Inorganic Chemistry: Tin (IV) Iodide

1819 Words4 Pages

Inorganic Chemistry
Experiment 3: Tin (IV) Iodide
Aim:
The primary objective of this procedure is to investigate the various properties of tin(IV) iodide.
Introduction:
Tin is a very important element as it is classified as a group 14 element. Examples of elements in this group include carbon, silicon and lead. It can be found in two different oxidation states, +2 and +4, almost identical to the element directly below it, which is lead (Pb). In terms of its properties, it is very similar to silicon. This is because they are on the metal/non-metal border. Another similarity that they share is that they are both excellent semiconductors. Tin is therefore an element very much of interest in the semiconductor industry.
Pre-Practical Questions: …show more content…

It must be performed under dry conditions. This is due to the fact that in the presence of water (or aqueous solution), tin is hydrolysed. The following reaction occurs:

SnI4 + 4H2O ----> Sn(OH)4 + 4HI

The bonding in this molecule must clearly be covalent as 1,1,1-trichloroethane must be non-polar in order to act as a solvent. The rule ‘like dissolves like’ must be taken into account. The electronegativity difference between tin and iodine is 0.7, which means that it is polar covalent. The electronegativity difference between tin and fluorine is 2.02, making it an ionic molecule. As neither molecule is polar, tin(IV) tetrafluoride would be more soluble in the non-polar solvent.

Procedure:
Part 1: …show more content…

This figure is slightly higher than its actual value, which should be approximately 81%. This was obtained as a result of finding the atomic mass of iodine. The atomic mass of the iodine was then multiplied by 4 as there were 4 atoms of iodine in tin(IV) iodide. This was then placed over the molecular mass of tin(IV) iodide and multiplied by 100 to be expressed as a percentage. This value could have been more accurate to its real value (i.e. approximately 81%) if the titration had been stopped as soon as the solution went yellow. At this point, the final colour of the solution is yellow due to the fact that some ICl2- (iodine dichloride) formed. This compound is formed from the reaction of iodine monochloride (ICl) with excess hydrochloric acid (HCl). In the preparation of the tin(IV) iodide, the crude % yield was found to be 61.02596306%. This yield could have been higher if the mixture in the round-bottomed flask was allowed to reflux for a longer time period. In addition to this, the crude yield could have been higher if the tinfoil was cut up into smaller pieces. This would have meant more surface area exposed to the mixture of iodine, acetic anhydride and glacial acetic acid. More surface area means more molecules have more space to react with the solution. This also means that the reaction would have occurred at a much

More about Inorganic Chemistry: Tin (IV) Iodide

Open Document