waecE-LEARNING
Chemistry Paper 2,Nov/Dec 2012  
Questions:   1 2 3 4 7 8   Main
General Comments
Weakness/Remedies
Strength







































































Question 4

  1. Question 4

    1. (i)         Define hybridization?

    (ii)        State the type(s) of hybridization exhibited by carbon in propene.
                (iii)      Mention the shape of the

      1.  s-orbital;
      2. p-orbital.
    1. Diamond and graphite are both allotropes of carbon.  Diamond is hard while graphite is soft.  Explain briefly this observation.              
    2. A monochloroalkane was analysed by converting all the chlorine into chloride ions and precipitating the chloride as silver chloride.  If 0.08226g of the chloroalkane gave 0.1837g of silver chloride,
    3.  calculate the molecular mass of the chloroakane;

    (ii)  determine the molecular formula of the chloroalkane;
    (iii) draw the structure of the chloroakane and name it.
                           [ H = 1.00; C = 12.0; Cl = 35.5; Ag = 108.0 ]

_____________________________________________________________________________________________________
OBSERVATION

 This question was not so popular with the candidates.  Only few candidates attempted this question.  In (a), majority of the candidates do not have good knowledge of hybridization.  In (b) some candidates could not explain why diamond is hard and graphite soft.  In (c), a fair attempt was made at the calculation of the molecular mass of the choroalkane and the drawing of its structure.
The expected answers include:
4.

(a)        (i)         Hybridization is defined as mixing of different atomic orbitals
to produce identical new orbitals (of the same shape and energy).
                                                                                                                       
                        OR
Hybridization is the mixing of two or more orbitals to give new sets of two or
more orbitals which are exactly equivalent.

            (ii)        sp3  and sp2
(iii)       I. s – orbital                 -           spherical 
                        II. p – orbital               -           dumb-bell/pear shape/ figure eight

(b)        In diamond each carbon atom is sp3 hybridized while in graphite each carbon atom is sp2 hybridized. Diamond has a giant covalent structure with a network of strong covalent bonds holding each atom tightly into crystals.

            In graphite, each carbon atom is covalently linked to three neighbouring atoms
in the same plane forming layers of carbon atoms held together by weak
van der Waals forces.                                                                         
(c)        (i)         R – Cl     Ag+    AgCl   
                        1 mole             1 mole
                        M(AgCl) = 108 + 35.5
    = 143.5gmol-1         
                        Moles of AgCl =  mass          
                                                     Molar mass
                                                  
                                                 = 1.280 x 10-3 moles  
I mole of RCl   = 1 mole of AgCl      
\mole of RCl  = 1.280 x 10-3 moles
                          Mass of RCl   = 0.0826 g
                          Molar mass     =    0.0826     
                                                      1.280 x 108 
                                                                         =  64.52 gmol-1     
 
ALTERNATIVE   (A)

            R – Cl     Ag+    AgCl   
            M(AgCl)  =  108 + 35.4  =  143.5 gmol -1
            Moles of AgCl   =    mass     =  0.1837  
                                         Molar mass   143.5
                                                            = 1.280 x 10-3
            143.5 g of AgCl    ≡ 35.5 g of Cl
             0.1837 g of AgCl ≡   35.5  x 0.1837    
                                                143.5
                                           =  0.0454 g
            Hence 0.0454 g of Cl  = 0.0826 g of RCl
                           35.5 g of Cl  = 0.0826  x 35.5 
                                                   0.0454
                                                =  64.52 gmol-1     

            ALTERNATIVE   B

            R – Cl     Ag+    AgCl
            1 mole              1 mole
            M(AgCl)  =  108 + 35.5    =   143 . 5
            0.1837 g of AgCl produced from 0.0826 g of RCl 
            ...         143.5 g of AgCl will be produced from:
                                                0.082   x 143.5  
                                                0.1837
                                          =    64.5 gmol-1     
           
            (ii)        Since the substance is an alkyl chloride/ chloroalkane
                        CnH2n+1Cl = 64.5      
                        =  12n +2n + 1 +35.5 = 64.5
                        14n = 64.5 – 36.5
                        = 28                
                               n =  = 2             
                        Molecular formulae = C2H5Cl

                                    H         H
                                                           
            (iii)       H         C         C         Cl       


                                    H         H

                        Chloroethane                                                  
Powered by Sidmach Technologies(Nigeria) Limited .
Copyright © 2015 The West African Examinations Council. All rights reserved.