Periodic Properties Essay Example for Free

Periodic Properties EssayThe halogens F, Cl, Br and I (At has not been included because of its scarcity and nuclear instability) are truly(prenominal) reactive non-metals that occur in the penultimate group of the periodic table, thusly they all in all require honorable one negatron to complete their valence shell. tout ensemble of the elements exists as diatomic molecules (F2, Cl2, Br2, I2) in which the atoms are joined by single covalent bonds. Going down a group of the periodic table, for successive elements on that point are more energy levels filled with electrons, so the outer electors are in higher energy levels and farther from the nucleus. Fluorine and chlorine are gases, atomic number 35 a liquid and ace a solid that forms a purple vapour on heating. The halogens are all quite electronegative elements. They require just one electron to complete their valence shell, hence they readily gain electrons to form the singly charged halide ions (F,Cl,Br,I). The ease wi th which they gain electrons gained is further from the nucleus and hence little strongly attracted. This means that, in contrast to the alkali metals, the reactivity of the halogens decreases going down the group.Method1) Test the solubility of Iodine1. A very small amount of iodine was put into pissing, cyclohexane and KI(aq) respectively2. The warp changes of the declarations and the solubility in each solvent were recorded2) Test iodine reacts with starch1. cardinal drops of I2-KI solution were put into a test electron tube-shaped structure2. A few drops of starch solution were added after that3. The tinct of solution was recorded3) Test the acid-base properties1. A few drops of chlorine water were put in a test surface, and it was tested with universal indicator paper2. This was repeated first victimisation water and and so using iodine solution instead of the chlorine water3. The color changes were recorded4) Displacements mingled with halogen elements1. 2cm depth o f each aqueous solution sodium chloride, potassium bromide and potassium iodide were put into 3 respective test tubes and labeled2. An equal volume of chlorine water was added into each test tube and the results were recorded3. A little hexane was added to form a separate upper layer of a non-polar solvent4. The mixtures were shook and the changes were recorded5. Step 1, 2, 3 and 4 were repeated first using water and then iodine solution instead of chlorine water5) Tests for halide ions Halide ions (Cl-, Br- and I-) with silver ions1. About 1cm depth of aqueous sodium chloride was put into a test tube2. A little aqueous silver nitrate was added and then the observations were recorded3. The test tube was placed in a sunny place, and left there for about 5 minutes and then it was observed again4. Step 1, 2 and 3 were repeated using aqueous potassium bromide, then aqueous potassium iodide instead of sodium chloride?Data Collection?1) The solubility of iodine in different solvent vividn essSolubilityWaterColorlessInsolubleCyclohexanePurpleSolubleEthanolYellowSolubleKI(aq)Yellow-brownSoluble2) Test iodine reacts with starchThe color of the solution is black.3) Test the acid-base propertiesCl2Br2I2-KIpH value43124) Displacements between halogen elementsThe color change of the solution after Cl2, Br2, I2 added into NaCl, KBr and KI respectivelyCl2Br2I2NaClNo changeNo changeBrownKBrPale yellow solutionNo changeBrownKIyellowyellowBrownThe color of the upper layers of the solution after hexane addedCl2Br2I2NaClNo changeNo changePurple deprivationKBrPale purpleNo changePurple redKIpurplePale purplePurple red5) Tests for halide ionsHalide ions (Cl-, Br- and I-) with silver ionsNaClWhite lessen is producedDarkens after it was placed in sunlightKBrCream precipitate is produced.KIYellow precipitate is produced.?Data Analysis?1) The solubility of iodine in different solventsThe solubility is larger in non-polar solvent (water, ethanol) and smaller in polar solvents.(cylohexa ne and KI) The purple color of iodine in cyclohexane is that because in non-polar solvents, iodine froms the violet solution.2) Test iodine reacts with starchAccording to the general knowledge we knew, the phenomenon of this response should be blue, but the color observed was black-green. That was because some of the starch hydrolysis in water and produced something could make the color darker.3) Test the acid-base properties1. Cl2The color of the universal indicator paper showed that Cl2 is strong acid.2. Br2The color of the universal indicator papers showed that Br2 is a kind of acid, but not very strong.3. I2The color of the universal indicator papers showed that I2 is a strong base. Actually, I2 is acid. The understanding is that the original color of I2 is red-brown, that made us cant see the phenomenon clearly.4) Displacements between halogen elementsAs what I mentioned above in background, the rule of displacements between halogen elements is that more reactive ones displa ce less reactive ones.Thats the reason why Br -cant displace Cl -, and I -cant displace Br and Cl-. When there was no reaction between 2 elements, the color we observed was the blend of original colors of the less reactive element and the solution containing the more reactive element. If theres a reaction between two elements, the color we can observe is the color of the displaced element.According to the information we got from Internet, we knew hexane is a kind of oil and is insoluble in water-solvent. That was the reason why we could differentiate the two layers of each solution very clearly.The colors of each solutions under layer were the original colors of the saline solutions. There were two kinds of events of the color of upper layer of each solution. For the solutions those do not have I ion, they were colorless. Thats because hexane is colorless and cannot react with Cl or Br -. Another instance is that the solutions include I -, when I meets hexane, it get out show the color of itself. That was why we could observe color of purple in this experiment.5) Test for halide ionsWhen halide ions dissolved into silver salts, then the precipitate is appear commonly.The white precipitate is AgCl AgNO3+NaClAgCl+NaNO3The off-white precipitate is AgBr AgNO3+KBrAgBr+NaNO3The pale yellow precipitate is AgI AgNO3+NaIAgI+NaNO3After 10 minutes under the sunshine, photodissociation happened on all of them, so the black precipitate on the bottoms of three test tubes are the products of photodissociation.1. Going down the group, the elements of this group have the same effective nuclear charge. atomic radius of these elements becomes bigger because of the increase of the number of energy levels. The attraction between nucleus and valence electrons gets weaker. Less energy is required to remove the first electron from one mole of gaseous atoms. The ionization energy going down the group decreases. The ability to attract electrons becomes weaker. The electronegati vity going down the group decreases.2. ingrained solvents always contain the element carbon. In radical solvents dont contain the element carbon. The most common solvent, water, is an example of an inorganic solvent. There are many more organic solvents than inorganic solvents. Compare with organic and inorganic solvent, the solubility of iodine is higher in organic solvent.3. The oxidizing power of the halogens decrease going down the group as the size of the atoms increase going down the group as the size of the atoms increases and the attraction between the nucleus and the electrons becomes less. In that case, going down the group, the elements become less powerful oxidising agents. This means that a higher halogen will displace a lower halogen from its salts. A lower halogen cannot displace a higher halogen from its salts.4. When starch reacts with iodine, the emblematic blue black color will appear. Thats a good way for us to identify starch and iodine.5. After photodissociat ion, the color of some precipitates will change. will become black. Thats the most obvious one. Other precipitates will become darken.1. Because we use solid iodine in the first experiment. If we add the solvent into the test tube first, the test tube will be wet and the solid iodine we put in later will attach on the surface wrong instead of fall into the liquid. For this reason we must add solid iodine first in experiment 1.2. According to the first experiment, we found that the solubility of iodine in pure water is very low. But the solubility of iodine in potassium iodide solution is relatively much higher. So we use I2-KI solution to increase the amount of iodine in order to let the phenomenon more obvious.REFERENCE1) Chemistry(for use with the International Baccalaureate Diploma Programme)3rd Edition John color Sadru DamjiFirst published in 2007 by IBID Press, Victoria, Page 77 to 78.2) http//www.epa.gov/ttn/atw/hlthef/hexane.html3) http//baike.baidu.com/view/373611.htm4) h ttp//baike.baidu.com/view/908645.htm