Experiment 11. Periodic Properties: Cation and Anion Analysis. Purpose: Background:

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Experiment 11 Periodic Properties: Cation and Anion Analysis Purpose: The purpose of this experiment is to develop a scheme to distinguish (classify) Group 2A (Alkaline Earth Metals) cations based on the solubility properties of salts formed with polyatomic anions and to determine the relative ability of halogens to oxidize halide anions. Background: In the nineteenth century, chemists attempted to classify and organize chemical elements in a useful way. The Russian chemist Dimitri Mendeleev is generally credited with the publication, in 1869, of the first widely recognized periodic table. He developed his table to illustrate the observed periodic trends in the physical and chemical properties of the then-known elements. Later it was observed that if the elements are arranged in order of increasing atomic number (number of protons in the nucleus) elements with similar properties tend to appear at somewhat regular intervals in the list. The standard form of the Periodic Table consists of a grid of elements laid out in 18 columns and 7 rows, with a double row of elements below that. The rows of the table are called periods and the columns are called groups. Each group contains one family of elements, with some of these having common names such as Alkali metals, Alkaline Earth Metals Halogens or Noble Gases. The table can also be deconstructed into four rectangular blocks presenting the buildup of the s, p, d, f electron subshells (orbitals): the s-block to the left, the p-block to the right, the d- block in the middle, and the f-block below that. Each block is named according to the subshell (orbital) in which the "last" electron notionally resides. The s-block, where the outer electron shell contains either one or two s electrons, comprises the first two groups (alkali metals and alkaline earth metals) as well as hydrogen and helium. The p-block, where 6 electrons are added to the 3 p orbitals after the s orbital shell is filled, comprises the last six groups: 13-18 (IUPAC) or 3(III)A 8(VIII)A (American). The elements in this group include the nonmetals and the metalloids. The d-block, where the 5 d orbitals (10 electrons total) are added, comprises groups 3 to 12 or 3(III)B to 2(II)B, and contains all of the transition metals. The f-block ( 7 orbitals,14 electrons total) comprises the lanthanides and actinides, which are usually offset below the rest of the Periodic Table. The similar physical and chemical properties of the families of elements are determined by their electronic structures around the nucleus, that is, each family has a similar s and p outer shell configuration. For example, the group 1A elements (alkali metals) are distinguished by

having a single s electron in the outer shell (2s 1, 3s 1 ); the group 2A elements (alkaline earth metals) have two s electrons (1s 2, 3s 2 ); and the group 3A elements (nonmetals) have two s electrons and a single p electron (2s 2 2p 1, 3s 2 3p 1 ). Elements, in their chemical reactions, attempt to gain or lose electrons to achieve a stable noble gas configuration in which the s orbitals are filled with 2 electrons and p orbitals are filled with 6 electrons. After helium, with just 2 s electrons, the elements try to achieve an octet configuration (the octet rule) consisting of 2 s electrons and 6 p electrons. The first 20 elements are most likely to follow the octet rule, after which the stability is governed by the filling of the 5 d orbitals (10 electrons). The periodic trends that arise from the arrangement of the elements in the periodic table exist because of the similar atomic structure of the elements within their respective group families or period and the periodic nature of the elements. The main periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Electronegativity: Electronegativity describes an atom's ability to attract and bind to electrons. Elements on the left side of the periodic table have less than a half-full valence shell, thus, the energy required to gain electrons is significantly higher compared to the energy required to lose electrons. As a result, the elements on the left side of the periodic table generally lose electrons in forming bonds. Conversely, elements on the right side of the periodic table are more energy-efficient in gaining electrons to create a complete valence shell of 8 electrons. The more inclined an atom is to gain electrons, the more likely that atom will pull electrons toward itself. Electronegativity increases moving to the right across a period of elements As you move down a group, electronegativity decreases. This is because the atomic number increases down a group and thus there is an increased distance between the valence electrons and the positively charged nucleus, or a greater atomic radius Ionization Energy: Ionization Energy is the amount of energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is considered the opposite of electronegativity. The ionization energy of the elements within a period generally increases from left to right. The closer the electronic configuration gets to the stable noble gas configuration the more difficult (more energy required) it is to remove an electron. The ionization energy of the elements within a group generally decreases from top to bottom. This is due to an electron being shielded from the influence of the nucleus.

Electron Affinity: Electron affinity describes the ability of an atom to accept an electron. Unlike electronegativity, electron affinity is a quantitative measure that measures the energy change that occurs when an electron is added to a neutral gas atom. When measuring electron affinity, the more negative the value, the more of an affinity to electrons that atom has. Electron affinity increases from left to right within a period. This is caused by the decrease in atomic radius. Electron affinity decreases from top to bottom within a group. This is caused by the increase in atomic radius. Atomic Radius: Atomic radius can be described one-half the distance between the nuclei of two like atoms held together by a single covalent bond. Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are being added to the same shell. But, at the same time, protons are being added to the nucleus, making it more positively charged. The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction. This means that the nucleus attracts the electrons more strongly, having the atom's shell pulled closer to the nucleus. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases. Going down a group, it can be seen that atomic radius increases. The valence electrons occupy higher levels due to the increasing quantum number (n). As a result, the valence electrons are further away from the nucleus as the number of neutrons increases. The dominance of electron shielding over nuclear charge prevents the outer electrons from being attracted to the nucleus; thus, they are loosely held and the resulting atomic radius is large. Ionic Radius: Ions may be larger or smaller than the neutral atom, depending on the ion's charge. When an atom loses an electron to form a cation, the lost electron no longer contributes to shielding the other electrons from the charge of the nucleus. As a result, the other electrons are more strongly attracted to the nucleus, and the radius of the atom gets smaller. Similarly, when an electron is added to an atom, forming an anion, the added electron shields the other electrons from the nucleus, with the result that the size of the atom increases. As with elemental radius, ionic radius increases going down a group as the number of energy levels increase. Ionic size becomes more complicated going across a period. As the atomic number increases from groups 1A to 3A, the cations show a decrease in ionic radius.

As the ions change to anions (greater electron affinity) in groups 5A to 7A, the ionic radii are considerably greater relative to the cations, but still show a decreasing trend in size. Cations (Alkaline Earth Metals) Table 1 lists the physical properties of the elements in group 2A of the periodic table. These elements are classified as metals and are generally referred to as the Alkaline Earth Metals. They have 2 electrons filling an s orbital and their removal results in a noble gas configuration for the resulting di-cation. M : M + 2e 2+ - Like other metals, the Alkaline Earth metals have low electronegativities and tend to form positively charged cations. The di-cation members of this group form ionic salts with many negatively charged ions. They can be distinguished from other cations and from each other by their reactions and solubility properties with the following polyatomic ions: Anions (Halogens): Sulfate (SO 4 ) 2- Carbonate (CO 3 ) 2- Oxalate (C 2 O 4 ) 2- Chromate (CrO 4 ) 2- Halogens exist naturally as diatomic covalently bonded molecules (Chlorine (Cl 2 ), Bromine (Br 2 ), Iodine (I 2 )). In ionic solutions the halogens typically exist as negatively charged halide anions (Cl -, Br -, I - ) forming many of the commonly known salts such as sodium chloride (NaCl). Table 2 lists the physical properties of the elements in group 7A of the periodic table. An important chemical property of the Group 7A halogens is that they can act as oxidizing agents in oxidation/reduction (redox) reactions, where the oxidation state of the resulting halide can range from -1 to +7. An oxidation takes place when an atom or molecule loses one or more electrons. A reduction occurs when an atom or molecule gains one or more electrons. The number of electrons gained must equal the number of electrons lost and the two reactions must always occur together. The species, which is being oxidized, gives (loses) its electron(s) to another substance, which is then reduced (gains electrons). Thus, the oxidized species acts as a reducing agent and the reduced species acts as an oxidizing agent. Oxidized Species Reduced Species loses electrons gains electrons increases its oxidation state decrease its oxidation state is a reducing agent is an oxidizing agent The Group 7A halogens are relatively easily reduced (good oxidizing agents) compared to non-halogens. The reaction for the reduction of a halogen to two halide anions can be written:

(1) The diatomic halogen, X 2, has 14 valence electrons and the two halide anions have a total of 16 valence electrons, thus, the halogen atoms have been reduced by gaining electrons. There are some powerful oxidizing agents which can oxidize the halide anions to a halogen molecule by removing halide electrons: (2) Notice that equations (1) and (2) above are the reverse of each other. Halogens are reduced to halide ions and halide ions are oxidized to halogens. It is understood that there are spectator cations present which balance the halide anion charge. What might be expected when two different halogen/halide pairs are mixed together? Suppose Cl and Br are chosen as the Group 7A pair. There are two combinations for mixing these: chlorine/bromide and bromine/chloride. (Note that the diatomic halogen elements have the ine suffix and the halide anions have the ide suffix.) Although balanced chemical reactions can be written for each oxidation/reduction combination, only one of these reactions will actually occur. or - - 2 2 Cl + 2Br 2Cl + Br - - 2 2 2Cl + Br Cl + 2Br (3) (4) If reaction (3) tales place, it means that chlorine (which is reduced to chloride in the reaction) is a stronger oxidizing agent than bromine because chlorine oxidizes bromide to form bromine. If reaction (4) takes place, it means that bromine is a stronger oxidizing agent than chlorine, because bromine oxidizes chloride to form chlorine. If equation (3) is read from right to left, it is identical to equation (4.), likewise, if reaction (4) is read from right to left it is identical to equation (3). Only one of these reactions can take place.

Table 1: Physical Properties of the Alkaline Earth Elements Physical Properties of the Alkaline Earth Elements (Group 2A) Property Magnesium Calcium Strontium Barium Electronic Configuration Atomic Radius (nm) Ionic Radius M 2+ (nm) (Ne)3s 2 (Ar)4s 2 (Kr)5s 2 (Xe)6s 2 0.136 0.174 0.191 0.198 0.065 0.099 0.113 0.135 Ionization Energy (kj mol -1 ) 1 st 738 2 nd 1451 589 1148 550 1061 503 965 Electronegativity 1.2 1.0 1.0 0.9 Table 2: Physical Properties of the Halogens Physical Properties of the Halogens (Group 7A) Property Chlorine Bromine Iodine Electronic Configuration Atomic Radius (nm) Ionic Radius X - (nm) (Ne)3s 2 3p 5 (Ar)3d 10 4s 2 4p 5 (Kr)4d 10 5s 2 5p 5 0.099 0.114 0.133 0.181 0.195 0.220 Ionization Energy (kj mol -1 ) 1254 1141 1007 Electron Affinity 348 324 295 Common -1, +1, +3, +5, +7 for all three Oxidation States Electronegativity 3.0 2.8 2.5

Part A: Alkaline Earth Cations The Experiment: In this part of the experiment the solubility properties of the group 2A Alkaline Earth metals will be classified according their reactions with the divalent polyatomic anions listed above. In order to express these reactions it will necessary to become proficient in writing equations involving positively charged cations and negatively charged anions, that is, ionic equations. Many ionic compounds do not dissociate appreciably in aqueous solution. Although composed of ions, they are confined in their crystal lattice. Solid ionic compounds are represented in chemical equations by leaving out the charges. For example copper sulfide is composed of copper cations and sulfide anions and in undissociated form would be written as CuS. The formula for a crystalline compound, which is present in its undissociated form, is usually followed by the notation (s) which means it is a solid. Any compound which is significantly ionized (dissociated) in water is written as the formula of the ion with a numerical superscript indicating its charge. For example, solid sodium chloride (NaCl, table salt) will dissolve in water forming sodium cations (Na + ) and chloride (Cl - ) anions. HO NaCl(s) Na + Cl 2 + - In aqueous solution ions are surrounded by molecules of water which stabilize the ions and keep them separated. These solvated ions are commonly written as Na + (aq) and Cl - (aq). In order to help keep track of ions in aqueous solution, enclose in parentheses the symbols of all ions derived from the same compound formula. For example, potassium bromide and lithium hydroxide when dissolved in water form solvated ions: HO 2 + - + - KBr(s) + LiOH(s) (K + Br ) + (Li + OH ) Inspect the following chemical reaction: (Ag + NO ) + (Na + Cl ) AgCl(s) + Na + NO + - + - + - 3 3 This equation should be interpreted as a reaction taking place when an aqueous solution of dissociated silver nitrate is mixed with an aqueous solution of dissociated sodium chloride. When the Ag + and the Cl - ions encounter each other in solution, they combine into an undissociated solid ionic compound. The ions are confined to their crystal lattice, and the AgCl, which is insoluble in water, precipitates out of the aqueous solution. Sometimes, a vertical arrow is written beside the compound to indicate this precipitation reaction. Notice that the nitrate and sodium ions are not involved in the overall precipitation reaction. They are referred to as spectator ions in the solution. To write a net ionic equation, cross out the spectator ion on both sides of the chemical equation and rewrite the resulting equation:

+ - Ag + Cl AgCl(s) In this experiment, aqueous nitrate solutions of Alkaline Earth cations will be mixed together with polyatomic anion. The results will be reported in terms of the relative amounts of precipitate formed, if any. Soluble Cloudy Light Precipitate Heavy Precipitate All the mixtures used in this experiment eventually form precipitates, so the results should be reported within the first minute after mixing and stirring. Also, the color of the precipitate and remaining solution should be noted. The results of the experiment will be to develop a plan for the identity of an unknown Alkaline Earth metal using just two of the anion reagents. The goal is to understand the solubility properties of cations sufficiently well to develop a logical analysis scheme. Pre-Lab Report & Notebook: Note: The Cation and Anion Analysis experiments can be done in either one or two laboratory sessions as time permits. The instructor will determine whether one or two lab reports are to be prepared. Download from the department data base to your hard drive or flash drive a copy of the lab report template and the data summary tables for both the Cation Analysis and the Anion Analysis experiments. http://chem.gmu.edu/templates Print the summary results tables for the Cation and Anion Analysis experiments. Prepare the Pre-lab report according to instructor s instructions. Materials and Equipment: Materials Equipment Drop Bottles for Cations 4 small test tubes (labelled) 0.1 M Mg(NO 3 ) 2 metal spatula or glass stirring rod 0.1 M Ca(NO) 2 0.1 M Ba(NO 3 ) 2 0.1 M Sr(NO 3 ) 2 Drop Bottles for Anions 1.0 M H 2 SO 4

Procedure: 1.0 M Na 2 (CO 3 ) 0.2 M (NH 4 ) 2 (C 2 O 4 ) 1.0 M K 2 (CrO 4 ) Unknown Cation Solution 1. Obtain 4 small test tubes and attach a label indicating the Alkaline Earth metal being tested 2. Each test tube must be cleaned and rinsed thoroughly with distilled water. Note: The test tubes need not be dried. The stirring rod should be rinsed thoroughly between tubes The aqueous solutions from each test can be flushed down the drain 3. Transfer 10 drops of each alkaline earth metal nitrate solution to one of the four small labelled test tubes 4. To each of the test tubes add 10 drops of 1-M sulfuric acid and mix well Note: Sulfuric acid is soluble in water and acts as a source of sulfate (SO 4 2- ) ions. 5. Note the results of the reaction, if any, within 1 minute of mixing the reagents 6. Flush the reaction mixtures down the drain and clean the tubes with copious amount of tap water, then with distilled water 7. Repeat step 3 through 6 testing the same Group 2 cations with 1-M sodium carbonate solution Na 2 CO 3 is a soluble salt and acts as a source of carbonate (CO 3 2- ) ions 8. Repeat step 3 through 6 testing the same Group 2 cations with 1-M ammonium oxalate solution (NH 4 ) 2 (C 2 O 4 ) is a soluble salt and acts as a source of oxalate (C 2 O 4 2- ) ions 9. Repeat step 3 through 6 testing the same Group 2 cations with 1-M potassium chromate solution K 2 (CrO 4 ) is a soluble salt and acts as a source of chromate (CrO 4 2- ) ions Unknown Alkaline Earth Metal: Devise a plan, based on the results above, for analyzing an unknown alkaline earth cation using just two of the anion reagents. Obviously, the unknown could be identified by simply following the steps used with the known cations, but the goal here is to understand the solubility properties of the cations sufficiently well to develop a more efficient logical analysis. 10. Obtain an unknown alkaline earth nitrate solution from the instructor 11. Transfer 10 drops of the unknown solution to one of the labelled test tubes representing the first of the two anions you identified in you analysis scheme. 12. Transfer 10 drops of the anion solution to the test tube and mix well 13. Note the results of the reaction, if any, within 1 minute of mixing the reagents 14. Repeat steps 11 through 13 for the second anion solution from your analysis scheme

15. Clean and dry the test tubes and police you bench work area Data Analysis: Use the printed Pre-lab as the laboratory notebook to record the experimental results of the experiment in the appropriate procedure results block. Summarize the measured and computed laboratory results in the printed copy of the Cation Analysis Results Summary Table. If required by the instructor, transfer the laboratory results to the electronic files and finalize the laboratory report. Discussion items for laboratory report: a) State which Group 2(II) alkaline earth cation forms the fewest insoluble salts. b) Which cation forms the most insoluble salts? c) Describe how the relative solubilities vary according to the position of the cation in the Periodic Table. d) Predict and describe your reasoning whether RaCrO 4 to be soluble or insoluble in water. e) Describe your reasoning for the scheme you devised to determine the identity of an unknown cation using just 2 reagents.

Part B: Oxidation/Reduction of Group 7A Halogens The Experiment In this part of the experiment the relative oxidizing strengths of the halogens chlorine, bromine, and iodine will be investigated. The determination of which halogen is the stronger oxidizing agent could be determined by consulting a table on standard electrode potentials. The halogen with the higher (more positive) standard electrode potential is the stronger reducing agent. In this experiment, however, the relative strength of a halogen acting as an oxidizing agent will be determined experimentally. The experiment takes advantage of the different colors and the different solubility characteristics of the halogens and halides in order to judge whether a redox reaction has occurred. Halogens (X 2 ) and halides (X - ) have different solubilities in water and in certain organic solvents. In polar water, the ionic halides are more soluble than are the halogens; in the relatively non-polar organic solvent, the non-polar halogens are more soluble than are the halides. Polar water and a non-polar solvent are immiscible, that is, they are insoluble in each other, and they will separate into two distinct liquid layers upon being mixed. Depending on the density of the organic solvent used, it may be either lighter or heavier than water and thus it may be either the upper or lower liquid layer. The instructor will indicate the solvent to be used and its density relative water. The halides are all colorless in aqueous solution in contrast to the halogens, each of which has a distinctive color when dissolved in an appropriate solvent. Chlorine (Cl 2 ) is a very pale yellow, Bromine (Br 2 ) is red-brown, and Iodine (I 2 ) is purple. Thus, although individual halides in aqueous solution cannot be distinguished, it is quite easy to differentiate their colored halogen counterparts in the organic solvent layer. Each halogen dissolved in an organic solvent will be mixed with each of the aqueous halide solutions. The resultant changes in the color of the organic layer will be observed. From these results a process for analyzing an unknown halide anion with a minimum number of halogens will be developed. Afterwards, the analysis scheme will be used to identity of an unknown halide. Example: Using the previous Cl/Br example above, suppose a pale-yellow Cl 2 is mixed with colorless Br -. The organic layer develops a red-brown color. This means that Br 2 (red-brown) has formed by the redox equation (3), and thus Cl 2 has oxidized Br -, that is Br - has lost electrons to chlorine forming Cl -. As a check for this result, mixing colorless Cl - with redbrown Br 2 (equation (4) would produce no change in the organic layer. Since Cl 2 mixed with Br - results in an oxidation of bromine (loss of electrons), but Br 2 mixed with Cl -, chlorine does not produce a reaction, chlorine is a stronger oxidizing agent than bromine.

Materials and Equipment: Materials Halogen Solutions Chlorine (Cl 2 ) Bromine (Cr 2 ) Iodine (I 2 ) Halide Solutions Chloride (Cl - ) Bromide (Br - ) Iodide (I - ) Organic Solvent Equipment 9 small test tubes (labelled) Procedure: 1. Set up 9 clean, distilled water washed test tubes (need not be dry) 2. Add 1 ml of a halogen solution to 3 test tubes 3. Repeat for the next halogen solution 4. Repeat for the next halogen solution 5. Label each test tube for its halogen content 6. Add 2 ml of the organic solvent to each test tube 7. Cork each test tube and shake thoroughly 8. Note the color of the organic layer color of halogen X 2 9. Add 1 ml of a halide solution (X - ) drop by drop to a halogen test tube 10. Shake gently during the addition; then cork and shake thoroughly 11. Observe the color of the organic layer the top layer 12. Repeat using the same halide solution on another halogen test tube 13. Repeat for the third halogen test tube 14. Repeat steps 9-13 for the second halide solution 15. Repeat steps 9-13 for the third halide solution 16. Obtain an unknown halide from the instructor Unknown Halide: 1. Set up 3 clean, distilled water washed test tubes (need not be dry) 2. Add 1 ml of the Cl 2 solution to the first test tube 3. Add 1 ml of the Br 2 solution to the second test tube 4. Add 1 ml of the I 2 solution to the third test tube 5. Add 2 ml of the organic solvent to each test tube 6. Cork each test tube and shake thoroughly 7. Add 1 ml of the unknown halide solution (X - ) drop by drop to the Cl 2 test tube

8. Shake gently during the addition; then cork and shake thoroughly 9. Observe the color of the organic layer the top layer 10. Repeat adding the unknown halide solution on the Br 2 test tube 11. Repeat adding the unknown halide solution on the I 2 test tube Waste disposal: All reaction solutions contain an organic solvent, thus, all test solutions must be disposed of in the designated waste jar in the hood. Data Analysis: Write balance chemical equations for each test performed which underwent a redox reaction. Use the printed Pre-lab as the laboratory notebook to record the experimental results of the experiment in the appropriate procedure results block. Summarize the measured and computed laboratory results in the printed copy of the Anion Analysis Results Summary Table. If required by the instructor, transfer the laboratory results to the electronic files and finalize the laboratory report. Discussion items for laboratory report: a) Formulate a statement about the relative oxidizing powers of the halogens (X 2 ) and their positions in the Periodic Table. Explain you reasoning. b) Write equations for the following reactions and provide an explanation as to whether the reactions will occur: Astatine (At 2 ) can oxidize iodide (I - ) Chloride (Cl - ) can reduce fluorine (F 2 ) c) Describe your reasoning for the scheme you devised to determine the identity of an unknown cation using just 2 reagents

Results Summary Tables PERIODIC PROPERTIES: CATION ANALYSIS Group II Nitrate Salt (SO 4 ) 2- (CO 3 ) 2- (C 2 O 4 ) 2- (CrO 4 ) 2- Mg 2+ Sol PPT Sol Sol Yellow Rxn (SO 4 ) 2- Write, type, or insert equation here, if precipitate forms Rxn (CO 3 ) 2- Rxn (C 2 O 4 ) 2- Rxn (CrO 4 ) 2- Ca 2+ Rxn (SO 4 ) 2- Rxn (CO 3 ) 2- Rxn (C 2 O 4 ) 2- Rxn (CrO 4 ) 2- Sr 2+ Rxn (SO 4 ) 2- Rxn (CO 3 ) 2- Rxn (C 2 O 4 ) 2- Rxn (CrO 4 ) 2- Ba 2+ Rxn (SO 4 ) 2- Rxn (CO 3 ) 2- Rxn (C 2 O 4 ) 2- Rxn (CrO 4 ) 2- Unknown # Rxn ( ) 2- Rxn ( ) 2-

PERIODIC PROPERTIES: ANION ANALYSIS Halogen Halide Cl - Br - I - Cl 2 Color: Color: Color: Rxn (Cl 2 Cl - Rxn (Cl 2 Br - ) Rxn (Cl 2 I - ) Br 2 Color: Color: Color: Rxn (Cl 2 Cl - Rxn (Cl 2 Br - ) Rxn (Cl 2 I - ) I 2 Color: Color: Color: Rxn (Cl 2 Cl - Rxn (Cl 2 Br - ) Rxn (Cl 2 I - ) Unknown Halide Halide Halogen Cl 2 Br 2 I 2 Unk # Color: Color: Color: Rxn w/cl 2 Rxn w/br 2 Rxn w/i2