1 Chemical Bonds 1. Important points about Lewis Dot: a. Duet Rule: 2 electrons needed to satisfy valence shell. i. What follows this rule? Hydrogen and Helium b. Octet Rule: 8 electrons needed to satisfy valence shell. i. What follows this rule? C, N, O and F are the only elements that pretty much always follow the rule. ii. What can exceed the octet? Any element whose valence shell is in n=3 or higher have the ability to carry more than 8 electrons in their valence. This is because at n = 3, there are additional d orbitals available for bonding. iii. What can have less than the octet? Boron is happy with 6 electrons in its valence. 2. What are the two types of electron pairs? a. Bonding Pair 2 electrons shared between two atoms in a molecule. These electrons are confined to the area between the two atoms sharing them. b. Lone Pair 2 unshared electrons that are found on one atom. These electrons take up more room than a bonding pair and tend to push bonding pairs closer together.
2 3. Steps for drawing a Lewis Dot structure a. Add up all valence electrons available using the periodic table. b. Draw a skeleton structure. c. Give all atoms in the compound the need duet or octet. d. Add up all electrons in use. Compare the number of electrons used to the number available. Remember you cannot exceed or go below the valence available. You must use the exact value. e. If you have used too many electrons you will need to make a double bond between two atoms. When you add the double bond pull a lone pair off each of the atoms participating in the double bond. Continue this process until you have used the correct number of electrons. f. If you have too few electrons, you will typically add them to the central atom. 4. Draw the Lewis Dot for a. CH 2 O Step 1 Count up valence electrons Based on the periodic table we can see that: C 4 valence, H 1 valence, O 6 valence Adding them all up we get:
3 Step 2 Build a skeleton If carbon is in a structure it is typically the central atom. Step 3 Give all atoms octet/duet Step 4 Add up electrons in Lewis Each bond counts for 2 electrons and each lone pair counts for 2 electrons. In this structure there are 3 bonds and 4 lone pairs: 3(2e - ) + 4(2e - ) = 14e - in use This value exceeds the 12e - available so we are going to have to make additional bonds. Step 5 replace lone pairs with bonds The only place we can add additional bonds is between carbon and oxygen (hydrogen does not make more than one bond). When we make this bond we will have to remove one lone pair from oxygen and one lone pair carbon. We needed to remove a lone pair from each because they share the bond and thus the bond counts as 2 electrons for both.
4 We now recount the number of electrons in use to see if it matches the number of electrons available. There are a total of 4 bonds and 2 lone pairs: 4(2e - ) + 2(2e - ) = 12e - in use As this value matches the original valence count this is a valid Lewis structure. 5. What do you do if there is more than one viable Lewis Structure? You need to look at the formal charges on the atoms in the compound and see which Lewis structure gives the best formal charges. a. How do you determine the formal charge? (# of valence electrons) (# of actual electrons in compound) = formal charge # of valence is the value you get from the periodic table. When counting the actual electrons remember that: lone pair = 2 e - bonding pair = 1e - b. What are favorable formal charges? 1. Values close to zero. 2. Negative formal charges on the most electronegative atom. 3. Positive formal charges on the least electronegative atom. 6. Draw the Lewis Dot with the best formal charges for a. OCN - Step 1 Count up valence electrons
5 Based on the periodic table we can see that: C 4 valence, N 5 valence, O 6 valence Adding them all up we get: Step 2 Build a skeleton If carbon is in a structure it is typically the central atom. Step 3 Give all atoms octet/duet Step 4 Add up electrons in Lewis Each bond counts for 2 electrons and each lone pair counts for 2 electrons. In this structure there are 2 bonds and 8 lone pairs: 2(2e - ) + 8(2e - ) = 20e - in use This value exceeds the 16e - available so we are going to have to make additional bonds. Step 5 replace lone pairs with bonds We needed to remove a lone pair from each because they share the bond and thus the bond counts as 2 electrons for both.
6 We now recount the number of electrons in use to see if it matches the number of electrons available. There are a total of 3 bonds and 6 lone pairs: 3(2e - ) + 6(2e - ) = 18e - in use We are still over our valence electron limit this means that we are going to have to make another bond. ` Once again, I need to add up the electrons in use there are four bonds and 4 lone pairs. 4(2e - ) + 4(2e - ) = 16e - We are using the correct number of valence electrons so this is a valid Lewis structure but is it the best? Couldn t we, just as easily, have drawn the structure as: or At this point we have 3 valid Lewis structures (valid meaning that octet/duets are satisfied and proper number of electrons have been used). Step 6 Compare formal charges
7 Looking at the formal charges, structure b is definitely not the correct structure, a formal charge of -2 on nitrogen would be too high energy. Comparing structure a and c the difference between their formal charges is the atom on which the negative charge is located. Remember that oxygen is more electronegative than N. This means that it would be preferential for the negative charge to be located on oxygen. This leads us to the conclusion that structure c is the best Lewis structure. Thus our final answer would be: When your compound has an overall charge it is important to remember to bracket the structure and place the charge outside. b. NO Step 1 Count up valence electrons Based on the periodic table we can see that: N 5 valence, O 6 valence Adding them all up we get: Step 2 Build a skeleton If carbon is in a structure it is typically the central atom. Step 3 Give all atoms octet/duet
8 Step 4 Add up electrons in Lewis Each bond counts for 2 electrons and each lone pair counts for 2 electrons. In this structure there is 1 bond and 6 lone pairs: 1(2e - ) + 6(2e - ) = 14e - in use This value exceeds the 11e - available so we are going to have to make additional bonds. Step 5 replace lone pairs with bonds We now recount the number of electrons in use to see if it matches the number of electrons available. There are a total of 2 bonds and 4 lone pairs: 2(2e - ) + 4(2e - ) = 12e - in use We are still over our valence electron limit but only by one electron this means that we cannot make a bond. We need to remove one electorn from either O or N and form a radical ( a radical is a compound that contains an unpaired electron). This means we can form or At this point we have 2 valid Lewis structures. Step 6 Compare formal charges
9 Looking at the formal charges, structure b is the best Lewis dot. Al formal charges are equal to zero. 7. What is resonance? Resonance is a situation in which there are more than one valid structure for a molecule such that the actual compound is some combination of all resonance structures. a. How do you know if you should pick the best structure or draw all resonance structures? In a case where all formal charges are equal you will have to draw each resonance structure out. When there is a case where there are variations in formal charges (as in the example above) you will pick the best structure. 8. Draw the Lewis Structure for a. NO 3 - Step 1 Count up valence electrons Based on the periodic table we can see that: N 5 valence, O 6 valence Adding them all up we get: Step 2 Build a skeleton If carbon is in a structure it is typically the central atom.
10 Step 3 Give all atoms octet/duet Step 4 Add up electrons in Lewis In this structure there is 3 bond and 10 lone pairs: 3(2e - ) + 10(2e - ) = 26e - in use This value exceeds the 24e - available so we are going to have to make additional bonds. Step 5 replace lone pairs with bonds We now recount the number of electrons in use to see if it matches the number of electrons available. There are a total of 4 bonds and 8 lone pairs: 4(2e - ) + 8(2e - ) = 24 e - in use We are using the correct number of electrons. But now we have to consider that there is no reason why the N has to be double bonded to one oxygen over another. This is a resonance case. That double bond is equally valid between the N and any of the oxygens. There is no diffenece in formal charges so all structures must be drawn:
11 9. How do you determine if a molecule is polar? Determining the electronegativity difference between the two atoms within the bond. Remember that electronegativity is the ability of an atom to pull electrons towards oneself. Generally: EN >1.8 indicates Ionic Bond 0.4< EN<1.8 indicates Polar Covalent EN< 0.4 indicates Non Polar Covalent 10. What is a handy way to remember electronegativity ordering? F >O>N>Cl>Br>I>S>C>H (pronounced FONClBrISCH ) Fluorine, Oxygen, Nitrogen, Chlorine, Bromine, Iodine, Sulfur, Carbon, Hydrogen (Fluorine being the most electronegative) The further apart 2 atoms are the greater the difference in their electronegativities and more polar bond they would form. Notice that metals are not part of this list as metals tend to lose electrons, meaning they would have very weak electronegativity. 11. What is the premise of VSEPR theory? Electron pairs orientate themselves in a manner which minimizes repulsion (i.e. maximizes distance between electron pairs).
12 12. VSEPR Chart
13 13. How do electronic and molecular structure differ? Electronic structure is the shape that a molecule would take based solely on the number of areas of electron density (also known as steric number) in this case we do not need to differentiate between bonding pairs and lone pairs. Meaning we are looking at how ALL of the electrons lone/bonding will be arranged Molecular structure is the arrangement that focuses on how the bonding electrons will be orientated in space. 14. What are the bond angles of the non-lone pair bearing shapes? a. Linear (2 areas of e - density, no lone pairs) 180 o b. Trigonal Planar (3 areas of e - density, no lone pairs) 120 o c. Tetrahedral (4 areas of e - density, no lone pairs) o d. Trigonal bipyramidal (5 areas of e - density, no lone pairs) 120 o and 90 o e. Octrahedral (6 areas of e -, no lone pairs) 90 o 15. Do lone pairs affect bond angles? Because lone pairs take up more room they essentially push the bonding pairs closer together, thus altering typical bond angles. For example in NH 3 (which has one lone pair) the bond angle is o and in H 2 O (which has 2 lone pairs) the bond angle is o. In both cases however, there are 4 areas of electron density around the central atom. 16. How can shape affect polarity? Symmetry can cancel out a polar bond. If identical atoms are attached to an atom and the molecule is symmetric, it will be non-polar overall.
14 Let s say that B is more electronegative than A, such that they form a polar bond. The following is a list of some examples of how symmetry can cancel out polarity (keep in mind, this list is not exhaustive). Basically what you are looking for is a symmetry in which the electrons are being pulled with equal strength but in opposing directions. If you picture the pull on electrons as a tug of war imagine that the opposing teams are identical in strength and are pulling at angles that perfectly counter each other. If this were the case the electrons would not be pulled in any one direction more strongly than the other thus they would stay where they were. 17. Determine the electronic and molecular structures around each of the central atoms and whether the molecule is polar. In order to do this we will first have to draw the Lewis Dot structure. a. CHF 3 Following the rules described in preceding problems we would get the structure: In this case, the central atom, carbon, has 4 areas of electron density. That means that electronic configuration around carbon would be tetrahedral configuration. Additionally, because there are no lone pairs on the carbon, the molecular structure would also be tetrahedral.
15 The molecular structure would look like: In a tetrahedral the bond angles are o. This would be a polar molecule as F is more electronegative than C. b. I 3 - In this case, the central I has 5 areas of electron density. This means that the electronic configuration would be trigonal bipyramidal. Because three of the areas of electron density are lone pairs, the molecular configuration would be a linear configuration. With all atoms in the bond being identical, and it being symmetric, this would be a nonpolar molecule. c. BrF 5 In this case, the central atom, bromine, has 6 areas of electron density. This means that the electronic configuration would be octahedral. As one of these areas corresponds to a lone pair the molecular configuration would be square pyramidal. Thus the molecule around bromine would look like:
16 This would be a polar molecule. Due to the lone pairs, there would not be a canceling of polar bonds due to symmetry. 18. Explain why CF 4 and XeF 4 are non-polar yet SF 4 is polar. In order to explain this, we will have to look at the shape as predicted by VSEPR. We ll start with our Lewis Dots. 19. Indicate the symmetry around all non-hydrogen atom The carbon has three areas of electron density, no lone pairs. This would mean that around the carbon the arrangement would be trigonal planar. The oxygen oxygen in the C-O-H bond has four areas of electron density (with 2 lone pairs) so the atoms would be arrange in a bent (or v-shaped) fashin around that oxygen. Putting all of this together the VSEPR structure would look like:
17 20. Consider CH 4 a. What orbitals are available for bonding for each atom involved in the molecule? Each hydrogen has a 1s orbital available for bonding. Carbon has one 2s and 3 2p orbitals available for bonding. This was determined by looking at the valence of each atom. Rememebr that only valence electrons participate in bonding. b. Based on this would you expect all the C-H bonds in methane (CH 4 ) to be identical? To answer this let s look at the valence electron configuration for carbon. In it s current format, it doesn t look like carbon can make more than 2 bonds, as it has only 2 unpaired electrons. What winds up happening is that one of the 2s electrons gets promoted into the 2p shell so that we get: When this happens, we are now able to bond carbon 4 times, as there are four unpaired electrons. Next let s consider this in conjunction with the 4 hydrogen atoms. From this we can see that each one of the hydrogens can pair up with one of the unpaired electrons from the carbon atom.
18 Looking at these results we can now say that based on this initial information we would expect 3 of the four bond to be identical. This is the case because 3 of the bonds are made from a 1s/2p combination and one bond is made from a 1s/2s combination. Different orbitals used = different bonds. c. What have experiments shown? Experiments have shown that all bonds are actually equivalent (i.e. identical to one another).. d. What does this indicate? This means that carbon cannot be using its atomic orbitals (2s and 2p). If all bonds are identical is must be using 4 equivalent orbitals to bond with the hydrogens. 21. This means that when an atom patricpiates in a bond, its atomic orbitals (regular valence shell orbitals) are combined to form a hybrid. This new hybrid orbital is what particpates in the bond. 22. How does the energy of the hybrids compare to the atomic orbitals? The hybrid orbital has an energy level somewhere between the energy level of the composite atomic orbitals. 23. The number of hybrid orbitals formed is equal to the number of atomic orbitals used.
19 24. What should you do first, when trying to determine the type of hybrid orbitals needed? Draw out your Lewis Dot structure. a. Why? It will help you to determine the number of areas of electron density (or steric number) of each of atom in the molecule. The type of hybrid used depends on the steric number. b. How many areas of electron density do the following account for? i. Lone Pairs - one ii. Single bond - one iii. Double Bond - one iv. Triple Bond one 25. Fill In the Following
20 26. Draw the following a. CH 4 Step 1 Lewis Dot Step 2 Determine the Steric Number for Atoms in Molecule Carbon Steric Number = 4 This means we will need to create a hybrid that has 4 orbitals. This means we need to combine 4 atomic orbitals. One s and three p orbitals which come together to form 4sp 3 hybrids. Hydrogen doesn t create hybrid orbitals because it only has a 1s electron available Thus, the molecule would look like: b. CH 2 O Step 1 Lewis Dot Step 2 Determine the Steric Number for Atoms in Molecule Carbon Steric Number = 3 This means that we need to create 3 hybrid orbitals. This means we need to combine 3 atomic orbitals. One s and two p orbitals which come together to form 3 sp 2 hybrids and one unhybridized p is left over.
21 Oxygen Steric Number = 3 This means that we need to create 3 hybrid orbitals. This means we need to combine 3 atomic orbitals. One s and two p orbitals which come together to form 3 sp 2 hybrids and one unhybridized p is left over. Hydrogen Does not form hybrids Thus the molecule would look like: 27. What are sigma and pi bonds? Sigma bonds are bonds that occur when orbitals have a head on overlap. These are all examples of sigma bonds. Pi bonds overlapping parallel p orbitals. These are two ways to represent the pi bond. 28. What are each type of bond composed of? Single = 1 sigma bond Double = 1 sigma bond + 1 pi bond Triple = 1 sigma bond + 2 pi bonds
22 29. Predict the shape, hybridization of central atom and polarity for the following a. OF 2 Shape Bent (4 areas of e - density and 2 lone pairs) sp 3 hybridization (Steric Number = 4, combination of 4 atomic orbitals) Polar b. TeF 4 Shape See Saw ( 5 areas of e - density and 1 lone pair) sp 3 d hybridization (Steric Number = 5, combination of 5 atomic orbitals) Polar c. BF 3 Shape Trigonal Planar (3 areas of e - density) sp 2 hybridization (Steric Number = 3, combination of 3 atomic orbitals) Non-polar (symmetry cancels out polar bonds)
23 30. Label the hybridization of C, O, and N in the following molecules. Also count total number of sigma bonds and total number of pi bonds. 31. Are all the atoms in the same plane? a. C 2 H 2 This would be a linear molecule (2 areas of e - density around each carbon, no lone pairs) b. CH 2 CCH 2 All of the atoms are not in the same plane. In order for the central carbon to double bond to each of the outer carbons they must form a pi bond. Pi bonds are made up of parallel p orbitals. This means that one of the carbons will have to bond with the p y and the other will bond with the p x. The sigma bonds occur on the z plan. 32. How can CO 3 2- help us understand the short falls of LE model? If you look at the Lewis Dot for CO 3 2- : This is a resonance structure. As you can see based on where we position the double bond, the hybridization of the oxygen atoms change. We cannot label it sp 2 or sp 3. Our
24 model is falling short here because we cannot describe resonance structures. This means that we need to revise out understanding a bit. 33. Orbitals are wave functions. One property of waves is that they can constructively or destructively interfere with each other. 34. What is constructive interference? When two waves constructively interfere the resultant wave has an increased amplitude. 35. What is destructive interference? When two waves destructively interfere they basically cancel each other out. 36. How does this relate to our understanding of orbital interactions? Just as waves have a positive phase and negative phase - so do orbitals. s orbital phases: p orbitals phase * a node is a location where there can be no electrons. When orbitals of the same phase interact with each other a bond forms. The electrons are located, primarily, between the two nuclei of each atom. When orbitals of different phases interact with each an anti-bond occurs. There is a node between the two nuclei of the atoms.
25 37. What is Molecular Orbital Theory? In this theory the atomic orbitals of different atoms are combined to make new bonding orbitals called molecular orbitals. 38. Facts about molecular orbitals? a. Molecular orbitals can hold up to 2e - with opposite spin. b. (Molecular Orbital) 2 = 90% probablility of finding e - within orbital. c. Only Molecular Orbitals exist after combining, no atomic orbitals. d. Molecular orbitals can be used to electron configuration for molecules. e. # of atomic orbitals used = # of molecular orbitals formed. f. When atomic orbitals interact they can do so constructively and destructively. g. Constructive interference produces bonding molecular orbitals that are lower in energy than the atomic orbitals contributors. h. Destructive interference produces anti-bonding molecular that are higher in energy than the atomic orbital contributors. i. Generally sigma bonds are lower in energy than pi bonds. 39. Why do bonds occur? The energy of the atoms combined is lower than the sum of all their energies independent from one another. 40. Fill in the following chart
26 Remember that degeneracy is when orbitals are at the same energy level. 41. What is a Molecular Orbital Diagram? A diagram that shows which atomic orbitals are being contributed to by the atoms and what molecular orbitals they will form. Additionally it indicates the relative energy level of all orbitals.
27 42. Draw the Molecular Orbital Diagram a. H 2 b. H 2 + When filling in the diagram always fill in lowest available molecular orbital first. In this case the molecule being formed is H 2 +. The positive charge indicates that one electron has been lost thus only one electron will be contributed. It does not matter which hydrogen you leave as the electron contributor. c. H 2 -
28 In this case the molecule being formed is H 2 -. The negative charge indicates that an extra electron has been gained thus three electrons are being contributed. It doesn t matter which hydrogen you add the electron to. 43. What is bond order? Bond order indicates the relative strength/stability of a bond. The formula for determining the bond order of a compound is: 44. Some facts about bond order: a. A bond order of zero indicates that the compound does not exist. b. The bond order typically corresponds to the type of bond present. i.e. Bond order = 1 indicates a single bond Bond order = 2 indicates a double bond Bond order = 3 indicates a triple bond 45. What do the bond orders for #41 tell us about relative stabilities of H 2, H 2 +,H 2 -?
29 H 2 = (2 0) / 2 = 1 H 2 + = (1-0)/2 = ½ H 2 - = (2-1)/(2) = ½ Based on these bond orders we can see that H 2 is the most stable compound. 46. Paramagnetic substances have unpaired electrons and are therefore attracted toward a magnetic field. Diamagnetic substances have paired electrons and are therefore repelled by a magnetic field. 47. What is the order of molecular orbitals for B 2, C 2 and N 2? σ 2s σ * 2s π 2p σ 2p π * 2p Remember that there are 2 degenerate orbitals in both the π and π *. Thus there are a total of 4 electrons that are contained within the π orbitals. 48. What is the order of molecular orbitals for O 2 and F 2? σ 2s σ * 2s σ 2p π 2p π * 2p 49. Draw the Molecular orbital diagrams for N 2 and O Why do B 2, C 2 and N 2 have a different MO order compared to O 2 and F 2? π 2p is lower in energy for B 2, C 2 and N 2 because of e - repulsion σ 2p has with σ 2s.
30 Additionally there is some s/p orbital mixing that changes energies. σ 2p is lower in energy because as protons are added the 2s orbital diminishes in size such that the e - repulsion is minimalized. 51. Write out the MO electron configuration for each. N 2 2- : N 2 2-, O 2 2-, F 2 2- O 2 2- : F 2 2- : a. Which of the following is predicted to be the most stable diatomic species? In order to determine this we will have to look at the bond orders of each molecule. You can look at all the electrons or just the valence electrons to determine the bond order you will get the same answer using either method. N 2 2- : using all electrons (10-6)/2 = 2, gives the same answer as looking only at the valence electrons (8-4)/2 = 2.
31 O 2 2- : (8-6)/2 = 1 F 2 2- : (8-8)/2 = 0 Based on these bond order values we can determine that N 2 2- is the most stable of all the species. We can also determine that F 2 2- doesn t exist as it has a bond order equal to zero. b. Indicate whether each is paramagnetic or diamagnetic. In order to determine this, you will need to determine if the electrons in the last molecular orbital are paired up or not. If they are paired up, it is diamagnetic; if not, paramagnetic. N paramagnetic O 2 2- : diamagnetic F 2 2- : paramagnetic 52. Use the MO model to predict the bond order and magnetism for Ne 2 and P 2. Ne 2: Because Ne comes after O and F on the periodic table, we will use their molecular orbital configuration. Bond Order = (8-8)/2 = 0 This compound does not exist. P 2 : Because P is in the same family as N, we will use the B 2, C 2, N 2 configuration. Bond Order = (8-2)/2 = 3 Diamagnetic 53. Complete the MO diagram for NO.
32 When creating the MO diagram for a molecule that has elements has B, C or N and O or F default to the B, C, and N MO order. Notice that when using molecular orbital theory, there is absolutely no issue with an odd number of electrons, unlike Lewis. 54. How do we reconcile these two theories (i.e. LE and MO models)? For simplicities sake we will view σ bonds as being localized (hybrid orbitals) and we will view π bonds through molecular orbital theory, in which the electrons are delocalized. To better understand this, we will consider benzene, C 6 H 6, a resonance structure. In the combined view of localized σ bonds and delocalized π, the new depiction looks like: