Question 2
- (i) What are Van der Waals forces?
(ii) List two factors which will favour the formation of each of the following bonds:
I. covalent bond;
II dative bond. [6 marks] - Consider the following pairs of bonds:
C – N and C – H, C – O and C – N, C – F and C – O.
(i) Which bond in each pair is more polar?
(ii) Arrange the bonds in increasing order of polarity. [4 marks] - (i) Consider the following pair of elements 4Be and 12Mg.
1. Give the electron configuration of each element.
II. Explain briefly why the elements have similar chemical properties.
(ii) Arrange the following species in order of decreasing size: Ne, Na+, Mg 2+.
Explain briefly your answer. [8 marks] - (i) Define transition elements.
(ii) Explain briefly why most transition elements have variable oxidation states.
(iii) Give two uses of metals. [7 marks]
Observation
It was an optional question. Few candidates responded to it.
In part (a), majority of the candidates defined van der Waals forces as weak intermolecular force.
Besides, they could not state two factors which would favour the formations of covalent and dative bond.
In part (b), majority of the candidates could not arrange the listed bonds in increasing order of polarity.
In part (c), majority of the candidates responded to the question satisfactorily. They used the concepts of periodic table to give the electron configuration of the elements. They gave satisfactorily explanation on why the elements have similar chemical properties.
In part (d), majority of the candidates defined transition elements correctly, but only few of them could state why transition elements have variable oxidation states.
The expected answers include:
- (i) Van der Waals forces are weak ( intermolecular) forces of attraction between molecules / atoms
(ii) I - high ionization energy
- high electron affinity
- low / close electronegativity difference between combining atoms
- presence of unpaired electrons
II - vacant / empty orbital in one atom
- presence of lone pair of electrons ( in another ) - (i) C – N
C – O
C – F
(ii) C – H < C – N < C – O < C – F - (i) I. 4Be – 1s2 2s2
12Mg – 1s2 2s2 2p6 3s2
II. 4Be and 12Mg have two / same valence electrons each, hence they have the same chemical properties.
(ii) Ne > Na+> Mg2+ Each of the species Ne, Na+, Mg2+ has ten or same number of electrons / isoelectronic, the smaller the nuclear charge, the bigger the size of the species . Ne has the smallest nuclear charge while Mg2+ has the highest hence order of decreasing size will depend on the nuclear charge. - (i) A transition element is an element whose atom has a partially filled d-subshell / which can give rise to ions / cations with an incomplete d-sub shell
OR is one with partially/ incompletely filled d-orbitals.
(ii) Most transition elements have variable oxidation states because they can use 3d or 4s electrons when forming bonds due to the closeness in energy of the 3d and 4s orbitals / electrons. Electrons can move easily between the 3d and 4s orbitals.
(iii) - automobile parts
- roofing sheets
- musical instruments
- electrical wires
- bridges
- reducing agent
- alloys e.g. cutlery, iron gates, coins, burglar proof, jewelleries
- aluminum / tin foil
- nails
- metal implements etc