Preparation of tin (IV) iodide SnI4
Key mandatory areas covered:
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Pure covalent bonding and ionic bonding can be considered as being at opposite ends of a
bonding continuum with polar covalent bonding lying between these two extremes.
 If the difference is large then the movement of bonding electrons from the element of lower
electronegativity to the element of higher electronegativity is complete resulting in the
formation of ions. Compounds formed between metals and non-metals are often, but not
always ionic.
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The spatial arrangement of polar covalent bonds can result in a molecule being polar or having no
overall polarity due to symmetry of polar bonds.
 Ionic compounds and polar molecular compounds tend to be soluble in polar solvents such
as water and insoluble in nonpolar solvents. Non-polar molecular substances tend to be
soluble in non-polar solvents and insoluble in polar solvents
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For a particular set of reaction conditions, the percentage yield provides a measure of the degree to
which the limiting reagent is converted into the desired product. It is possible to calculate the
percentage yield using the equation below. Percentage yield = (actual yield / theoretical yield) x 100
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Percentage yield is the quantity of the desired product formed under the prevailing reaction
conditions whilst the ‘theoretical yield’ is the quantity of desired product which would be obtained,
assuming full conversion of the limiting reagent, as calculated from the balanced equation.
Percentage yields can be calculated from mass of reactant(s) and product(s) using a balanced
equation. Given costs for the reactants, a percentage yield can be used to calculate the feedstock’s
cost for producing a given mass of product.
Key skills developed
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Vacuum filtration using Buchner funnel
Using melt-temp to analyse purity of product
Filtration
Manipulation of non-polar solvent temperature to recrystallize product.
Recrystallization
Synthesis of chemicals
Since each chemical synthesis is unique, there is no absolutely formulaic procedure by which all
compounds can be produced. Most synthesis procedures do share a number of commonalities
which will be discussed on this page.
If the reagents that are being used in the reaction are not properly mixed (they are not
homogeneously incorporated – in the same phase) the rate of reaction will be drastically slowed.
The solvent that is being used not only has to be able to support the reaction without interference;
it must also dissolve the starting materials. The product may or may not be soluble in the solvent.
If the product is insoluble in the solvent it will precipitate and therefore be easily separated from
the residual starting materials.
Many reactions produce heat; care must be taken when dealing with such situations. Frequently,
stirring can be utilized as a means of dispersing the heat. Also, if the heating problem is
anticipated, the reagents can be combined slowly to avoid excessive heating. Also keep in mind
that although a reaction may be exothermic, it may be necessary to heat it to increase the rate at
which the reaction occurs. For example, the synthesis of SnI 4 from tin and I2 requires heat to begin
the reaction. Once begun, the reaction produces heat and the external source of heat must be
removed.
After the synthetic procedure has been carried out, the product must be separated from byproducts and the reagents that were used to make the product. In some cases, the product will
precipitate from solution and can be isolated simply by filtration. This is the case for tin(iv) iodide
synthesis. If the product is soluble in the solvent, it will be necessary to remove some or all of the
solvent. Unfortunately, removal of the solvent usually leaves behind not only the product but also
any un-reacted starting materials or by-products. These undesired compounds must then be
separated from the product.
If the product is a solid, recrystallization can be used to purify it. A solvent can be chosen that will
dissolve the product but will not dissolve the unwanted starting materials and by-products, or, the
solvent may dissolve the impurities but not the product. If the product is a liquid; it can
be distilled to remove impurities that have boiling points different from that of the product. There
are also chromatographic techniques that can be used for both solids and liquids.
After the substance has been isolated it must be identified. Even though the synthetic procedure
was devised to allow you to prepare a certain compound, the substance obtained in the reaction
may not actually be that compound. In order to identify the substance you generally need to
perform several measurements of physical and chemical properties. For example, if you are trying
to prepare SnI4, you can measure the melting point - it should be 144°C, according to the
literature. Usually the comparison of several measurements with those reported in the literature
are needed in order to be sure that you have indeed prepared the correct compound.
Overview of bonding and properties
Most of chemistry deals not with individual atoms but with combinations of atoms of different
elements. These combinations of atoms are called compounds and have completely different
chemical and physical properties than the constituent elements. For example, a common general
chemistry laboratory experiment involves the addition of tin metal to iodine. Both of these are in
their elemental forms, tin being a metal and iodine being a molecular solid. As shown in Figure 25,
tin is grey, while iodine is purple (you may have used tincture of iodine as an antiseptic for cuts).
When the two elements are heated, an orange solid forms. This orange solid is tin tetraiodide and
consists of tin atoms bonded to iodine atoms in a compound that has the formula SnI 4. This
compound melts at 144 °C and will dissolve in many organic solvents, whereas metallic tin melts
over 300 °C and is insoluble in organic solvents. If we take another metal such as sodium and
allow it to react with elemental iodine, not only will we get a much more violent reaction, but we will
also obtain a white product that is not very soluble in organic solvents but highly soluble in water.
Moreover, this compound, sodium iodide, has a melting point of 651 °C.
Elemental tin, iodine and SnI4.
Generally, this kind of difference in physical properties is indicative of something quite different
about the compounds at the atomic level. In fact, compounds are placed into two general
categories--ionic and covalent. Ionic compounds are formed between metals and non-metals,
while covalent compounds are formed between two non-metals, or, as is the case with tin
tetraiodide, between an element that is a borderline metal (notice the position of tin on the periodic
chart) and a non-metal like iodine. In ionic compounds, the atoms are present as ions; that is, as
charged particles. For example, in NaI, the sodium is present as sodium ions and the iodine as
iodide ions. The sodium ion, you will remember, has 12 protons in the nucleus, but only 11
extranuclear electrons. Consequently, it has one more positive charge than negative change and
therefore has an overall plus one charge (Na +). The iodide ion has just the opposite situation: the
number of electrons is greater than that of protons by one and is negatively charged (I -). In some
ionic compounds there are plus two or plus three cations and minus two or minus three anions.
Magnesium carbonate contains a plus two magnesium ion (Mg2+) and a minus two carbonate ion
(CO32-).
In covalent compounds there are no charged ions; the atoms are held together by covalent bonds.
These covalent bonds are believed to be a result of the sharing of electron density between two
atoms. Covalent compounds either contain molecules or a large network of atoms linked together.
A representation of both compounds is shown in Figure 26.
. Two types of covalent compounds (a) molecular (NH3) and (b) covalent network (a portion of the
diamond structure).
Recrystallization
The principle behind recrystallization is that the amount of solute that can be dissolved by a
solvent increases with temperature. In recrystallization, a solution is created by dissolving a solute
in a solvent at or near its boiling point. At this high temperature, the solute has a greatly increased
solubility in the solvent, so a much smaller quantity of hot solvent is needed than when the solvent
is at room temperature. When the solution is later cooled, after filtering out insoluble impurities, the
amount of solute that remains dissolved drops precipitously. At the cooler temperature, the
solution is saturated at a much lower concentration of solute. The solute that can no longer be
held in solution forms purified crystals of solute, which can later be collected.
Recrystallization works only when the proper solvent is used. The solute must be relatively
insoluble in the solvent at room temperature but much more soluble in the solvent at higher
temperature. At the same time, impurities that are present must either be soluble in the solvent at
room temperature or insoluble in the solvent at a high temperature. For example, if you wanted to
purify a sample of Compound X which is contaminated by a small amount of Compound Y, an
appropriate solvent would be one in which all of Compound Y dissolved at room temperature
because the impurities will stay in solution and pass through filter paper, leaving only pure crystals
behind. Also appropriate would be a solvent in which the impurities are insoluble at a high
temperature because they will remain solid in the boiling solvent and can then be filtered out.
When dealing with unknowns, you will need to test which solvent will work best for you. According
to the adage "Like dissolves like," a solvent that has a similar polarity to the solute being dissolved
will usually dissolve the substance very well. In general, a very polar solute will easily be dissolved
in a polar solvent and will be fairly insoluble in a non-polar solvent. Frequently, having a solvent
with slightly different polarity characteristics than the solute is best because if the polarity of the
two is too closely matched, the solute will likely be at least partially dissolved at room temperature.
How to calculate % Yield
To accurately calculate the yield, the equation needs to be balanced. Next, identify the limiting
reagent. Then the theoretical yield of the product can be determined and, finally, compared to the
actual yield. Then, percent yield can be calculated.
Chemists carry out chemical reactions to make new products. The ideal reaction for a chemist is
one where all the reactants are converted into the desired product. In reality, this does not always
occur. One method of assessing a reaction is to calculate the percentage yield. The following
examples will explain how this works. Consider a simple reaction such as methane reacting with
oxygen:
CH4 + 2O2  CO2 + 2H2O
If 16g of methane was reacted with oxygen, we would expect 44g of carbon dioxide to form. (1
mole of methane, gfm 16g  1 mole of carbon dioxide, gfm 44g)
The quantity of product, 44g, is called the theoretical yield.
The actual yield is the quantity obtained when the experiment is carried out. Once the actual yield
known, the percentage yield can be calculated.
Determination of Melting Point
Determining the melting point of a compound is one way to test if the
substance is pure. A pure substance generally has a melting range (the
difference between the temperature where the sample starts to melt and
the temperature where melting is complete) of one or two degrees.
Impurities tend to depress and broaden the melting range so the purified
sample should have a higher and smaller melting range than the
original, impure sample.
1. Fill a capillary tube with crystals about 3 mm high. Put the capillary
tube (open end down) into the crystals and tap it on the bottom of the
crystallization dish to get the crystals into the tube. Force the crystals
to slide to the bottom of the tube using one of the following methods:
tap the tube (open end up) on the lab bench; drop the capillary tube
through a 2-3 foot piece of glass tubing; or rub the capillary tube
along a piece of wire gauze.
2. Place the capillary tube in the MEL-TEMP melting point apparatus.
Set the MEL-TEMP at a high enough level to make a rapid
determination of melting point. Observe the melting process though
the magnifying lens.
3. Once a melting point range is determined, prepare another capillary
tube (tubes should only be used once and then discarded) and set the
MEL-TEMP to the appropriate power level, based on the power
level/temperature chart. This time, make sure that the increase in
temperature is no more than 2oC per minute. Again, observe through
the lens.
Questions
Q1) a) Write the balanced chemical equation for the reaction of tin with iodine.
b) What type of bonding does tin and iodine have?
c) How did you know all the iodine had reacted?
d) Tin is the reactant in excess, which means some tin will remain after all the iodine has reacted.
How is the unreacted tin removed from the reaction mixture?
Q2) Toluene is a non-polar solvent and is used to dissolve tin iodide. By drawing a diagram show
how tin iodide can dissolve in toluene, even though it has polar covalent bonds.
Q3) a) Write down the mass of tin iodide produced from your synthesis in the lab.
b) Considering we used 2.5g of iodine, calculate the percentage yield obtained from your
synthesis.
c) State two reasons the actual yield is lower than the theoretical yield.
Q4) a) Write down the melting range of your synthesised tin iodide. This is a method to
determine the purity of product. (144 °C)
b) What does this fairly low melting temperature suggest about the bonding in tin(iv) iodide?