Title:
Tin (IV) Iodide
Intro:
Tin is an element on the metal/non-metal border and like silicon is of interest due to its importance in the semi-conductor industry.
Procedure:
Before beginning the experiment all the glassware was dried thoroughly.
25cm3 of acetic acid anhydride was added to 25cm3 glacial acetic acid in a 100cm3 round-bottomed flask. To this mixture, 0.5g of finely chopped tin-foil was added along with 2.3g of iodine. The round-bottomed flask was fitted with a condenser and a drying tube which consisted of CaO chips and cotton wool. The mixture was refluxed in an oil bath on a stirrer hot-plate. Refluxing was completed with the disappearance of the violet colour. The remaining red mixture was cooled in an ice-bath. The tin(IV) iodide was filtered off with suction from a buchner funnel. The product was recrystallised using 1,1,1-trichloroethane as a solvent.
Pre-practical questions:
1.
Tin belongs to group 14 – IVA.
C – non-metalSi – non-metal Ge – non-metalPb – metalSn – metal.
2.
The reaction must be performed under dry conditions, else the tin got to tin hydroxide Sn(OH)2 SnO2
3.
From general rule, like dissolves like. 1,1,1-trichloroethane is a non-polar solvent so SnI4 must also be non-polar to dissolve. There the bonding that exists between 1,1,1-trichloroethane and SnI4 is covalent. The electro negativity between Sn and I is very low (0.7) so the bonding is just about covalent while the electro negativity difference between Sn and F is greater than 1.7. Hence the tetraflouride would be soluble in the organic solvent.
Post-practical questions:
1.
No. The divalent state is stable if promotion energies are examined. Bond energies decrease in general. Take for example the reaction MX2 + X2 = MX4
The MX bond energies are decreasing. However, a stage may be reached when the energy becomes too small to compensate for M2 M4 promotion energy.
MX2 becomes more stable
E.g.PbCl2 + Cl2 PbCl4
PbCl4 decomposes except at low temperatures while PbBr4 and PbI4 don’t exist possibly because of the reducing power of Br – and I -.
2.
The bond energies for Sn-I is 187 KJ mol –1 (relatively low energy level).
Therefore a good deal of energy is not needed to form these bonds. When the vapours pass up the condenser, the cold sides of the condenser cause the condenser cause the crystals to be formed.
3.
SnCl4 / SnI4(1) / Liquid (r.t.) / Solid (r.t.)
(2) / Colourless / Orange powder
(3) / b.p. 114.1C / b.p. 384C
The high boiling point can be put down to the fact that iodine atoms are held more tightly to the Sn than Cl atoms. Hence a higher boiling point would be required to split the molecules apart. The physical difference can be attributed to the fact that chlorine is a gas and Iodine is a solid at room temperature. Hence SnCl4 is a liquid and SnI4 is a solid.
4.
Electron position in Sn
In this configuration the 5p and 4d sublevels are very close together – like the d sublevels in transition metals. It is the excitation of e- in the d sublevels that give use to the different colours of the compounds. This is probably the case with Sn. The electrons in the 4d orbital can be promoted to the 4p orbital. When they fall back down they give off a photon of light. In this case, it is a frequency which produces an orange colour.
5.
For the reaction with water – no gas was evolved – a grainy substance fell out of solution and was yellow in colour. A brown colour remained.
SnI4(aq) + 2 H2O(l) SnO2 (ppt) + 4HI(aq)
For the KI reaction no precipitate was formed – no gas was evolved – the reaction went from light brown to a very dark brown solution.
SnI4 (aq) + 2 KI(aq) SnI62-(aq) + 2K2+
(light brown) (dark brown)
Iodine has a different oxidation state causing the change in colour.
6.
IO3-(aq) + 6 H +(aq) 2I-(aq) + 3Cl-(aq)3ICl(aq) + 3 H2O(l)
IO3- 3ICl
+5 +1
2I- 3ICl
-1 +1
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Paul Walsh – 2001