Experiment 5 Molecular Geometry & Polarity

Transcription

1 Experiment 5 Molecular Geometry & Polarity Name Date Section Purpose: To determine molecular geometry from a Lewis structure. To determine molecular polarity from bond polarity and geometry. To use polarity to understand intermolecular forces and its impact on physical properties and solubility. Background Information: Molecular formulas, and more specifically structural formulas with hydrocarbons, distinguish one substance from another. As you should know by now, H 2 O (water) is different than H 2 O 2 (hydrogen peroxide) not only in the number of oxygen atoms, but also in many physical and chemical properties. A simple Lewis structure provides a picture of atom connectivity and can help explain these differences. Additionally, the Lewis structures can be redrawn to provide a three dimensional representation of the molecules using the VSEPR theory (valence shell electron pair repulsion). Using atomic electronegativities with these molecular geometries, one can then determine polarity. It is this concept of polarity that can provide further insight into physical properties such as melting point and boiling point as well as solubility and chemical reactivity. Lewis structures can be created from a simple understanding of the number of valence electrons (outer shell) on an atom and the concepts of covalent bonds and the octet rule. The number of valence electrons on an atom can be determined from its position on the periodic table and group number. Taking carbon as an example, it belongs to group IVA on the periodic table and has four valence electrons. A Lewis symbol can be represented for carbon by imagining a box around the atomic symbol with dots placed around this box to represent the valence electrons with no more than two per side, avoiding pairing until necessary. From a formula such as CH 4, one can provide the Lewis symbols for each atom in the formula, as in the figure below. These individual symbols can then be connected by sharing one electron dot on one atom with an electron dot on another atom to form a covalent bond. This pattern is continued until all atoms in the formula are bonded to one or more atoms such that any atom has no more than eight total electrons (only two for hydrogen) as shown for CH 4 below. Figure 1: Lewis Structure & VESPR Structure of Methane If an atom has more than four valence electrons, as in nitrogen and oxygen (5 and 6 respectively), its Lewis symbol will contain paired electrons (dots). These paired electrons are known as lone pairs (l.p.) and are not bonded to another atom within a molecule. In completion of this experiment, you will see this in the NH 3 and H 2 O molecules. Once a correct Lewis structure has been determined, the VSEPR theory can be applied to obtain a three dimensional representation of the molecule. For a given central atom, such as carbon in CH 4, the surrounding bonded atoms and any lone pair electrons will repel each other and maximize their separation. The visualization of this concept is easier for some molecules than others. The following table outlines the

2 assignment of electron group arrangement and ultimately molecular geometry based on the number of bonded atoms and lone pairs on a central atom. The C in CH 4 has four bonded atoms (hydrogens) and no lone pairs, so the electron arrangement and geometry will be tetrahedral as shown in the Figure 1. Figure 2: s around Center Atom (or Atom of Interest) in VSEPR Theory Electron Groups Atoms & Lone Pairs (l.p.) Electron Group Arrangement Molecular Geometry 2 2 atoms, 0 l.p. Linear Linear atoms, 0 l.p. Trigonal planar Trigonal planar 2 atoms, 1 l.p. Trigonal planar Bent or V-shaped 4 atoms, 0 l.p. Tetrahedral Tetrahedral 3 atoms, 1 l.p. Tetrahedral Pyramidal 2 atoms, 2 l.p. Tetrahedral Bent or V-shaped To determine molecular polarity, one must first understand bond polarity. Due to differences in electronegativities, any bond between two different atoms will have a dipole moment or a partial separation of charge. The more electronegative an atom, the greater is its pull on the bonded electrons. The electronegativity of the elements follow a periodic trend, increasing from left to right across a row and decreasing down a column. Electronegativity values for selected elements are provided in the table below. Considering the H Cl bond, chlorine will have a partial negative (δ ) charge while hydrogen will be partially positive (δ+). This creates a dipole vector that points towards the negative atom by convention. δ+ H Cl δ + Figure 3: Electronegativity (χ) of Select Elements Element χ Element χ Element χ Element χ Element χ H 2.20 Li 0.98 C 2.55 N 3.04 O 3.44 F 3.98 Na 0.93 Si 1.90 P 2.19 S 2.58 Cl 3.16 K 0.82 Ge 2.01 As 2.18 Se 2.55 Br 2.96 Molecular polarity is determined by summing the individual bond dipole vectors in a molecule. Consider the hydrogen cyanide, HCN, molecule. Following the Lewis structure and VSEPR theory approach above, you should be able to conclude the H C N: molecule will have a linear geometry. A dipole vector between H C points towards the carbon given it is more electronegative than hydrogen. Likewise a vector between C N points toward nitrogen. Given the linear geometry and that both vectors point towards the nitrogen end of the molecule, the two vectors will add to give a net dipole moment, rendering the hydrogen partially positive and the nitrogen partially negative. Thus HCN is a polar molecule. In general symmetrical molecules (those having only one type of atom bonded to a central atom) and no lone pairs on the center atom will be non-polar. For this reason CH 4 and hydrocarbons in general are non-polar, since the individual bond dipole vectors cancel each other. Molecules with lone pairs on the central atom tend to have bent or pyramidal geometries and will be polar. Molecules with oxygen or nitrogen follow this pattern and are polar. The polarity of molecules is used to determine the intermolecular forces that influence physical properties. There are three major classifications of intermolecular forces and these are hydrogen-bonding, dipole-dipole, and London dispersion in order of decreasing strength. Molecules containing O H and N H are very polar and possess relatively strong hydrogen-bonding forces between molecules. Non-polar molecules, such as the hydrocarbons, have the weakest London dispersion forces between molecules. Polar molecules that do not contain O-H or N-H bonds will have the intermediate strength dipole-dipole forces between molecules.

3 The stronger the intermolecular forces between molecules, the higher the melting point and boiling points they will have compared to molecules with similar molecular mass. CH 4 is non-polar with London dispersion forces between molecules and is a gas at room temperature (b.p. = 164 ) while H 2 O, with H-bonding forces, is a liquid at room temperature (b.p. = 100 ). Likewise, intermolecular forces influence viscosity and surface tension of liquids. The stronger the forces the more viscous the liquid will be. Solubility of one substance in another will also be influenced by these intermolecular forces. In general, polar molecules are more soluble in other polar substances having similar intermolecular forces, likes dissolve likes. As you likely know, oils are not soluble in water. Oils are hydrocarbons with London dispersion intermolecular forces, while water has strong hydrogen-bonding forces. The oil molecules prefer interacting with other oil molecules and not with water; therefore, oil will not dissolve in water. During this experiment you will draw Lewis and VSEPR structures of several molecules and use molecular model kits to help visualize the three dimensional shapes. You will examine several physical properties, making comparisons between polar and non-polar molecules. Lastly you will compare the solubility of I 2 (non-polar) in water and toluene, C 6 H s CH 3. Materials: Molecular Model Kit Watch Glasses Benzoic Acid Acetone Toluene Mel-Temp Instrument 10mL Graduated Cylinder Napthalene Hexane Tape Capillary Melting Tubes Large Test Tube Isopropyl Alcohol Iodine Filter Paper Safety: Eye Safety Poisonous Chemical Fire Safety -Make sure you understand the lab procedure before you begin -Review the location of the safety equipment -Eye goggles must be worn at all times during this experiment -Benzoic Acid, isopropyl alcohol, acetone, hexane, iodine, and toluene are poisonous if ingested -Hexane and toluene are volatile and extremely flammable Procedure: There are four parts to this experiment which do not have to be done in order. Part I: Measure the melting point for benzoic acid and napthalene. 1. Place a small amount of each benzoic acid in melting point capillary tubes. 2. Place the melting point tubes in the Mel-temp instrument and use as instructed. 3. Record in Table 1 of the data sheet the temperatures for the onset of melting and point of complete melt. 4. Repeat steps 1-3 for napthalene. Part II: Compare evaporation rates of acetone, isopropyl alcohol, and hexane. 1. Lay three pieces of filter paper on watch glass. Place two drops of each liquid in the same spot on separate pieces of filter paper. Start timing. 2. Hang the filter papers with small piece of tape from mini-fume hood. Stop timing when the spot evaporates. Record in Table 2 of the data sheet the time elapsed for each liquid to evaporate. Part III: Compare the solubility of iodine (I 2 ) in water and toluene. 1. Place two small crystals of I 2 in a large test tube. 2. Add 2 ml distilled water and stir with a glass rod for several minutes. 3. Add 2 ml toluene and stir for several minutes. 4. Record all observations in Table 3 of the data sheet. Part IV: Complete Table 4 in the data sheet. 1. Draw the Lewis structure of the molecule. 2. Determining the electron-group arrangement and molecular geometry by using Figure 2 and the molecular model kit (build the molecules to help visualize the geometry and polarity). 3. Draw the VESPR structure of the molecule. 4. Determine if the molecule is polar (insert polar or non-polar in the data sheet).

4 Data Sheet: Name Date Section Part I: Table 1 (Melting Points) Compound Polarity Onset Temperature Benzoic Acid, C 6 H 5 CO 2 H Naphthalene, C 10 H 8 Full Melt Temperature Average Temperature Part II: Table 2 (Evaporation Rates) Compound Intermolecular Force(s) Time (min) Acetone, CH 3 C(O)CH 3 Isopropyl alcohol, CH 3 CH(OH)CH 3 Hexane, CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 Part III: Mixture I 2 + water Table 3 (Solubility) Observations I 2 + water + toluene Part IV: Table 4 (Lewis & VESPR Structures) Formula Lewis structure VSEPR Structure Geometry and Polarity CO 2 COH 2 PBr 3

5 Formula Lewis structure VSEPR structure Geometry and Polarity H 2 O CHCl 3 NH 3 *CO C O CH 3 OH C O CH 3 CH 3 *NH 4 + *The atoms in these molecules may not form the typical number of bonds

6 Post Lab Questions: Name Date Section 1. a. Provide a vector representation of the bond polarity in H 2 O. b. Which is more polar, CHCl 3 or CH 4? Explain. 2. Briefly explain the observed trend in melting points of benzoic acid and naphthalene in terms of intermolecular forces and molecular polarity. 3. Draw the Lewis structures for the molecules in part two (acetone, isopropyl alcohol, and hexane). Using these structures, chemically explain your results for part II. 4. Why would you expect ethylene glycol, HOCH 2 CH 2 OH, to be more viscous than hexane, C 6 H 14? 5. In terms of molecular polarity and intermolecular forces, why is toluene (C 6 H 5 CH 3 ) not soluble in water?

7 Experiment 5 Molecular Geometry & Polarity Name Date Section Pre-Lab Questions: 1. a) What is the main purpose of this experiment? b) Explain main safety precautions for this experiment. 2. a) Complete the table. Group Element Number in Periodic Table H Valence Electrons Bonds it Usually Forms Lone Pairs it Usually has C P S F b) Provide the Lewis dot symbols for an atom of nitrogen and an atom of oxygen. 3. a) What is the octet rule in terms of Lewis structures? b) Are there any exceptions to the octet rule? 4. Using Figure 2, what geometry is expected for the following Lewis structures? a) b) 5. a) What are the three major types of intermolecular forces? b) Using a reliable resource, determine the Lewis structure of the following molecules and sketch them. i) benzoic acid, C 6 H 5 CO 2 H ii) naphthalene, C 10 H 8 iii) toluene, C 6 H 5 CH 3