CHEMISTRY PAPER – COMPREHENSION-I

CHEMISTRY

 

Comprehension-I

 

Read the paragraph and answer the questions (18 – 20) given below:

 

Carbene is an important intermediate and can be formed in many reactions. One of these is             elimination of gem dihalide by active metal. Carbon is divalent in carbene and highly reactive. The two non–bonding electrons decide the shape and the bond angle of this intermediate. One of these is called singlet carbene (single orientation in magnetic field) and other is called triplet carbene (three orientations in magnetic field). Carbon of carbene is electron deficient and will try to get stabilized by reactions such as insertion reactions in alkanes and cyclo–addition reactions in alkenes.

 

18. In which of the following singlet carbene is most stable?

(A)                                                           (B)

(C)                                                          (D)

 

19. Which carbene will have greater bond angle?

(A)  Singlet                                                        (B)  Triplet

(C)  Both have similar bond angle                       (D)  Bond angle cannot be measured

 

20. When triplet carbene reacts with cis2–butene, how many products are possible?

(A)  1                                                                (B)  2

(C)  3                                                                (D)  4

 

18. B
19. B
20. D

 

 

Comprehension-II

 

Read the paragraph and answer the questions (21 – 23) given below:

 

Emission of electrons from metal surface under the influence of light radiation of appropriate  leaving the metal positively charge is called photoelectric effect.

Work function may be defined as the minimum amount of energy required to eject electron from metal surface, also known as threshold energy.

According to Einstein,

Maximum K.E. of ejected electrons = absorbed energy – work function

 

21. The wavelength and frequency of a proton having an energy of 1 eV_o

(A)  22.42 × 10^–7 M, 3.41 × 10¹⁰ sec                     (B)  2.2 × 10^–7 M, 4.2 × 10¹⁴ sec^–1

(C)  12.42 × 10^–7 M, 2.41 × 10¹⁴ sec                     (D)  2.2 M, 4.2 sec^–1

 

22. Energy required to stop the ejection of ejections from Cu plate is 0.24 eV. The work function when radiation of strikes the plate

(A)  7.835 × 10^–19 J                                             (B)  8.219 × 10^–19 J

(C)  7.45 × 10^–19 J                                               (D)  7.45 × 10^–21 J

 

23. Which is/are the correct graph?

(A)		(B)
(C)		(D)	None of these

 

21. C
22. C
23. B

 

 

Comprehension-III

 

Read the paragraph and answer the questions (24 – 26) given below:

 

K₂Cr₂O₇ is good oxidising agent in acidic medium.

In alkaline solution, orange colour of  changes to yellow colour due to formation of  and again yellow colour changes to orange colour in acid medium.

Then  and  exist in equilibrium are interconvertable by altering pH of solution. When heated with H₂SO₄ and soluble chloride K₂Cr₂O₇ gives vapours of CrO₂Cl₂. CrO₂Cl₂ when passed into aqueous NaOH solution yellow colour solution of  is obtained. This on reaction with lead acetate gives yellow ppt of PbCrO₄.

When H₂O₂ is added to an acidified solution of dichromate ion a complicated reaction occurs. The products depend on the pH and concentration of Cr.

A deep blue–violet coloured peroxo compound CrO(O₂)₂ called chromic peroxide is formed. This decomposes rapidly in aqueous solution into aqueous solution into Cr³⁺ and oxygen.

 

24. Which of the following statements is not correct when a mixture of NaCl and K₂Cr₂O₇ is gently warmed with conc. H₂SO₄?

(A)  A deep red vapour is evolved

(B)  The vapour when passed through NaOH solution gives a yellow solution of Na₂CrO₄.

(C)  Chlorine gas is formed

(D) Chromyl chloride is formed

 

25. The equivalent mass of KIO₃ in the reaction

 

(A)  Molecular mass                                           (B)

(C)                                                                (D)

 

26. When H₂O₂ is added to an acidified solution of K₂Cr₂O₇

(A)  Solution turns green due to formation of Cr₂O₃.

(B)  Solution turns yellow due to formation of K₂CrO₄

(C)  A deep blue–violet coloured compound CrO(O₂)₂ is formed

(D)  Solution gives green ppt of Cr(OH)₃

 

24. C
25. C
26. C

 

 

Comprehension-IV

 

Read the paragraph and answer the questions (27 – 29) given below:

 

Volumetric titrations involving KMnO₄ are carried out only in presence of dilute H₂SO₄ but not in the presence of HCl or HNO₃. This is because oxygen produced from KMnO₄ + dil. H₂SO₄ is used only for oxidizing the reducing agent. Moreover, H₂SO₄ does not give any oxygen of its own to oxidize the reducing agent. In case HCl is used, the oxygen produced from KMnO₄ + HCl is partly used up to oxidize HCl and in case HNO₃ is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. KMnO₄ in various mediums gives following products–

 

27. (1 mole) in neutral medium disproportionates to,

(A)           (B)

(C)          (D)

 

28. Number of electrons transferred in each case when KMnO₄ acts as an oxidizing agent to give MnO₂, Mn⁺⁺, Mn(OH)₃ and are respectively–

(A)  3, 5, 4, 1                                                     (B)  4, 3, 1, 5

(C)  1, 3, 4, 5                                                     (D)  5, 4, 3, 1

 

29. The number of moles of KMnO₄ that will be needed to react with one mole of sulphite in an acidic solution is–

(A)                                                                (B)

(C)                                                                (D)  1

 

27. A
28. A
29. A

 

 

Numerical Based Questions

 

30. Calculate the amount of propene (in g) formed from 34 g of 1-iodo propane by reaction with alcoholic KOH if yield is 36%. Give your answer upto two decimal points.

170 g of 1-iodo propane gives = 42 g of propene.

34 g of 1-iodo propane gives

= 8.4 g if yield is 100%

Since yield is 36%

Amount of propene formed

= 3.024 g

30. 3.02

 

31. It required 0.7 g of hydrocarbon (A) to react completely with 2.0 g of Br₂. On treatment of ‘A’ with HBr it gave monobromo alkane (B). The same compound (B) was obtained when (A) was treated with HBr in presence of peroxide. Calculate the molecular weight of ‘B’ (in g mol^–1).
32. 137
33. 2.0 g of Br₂ requires = 0.7 g of hydrocarbon

160 g of Br₂ requires

The hydrocarbon ‘A’ is C₄H₈, i.e. butene.

Since, it gives same product with HBr in presence as well as absence of peroxide, therefore, it is But-2-ene

 

The molecular weight of 2-bromobutane is C₄H₉Br(4 × 12 + 9 + 80) = 48 + 9 + 80 = 137 g mol^-1

 

32. The H–O-H, bond angle in water molecule is 105°, the H–O bond distance being 0.94. The dipole moment for the molecule is 1.85 D. The charge on the oxygen atom [cos 105° = – 0.258] is ……………………. ´ 10^–10 esu cm. Give your answer upto two decimal points
33. 3.23

Since, H₂O has two vectors of

O–H bond acting at 105°, let dipole moment of O–H be ‘a’

i.e.

1.52´10^–18 =0.94´10^–8

d = 1.617´10^–10 esu

Since ‘O’ atom acquires 2d  charges. One d charge from each bond. Thus charge on ‘O’ atom
= 2d

= 2 ´ 1.617´10^–10

= 3.23´10^–10 esu cm

 

33. Find the quantum numbers of excited state of electrons in He⁺ ion which on transition to ground state and first excited state emit two photons of wavelengths, 30.4 nm and 108.5 nm respectively.
34. 5
35. For transition to ground state, n₁ = 1, n₂ = ?

 

This gives, n₂ = 2

For n₁ = 2 (first excited state), n₂ = ?

 

This gives, n₂ = 5

 

34. 1 ml solution containing (NH₄)₂ SO₄ was treated with excess of NaOH. The NH₃ released was absorbed in 50 ml of 0.1 N HCl solution. The solution after passage of NH₃ in it requires 20 ml of 0.1 N NaOH for complete neutralization. Calculate the strength of (NH₄)₂SO₄ solution (in g/L).
35. 198