Lewis Model of Electronic Structure

Transcription

1 Activity 21 Lewis Model of Electronic Structure Why? The Lewis model of molecular electronic structure describes how atoms bond to each other to form molecules. It determines the number of bonds formed between pairs of atoms in a molecule and the number of electrons that exist as lone or nonbonding pairs. This information makes it possible for you to predict the geometry of molecules (e.g., 2 is linear but S 2 and 2 are bent) and relative bond strengths and bond lengths. Learning bjective Understand how to draw Lewis structures and interpret them in terms of molecular properties Success riteria onstruct realistic Lewis structures Identify relative bond strengths and lengths from Lewis structures Prerequisite Activity 3: Molecular Representations Information Molecules exist because they are more stable than the separated atoms. By more stable, we mean that they have lower energy in the same way that a skateboarder at the bottom of a hill has less energy or is more stable than one at the top. G.N. Lewis recognized, from the very low chemical reactivity of the noble gases, that a configuration with eight electrons in a shell produces a very stable situation. e therefore proposed that molecules form so that atoms can transfer or share electrons and produce this very stable octet structure. Lewis structures are used to model how the electrons are arranged to produce these stable eightelectron configurations. In these diagrams, dots are used to represent electrons; a line between atoms represents a single covalent bond formed by a pair of electrons; other dots represent nonbonding electrons; and charges are written to identify a formal distribution of charge. The bonds show how the atoms in a molecule are connected to each other. A Lewis diagram does not show bond lengths, bond angles, the arrangement of atoms in three-dimensional space, or the actual charges on atoms. Some molecules require more than one Lewis structure to describe them. These multiple structures are called resonance structures. In some situations, atoms in period 2 can have fewer than 9 electrons and atoms in periods 3 and higher have more than 8 electrons. Activity 21 Lewis Model of Electronic Structure 131

2 Some atoms (e.g.,, N,, and S) form double bonds, which are represented by double lines. Some atoms (e.g., and N) form triple bonds, which are represented by triple lines. A double bond is stronger and shorter than single bond, and a triple bond is strongest and shortest of the three. ow does one determine and draw a Lewis structure? irst determine whether the molecule is ionic or covalent. If it is ionic, draw each ion separately. or covalent molecules and polyatomic ions, follow the methodology given below. Model 1: Methodology for onstructing Lewis Structures A Simple Example Methodology Step 1: ount the valence electrons from all the atoms. Add electrons for negative ions, subtract electrons for positive ions. The number of valence electrons can be determined from the atom s position in the Periodic Table. Table 21.1 Example: Single Bond or hydrochloric acid, has 1 and has 7 for a total of 8 valence electrons. Step 2: Assemble the bonding framework. Decide which atoms are connected to each other and use a pair of electrons, represented by a line, to form a bond between each pair of atoms that are bonded together. Step 3: Arrange the remaining electrons so that each atom has 8 electrons around it (the octet rule). If necessary, place additional pairs of electrons between the atoms to form additional bonds. Step 4: heck for exceptions to the octet rule. or only 2 electrons are needed (the duet rule). Be and B are often electron deficient and may have only 4 or 6 electrons. Atoms in the second period cannot have more than 8 electrons. The third and higher period elements sometimes have more than 8 electrons. Step 5: Determine the formal charges () on the atoms. satisfies the octet rule. satisfies the duet rule. = number of atomic valence electrons number of lone pair electrons 0.5(number of shared electrons). Evaluate whether the formal charges () on the atoms are reasonable. The structure is reasonable if the charges are zero or small, and the negative charges reside on the most electronegative atoms. ther structures with larger formal charges are higher energy and not representative of the lowest energy structure of the molecule. Step 6: Draw resonance structures. () = (2) = 0 () = (2) = 0 Reasonable because is zero. none in this case 132 oundations of hemistry

3 Key Questions 1. ow can you determine the number of valence electrons that an atom has? 2. a) ow many valence electrons does have and how many more does it need to fill the first shell? b) ow many valence electrons does have and how many more does it need to fill a shell? 3. rom the Lewis perspective, why do hydrogen and chlorine combine to form the molecule? 4. ow is formal charge determined? 5. ow is formal charge used to identify unreasonable Lewis structures? Exercises 1. Write the number of valence electrons for and for. Draw the Lewis structure for. ydrogen has one valence electron; fluorine has seven. The Lewis structure for is shown. 2. Write the number of valence electrons for. Determine how many more electrons needs to form a molecule, and draw the Lewis structure for a molecule composed of and. arbon has four valence electrons and needs four more to fill its outer shell. The Lewis structure of methane, 4, is shown. Activity 21 Lewis Model of Electronic Structure 133

4 3. Write the number of valence electrons for S. Determine how many more electrons S needs to form a molecule, and draw the Lewis structure for a molecule composed of S and. Sulfur has six valence electrons and needs two more to fill its outer shell. The Lewis structure of hydrogen sulfide, 2 S, is shown. S 4. Draw Lewis structures for 2, N 3, P 3, 2 6, and Na. (Na is ionic, so draw the two ions separately with no line connecting them and indicate the charge on the ion with a + or - sign.) N P Na Need for Double Bonds Table 21.2 Methodology Step 1: ount the number of valence electrons from all the atoms. Example: Multiple Bonds or carbon dioxide, has 4 valence electrons and has 6 valence electrons for a total of 16 valence electrons. Step 2: Assemble the bonding framework. Step 3: Arrange the remaining electrons so that each atom has 8 electrons around it (the octet rule). If necessary, place additional pairs of electrons between the atoms to form additional bonds to satisfy the octet rule. Step 4: heck for exceptions to the octet rule. No exceptions present. 134 oundations of hemistry

5 Table 21.2, continued Methodology Example: Multiple Bonds () = (8) = 0 Step 5: Evaluate whether the formal charges on the atoms are reasonable. Step 6: Draw resonance structures. () = (4) = 0 Reasonable because is zero. none in this case Key Questions 6. ow many valence electrons are there in the carbon dioxide molecule? 7. Why is it sometimes necessary in constructing Lewis structures to put double or even triple bonds between atoms? Exercises 5. Draw the Lewis structures for 2, N 2, 2 4, and 2 2. N N Activity 21 Lewis Model of Electronic Structure 135

6 I 4 Exception to the ctet Rule Table 21.3 Methodology Step 1: ount the number of valence electrons from all the atoms, and, because it is a 1 anion, add one. Example: Third Period Element or I 4, there are 36 valence electrons ( ). Step 2: Assemble the bonding framework. I Step 3: Arrange the remaining electrons so that each atom has 8 electrons around it. Step 4: heck for exceptions to the octet rule. After satisfying the octet rule for each atom, 4 electrons remain. These are placed as nonbonding electrons on the fifth period element, iodine. : : : : I I : : : : - Step 5: Evaluate whether the formal charges on the atoms are reasonable. Note: The negative formal charge is on I, which is less electronegative than. This result is correct and is an exception to the general rule given earlier. Step 6: Draw resonance structures. (I) = 7 8 = 1 () = 7 7 = 0 none in this case 136 oundations of hemistry

7 Key Questions 8. ow does the example of I 4 differ from the examples of and 2? Exercise 6. Draw the Lewis structure for P 5. P N 2 The Need for Multiple Lewis Structures Table 21.4 Methodology Step 1: ount the number of valence electrons from all the atoms, and, because it is a 1 ion, add one. Example: Resonance or N 2 there are 18 valence electrons (5 + (2 6) + 1) Step 2: Assemble the bonding framework. N Step 3: Arrange the remaining electrons so that each atom has 8 electrons around it. N - Step 4: heck for exceptions to the octet rule. none Step 5: Evaluate whether the formal charges on the atoms are reasonable. (N) = 5 5 = 0 (1) = 6 7 = 1 (2) = 6 6 = 0 Step 6: In Step 3 the double bond could just as well have been drawn on the left side of the nitrogen. This second structure is called a resonance structure. In fact, experimental measurements of the bond lengths show that both N- bonds are the same and in between that of a single bond and that of a double bond; therefore the second structure is needed to adequately describe the bonding. Draw the resonance structure to show that both bonds are equivalent. N - Activity 21 Lewis Model of Electronic Structure 137

8 Key Questions 9. Why is it not possible to describe N 2 by a single Lewis structure? Exercise 7. Draw the Lewis structure for ozone 3. (+) (+) (-) (-) Model 2: Bond Strengths and Lengths Table 21.5 Bond Average Dissociation Energy in kj/mole Average Length in pm = Key Questions 10. The energy it takes to dissociate or break a bond is a measure of the bond strength. In view of the data in Model 2, how does the Lewis structure help you identify the strongest bonds in a molecule? 11. In view of the data in Model 2, how does the Lewis structure help you identify the shortest bonds in a molecule? 138 oundations of hemistry

9 Exercises 8. Identify which bond in acetic acid is the shortest and strongest. Model 3: Which Lewis Structure is Better? The two structures below represent the thiocyanate ion. Which one has a lower energy and is therefore the better description of the lowest energy state of this ion? igure S N N S Key Questions 12. What are the formal charges on the atoms in the thiocyanate ion in Model 3? Write the formal charge in parentheses above each atom in these structures. Structure A: Sulfur has a formal charge of -1, while carbon and nitrogen have a formal charge of zero, as shown. A (-) S N - Structure B: Sulfur has a formal charge of -1, carbon has a formal charge of -1, and nitrogen has a formal charge of +1, as shown. B (-) (+) (-) N S ow do the formal charges help you identify the lower energy and better structure? Activity 21 Lewis Model of Electronic Structure 139

10 14. Which structure in Model 3 has the lower energy and is therefore the better description of the lowest energy state of this ion? Exercises 9. In view of the formal charges, why is = = a better representation of carbon dioxide than? Don t forget to add the nonbinding electrons In structure A, the formal charges on all of the atoms are zero. In structure B, one of the oxygens has a formal charge of +1; one has a formal charge of -1. Structure A has fewer non-zero formal charges and is a better representation of 2. A B + - Information In assembling the bonding framework of a molecule, you may find it difficult to identify which atoms are bonded to each other. Sometimes this information cannot be deduced from first principles: you simply need to know the molecular structure. But here are some guidelines that are often helpful. It is useful to think in terms of outer atoms and inner atoms. An outer atom bonds to only one other atom while an inner atom bonds to more than one other atom. ydrogen atoms are always outer atoms because they can form only one bond. uter atoms other than hydrogen are usually the ones with the highest electronegativities. The bonding framework is often indicated by the order in which the atoms are written in a molecular formula. or example, in S the carbon atom is the inner atom. Parentheses are often used in a molecular formula to indicate the bonding framework. or example in ( 3 ) 2, the hydrogen atoms are bonded to carbon to form two methyl radicals, and oxygen is an outer atom bonded to an inner carbon atom. Multiple atoms of the same element are usually the outer atoms around a single atom of another element. or example, in P 6 phosphorous is the inner atom. Sometimes it is helpful to know the chemical properties of a molecular formula. or example, in N 3 the hydrogen atom might be bonded to the nitrogen or to oxygen. If you know that nitric acid is an oxyacid, you can locate the hydrogen as bonded to an oxygen atom. inally, the most likely structure is the one which has the most reasonable formal charges on the atoms. By reasonable, we mean that the formal charges should be small or zero, and the negative charges should be located on the most electronegative atoms. 140 oundations of hemistry

11 Exercises 10. Draw Lewis structures for the following molecules. A Lewis structure includes the formal charge on each atom if it differs from zero and any resonance structures that are significant. To be significant, a resonance structure does not increase the formal charge on any of the atoms. Indicate the formal charge on an atom by writing it in parentheses next to the atom, e.g., (-1). Se 2 I 3 S Se I I - I - S N 3 Si 4 N + - N+ - Si 11. Use Lewis structures to arrange the following compounds in order of increasing carbon-carbon bond strength. Explain Use Lewis structures to arrange the following compounds in order of increasing carbon-carbon bond length. Explain Activity 21 Lewis Model of Electronic Structure 141

12 13. Explain which N N bond is more stable, the one in nitrogen N 2 or the one in hydrazine N 2 4. The N-N triple bond in nitrogen is stronger and more stable than the N-N single bond in hydrazine. N N Nitrogen N N ydrazine 14. Explain which bond is shorter, the one in methanol 3 or the one in formaldehyde 2. The = double bond in formaldehyde is shorter than the - single bond in methanol. Methanol ormaldehyde Problems 1. Use Lewis structures to identify which of the following compounds you would most likely be successful at synthesizing: Si 4, 4, S 6, or 6. Explain. Si S The 4 and 6 Lewis structures show expanded octets, which are never observed for the second period element oxygen. Si 4 has a complete octet around the central atom and S 6 is able to accommodate additional electrons around the central sulfur atom. Si 4 and S 6 are the most stable compounds of the four and would be easier to synthesize than 4 and oundations of hemistry

13 2. Two Lewis structures are needed to describe the bonding in formamide, N 2. Write these two resonance structures. ne should have no formal charges on the atoms, and the other will have a formal charge of +1 on N and 1 on. Which structure that you have drawn do you expect to be the better description of formamide? Exercise 10 states that a significant resonance structure doesn t increase the formal charge on any of the atoms. Using that criterion, the structure on the left would be the better description of formamide because all of the atoms in the structure have formal charges of zero. The structure on the right is less significant because of the negative formal charge on the oxygen and the positive formal charge on the nitrogen. - N N + 3. or the two resonance structures of formamide in Problem 2, explain why each of the following statements is either correct or incorrect. a) The molecule is not oscillating back-and-forth between the two structures, rather the molecule is an average or superposition of the two structures. b) The expected bond length is between that for a normal double bond and that for a normal single bond. c) The nitrogen in the molecule has a lower electron density associated with it than is found for the free nitrogen atom. d) Both resonance structures have the same energy. Activity 21 Lewis Model of Electronic Structure 143

14 Reflection Develop a checklist that you can use to assure that you have written a correct Lewis structure. 144 oundations of hemistry