Chemistry Paper 2 (Essay) ,May/June 2009

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Questions:		1	2	3	4	5	6	7	8		Main

General Comments
Weakness/Remedies
Strength

Question 4

(a)   Atoms of four non-metallic elements in the same group of the periodic table are arranged in order of increasing atomic radius as R < T < W < X.
            Which of the following elements
(i)   would readily lose electron(s) from the outermost shell;
(ii)     is most electronegative;
(iii)    would T displace from aqueous solution;

(iv)    is at the top of the group?                                [5marks]


(b) The following table shows the electronic configuration of two elements Y and Z.

Element	Electronic Configuration
Y	1s2 2s2 2p5
Z	1s2 2s2 2p6 3s1

(i) Name the

group to which Y belongs.
period to which Z belongs.

(ii)   What is the number of protons present in an atom of Z?
(iii)    How many unpaired electrons are in an atom of Y?
(iv)    Write the formula of the compound formed between Y and Z.                   [5 marks]

CaCI2 ;
NH4CI;
CCI4.                                                                                                          [5 marks]
(d)      Describe how the conductance of a molar solution of ammonia compares to that of sodium hydroxide solution.                                                                    [5 marks]

(e)      State the type of reaction illustrated by each of the following equations:
1. CH3CH2OH(1) + CH3COOH(aq) ⇌ CH3COOC2H5(l) + H2O(l) ;
2. H +(aq) + OH- (aq) → H2O(l).
                                                                                                        [2 marks]

(f)      Determine the volume of the residual gas when 20.0 cm3 of hydrogen was sparked with
          15.0cm3of oxygen and the resulting mixture cooled to room temperature.            [3 marks]

_____________________________________________________________________________________________________

OBSERVATION

This is another popular question among candidates and the performance was good.

In part (a), candidates knew that X would readily loose electron(s) from the outermost shell, R was most electropositive, W and X would be displaced by T in aqueous solution and R was at the top of the group.

In part (b), candidates correctly stated that Y belongs to group VII and Z belongs to period III. They also knew that eleven protons were present in an atom of Z, one unpaired electron was in Y and the formula of the compound formed between the two elements was ZY.

In part (c), candidates named the type of bond(s) in each of the given compounds but some of them lost marks in (ii) because they either mentioned dative or co-ordinate covalent bond as the only bond present in NH4Cl.

The required answers were as follows:

CaCl2 - ionic/electrovalent;
1. NH4Cl - covalent/co-ordinate covalent and ionic.
1. CCl4 - covalent.

1. In (d), candidates could not describe how the conductance of a molar solution of ammonia compares to that of sodium hydroxide solution.
  
  The expected response from candidates was as described below:
  1. pour some quantity of each solution into a separate beaker.
  2. insert electrodes into the solutions and close the circuit.
  3. take the readings on the ammeter.
  4. rinse the electrodes properly with distilled water and dip them into the second solution. Close the circuit and take the reading on the ammeter again.
  1. the solution with the higher reading is NaOH(aq) while the one with lower reading is ammonia.
    OR
    NaOH solution is a strong electrolyte and completely ionizes to produce a lot of ions. NH3 solution is a weak electrolyte and produces very few ions. The ions produced by NaOH far outnumber those produced by NH3 solution. Hence NaOH solution conducts more than NH3 solution.
  In (e)(ii), candidates correctly stated esterification as the reaction illustrated by the equation. However, some of them lost the mark in (e)(ii) because they confused neutralization for hydrolysis.
  
  In (f), candidates determined the volume of the residual gas as follows:
  
                       2H2(g) + O2(g) → 2H2O(1)
  
                       from the equation
  
                       2 vols of H2 reacts with 1 vol of O2/20cm3 of H2(s) reacts with 10 cm3 of O2(g)
  
  residual gas    =    (15 – 10)cm3 = 5cm3.