê ×¤üµÖê ÝÖµÖê Ûêú †®ÖãÃÖÖ¸ü ‹Ûú †×³Ö×ÛÎúµÖÖ ÀÖéÓÜÖ»ÖÖ ¤êüŸÖÖ

Æîü … ×®Ö´®Ö †×³Ö×ÛÎúµÖÖ†Öë ´Öë A, B, C, D †Öî ü E Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ ×»Ö×ÜÖ‹ : 5

An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of

reactions as shown below. Write the structures of A, B, C, D and E in the following

reactions :

†£Ö¾ÖÖ/OR

25. (a) •Ö²Ö ‹×®Ö»Öß®Ö ×®Ö´®Ö †×³ÖÛúÖ¸üÛúÖë Ûêú ÃÖÖ£Ö †×³Ö×ÛÎúµÖÖ Ûú¸üŸÖÖ Æîü ŸÖ²Ö ´ÖãÜµÖ ˆŸ¯ÖÖ¤üÖë Ûúß ÃÖÓ ü“Ö®ÖÖ‹Ñ ×»Ö×ÜÖ‹ :

(i) Br2 •Ö»Ö

(ii) HCl

(iii) (CH3CO)2O / pyridine

(b) ×®Ö´®Ö ÛúÖê ˆ®ÖÛêú Œ¾Ö£Ö®ÖÖÓÛú Ûêú ²ÖœÌüŸÖê ÛÎú´Ö ´Öë ×»Ö×ÜÖ‹ :

C2H5NH2, C2H5OH, (CH3)3N

(c) ×®Ö´®Ö µÖÖî×ÝÖÛúÖë Ûêú µÖãÝ´Ö ´Öë ¯ÖÆü“ÖÖ®Ö Ûú¸ü®Öê Ûêú ×»ÖµÖê ‹Ûú ÃÖÖ´ÖÖ®µÖ ¸üÖÃÖÖµÖ×®ÖÛú •ÖÖÑ“Ö ¤üß×•Ö‹ :

(CH3)2NH †Öî ü (CH3)3N 5

56/1/1/D 11 [P.T.O.

(a) Write the structures of main products when aniline reacts with the following

reagents :

(i) Br2 water

(ii) HCl


(b) Arrange the following in the increasing order of their boiling point :


(c) Give a simple chemical test to distinguish between the following pair of

compounds :

(CH3)2NH and (CH3)3N

26. •Ö»ÖßµÖ ×¾Ö»ÖµÖ®Ö ´Öë ´Öê×£Ö»Ö ‹êÃÖß™êü™ü Ûêú •Ö»Ö-†¯Ö‘Ö™ü®Ö Ûêú ×»ÖµÖê ×®Ö´®Ö ¯Ö×¸üÞÖÖ´Ö ¯ÖÏÖ¯ŸÖ ÆãüµÖê £Öê :

t/s 0 30 60

[CH3COOCH3]/mol L–1 0.60 0.30 0.15

(i) ×¤üÜÖ»ÖÖ‡‹ ×Ûú µÖÆü ‹Ûú ”û¤Ëü´Ö ¯ÖÏ£Ö´Ö ÛúÖê×™ü †×³Ö×ÛÎúµÖÖ Ûêú †®ÖãÃÖÖ¸ü Æîü ŒµÖÖë×Ûú •Ö»Ö ÛúÖ ÃÖÖÓ¦üÞÖ ×Ã£Ö¸ü ¸üÆüŸÖÖ

Æîü …

(ii) ÃÖ´ÖµÖ 30 ÃÖê 60 ÃÖêÛúÞ›ü Ûêú ²Öß“Ö †×³Ö×ÛÎúµÖÖ Ûúß †ÖîÃÖŸÖ ¤ü¸ü ¯Ö×¸üÛú×»ÖŸÖ Ûúß×•Ö‹ …

(×¤üµÖÖ ÝÖµÖÖ Æîü log 2 = 0.3010, log 4 = 0.6021) 5

For the hydrolysis of methyl acetate in aqueous solution, the following results were

obtained :

t/s 0 30 60

[CH3COOCH3]/mol L–1 0.60 0.30 0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water

remains constant.

(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(Given log 2 = 0.3010, log 4 = 0.6021)

†£Ö¾ÖÖ/OR

56/1/1/D 12

26. (a) ‹Ûú †×³Ö×ÛÎúµÖÖ A + B → P Ûêú ×»ÖµÖê ¤ü¸ü ¤üß ÝÖ‡Ô Æîü

¤ü¸ü = k[A] [B]2

(i) µÖ×¤ü B ÛúÖ ÃÖÖÓ¦üÞÖ ¤üÖê ÝÖã®ÖÖ Ûú¸ü ×¤üµÖÖ •ÖÖµÖê ŸÖÖê †×³Ö×ÛÎúµÖÖ Ûúß ¤ü¸ü ÛîúÃÖê ¯ÖÏ³ÖÖ×¾ÖŸÖ ÆüÖêÝÖß ?

(ii) µÖ×¤ü A ²ÖÆãüŸÖ †×¬ÖÛú ´ÖÖ¡ÖÖ ´Öë ´ÖÖî•Öæ¤ü ÆüÖê ŸÖÖê †×³Ö×ÛÎúµÖÖ Ûúß Ûãú»Ö ÛúÖê×™ü ŒµÖÖ Æîü ?

(b) ‹Ûú ¯ÖÏ£Ö´Ö ÛúÖê×™ü †×³Ö×ÛÎúµÖÖ 50% ¯Öæ üß ÆüÖê®Öê ´Öë 30 ×´Ö®Ö™ü »ÖêŸÖß Æîü … ‡ÃÖ †×³Ö×ÛÎúµÖÖ ÛúÖê 90% ¯ÖæÞÖÔ ÆüÖê®Öê ´Öë

•ÖÖê ÃÖ´ÖµÖ »ÖÝÖêÝÖÖ ˆÃÖÛúÖ ¯Ö×¸üÛú»Ö®Ö Ûúß×•Ö‹ …

(log 2 = 0.3010) 5

(a) For a reaction A + B → P, the rate is given by

Rate = k[A] [B]2

(i) How is the rate of reaction affected if the concentration of B is doubled ?

(ii) What is the overall order of reaction if A is present in large excess ?

(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time

required for 90% completion of this reaction.

(log 2 = 0.3010)

_________

Qu

es. Answers Marks

1 BaCl2 because it has greater charge / +2 charge ½ +½

2 X2Y3 1

3 3 1

4 2, 5 - dinitrophenol 1

5 CH3-CH2-Br

Because it is a primary halide / (10) halide

½ +½

6. When vapour pressure of solution is higher than that predicted by Raoult’s law /

the intermolecular attractive forces between the solute-solvent/(A-B) molecules are weaker than

those between the solute-solute and solvent-solvent molecules/A-A or B-B molecules.

Eg. ethanol-acetone/ethanol-cyclohexane/CS2-acetone or any other correct example

ΔmixH is positive

OR

(a)Azeotropes are binary mixtures having the same composition in the liquid and vapour phase

and boil at a constant temperature.

(b) Minimum boiling azeotrope

eg - ethanol + water or any other example

1

½

½

1

½

½

7.

(i)Ag+

(aq) + e- Ag (s)

Reaction with higher E0

value / ∆ G0

negative

(ii) Molar conductivity of a solution at infinite dilution or when concentration approaches

zero

Number of ions per unit volume decreases

½

½

½

½

8. Elements which have partially filled d-orbital in its ground states or any one of its oxidation

states.

1) Variable oxidation states

2) Form coloured ion

Or any other two correct characteristics

1

½ +½

9. 1) Diamminedichloridoethylenediaminechromium(III) chloride

2) [Co(NH3)5(ONO)]2+

1+ 1


DELHI -2015

SET -56/1/3/D

10 (i)LiAlH4 / NaBH4 /H2, Pt

(ii)KMnO4 , KOH

1

1

11 (i)Hexamethylene diamine NH2 (CH2)6 NH2 and

adipic acid HOOC- (CH2)4- COOH

(ii)3 hydroxybutanoic acid CH3CH(OH)CH2COOH and

3 hydroxypentanoic acid CH3CH2CH(OH)CH2COOH

(iii)Chloroprene H2C=C(Cl)CH=CH2

IUPAC names are accepted

Note : ½ mark for name /s and ½ mark for structure / s

½

½

½

½

½

½

12 (i)CH3CH2CH3

(ii) C6H5COONa + CHI3

(iii)CH4

1

½, ½

1

13

13

(i) C6H5OH + NaOH C6H5ONa CH3X C6H5OCH3

Or

C6H5OH + Na C6H5ONa CH3X C6H5OCH3

(ii)CH3CH(OH)CH3 CrO3 or Cu/573K CH3COCH3 (i)CH3MgX (CH3)2C(OH)CH3

(ii)H2O

(iii)C6H5NH2 NaNO2 + HCl C6H5N2Cl H2O warm C6H5OH

273K

OR

a)

b)

1

1

1

½

½

1

1

(Acetyl chloride instead of acetic anhydride may be used)

14 (i)Maltose

(ii) fibrous proteins: parallel polypeptide chain , insoluble in water

Globular proteins: spherical shape, soluble in water, (or any 1 suitable difference)

(iii) Vitamin D

1

1

1

15 (i)Larger surface area, higher van der Waals’ forces , higher the boiling point

(ii)Rotation due to one enantiomer is cancelled by another enantiomer

(iii) - NO2 acts as Electron withdrawing group or –I effect

1

1

1

16.

∆Tf = i Kf m

∆Tf = i Kf mb x1000

Mb x ma

1.62 K = i x 4.9K kg mol-1

x 3.9 g x 1000

122 gmol-1

49 kg

i = 0.506

Or by any other correct method

As i<1 , therefore solute gets associated.

½

1

½

1

17 (i) Zinc being low boiling will distil first leaving behind impurities/ or on electrolysis the pure

metal gets deposited on cathode from anode.

(ii)Silica acts as flux to remove iron oxide which is an impurity as slag or FeO + SiO2 FeSiO3

(iii)Wrought iron

1

1

1

18 d = z x M

a3 NA

z = d a3 NA

M

z = 2.7 g cm-3

x 6.022 x1023

mol-1

x ( 4.05 x 10-8

cm)3

27 g mol-1

= 3.999 ≈ 4

Face centered cubic cell/ fcc

½

1

½

1

19 (i) 5f orbital electrons have poor shielding effect than 4f

(ii)due to d-d transition / or the energy of excitation of an electron from lower d orbital to higher

d-orbital lies in the visible region /presence of unpaired electrons in the d-orbital.

(iii) 2 MnO4- + 6 H

+ + 5 NO2

- 2 Mn

2+ + 3 H2O + 5 NO3

-

1

1

1

20 (i)

(ii)t2 g3 e g

1

(iii) sp3 , diamagnetic

1

1

½+ ½

21 The cell reaction : Fe(s) + 2H+ (aq) Fe

2+ (aq) + H2(g)

Eocell = E

oc - E

oa

= [0-(-0.44)]V=0.44V

Ecell = Ecell - 0.059 log [ Fe2+

]

2 [ H+]2

Ecell = 0.44 V - 0.059 log ( 0.001 )

2 ( 0.01 ) 2

= 0.44 V - 0.059 log ( 10 )

2

= 0.44 V - 0.0295 V

=≈ 0.410 V

1

1

1

22 (i) mutual coagulation

(ii)strong interaction between dispersed phase and dispersion medium or solvated layer

(iii)CO acts as a poison for catalyst or iron

1

1

1

23 (i) Concern for students health, Application of knowledge of chemistry to daily life,

empathy , caring or any other

(ii)Through posters, nukkad natak in community, social media, play in assembly or any other

(iii)Tranquilizers are drugs used for treatment of stress or mild and severe mental disorders .. Eg:

equanil (or any other suitable example)

(iv) Aspartame is unstable at cooking temperature.

½, ½

1

½ , ½

1

o

24

24.

(a)

k = 2.303 log [ A0 ]

t [A]

k = 2.303 log 0.60

30 0.30

k = 2.303 x 0.301 = 0.023 s-1

30

k = 2.303 log 0.60

60 0.15

k = 2.303 x 0.6021 = 0.023 s-1

60

As k is constant in both the readings, hence it is a pseudofirst order reaction.

ii)

Rate = - Δ[R]/Δt

= -[0.15-0.30]

60-30

= 0.005 mol L-1

s-1

OR

a) (i) Rate will increase 4 times of the actual rate of reaction.

(ii) Second order reaction

b) t

1/2 = 0.693

k

30min =

0.693

k

1

½

½

1

½

½

1

1+1

½

k = 0.0231min

-1

k = 2.303 log [ A0 ]

t [A]

t = 2.303 log 100

0.0231 10

t = 2.303 min

0.0231

t = 99.7min

½

½

½

1

25

25

(a) (i) Due to decrease in bond dissociation enthalpy from HF to HI , there is an increase in acidic

character observed.

(ii)Oxygen exists as diatomic O2 molecule while sulphur as polyatomic S8

(iii)Due to non availability of d orbitals

(b)

OR

(i) White Phosphorus because it is less stable due to angular strain

(ii)Nitrogen oxides emitted by supersonic jet planes are responsible for depletion of ozone layer.

Or NO+O3 NO2+ O2

(iii)due to small size of F, large inter electronic repulsion / electron- electron repulsion among the

lone pairs of fluorine

1

1

1

1

1

½ , ½

1

1

(iv)Helium

(v) XeF2 + PF5 [XeF]+ [PF6]

- 1

1

26.

26.

A = B = C = D = E =

OR

O

a. i)

ii) iii)

b. ( CH3)3N < C2H5NH2 < C2H5OH

c. By Hinsberg test secondary amines ( CH3)2NH shows ppt formation which is insoluble

KOH

while

in tertiary amines ( CH3)3N do not react with benzene sulphonyl choride

1x5=

5

1

1

1

1

1

56/2/2/F 1 P.T.O.




Code No.

amob Z§. Roll No.


CHEMISTRY (Theory)












attempting it.





56/2/2/F

SET-2

56/2/2/F 2

gm_mÝ` ZXe :











1 mark each.


each.


3 marks each.



each.


1. XE JE `m¡JH$ H$m AmB©. y.nr.E.gr. Zm_ bIE : 1

HO CH2 CH = C CH3

CH3

56/2/2/F 3 P.T.O.


HO CH2 CH = C CH3

CH3














(i) CH3 CHO ?

CH3 CH CH3

|

OH

(ii) CH3 COOH ?

CH3 COCl


(i) CH3 CHO ?

CH3 CH CH3 |

OH

(ii) CH3 COOH ?

CH3 COCl

56/2/2/F 4


Cu2+ (aq) + 2e Cu(s) E0 = + 0·34 V

H+ (aq) + e 2

1 H2(g) E

0 = 0·00 V






Cu2+ (aq) + 2e Cu(s) E0 = + 0·34 V

H+ (aq) + e 2

1 H2(g) E

0 = 0·00 V




one application.


AWdm

EµOAmoQ´>mon H$mo n[a^mfV H$sOE & amCëQ> Ho$ Z`_ go G$UmË_H$ dMbZ Ûmam ~ZZo dmbm EµOAmoQ´>mon H$g àH$ma H$m hmoVm h¡ ? EH$ CXmhaU XrOE & 2

56/2/2/F 5 P.T.O.



OR







states and why ?

10. (i) ZåZbIV H$m°åßboŠg H$m AmB©.`y.nr.E.gr. Zm_ bIE :

[Cr (en)3]Cl3




[Cr (en)3]Cl3





0

)Zn/2Zn(E = 0.76 V, 0

)2

H/H(E = 0.00 V



0

)Zn/2Zn(E = 0.76 V, 0

)2

H/H(E = 0.00 V

56/2/2/F 6







water.


charcoal.












h¡ &



3–2

4MnO + 4H+ 3

56/2/2/F 7 P.T.O.



Mn2O7.


properties.


3–2

4MnO + 4H+


(i) CH3 CH = CH2 –

22

62

OH/OH3)ii

HB)i ?



?


(i) CH3 CH = CH2 –

22

62

OH/OH3)ii

HB)i ?



?




4 × 108 cm.

56/2/2/F 8







(i) Q>oarbrZ

(ii) ~¡Ho$bmBQ>

(iii) ~wZm-S


polymers :

(i) Terylene

(ii) Bakelite

(iii) Buna-S


(b) ã`wQ>¡Z-1-Am°b àH$meH$s` ZpîH«$` (Y«wdU AKyU©H$) h¡ naÝVw ã`wQ>¡Z-2-Am°b

àH$meH$s` gH«$` (Y«wdU KyU©H$) h¡ & Š`m| ?

(c) `Ún ŠbmoarZ BboŠQ´>m°Z H$mo AmH$f©V H$aZo dmbm J«wn h¡ \$a ^r `h BboŠQ´>m°ZñZohr Eamo_¡Q>H$ àVñWmnZ A^H«$`mAm| _| Am°Wm- VWm n¡am- ZXeH$ h¡ & Š`m| ? 3



active ?



reactions. Why ?

56/2/2/F 9 P.T.O.





AWdm




(iii) EQ>mS>© A^H«$`m How do you convert the following :




OR







(iii) Cg dQ>m_Z H$m Zm_ bIE OgH$s H$_r Ho$ H$maU _gy‹S>m| _| IyZ AmZo bJVm h¡ & 3




acids ?


gums.

56/2/2/F 10


C„oI H$sOE &




Nickel.








nwZ: ZarjU H$adm`m Am¡a ~ƒm| Ho$ ñdmñÏ` _| AZ wn_ gwYma nm`m J`m &

Cn w©º$ àH$aU H$mo n‹T>Zo Ho$ ~mX, ZåZbIV àíZm| Ho$ CÎma XrOE : 4 (i) lr Mmon ‹S>m Ûmam H$Z _yë`m| (H$_-go-H$_ Xmo) H$mo Xem©`m J`m h¡ ?













56/2/2/F 11 P.T.O.





a doctor ?



(i)






(ii) XeF2 5

AWdm


(i) XeF4

(ii) H2S2O7






(i) Bond angle in




56/2/2/F 12



(ii) XeF2

OR


(i) XeF4

(ii) H2S2O7






25. (a) àË`oH$ Ho$ bE Cn wº$ CXmhaU XoVo hþE ZåZbIV A^H«$`mAm| H$mo àXe©V H$sOE : (i) A_moZrH$aU (ii) H$pßb¨J (`w½_Z) A^H«$`m (iii) Eo_rZm| H$m EogrQ>brH$aU


AWdm




(i) HBF4 /

(ii) Cu / HBr

56/2/2/F 13 P.T.O.


5

(a) Illustrate the following reactions giving suitable example in each

case :

(i) Ammonolysis






OR


chloride (C6 H5–


(i) HBF4 /

(ii) Cu / HBr


56/2/2/F 14


(i) gH«$`U D$Om©

(ii) Xa pñWam§H$



(X`m J`m : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

AWdm


ln [R]

t (s)








(ii) Rate constant


Calculate t1/2

for the reaction.

(Given : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

OR

56/2/2/F 15 P.T.O.



ln [R]

t (s)






1





1


4 Rhombic sulphur 1

5 X2Y3 1

6 (i) CH3MgBr/ H3O+


1

1

7 a) Cu2+






½ , ½

1

1

8

8



∆mixH= negative

OR






1

½

½

1

½

½



electrons.

1

½ ,½


SET -56/2/2 F

2

10 i) tris-(ethane-1,2-diamine)chromium(III) chloride

ii) K3[ Cr(C2O4)3]

1

1

11 E cell = E

0 cell –

log

E cell = 0.76 V -

V log

( )

E cell = 0.76 – 0.0295 V log 10

= 0.7305 V

1

1

1




1

1

1

13

i)


ii) t2g4

iii) dsp 2

, diamagnetic

1

1

½ , ½



iii) 3MnO42-

+ 4H+ 2MnO4

- + MnO2 + 2H2O

1

1

1

15 i) CH3CH2CH2OH

ii)

iii) CH3CHO

1

1

1

3

16 d=

6.23 g cm-3

=

( )

z=4

fcc

½

½

1

1

17 ∆ Tf = i. Kf m

= i Kf wB x 1000

MB x wA

2K= 2 x 1.86K kg/mol x wB x 1000

58.5 g/mol x 37.2 g

wB = 1.17g

1

1

1

18

i)

ii)


iii)


1

1

1




1

1

1

20

i) C6H5COOH PCl5 C6H5COCl H2/Pd C6H5CHO

BaSO4


/H2SO4 CH3CHO

iii) CH3COOH NaOH


CH4

OR

1

1

1

4

20

i)

ii)

iii)

1

1

1

21 i) Fructose



iii) Vitamin C

1

1

1





1

1

1




relevant answer)



½ , ½

1

1

½+ ½

24

a) i)Because of lone pair in NH3 , lone pair- bond pair repulsion decreases the bond angle



b)

1

1

1

1,1

5

24

OR

a)





1,1

1

1

1

25

a) i)ammonolysis

ii)

(any one)

1

1

6

25



(soluble in alkali)




OR

a) i)

ii)

1

1

1

1

1

½,½,

½

7

b) i) A- B- C-


½ ,½,

½

26

26

a)i) Activation energy- Extra energy required by reactants to form activated complex.


b)

k= 2.303 log [ A0 ]

t [A]

k = 2.303 log 100

10 min 75

k = 2.303 x 0.125

10 min

k = 0.02879 min-1

t1/2 =

=

t1/2 = 24.07min

OR


b)

t =

log

t99% =

log

t =

x 2

1

1

½

½

1

1

1,1,1

½

8

t90% =

log

=

t99% = 2 x t90%

½

1

56/1/2/D 1 [P.T.O.


Roll No.













56/1/2/D
















SET – 2

56/1/2/D 2











C|

C|

B

H3

–

r

CH3



C|

C|

B

H3

–

r

CH3

2. BaCl2 †Öî ü KCl ´Öë ÃÖê ÛúÖî®Ö ŠúÞÖÖŸ´ÖÛú “ÖÖ•ÖÔ ¾ÖÖ»Öê ÛúÖê»ÖÖ‡›üß ÃÖÖò»Ö ÛúÖ ÃÛÓú¤ü®Ö †×¬ÖÛú ¯ÖÏ³ÖÖ¾Ö¿ÖÖ»Öß œÓüÝÖ ÃÖê

Ûú¸êüÝÖÖ ? ÛúÖ¸üÞÖ ¤üß×•Ö‹ … 1



3. ˆÃÖ µÖÖî×ÝÖÛú ÛúÖ ÃÖæ¡Ö ŒµÖÖ ÆüÖêÝÖÖ ×•ÖÃÖ´Öë Y ŸÖ¢¾Ö ccp •ÖÖ»ÖÛú ²Ö®ÖÖŸÖÖ Æîü †Öî ü X “ÖŸÖãÂ±ú»ÖÛúßµÖ ×¸ü×ŒŸÖ ÛúÖ 1/3¾ÖÖÑ

³ÖÖÝÖ ‘Öê üŸÖÖ Æîü ? 1




56/1/2/D 3 [P.T.O.







(i) CH3 – CO – CH3 ?

→ CH3 – C|

O

H

H

– CH3

(ii) C6H5 – CH2 – CH3

?→ C6H5 – COO–K+

7. ¸üÖˆ»™ü ×®ÖµÖ´Ö ÃÖê ¬Ö®ÖÖŸ´ÖÛú ×¾Ö“Ö»Ö®Ö ÃÖê ŒµÖÖ ŸÖÖŸ¯ÖµÖÔ Æîü ? ‹Ûú ˆ¤üÖÆü¸üÞÖ ¤üß×•Ö‹ … ¬Ö®ÖÖŸ´ÖÛú ×¾Ö“Ö»Ö®Ö Ûêú ×»ÖµÖê

∆mixH ÛúÖ ×“ÖÅ®Öü ŒµÖÖ Æîü ? 2



†£Ö¾ÖÖ/OR





56/1/2/D 4


Ag+(aq) + e– → Ag(s) E° = +0.80 V

H+(aq) + e– → 1

2H2(g) E° = 0.00 V






chloride solution :

Ag+(aq) + e– → Ag(s) E° = +0.80 V

H+(aq) + e– → 1

2H2(g) E° = 0.00 V







10. (i) ÛúÖò ¯»ÖêŒÃÖ [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine) ÛúÖ †Ö‡Ô.µÖæ.¯Öß.‹.ÃÖß. ®ÖÖ´Ö

×»Ö×ÜÖµÖê …







56/1/2/D 5 [P.T.O.






if ∆o < P.


(At.no. of Ni = 28)



E°(Fe2+ | Fe) = –0.44 V E°(H+ | H2) = 0.00 V 3



E°(Fe2+ | Fe) = –0.44 V E°(H+ | H2) = 0.00 V











(ii) PHBV



(i) Nylon-6, 6

(ii) PHBV

(iii) Neoprene

56/1/2/D 6


(i) CH3 – C|C

=

H3

O (i) H2N – NH3


?


→ ? + ?


→∆

? 3


(i) CH3 – C|C

=

H3

O (i) H2N – NH3


?


→ ? + ?


→∆

?









†£Ö¾ÖÖ/OR


2CH3CH2OH H+




2CH3CH2OH H+



56/1/2/D 7 [P.T.O.





Glucose ?







Give reasons :









or dissociated).








56/1/2/D 8

21. ´ÖÖê»Ö¸ü ¦ü¾µÖ´ÖÖ®Ö 27 g mol–1 Ûêú ÃÖÖ£Ö ‹Ûú ŸÖŸ¾Ö ÛúÖê ü »Ö´²ÖÖ‡Ô 4.05 × 10–8 cm ¾ÖÖ»Öß ‹Ûú ŒµÖæ×²ÖÛú µÖæ×®Ö™ü

ÃÖê»Ö ²Ö®ÖÖŸÖÖ Æîü … µÖ×¤ü ‘Ö®ÖŸ¾Ö 2.7 g cm–3 ÆüÖê, ŸÖÖê ŒµÖæ×²ÖÛú µÖæ×®Ö™ü ÃÖê»Ö ÛúÖ Ã¾Öºþ¯Ö ŒµÖÖ Æîü ? 3







2MnO–

4 + 6H+ + 5NO–

2 →





2MnO–

4 + 6H+ + 5NO–

2 →













56/1/2/D 9 [P.T.O.








students.










reactions :

†£Ö¾ÖÖ/OR

56/1/2/D 10


(i) Br2 •Ö»Ö

(ii) HCl







reagents :

(i) Br2 water

(ii) HCl





compounds :



t/s 0 30 60

[CH3COOCH3]/mol L–1 0.60 0.30 0.15


Æîü …



56/1/2/D 11 [P.T.O.


obtained :

t/s 0 30 60

[CH3COOCH3]/mol L–1 0.60 0.30 0.15


remains constant.


(Given log 2 = 0.3010, log 4 = 0.6021)

†£Ö¾ÖÖ/OR


¤ü¸ü = k[A] [B]2





(log 2 = 0.3010) 5


Rate = k[A] [B]2





(log 2 = 0.3010)






(i) Cl F3

(ii) XeF4 5

56/1/2/D 12




and sulphur.



(i) Cl F3

(ii) XeF4

†£Ö¾ÖÖ/OR





Æîü ?


XeF2 + PF5 → 5



layers ?




XeF2 + PF5 →

_________

Qu


1 CH3-CH2-Br


½ +½


3 X2Y3 1

4 3 1


6. (i)LiAlH4 / NaBH4 /H2, Pt

(ii)KMnO4 , KOH

1

1

7.

7.

When vapour pressure of solution is higher than that predicted by Raoult’s law /




ΔmixH is positive

OR





1

½

½

1

½

½

8. (i)Ag+

(aq) + e- Ag (s)


value / ∆ G0

negative


zero


½

½

½

½


states.




1

½ +½

10 1) Diamminedichloridoethylenediaminechromium(III) chloride

2) [Co(NH3)5(ONO)]2+

1+ 1


DELHI -2015

SET -56/1/2/D

11 (i)

(ii)t2 g3 e g

1


1

1

½+ ½


2+ (aq) + H2(g)

Eocell = E

oc - E

oa

= [0-(-0.44)]V=0.44V


]

2 [ H+]2

Ecell = 0.44 V - 0.059 log ( 0.001 )

2 ( 0.01 ) 2

= 0.44 V - 0.059 log ( 10 )

2

= 0.44 V - 0.0295 V

=≈ 0.410 V

1

1

1



(iii)CO acts as a poison for catalyst.

1

1

1








½

½

½

½

½

½

15 (i)CH3CH2CH3


(iii)CH4

1

½, ½

1

o

16.

16.


Or



(ii)H2O


273K

OR

a)

b)


1

1

1

½

½

1

1

17 (i)Maltose



(iii) Vitamin D

1

1

1




1

1

1

19

∆Tf = i Kf m

∆Tf = i Kf mb x1000

Mb x ma

1.62 K = i x 4.9K kg mol-1

x 3.9 g x 1000

122 gmol-1

49 kg

i = 0.506



½

1

½

1




(iii)Wrought iron

1

1

1

21 d = z x M

a3 NA

z = d a3 NA

M

z = 2.7 g cm-3

x 6.022 x1023

mol-1

x ( 4.05 x 10-8

cm)3

27 g mol-1

= 3.999 ≈ 4


½

1

½

1




(iii) 2 MnO4- + 6 H

+ + 5 NO2

- 2 Mn

2+ + 3 H2O + 5 NO3

-

1

1

1







½, ½

1

½ , ½

1

24

24.

A = B = C = D = E =

OR

O

a. i)

ii) iii)


c. By Hinsberg test secondary amines ( CH3)2NH shows ppt formation which is

tertiary insoluble in KOH , tertiary amines ( CH3)3N do not react with benzene sulphonyl choride

1x5=

5

1

1

1

1

1

25

(a)

k = 2.303 log [ A0 ]

t [A]

k = 2.303 log 0.60

30 0.30

k = 2.303 x 0.301 = 0.023 s-1

30

k = 2.303 log 0.60

60 0.15

k = 2.303 x 0.6021 = 0.023 s-1

60

1

½

½

1

25.


ii)

Rate = - Δ[R]/Δt

= -[0.15-0.30]

60-30

= 0.005 mol L-1

s-1

OR

a) (i) Rate will increase 4 times of the actual rate of reaction.


b) t

1/2 = 0.693

k

30min =

0.693

k

k = 0.0231min-1

k = 2.303 log [ A0 ]

t [A]

t = 2.303 log 100

0.0231 10

t = 2.303 min

0.0231

t = 99.7min

½

½

1

1+1

½

½

½

½

1

26.

26.


character observed.



(b)

OR



Or NO+O3 NO2+ O2



(iv)Helium


-

1

1

1

1

1

½ , ½

1

1

1

1

56/2/1/F 1 P.T.O.




Code No.

amob Z§. Roll No.


CHEMISTRY (Theory)












attempting it.





56/2/1/F

SET-1

56/2/1/F 2

gm_mÝ` ZXe :











1 mark each.


each.


3 marks each.



each.






56/2/1/F 3 P.T.O.



3. XE JE `m¡JH$ H$m AmB©. y.nr.E.gr. Zm_ bIE : 1

HO CH2 CH = C CH3

CH3


HO CH2 CH = C CH3

CH3







6. (i) ZåZbIV H$m°åßboŠg H$m AmB©. y.nr.E.gr. Zm_ bIE :

[Cr (en)3]Cl3




[Cr (en)3]Cl3



56/2/1/F 4


AWdm

EµOAmoQ´>mon H$mo n[a^mfV H$sOE & amCëQ> Ho$ Z`_ go G$UmË_H$ dMbZ Ûmam ~ZZo dmbm EµOAmoQ´>mon H$g àH$ma H$m hmoVm h¡ ? EH$ CXmhaU XrOE & 2



OR




(i) CH3 CHO ?

CH3 CH CH3

|

OH

(ii) CH3 COOH ?

CH3 COCl


(i) CH3 CHO ?

CH3 CH CH3 |

OH

(ii) CH3 COOH ?

CH3 COCl


Cu2+ (aq) + 2e Cu(s) E0 = + 0·34 V

H+ (aq) + e 2

1 H2(g) E

0 = 0·00 V



56/2/1/F 5 P.T.O.




Cu2+ (aq) + 2e Cu(s) E0 = + 0·34 V

H+ (aq) + e 2

1 H2(g) E

0 = 0·00 V




one application.





states and why ?






56/2/1/F 6


(i) Q>oarbrZ

(ii) ~¡Ho$bmBQ>

(iii) ~wZm-S


polymers :

(i) Terylene

(ii) Bakelite

(iii) Buna-S



(iii) Cg dQ>m_Z H$m Zm_ bIE OgH$s H$_r Ho$ H$maU _gy ‹S>m| _| IyZ AmZo bJVm h¡ & 3




acids ?


gums.


C„oI H$sOE &




Nickel.



56/2/1/F 7 P.T.O.



0

)Zn/2Zn(E = 0.76 V, 0

)2

H/H(E = 0.00 V



0

)Zn/2Zn(E = 0.76 V, 0

)2

H/H(E = 0.00 V







water.


charcoal.










56/2/1/F 8


(b) ã`wQ>¡Z-1-Am°b àH$meH$s` ZpîH«$` (Y«wdU AKyU©H$) h¡ naÝVw ã`wQ>¡Z-2-Am°b àH$meH$s` gH«$` (Y«wdU KyU©H$) h¡ & Š`m| ?

(c) `Ún ŠbmoarZ BboŠQ´>m°Z H$mo AmH$f©V H$aZo dmbm J«wn h¡ \$a ^r `h BboŠQ´>m°ZñZohr

Eamo_¡Q>H$ àVñWmnZ A^H«$`mAm| _| Am°Wm- VWm n¡am- ZXeH$ h¡ & Š`m| ? 3



active ?



reactions. Why ?





AWdm




(iii) EQ>mS>© A^H«$`m How do you convert the following :




OR





56/2/1/F 9 P.T.O.



h¡ &



3–2

4MnO + 4H+ 3



Mn2O7.


properties.


3–2

4MnO + 4H+


(i) CH3 CH = CH2 –

22

62

OH/OH3)ii

HB)i ?



?


(i) CH3 CH = CH2 –

22

62

OH/OH3)ii

HB)i ?



?

56/2/1/F 10




4 × 108 cm.






nwZ: ZarjU H$adm`m Am¡a ~ƒm| Ho$ ñdmñÏ` _| AZwn_ gwYma nm`m J`m &

Cn w©º$ àH$aU H$mo n‹T>Zo Ho$ ~mX, ZåZbIV àíZm| Ho$ CÎma XrOE : 4

(i) lr Mmon ‹S>m Ûmam H$Z _yë`m| (H$_-go-H$_ Xmo) H$mo Xem©`m J`m h¡ ?













56/2/1/F 11 P.T.O.





a doctor ?



(i) gH«$`U D$Om©

(ii) Xa pñWam§H$



(X`m J`m : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

AWdm


ln [R]

t (s)






56/2/1/F 12



(ii) Rate constant


Calculate t1/2

for the reaction.

(Given : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

OR



ln [R]

t (s)







(i)






(ii) XeF2 5

AWdm

56/2/1/F 13 P.T.O.


(i) XeF4

(ii) H2S2O7






(i) Bond angle in






(ii) XeF2

OR


(i) XeF4

(ii) H2S2O7






56/2/1/F 14

26. (a) àË`oH$ Ho$ bE Cn wº$ CXmhaU XoVo hþE ZåZbIV A^H«$`mAm| H$mo àXe©V H$sOE :

(i) A_moZrH$aU

(ii) H$pßb¨J (`w½_Z) A^H«$`m

(iii) Eo_rZm| H$m EogrQ>brH$aU


AWdm




(i) HBF4 /

(ii) Cu / HBr


5 (a) Illustrate the following reactions giving suitable example in each

case :

(i) Ammonolysis






56/2/1/F 15 P.T.O.

OR


chloride (C6 H5–


(i) HBF4 /

(ii) Cu / HBr


1



2 Rhombic sulphur 1


4 X2Y3 1



1

6. i) tris-(ethane-1,2-diamine)chromium(III) chloride

ii) K3[ Cr(C2O4)3]

1

1

7.

7.



∆mixH= negative

OR






1

½

½

1

½

½

8. (i) CH3MgBr/ H3O+


1

1

9. a) Cu2+






½ , ½

1

1


SET -56/2/1 F

2



electrons.

1

½ ,½

11 ∆ Tf = i. Kf m

= i Kf wB x 1000

MB x wA

2K= 2 x 1.86K kg/mol x wB x 1000

58.5 g/mol x 37.2 g

wB = 1.17g

1

1

1

12

i)

ii)


iii)


1

1

1

13 i) Fructose



iii) Vitamin C

1

1

1





1

1

1

3

15 E cell = E

0 cell –

log

E cell = 0.76 V -

V log

( )

E cell = 0.76 – 0.0295 V log 10

= 0.7305 V

1

1

1




1

1

1

17

i)


ii) t2g4

iii) dsp 2

, diamagnetic

1

1

1/2 , ½




1

1

1

19

19.

i) C6H5COOH PCl5 C6H5COCl H2/Pd C6H5CHO

BaSO4


/H2SO4 CH3CHO

iii) CH3COOH NaOH


CH4

OR

i)

ii)

iii)

1

1

1

1

1

1

4



iii) 3MnO42-

+ 4H+ 2MnO4

- + MnO2 + 2H2O

1

1

1

21 i) CH3CH2CH2OH

ii)

iii) CH3CHO

1

1

1

22 d=

6.23 g cm-3

=

( )

z=4

fcc

½

½

1

1




relevant answer)



½ , ½

1

1

½+ ½

24

a)i) Activation energy- Extra energy required by reactants to form activated complex.


b)

k= 2.303 log [ A0 ]

t [A]

k = 2.303 log 100

10 min 75

k = 2.303 x 0.125

10 min

1

1

½

½

5

24.

k = 0.02879 min-1

t1/2 =

=

t1/2 = 24.07min

OR


b)

t =

log

t99% =

log

t =

x 2

t90% =

log

=

t99% = 2 x t90%

1

1

1,1,1

½

½

1

25

a) i)Because of lone pair in NH3 , lone pair- bond pair repulsion decreases the bond angle



b)

OR

1

1

1

1,1

6

25.

a)





1,1

1

1

1

26

a) i)ammonolysis

ii)

(any one)


1

1

1

7

Sr. Name Sr. Name

26.


(soluble in alkali)




OR

a) i)

ii)

b) i) A- B- C-


1

1

1

1

½,½,

½

½ ,½,

½

8

No. No.