The Oxidation States of Tin. Preparation of Tin (IV) Iodide and Tin (II) Iodide
Taylor Filewood
Station 4C, Tuesdays 1:30-5:20pm
June 26th, 2018
Chem 236W

Metals are stable in different states based on the bonding that they have and whether or not a lattice is formed. Tin is metallic and stable in both of its states, II and IV.1 This is because there are stronger covalent bonds that are occurring and since lattices are being formed when these bonds are occurring. When trying to determine if a compound is stable or not, the best thing to do is determine its melting point.2 This will tell you if a lattice is formed or if just weak Van der Waals forces are occurring.3 The higher the melting point is, the stronger the bonds will be which means there is a lattice that has formed. The melting point also tells you if your compound is pure or not based on how close it is to the theoretical melting point and how large the range is. The larger the range and the farther it is from the theoretical melting point, the more impure it will be.2
In this lab we are going to synthesize Tin (IV) Iodide and Tin (II) Iodide and then determine the melting points of each. This will allow us to determine whether or not a lattice is formed and if we truly formed the product that we wanted to produce. A reflux system will be used for each with different atmospheres, Tin (II) Iodide with nitrogen and Tin (IV) Iodide with oxygen, and use different solvents, glacial acetic acid with acetic anhydride for Tin (IV) Iodide and 2M Hydrochloric acid for Tin (II) Iodide, to cause the different products to form. Vacuum filtration will also be used to wash and dry the products made.
Tin (IV) Iodide:
A reflux system was set up on May 22nd, 2018 that was used to synthesize Tin (IV) Iodide. A CaCl2 drying tube was used to prevent any of the reaction from leaving the system and it was filled about half way with CaCl2. 0.12g of tin was grinded until somewhat shiny and added to a 50mL round bottom flask. The tin started off as a greenish brown color. Then 0.48g of iodine, which was a silver, grey charcoal color, was also added to the same round bottom flask. Nothing occurred in the round bottom flask. 4.0mL of glacial acetic acid, which was a clear liquid, and 4.0mL of acetic anhydride, which was also a clear liquid, were then added to the round bottom flask. This caused a yellow brown solution to form before the reflux began. At 2:28pm the solution started to stir and heat up which resulted in the solution becoming a brown almost dark purple color. Once the solution had heated to about 135? the reflux then started. This was at 2:44pm. A purple vapor then began to form as the solution continued to stir and maintained a temperature around 150?. At 3:01pm some of the purple vapor disappeared but the solution became more of a dark brown orangey color. The water condenser had some condensation on the bottom and the dark brown orangey solution was dripping back into the round bottom flask as the reflux continued. The purple vapor still remained at 3:17pm and the solution remained the same color as well. The solution also was bubbling the entire time the reflux was going. The reflux was stopped after about an hour, at 3:44pm. The purple vapor still remained so that means that too much iodine was used, but some precipitate still formed. The solution was still the dark brown orangey color. The solution was transferred into a 50mL Erlenmeyer flask using a pipette that was preheated with a solution of 5mL of glacial acetic acid and 5mL of acetic anhydride. It was then cooled to room temperature using an ice water bath, which caused a darker orange paste to form. Vacuum filtration was then used to dry the paste until a lighter brown orangey powder formed. Then the product was placed in a glass vial and put into a vacuum desiccator to dry for a week. On May 29th, 2018 at 2:50pm, the product, SnI4, had been drying for a week and had become a lighter orange color and more of a powder then it was the week before.
Tin (II) Iodide:
On May 29th, 2018, Tin (II) Iodide was synthesized using a reflux system that was flushed with nitrogen. 1.0g of tin, which was a green, brown color, was grinded until somewhat shiny and added to a 25mL round bottom flask that was purged with nitrogen for 5 minutes, from 1:56pm until 2:01pm. 1.39g of iodide, which was a silver, grey solid, was also added to the same round bottom flask. Nothing occurred when these two solids were put together nor when 10mL of 2M HCl, which is a clear liquid, was added to the same round bottom flask. The solution was set up to the water condenser to begin its reflux at 2:13pm. As the solution was being stirred and heated, it turned to a dark red color right as the reflux began. At 2:17pm the solution became pale yellow and then at 2:20pm it became an even darker yellow. The solution remained this color and began to bubble as the reflux continued. At 2:44pm the reflux was stopped and had gone on for about 30minutes. The solution had become a yellow, transparent liquid and there was some condensation present on the bottom of the water condenser attached to the round bottom flask. The solution was then transferred to a new 25mL round bottom flask, which had been purged with nitrogen for about 5 minutes, from 2:35pm until 2:40pm, to cool to room temperature in an ice, water bath. A nitrogen line was also connected to keep the atmosphere nitrogen based. As this occurred, a bright neon orange precipitate began to form. Once fully formed, the precipitate was washed using 3mL of water with a drop of 2M HCl and then dried using vacuum filtration and an orange flaky paste began to form. The product was then placed in a glass vial inside a vacuum desiccator to dry further. On May 31st, 2018 at 1:40pm, the product, SnI2, had been drying for two days and had become more of a powder, but maintained its bright neon orange color.
Compound Drawn Chemical Equation Theoretical and Percent Yield Theoretical Melting Point Determined Melting Point
Sn + 2I2 ? SnI4 Theoretical Yield: 0.592g
Percent Yield: 20.26% 144? 144.8? – 146.2?
Sn + I2 ? SnI2 Theoretical Yield: 2.056g
Percent Yield: 46.57% 320? 311.8? – 321.1?

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Calculation for Theoretical and Percent Yield:
0.12g Sn × (mol Sn ÷ 118.71g Sn) = 0.0010 mol Sn
0.48g I2 × (mol I2 ÷ 253.8089g I2) = 0.0019 mol I2
To compare as a 1:1 ratio we must divide the number of I2 moles by 2 which gives 0.000946 moles of I2.
So, this means that I2 is the limiting reagent so 0.000946 moles of SnI4 is produced.
0.00946mol SnI4 × (626.328g SnI4 ÷ mol SnI4) = 0.592g SnI4 is the theoretical yield.
Actual Yield: 0.12g
Percent Yield: (0.12g ÷ 0.592g) × 100% = 20.26%
1.0g Sn × (mol Sn ÷ 118.71g Sn) = 0.00842mol Sn
1.39g I2 × (mol I2 ÷ 253.8089g I2) = 0.00548 mol I2
Since already is in a 1:1 ratio based off the balanced equation, the limiting reagent is I2. So that means that 0.00548 moles of SnI2 should be produced.
0.00548 mol SnI2 × (372.5189g SnI2 ÷ mol SnI2) = 2.04g SnI2 is the theoretical yield.
Actual Yield: 0.95g
Percent Yield: (0.95g ÷ 2.04g) × 100% = 46.57%
The two products synthesized in the lab were SnI4 and SnI2. The melting point that was found for both tells us how pure the compound was and how strong the bonds were.4 For SnI4, the melting point range was small and it was close to the theoretical melting point which means that the compound was purer and contained almost all SnI4. The reason it may not have been completely pure is because there could have been a side reaction with oxygen.2 This is why the glacial acetic acid and acetic anhydride was used as a solvent. Both prevent water from entering the system by reacting with it, but that was the acetic anhydride’s main purpose as the solvent whereas glacial acetic acid was a good solvent since it allows iodine to dissolve in it to react with the tin. It also removed the oxide layer on the tin to allow iodine to react with the tin as well. These two solvents reduced the chance of tin reacting with the oxygen and prevented the water from causing SnI4 from going through hydrolysis.
For SnI2 the melting point range was very large and it was not as close to the theoretical one as SnI4 was. This is because SnI2 needed to be done in a nitrogen atmosphere to prevent any oxygen from reacting with the tin. Both SnI4 and SnI2 have this risk but it is much greater for SnI2 since tin is in the II state which attracts oxygen more since it has the tendency to form SnO2. The side reaction that would have occurred would be, Sn4+ + O2- ? SnO2. The oxygen would have caused the tin to oxidize to the (IV) state and then react with the oxygen present in the system. This would have affected how much product was produced and how pure the SnI2 was. When this reaction could have occurred was if oxygen had entered the system when the solution was being transferred and if it was not done fast enough so the product could have not all been SnI2. Also, the tin being oxidized to the IV state during the reaction would have caused the melting point to be lower as well since SnI4 could have also been formed since both states of tin may have been present at that time. SnI2 also uses a different solvent, which was 2M HCl, and not the glacial acetic acid with acetic anhydride. This is because SnI2 does not have the risk of going through hydrolysis if water enters the system since there is a nitrogen atmosphere. This atmosphere prevents any reaction with oxygen and water so all that is needed is the 2M HCl since it removes the oxide layer from the tin to allow it to react with the iodine. The majority of the product that was produced was SnI2 since the melting point was much closer to the theoretical one for SnI2 then the theoretical one for SnI4 and the product was in a nitrogen atmosphere for the majority of the time the reaction was going on so there would be more SnI2 present.
This melting point also tells us about the structure of the two compounds. The higher the melting point is, the stronger the bonds will be which indicates that a lattice has formed.3 For SnI4 there is not enough cations to anions to form a lattice which means only weaker Van Der Waals bonds will form. This agrees with the melting point determined since it is smaller and it will not take as much energy to break the bonds as it would for a lattice.3 For SnI2, there is enough cations to anions to form a lattice. The bonds created from this are much stronger and are covalent which results in the higher melting point. It will take more energy to break the bonds for a lattice since the entire lattice must be broken. The melting point obtained therefore matches this theory since it was much higher then the melting point for SnI4, which only has the weak forces holding it together.
Both reactions are redox reactions. The half reactions for SnI2 are, Sn(s) ? Sn2+ + 2é and I2 + 2é ? 2I-. The half reactions for SnI4 are, Sn(s) ? Sn4+ + 4é and 2I2 + 4é ? 4I-. For both of the reactions, tin is being oxidized and iodide is being reduced. This makes the reactions and the way the reactions occur similar which is why the different conditions are needed to produce the different products and to prevent, as much as possible, the side reactions from occurring.
The resultant products for both SnI2 and SnI4 showed that even with different reaction conditions, side reactions and the oxidation of tin can still occur. The melting points also showed the impurities of the compounds based on range and closeness to the theoretical value. The melting point for SnI4 was 144.8? – 146.2?, meaning the product was close to pure due to the small range and closeness to the theoretical value. For SnI2 it was 311.8? – 321.1?, which means it was less pure due to the large range. The percent yield also showed how pure the compound was and how the side reactions caused less product to be produced. The yield for SnI4 was 20.26% and the yield for SnI2 was 46.57%. Determining the yields and the melting points from the products allowed us to see how pure the product was and the different bonds that formed within the compounds. Overall, the nitrogen atmosphere did work, but when transferring the solution, the atmosphere would be oxygen which is what caused side reactions and the oxidation of tin to occur for SnI2. For SnI4, water entering the system or due to the fact that too much iodine was added, caused the yield to be a bit lower. However, the melting point
determined still gave the information wanted about the compounds SnI2 and SnI4.


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