Chemistry Paper 2B Nov/Dec 2015

Question 4


  1. (i)   Define the term allotropy.

(ii)        State the type of bond present in diamond and graphite.

  1. Explain briefly why:
  1. diamond d is hard while graphite is soft;
  2. graphite is a good conductor of electricity while diamond

is not.                                                                               [10 marks]

(b)  Under two different conditions carbon burns in air to form two gaseous products.
(i)  State the condition under which each product is formed.
(ii) Write a balanced equation for the formation of each product formed in 4(b)(i)

  1. Name the product formed under each condition stated in 4(b)(i).           [8 marks]

(c)  When excess lead trioxonitrate (V) solution was added to a solution of sodium
chloride, the precipitate formed weighed 5.56 g.

  1. Write a balanced equation for the reaction.
  2. Calculate the mass of sodium chloride in the solution.

[Na = 23.0, Cl = 35.5, Pb = 207.0]                                                            [7 marks]

In part (a) (i), majority of the candidates defined allotropy correctly, but some could not.

 

In part (a) (ii), majority of the candidates could not state the type of bond present in diamond and graphite as covalent, only a few could.

In part (a) (iii), majority of the candidates could not explain correctly why diamond is hard, and graphite soft.

In part (b), majority of the candidates stated the different conditions under which carbon burns in air to form two gaseous products.  They equally wrote and balanced the equation for the formation of each product, and named the product formed under each condition.  However, some of the candidates wrote the equations, but could not state the conditions.

In part (c), only few candidates could write the equation for the reaction between excess lead trioxonitrate (V) solution and sodium chloride.  Majority of the candidates could not solve question (c) (ii) correctly.

 

Observation

 

(a)        (i)         Allotropy is the existence of difference forms of the same element in the same
(physical) state.
           

  1. Covalent bond

           

  1. I. Diamond – strong covalent bonds between all atoms with tetrahedral

    structure which is rigid and difficult to break.

Graphite – weak forces (of attraction) between layers making layers to slide and hence soft.

II.  In diamond  all the four valence electrons are bonded but in graphite three       our of the four are bonded thus there is one free mobile/delocalized electron which conducts electricity.

  (b)     (i)        Sufficient/plentiful supply/excess of air Limited/small amount of air
(ii)        Cs + O2(g)            CO2(g)
2C(s) + O2(g)          2CO(g)

  1. Carbon (IV) oxide

Carbon (II)  oxide

 (c)       (i)         Pb(NO3)2+2NaCl            PbCl2+NaNO3

            (ii)        M(NaCl)                      = 23+35.5           + 58.5

                        M(PbCl)                      = 207 + 2(35.5)  = 278

                        278g PbCl2                 = 2x58.5g NaCl

                        5.56g PbCl2                        =  5.56×2 × 58.5  = 2.34g NaCl
278