Structures and Physical Properties
Structures of the elements
The trend from non-metal to metal as you go down the Group is clearly seen in the structures of the elements themselves.
Carbon at the top of the Group has giant covalent structures in its two most familiar allotropes – diamond and graphite.
Diamond has a three-dimensional structure of carbon atoms each joined covalently to 4 other atoms. The diagram shows a small part of that structure.
Exactly this same structure is found in silicon and germanium and in one of the allotropes of tin – “grey tin” or “alpha-tin”.
The common allotrope of tin (“white tin” or “beta-tin”) is metallic and has its atoms held together by metallic bonds. The structure is a distorted close-packed arrangement. In close-packing, each atom is surrounded by 12 near-neighbours.
By the time you get to lead, the atoms are arranged in a straightforward 12-co-ordinated metallic structure.
There is therefore a clear trend from the typical covalency found in non-metals to the metallic bonding in metals, with the change-over obvious in the two entirely different structures found in tin.
Physical properties of the elements
Melting points and boiling points
If you look at the trends in melting and boiling points as you go down Group 4, it is very difficult to make any sensible comments about the shift from covalent to metallic bonding. The trends reflect the increasing weakness of the covalent or metallic bonds as the atoms get bigger and the bonds get longer.
The low value for tin’s melting point compared with lead is presumably due to tin forming a distorted 12-co-ordinated structure rather than a pure one. The tin values in the chart refer to metallic white tin.
There is a much clearer non-metal / metal difference shown if you look at the brittleness of the elements.
Carbon as diamond is, of course, very hard – reflecting the strength of the covalent bonds. However, if you hit it with a hammer, it shatters. Once you apply enough energy to break the existing carbon-carbon bonds, that’s it!
Silicon, germanium and grey tin (all with the same structure as diamond) are also brittle solids.
However, white tin and lead have metallic structures. The atoms can roll over each other without any permanent disruption of the metallic bonds – leading to typical metallic properties like being malleable and ductile. Lead in particular is a fairly soft metal.
Carbon as diamond doesn’t conduct electricity. In diamond the electrons are all tightly bound and not free to move.
|White tin and lead are normal metallic conductors of electricity.|
There is therefore a clear trend from the typically non-metallic conductivity behaviour of carbon as diamond, and the typically metallic behaviour of white tin and lead.
Trying to explain the trends
The main characteristic of metals is that they form positive ions. What we need to do is to look at the factors which increase the likelihood of positive ions being formed as you go down Group 4.
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is usually measured on the Pauling scale, where the most electronegative element (fluorine) is given an electronegativity of 4.
The lower the electronegativity of an atom, the less strongly the atom attracts a bonding pair of electrons. That means that this atom will tend to lose the electron pair towards whatever else it is attached to. The atom we are interested in will therefore tend to carry either a partial positive charge or form a positive ion.
Metallic behaviour is usually associated with a low electronegativity.
Note: If you aren’t sure about electronegativity you really ought to read about it before you go any further.Use the BACK button on your browser to return quickly to this page.
|So what happens to electronegativity in Group 4? Does it decrease as you go down the Group, suggesting a trend towards metallic behaviour?|
Well! It certainly falls from carbon to silicon, but from there on it is a complete mess!
There therefore seems to be no relationship between the non-metal to metal trend and the electronegativity values. Assuming the electronegativity values are correct, I am completely at a loss to understand this!
If you are thinking about the formation of positive ions, the obvious place to start looking is how ionisation energies change as you go down Group 4.
Ionisation energies are defined as the energy needed to carry out each of the following changes. They are quoted in kJ mol-1.
First ionisation energy:
Second ionisation energy:
. . . and so on.
None of the Group 4 elements form 1+ ions, so looking at the first ionisation energy alone isn’t very helpful. Some of the elements do, however, form 2+ and (to some extent) 4+ ions.
The first chart shows how the total ionisation energy needed to form the 2+ ions varies as you go down the Group. The values are all in kJ mol-1.
You can see that the ionisation energies tend to fall as you go down the Group – although there is a slight increase at lead. The main trend is because:
- The atoms are getting bigger because of the extra layers of electrons. The further the outer electrons are from the nucleus, the less they are attracted – and so the easier they are to remove.
- The outer electrons are screened from the full effect of the nucleus by the increasing number of inner electrons.
- These two effects outweigh the effect of the increasing nuclear charge.
What is clear looking at these two charts is that you have to put in large amounts of ionisation energy to form 2+ ions, and huge amounts to form 4+ ions.
However, in each case there is a fall in ionisation energy as you go down the Group which makes it more likely that tin and lead could form positive ions – however, there is no indication from these figures that they are likely to form positive ions.
The ionisation energies of carbon at the top of the Group are so huge that there is no possibility of it forming simple positive ions.
OXIDATION STATE TRENDS IN GROUP 4
Some examples of the trends in oxidation states
The overall trend
The typical oxidation state shown by elements in Group 4 is +4, found in compounds like CCl4, SiCl4 and SnO2.
However, as you go down the Group, there are more and more examples where the oxidation state is +2, such as SnCl2, PbO, and Pb2+.
With tin, the +4 state is still more stable than the +2, but by the time you get to lead, the +2 state is the more stable – and dominates the chemistry of lead.
An example from carbon chemistry
The only common example of the +2 oxidation state in carbon chemistry occurs in carbon monoxide, CO. Carbon monoxide is a strong reducing agent because it is easily oxidised to carbon dioxide – where the oxidation state is the more thermodynamically stable +4.
For example, carbon monoxide reduces many hot metal oxides to the metal – a reaction which is used, for example, in the extraction of iron in a blast furnace.
Examples from tin chemistry
By the time you get down the Group as far as tin, the +2 state has become increasingly common, and there is a good range of both tin(II) and tin(IV) compounds. However, tin(IV) is still the more stable oxidation state of tin.
That means that it will be fairly easy to convert tin(II) compounds into tin(IV) compounds. This is best shown in the fact that Sn2+ions in solution are good reducing agents.
For example, a solution containing tin(II) ions (for example, tin(II) chloride solution) will reduce a solution of iodine to iodide ions. In the process, the tin(II) ions are oxidised to tin(IV) ions.
Tin(II) ions also reduce iron(III) ions to iron(II) ions. For example, tin(II) chloride solution will reduce iron(III) chloride solution to iron(II) chloride solution. In the process, the tin(II) ions are oxidised to the more stable tin(IV) ions.
Tin(II) ions will also, of course, be easily oxidised by powerful oxidising agents like acidified potassium manganate(VII) solution (potassium permanganate solution). This reaction could be used as a titration to find the concentration of tin(II) ions in a solution.
Examples from lead chemistry
With lead, the situation is reversed. This time, the lead(II) oxidation state is the more stable, and there is a strong tendency for lead(IV) compounds to react to give lead(II) compounds.
Lead(IV) chloride, for example, decomposes at room temperature to give lead(II) chloride and chlorine gas:
. . . and lead(IV) oxide decomposes on heating to give lead(II) oxide and oxygen.
Lead(IV) oxide also reacts with concentrated hydrochloric acid, oxidising some of the chloride ions in the acid to chlorine gas. Once again, the lead is reduced from the +4 to the more stable +2 state.
Trying to explain the trends in oxidation states
There’s nothing surprising about the normal Group oxidation state of +4.
All of the elements in the group have the outer electronic structure ns2npx1npy1, where n varies from 2 (for carbon) to 6 (for lead). The oxidation state of +4 is where all these outer electrons are directly involved in the bonding.
As you get closer to the bottom of the Group, there is an increasing tendency for the s2 pair not to be used in the bonding. This is often known as the inert pair effect – and is dominant in lead chemistry.
However, just giving it a name like “inert pair effect” explains nothing. You need to look at two different explanations depending on whether you are talking about the formation of ionic or covalent bonds.
The inert pair effect in the formation of ionic bonds
If the elements in Group 4 form 2+ ions, they will lose the p electrons, leaving the s2 pair unused. For example, to form a lead(II) ion, lead will lose the two 6p electrons, but the 6s electrons will be left unchanged – an “inert pair”.
You would normally expect ionisation energies to fall as you go down a Group as the electrons get further from the nucleus. That doesn’t quite happen in Group 4.
This first chart shows how the total ionisation energy needed to form the 2+ ions varies as you go down the Group. The values are all in kJ mol-1.
Notice the slight increase between tin and lead.
This means that it is slightly more difficult to remove the p electrons from lead than from tin.
However, if you look at the pattern for the loss of all four electrons, the discrepancy between tin and lead is much more marked. The relatively large increase between tin and lead must be because the 6s2 pair is significantly more difficult to remove in lead than the corresponding 5s2 pair in tin.
Again, the values are all in kJ mol-1, and the two charts are to approximately the same scale.
The reasons for all this lie in the Theory of Relativity. With the heavier elements like lead, there is what is known as a relativistic contraction of the electrons which tends to draw the electrons closer to the nucleus than you would expect. Because they are closer to the nucleus, they are more difficult to remove. The heavier the element, the greater this effect.
This affects s electrons much more than p electrons.
In the case of lead, the relativistic contraction makes it energetically more difficult to remove the 6s electrons than you might expect. The energy releasing terms when ions are formed (like lattice enthalpy or hydration enthalpy) obviously aren’t enough to compensate for this extra energy. That means that it doesn’t make energetic sense for lead to form 4+ ions.
The inert pair effect in the formation of covalent bonds
You need to think about why carbon normally forms four covalent bonds rather than two.
Using the electrons-in-boxes notation, the outer electronic structure of carbon looks like this:
There are only two unpaired electrons. Before carbon forms bonds, though, it normally promotes one of the s electrons to the empty p orbital.
That leaves 4 unpaired electrons which (after hybridisation) can go on to form 4 covalent bonds.
It is worth supplying the energy to promote the s electron, because the carbon can then form twice as many covalent bonds. Each covalent bond that forms releases energy, and this is more than enough to supply the energy needed for the promotion.
One possible explanation for the reluctance of lead to do the same thing lies in falling bond energies as you go down the Group. Bond energies tend to fall as atoms get bigger and the bonding pair is further from the two nuclei and better screened from them.
For example, the energy released when two extra Pb-X bonds (where X is H or Cl or whatever) are formed may no longer be enough to compensate for the extra energy needed to promote a 6s electron into the empty 6p orbital.
This would would be made worse, of course, if the energy gap between the 6s and 6p orbitals was increased by the relativistic contraction of the 6s orbital.