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-metal
Si – non-metal
metal.
Ge – non-metal
Pb – metal
Sn –
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 1,1,1trichloroethane 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.
(1)
(2)
(3)
SnCl4
Liquid (r.t.)
Colourless
b.p. 114.1C
SnI4
Solid (r.t.)
Orange powder
b.p. 384C
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
--------------------------Paul Walsh – 2001
pwalsht@maths.tcd.ie