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