CH342 Handin Homework 8 Answers. FC = 7-8/2 = +3 FC = 7-6/2-2 = +2 FC = 7-4/2-4 = +1 FC = 7-2/2-6 = 0 all sp 3 hybridized

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1 CH342 Handin Homework 8 Answers 1. Draw the Lewis structures and determine the formal charge and hybridization for the chlorine in 4, 3, 2, and. FC = 78/2 = +3 FC = 76/22 = +2 FC = 74/24 = +1 FC = 72/26 = 0 all sp 3 hybridized 2. Can you decrease the formal charge on chlorine by forming a double bond in 4, 3, 2, and? Show the Lewis dot structure for each ion (only one resonance form is necessary). Which species is most likely to form a double bond? FC = 710/2 = +2 FC = 78/22 = +1 FC = 76/24 = 0 no double bond Which most likely?: The species with the highest formal charge on is the most likely to form a double bond. A higher formal charge would make the hungrier for electrons, which it could gain by sharing in a double bond. So 4 is most likely to double bond. 3. To form a double bond, these oxyanions would be expected to use dorbitals in some way. Would the expected hybridization for the chlorine change if double bonding occurs? What orbital on chlorine would be used for pibonding? There are roughly three possibilities: (1.) The hybridization for the formation of the sigma bonds could remain sp 3 if the atom used a dorbital for pi bonding, Figure 1. (2.) r, the could use sp 2 d hybridization for the atom and have a p orbital left over for pibonding, Figure 2. (3.) r, alternatively, the could use sp 3 d hybrids for both bonds in the double bond. In this model the two bonds to the double bonded oxygen would be equivalent and each bond of the pair is made from the overlap of an sp 3 d orbital on the and an sp 3 hybrid on the oxygen, Figure 3. 1

2 Figure 1. sp 3 + d Figure 2. sp 2 d + p Figure 3. sp 3 d the atom is at the center of the trigonal bipyramid 4. Draw the molecular orbital diagram for and. Determine the bond order from the M filling. Which species, or, is predicted to have the stronger bond and why? E E 2p π 3p π 2s 3s 2s 3s 2p π 3p π : B = = 1½ : B = = 1 Therefore, is predicted to have a stronger bond than. However, is an oddelectron species, so it is a free radical. Free radicals are very reactive. However, note that hypochlorite ion,, is the active ingredient in bleach. So is also reactive; it is a good oxidizing agent. Notice that the pi bonding and antibonding orbitals are all filled in, so there is no net pibonding, as predicted in Question 2.

3 5. Using the data from the Molecular Structure Calculations online database ( determine the charge on the chlorine, the bond order, the chlorine hybridization, and the bond angle for the chlorine oxyanions. Use the following table to report your results: oxyanion q() Bondrder hybrid Bond angle Hybridization predicted from bond angles s 0.44 p s 0.59 p sp s 0.69 p 3 d sp sp 2.93 d sp 3 6. Do you see evidence for the formation of a partial double bond in any of the oxyanions? Do you see evidence that any of the oxyanions have significant dorbital character for the? Compare these results to your answer for question 3. The strongest case for double bonding is 4. For 4 with one double bond as shown in Question 2, there are four resonance structures with an average qualitative bond order for each bond of Therefore, you might even argue that 4 forms a second partial double bond. The average qualitative bond order for 3 in Question 2 is So, you might argue for a small amount of double bond character in 3. n the other hand none of the ions show a significant contribution from d orbitals. Compared to question 3, above, there is a ~6% contribution from dorbitals that is consistent with models 2 or The bond angles around an atom can be used to predict the hybridization of that atom. For example, ~109 bond angles are indicative of sp 3 hybridization. Use the bond angle from the database calculations to predict the corresponding hybridization for chlorine in each oxyanion. Record this prediction in the last column of your table. Do any of these predicted hybrizations suggest any pibonding? Bond angles close to 120 correspond to sp 2 hybridization. Then sp 2 hybridization leaves a p orbital left over for pibonding. Looking at the table, 2 has almost 120 bond angles. But where then, you might ask, do the nonbonding electrons go? For 2, you can picture one lone pair in the third sp 2 hybrid, just like 3 and S 2, and the remaining lone pair in the pure p orbital. This porbital is the porbital that would be used for forming a double bond, but the oxygen atoms already have sufficient electrons, so no double bonding occurs. (A similar situation occurs in, which has all the pibonding and piantibonding orbitals filled). 8. Questions 14 are based on the clearcut predictions of the simple bonding theories that are presented in General Chemistry text books. Problems 57 show that the real world is not quite so simple. There are shades of grey, if you like, and the simple theories in the text book aren t the end of the story. Now let s say your younger sister is taking high school chemistry and she asks

4 you what the Lewis dot structures are for 4, 3, 2, and. She is just learning about the minimization of formal charge and resonance structures. You want to be as honest as possible without having to get into too much detail. What would you choose for the best Lewis structures? I would give the generalization that halogens don t like to double bond and suggest the structures in Question 1 are the best firstlevel approximation. Another generalization is to say that you don t use dorbitals (expand the octet) unless absolutely necessary. The firstlevel approximation predicted hybridization is also sp 3. Summary for Questions 18 In Question 3, we proposed three different hybridization schemes that might be used to explain the formation of some double bond character. You might be wondering which is the best scheme as compared to the results listed in the database. There are no explicit pibonds listed in the Best Lewis structure section for 4. So perhaps Figure 3 isn t so bizarre after all. 1 In fact none of the three schemes really explain the partial double bond character very well; they all overemphasize the contribution of the dorbitals. At some point, it is best to just move on to the more careful and complete theory, which is M theory in this case. 9. The HM for CH 4 at the HF/321G(*) level from Spartan is given below: M: 5 Eigenvalues: (ev): T2 1 H1 S H1 S C1 S C1 S C1 PX C1 PY C1 PZ C1 S C1 PX C1 PY C1 PZ H2 S H2 S H3 S H3 S H4 S H4 S The HM is M 5 since there are 10 total electrons with two electrons per M. At the HF/3 21G(*) there are no polarization functions on either C or H (3 rd row elements are given polarization functions). Since the Gaussian orbitals are at the 321G(*) level, the valence shells are split into an inner and outer part.: Ψ M5 = Ψ C, 2px (inner) Ψ C, 2py (inner) Ψ C, 2px (outer) Ψ C, 2py (outer) Ψ H2,1s (inner) Ψ H2,1s (outer) Ψ H3,1s (inner) Ψ H3,1s (outer) Ψ H4,1s (inner) Ψ H4,1s (outer)

5 Write the HM in a similar fashion for formaldehyde at the HF/321G(*), or equivalently HF/321G, level. Compare the relative importance of the inner and outer portions of the split valence shell orbitals on carbon between methane and formaldehyde ANSWER The HM for formaldehyde is: M: 8 Eigenvalues: (ev): B1 1 C1 S C1 S C1 PX C1 PY C1 PZ C1 S C1 PX C1 PY C1 PZ H2 S H2 S H3 S H3 S S S PX PY PZ S PZ PX PY Ψ M8 = Ψ C, 2py (inner) Ψ C, 2py (outer) Ψ H2,1s (inner) Ψ H2,1s (outer) Ψ H3,1s (inner) Ψ H3,1s (outer) Ψ, 2px (inner) Ψ, 2py (outer) The inner portion on C is much more important in formaldehyde.

6 Mutagenicity of Chlorofuranone Derivatives Answers The best correlation with the activity is with the LUM energy. The correlation with the PM3_Dipole is also very good, but since the correlation coefficient between the PM3_Dipole and the LUM energy is 82 using both descriptors together will probably not be a good idea. vol should be removed from the data set. The fit with SlogP is the best with R 2 = : activity = * PM3_LUM * SlogP The fit with the PM3 Heat of formation is next with R 2 = : activity = * PM3_LUM * PM3_HF The fit with the PEE_VSA_Hydrophobic surface area is third with R 2 = : activity = * PEE_VSA_HYD * PM3_LUM The correlation plot with SlogP is quite good:

7 predicted activity The predicted value is , which is not very good activity. activity = * PM3_LUM * SlogP activity = * * =

Hybrid Molecular Orbitals

Hybrid Molecular Orbitals Last time you learned how to construct molecule orbital diagrams for simple molecules based on the symmetry of the atomic orbitals. Molecular orbitals extend over the entire molecule