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Chemistry Mains – Answers

1.a)Ave. Mol. wt = 92

b) Ammineaquanitropyridineplatinum(II)nitrate

c) TeF₅-hybridisation sp³d² and shape is square pyramidal.

 

2.a) 2.92

b)Cu²⁺ has configuration ….3d⁹4s⁰. So it seems there is no completely vacant ‘3d’ orbital to take part in dsp² hybridisation. But in the presence of strong ligand(NH₃ is moderately strong field ligand) the unpaired electron from the 3d orbital is promoted to vacant 4p orbital and the vacant 3d orbital is available for hybridisation. So Cu²⁺ can undergo dsp² hybridisation after promotion of unpaired electron.

 

3.       A: Ca₂B₆O₁₁

         B: Na₂B₄O₇

         C: NaBO₂

         D: B₂O₃

         E: Cr₂(SO₄)₃

         F: Na₂[(OH)₂B(O-O)₂B(OH)₂].6H₂O

 

4.a) 58.94 gm/cc

b)The emf of the cell is decreased by 0.3245 V

 

5.

 

6.  a) 1.35 * 10⁵

     b) 0.4114 M

 

7.  b)  A: PhS-CH₂-CH=CMe₂

         B: CH₂=CH-CMe=CH₂ 

    c) A,B and C are ortho, meta and para Chlorotoluene.

         X is 0-toluidine, Y is m-toluidine and Z is p-toluidine.

 

8. A: CH₃I

B: C₂H₆

9. A: CH₂(COOH)₂

B: CH₃COOH

C: CH₃COCH₃

D: CH₃-CH₂-CH(CH₃)-CH₂-CH=C(CH₃)-CH₃

E: CH₃-CH₂-CH(CH₃)-CH₂-CHO

F: CH₃-CH₂-CH(CH₃)-CH₂-CH₂OH

G: CH₃-CH₂-C(CH₃)=CH-CH₃

H: CH₃CH₂COCH₃

I  : CH₃CHO

    10.  pT=14.085