2.2.2 (n,o) Simple Molecular Lattices
Syllabus
(n) explanation of the solid structures of simple molecular lattices, as covalently bonded molecules attracted by intermolecular forces, e.g. I2, Ice
(o) explanation of the effect of structure and bonding on the physical properties of covalent compounds with simple molecular lattice structures including melting and boiling points, solubility and electrical conductivity.
What does this mean?
Ice
Even simple molecular substances can freeze.
So, they must form some sort of regular arrangement of molecules as they crystalise.
Exactly what type of lattice they adopt will depend entirely on the shape of the molecules involved and the intermolecular forces involved.
As we saw in 2.2.2 (m)- Anomalous Properties of Water - ice is held together with hydrogen bonds, leading to it floating on water.
It is a V-shaped molecule but with two lone-pairs in the tetrahedral position.
So each lone-pair on the O-atom of a water molecule hydrogen-bonds to neighbouring H-atom at 109.5o.
And the H-atoms on the same water molecule hydrogen-bond to lone-pairs on neighbouring O-atoms, also at 109.5o.
This makes it very similar in structure to diamond, which you're probably familiar with from GCSE.
And since Hydrogen bonds are unusually strong intermolecular forces the melting and boiling points of water are higher than for similar size molecules that cannot Hydrogen-bond.
Discussing the solubility of water in water seems somewhat pointless but its conductivity may be asked about.
You should know that water has very polar-bonds - which makes these bonds less than 100% covalent.
But water has a very small ionic nature - in other words it is genuinely very poor at ionising.
H2O → H+ + OH-
In normal conditions the concentration of H+ & OH- ions in pure water is 1 x 10-7 mol/dm3 - in other words, next to nothing.
So, it shouldn't be a surprise that pure water is such a poor electrical conductor - there are almost no ions to carry charge and no free electrons because all the valence electrons are held in bonds.
Iodine
Uniquely for a halogen Iodine (I2) is a solid at room temperature.
Something must be holding its lattice together and, since I2 molecules are non-polar by definition, it can only be van der Waals forces.
But Iodine is a large atom with many electrons so it can produce stronger van der Waals than other diatomic halogen molecules.#
The first representation of the lattice is realistic but not very helpful and fundamentally impossible to draw well in an exam.
The examiner will be looking for a more open lattice as you are used to drawing for NaCl.
You can see that the molecules are not all in the same orientation but the examiner won't be looking for that, so you can ignore it.
What (s)he'll be looking for is the alternating pattern of layers, with every other point unoccupied.
The solubility of Iodine in water is poor because the only sort of intermolecular force it can produce are van der Waals.
These are much weaker than the Hydrogen bonds that water produces.
So, for Iodine to dissolve there would be a large energy requirement to break Hydrogen bonds, but a poor energy return when Iodine formed van der Waals with water.
More energy in than out generally makes substances insoluble in water.
In the case of Iodine - it does dissolve in water, just not well.
The photo shows a yellow-brown solution of Iodine in water.
But what is the purple layer?
Being poorly soluble in water doesn't make Iodine poorly soluble in everything, of course.
In the photo above, the top layer is cyclohexane (a non-polar solvent, unlike water) & we see that it dissolves well in that - another non-polar substance which only forms van der Waals forces - so most of the Iodine has moved into the top layer where it forms a purple solution.
The energy input and output are more similar when a non-polar substance dissolves in a polar solvent.
And since Iodine has no free electrons and does not readily ionise to make I+ and I- ions - it is a poor electrical conductor.
You may know that Iodine is one of the few substances to sublime - go from solid to gas without passing through a liquid phase.
No one will ask you why though.
But since you're interested...
If you want to think about the van der Waals being strong enough to hold the solid together but still fairly easily overcome. Then given enough energy for them to just break there's really nothing to stop individual molecules flying off into the gas-phase.
Of course, you could make the same argument about any lattice held together by van der Waals - and Bromine forms a liquid readily enough.
But then so would Iodine at different pressures.
Which is why no one will ask.
A typical phase diagram looks like this.
At higher pressures a solid changes to a liquid before vapourising.
Its just that Iodine happens to follow the green path at normal pressures.
You must be glad you asked!
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