Periodic Properties Essay Example for Free

Periodic Properties Essay The halogens F, Cl, Br and I (At has not been included because of its scarcity and nuclear instability) are very reactive non-metals that occur in the penultimate group of the periodic table, hence they all require just one electron to complete their valence shell. All 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 there 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, bromine a liquid and iodine 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 (Fà ¯Ã‚ ¿Ã‚ ½,Clà ¯Ã‚ ¿Ã‚ ½,Brà ¯Ã‚ ¿Ã‚ ½,Ià ¯Ã‚ ¿Ã‚ ½). The ease with which they gain electrons gained is further from the nucleus and hence less strongly attracted. This means that, in contrast to the alkali metals, the reactivity of the halogens decreases going down the group. Method 1) Test the solubility of Iodine: 1. A very small amount of iodine was put into water, cyclohexane and KI(aq) respectively 2. The color changes of the solutions and the solubility in each solvent were recorded 2) Test iodine reacts with starch: 1. Three drops of I2-KI solution were put into a test tube 2. A few drops of starch solution were added after that 3. The color of solution was recorded 3) Test the acid-base properties: 1. A few drops of chlorine water were put in a test surface, and it was tested with universal indicator paper 2. This was repeated first using water and then using iodine solution instead of the chlorine water 3. The color changes were recorded 4) Displacements between halogen elements: 1. 2cm depth of each aqueous solution: sodium chloride, potassium bromide and potassium iodide were put into 3 respective test tubes and labeled 2. An equal volume of chlorine water was added into each test tube and the results were recorded 3. A little hexane was added to form a separate upper layer of a non-polar solvent 4. The mixtures were shook and the changes were recorded 5. Step 1, 2, 3 and 4 were repeated first using water and then iodine solution instead of chlorine water 5) Tests for halide ions [Halide ions (Cl-, Br- and I-) with silver ions]: 1. About 1cm depth of aqueous sodium chloride was put into a test tube 2. A little aqueous silver nitrate was added and then the observations were recorded 3. The test tube was placed in a sunny place, and left there for about 5 minutes and then it was observed again 4. 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 Color Solubility Water Colorless Insoluble Cyclohexane Purple Soluble Ethanol Yellow Soluble KI(aq) Yellow-brown Soluble 2) Test iodine reacts with starch: The color of the solution is black. 3) Test the acid-base properties: Cl2 Br2 I2-KI pH value 4 3 12 4) Displacements between halogen elements: The color change of the solution after Cl2, Br2, I2 added into NaCl, KBr and KI respectively Cl2 Br2 I2 NaCl No change No change Brown KBr Pale yellow solution No change Brown KI yellow yellow Brown The color of the upper layers of the solution after hexane added Cl2 Br2 I2 NaCl No change No change Purple red KBr Pale purple No change Purple red KI purple Pale purple Purple red 5) Tests for halide ions: Halide ions (Cl-, Br- and I-) with silver ions: NaCl White precipitate is produced Darkens after it was placed in sunlight KBr Cream precipitate is produced. KI Yellow precipitate is produced. ?Data Analysis? 1) The solubility of iodine in different solvents: The solubility is larger in non-polar solvent (water, ethanol) and smaller in polar solvents.(cylohexane 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 starch: According to the general knowledge we knew, the phenomenon of this reaction 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 properties: 1. Cl2: The color of the universal indicator papers showed that Cl2 is strong acid. 2. Br2: The color of the universal indicator papers showed that Br2 is a kind of acid, but not very strong. 3. I2: The color of the universal indicator papers showed that I2 is a strong base. Actually, I2 is acid. The reason is that the original color of I2 is red-brown, that made us cant see the phenomenon clearly. 4) Displacements between halogen elements: As what I mentioned above in background, the rule of displacements between halogen elements is that more reactive ones displace less reactive ones. Thats the reason why Br -cant displace Cl -, and I -cant displace Br and Cl-. When there was no reaction between two 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 instances 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 will show the color of itself. That was why we could observe color of purple in this experiment. 5) Test for halide ions: When halide ions dissolved into silver salts, then the precipitate is appear commonly. The white precipitate is AgCl: AgNO3+NaClà ¯Ã‚ ¿Ã‚ ½AgCl+NaNO3 The off-white precipitate is AgBr: AgNO3+KBrà ¯Ã‚ ¿Ã‚ ½AgBr+NaNO3 The pale yellow precipitate is AgI: AgNO3+NaIà ¯Ã‚ ¿Ã‚ ½AgI+NaNO3 After 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 electronegativity going down the group decreases. 2. Organic solvents always contain the element carbon. Inorganic 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 typical blue black color will appear. Thats a good way for us to identify starch and iodine. 5. After photodissociation, 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 inside 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. REFERENCE 1) à ¯Ã‚ ¿Ã‚ ½Chemistryà ¯Ã‚ ¿Ã‚ ½(for use with the International Baccalaureate Diploma Programme) [3rd Edition] John Green Sadru Damji First published in 2007 by IBID Press, Victoria, Page 77 to 78. 2) http://www.epa.gov/ttn/atw/hlthef/hexane.html 3) http://baike.baidu.com/view/373611.htm 4) http://baike.baidu.com/view/908645.htm