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CHEMISTRY PAPER – COMPREHENSION-I

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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.

 

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

(A)                                                           (B)

(C)                                                          (D)

 

  1. Which carbene will have greater bond angle?

(A)  Singlet                                                        (B)  Triplet

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

 

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

(A)  1                                                                (B)  2

(C)  3                                                                (D)  4

 

  1. B
  2. B
  3. 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

 

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

(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

 

  1. 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

 

  1. Which is/are the correct graph?
(A) (B)
(C) (D) None of these

 

  1. C
  2. C
  3. 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.

 

  1. 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

 

  1. The equivalent mass of KIO3 in the reaction

 

(A)  Molecular mass                                           (B)

(C)                                                                (D)

 

  1. 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

 

  1. C
  2. C
  3. 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–

 

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

(A)           (B)

(C)          (D)

 

  1. 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

 

  1. 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

 

  1. A
  2. A
  3. A

 

 

Numerical Based Questions

 

  1. 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

  1. 3.02

 

  1. 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 mol1).
  2. 137
  3. 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

 

  1. 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 ……………………. ´ 1010 esu cm. Give your answer upto two decimal points
  2. 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´1018 =0.94´108

d = 1.617´1010 esu

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

= 2 ´ 1.617´1010

= 3.23´1010 esu cm

 

  1. 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.
  2. 5
  3. For transition to ground state, n1 = 1, n2 = ?

 

This gives, n2 = 2

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

 

This gives, n2 = 5

 

  1. 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).
  2. 198

 

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