2.3 Questions and Answers - Chemteach - University of Canterbury - New Zealand

2.4 Questions and Answers Archive

  1. Lewis diagrams
  2. 2004 2.4 exam questions
  3. Intermolecular forces

Q1. When you draw the Lewis dot diagram (below), you get one lone electron. Can you explain how this can exist?

 ..    .. 
.N::O:

A. In any molecule where there is an uneven number of total valence electrons there must be a single unpaired electron somewhere in the structure which will also mean an atom (in this case N) will not have a full “octet”. This is acceptable in some molecules (after all we are happy to allow BCl3 to exist when it only has 6 electrons around the central atom). NO2  is another well known example with a single electron but in this case the molecules readily dimerise to from N2O4 which eliminates the issue.

I am sure that there is a fuller explanation relating to molecular orbital theory but I suspect you do not require this. Back to top.

Q2. I sent you the below question last week.  I also pass on a question asked of me by a Year 12 student earlier in the week.  The Chemistry 2.4 2004 paper included the following question:

Oxide

Melting point (°C)

Conductivity of solid

Hardness of solid

CO2

sublimes at –78

poor

brittle

SiO2

1700

poor

very hard

Discuss the structure and bonding within carbon dioxide and silicon dioxide, and relate these to the properties shown in the table above. The model answers referred to as all valence electrons are involved in forming covalent bonds there are no free moving charges and so no electrical conduction.   Every oxygen in SiO2 has two lone pairs of electrons.  Maybe molecular orbital theory deals with this, my old textbooks never went as complicated as SiO2 for molecular orbital theory, but other than this the model answer is in conflict with current theory.

Also Question six of AS 2.4 2005, the question asks to Discuss the properties of iodine and potassium iodide, in terms of the structure and bonding within each solid and one of the properties is iodine solubility in cyclohexane.

The answer said since both molecules have similar weak intermolecular forces, then weak forces will exist between the two different molecules.  In the evidence section, it states must refer to similar strengths of intermolecular forces.

The answer seems to require specific vague information.  It explains nothing to me and certainly sheds no light as to why a non polar solid would not dissolve in a polar solvent.

A. Wrt the conduction question I suspect the response relating to all valence electrons being involved in bonding was simply a careless slip that was meant to refer to the central Si atom only. The key point at Yr 12 is if they can recognize that a substance does not conduct unless it has charged particles that are free to move –either delocalized electrons (as in metals or graphite) or mobile ions as in molten or aqueous salt solutions. I'm sure we are not expecting any information relating to band theory or the impact of small amounts of impurities into pure silica. 

Wrt the solubility question – Chemical changes occur because they are thermodynamically favourable. When two components are mixed there is generally a favorable increase in entropy arising from the increased “disorder or randomness” of the mixture but we also have to consider the enthalpy factor which is favoured by a reduction in energy. If substance A is to dissolve in substance B the existing attractive forces between the B molecules (and also between the A molecules) have to be broken which is an endothermic process (ie requires energy input). This energy requirement may be partially or fully offset by any new attractions formed between the A and B molecules. If A and B do not form some type of attractive forces then the mixing process is unlikely to be energetically favorable. Therefore, non-polar solutes will generally dissolve in non polar solvents because you are replacing existing attractions with new attractions of similar type (instantaneous dipole) and strength. Similarly polar solutes will generally dissolve in non-polar solvents because the existing attractive forces between the polar molecules will simply be replaced by equivalent attractive forces between the two polar molecules which will still have permanent dipole –permanent dipole attractions between the components of the mixture. Whereas if a polar solute is inserted between polar molecules the original attractions between the polar molecules has to be broken but is not compensated for by any new attractive force (remembering that the instantaneous dipole attractions exists both before and after the mixing process).

Having just written all this I now realize why the examiners report tries to summarise it by relating the requirement to “similar strengths of intermolecular forces”.

Some simple test tube reactions investigating temperature changes associated with mixing of water and alcohol, water and  ethyl ethanoate and water and cyclohexane can produce some interesting discussion about the nature and strength of the enthalpy changes involved but of course solubility s not determined by enthalpy alone because of the entropy contribution. But let's leave true thermodynamics to tertiary study. I'm happy to tell my students that the reality of teenagers lazing around in a very untidy bedroom is simply the consequence of the first two laws of thermodynamics.

Hope this off the top of my head ramble is of some use. I guess my main point is that at Yr 12 we should be fairly understanding as to the expected depth of understanding of particular chemical concepts. Back to top.

Q3. There appears to be considerable variation (in different texts etc) as to which intermolecular forces of attractions, the "van der Waals forces" generic term actually encompasses.  We avoid the debate by using the phrase temporary (or instantaneous) dipole attractions in an analogous manner to the permanent dipole attractions used in this resource. Apart from avoiding the debate about what van der Waals forces are it gives the students a much clearer picture of what type of forces are involved in each case.

A. If you visit the Activities folder on the Chemteach website ( www.chemteach.ac.nz ) there are two activities that you can download that give students practice with inter and intramolecular forces. Back to top.

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